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\title[Rationally simply connected complete intersections]
{Low degree complete intersections are rationally simply connected}

\author[de Jong]{A. J. de Jong}
\address{Department of Mathematics \\
Columbia University \\
New York, NY 10027}
\email{dejong@math.columbia.edu}

\author[Starr]{Jason Michael Starr}
\address{Department of Mathematics \\
  Stony Brook University \\ Stony Brook, NY 11794}
\email{jstarr@math.sunysb.edu} 

\date{\today}

\begin{document}

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\begin{abstract}
We prove smooth, 
low degree complete intersections in projective space are rationally
simply connected in a strong sense.  Using a result of Hassett, we deduce
weak approximation for smooth, low degree complete intersections defined 
over the function field of a curve.
\end{abstract}

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\section{Main results} \label{sec-intro}
\marpar{sec-intro}

\mni
There is a well-developed algebro-geometric analogue of path
connectedness: 
A nonempty, projective, complex variety is \emph{rationally
  connected} if each 
pair of closed points is contained in a rational curve, cf. ~\cite{K},
~\cite{De}.  To give a flavor of this notion, we mention that a
smooth complete intersection
of $c$ hypersurfaces in $\PP^n$ of degrees $d_1,\dots,d_c$ is
rationally connected if and only if 
$$
n \geq \sum_{i=1}^c d_i.
$$

\mni
A path connected
space is simply connected if the space of based paths is
path connected.  Barry Mazur suggested an algebro-geometric 
analogue of simple
connectedness: 
a rationally connected variety should be
\emph{rationally simply connected} if the space of based,
2-pointed rational curves of suitably positive class is rationally
connected.  

\mni
We must assume the 
curve class is suitably positive or else the space may be empty or
atypical in some other way.  Similarly we must assume the two
basepoints are general.
Finally, since the space of
2-pointed rational curves is typically not compact, it should be
compactified.  We compactify with the Kontsevich moduli space of
stable maps.
For a projective scheme $X$ and nonnegative
integers $g$, $m$ and $e$, the \emph{Kontsevich space} $\Kgnb{g,m}(X,e)$
parametrizes data $(C,p_1,\dots,p_m,h)$ of
\begin{enumerate}
\item[(i)]
a proper, connected, at-worst-nodal, arithmetic genus $g$ curve $C$,
\item[(ii)]
an ordered collection $p_1,\dots,p_m$ of distinct smooth points of
$C$, 
\item[(iii)]
and a morphism $h:C\rightarrow X$ whose image has degree $e$
\end{enumerate}
such that $(C,p_1,\dots,p_m,h)$ has only finitely many automorphisms,
cf. ~\cite{FP}.  Oftentimes this is refined by using a \emph{curve
  class} $\beta$ in $X$ in place of the degree $e$.  
There is an evaluation morphism
$$
\text{ev}:\Kgnb{g,m}(X,e) \rightarrow X^m, \ \ (C,p_1,\dots,p_m,h)
\mapsto (h(p_1),\dots,h(p_m)).
$$
In particular, a general fiber of 
$$
\text{ev}:\Kgnb{0,2}(X,e) \rightarrow X\times X
$$
is an algebraic analogue of the space of based paths in topology.  

\begin{thm} \label{thm-main}
\marpar{thm-main}
Let $X$ be a smooth complete intersection in $\PP^n$ of type
$\underline{d} = (d_1,\dots, d_c)$.  For simplicity, assume all
$d_i\geq 2$.  The variety $X$ 
is rationally simply connected if 
$$
n+1 \geq \sum_{i=1}^c d_i^2
$$
with the one exception $n=3$ and $\underline{d} = (2)$, i.e., a
quadric surface.  Also $\PP^n$ is rationally simply connected for
$n\geq 2$.

\mni
To be precise, for every $e\geq 2$ there is a canonically defined
irreducible component $M_{e,2}\subset \Kgnb{0,2}(X,e)$ such that the
restriction 
$$
\text{ev}:M_{e,2} \rightarrow X \times X
$$ 
is dominant with rationally connected general fiber.  
\end{thm}

\mni
A general point of $M_{e,2}$ parametrizes embedded, smooth rational curves.
Moreover, if $X$ is general then $\Kgnb{0,2}(X,e)$ is irreducible,
i.e., $M_{e,2}= \Kgnb{0,2}(X,e)$, cf. \cite{HRS2}.  

\mni
What of $m$-pointed curves with $m> 2$?  The fiber of
$$
\text{ev}:\Kgnb{0,m}(X,e) \rightarrow X^m
$$
over a general point $(p_1,\dots,p_m)$
is an algebraic analogue of the space of based $m$-pointed paths to a path
connected topological space $X$
$$
\mc{P}_{p_1,\dots,p_m}  = 
\{(\gamma,t_1,\dots,t_m)| \gamma:[0,1]\rightarrow X \text{ continuous
}, 
$$
$$
0=t_1 < t_2 <
\dots <t_m = 1, \ \gamma(t_1) = p_1,\dots, \gamma(t_m) = p_m\}.
$$
Of course this is homotopy equivalent to the $(m-1)$-fold product of
the space of based paths in $X$.  So if $X$ is simply connected, then
$\mc{P}_{p_1,\dots,p_m}$ is path connected.  

\mni
Unfortunately the
analogue in algebraic geometry is unclear (and quite possibly false):
rational connectedness of the space of
based $m$-pointed rational curves
does not obviously follow from rational connectedness of the space of
based $2$-pointed rational curves.  However, the methods for proving
rational connectedness of the space of based $2$-pointed rational
curves often do prove rational connectedness of the space of based
$m$-pointed rational curves.  

\begin{thm} \label{thm-main2}
\marpar{thm-main2}
Let $X$ be a smooth complete intersection in $\PP^n$ of type
$\underline{d} = (d_1,\dots,d_c)$.  For simplicity, assume all
$d_i\geq 2$.  The variety $X$ is strongly rationally simply connected
if
$$
n+1 \geq \sum_{i=1}^d (2d_i^2 - d_i).
$$
Also linear varieties of dimension $\geq 2$ and
smooth quadric hypersurfaces of dimension $\geq 3$ are rationally
simply connected.  

\mni
To be precise, for every integer $m\geq 2$ and every
integer $e\geq 4m-6$ there is a canonically defined irreducible component
$M_{e,m} \subset \Kgnb{0,m}(X,e)$ such that the restriction
$$
\text{ev}: M_{e,m} \rightarrow X^m
$$
is dominant with rationally connected general fiber. 
\end{thm}

\mni
The irreducible component $M_{e,m}$ above is the unique one dominating the
irreducible component $M_{e,1}$ from Theorem ~\ref{thm-main}, in particular
a general point parametrizes a smooth, embedded curve.
The inequality in Theorem ~\ref{thm-main2} is worse than the
inequality in Theorem ~\ref{thm-main}.  We expect Theorem
~\ref{thm-main2} holds whenever
$$
n \geq \sum_{i=1}^c d_i^2.
$$
This would be the analogue of the theorem above that a smooth
complete intersection is rationally connected when
$$
n \geq \sum_{i=1}^c d_i.
$$
The methods of this paper combined with \cite{Sr1c} prove the following.

\begin{thm} \label{thm-main3}
\marpar{thm-main3}
A \emph{general} degree $d$ hypersurface in $\PP^n$ is strongly
rationally simply connected if 
$$
n\geq d^2, 
$$
i.e., the locus parametrizing strongly rationally simply
connected hypersurfaces is a dense Zariski open subset of the
parameter space of all degree $d$ hypersurfaces.
\end{thm}

\mni
\textbf{Weak approximation.}
There are applications of these theorems to \emph{weak
  approximation} for varieties defined over the function field of a
curve.  
There are equivalent algebraic and geometric formulations of weak
approximation.  

\mni
\textbf{Weak approximation: Algebraic formulation.}  First let $R$ be a
semilocal Dedekind domain with  
maximal ideals $\mf{m}_1,\dots,\mf{m}_n$ and fraction field $K$.
A smooth $K$-algebraic space (or $K$-scheme if you prefer)
$X$ is said to satisfy \emph{weak approximation
with respect to $R$, and $\mf{m}_1,\dots,\mf{m}_n$}
if for every smooth $R$-algebraic space 
$X_R$ with generic fiber $X_K\cong X$,
the image of the induced map of rational points
$$
X_R(R) \rightarrow \prod_{i=1}^n X_R(\widehat{R}_{\mf{m}_i})
$$
is dense for the adic topology.  In other words, for every sequence
$(\widehat{\sigma}_1,\dots,\widehat{\sigma}_n)$ of elements
$\widehat{\sigma}_i \in X_R(\widehat{R}_{\mf{m}_i})$, and for every
integer $N$, there exists an element $\sigma \in X_R(R)$ such that for
every $i=1,\dots,n$,
$$
\sigma \cong \widehat{\sigma}_i (\text{ modulo } \mf{m}_i^N).
$$

\mni
More generally, let $R$ be a Dedekind domain with fraction field $K$,
not necessarily semilocal.
A smooth
$K$-algebraic space $X$ is said to satisfy \emph{weak approximation
with respect to $R$} if it satisfies weak approximation with respect to
every semilocal localization of $R$.  

\mni
\textbf{Weak approximation: Geometric formulation.} 
Let $B$ be a connected, smooth curve over an algebraically
closed field, let $Z$ be a proper closed subset of $B$,
 and let $K$ be the function field of $B$.  A smooth
algebraic space $X$ over $K$ is said to satisfy \emph{weak
  approximation with respect to $B$ and $Z$} if for every smooth morphism
$$
\pi:\mc{X} \rightarrow B
$$
with generic fiber
$\mc{X}\otimes_{\OO_B} K \cong X$
and for every morphism of
$B$-schemes $\sigma_Z:Z\rightarrow \mc{X}$, there exists an open
subset $U$ of $B$ containing $Z$ and a commutative diagram of
$B$-schemes
$$
\begin{CD}
Z @> \sigma_Z >> \mc{X} \\
@V \subset VV  @VV = V \\
U @> \sigma_U >> \mc{X}
\end{CD}
$$
In other words, there exists a $B$-morphism
$\sigma_U:U\rightarrow \mc{X}$ such that $\sigma_U|_Z = \sigma_Z$.

\mni
More generally, let $B$ be a connected, smooth curve over an
algebraically closed field and let $K$ be the function field of $B$.
A smooth algebraic space $X$ over $K$ is said to satisfy \emph{weak
  approximation with respect to $B$} if for every proper closed subset
$Z$ of $B$, it satisfies weak approximation with respect to $B$ and $Z$.

\mni
The connection between the algebraic formulation and geometric formulation
comes by setting $R$ to
be the coordinate ring of any open affine subset of $B$ containing
$Z$.  Although nontrivial, the geometric formulation holds for
arbitrary proper closed subschemes $Z$ if and only if it holds for
\emph{reduced} proper closed subschemes, i.e., finite sets of closed
points of $B$, cf. ~\cite[Proposition 11]{HT06}.

\mni  
\textbf{Weak approximation and rational simple connectedness.}
The strong version of rational simple connectedness is related to weak
approximation by the following beautiful theorem of Hassett.

\begin{thm}~\cite{HassettWA} \label{thm-Hassett}
\marpar{thm-Hassett}
Let $K$ be the function field of a curve over an algebraically closed
field of characteristic $0$.  A smooth, projective $K$-scheme $X$
satisfies weak approximation if the geometric generic fiber
$X\otimes_K \overline{K}$ is strongly rationally connected. 

\mni
More precisely, $X$ satisfies weak approximation if 
for
every sufficiently positive and divisible
integer $m$ there exists a Galois-invariant curve class $\beta$
in $X\otimes_K \overline{K}$ and a Galois-invariant, irreducible,
closed subset
$M_{\beta,m}$ of $\Kgnb{0,m}(X\otimes_K\overline{K},\beta)$
such that some point of $M_{\beta,m}$ parametrizes an irreducible
curve and the restriction to $M_{\beta,m}$ of the evaluation map 
$$
\text{ev}|_M:M_{\beta,m}\subset \Kgnb{0,m}(X\otimes_K
\overline{K},\beta) \rightarrow (X\otimes_K\overline{K})^m
$$
is dominant with rationally connected geometric generic fiber.  
\end{thm}

\begin{cor} \label{cor-main3}
\marpar{cor-main3}
Let $K$ be the function field of a curve over an algebraically closed
field of characteristic $0$.  Let $X\subset \PP^n_K$ be a smooth
complete intersection of type $\underline{d} = (d_1,\dots,d_c)$.  
If
$$
n+1 \geq \sum_{i=1}^c (2d_i^2-d_i)
$$
then $X$ satisfies weak approximation.  

\mni
If
$
n\geq d^2
$
and if $X \subset \PP^n_K$ is a ``general'' hypersurface of degree $d$,
then $X$ satisfies weak approximation.  To be precise, there exists a
Zariski dense open subset $U$ in the parameter space
$\PP^N_{\overline{\QQ}}$, $N=
\binom{n+d}{d}-1$, of degree $d$ hypersurfaces
in $\PP^n$ defined over $\overline{\QQ}$.  If the $K$-valued point of
$\PP^{N}_{\overline{\QQ}}$ associated to $X$ lies in
$U$, then $X$ satisfies weak approximation.
\end{cor}

\mni
At least conjecturally, rational simple connectedness should imply
existence of rational points for varieties defined over the function
field of a surface.  

\begin{prin}[de Jong] \label{prin-deJong}
\marpar{prin-deJong}
A proper, smooth variety defined over the function field of a surface
over an algebraically closed field of characteristic $0$ has a
rational point if the base-change to the algebraic closure of the
function field satisfies a strong version of rational connectedness and
a certain Brauer
obstruction vanishes.
\end{prin}

\mni
In a forthcoming paper, we will present 
de Jong's strategy for proving
Principle~\ref{prin-deJong} and prove 
Principle~\ref{prin-deJong}
under some additional hypotheses. 

\mni
\textbf{Acknowledgments:}  We are grateful to Izzet Coskun, Joe Harris
and Brendan Hassett for useful conversations.

\subsection{Overview} \label{subsec-ov}
\marpar{subsec-ov}

\mni
In fact, the methods of this paper are not particular to complete
intersections, although this is the application we have in mind.  
Theorems ~\ref{thm-main} and ~\ref{thm-main2} are special cases of
the following theorems about Fano manifolds whose Chern classes
satisfy certain inequalities, together with linear sections of these
Fano manifolds.  

\begin{thm} \label{thm-trc1}
\marpar{thm-trc1}
Let $(X,\OO(1))$ be a smooth, projective variety.  Let $\alpha$ be an
$\OO(1)$-degree $1$ curve class.  Assume $c_1(T_X) = \langle c_1(T_X),
\alpha \rangle c_1(\OO(1))$, i.e., $c_1(T_X)$ is a multiple of
$c_1(\OO(1))$.  

\mni
Assume $(X,\OO(1))$ is neither a linear variety nor a
quadric hypersurface.  Assume that
$$
\text{ch}_2(T_X) = \frac{\langle 2\text{ch}_2(T_X),\Pi \rangle}{2}
c_1(\OO(1))^2 
$$
for some integer $\langle 2\text{ch}_2(T_X),\Pi \rangle$, i.e.,
$2\text{ch}_2(T_X)$ is a multiple of $c_1(\OO(1))^2$.  
Assume that
$$
\langle c_1(T_X),\alpha \rangle > 0,
$$
$$
\langle 2\text{ch}_2(T_X),\Pi \rangle \geq 0
$$
and
$$
\text{dim}(X) \geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
-1) + 4.
$$

\mni
Assume there exists a smooth, projective variety $Y$ such that $X$ is
a codimension $c$ linear section of $Y$.  Assume
$$
c \geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle - 1) -
\text{dim}(X) + 5,
$$
$$
c > \text{dim}(X) - 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
- 1) - 2\langle 2\text{ch}_2(T_X),\Pi \rangle + 2,
$$
and 
$$
\text{CH}^p(Y) = \ZZ\{c_1(\OO(1))^p\}, \text{ for } 0\leq p \leq
2(\text{dim}(X) - \langle c_1(T_X),\alpha - 1) + 4.
$$
By Barth's theorems, \cite{Barth70}, this last condition holds if
$$
c \geq 4(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -1) -
\text{dim}(X) + (h^0(X,\OO(1)) - \text{dim}(X) -2) + 9.
$$

\mni
Then, first, there exists a unique irreducible component $M_{\alpha,1}$ of
$\Kgnb{0,1}(X,\alpha)$ dominating $X$, and a general fiber of
$\Kgnb{0,1}(X,\alpha)$ over $X$ is geometrically irreducible and
geometrically rationally connected.
Second, for every integer $e \geq 2$ there is a canonically defined
irreducible component $M_{e\cdot \alpha,2}$ of $\Kgnb{0,2}(X,e
\alpha)$, constructed in
Notation
~\ref{notat-cic2}, for which
the fiber of the restriction
$$
\text{ev}|_M:M_{e\cdot \alpha,2} \subset \Kgnb{0,2}(X,e\alpha) \rightarrow X^2
$$
over a general point of $X^2$
is geometrically irreducible and geometrically rationally connected.

\mni
The same conclusion holds for every linear variety of dimension $\geq
2$
and every smooth quadric variety of
dimension $\geq 3$.
\end{thm}

\begin{thm} \label{thm-trc2}
\marpar{thm-trc2}
Let $(X,\OO(1))$ be a smooth, projective variety.  Let $\alpha$ be an
$\OO(1)$-degree $1$ curve class.  Assume $c_1(T_X) = \langle c_1(T_X),
\alpha \rangle c_1(\OO(1))$, i.e., $c_1(T_X)$ is a multiple of
$c_1(\OO(1))$.  

\mni
Assume $(X,\OO(1))$ is neither a linear variety nor a
quadric hypersurface.  Assume that
$$
\text{ch}_2(T_X) = \frac{\langle 2\text{ch}_2(T_X),\Pi \rangle}{2}
c_1(\OO(1))^2 
$$
for some integer $\langle 2\text{ch}_2(T_X),\Pi \rangle$, i.e.,
$2\text{ch}_2(T_X)$ is a multiple of $c_1(\OO(1))^2$.
Assume that
$$
\langle c_1(T_X),\alpha \rangle > 0,
$$
$$
2\langle 2\text{ch}_2(T_X),\Pi \rangle \geq \langle c_1(T_X),\alpha \rangle
$$
and
$$
\text{dim}(X) \geq 4(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
-1) + 4.
$$

\mni
Assume there exists a smooth, projective variety $Y$ such that $X$ is
a codimension $c$ linear section of $Y$.  Assume
$$
c \geq 4(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle - 1) -
\text{dim}(X) + 4,
$$
$$
c > \text{dim}(X) - 2(2\langle 2\text{ch}_2(T_X),\alpha \rangle -
\langle c_1(T_X),\alpha \rangle) - 4(\text{dim}(X) - \langle
c_1(T_X),\alpha \rangle -1) + 6,
%% \text{dim}(X) - 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
%% - 1) - 2\langle 2\text{ch}_2(T_X),\Pi \rangle + 2,
$$
and 
$$
\text{CH}^p(Y) = \ZZ\{c_1(\OO(1))^p\}, \text{ for } 0\leq p \leq
2(\text{dim}(X) - \langle c_1(T_X),\alpha - 1) + 4.
$$
By Barth's theorems, \cite{Barth70}, this last condition holds if
$$
c \geq 4(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -1) -
\text{dim}(X) + (h^0(X,\OO(1)) - \text{dim}(X) -2) + 9.
$$

\mni
Then, first, there exists a unique irreducible component $M_{\alpha,1}$ of
$\Kgnb{0,1}(X,\alpha)$ dominating $X$, and a general fiber of
$\Kgnb{0,1}(X,\alpha)$ over $X$ is geometrically irreducible and
geometrically rationally connected.
Next, for every integer $m\geq 2$, for every integer $e\geq 4m-6$,
there is a canonically defined irreducible component
$M_{e\cdot\alpha,m}$ of $\Kgnb{0,m}(X,e\alpha)$, constructed in
Notation ~\ref{notat-cic2}, for which the fiber of the restriction
$$
\text{ev}|_M:M_{e\cdot\alpha, m} \subset \Kgnb{0,m}(X,e\alpha)
\rightarrow X^m
$$
over a general point of $X^m$ is geometrically irreducible and
geometrically rationally connected.

\mni
The same conclusion holds for every linear variety of dimension $\geq
2$ and every smooth quadric variety of dimension $\geq 3$.
\end{thm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%
%% Section: Elementary results about complete intersections
%% 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Elementary results about complete 
intersections} \label{sec-eci}
\marpar{sec-eci}

\mni
We gather here some elementary, well-known results about complete
intersections which we use.  The first result we use is the
computation of the Chern character of a complete intersection.

\begin{lem} \label{lem-Chern}
\marpar{lem-Chern}
For a smooth, complete intersection $X$ of type $\underline{d} =
(d_1,\dots,d_c)$ in $\PP^n$, the Chern character equals
$$
\text{ch}(T_X) = n-c +  \sum_{k=1}^{n-c} \lt(n+1 - \sum_{i=1}^c
d_i^k\rt) \frac{c_1(\OO(1))^k}{k!}.
$$
\end{lem}

\begin{proof}
There is a short exact sequence
$$
\begin{CD}
0 @>>> T_X @>>> T_{\PP^n}|_X @>>> \oplus_{i=1}^c \OO_{\PP^n}(d_i)|_X
@>>> 0.
\end{CD}
$$
Therefore, by the axioms of the Chern character,
$$
\text{ch}(T_X) = \text{ch}(T_{\PP^n})_X - \sum_{i=1}^c
e^{d_ic_1(\OO(1))} = \text{ch}(T_{\PP^n}|_X) - c -
\sum_{k=1}^{n-c}\sum_{i=1}^c d_i^k \frac{c_1(\OO(1))^k}{k!}.
$$
Also, by the Euler sequence,
$$
\begin{CD}
0 @>>> \OO_{\PP^n} @>>> \OO_{\PP^n}(1)^{\oplus(n+1)} @>>> T_{\PP^n}
@>>> 0
\end{CD}
$$
and the axioms of the Chern character,
$$
\text{ch}(T_{\PP^n}) = (n+1)e^{c_1(\OO(1))} - 1 = n + \sum_{k=1}^n
(n+1) \frac{c_1(\OO(1))^k}{k!}.
$$
Combined, these formulas give the formula for $\text{ch}(T_X)$.   
\end{proof}

\mni
The second result we use is that every complete intersection is a
linear section of a smooth complete intersection of arbitrarily large
dimension.  

\begin{lem} \label{lem-ext1}
\marpar{lem-ext1}
Let
$Y\subset \PP^N$ 
be a smooth, quasi-projective variety, let $\Pi$ be a linear subspace of
$\PP^N$ intersecting $Y$ transversally in a smooth subvariety $Y_\Pi$,
let $d$ be a positive integer, and let $F$ be a global section
of $\OO_Y(d)$ whose zero locus $Z=\mathbb{V}(F)$ does not contain
$Y_\Pi$.  Denote the ideal sheaf of $\Pi$ in $\PP^N$ by $\mc{I}_\Pi$.

\mni 
If 
$$
\text{codim}(\Pi\subset \PP^N) \geq \text{dim}(\text{Sing}(\Pi\cap
Z))+ 1, 
$$
then there exists a global section $G$ of $\mc{I}_\Pi\cdot
\OO_{\PP^N}(d)$ such that the zero locus $Z'=\mathbb{V}(F +
G|_Y)$ is smooth.  Thus $Z'$ is a smooth Cartier divisor on $Y$
such that $Z'\cap \Pi = Z\cap \Pi$.
\end{lem}

\begin{proof}
The base locus of the
linear subsystem $L$ of $|\OO_Y(d)|$ spanned by $F$ and the image
of $|\mc{I}_\Pi\cdot \OO_{\PP^N}(d)|$ is contained in $Z\cap \Pi$.  
By Bertini's theorem, cf. ~\cite[Th\'eor\`eme I.6.3.2]{Jou}, a 
general member $Z'$ of $L$ is smooth away from $Z\cap \Pi$.
Thus to prove a general member is everywhere smooth 
it suffices to prove there exists at least one member that is smooth
at every point of $Z\cap \Pi$.  

\mni
Denote the ``bad locus'' $\text{Sing}(Z\cap \Pi)$ by $B$.  
Denote by $a$ the codimension $\text{codim}(Y_\Pi\subset Y)$.  Denote by
$b$ the dimension $\text{dim}(B)$.  Choose homogeneous
coordinates $[X_0,\dots,X_N]$ on $\PP^N$
so that  
$$
\Pi = \mathbb{V}(X_{N+1-a},\dots, X_N).
$$
Because $a>b$, there exist elements $G_{N+1-a},\dots,G_N\in
H^0(\PP^N,\OO(d-1))$ so that 
$$
B \cap \mathbb{V}(G_{N+1-a},\dots,G_N) = \emptyset.
$$
For each
scalar $\lambda$,
denote
$$
F_\lambda := F + \lambda\sum_{i=1}^a X_{N-a+i} G_{N-a+i}
$$
and denote the zero locus of $F_\lambda$ by $Z_\lambda$.  
The claim is that $Z_\lambda$ is smooth at every point of $Z\cap \Pi$ for a
general choice of the scalar $\lambda$.

\mni
Associated to $F_\lambda$ there is a derivative map
$$
dF_\lambda:T_Y|_{Z_\lambda} \rightarrow \OO_Y(d)|_{Z_\lambda}.
$$
The singular locus of $Z_\lambda$ is the support of the cokernel of
$dF_\lambda$.  Thus $Z_\lambda$ is smooth at every point of $Z\cap \Pi$ if
and only if the induced map
$$
dF_\lambda:T_Y|_{Z\cap\Pi} \rightarrow \OO_Y(d)|_{Z\cap\Pi}
$$
is surjective.  The restriction to the subsheaf $T_{Y_\Pi}|_{Z\cap\Pi}$ is
precisely
$$
d(F|_{Y_\Pi}):T_{Y_\Pi}|_{Z\cap\Pi} \rightarrow \OO_{Y_\Pi}(d)|_{Z\cap\Pi}.
$$
Denote the cokernel by
$$
C := \text{Coker}(d(F|_{Y_\Pi})).
$$
This is a locally principal coherent sheaf supported on $B$.  There is
an induced map
$$
\widetilde{dF}_\lambda:N_{Y_\Pi/Y}|_{Z\cap\Pi} \rightarrow C.
$$

\mni
The map $\widetilde{dF}_\lambda$
is \emph{linear} in $\lambda$. To be precise there exists a map
$$
\widetilde{G}:N_{Y_\Pi/Y}|_{Z_\Pi} \rightarrow C
$$
such that
$$
\widetilde{dF}_\lambda = \widetilde{dF}_0 + \lambda \widetilde{G}.
$$
The sequence $(X_{N+1-a},\dots,X_N)$ is an ordered basis for
$\mc{I}_\Pi(1)$, i.e., the induced map $\OO^{\oplus a} \rightarrow
\mc{I}_\Pi(1)$ is an isomorphism.  Transposing, twisting by $\OO(1)$
and restricting to $Y$ gives an isomorphism 
$N_{Y_\Pi/Y} \cong \OO_{Y_\Pi}(1)^{\oplus a}$. 
Inverting the isomorphism and composing with $\widetilde{G}$ gives the map
$$
\widehat{G}:\OO_{Y_\Pi}(1)^{\oplus a}
\rightarrow \OO_{Y_\Pi}(d)
$$
whose restriction to the $i^\text{th}$ summand $\OO_{Y_\Pi}(1)$ of
$\OO_{Y_\Pi}(1)^{\oplus a}$ is the restriction to $Y_\Pi$ of the
multiplication map
$$
G_{N-a+i}:\OO(1) \rightarrow \OO(d).
$$

\mni
For scalars $\lambda, \mu$, consider the map
$$
\mu\widetilde{dF} +\lambda \widetilde{G}:N_{Y_\Pi/Y}|_{Z_\Pi} \rightarrow
C.
$$
For $\mu=0$ and $\lambda = 1$, the map is surjective because 
$$
B\cap \mathbb{V}(G_{N+1-a},\dots,G_N) = \emptyset.
$$
Therefore, by upper-semicontinuity, the map is surjective for general
choice of $(\lambda,\mu)$.  In particular, there exists $\mu\neq 0$
such that the map is surjective.  But then the map is also surjective
for $(\lambda/\mu,1)$.  Thus $Z_{\lambda/\mu}$ is smooth at every
point of $Z_\Pi$.  
\end{proof}

\begin{defn} \label{defn-ext1}
\marpar{notat-ext1}
Let $P$ be a pure-dimensional variety.  A sequence $Y_1,\dots, Y_r$
of locally closed,
pure-dimensional subvarieties
intersect \emph{dimensionally transversally} if either
$Y_1\cap \dots \cap Y_r$ is empty or else it is
pure-dimensional of dimension
$$
\text{dim}(Y_1\cap dots \cap Y_r) =
\text{dim}(Y_1)+\dots +\text{dim}(Y_r)-(r-1)\text{dim}(P).
$$
\end{defn}

\begin{lem} \label{lem-ext2}
\marpar{lem-ext2}
Let $(Y,\Pi,r,\underline{d},F_1,\dots,F_r)$ be a datum of a smooth,
quasi-projective variety
$Y\subset \PP^N$, a linear subspace $\Pi$ 
of $\PP^N$ intersecting $Y$ transversally in a smooth
subvariety $Y_\Pi$, a positive integer $r$,
a sequence of positive integers 
$\underline{d}=(d_1,\dots,d_r)$, 
and for $i=1,\dots,r$, a global section $F_i$ of $\OO_Y(d_i)$
with zero locus $Y_i = \mathbb{V}(F_i)$
such that
$Y_1,\dots, Y_r$ and $\Pi$ intersect dimensionally transversally.
Denote the ideal sheaf of $\Pi$ in $\PP^N$ by $\mc{I}_\Pi$.

\mni
If
$$
\text{codim}(\Pi\subset \PP^N) \geq (2^r-1)\text{dim}(Y_0\cap \Pi) -
2^r + r + 1
$$
then for $i=1,\dots,r$ 
there exist global sections $G_i$ of $\mc{I}_\Pi\cdot
\OO_{\PP^N}(d_i)$ giving Cartier divisors $Y'_i = \mathbb{V}(F_i +
G_i|_Y)$ such that for every
$k=1,\dots,r$ the intersection $Y'_1\cap \dots \cap Y'_k$ is smooth.
\end{lem}

\begin{proof}
This is proved by induction on $r$.  The base case, $r=1$ is Lemma
~\ref{lem-ext1}.  Thus, by way of induction, assume $r>1$ and assume
the result is known for $r-1$.  By Bertini's theorem,
cf. \cite[Th\'eor\`eme I.6.3]{Jou}, for
a general linear space $\Lambda$ of dimension
$$
\text{dim}(\Lambda) = \text{dim}(\Pi) + \text{dim}(Y\cap \Pi)
$$
containing $\Pi$, the intersection $Y\cap \Lambda$ is transverse and
smooth.  The singular locus of $Y_1\cap \Pi$ has dimension 
$$
\text{dim}(\text{Sing}(Y_1\cap \Pi)) \leq
\text{dim}(Y_1\cap \Pi) =\text{dim}(Y\cap \Pi) - 1.
$$
Thus
$$
\text{codim}(\Pi\subset \Lambda) > \text{dim}(\text{Sing}(\Pi\cap Y_1)).
$$
Thus
Lemma ~\ref{lem-ext1} implies there exists a global section $G_1$ of
$\mc{I}_\Pi \cdot \OO_{\PP^N}(d)$ such that the zero locus
$Y'_{1,\Lambda} = \mathbb{V}(F_1|_\Lambda + G_1|_\Lambda)$ is smooth.  
Since $Y'_{1,\Lambda}$ is
smooth, a second application of Lemma ~\ref{lem-ext1} implies there
exists $G$ as above such that also $Y'_1 = \mathbb{V}(F_1+G_1|_Y)$ is
smooth. 

\mni
For each of $i=2,\dots,r$, after replacing $F_i$ by $F_i + G_i|_Y$ for a
general member $G_i$ of $\mc{I}_\Pi\cdot \OO_{\PP^N}(d_i)$, $Y_1,
\dots, Y_r$ and $\Lambda$ intersect dimensionally transversally.

\mni
Now form the new datum $(Z,\Lambda,r-1, \underline{e},H_1,\dots,H_{r-1})$
where
$Z=Y'_1$,
$\underline{e} = (d_2,\dots,d_r)$, and for $i=1,\dots,r-1$, $H_i =
F_{i+1}|_Z$.
For each $i=1,\dots,r-1$, denote $Z_i = \mathbb{V}(H_i) = Z'_1\cap
Y_{i+1}$.     
The claim is that this datum also satisfies the hypotheses of the lemma.
One hypothesis that requires verification is:
$$
\text{dim}(\PP^N)-\text{dim}(\Lambda) \geq
(2^{r-1}-1)\text{dim}(Z\cap \Lambda) - 2^{r-1} + r.
$$

\mni
First observe that
$$
\text{dim}(Y_1\cap \Lambda) = \text{dim}(Y\cap \Lambda) - 1 = $$
$$
(\text{dim}(Y\cap \Pi) + \text{dim}(\Lambda) - \text{dim}(\Pi)) - 1
= 2\text{dim}(Y\cap \Pi) - 1.
$$
By construction of $\Lambda$ and the hypothesis on $\text{dim}(\Pi)$,
$$
\text{dim}(\PP^N)-\text{dim}(\Lambda) =
\text{dim}(\PP^N)-\text{dim}(\Pi) - \text{dim}(Y\cap \Pi) \geq
$$
$$
[(2^r-1)\text{dim}(Y\cap \Pi) - 2^r + r + 1] -
\text{dim}(Y\cap \Pi) =
$$
$$
(2^r-2)\text{dim}(Y\cap \Pi) - 2^r + r + 1 =
(2^{r-1}-1)(2\text{dim}(Y\cap \Pi)) - 2^r + r +1.
$$
Using the identity above, this is
$$
(2^{r-1}-1)\text{dim}(Y_1\cap \Lambda) + 2^{r-1}-1 - 2^r +r + 1.
$$
Using the elementary identity that
$$
2^{r-1} -1 -2^r +r +1 = -(2^r-2^{r-1}) + r = -2^{r-1} + r, 
$$
it follows that
$$
\text{dim}(\PP^N) - \text{dim}(\Lambda) \geq
(2^{r-1}-1)\text{dim}(Z\cap \Lambda) - 2^{r-1}+r.
$$
So the datum satisfies the hypotheses of the lemma.

\mni
By the induction assumption, for each $i=1,\dots,r-1$, there exist
elements $J_i$ of $\mc{I}_\Lambda \cdot \OO_{\PP^N}(d_{i+1})$ giving
Cartier divisors $Z'_i = \mathbb{V}(H_i + J_i|_Z)$ 
such that for every $k=1,\dots, r-1$, the
intersection
$Z'_1\cap \dots \cap Z'_k$ is smooth.  In other words, setting $G_{i+1} =
J_i$, 
for the Cartier
divisors $Y'_{i+1} = \mathbb{V}(F_{i+1} + G_{i+1}|_Y)$, 
$$
Y'_1\cap Y'_2 \cap \dots \cap Y'_{k+1} = Z'_1 \cap \dots \cap Z'_k
$$
is smooth.
Thus $(G_1,G_2,\dots,G_r)$ satisfy the conditions of the lemma.
Therefore the lemma is proved by induction on $r$.
\end{proof}

\begin{lem} \label{lem-ext3}
\marginpar{lem-ext3}
Let $X$ be a complete intersection, not necessarily smooth, in $\PP^n$
of type $\underline{d} = (d_1,\dots,d_r)$.  For every integer
$$
c \geq (2^r-1)n-2^r + r +1,
$$
for every linear embedding $\PP^n\subset \PP^{n+c}$, there exists a
smooth complete intersection $X'$ in $\PP^{n+c}$ of type
$\underline{d}$ such that $X=\PP^n\cap X'$.
\end{lem}

\begin{proof}
This follows from Lemma ~\ref{lem-ext2} by setting $\PP^N=\PP^{n+c}$,
$\Pi = \PP^n$, $Y = \PP^N$, and, for $i=1,\dots,r$, letting $F_i$ be a
general section of $\mc{I}_X\cdot \OO_{\PP^N}(d_i)$.  
\end{proof}

\begin{rmk} \label{rmk-ext3}
\marpar{rmk-ext3}
The inequality in Lemma ~\ref{lem-ext3} is certainly not the best
possible, but it suffices for our purposes.
\end{rmk}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%
%% Section: Canonical irreducible components
%% 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Canonical irreducible components} \label{sec-cic}
\marpar{sec-cic}

\mni
The space of rational curves of a given class in a rationally
connected variety may be reducible.  However, often there is a
distinguished irreducible component.  In this section we present a
construction associating to a distinguished component for one curve
class $\alpha$ and a positive integer $e$, a
distinguished component for the curve class $e\alpha$.  In fact, there
are two constructions: the first uses multiple covers and the second
uses reducible curves. 
The main result of this section, Lemma ~\ref{lem-cic1}, proves these
constructions agree under a suitable hypothesis.

\begin{lem}\cite[Theorem II.7.6]{K} \label{lem-cic0.1}
\marpar{lem-cic0.1}
Let $f:C\rightarrow X$ be a stable map such that every component of
$C$ is contracted or free.  Then $\Kgnb{0,0}(X)$ is smooth at $[(C,f)]$,
and there exist deformations of $(C,f)$ smoothing all the nodes of
$C$.  A general such deformation is free.
\end{lem}

\mni
The following criterion from \cite[\S 4]{EGA4} will be useful.

\begin{lem} \label{lem-cic0.15}
Let
$$
i:N\rightarrow M, \ \ \ e:M\rightarrow X
$$
be morphisms of irreducible schemes.  If $i$ maps the generic point of
$N$ to a normal point of $M$, and if $e\circ i$
is dominant with irreducible geometric generic fiber, then also $e$ is
dominant with irreducible geometric generic fiber.
\end{lem}

\begin{proof}
If $e\circ i$ is dominant, clearly $e$ is dominant.  By
\cite[Proposition 4.5.9]{EGA4}, to prove the geometric generic fiber
of $e$ is irreducible, it is equivalent to prove the field extension
$K(M)/K(X)$ is a \emph{primary} extension, i.e., $K(X)$ is separably
closed in $K(M)$.  
Denote by $\eta_i$ the image under $i$ of the generic point of $N$.
Because the $K(X)$-algebra $\OO_{M,\eta_i}$ is normal, 
the separable closure of $K(X)$
in $K(M)$ equals the separable closure of $K(X)$ in $\OO_{M,\eta_i}$.
Because the separable closure 
is a field, it maps isomorphically to its image in the
residue field $\kappa(\eta_i)$ of $\OO_{M,\eta_i}$.  Thus the
separable closure of $K(X)$ in $K(M)$ is contained in 
the separable closure of
$K(X)$ in $\kappa(\eta_i)$.  Of course $\kappa(\eta_i)/K(X)$ is a
subextension of $K(N)/K(X)$.  Because the geometric generic fiber of
$e\circ i$ is irreducible, $K(X)$ is separably closed in $K(N)$, thus
separably closed in $\kappa(\eta_i)$.  Therefore also $K(X)$ is separably
closed in $K(M)$.
\end{proof}

\mni
The first construction of distinguished irreducible components uses
multiple covers.

\begin{lem} \label{lem-cic0.2}
\marpar{lem-cic0.2}
Let $M_{\alpha,0}$ be an irreducible component of
$\Kgnb{0,0}(X,\alpha)$
whose general point parametrizes a smooth, free curve.  For every
positive integer $e$ there exists a unique irreducible component
$M_{e\cdot \alpha,0}$ of $\Kgnb{0,0}(X,e\alpha)$ 
parametrizing a degree $e$ cover of a smooth, free curve parametrized by
$M_{\alpha, 0}$.  

\mni
For every
positive integer $e$ denote by $M_{e\cdot \alpha,1}$ the unique
irreducible component of $\Kgnb{0,1}(X,e\alpha)$ dominating
$M_{e\cdot\alpha,0}$.  If the geometric generic fiber of
$$
\text{ev}|_M:M_{\alpha,1} \rightarrow X
$$
is irreducible, then for every $e$ the geometric generic fiber of 
$$
\text{ev}|_M:M_{e\cdot\alpha,1} \rightarrow X
$$
is irreducible.
\end{lem}

\begin{proof}
The subvariety $S$ of $\Kgnb{0,0}(X,e\alpha)$ parametrizing degree $e$
covers of smooth, free curves parametrized by $M_{\alpha,0}$ fibers
over $M_{\alpha,0}$ with fibers isomorphic to $\Kgnb{0,0}(\PP^1,e)$.  
Because $\Kgnb{0,0}(\PP^1,e)$ is smooth and geometrically connected,
$S$ is irreducible.  Moreover, because a multiple cover of a free
curve satisfies the hypothesis of Lemma ~\ref{lem-cic0.1},
$\Kgnb{0,0}(X,e\alpha)$ is smooth at every point of $S$.  Therefore
there exists a unique irreducible component $M_{e\cdot \alpha,0}$ that
intersects $S$.

\mni
The proof of the second part uses Lemma ~\ref{lem-cic0.15}.  The role
of $e$ is played by
$$
\text{ev}|_M:M_{e\cdot\alpha,1} \rightarrow X.
$$
The role of $i:N\rightarrow M$ is played by the inverse image $S_1$ of
$S$ in $M_{e\cdot\alpha,1}$.  By the argument above, $S_1$ is in the
normal locus of $M_{e\cdot\alpha,1}$.  By Lemma ~\ref{lem-cic0.15}, to
prove the geometric generic fiber of $e$ is irreducible, it suffices
to prove that $i\circ e$ is dominant with irreducible geometric
generic fiber.

\mni
The morphism $e\circ i$ factors
as
$$
S_1 \rightarrow M_{\alpha,1} \rightarrow X
$$
where the first morphism ``forgets'' about the multiple covering.  The
first morphism is dominant and its geometric generic fiber is
isomorphic to a base change of the geometric generic fiber of
$$
\text{ev}:\Kgnb{0,1}(\PP^1,e) \rightarrow \PP^1,
$$
which is irreducible.  Thus both 
$$
S\rightarrow M_{\alpha,1} \text{ and } M_{\alpha,1} \rightarrow X
$$
are dominant morphisms whose geometric generic fiber is irreducible.
Therefore the composite $e\circ i$ is also 
dominant with irreducible geometric generic fiber.
\end{proof}

\mni
The second construction of distinguished irreducible components uses
reducible curves.

\begin{lem} \label{lem-cic0.3}
\marpar{lem-cic0.3}
For $i=1,2$, let $M_i$ be an irreducible component of
$\Kgnb{0,1}(X,\beta_i)$ whose general point parametrizes a smooth,
free curve.  Assume that for at least one of $i=1,2$, the geometric
generic fiber of
$$
\text{ev}|_{M_i}:M_i \subset \Kgnb{0,1}(X,\beta_i) \rightarrow X
$$
is irreducible.
Then there exists a unique irreducible component $M$ of
$\Kgnb{0,0}(X,\beta_1+\beta_2)$ parametrizing a reducible curve
$C=C_1\cup C_2$ whose component $C_i$ is a smooth, free curve
parametrized by $M_i$.  Moreover a general point of $M$ parametrizes a
smooth, free curve.  And, if $\text{ev}_{M_i}$ has geometrically
irreducible generic fiber for both $i=1$ and $2$, then for the unique
irreducible component $M'$ of $\Kgnb{0,1}(X,\beta_1+\beta_2)$
dominating $M$, the restriction
$$
\text{ev}_{M'}:M'\subset \Kgnb{0,1}(X,\beta_1+\beta_2) \rightarrow X
$$
has geometrically irreducible generic fiber.
\end{lem}

\begin{proof}
Denote by $M_i^\circ$ the open subset of $M_i$ parametrizing smooth,
free curves.  Then, by \cite[Corollary II.3.5.4.2]{K}, the evaluation
morphism
$$
\text{ev}|_{M_i^\circ}:M_i^\circ \subset \Kgnb{0,1}(X,\beta_i)
\rightarrow X
$$
is smooth.  Therefore the fiber product
$$
F := M_1^\circ \times_X M_2^\circ
$$
is smooth, and projection onto each factor is smooth.  Moreover, the
geometric generic fiber of projection onto $M_2^\circ$, resp. onto
$M_1^\circ$, equals the geometric generic fiber of 
$$
\text{ev}|_{M_i^\circ}: M_i^\circ \rightarrow X.
$$
By hypothesis, at lease one of the geometric generic fibers is
irreducible.  Therefore the geometric generic fiber of one of the two
projections of $F$ is irreducible.  Since the base of each projection is
irreducible, it follows that $F$ is irreducible.  

\mni
There is a boundary morphism
$$
\Delta:\Kgnb{0,1}(X,\beta_1)\times_X \Kgnb{0,1}(X,\beta_2) \rightarrow
\Kgnb{0,0}(X,\beta_1+\beta_2)
$$
associating to a pair $((C_1,p_1,f_1),(C_2,p_2,f_2))$ with
$f_1(p_1)=f_2(p_2)$ the reducible curve $C=C_1\cup C_2$ obtained by
gluing $p_1$ to $p_2$, and the unique morphism $f:C\rightarrow X$
whose restriction to $C_i$ equals $f_i$ for each $i$.  
As $F$ is irreducible, also 
the image
$\Delta(F)$ is also irreducible.  By Lemma ~\ref{lem-cic0.1},
$\Kgnb{0,0}(X,\beta_1+\beta_2)$ is smooth at every point of
$\Delta(F)$.  Therefore there exists a unique irreducible component
$M$ of $\Kgnb{0,0}(X,\beta_1+\beta_2)$ intersecting $\Delta(F)$.  By
Lemma ~\ref{lem-cic0.1}, the generic point of $M$ parametrizes a
smooth, free curve.

\mni
Next assume that $\text{ev}|_{M_i}$ has geometrically irreducible
generic fiber for each of $i=1,2$.  Denote by $(M_1^o)'$ the open subset
of $\Kgnb{0,2}(X,\beta_1)$ parametrizing smooth,
$2$-pointed curves whose associated $1$-pointed curve (forgetting the
first point) is a free curve parametrized by $M_1$.  The projection
$$
(M_1^o)' \rightarrow M_1^\circ, \ \ [(C_1,p_0,p_1,f_1)] \mapsto [(C_1,p_1,f_1)]
$$
is smooth with geometrically irreducible
fibers.  Therefore also the fiber product
$$
F' := (M_1^o)'\times_{M_1^\circ} F = 
$$
$$
\{([(C_1,p_0,p_1,f_1)],[(C_2,p_2,f_2)])| [(C_i,p_i,f_i)] \in M_i^\circ
\text{ for } i=1,2, \text{ and } f_1(p_1) = f_2(p_2) \}
$$
is smooth and irreducible.  There is a boundary map
$$
\Delta:\Kgnb{0,2}(X,\beta_1)\times_X \Kgnb{0,1}(X,\beta_2) \rightarrow
\Kgnb{0,1}(X,\beta_1+\beta_2), 
$$
$$
([(C_1,p_0,p_1,f_1)],[(C_2,p_2,f_2)]) \mapsto [(C ,p_0,f)]
$$
where $C$ and $f$ are as above.  Again by Lemma ~\ref{lem-cic0.1},
$\Kgnb{0,1}(X,\beta_1+\beta_2)$ is smooth at every point of
$\Delta(F')$, so there is a unique irreducible component $M'$
intersecting $\Delta(F')$.  By Lemma ~\ref{lem-cic0.1}, the generic
point of $M'$ parametrizes a smooth curve.  Thus the image of $M'$ in
$\Kgnb{0,0}(X,\beta_1+\beta_2)$ is an irreducible component
intersecting $\Delta(F)$, i.e., it is precisely $M$.  

\mni
To prove the restriction
$$
\text{ev}_{M'}:M'\subset \Kgnb{0,1}(X,\beta_1+\beta_2) \rightarrow X
$$
has geometrically irreducible generic fiber, we use Lemma
~\ref{lem-cic0.15} with $\text{ev}_{M'}$ in the role of $e$, and with
$$
\Delta:F'\rightarrow M'
$$
in the role of $i$.  Thus it suffices to prove that the geometric
generic fiber of
$$
\text{ev}\circ \Delta:F'\rightarrow X
$$
is irreducible.
By
construction, this
is a composition of the dominant
morphisms
$$
F' \rightarrow (M_1^o)' \rightarrow M_1^o \rightarrow X.
$$
Each of these has irreducible geometric generic fibers by the hypotheses
that $\text{ev}|_{M_i}$ have geometrically irreducible fibers.  Thus
the composition also has irreducible geometric generic fibers.
\end{proof}

\mni
The following lemma shows that in good
cases the constructions from Lemmas ~\ref{lem-cic0.2} and \ref{lem-cic0.3}
agree.

\begin{lem} \label{lem-cic1}
\marpar{lem-cic1}
Let $M_{\alpha,0}$ be an irreducible component of
$\Kgnb{0,0}(X,\alpha)$ whose general point parametrizes a smooth, free
curve.  Denote by $M_{\alpha,1}$ the unique irreducible component of
$\Kgnb{0,1}(X,\alpha)$ dominating $M_{\alpha,0}$.  Assume that the
generic fiber of the restriction
$$
\text{ev}|_{M}:M_{\alpha,1} \rightarrow X
$$
is geometrically irreducible.

\mni
For every positive integer $e$ there is a unique irreducible component
$M_{e\cdot \alpha,0}$ of $\Kgnb{0,0}(X,e\alpha)$ parametrizing
(among others) a reducible curve whose (non-contracted) components are
all multiple covers of free curves parametrized by $M_{\alpha,0}$.

\mni
A general point of $M_{e\cdot \alpha,0}$
parametrizes a smooth, free curve.  Denoting by $M_{e\cdot
  \alpha,1}$ the unique irreducible component of
$\Kgnb{0,1}(X,e\alpha)$ dominating $M_{e\cdot \alpha,0}$, the
restriction 
$$
\text{ev}|_{M}:M_{e\cdot \alpha,1} \rightarrow X
$$
is dominant with irreducible geometric generic fiber.  
  
% Then for every
% integer $e$ there is an associated irreducible component $M_{e\cdot
% \alpha,1}$ of $\Kgnb{0,1}(X,\alpha)$ whose general point is a smooth
% rational curve and such that
% $$
% \text{ev}|_{M}:M_{e\cdot \alpha,1} \rightarrow X
% $$
% is dominant with irreducible generic fiber.
\end{lem}

\begin{proof}
The very last part is just the restatement of the second half of Lemma
~\ref{lem-cic0.2}.  

\mni
Let $(D,g)$ be a stable map all of whose non-contracted components $D_i$ are
multiple covers of free curves $C_i$ parametrized by $M_{\alpha,0}$.  By
Lemma ~\ref{lem-cic0.1}, $\Kgnb{0,0}(X)$ is smooth at $[(D,g)]$.  In
particular, there is a unique irreducible component $M$ containing
$[(D,g)]$.  Let $e$ be the degree of $D_i$ over $C_i$.  
Since the space of multiple covers $\Kgnb{0,r}(\PP^1,e)$ is
irreducible, we can deform $D_i$ to become a reducible cover of $C_i$
all of whose non-contracted components map isomorphically to $C_i$.
So there is a deformation of   
$(D,g)$ in $M$ which specializes to a stable map $(C,f)$ all of whose
non-contracted components are equal to free curves $C_i$ parametrized
by $M_{\alpha,0}$.  Therefore it suffices to prove there exists a
unique irreducible component $M$ parametrizing a stable map $(C,f)$
all of whose non-contracted components are free curves parametrized by
$M_{\alpha,0}$.  

\mni
For each integer $e$, if
$M_{e\cdot \alpha,1}$ is an irreducible component satisfying the
hypotheses, then the image
$M_{e\cdot \alpha,0}$ of $M_{e\cdot\alpha,1}$ under the
forgetful morphism
$$
\Kgnb{0,1}(X,e\alpha) \rightarrow \Kgnb{0,0}(X,e\alpha)
$$
is an irreducible component of $\Kgnb{0,0}(X,e\alpha)$, and
$M_{e\cdot\alpha,1}$ is the unique component of
$\Kgnb{0,1}(X,e\alpha)$ 
dominating $M_{e\cdot \alpha,0}$.  

\mni
The lemma is established by induction on $e$.  The base case $e=1$ is
tautological.  
Thus, by way of induction, assume $e>1$ and assume the lemma is true
for $e-1$.  
The irreducible components $M_{\alpha,1}$ and $M_{(e-1)\cdot
  \alpha,1}$ satisfy the hypotheses for $M_1$ and $M_2$ in Lemma
~\ref{lem-cic0.3}.  Therefore there exists a unique irreducible
component $M_{e\cdot\alpha,0}$ of $\Kgnb{0,0}(X,e\alpha)$
parametrizing a reducible curve $C=C_1\cup C_2$ whose component $C_1$,
resp. $C_2$,
is a smooth, free curve parametrized by $M_{\alpha,1}$,
resp. $M_{(e-1)\cdot\alpha, 1}$.  Moreover a general point of
$M_{e\cdot \alpha,0}$ parametrizes a smooth,free curve and the
generic fiber of $M_{e\cdot\alpha,1}$ over $X$ is geometrically
irreducible.  By the induction hypothesis, there is a reducible curve
$C_2$ parametrized by  $M_{(e-1)\cdot\alpha,1}$, all of whose
non-contracted components are free curves parametrized by
$M_{\alpha,0}$.  Because they are free curves, for a general
deformation of $C_2$ as above, the marked point is a general point of
$X$.  Because $M_{\alpha,1}$ dominates $X$, there exists a smooth,
free curve $C_1$ parametrized by  $M_{\alpha,1}$ whose marked point
agrees with the  marked point of $C_2$.  Thus, by the construction of
$M_{e\cdot \alpha,0}$,  $C=C_1\cup C_2$ is a
point of $M_{e\cdot \alpha,0}$, and all of its non-contracted
components are free curves parametrized by $M_{\alpha,0}$.

\mni
It remains to prove that $M_{e\cdot\alpha,0}$ is the \emph{only} 
irreducible component of $\Kgnb{0,0}(X,e\alpha)$ parametrizing a curve
all of whose non-contracted components are free curves parametrized by
$M_{\alpha,0}$.  Let $C$ be such a curve.  There exists an irreducible
component $C_1$ of $C$ meeting the remainder $C_2$ of $C$ in a single
node.  Necessarily $C_1$ is not contracted, since it does not contain
$3$ special points.  Therefore $C_1$, marked by the node, is a free
curve in $M_{\alpha,1}$.  Moreover, $C_2$ is a curve parametrized by
$\Kgnb{0,0}(X,(e-1)\alpha)$ all of whose non-contracted components are
free curves parametrized by $M_{\alpha,0}$.  By the induction
hypothesis, $C_2$, marked by the node, is parametrized by
$M_{(e-1)\alpha,1}$.  Thus $C$ is in the image
$\Delta(M_{\alpha,1}\times_X  M_{(e-1)\alpha,1})$.  By the
construction in the proof of Lemma ~\ref{lem-cic0.3}, $C$ is contained
in $M_{e\cdot\alpha,0}$.  By Lemma ~\ref{lem-cic0.1},
$\Kgnb{0,0}(X,e\alpha)$ is smooth at $[C]$, therefore
$M_{e\cdot\alpha,0}$ is the unique irreducible component containing $C$.
\end{proof}

\mni
Here is a useful ``recognition criterion'' for the irreducible
component $M_{e\cdot \alpha,0}$.

\begin{lem} \label{lem-cic1.5}
\marpar{lem-cic1.5}
Let $C$ be a smooth point of $\Kgnb{0,0}(X,e\alpha)$.  Assume
that every non-contracted component $C_i$ of $C$ is in one of the sets
$M_{e_i\cdot\alpha,0}$.  Moreover, assume that at most one of the $C_i$
is not a free curve.  Then $M_{e\cdot \alpha,0}$ is the unique
irreducible component of $\Kgnb{0,0}(X,e\alpha)$ containing $C$.  
\end{lem}

\begin{proof}
Let $C_i$ be an irreducible component of $C$ such that for every other
component $C_j\neq C_i$, $C_j$ is free.  Then, for every deformation
of $C_i$, there exists a deformation of the remaining irreducible
components $C_j$ giving rise to a deformation of $C$.  Because $C$ is
a smooth point of $\Kgnb{0,0}(X,e\alpha)$, the unique irreducible
component containing $C$ also contains this deformation.  Because
$C_i$ is in one of the sets $M_{e_i\cdot\alpha,0}$, a general
deformation of $C_i$ is a free curve.  Thus the unique irreducible
component containing $C$ also contains a curve $C^\text{new}$ as above
all of whose components are free.  
Thus, without
loss of generality, assume $C$ has been deformed so that every
irreducible component of $C$ is a free curve parametrized by some
$M_{e'\cdot\alpha}$.  

\mni
The proof is by induction on the number of components of $C$.  If $C$
is irreducible, then the result is tautological.  Thus assume the
number of components $c$ is positive, and the result is known for all
smaller values of $c$.  Let $C_i$ be a leaf of $C$ and let $C'$ denote
the union of all components $C_j\neq C_i$.  Mark each of $C_i$ and
$C'$ by their intersection point $p$.
By the induction
hypothesis, the unique irreducible component of
$\Kgnb{0,0}(X,(e-e_i)\cdot\alpha)$ containing $C'$ is
$M_{(e-e_i)\cdot\alpha,0}$.  Thus the curve $C$ is in the image of the
boundary map
$$
\Delta:M_{e_i\cdot\alpha,1}\times_X M_{(e-e_i)\cdot\alpha,1}
\rightarrow \Kgnb{0,0}(X,e\cdot \alpha).
$$
Moreover, $C$ is in the unique component $M'$ dominating
$M_{(e-e_i)\cdot\alpha,1}$.  By the same argument as in the previous
proof, the unique component of $\Kgnb{0,0}(X,e\cdot\alpha)$ containing
the image of $M'$ contains a reducible curve, every (non-contracted)
component of which is a smooth, free curve in $M_{\alpha,0}$.
Therefore, by Lemma ~\ref{lem-cic1}, the unique component containing
the image of $M'$ is $M_{e\cdot\alpha,0}$.
\end{proof}

\mni
It is useful to extend this to $m$-pointed maps. 

\begin{notat} \label{notat-cic2}
\marpar{notat-cic2}
For every integer $e\geq 1$ and every integer $m\geq 0$, denote by
$M_{e\cdot \alpha,m}$ the unique irreducible component of
$\Kgnb{0,m}(X,e\cdot\alpha)$ whose image under the forgetful morphism
$$
\Kgnb{0,m}(X,e\cdot \alpha) \rightarrow \Kgnb{0,0}(X,e\cdot \alpha)
$$
equals $M_{e\cdot\alpha,0}$.
\end{notat}

\mni
Finally, one can consider Behrend-Manin stable maps whose components
are free curves parametrized by some $M_{e\cdot\alpha,m}$.

\begin{notat} \label{notat-cic3}
\marpar{notat-cic3}
Let $\tau$ be a genus $0$, stable $A$-graph for $X$
in the sense of \cite{BM}, all of whose curve classes are multiples of
$\alpha$.  Denote by $M_\tau$ the closure in the Behrend-Manin stack
$\M(X,\tau)$ of the locus
parametrizing stable maps each of whose (non-contracted)
irreducible components is a
smooth, free curve parametrized by some $M_{e\cdot\alpha,m}$.  
\end{notat}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%
%% Section: Pointed Lines
%% 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Pointed lines} \label{sec-pl}
\marpar{sec-pl}

\mni
In order to apply Lemma \ref{lem-cic1}, we must first have an 
irreducible component $M_{\alpha,1}$ dominating $X$ whose geometric
generic fiber over $X$ is irreducible.  
When $\alpha$ is the class of a
line, i.e., the $\OO(1)$-degree of $\alpha$ is $1$, there are
conditions on the dimension of $X$ and the intersection number
$\langle c_1(T_X),\alpha \rangle$ guaranteeing the existence of
such a component $M_{\alpha,1}$, which is moreover the unique
component dominating $X$.
The conditions are stronger than might be expected (by doing a
parameter count, for instance).
However,
if we assume that $X$ is a linear section of a smooth,
projective variety of sufficiently high dimension
(which is true for every smooth complete intersection),
then the strong conditions may be replaced by weaker conditions.

\mni
Let $X$ be a smooth, projective variety with a very ample invertible
sheaf $\OO(1)$.  All degrees are relative to $\OO(1)$.  Let $\alpha$ be
a curve class on $X$.
Denote by
$$
\begin{CD}
\mc{C} @> f >> X \\
@V \pi VV \\
\Kgnb{0,1}(X,\alpha)
\end{CD}
$$ 
the universal family of stable maps over $\Kgnb{0,1}(X,\alpha)$.  
If $\alpha$ is ``indecomposable'', then we know quite a bit about
$\Kgnb{0,1}(X,\alpha)$ and the evaluation morphism $\text{ev}$.

\begin{lem} \label{lem-pl1}
\marpar{lem-pl1}
Assume there is no nontrivial decomposition $\alpha =
\alpha'+\alpha''$ of $\alpha$ as a sum of effective rational curve
classes, e.g., the $\OO(1)$-degree of $\alpha$ equals $1$.
\begin{enumerate}
\item[(i)]
A general fiber of
$$
\text{ev}: \Kgnb{0,1}(X,\alpha) \rightarrow X
$$
is smooth and projective.  
\item[(ii)]
Every (nonempty) connected component $M$ of a general fiber of $\text{ev}$
has dimension 
$$
\text{dim}(M) = \langle c_1(T_X),\alpha \rangle - 2.
$$
In particular, if $\text{ev}$ is dominant then $\langle
c_1(T_X),\alpha \rangle \geq 2$.
\item[(iii)]
The first Chern class of $M$ is
$$
c_1(T_M) = \pi_*f^*[\text{ch}_2(T_X) + \frac{1}{2\langle
  c_1(T_X),\alpha \rangle}c_1(T_X)^2] - \psi,
$$
where $-\psi$ is the first Chern class of the normal bundle of the
marked section in the universal curve over $M$.  In particular, if the
$\OO(1)$-degree of $\alpha$ equals $1$, this reduces to
$$
c_1(T_M) = \pi_*f^*[\text{ch}_2(T_X) + \frac{1}{2\langle
  c_1(T_X),\alpha \rangle}c_1(T_X)^2 - c_1(\OO(1))^2].
$$
\end{enumerate}
\end{lem}

\begin{proof}
The first part follows from \cite[1.1]{KMM92c}. The second part
follows from \cite[Theorem II.1.7.2]{K}.  The third part follows from 
\cite{dJS3}.  
\end{proof}

\mni
Because of Lemma ~\ref{lem-pl1}(ii), $\text{ev}$ is dominant only if
$$
\langle c_1(T_X),\alpha \rangle \geq 2.  
$$
Moreover, if there is
equality, a general fiber is finite.  If the degree is not $1$,
then the fiber is disconnected.  However, if the inequality is strict,
i.e., if
$$
\langle c_1(T_X)\alpha
\rangle \geq 3,
$$
we may well expect a general fiber of $\text{ev}$ to be irreducible.
The next lemmas prove dominance of $\text{ev}$ and irreducibility of a
general fiber of $\text{ev}$ under inequalities stronger than this
naive one.  After that, we prove the naive inequality suffices if $X$
is a linear section of a smooth projective variety of large dimension.

\begin{lem} \label{lem-pl1.5}
\marpar{lem-pl1.5}
Let $\alpha$ be an $\OO(1)$-degree $1$ rational curve class.  Assume
that
$$
2 \langle c_1(T_X), \alpha \rangle \geq \text{dim}(X) + 3.
$$
If a general fiber of 
$$
\text{ev}:\Kgnb{0,1}(X,\alpha) \rightarrow X
$$
is nonempty, then it is irreducible.
\end{lem}

\begin{proof}
Let $M'$ and $M''$ be irreducible components of the fiber of
$\text{ev}$ over a general point $p$ of $X$.  
Denote by $\Pi'$, resp. $\Pi''$, the subvarieties of $X$ swept out by
curves parametrized by $M'$, resp. $M''$.  By Lemma
~\ref{lem-pl1}(ii),
$$
\text{dim}(\Pi') = \text{dim}(\Pi'') = \langle c_1(T_X),\alpha \rangle -
1.
$$
Moreover they intersect at $p$.  Because $X$ is smooth at $p$,
$$
\text{dim}_p(\Pi' \cap \Pi'') \geq \text{dim}_p(\Pi') + \text{dim}_p(\Pi'')
- \text{dim}_p(X) = 2\langle c_1(T_X),\alpha \rangle -2 - \text{dim}(X).
$$
By hypothesis, this is $\geq 1$.  In particular, there exists $q\neq
p$ such that $\Pi'$ and $\Pi''$ intersect at $q$.  There is a
unique line $L$ containing $p$ and $q$.  Thus 
$$
[L]\in M'\cap M''.
$$
By Lemma ~\ref{lem-pl2}(i), the fiber of $\text{ev}$ over $p$ is
smooth.  In particular, every connected component is irreducible.
Therefore
$$
M' = M'',
$$
i.e., the fiber is irreducible.
\end{proof}

\mni
Because of Lemma ~\ref{lem-pl1.5}, irreducibility follows from
dominance of $\text{ev}$, if the index $\langle c_1(T_X),\alpha
\rangle$ is sufficiently large.  The next lemma shows that if the
index is sufficiently large, then $\text{ev}$ is dominant.

\begin{lem} \label{lem-pl2}
\marpar{lem-pl2}
Assume that for every effective rational curve class $\alpha'\neq
\alpha$,
$$
\langle c_1(T_X), \alpha' \rangle \geq \text{dim}(X) + 2.
$$
Then a
general fiber of 
$$
\text{ev}: \Kgnb{0,1}(X,\alpha) \rightarrow X
$$
is nonempty.  In particular, if $c_1(T_X) = \langle c_1(T_X),\alpha
\rangle c_1(\OO(1))$ for an $\OO(1)$-degree $1$ class $\alpha$, i.e.,
if $c_1(T_X)$ is a multiple of $c_1(\OO(1))$, 
the inequality holds if
$$
2\langle c_1(T_X),\alpha \rangle \geq \text{dim}(X) + 2.
$$
\end{lem}

\begin{proof}
This follows from \cite[Theorem V.1.6]{K}.
\end{proof}

\mni
Together, Lemmas ~\ref{lem-pl1.5} and \ref{lem-pl2} imply that
$\text{ev}$ is dominant, resp. $\text{ev}$ is dominant and a general
fiber is irreducible, if 
$$
2\langle c_1(T_X),\alpha \rangle \geq \text{dim}(X) +2, \text{ resp. }
\geq \text{dim}(X) + 3.
$$
However, as observed above, we expect this to hold under the weaker inequality
$$
\langle c_1(T_X),\alpha \rangle \geq 2, \text{ resp. } \geq 3.
$$
The next lemma proves the weaker inequality suffices provided $X$ is a
linear section of sufficiently high codimension of a smooth,
projective variety $Y$.

\begin{lem} \label{lem-pl2a}
\marpar{lem-pl2a}
Let $\alpha$ be an $\OO(1)$-degree $1$ curve class.
Assume that
$$
\langle c_1(T_X), \alpha \rangle \geq 2, \text{ resp. } \geq 3.
$$
Assume there exists a smooth projective variety $Y$ such that $X$ is a
codimension $c$ linear section of $Y$ and
$$
\text{ev}:\Kgnb{0,1}(Y,\alpha) \rightarrow Y
$$
is dominant, resp. dominant with connected generic fiber.  Then
$$
\text{ev}: \Kgnb{0,1}(X,\alpha) \rightarrow X
$$
is dominant, resp. dominant with connected generic fiber.  In the
latter case, the unique irreducible component $M_{\alpha,1}$
dominating $X$ satisfies the hypotheses of Lemma ~\ref{lem-cic1}.

\mni
In particular, if $c_1(T_X) = \langle c_1(T_X),\alpha
\rangle c_1(\OO(1))$,
i.e., if $c_1(T_X)$ is a multiple of $c_1(\OO(1))$, and if $X$ is a
codimension $c$ linear section of a smooth projective variety $Y$, then
$$
\text{ev}:\Kgnb{0,1}(X,1) \rightarrow X
$$
is surjective, resp. surjective with connected generic fiber, if
$$
\langle c_1(T_X), \alpha \rangle \geq 2, \text{ resp. } \geq 3
$$
and
$$
c \geq \text{dim}(X) - 2\langle c_1(T_X),\alpha \rangle + 2, \text{
  resp. } \geq 
\text{dim}(X) - 2\langle c_1(T_X),\alpha \rangle + 3. 
$$
In the latter case, the unique irreducible component $M_{\alpha,1}$
dominating $X$ satisfies the hypotheses of Lemma ~\ref{lem-cic1}
\end{lem}

\begin{proof}
By hypothesis, the morphism
$$
\text{ev}:\Kgnb{0,1}(Y,1) \rightarrow Y
$$
is dominant.  Thus for some $\OO(1)$-degree $1$ curve class $\alpha$
on $Y$, the morphism
$$
\text{ev}:\Kgnb{0,1}(Y,\alpha) \rightarrow Y
$$
is dominant.
By the Lefschetz hyperplane theorem, the curve class $\alpha$ on $Y$
is the image of a curve class $\alpha$ on $X$.
By Lemma ~\ref{lem-pl1}, a general fiber of $\text{ev}$ has dimension
$$
\langle c_1(T_Y),\alpha \rangle - 2 = c + \langle c_1(T_X), \alpha
\rangle - 2.
$$
By the hypothesis that $\langle c_1(T_X),\alpha \rangle \geq 2$, this
implies a general fiber has dimension $\geq c$.  By upper
semicontinuity of fiber dimension, every fiber has dimension $\geq
c$.  

\mni
Let $p$ be a point of $X$.
Of course $Y$ is embedded in a projective space $\PP^N$ and $X$ is the
intersection of $Y$ with a codimension $c$ linear subspace $\Lambda$
containing
$p$.  The variety of lines in $\PP^N$ containing $p$ is isomorphic to
$\PP^{N-1} = \PP^N/\{p\}$.  By the previous paragraph, the variety of
lines in $Y$ containing $p$ is a subvariety of $\PP^{N-1}$ of
dimension $\geq c$.  The variety of lines in $X$ containing $p$ is the
intersection of this subvariety with the codimension $c$ linear
subspace $\Lambda/\{p\}$.  Because every $c$-dimensional variety in
projective space intersects every codimension $c$ linear subspace,
there exists a line in $X$ containing $p$, i.e.,
$$
\text{ev}:\Kgnb{0,1}(X,1) \rightarrow X
$$
is surjective.  

\mni
Next assume that
$$
\langle c_1(T_X),\alpha \rangle \geq 3.
$$
By \cite[Theorem II.1.7.2]{K}, every irreducible component of
$\Kgnb{0,1}(Y,\alpha)$ has dimension 
$$
\geq \langle c_1(T_Y),\alpha \rangle + \text{dim}(Y) - 2.
$$
By the argument above, the fiber of
$$
\text{ev}_X:\Kgnb{0,1}(X,\alpha) \rightarrow X
$$
over a point $p$ is an intersection of $c$ ample divisors in the fiber of
$$
\text{ev}_Y:\Kgnb{0,1}(Y,\alpha) \rightarrow Y
$$
over $p$.  If $p$ is a general point of $X$, then by Lemma
~\ref{lem-pl1}(ii),
$$
\text{dim}(\text{ev}_X^{-1}(p)) \leq \langle c_1(T_X),\alpha \rangle -
2.
$$
Therefore 
$$
\text{dim}(\text{ev}_Y^{-1}(p))
\leq \langle c_1(T_X),\alpha \rangle - 2 + c = \langle c_1(T_Y),
\alpha \rangle -2 =
\text{dim}(\Kgnb{0,1}(Y,\alpha) - \text{dim}(Y).
$$
This implies that $\text{ev}_Y$ is a flat, local complete intersection
morphism at every point of $\text{ev}_Y^{-1}(p)$, cf. ~\cite[Theorem
II.1.7.3]{K}.  In particular, 
every irreducible component 
of $\text{ev}_Y^{-1}(p)$ is contained in an
irreducible component $M$ of $\Kgnb{0,1}(Y,\alpha)$
that dominates $Y$.  By hypothesis, a general fiber of $\text{ev}_Y$
is irreducible.  Thus, in fact, there is only one irreducible
component $M$.  The restriction
$$
\text{ev}|_M:M\rightarrow Y
$$
is a projective morphism whose geometric generic fiber is connected.
By \cite[Corollary III.11.5]{H}, every fiber is connected.  
Therefore the fiber
$\text{ev}_Y^{-1}(p)$ is connected.  Also, as mentioned above, it is a
local complete intersection variety of 
dimension
$$
\geq \langle c_1(T_X) ,\alpha \rangle - 2 + c \geq c + 1.
$$
Thus an intersection with $c$ ample divisors is still connected,
cf. \cite[Corollary III.7.9]{H}.
Therefore $\text{ev}_X^{-1}(p)$ is connected.  

\mni
The last part of the lemma follows from the first parts together with
Lemmas ~\ref{lem-pl1.5} and ~\ref{lem-pl2}.
\end{proof}

\mni
The main application is to complete intersections.

\begin{cor} \label{cor-pl3}
\marpar{cor-pl3}
Let $X$ be a smooth complete intersection in $\PP^n$ of multidegree
$\underline{d} = (d_1,\dots,d_c)$ and thus of index 
$$
\langle c_1(T_X), \alpha \rangle =n+1 - \sum
d_i.
$$
A general fiber of $\text{ev}$ is nonempty, resp. nonempty and
irreducible, 
if
$$
\langle c_1(T_X), \alpha \rangle \geq 2, \text{ resp. } \geq 3.
$$
In the latter case, the unique irreducible component $M_{\alpha,1}$ of
$\Kgnb{0,1}(X,\alpha)$ dominating $X$ satisfies the hypotheses of
Lemma ~\ref{lem-cic1}.
\end{cor}

\begin{proof}
This follows from Lemma ~\ref{lem-pl2a}: for every sufficiently positive
integer $c$ there exists a smooth projective variety $Y$ such that $X$
is a codimension $c$ linear section of $Y$ by Lemma ~\ref{lem-ext3}.  
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%
%% Section: Minimal pointed curves. Existence and connectedness
%% 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Minimal pointed curves. Existence and connectedness} 
\label{sec-mpc}
\marpar{sec-mpc}

\mni
The proofs of Theorems ~\ref{thm-main} and \ref{thm-main2} are
essentially induction arguments.  A key role is played by rational
curves having minimal curve class among those
that contain $1$, resp. $2$ and $4$, general
fixed points of $X$.  More generally, a curve having minimal curve
class among those containing $m$ general fixed points of $X$ is called
a \emph{minimal pointed curve}.
The case of $1$ point, i.e., pointed lines, was
considered in Section ~\ref{sec-pl}.  The higher cases are considered
in this section and Section ~\ref{sec-mpc2}.  

\mni
There are a few features
of the space of minimal pointed curves that are important.  First of
all, in order to prove irreducibility and rational connectedness of a
general fiber of 
$$
\text{ev}:\Kgnb{0,m}(X,\beta) \rightarrow X^m
$$
for large $m$ and $\beta$, it is crucial to first prove irreducibility
and rational connectedness of a general fiber of
$$
\text{ev}:\Kgnb{0,2}(X,2\alpha) \rightarrow X^2.
$$
Secondly, in order to make the induction argument work, we need to
prove existence of some special rational surfaces in $X$.  Although it
is possible one can prove this directly, we find it useful to think of
such rational surfaces as minimal degree free curves in a space of
minimal pointed curves.  For this reason, it is important to study the
canonical bundle and uniruledness of the spaces of minimal pointed
curves.  This is done in Section ~\ref{sec-mpc2}.  Finally, although
we are primarily concerned with the cases $m=2$ and $m=4$, the methods
apply to arbitrary $m$.  Thus the results are proved for arbitrary $m$.

\mni
As above, $X$ is a smooth, projective variety with a very ample
invertible sheaf $\OO(1)$.  The goal in this section is to understand
curve classes $\beta$ such that the geometric generic fiber of 
$$
\text{ev}:\Kgnb{0,m}(X,\beta) \rightarrow X^m
$$
is smooth, resp. smooth and automorphism free.  There is one simple
criterion for smoothness of a general fiber of $\text{ev}$.

\begin{lem} \label{lem-mpc1}
\marpar{lem-mpc1}
If every point in a general fiber of $\text{ev}$ parametrizes a curve
whose irreducible components are all free, then a (non-empty)
general fiber of
$\text{ev}$ is smooth of the expected dimension
$$
\langle c_1(T_X),\beta \rangle -(m-1)\text{dim}(X) +m-3
$$
and the intersection with the boundary $\Delta$
is a simple normal crossings divisor.
\end{lem}

\begin{proof}
This is trivial if a general fiber is empty.  Thus assume $\text{ev}$
is dominant.  

\mni
The open substack $U$ of $\Kgnb{0,m}(X,\beta)$ parametrizing unions of free
curves is smooth and $U\cap \Delta$ is a simple normal
crossings divisor,
cf. \cite[Theorem II.7.6]{K}.  By
generic smoothness the intersection of $U$ with a general fiber of
$\text{ev}$ is smooth and the intersection with $U\cap \Delta$ is a
simple normal crossings divisor.
If $U$ contains a general
fiber of $\text{ev}$, the fiber is smooth and the intersection with
$\Delta$ is a simple normal crossings divisor.
\end{proof}

\mni
There is a simple criterion on the curve class $\beta$ insuring the
hypothesis of Lemma ~\ref{lem-mpc1}.  The relevant definitions are the
following, and the criterion is the next lemma.

\begin{defn} \label{defn-mpc2}
\marpar{defn-mpc2}
A curve class $\beta$ is \emph{$m$-dominant} if the evaluation
morphism
$$
\text{ev}:\Kgnb{0,m}(X,\beta) \rightarrow X^m
$$
is dominant.  

\mni
An $m$-dominant 
curve class $\beta$ is \emph{$m$-minimal} if for every
partition
$$
m = m_1 + \dots + m_r
$$
and for every collection $(\beta_1,\dots,\beta_r)$ of $m_i$-dominant curve
classes $\beta_i$ satisfying
$$
\sum_{i=1}^r \beta_i \leq \beta,
$$
in fact
$$
\sum_{i=1}^r \beta_i = \beta.
$$
In particular, this implies that $\beta$ is minimal among $m$-dominant
curve classes.
\end{defn}

\begin{lem} \label{lem-mpc3}
\marpar{lem-mpc3}
If $\beta$ is $m$-minimal, then every point in a general fiber of 
$$
\text{ev}:\Kgnb{0,m}(X,\beta) \rightarrow X^m
$$
is a union of free curves.  Therefore a general fiber of $\text{ev}$
is smooth of the expected dimension
$$
\langle c_1(T_X),\beta \rangle -(m-1)\text{dim}(X) +m-3
$$
and intersects $\Delta$ in a simple normal crossings divisor.
Moreover, every point in a general fiber parametrizes an
automorphism-free stable map.
\end{lem}

\begin{proof}
Let $p=(p_1,\dots,p_m)$ be \emph{the} geometric generic point of $X^m$.
Then for every subset $I \subset\{1,\dots,m\}$, setting $l:= \#I$, the
$l$-tuple $(p_i|i\in I)$  
maps to the geometric
generic point of $X^l$.

\mni
By Lemma ~\ref{lem-mpc1}, to prove the lemma it suffices to prove that
every point
$(C,q_1,\dots,q_r,f)$ in the fiber of 
$$
\text{ev}:\Kgnb{0,m}(X,\beta) \rightarrow X^m
$$
over $p$ is a union of free curves.
Let $C'$ be the maximal subcurve of $C$ whose irreducible
components are free curves.  Let $C'_1,\dots,C'_r$ denote the
connected components of $C'$ containing at least one of the marked
points.  Denote by $C''$ the union of all irreducible components of
$C$ not among $C'_1,\dots,C'_r$.  

\mni
There are two observations.  First of all, for every $i=1,\dots,m$,
every irreducible curve in $X$ containing $p_i$ is free by
\cite[1.1]{KMM92c}.  Therefore every point $q_i$ is contained in one
of the components $C'_1,\dots,C'_r$.  Secondly, let $C_j$ be a
contracted irreducible component of $C$ whose image equals $p_i$. 
Every irreducible component of $C$ intersecting $C_j$ is either
contracted -- and thus trivially free -- or else maps to an irreducible
curve in $X$ containing $p_i$ -- and thus free by the first
observation.  Therefore every connected component $C'_i$ contains at
least one irreducible component that is not contracted.  

\mni
Let $\beta_i$ denote the curve class of $C'_i$.  
Denote by $m_i$ the number of elements of
$\{q_1,\dots,q_r\}$ contained in $C'_i$.  Because every subset of
$\{p_1,\dots,p_m\}$ is general, $\beta_i$ is $m_i$-dominant.  And
$$
\sum_{i=1}^r \beta_i \leq \beta.
$$
Therefore, since $\beta$ is $m$-minimal, 
$$
\sum_{i=1}^r \beta_i = \beta.
$$
In particular, every irreducible component of $C''$ is contracted.
Because contracted curves are free, every irreducible component of $C$
is free.

\mni
Finally, to prove that $C$ is automorphism-free, it suffices to prove
that every non-contracted component $C_i$ has degree $1$ over its
image.  But if the degree is $>1$, then $C_i$ can be replaced by a
curve with smaller curve class also meeting all marked points
contained in $C_i$: namely the normalization of the image of $C_i$ (if
any ``special'' points are mapped to the same point in the image, one
may also need to attach some contracted components).  
Thus $C$ can be replaced by an $m$-dominating curve with smaller curve
class.  This contradicts that $\beta$ is $m$-minimal.  
\end{proof}

\mni
What is the $\OO(1)$-degree of an $m$-minimal curve?  Assuming
$$
m \leq h^0(X,\OO(1))
$$
then $m$ general points on $X$ impose independent conditions on linear
forms, i.e., they are in \emph{linear general position}.  The minimum
$\OO(1)$-degree of a rational curve in $\PP^N$ 
containing $m$ points in linearly
general position is $m-1$.  However, there seem to be very few
varieties $X$ with an $m$-dominating class of $\OO(1)$-degree $m-1$.

\begin{ques}[Coskun, Harris] \marpar{ques-mpcCH}
\label{ques-mpcCH}
Let $(X,\OO(1))$ be a smooth projective variety, and let $m$ be an
integer satisfying $2\leq m \leq h^0(X,\OO(1))$.  Assume that every
general $m$-tuple of points on 
$X$ is contained in a curve in $X$ of $\OO(1)$-degree $m-1$.  Is it true that
$(X,\OO(1))$ is a linear variety, a quadric hypersurface, or the
projection of a rational
normal scroll or a Veronese surface? 
\end{ques}

\mni
There is some evidence for a positive answer.  This gives one
criterion for a curve class to be $m$-minimal.

\begin{lem}\label{lem-mpcCH}
\marpar{lem-mpcCH}
Let $(X,\OO(1))$ be a smooth, projective variety and let $m$ be a positive
integer.  Assume that $c_1(T_X) = \langle
c_1(T_X),\alpha \rangle
c_1(\OO(1))$ for some $\OO(1)$-degree $1$ curve class $\alpha$, i.e.,
$c_1(T_X)$ is a multiple of $c_1(\OO(1))$.

\mni
If there exists an $m$-dominating class $\beta$ of
$\OO(1)$-degree $<m$, then $(X,\OO(1))$ is isomorphic to either
$(\PP^n,\OO_{\PP^n}(1))$ or a quadric hypersurface
$(Q,\OO_{\PP^{n+1}}(1)|_Q)$.  

\mni
If $(X,\OO(1))$ is neither a linear variety nor a quadric
hypersurface, then every $\OO(1)$-degree $m$, $m$-dominating class is
$m$-minimal.  

\mni
If $(X,\OO(1))$ is a linear variety, $[\text{line}]$ is $1$-minimal.
If $(X,\OO(1))$ is a quadric hypersurface, $m[\text{line}]$ is
$m$-minimal for $m=1,2$.  
\end{lem}

\begin{proof}
\mni
For a general $m$-tuple of points on $X$, every irreducible curve $C$
containing the points satisfies
$$
h^1(C,N_{C/X}(-m)) = 0
$$
and therefore
$$
\chi(C,N_{C/X}(-m)) \geq 0.
$$
By Riemann-Roch, this implies
$$
\langle c_1(T_X), C \rangle -2 -(m-1)(\text{dim}(X)-1) =
\text{deg}(N_{C/X}(-m)) + \text{rank}(N_{C/X}(-m))
\geq 0. 
$$
Substituting in 
$$
c_1(T_X) = \iota_X c_1(\OO(1))
$$
and
$$
\langle c_1(\OO(1)), C \rangle \leq m-1
$$
gives,
$$
\iota_X \geq \text{dim}(X)
$$
As is well-known, this implies either $X\cong \PP^n$ or $X$ is a
quadric hypersurface.

\mni
For linear varieties and quadric hypersurfaces, lines are obviously 
$1$-minimal.  For a general pair of points
on a quadric hypersurface, the join of the pair is not contained in
the hypersurface.  Thus conics are $2$-minimal on a quadric hypersurface.

\mni
Assume $(X,\OO(1))$ is neither a linear variety nor a quadric
hypersurface.  Let $\beta$ be an $m$-minimal curve class of
$\OO(1)$-degree $m$.  Let 
$$
m = m_1 + \dots + m_r
$$
be a partition of $m$ and let $(\beta_1,\dots,\beta_r)$ be a sequence
of $m_i$-dominant classes $\beta_i$ such that
$$
\sum_{i=1}^r \beta_i \leq \beta.
$$
By the first part of the lemma, the $\OO(1)$-degree of $\beta_i$
is $\geq m_i$.  Thus the sum of the $\OO(1)$-degrees of the classes
$\beta_i$ equals the $\OO(1)$-degree of $\beta$.  Since $\OO(1)$ is
ample, this implies
$$
\sum_{i=1}^r \beta_i = \beta.
$$
Therefore $\beta$ is $m$-minimal.
\end{proof}

\mni
When is an $\OO(1)$-degree $m$ curve class $m$-dominant?  By Lemma
~\ref{lem-mpc1}
if $c_1(T_X)=\langle c_1(T_X),\alpha \rangle
c_1(\OO(1))$ and if $m\alpha$ is $m$-dominant, then 
$$
\text{dim}(X) \geq m(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -
1) + 3.
$$
If $\text{ev}$ is not generically 1-to-1, then a general fiber of
$\text{ev}$ is irreducible only if
$$
\text{dim}(X) \geq m(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -
1) + 4.
$$
Assuming these inequalities hold, is $m\alpha$ an $m$-dominant class,
resp. $m$-dominant with irreducible geometric generic fiber over $X^m$?

\mni
For $m=1$,
this is precisely the issue addressed by Lemma ~\ref{lem-pl2a}.  For
$m>1$, the question can be reduced to the case of $m_i < m$.  

\mni
Let
$m_1,m_2$ be positive integers such that $m_1 + m_2 = m$.  For
$i=1,2$, let $\beta_i$ be an $\OO(1)$-degree $m_i$ curve class.
Denote
$$
\beta = \beta_1 + \beta_2.
$$
For
$i=1,2$, let $p_i$ be a general point of $X^{m_i}$.  Denote
$$
p = (p_1,p_2) \in X^m.
$$
For $i=1,2$, let $\Pi_i$ be the subvariety of $X$ (possibly empty)
swept out by curves in the fiber of
$$
\text{ev}:\Kgnb{0,m_i}(X,\beta_i) \rightarrow X^{m_i}
$$
over $p_i$.  

\begin{lem} \label{lem-mpcdom}
\marpar{lem-mpcdom}
Let $(X,\OO(1))$ be a smooth projective variety and let $\beta$ be an
$\OO(1)$-degree $m$ curve class on $X$.  If for every general point
$p\in X^m$ 
there exists a
decomposition,
$$
m = m_1 + m_2,  \ \ \beta = \beta_1 + \beta_2
$$
for which $\Pi_1$ intersects $\Pi_2$, then
$\beta$ is $m$-dominating.

\mni
Conversely, if $\beta$ is $m$-dominating and if
$$
\text{dim}(X)\geq m(\text{dim}(X) - 1) - \langle c_1(T_X),\beta
\rangle + 4
$$
then there exists a nontrivial decomposition as above such that $\Pi_1$
intersects $\Pi_2$.  
\end{lem}

\begin{proof}
The first part is obvious by taking a union of two intersecting
curves.  The second part follows from the bend-and-break lemma,
cf. \cite[Corollary II.5.5]{K}.
\end{proof}

\mni
These results imply a consequence for the canonical irreducible
components of Section ~\ref{sec-cic}.

\begin{lem} \label{lem-mpcM}
\marpar{lem-mpcM}
Assume $(X,\OO(1))$ is neither a linear variety nor a quadric
hypersurface and $\text{Pic}(X) = \ZZ\{c_1(\OO(1))\}$; in particular,
$c_1(T_X) = \langle c_1(T_X),\alpha \rangle
c_1(\OO(1))$ 
for some $\OO(1)$-degree $1$ curve class $\alpha$.

\mni
Assume there exists a unique irreducible component $M_{\alpha,1}$ of
$\Kgnb{0,1}(X,\alpha)$ such that the restriction
$$
\text{ev}|_M:M_{\alpha,1} \rightarrow X
$$
is dominant, and further assume the general fiber is irreducible.
Assume
$$
\text{dim}(X) \geq m(\text{dim}(X)-\langle c_1(T_X),\alpha \rangle
- 1) + 4
$$
If there exists an irreducible component $M_m$ of
$\Kgnb{0,m}(X,m\alpha)$ such that the restriction
$$
\text{ev}|_M:M_m\rightarrow X^m
$$
is dominant, then $M_m$ equals the component $M_{m\cdot\alpha,m}$ from
Notation \ref{notat-cic2}.
\end{lem}

\begin{proof}
\mni
The lemma will be proved by induction on $m$.  For $m=1$, it is tautological.
Thus assume $m>1$ and the lemma is true for all smaller values
$m'<m$.  
First of all, by Lemmas ~\ref{lem-mpc1}, \ref{lem-mpc3} and
\ref{lem-mpcCH}, $m\alpha$ is $m$-minimal and a general point of $M_m$
parametrizes a smooth curve.  Therefore the image of $M_m$ in
$\Kgnb{0,0}(X,m\alpha)$ is an irreducible component $M_0$, and $M_m$
is the unique irreducible component of $\Kgnb{0,m}(X,m\alpha)$
dominating $M_0$.  

\mni
Let $p=(p_1,\dots,p_m)$ be a general point of $X^m$.  
By the
second part of Lemma ~\ref{lem-mpcdom}, there exists a decomposition
$$
m = m_1 + m_2
$$
and a reducible curve $C=C_1\cup C_2$ parametrized by $M_m$ such that
$$
(C_1\cap \{p_1,\dots,p_m\}) \sqcup (C_1 \cap \{p_1,\dots,p_m\})
$$
is a partition of $\{p_1,\dots,p_m\}$ into subsets of size $m_1$ and
$m_2$ respectively.  The sum of the $\OO(1)$-degrees of $C_1$ and
$C_2$ equals the $\OO(1)$-degree of $C$, $m$.  On the other hand, by
Lemma ~\ref{lem-mpcCH}, the $\OO(1)$-degree of $C_i$ is $\geq m_i$.
Therefore the $\OO(1)$-degree of $C_i$ is precisely $m_i$.

\mni
Because $p$ is general, also every subset of $\{p_1,\dots,p_m\}$ is
general.  
Therefore the component $M'_{m_i}$ of $\Kgnb{0,m_i}(X,m_i\alpha)$
parametrizing $[C_i]$ dominates $X^{m_i}$.  
Moreover, because $X$ is neither linear nor a quadric
hypersurface,
$$
\text{dim}(X) - \langle c_1(T_X),\alpha \rangle - 1 \geq 0.
$$
Therefore, since
$$
\text{dim}(X) - 4 \geq m(\text{dim}(X)-\langle c_1(T_X),\alpha \rangle
- 1),
$$
also for $i=1,2$,
$$
\text{dim}(X) - 4 \geq m_i(\text{dim}(X)-\langle c_1(T_X),\alpha \rangle
- 1).
$$
Thus $m_i$ and $M_{m_i}$ satisfy all the hypotheses of the Lemma.
Because $m_i < m$,
by the induction
hypothesis, $M'_{m_i}$ equals $M_{m_i\cdot \alpha,m_i}$.  

\mni
By
Lemmas ~\ref{lem-mpc3} and ~\ref{lem-mpcCH}, every irreducible
component of $C_i$ is a free curve.   Therefore $M_0$ contains a point
in the image of
$$
\Delta:M_{m_1\cdot \alpha,1}\times_X M_{m_2\cdot \alpha,1} \rightarrow
\Kgnb{0,0}(X,m\alpha) 
$$
which is a smooth point of $\Kgnb{0,0}(X,m\alpha)$.  By Lemma
~\ref{lem-cic1}, because both morphisms
$$
\text{ev}:M_{m_i\cdot \alpha,1} \rightarrow X
$$
are dominant with irreducible generic fiber, $M_{m_1\cdot
  \alpha,1}\times_X M_{m_2\cdot \alpha,1}$ is irreducible (by the same
argument as in the proof of Lemma ~\ref{lem-cic0.3}).
Therefore
$M_0$ is the unique irreducible component of $\Kgnb{0,0}(X,m\alpha)$
containing the image of $M_{m_1\cdot \alpha,1}\times_X M_{m_2\cdot
  \alpha,1}$.  

\mni
By Lemma ~\ref{lem-cic1}, each of $M_{m_i\cdot \alpha,1}$ parametrizes
a reducible curve $C_{m_i}$ whose (non-contracted) components are all smooth,
free curves parametrized by $M_{\alpha,1}$.  Because they are free,
they deform so that the marked point $p_i$ is a general point of $X$.  In
particular, there exists a pair of curves
$((C_{m_1},p_1),(C_{m_2},p_2))$ as above so that $p_1$ and $p_2$
coincide in $X$, i.e., the pair is contained in the fiber product
$$
M_{m_1\cdot \alpha,1}\times_X M_{m_2\cdot \alpha,1}.
$$ 
The image under $\Delta$ parametrizes a reducible curve
$C=C_1\cup C_2$.  Every (non-contracted) component of $C$ is either a
(non-contracted) component of $C_1$ or of $C_2$.  Thus, by hypothesis,
it is a smooth, free curve parametrized by $M_{\alpha,1}$.  Since $M_0$
contains such a curve, by Lemma ~\ref{lem-cic1}, $M_0$ equals
$M_{m\cdot \alpha,0}$.  Since $M_m$ is the unique irreducible
component dominating $M_0$, $M_m$ equals $M_{m\cdot \alpha,m}$.  The
lemma is proved by induction on $m$.  
\end{proof}

\mni
Lemma ~\ref{lem-mpcdom} provides the induction step in an inductive
proof of the existence of an $m$-dominating curve class of
$\OO(1)$-degree $m$.  For the induction step to apply, it is necessary
that certain cycles in $X$ have nonempty intersection.  The simplest
way to insure this is to require the relevant Chow groups of $X$ to
be generated by the appropriate power of $c_1(\OO(1))$.  

\begin{lem} \label{lem-mpcdom2}
\marpar{lem-mpcdom2}
Assume that $c_1(T_X) = \langle c_1(T_X),\alpha \rangle c_1(\OO(1))$
for an $\OO(1)$-degree $1$ curve class $\alpha$, i.e., $c_1(T_X)$ is a
multiple of $c_1(\OO(1))$, resp. assume $\text{Pic}(X)$ is generated
by $c_1(\OO(1))$.  
Assume that $(X,\OO(1))$ is neither a
linear space nor a quadric hypersurface.
Let $m$ be an integer $m\geq 2$.  Assume
that
$$
\text{dim}(X) \geq m(\text{dim}(X)-\langle c_1(T_X),\alpha \rangle-1)
+ m + 2, 
$$
$$
\text{resp. } \geq 2(m-1)(\text{dim}(X)-\langle c_1(T_X),\alpha
\rangle-1) +2m-6.
$$
Assume that for every pair of nonempty closed subsets $\Pi_1,\Pi_2$, 
resp. for every $m$-tuple of nonempty closed subsets
$\Pi'_1,\dots,\Pi'_m$, 
of $X$ with
pure dimensions
$$
\text{dim}(\Pi_1) = \langle c_1(T_X),\alpha \rangle -1,
$$
and
$$
\text{dim}(\Pi_2) = \langle c_1(T_X),\alpha \rangle -
(m-2)(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle) + m-3, 
$$
respectively
$$
\text{dim}(\Pi'_i) = \langle c_1(T_X), \alpha \rangle - 1, \text{
  for } i=1,\dots,m,
$$
the intersection $\Pi_1\cap \Pi_2$ is nonempty, resp. $\Pi'_1\cap
\dots \cap \Pi'_m$ is nonempty and
connected.  In particular this holds if
$$
\text{CH}^p(X) = \ZZ\{c_1(\OO(1))^p\}, \text{ for } 0\leq p \leq
\text{dim}(X) +1 - \langle c_1(T_X), \alpha \rangle,
$$
respectively
$$
\text{for } 0
\leq p \leq m(\text{dim}(X) + 1 - \langle c_1(T_X), \alpha \rangle).
$$
Then $m\alpha$ is $m$-dominating, resp. $m\alpha$ is $m$-dominating
and a general fiber of $\text{ev}$ is connected.
\end{lem}

\begin{proof}
First of all, because $X$ is neither linear nor a quadric
hypersurface,
$$
\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -1 \geq 0.
$$
This means that the various inequalities above for $m$ imply the
analogous inequalities for $m'<m$.  This is the basis for the
induction argument.

\mni
There are two directions: nonemptiness and connectedness.
Nonemptiness of a general fiber of
$$
\text{ev}:\Kgnb{0,m}(X,m\alpha) \rightarrow X^m
$$
will be proved by induction on $m$.  For $m=1$, connectedness follows
from Lemma ~\ref{lem-pl2a}.  Thus, assume $m>1$ and the result is
known for $m-1$.  By the induction hypothesis, the fiber of 
$$
\text{ev}:\Kgnb{0,m-1}(X,(m-1)\alpha) \rightarrow X^{m-1}
$$
over a general point $(p_1,\dots,p_{m-1})$
is nonempty.  By Lemmas ~\ref{lem-mpc1}, \ref{lem-mpc3} and
\ref{lem-mpcCH}, the fiber has pure dimension
$$
\langle c_1(T_X),\alpha \rangle - (m-2)(\text{dim}(X)-\langle
c_1(T_X),\alpha \rangle) + m - 4.
$$
Therefore the subvariety $\Pi''$ swept out by the curves parametrized
by the fiber has dimension one greater,
$$
\text{dim}(\Pi'') = 
\langle c_1(T_X),\alpha \rangle - (m-2)(\text{dim}(X)-\langle
c_1(T_X),\alpha \rangle) + m - 3.
$$
Fix a general point $p_m$ in $X$.  By Lemma ~\ref{lem-pl2a}, a general
fiber of
$$
\text{ev}:\Kgnb{0,1}(X,\alpha) \rightarrow X
$$
is connected.  By Lemma ~\ref{lem-pl1}, the fiber has dimension
$\langle c_1(T_X),\alpha \rangle - 2$.  Therefore the subvariety
$\Pi'$ swept out by curves in the fiber has dimension one greater,
$$
\text{dim}(\Pi') = \langle c_1(T_X),\alpha \rangle - 1.
$$
By hypothesis, $\Pi'$ and $\Pi''$ intersect.  Therefore there exists a
curve $C''$ of class $(m-1)\alpha$ containing $p_1,\dots,p_{m-1}$ and
a curve $C'$ of class $\alpha$ containing $p_m$ such that $C'$
intersects $C''$.  The union $C=C'\cup C''$ is a curve of class
$m\alpha$ containing $p_1,\dots,p_{m-1},p_m$.  Therefore $m\alpha$ is
$m$-dominating.  

\mni
Next we address connectedness.  Let $M$ be an irreducible component of
a the fiber of
$$
\text{ev}:\Kgnb{0,m}(X,m\alpha) \rightarrow X^m
$$
over a general point $p=(p_1,\dots,p_m)$ of $X^m$.  By repeatedly
applying the second part of Lemma ~\ref{lem-mpcdom}, $M$ parametrizes
some curves $C$ every irreducible component of which has class
$\alpha$ and contains precisely one of the points
$\{p_1,\dots,p_m\}$.  Moreover, the locus $\Lambda$
of such curves has dimension
$$
m\langle c_1(T_X),\alpha \rangle - (m-1)\text{dim}(X) - 2.
$$

\mni
Now consider the pointed lines parametrized by this locus.  The total
space of the family is a $\PP^1$-bundle over the locus, together with
a section contracted to a point $p_i$.  For such a $\PP^1$-bundle over
a $1$-dimensional base, any two effective curves that do not intersect
the contracted section do intersect each other.  From this it follows
that there is a nonempty sublocus $\Lambda'$ of dimension
$$
m\langle c_1(T_X),\alpha \rangle - (m-1)\text{dim}(X) - m
$$
parametrizing curves such that every component is a curve of class
$\alpha$ containing precisely one of the points $\{p_1,\dots,p_m\}$,
and such that all components intersect in a common point $q$.  For
$i=1,\dots,m$, denote by $\Pi'_i$ the closed subvariety of $X$ swept
out by all $\alpha$-curves containing $p_i$.  Then clearly $\Lambda'$
is a $\Kgnb{0,m}$-bundle over the intersection
$$
\Pi' := \Pi'_1 \cap \dots \cap \Pi'_m.
$$
By Lemmas ~\ref{lem-pl1} and \ref{lem-pl2}, each $\Pi'$ is nonempty of
dimension 
$$
\text{dim}(\Pi'_i) = \langle c_1(T_X), \alpha \rangle -1.
$$
By hypothesis, the intersection $\Pi'$ of an $m$-tuple of such
subvarieties is nonempty and connected.  Therefore $\Lambda'$ is
connected.  Therefore $\text{ev}^{-1}(p)$ is connected.
\end{proof}

\mni
The following two lemmas form the analogue for higher $m$ of Lemma
~\ref{lem-pl2a}.  

\begin{lem} \label{lem-mpcdom3}
\marpar{lem-mpcdom3}
Let $(X,\OO(1))$ be a smooth, projective variety that is neither a
linear space nor a quadric hypersurface.
Let $m$ be a
positive integer $\leq h^0(X,\OO(1))$.
Let $\beta$ be an $\OO(1)$-degree $m$ curve class
on $X$.
Assume that
$$
\text{dim}(X) \geq m(\text{dim}(X) - 1) - \langle c_1(T_X),\beta
\rangle + 3, \text{ resp. } \geq m(\text{dim}(X)-1) -\langle
c_1(T_X),\beta \rangle + 4.
$$
In particular, if $\beta = m\alpha$, assume
$$
\text{dim}(X) \geq m(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -
1) +3, \text{ resp. } \geq m(\text{dim}(X) - \langle c_1(T_X),\alpha
\rangle - 1) +4.
$$
Assume there exists a smooth projective variety $Y$ such that $X$ is a
codimension $c$ linear section of $Y$ and
$$
\text{ev}_Y:\Kgnb{0,m}(Y,\beta) \rightarrow Y^m
$$
is dominant, resp. dominant with connected generic fiber.  
Then 
$$
\text{ev}_X:\Kgnb{0,m}(X,\beta) \rightarrow X^m
$$
is dominant, resp. dominant with connected generic fiber.
\end{lem}

\begin{proof}
The proof is very similar to the proof of Lemma ~\ref{lem-pl2a}.
Embed $Y$ in $\PP^N$ and let $X = Y \cap
\Lambda$ for a linear 
subspace $\Lambda$ of codimension $c$.   For an $m$-tuple of
points $p_1,\dots,p_m$, 
in $X$, a $\beta$-curve in $Y$ containing $p_1,\dots,p_m$ is contained
in $X$ if and only if its span is contained in $p_1,\dots,p_m$.  Using
Lemmas ~\ref{lem-mpc1}, \ref{lem-mpc3} and \ref{lem-mpcCH}, together
with the argument from the proof of Lemma ~\ref{lem-pl2a}, the
morphism $\text{ev}_Y$ is a flat, local complete intersection morphism
at every point of the fiber $\text{ev}_X^{-1}(p)$ over a general point
$p$ of $X^m$.  The claim is that $\text{ev}_X^{-1}(p)$ is an
intersection of $c$ big, base-point-free divisors in
$\text{ev}_Y^{-1}(p)$.  This will immediately imply that every
component of $\text{ev}_Y^{-1}(p)$ meets $\text{ev}_X^{-1}(p)$, in
particular $\text{ev}_X^{-1}(p)$ is not empty.
Moreover, if a
general fiber of $\text{ev}_Y$ is connected, 
by the same argument from the proof of Lemma ~\ref{lem-pl2a},
$\text{ev}_Y^{-1}(p)$ is a connected, locally complete intersection.
Then by \cite[Corollary III.7.9, Corollary III.11.5]{H}, 
$\text{ev}_X^{-1}(p)$ is also
connected.

\mni
To see that $\text{ev}_X^{-1}(p)$ is an intersection of big,
base-point-free divisors in $\text{ev}_Y^{-1}(p)$, fix a hyperplane
$H\subset \PP^N$ containing $\text\{p_1,\dots,p_m\}$.  Associated to every
$\OO(1)$-degree $m$ curve $C$ containing $\{p_1,\dots,p_m\}$, the span
of $C$ is an $m$-plane containing $\text{span}(p_1,\dots,p_m)$.  In
other words, it is a point in $\PP^{N-m} \cong
\PP^N/\text{span}(p_1,\dots,p_m)$.  The hyperplane $H$ corresponds to
a hyperplane in $\PP^{N-m}$.  Therefore the set of curves $C$
contained in $H$ forms a divisor in $\text{ev}_Y{-1}(p)$.  Varying
$H$, it is clear that this divisor class is big and base-point-free.
\end{proof}

\begin{cor} \label{cor-mpcdom4}
\marpar{cor-mpcdom4}
Assume that $c_1(T_X) = \langle c_1(T_X),\alpha \rangle c_1(\OO(1))$
for an $\OO(1)$-degree $1$ curve class $\alpha$, i.e., $c_1(T_X)$ is a
multiple of $c_1(\OO(1))$, resp. assume $\text{Pic}(X)$ is generated
by $c_1(\OO(1))$.  
Assume that $(X,\OO(1))$ is neither a
linear space nor a quadric hypersurface.
Let $m$ be an integer $m\geq 2$.  
Assume that
$$
\text{dim}(X) \geq m(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -
1) + 3, \text{ resp. } \geq m(\text{dim}(X) - \langle c_1(T_X),\alpha
\rangle - 1) + 4.
$$
Assume there exists a smooth, projective variety $Y$ such that $X$ is
a codimension $c$ linear section of $Y$.  
If
$$
c \geq m(\text{dim}(X)-\langle c_1(T_X),\alpha \rangle -1 ) -
\text{dim}(X)+m-2, 
$$
respectively
$$
c \geq 2(m-1)(\text{dim}(X)-\langle c_1(T_X),\alpha
\rangle-1) -\text{dim}(X) +2m - 6
$$
and if
$$
\text{CH}^p(Y) = \ZZ\{c_1(\OO(1))^p\}, \text{ for } 0 \leq p \leq
\text{dim}(X) + 1 - \langle c_1(T_X),\alpha \rangle,
$$
respectively
$$
\text{for } 0 \leq p \leq m(\text{dim}(X)+1 - \langle c_1(T_X), \alpha
\rangle),
$$
then 
$$
\text{ev}_X:\Kgnb{0,m}(X,m\alpha) \rightarrow X^m
$$
is dominant, resp. dominant with connected fibers.  By Barth's
theorems, \cite{Barth70}, this last condition holds if 
$$
c \geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -1)
-\text{dim}(X) + (h^0(X,\OO(1))-\text{dim}(X)-2) + 5,
$$
respectively
$$
c \geq 2m(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle - 1) -
\text{dim}(X) + (h^0(X,\OO(1))-\text{dim}(X)-2) + 4m+1.
$$
\end{cor}

\begin{proof}
This follows from Lemmas ~\ref{lem-mpcdom2} and \ref{lem-mpcdom3}.
\end{proof}

\mni
The main application is to complete intersections.

\begin{cor} \label{cor-mpcdom5}
\marpar{cor-mpcdom5}
Let $X$ be a smooth complete intersection in $\PP^n$ of multidegree
$\underline{d} = (d_1,\dots,d_c)$.  For simplicity, assume all
$d_i\geq 2$ and assume $\underline{d} \neq (2)$.  
Let $m$ be an integer $1\leq m
\leq n$.  
The fiber of
$$
\text{ev}:\Kgnb{0,m}(X,m) \rightarrow X^m
$$
over a general point in $X^m$ is nonempty, resp. nonempty and
irreducible, if 
$$
\text{dim}(X) \geq m(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -
1) +3, \text{ resp. } \geq m(\text{dim}(X) - \langle
c_1(T_X),\alpha \rangle - 1) + 4,
$$
in other words, if
$$
n \geq \sum_{i=1}^c (md_i-m+1) - m + 3, \text{ resp. } \geq
\sum_{i=1}^c(md_i - m + 1) - m + 4.
$$
\end{cor}

\begin{proof}
This follows from Corollary ~\ref{cor-mpcdom4} since complete
intersections are arbitrarily extendable by Lemma ~\ref{lem-ext3}.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%
%% Section: Minimal pointed curves. Canonical bundle and uniruledness
%% 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Minimal pointed curves. 
Canonical bundle and uniruledness} \label{sec-mpc2}
\marpar{sec-mpc2}

\mni
The rational surfaces we will need in Section ~\ref{sec-q} arise from
minimal free curves in the space of minimal pointed curves.  We study
these curves using the formula for the canonical bundle from
\cite{dJS3} together with the method of Koll\'ar-Miyaoka-Mori,
cf. \cite{Miyaoka-Mori86}, \cite[\S V.1]{K}.  As in Sections
~\ref{sec-pl} and \ref{sec-mpc},  the existence of the curves follows
from hypotheses on $\text{dim}(X)$, $\langle c_1(T_X),\alpha
\rangle$, and also the second graded piece of the Chern character
$\langle \text{ch}_2(T_X),\Pi \rangle$.  The hypotheses are stronger
than the weak hypotheses suggested by a parameter count.  However,
under when $X$ is a linear section of sufficiently high
codimension in a smooth, projective variety $Y$, the weaker hypotheses
suffice (and, in fact, are also necessary conditions).

\mni
Because of the criterion in Lemma ~\ref{lem-mpcCH}, we shall focus on
the case of $m$-minimal classes of $\OO(1)$-degree $m$.  Moreover the
varieties we will focus on have another feature formalized in the
following definition/hypothesis.

\begin{defn} \label{defn-mcpU}
\marpar{defn-mcpU}
Let $m$ and $N$ be integers satisfying $1\leq m \leq N$.  For a
linearly general $m$-tuple
$p=(p_1,\dots,p_m)$ in $(\PP^N)^m$, the \emph{linearly
  nondegenerate locus} $U_p$ is 
the maximal open substack of the corresponding fiber of
$$
\text{ev}:\Kgnb{0,m}(\PP^N,m) \rightarrow (\PP^N)^m
$$
parametrizing stable maps for which no irreducible component
is mapped into the linear subspace $\text{span}(p_1,\dots,p_m)$.  By
B\'ezout's theorem, this is equivalent to the condition that no
non-contracted component is mapped into the linear subspace
$\text{span}(p_1,\dots, p_m)$.  
\end{defn}

\begin{lem} \label{lem-mpcU}
\marpar{lem-mpcU}
Let $(X,\OO(1))$ be a smooth, projective variety.
Let $\alpha$ be an
$\OO(1)$-degree $1$ curve class.
Let $m$ be a positive
integer $\leq h^0(X,\OO(1))$.  
If $m\geq 2$, assume $(X,\OO(1))$ is
not a linear variety.  If $m\geq 3$, assume $(X,\OO(1)$ is not a
quadric hypersurface.  
Assume $c_1(T_X) = \langle
c_1(T_X), \alpha \rangle c_1(\OO(1))$, i.e., $c_1(T_X)$ is a multiple
of $c_1(\OO(1))$.  
Assume $m\alpha$ is an
$m$-dominating class.  Assume either
\begin{enumerate}
\item[(i)]
$$
\text{dim}(X) > \langle c_1(T_X),\alpha \rangle + 1,
$$
\item[(ii)]
$$
\text{dim}(X) = \langle c_1(T_X),\alpha \rangle + 1,
$$
and 
$$
h^0(X,\OO(1)) \geq \text{dim}(X) + m  - 2,
$$
\item[(iii)]
$(X,\OO(1))$ is a quadric hypersurface and $m \leq 2$, or
\item[(iv)]
$(X,\OO(1))$ is linear and $m=1$.
\end{enumerate}
Then the fiber of
$$
\text{ev}:\Kgnb{0,m}(X,m\alpha) \rightarrow X^m
$$
over a general point $p$ of $X^m$ is contained in the nondegenerate
locus $U_p$.
\end{lem}

\begin{proof}
Embed $X$ in $\PP^N$ by the complete linear system $|\OO(1)|$, where
$N=h^0(X,\OO(1)) - 1$.  
First consider the case when
$$
h^0(X,\OO(1)) \geq \text{dim}(X) + m.
$$
The codimension of $X$ is $c=h^0(X,\OO(1)) -
\text{dim}(X) - 1 \geq m-1$.  Also the degree of $X$ is $\geq
\text{codim}(X) +1 \geq m$.  Therefore $m$ general points
$p_1,\dots,p_m$
of $X$ are
contained in a codimension $c$ linear section of $X$, which is
necessarily zero dimensional.  For every curve $C$ in $X$ containing
$p_1,\dots,p_m$, $\text{span}(C)$ contains
$\text{span}(p_1,\dots,p_m)$, which is contained in the zero
dimensional linear section of $X$.  Since $\text{dim}(C)$ is positive,
$\text{span}(C)$ must properly contain $\text{span}(p_1,\dots,p_m)$.
In particular, if $C$ is a degree $m$ rational curve, $C$ is linearly
nondegenerate.  

\mni
Next assume that
$$
\text{dim}(X) + m > h^0(X,\OO(1)).
$$
The claim is that
$$
h^0(X,\OO(1)) \geq \text{dim}(X) - m(\text{dim}(X) -\langle
c_1(T_X),\alpha \rangle - 1) + m -2.
$$
Indeed, $h^0(X,\OO(1)) \geq \text{dim}(X) + 1$ always.  So in case
(i),
$$
h^0(X,\OO(1)) \geq \text{dim}(X) + 1 \geq \text{dim}(X) -2 \geq
\text{dim}(X) -2 - m(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -
2).
$$
In case (ii), by hypothesis,
$$
h^0(X,\OO(1)) \geq \text{dim}(X) + m -2 = \text{dim}(X) -
m(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -1) + m -2.
$$
In case (iii), since $m\leq 2$,then
$$
h^0(X,\OO(1)) = \text{dim}(X) + 2 \geq \text{dim}(X) -m(-1) + m -2 =
\text{dim}(X) - m (\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -1)
+ m -2.
$$
And case (iv) is clear.  

\mni
Let $Z$ be the intersection of $X$ with a general $\PP^{m-1}$ in
$\PP^N$.  Then
$$
\text{dim}(Z) = \text{dim}(X) + m - h^0(X,\OO(1)),
$$
and
$$
\langle c_1(T_Z),\alpha \rangle = \langle c_1(T_X),\alpha \rangle + m
- h^0(X,\OO(1)).
$$
Thus,
$$
\text{dim}(Z) - \langle c_1(T_Z),\alpha \rangle -1 = \text{dim}(X) -
\langle c_1(T_X),\alpha - 1.
$$
Plugging this in,
$$
\text{dim}(Z) - m(\text{dim}(Z)-\langle c_1(T_Z),\alpha \rangle -1) -
3 = \text{dim}(X) +m - h^0(X,\OO(1)) - m(\text{dim}(X)-\langle
c_1(T_X),\alpha \rangle - 1) - 3.
$$
Because 
$$
h^0(X,\OO(1)) \geq \text{dim}(X) - m(\text{dim}(X) - \langle
c_1(T_X),\alpha \rangle - 1) - 3,
$$
it follows that
$$
\text{dim}(Z) - m(\text{dim}(Z) - \langle c_1(T_Z),\alpha \rangle - 1)
- 3
$$
is negative.  By Lemma ~\ref{lem-mpc1}, there is no $\OO(1)$-degree
$m$ curve class on $Z$ that is $m$-dominating.  Let $p_1,\dots,p_m$ be
$m$ general points of $Z$.  These span a $\PP^{m-1}$, i.e.,
$\text{span}(p_1,\dots,p_m) = \text{span}(Z)$.
Let $C$ be a
degree $m$ curve in $X$ containing $p_1,\dots,p_m$.  Then
$\text{span}(C)$ contains $\text{span}(p_1,\dots,p_m) =
\text{span}(Z)$.  Because there is no $\OO(1)$-degree $m$ curve class
on $Z$ that is $m$-dominating, the curve $C$ cannot be contained in
$Z$, i.e., it cannot be contained in $\text{span}(Z)$.  Therefore
$\text{span}(C)$ properly contains $\text{span}(Z) \cong \PP^{m-1}$,
i.e., $C$ is a linearly nondegenerate curve.
\end{proof}

\mni
Because of Lemma ~\ref{lem-mpcU}, we introduce the following
hypothesis.

\begin{hyp} \label{hyp-mpcU}
\marpar{hyp-mpcU}
Let $(X,\OO(1),m)$ be a smooth, projective variety together with a
positive integer $m$.
Let $\alpha$ be an
$\OO(1)$-degree $1$ curve class.
Assume
$m \leq h^0(X,\OO(1))$.  
Assume $c_1(T_X) = \langle
c_1(T_X), \alpha \rangle c_1(\OO(1))$, i.e., $c_1(T_X)$ is a multiple
of $c_1(\OO(1))$.  
If $m\geq 2$, assume
$(X,\OO(1))$ is not a linear variety.  If $m\geq 3$, assume
$(X,\OO(1))$ is not a quadric hypersurface.  If $\text{dim}(X) =
\langle c_1(T_X),\alpha \rangle + 1$, assume $h^0(X,\OO(1)) \geq
\text{dim}(X) + m -2$.  
Assume $m\alpha$ is an
$m$-dominating class.  
By Lemma ~\ref{lem-mpcU}, these assumptions imply 
the fiber $M_p$ of
$$
\text{ev}:\Kgnb{0,m}(X,m\alpha) \rightarrow X^m \subset (\PP^N)^m
$$
over a general point $p$ of $X^m$
is contained in the linearly nondegenerate locus $U_p$.
\end{hyp}

\mni
The varieties we consider satisfy Hypothesis ~\ref{hyp-mpcU}.  The hypothesis
implies consequences for the
divisor theory of $M_p$.  
As usual, denote by 
$$
(\pi:\mc{C} \rightarrow U, (s_i:U\rightarrow \mc{C}|i=1,\dots,m),
g:\mc{C} \rightarrow \PP^N)
$$
the universal family of stable maps over $U$.

\begin{lem} \label{lem-mpc4}
\marpar{lem-mpc4}  
The stack $U_p$ is
automorphism-free, i.e., $U_p$ is a quasi-projective scheme.  
Moreover, there exists a big, base-point-free Cartier divisor
class $\lambda$ on $U$ such that
$$
\pi_*g^* c_1(\OO_{\PP^N}(1))^2 = m \lambda,
$$
$$
\pi_*(s_i(U)\cdot s_i(U)) = -\lambda, \ \ i=1,\dots,m.
$$
If $m>1$, also
$$
\frac{1}{m-1} \sum_{\{A,B\},A\cup B = \{1,\dots,m\}} \#A \cdot \#B
\Delta_{\{A,B\}} = m \lambda.
$$
\end{lem}

\begin{proof}
Denote by $E$ the locally free sheaf
$$
E := \pi_* g^* \OO_{\PP^N}(1).
$$
Denote by $F$ the vector space
$$
F := \oplus_{i=1}^m \OO_{\PP^N}(1)\otimes \kappa(p_i)
$$
Because $g\circ s_i$ is a constant morphism with image $p_i$, there is
an evaluation morphism
$$
E\rightarrow F\otimes \OO_U.
$$
This is surjective, because it factors the surjective evaluation
$$
H^0(\PP^N,\OO(1))\otimes \OO_U \rightarrow E \rightarrow F\otimes
\OO_U.
$$
Denote the kernel by
$$
\mc{L} := \text{Ker}(E \rightarrow F\otimes \OO_U).
$$
This is an invertible sheaf on $U$.  Denote its divisor class by
$$
\lambda := c_1(\mc{L}).
$$
It is clear that $\lambda$ is base-point-free.  The complete linear
system defines a morphism
$$
\phi_\lambda: U_p \rightarrow \PP^{N-m} =
\PP^N/\text{span}(p_1,\dots,p_m)
$$
that associates to a stable map $(C,(q_1,\dots,q_m),f)$ the point
$$
\text{span}(f(C)) \in \PP^N/\text{span}(p_1,\dots,p_m).
$$
It is clear that the only locus contracted by this morphism is the
locus of curves with a contracted component (not mapping into
$\text{span}(p_1,\dots,p_m)$!) containing at least $4$ double points.
Therefore $\phi_\lambda$ is big.  

\mni
By adjunction there is a natural homomorphism
$$
\pi^* E \rightarrow g^* \OO_{\PP^N}(1).
$$
Restricting to $\mc{L}$ gives a morphism of invertible sheaves
$$
\pi^* \mc{L} \rightarrow g^* \OO_{\PP^N}(1),
$$
or equivalently, a global section of the twist
$$
\sigma:\OO_{\mc{C}} \rightarrow \pi^*\mc{L}^\vee \otimes
g^*\OO_{\PP^N}(1).
$$
The zero locus of this section is precisely 
$$
\text{Zero locus}(\sigma) = g^{-1}\text{span}(p_1,\dots,p_m).
$$
Of course there is an inclusion
$$
\cup_{i=1}^m s_i(U) \subset g^{-1}\text{span}(p_1,\dots,p_m).
$$
Now a curve of degree $m$ spans at most a $\PP^m$.  Thus
$\text{span}(p_1,\dots,p_m)$ is a hyperplane in the span of the
curve.  By B\'ezout's theorem, if the intersection has degree $\geq
m+1$, then some irreducible component of the curve is contained in
$\text{span}(p_1,\dots,p_m)$.  By hypothesis, every curve parametrized
by $U$ has no irreducible component contained in
$\text{span}(p_1,\dots,p_m)$.  Thus
$$
\text{Zero locus}(\sigma) = \cup_{i=1}^m s_i(U)
$$
as closed subschemes of $\mc{C}$.  This implies an isomorphism of
invertible sheaves
$$
\OO_{\mc{C}}(\sum_{i=1}^m s_i(U)) \cong \pi^*\mc{L}^\vee g^*\OO(1).
$$

\mni
Applying $s_i^*$ to each side of this isomorphism gives
$$
s_i^*\OO_{\mc{C}}(s_i(U)) \cong \mc{L}^\vee,
$$
or equivalently,
$$
-\pi_*[s_i(U)\cdot s_i(U)] = \lambda
$$
for every $i=1,\dots,m$.  Combined with \cite[Lemma 5.8]{dJS3}, this
implies the divisor class relations
$$
\frac{1}{m-1}\sum_{(A,B), 1\in A, A\cup B = \{1,\dots,m\}} \#A\cdot
\#B \Delta_{(A,B)} = m \lambda
$$
and
$$
\frac{1}{m-1} \sum_{\{A,B\}} \#A\cdot \#B \Delta_{\{A,B\}} = m\lambda.
$$
Also,
$$
\pi_*g^*c_1(\OO_{\PP^N}(1))^2 = \pi_*[\sum_{i=1}^m s_i(U)
+\pi^*\lambda]^2 = \sum_{i=1}^m \pi_*[s_i(U)\cdot s_i(U)] +
2\sum_{i=1}^m s_i^*\pi^* \lambda
$$
which reduces to
$$
\pi_*g^*c_1(\OO_{\PP^N}(1))^2 = m\lambda.
$$
\end{proof}

\mni
When $m$ is small, this allows us to give a useful formula for the
first Chern class of a general fiber of the evaluation morphism
$\text{ev}$. 

\begin{lem} \label{lem-mpc5}
\marpar{lem-mpc5}
If $(X,\OO(1),m)$ satisfies Hypothesis ~\ref{hyp-mpcU},
then the fiber $M_p$ of
$$
\text{ev}:\Kgnb{0,m}(X,m\alpha) \rightarrow X^m
$$
over a general point $p$ of $X^m$ is a
smooth, projective variety of the expected dimension 
$$
\text{dim}(X) - m (\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -
1) - 3
$$
whose
first Chern class equals
$$
c_1(T_M) = \pi_*g^*[\text{ch}_2(T_X) - \frac{(m-2)\langle
  c_1(T_X),\alpha\rangle + 2m}{2m}c_1(\OO(1))^2]
+ 2\Delta.
$$

\mni
In particular, for $m=1$ this is the formula from Lemma
~\ref{lem-pl1}
$$
c_1(T_M) = \pi_* g^*[\text{ch}_2(T_X) + \frac{\langle c_1(T_X),\alpha
  \rangle - 2}{2} c_1(\OO(1))^2],
$$
for $m=2$ this is
$$
c_1(T_M) = \pi_* g^*[\text{ch}_2(T_X) + c_1(\OO(1))^2],
$$
and for $m=3$ this is
$$
c_1(T_M) = \pi_* g^*[\text{ch}_2(T_X) -\frac{\langle c_1(T_X),\alpha
  \rangle - 6}{6} c_1(\OO(1))^2].
$$
\end{lem}

\begin{proof}
For $m=1$, this follows from Lemma ~\ref{lem-pl1}. Thus assume $m>1$.

\mni
First of all, by Lemma ~\ref{lem-mpcCH}, $m\alpha$ is $m$-minimal.
Thus, by Lemma ~\ref{lem-mpc3}, $M_p$ is smooth and
automorphism-free. 
By
\cite[Theorem 1.1]{dJS3}, the first Chern class of $M_p$ equals
$$
c_1(T_M) = \pi_*g^*[\text{ch}_2(T_X) + \frac{1}{2\langle
  c_1(T_X),m\alpha \rangle} c_1(T_X)^2] + 2 \Delta 
$$
$$
-\frac{1}{r-1}\sum_{\{A,B\}} \#A\cdot \#B \Delta_{\{A,B\}} -
\frac{1}{2\langle c_1(T_X),m\alpha \rangle}
\sum_{\{\beta',\beta''\}, \beta'+\beta'' = m\alpha} 
\langle c_1(T_X),\beta'\rangle \langle
c_1(T_X), \beta'' \rangle \Delta_{\{\beta',\beta''\}}.
$$
Using the fact that every curve class is a multiple of $\alpha$, this
reduces to
$$
\pi_*g^*[\text{ch}_2(T_X) + \frac{1}{2\langle
  c_1(T_X),m\alpha \rangle} c_1(T_X)^2] + 2 \Delta 
$$
$$
-\frac{1}{r-1}\sum_{\{A,B\}} \#A\cdot \#B \Delta_{\{A,B\}} -
\frac{\langle c_1(T_X),\alpha\rangle}{2m}
\sum_{\{m',m''\}m'+m''=m} m' m''
\Delta_{\{ m'\alpha,m''\alpha \}}.
$$
Next, because $m\beta$ is $m$-minimal,
$$
\sum_{\{A,B\}, \#A = m',\#B=m''} \Delta_{\{A,B\}} =
\Delta_{\{m'\alpha,m''\alpha\}}.
$$
So the formula reduces further to
$$
\pi_*g^*[\text{ch}_2(T_X) + \frac{1}{2\langle
  c_1(T_X),m\alpha \rangle} c_1(T_X)^2] + 2 \Delta 
$$
$$
-\frac{1}{m-1}\sum_{m'+m''=m} m'm'' \Delta_{\{m'\alpha,m''\alpha\}} -
\frac{\langle c_1(T_X),\alpha\rangle}{2m}
\sum_{m'+m''=m} m' m''
\Delta_{\{m'\alpha,m''\alpha\}}.
$$
Finally, by Lemma ~\ref{lem-mpc4}, there is a divisor class relation
$$
\frac{1}{m-1} \sum_{m'+m''=m} m' m'' \Delta_{\{m'\alpha,m''\alpha\}} =
m\lambda = \pi_* g^* c_1(\OO(1))^2.
$$
Substituting this in and simplifying gives the formula.

\mni
The formulas in the cases $m=2$ and $3$ follow from the formula above
together with the relations from Lemma ~\ref{lem-mpc4},
$$
\Delta_{1,1} = \frac{1\cdot 1}{1} \Delta_{1,1} = 2\lambda = \pi_* g^*
c_1(\OO(1))^2, 
$$
$$
\Delta_{2,1} = \frac{2\cdot 1}{2} \Delta_{2,1} = 3\lambda = \pi_* g^*
c_1(\OO(1))^2.
$$
When $m>3$, the boundary $\Delta$ is not of a single type, i.e.,
$\Delta \neq \Delta_{i,j}$.  Thus there is not a similar formula for
$m>3$.   
\end{proof}

\mni
Because of the formula in the last lemma, $\text{ch}_2(T_X)$ plays a
significant role in the geometry of a general fiber of $\text{ev}$.
It is useful to make the following hypothesis, which holds for all
complete intersections.

\begin{hyp} \label{hyp-mpcU2}
\marpar{hyp-mpcU2}
Assume $(X,\OO(1),m)$ satisfies Hypothesis ~\ref{hyp-mpcU}.  Further
assume that
$$
\text{ch}_2(T_X) = \frac{\langle 2\text{ch}_2(T_X),\Pi\rangle}{2}
c_1(\OO(1))^2
$$
for some integer $\langle 2\text{ch}_2(T_X),\Pi \rangle$.
If $X$
contains the class of a linear $2$-plane $\Pi$, then the integer is
the intersection number indicated.  
\end{hyp}

\mni
Koll\'ar, Miyaoka and Mori proved uniruledness of varieties with
positive first Chern class.  In our setting, this gives the following.

\begin{cor} \label{cor-mpc6}
\marpar{cor-mpc6}
Assume that $(X,\OO(1),m)$ satisfies Hypothesis ~\ref{hyp-mpcU2}.
If
$$
m\langle 2\text{ch}_2(T_X),\Pi \rangle - (m-2) \langle c_1(T_X),\alpha
\rangle - 2m > 0
$$
then the fiber $M_p$ of
$$
\text{ev}:\Kgnb{0,m}(X,m) \rightarrow X^m
$$ 
over a general point $p$ of $X^m$
is uniruled.
If
$$
(m-1)\text{dim}(X) +m \langle 2\text{ch}_2(T_X),\Pi \rangle -
2(m-1)\langle c_1(T_X),\alpha \rangle - 3m + 2 > 0
$$
then $M_p$ is uniruled by rational curves of $\lambda$-degree $1$.
\end{cor}

\begin{proof}
When the first inequality holds, by Lemma ~\ref{lem-mpc5},
$c_1(T_X)$ is the sum of a nef, big
divisor and an effective divisor.  Therefore, by
\cite{Miyaoka-Mori86}, $M$ is uniruled.  

\mni
Moreover, applying
bend-and-break, the minimal class $\gamma$ of a free rational curve in $X$
satisfies the inequality from \cite[Theorem V.1.6.1]{K},
$$
\langle c_1(T_M),\gamma \rangle \leq \text{dim}(M) + 1.
$$
Every free rational curve class in $X$ satisfies
$$
\langle \Delta, \gamma \rangle \geq 0
$$
since such curves deform out of $\Delta$.  If $\langle \lambda, \gamma
\rangle \geq 2$ and if the second inequality holds, then 
$$
\langle c_1(T_M), \gamma \rangle > \text{dim}(M) + 1.
$$
Therefore, if the second inequality holds, $M$ is uniruled by rational
curves of $\lambda$-degree $1$.
\end{proof}

\begin{lem} \label{lem-mpc7}
\marpar{lem-mpc7}
Assume $(X,\OO(1),m)$ satisfies Hypothesis ~\ref{hyp-mpcU}.
Assume that the fiber $M_p$ of
$$
\text{ev}:\Kgnb{0,m}(X,m\alpha) \rightarrow X^m
$$
over a general point $p$ of $X^m$ 
is uniruled by rational curves of $\lambda$-degree $1$.  
A general point of $M_p$ parametrizes a curve which is
either a hyperplane section of a degree $m$
rational normal surface scroll $\Sigma \subset X$ or
a hyperplane section of a Veronese surface $\Sigma \subset X$ (which
can only occur if $m=4$).
\end{lem}

\begin{proof}
First of all, by the proof of Lemma ~\ref{lem-mpc3}, a general fiber
$M$ is smooth and automorphism-free, a general point of $M$
parametrizes a smooth curve, and the boundary $\Delta\cap M$ is a
simple normal crossings divisor.

\mni
Let $D$ be a free rational curve in $M$ of $\lambda$-degree $1$.
Denote by $\pi:\mc{C}_D \rightarrow D$ the restriction of the
universal curve over $D$.  Denote by $(\Sigma,\OO(1))$ the linearly
normal
surface obtained by contracting all curves in $\mc{C}_D$ of
$g^*\OO(1)$-degree $0$.  In other words, if $\mc{C}_D \rightarrow
\PP^N$ is the morphism induced by the complete linear system of
$g^*\OO(1)$, then $\Sigma$ is the linearly normal
image of $\mc{C}_D$ in $\PP^N$.

\mni
By
Lemma ~\ref{lem-mpc4}, the surface $\Sigma$ 
has
$\OO(1)$-degree $m$ since $D$ has $\lambda$-degree $1$.
Because the invertible sheaf $\mc{L}|_D$ has degree $1$,
$$
h^0(\Sigma,\OO(1)) = h^0(D,\pi_*g^*\OO(1)|_D) = m+2.
$$
Therefore $\Sigma$ spans a linear $\PP^{m+1}$.  A surface of degree $m$
spanning $\PP^{m+1}$ is a surface of \emph{minimal degree}.  These
were classified by Del Pezzo in 1885, \cite{DP1885}.  For a modern
account and generalization, see \cite{EHmin}.  In particular, there
are three possibilities for $\Sigma$: $\Sigma$ is a cone over a
rational normal curve, $\Sigma$ is the Veronese surface, or $\Sigma$
is a (smooth) rational normal surface scroll.  In each of these cases,
$D$ is a pencil of hyperplane sections of $\Sigma$.  

\mni
If $\Sigma$ is a cone over a rational normal curve, then
the hyperplane containing the vertex
of the cone gives a point of $D$ that intersects the boundary $\Delta$
with multiplicity $>1$.  Assuming $D$ is general in its deformation
class, $D$ intersects any specified divisor transversally, cf. the proof of
\cite[Proposition II.3.7]{K}. Therefore $\Sigma$ is not a cone over a
rational normal surface scroll.

\mni
The final claim is that the morphism $\Sigma \rightarrow X$ is an
embedding.  If not, the image spans a $\PP^m$.  But then
$\text{span}(p_1,\dots,p_m) \cap \Sigma$ is a hyperplane section of
$\Sigma$, thus a curve in $\Sigma$.  In particular, it is strictly
larger than $\{p_1,\dots,p_m\}$.  
Therefore a curve parametrized by
$D$ intersects $\text{span}(p_1,\dots,p_m)$ in a subscheme of degree
$\geq m+1$.  By B\'ezout's theorem, the curve has an irreducible
component contained in $\text{span}(p_1,\dots,p_m)$.  This contradicts
the hypothesis that $D$ is contained in the open substack $U$ of Lemma
~\ref{lem-mpc4}.  Therefore $\Sigma \rightarrow X$ is an embedding.
\end{proof}

\mni
Using the previous result, we can compute the dimension of the space
of free $\lambda$-degree $1$ curves in $M_p$.  In particular, since the
dimension must be positive if such curves exist,
the dimension inequality gives
a necessary condition for the existence of free
$\lambda$-degree $1$ curves in $M_p$.

\begin{lem} \label{lem-mpc8}
\marpar{lem-mpc8}
Assume $(X,\OO(1),m)$ satisfies Hypothesis ~\ref{hyp-mpcU2}.
Assume that the
fiber $M_p$ of
$$
\text{ev}:\Kgnb{0,m}(X,m\alpha) \rightarrow X^m
$$
over a general point $p$ of $X^m$ 
is uniruled by rational curves of $\lambda$-degree $1$.  
With one exception,
the space of $\lambda$-degree $1$ curves in $M_p$
containing a general point of $M_p$ has dimension
$$
\frac{m}{2}\langle 2\text{ch}_2(T_X),\Pi \rangle - \frac{m-2}{2} \langle
c_1(T_X),\alpha \rangle + m-2.
$$
The one exception occurs when $m=4$ and the $\lambda$-degree $1$ curves
sweep out Veronese surfaces.  In this case the dimension equals
$$
2\langle 2\text{ch}_2(T_X),\Pi \rangle - \langle
c_1(T_X),\alpha \rangle.
$$
\end{lem}

\begin{proof}
For a free curve $D$ in a smooth projective variety $M$, the dimension
of the space of deformations of $D$ containing a general point $m$
equals
$$
\langle c_1(T_M), D \rangle - 2,
$$
cf. \cite[Theorem II.1.7, Corollary II.3.5.3]{K}.  The formula for
$c_1(T_M)$ is given in Lemma ~\ref{lem-mpc5}.  By hypothesis, the
$\lambda$-degree is $1$.  Thus the only missing data to compute
$\langle c_1(T_M),D \rangle$ is the intersection number with
$\Delta$.  The two possibilities for $D$ are given in Lemma
~\ref{lem-mpc7}: either $D$ is a pencil of hyperplane sections of a
rational normal surface scroll, or $D$ is a pencil of hyperplane
sections of a Veronese surface.  In the first case, the intersection
number with $\Delta$ equals $m$:  the hyperplane becomes reducible
precisely when it contains a line of ruling through one of the $m$
points $p_1,\dots,p_m$.  In the second case, the intersection number
with $\Delta$ is $3$ (even though $m=4$): a pencil of conics in
$\PP^2$ has $3$ reducible members.  Substituting this in gives the two
formulas above.  
\end{proof}

\mni
Assuming $X$ is a linear section of sufficiently high codimension in a
smooth projective variety $Y$, Lemma ~\ref{lem-mpc8} implies the
necessary condition above for the existence of a $\lambda$-degree $1$
free curve is also a sufficient condition.

\begin{lem} \label{lem-mpc9}
\marpar{lem-mpc9}
Assume $(X,\OO(1),m)$ satisfies Hypothesis ~\ref{hyp-mpcU2}.
If $m\neq 4$,
assume that
$$
\frac{m}{2}\langle 2\text{ch}_2(T_X),\Pi \rangle - \frac{m-2}{2} \langle
c_1(T_X),\alpha \rangle + m-2 \geq  0.
$$ 
If $m = 4$, assume that
$$
2\langle 2\text{ch}_2(T_X),\Pi \rangle - \langle
c_1(T_X),\alpha \rangle \geq 0.
$$
Assume there exists a smooth projective variety $Y$ such that $X$ is a
codimension $c$ linear section of $Y$.  If
$$
c > 2(m-1)\langle c_1(T_X),\alpha \rangle - m \langle
2\text{ch}_2(T_X), \Pi \rangle - (m-1)\text{dim}(X) +3m - 2
$$
then a general fiber $M$ of
$$
\text{ev}: \Kgnb{0,m}(X,m\alpha) \rightarrow X^m
$$
is uniruled by rational curves of $\lambda$-degree $1$.  
\end{lem}

\begin{proof}
First of all, the hypotheses of Corollary ~\ref{cor-mpc6} for $X$
imply the hypotheses for $Y$.  There is a formula
$$
(m-1)\text{dim}(Y) +m \langle 2\text{ch}_2(T_Y),\Pi \rangle -
2(m-1)\langle c_1(T_Y),\alpha \rangle -3 m + 2  =
$$
$$
c + (m-1)\text{dim}(X) +m \langle 2\text{ch}_2(T_X),\Pi \rangle -
2(m-1)\langle c_1(T_X),\alpha \rangle - 3 m + 2. 
$$
Therefore the inequality on $c$ implies that
$$
(m-1)\text{dim}(Y) +m \langle 2\text{ch}_2(T_Y),\Pi \rangle -
2(m-1)\langle c_1(T_Y),\alpha \rangle -3 m + 2  > 0.
$$
By Corollary ~\ref{cor-mpc6}, a general fiber of 
$$
\text{ev}:\Kgnb{0,m}(Y,m\alpha) \rightarrow Y^m
$$
is uniruled by $\lambda$-degree $1$ rational curves.  By Lemma
~\ref{lem-mpc8}, the dimension of the space of such rational curves is
at least
$$
a_Y = \frac{m}{2}\langle 2\text{ch}_2(T_Y),\Pi \rangle - \frac{m-2}{2} \langle
c_1(T_Y),\alpha \rangle + m-2
$$
if $m\neq 4$, and is at least
$$
a_Y = 2\langle 2\text{ch}_2(T_Y),\Pi \rangle - \langle
c_1(T_Y),\alpha \rangle 
$$
if $m=4$.  
Of course there are formulas
$$
\frac{m}{2}\langle 2\text{ch}_2(T_Y),\Pi \rangle - \frac{m-2}{2} \langle
c_1(T_Y),\alpha \rangle + m-2 =
$$
$$
c + \frac{m}{2}\langle 2\text{ch}_2(T_X),\Pi \rangle - \frac{m-2}{2} \langle
c_1(T_X),\alpha \rangle + m-2,
$$
respectively
$$
2\langle 2\text{ch}_2(T_Y),\Pi \rangle - \langle
c_1(T_Y),\alpha \rangle  =
$$
$$
c+ 2\langle 2\text{ch}_2(T_X),\Pi \rangle - \langle
c_1(T_X),\alpha \rangle.
$$
In other words,
$$
a_Y = c + a_X.
$$

\mni
By upper semicontinuity, for every point of
$\Kgnb{0,m}(Y,m\alpha)$ contained in an irreducible component
dominating $Y^m$, the dimension of the space of $\lambda$-degree $1$
curves is at least $a_Y$.  Let $[C]\in \Kgnb{0,m}(X,m\alpha)$ be a
general point of a general fiber of
$$
\text{ev}:\Kgnb{0,m}(X,m\alpha) \rightarrow X^m.
$$
By Lemma ~\ref{lem-mpc3}, $C$ is a smooth curve and
$$
h^1(C,T_X|_C(-m)) = 0.
$$
There is a short exact sequence
$$
\begin{CD}
0 @>>> T_X @>>> T_Y|_X @>>> \OO(1)^{\oplus c} @>>> 0
\end{CD}
$$
giving rise to a short exact sequence
$$
\begin{CD}
0 @>>> T_X|_C @>>> T_Y|_C @>>> \OO(1)|_{C}^{\oplus c} @>>> 0.
\end{CD}
$$
Since $C$ has $\OO(1)$-degree $m$, $\OO(1)|_C \cong \OO_C(m)$.
Therefore twisting this sequence by $\OO_C(-m)$ gives
$$
\begin{CD}
0 @>>> T_X|_C(-m) @>>> T_Y|_C(-m) @>>> \OO_C^{\oplus c} @>>> 0.
\end{CD}
$$
Since
$$
h^1(C,\OO_C) = 0
$$
by the long exact sequence of cohomology, also
$$
h^1(C,T_Y|_C(-m)) = 0
$$
Thus $C$ deforms to a curve in $Y$ passing through $m$ general
points.  Therefore the dimension of the space of $\lambda$-degree $1$
rational curves containing $C$, for $Y$, is at least $a_Y$.  

\mni
Denote by $M_X$ the fiber of
$$
\text{ev}:\Kgnb{0,m}(X,m\alpha) \rightarrow X^m
$$
over a general point $(p_1,\dots,p_m)$
and denote by $M_Y$ the fiber of
$$
\text{ev}:\Kgnb{0,m}(Y,m\alpha) \rightarrow Y^m
$$
over the same point $(p_1,\dots,p_m)$.
Recall there is a morphism $M_Y \rightarrow \PP^{N-m}$ defined as follows.
Associated to a curve $C$ in $\PP^N$ containing $p_1,\dots,p_m$,
there is the point $v=\text{span}(C)$ in the projective space $\PP^{N-m}
= \PP^N/\text{span}(p_1,\dots,p_m)$.  
Each $\lambda$-degree $1$
curve in $M_Y$ maps to 
a line in $\PP^{N-m}$.  Because $X$ is a
codimension $c$ linear section of $Y$, there is a codimension $c$
linear subspace $\Lambda$ of $\PP^{N-m}$ such that $M_X$ equals
$M_Y\times_{\PP^{N-m}} \Lambda$.  Let $[C]$ be a general point of
$M_X$ mapping to a point $v$ of $\PP^{N-m}$.  By the previous
paragraph, the space of $\lambda$-degree $1$ curves in $M_Y$
containing $[C]$ is at least $a_Y$.  These map to lines in $\PP^{N-m}$
containing $v$, i.e., these correspond to points in the quotient
projective space $\PP^{N-m-1} = \PP^{N-m}/v$.  Of course $v\in
\Lambda$ and $\Lambda/v$ is a codimension $c$ linear subspace of
$\PP^{N-m-1}$.  Therefore, as long as
$$
a_Y \geq c
$$
there exists a $\lambda$-degree $1$ curve in $M_Y$ containing $[C]$
whose image is containing in $\Lambda$, i.e., there exists a
$\lambda$-degree $1$ curve in $M_X$.  Because $a_Y = c + a_X$, the
inequality above is precisely,
$$
a_X \geq 0
$$
which is the inequality in the statement of the lemma.     
\end{proof}

\mni
The main application is to complete intersections.

\begin{cor} \label{cor-mpc10}
\marpar{cor-mpc10}
Let $X$ be a smooth complete intersection in $\PP^n$ of type
$\underline{d} = (d_1,\dots,d_c)$.  Without loss of generality, assume
all $d_i > 1$. Also assume $\underline{d} \neq (2)$.  If
$\underline{d} = (3)$, assume $m\leq 4$, and if $\underline{d}=
(2,2)$, assume $m\leq 5$.  

\mni
If $m\neq 4$,
assume
$$
n \geq \sum_{i=1}^c (\frac{m}{2}d_i^2 - \frac{m-2}{2}d_i) + 1-m.
$$
If $m=4$, assume
$$
n \geq \sum_{i=1}^c (2d_i^2 - d_i) - 1.
$$
Then $(X,\OO(1),m)$ satisfies Hypothesis ~\ref{hyp-mpcU2}
and a
general fiber of
$$
\text{ev}:\Kgnb{0,m}(X,m[\text{line}]) \rightarrow X^m
$$
is uniruled by $\lambda$-degree $1$ rational curves.
\end{cor}

\begin{proof}
First of all, the inequalities above implies the inequality in Corollary
~\ref{cor-mpcdom5}.  Thus, Corollary ~\ref{cor-mpcdom5} together with
Lemma~\ref{lem-mpcCH} implies $m[\text{line}]$ is an $m$-minimal class.
If $\text{dim}(X) = \langle c_1(T_X),\alpha \rangle + 1$, then
$\underline{d} = (3)$ or $\underline{d}=(2,2)$.  In the first case,
$h^0(X,\OO(1)) \geq \text{dim}(X) + 2 = \text{dim}(X) + 4 - 2$.  In
the second case, $h^0(X,\OO(1)) \geq \text{dim}(X) + 3 = \text{dim}(X)
+ 5-2$.  Thus, in every case $(X,\OO(1),m)$ satisfies Hypothesis
~\ref{hyp-mpcU}.  Moreover,
$$
\text{ch}_2(T_X) = \frac{\langle 2 \text{ch}_2(T_X),\Pi \rangle}{2}
c_1(\OO(1))^2,
$$
for the integer
$$
\langle 2\text{ch}_2(T_X),\Pi \rangle = n+1 - \sum_{i=1}^c d_i^2.
$$
Therefore $X$ satisfies Hypothesis ~\ref{hyp-mpcU2}.

\mni
Because a complete intersection is infinitely extendable, Lemma
~\ref{lem-ext3}, 
the second
hypothesis of Lemma~\ref{lem-mpc9} holds.  Substituting
in
$$
\langle 2 \text{ch}_2(T_X),\Pi \rangle = n+1 -\sum_{i=1}^c d_i^2
$$
and
$$
\langle c_1(T_X), [\text{line}] \rangle = n+1 - \sum_{i=1}^c d_i,
$$
the first hypothesis of Lemma ~\ref{lem-mpc9} is precisely 
$$
n \geq \sum_{i=1}^c (\frac{m}{2}d_i^2 - \frac{m-2}{2}d_i) + 1-m
$$
if $m\neq 4$,
respectively
$$
n \geq \sum_{i=1}^c (2d_i^2 - d_i) - 1
$$
if $m=4$.  Thus, by Lemma ~\ref{lem-mpc9}, a general fiber of
$$
\text{ev}:\Kgnb{0,m}(X,m[\text{line}]) \rightarrow X^m
$$
is uniruled by $\lambda$-degree $1$ rational curves.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%
%% Section: Twisting surfaces. Existence
%% 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Twisting surfaces. Existence} \label{sec-q}
\marpar{sec-q}

\mni
Every rational ruled surface
$$
\pi:\Sigma \rightarrow \PP^1
$$
is isomorphic to the projective bundle 
$$
\PP(\OO_{\PP^1} \oplus \OO_{\PP^1}(-h)) \rightarrow \PP^1
$$
for some nonnegative integer $h$.  We will call the integer $h$ the
\emph{Hirzebruch type} of the ruled surface, 
or \emph{H-type} for short.  We are mostly
interested in the case that $h=0$, i.e., $\Sigma = \PP^1 \times
\PP^1$.  But for some arguments it is convenient to have a more
general setup.

\mni
Denote by $F$ the divisor class of a fiber of $\pi$ in $\Sigma$.
Let $E$ be the divisor class of a section of $\pi$ with minimal
self-intersection, i.e., $E$ is the directrix.  If this section is
unique, also denote by $E$ the unique effective curve in this divisor
class.  The self-intersection of the section is
$$
(E\cdot E)_\Sigma = -h.
$$
Denote by $F'$ the divisor class 
$$
F' := E - (E\cdot E)_\Sigma F = E + hF
$$
i.e., $F'$ is the unique divisor class of a section such that
$$
(F'\cdot E)_\Sigma = 0.
$$
 
\mni
Associated to every morphism from a rational ruled surface to $X$,
$$
f:\Sigma \rightarrow X,
$$
there is a morphism
$$
(\pi,f):\Sigma \rightarrow \PP^1 \times X.
$$
In other words, consider $f$ as a family of morphisms from $\PP^1$ --
the fibers of $\pi$ -- to $X$ parametrized by the base of $\pi$, which
is also $\PP^1$.  If $(\pi,f)$ is finite, the \emph{vertical normal
  bundle} is defined to be
$$
N_{(\pi,f)} := \text{Coker}(d(\pi,f):T_\Sigma \rightarrow
(\pi,f)^*T_{\PP^1\times X}).
$$
Although $N_{(\pi,f)}$ need not be locally free, it is $\pi$-flat.

\begin{lem} \label{lem-q0}
\marpar{lem-q0}
The sheaf $N_{(\pi,f)}$ is $\pi$-flat. 
\end{lem}

\begin{proof}
By the local flatness criterion,
\cite[Theorem 22.5]{Ma}, it suffices to prove that for every geometric
point $t$ of 
$\PP^1$, the induced morphism
$$
d(f_t):\Sigma_t \rightarrow f_t^* T_X
$$
is injective, where $f_t$ is the restriction of $f$ to $\Sigma_t =
\pi^{-1}(t)$.
By hypothesis, $f_t$ is finite.  Since the
characteristic is $0$, this implies $f_t$ is generically unramified
and thus $d(f_t)$ is injective.
\end{proof}

\mni
There is a slightly technical lemma which is useful.  Let
$$
\pi:\Sigma \rightarrow \PP^1
$$
be a ruled surface.  
Let $n$ be a positive integer.
Let $N$ be a
coherent sheaf on $\Sigma$.    
There is a cup-product map
$$
H^0(\Sigma,\OO_\Sigma(F'+nF))\times H^1(\Sigma,N(-F'-nF))
\rightarrow H^1(\Sigma,N),
$$
or equivalently a map
$$
c:H^0(\Sigma,\OO(F'+nF)) \rightarrow
\text{Hom}(H^1(\Sigma,N(-F'-nF)), H^1(\Sigma,N)).
$$

\begin{lem} \label{lem-qtech}
\marpar{lem-qtech}
Let $n$ be a positive integer.
Let $N$ be a $\pi$-flat coherent sheaf on $\Sigma$.
Assume there
exists an $N$-regular section $\alpha$ of $H^0(\Sigma,\OO_\Sigma(F'+nF))$,
for
which $c(\alpha)$ is injective.  Also assume there exists an $N$-regular
divisor $D\in |\OO_{\Sigma}(F'+nF)|$ 
such that $N|_D$ is
generated by global sections.  Then $N$ is generated by global
sections and $h^1(\Sigma, N(-F'-nF)) = 0$.
\end{lem}

\begin{proof}
Let $U$ be the maximal open subscheme of $\Sigma$ on which
$N$ is generated by global sections.  The goal is to prove that $U$
equals all of $\Sigma$.  

\mni
If $c(\alpha)$ is injective for one $N$-regular $\alpha$, then it is
injective for a general choice of $\alpha$.  Similarly, if $N|_D$ is
generated by global sections for one $N$-regular $D$, it is generated by
global sections for a general choice of $D$.  Thus there exists
$\alpha$ and $D$ as above so that $D$ is the zero locus of $\alpha$.
Also, we may assume that $D$ is irreducible (hence smooth).

\mni
Consider the short exact sequence
$$
\begin{CD}
0 @>>> N(-D) @>>> N @>>> N|_D @>>> 0.
\end{CD}
$$
There is a long exact sequence of cohomology, some of whose terms are
$$
\begin{CD}
H^0(\Sigma,N) @>>> H^0(D,N|_D) @>>> H^1(\Sigma, N(-n,-1)) 
@> c(\alpha) >> H^1(\Sigma, N). 
\end{CD}
$$
By the hypothesis that $c(\alpha)$ is injective, the first map is
surjective.  By the hypothesis that $N|_D$ is generated by global
sections, $D$ is contained in $U$.  
Since $D$ is an ample divisor, the
complement of $U$ consists of finitely many closed points of
$\Sigma$.  

\begin{claim} \label{claim-qtecha}
\marpar{claim-qtecha}
Both
$$
R^1\pi_*N \text{  and  } R^1\pi_* N(-D) 
$$
are zero.  
\end{claim}

\mni
Because $N$ is $\pi$-flat, by
cohomology and base change  \cite[Theorem III.12.11]{H},
it suffices to prove that for
every fiber $F$ of $\pi$,
$$
h^1(F,N|_F) = h^1(F,N(-D)|_F) = 0.
$$
Now $N|_F$ is a coherent sheaf on $F\cong \PP^1$ whose global sections
generate $N$ on the dense open $F\cap U$.  
Every coherent sheaf on $\PP^1$ that is generically
generated is globally generated.  And for a globally generated
coherent sheaf $N|_F$ on $\PP^1$,
$$
h^1(F,N|_F) = h^1(F,N|_F(-1)) = 0.
$$
This is easy to see using Grothendieck's lemma on splitting of vector
bundles together with the cohomology of invertible sheaves on
$\PP^1$.  Therefore
$$
R^1\pi_* N = R^1\pi_* N(-D) = (0),
$$
i.e., Claim ~\ref{claim-qtecha} is valid.

\mni
Because of Claim ~\ref{claim-qtecha}, the Leray spectral sequence
associated to $\pi$ proves that $c(\alpha)$ is a morphism
$$
H^1(\PP^1,\pi_*N(-D)) \rightarrow
H^1(\PP^1,\pi_*N).
$$
Because $c(\alpha)$ is injective
$$
h^1(\PP^1,\pi_* N(-D)) \leq h^1(\PP^1,\pi_* N).
$$

\mni
There is an isomorphism
$$
\pi_* N(-D) \cong [\pi_*(N(-F'))]\otimes \OO(-n),
$$
where $F'$ is a general member in its linear equivalence class.
In particular, $F'$
is contained in $U$, thus $N|_{F'}$ is generated by global sections.
Since there is a short exact sequence
$$
\begin{CD}
0 @>>> N(-F') @>>> N @>>> N|_{F'} @>>> 0
\end{CD}
$$
and since
$$
R^1\pi_* N(-F') \cong (R^1\pi_* N(-D))(n)
$$
is zero by Claim ~\ref{claim-qtecha},
there is a short exact sequence
$$
\begin{CD}
0 @>>> \pi_* N(-F') @>>> \pi_* N @>>> N|_{F'}
@>>> 0.
\end{CD}
$$
Since $N|_{F'}$ is generated by global sections,
$$
h^1(\PP^1,N|_{F'}) = 0.
$$
Thus the long exact sequence of cohomology gives
$$
h^1(\PP^1,\pi_* N(-F')) \geq h^1(\PP^1,\pi_*N).
$$

\mni
Since $\pi_* N(-F')$ is coherent, by Grothendieck's lemma
it has the form
$$
\pi_* N(-F') \cong (\text{Torsion}) \oplus
\OO(a_1)\oplus \dots \oplus \OO(a_r)
$$ 
for some
integers $a_1,\dots,a_r$.  If $h^1$ is positive, some of the $a_i$ are
negative.  Then, of course, twisting by $\OO(-n)$ makes those $a_i$
more negative. Therefore, if
$$
h^1(\PP^1,\pi_* N(-F')) \neq 0,
$$
then
$$
h^1(\PP^1,[\pi_* N(-F')]\otimes \OO(-m)) >
h^1(\PP^1,\pi_* N(-F')),
$$
which implies
$$
h^1(\PP^1,\pi_* N(-D)) > h^1(\PP^1,\pi_*N).
$$
This contradicts the previous paragraph.  Therefore
$$
h^1(\PP^1,\pi_* N(-F')) = 0.
$$
By the previous inequalities this implies
$$
h^1(\PP^1,\pi_* N) = 0,
$$
which in turn implies
$$
h^1(\PP^1,\pi_* N(-D)) = 0.
$$
By Claim ~\ref{claim-qtecha} and the Leray spectral sequence for $\pi$,
$$
h^1(\Sigma,N(-D)) = h^1(\PP^1,\pi_* N(-D)) = 0.
$$
Thus it only remains to prove $N$ is globally generated.

\mni
Since $\pi_* N(-D)$ is a coherent sheaf on $\PP^1$
with vanishing $h^1$, the twist
$[\pi_* N(-D)]\otimes \OO(n)$ is globally generated,
i.e., $\pi_* N(-F')$ is globally generated.
As the first and last terms in the
short exact sequence
$$
\begin{CD}
0 @>>> \pi_* N(-F') @>>> \pi_* N @>>> N|_{F'}
@>>> 0
\end{CD}
$$
are globally generated, also
$\pi_* E$
is globally generated.  As established above, 
$E$ is $\pi$-relatively globally generated. 
Therefore $E$ is globally generated.  
\end{proof}

\mni
The geometric question behind these definitions is this: Given a
rational ruled surface in $X$, and given a deformation in $X$ of a
rational curve in the surface, is there a corresponding deformation of
the surface containing the deformation of the curve?  The following
lemma gives one answer.

\begin{lem} \label{lem-qmove}
\marpar{lem-qmove}
Let $\pi:\Sigma\rightarrow \PP^1$ be a ruled surface.  Let $f:\Sigma
\rightarrow X$ be a morphism such that $(\pi,f)$ is finite.  Let $n$
be a positive integer. 
Assume
that $N_{(\text{pr}_1,f)}$ is globally generated and
$h^1(\Sigma,N_{(\text{pr}_1,f)}(-F'-nF))=0$.  Every free curve in $\Sigma$
maps to a free curve in $X$.  Moreover, for every reduced curve
$D$ in $|\OO_{\Sigma}(F'+nF)|$, for every infinitesimal deformation of
$D$ in $X$, there exists an infinitesimal deformation of $\Sigma$ in $X$
containing the deformation of $D$.
\end{lem}

\begin{proof}
For a free curve $C$ in $\Sigma$, both $N_{C/\Sigma}$ and
$N_{(\text{pr}_1,f)}|_C$ are globally generated.  Therefore $N_{C/X}$
is globally generated, i.e., $C$ is a free curve in $X$. 

\mni
In particular, since a reduced curve $D$ in $|\OO_{\Sigma}(F'+nF)|$ is
free in $\Sigma$, it is free in $X$.  Thus the deformation space of
the morphism $(D,f|_D:D\rightarrow X)$ (allowing both $D$ and $f$ to vary) is
smooth.  Because 
$$
h^1(D,N_{(\text{pr}_1,f)}|_D) = h^1(\Sigma, N_{(\text{pr}_1,f)}(-D)) =
0,
$$
and by the long exact sequence of cohomology associated to the short
exact sequence 
$$
\begin{CD}
0 @>>> N_{(\text{pr}_1,f)}(-D) @>>> N_{(\text{pr}_1,f)} @>>>
N_{(\text{pr}_1,f)}|_D @>>> 0,
\end{CD}
$$
it follows that $h^1(\Sigma,N_{(\text{pr}_1,f)}) = 0$.  Thus
the deformation space of $(\pi,f):\Sigma \rightarrow \PP^1\times X$
(allowing both $\Sigma$ and $(\pi,f)$ to vary) is also smooth.  It
follows that the deformation space of the datum,
$$
(\Sigma,D\subset \Sigma,(\pi,f):\Sigma \rightarrow \PP^1\times X),
$$
(allowing $\Sigma$ and $(\pi,f)$ to vary, and allowing $D$ to vary as a
divisor in $\Sigma$) is also smooth.

\mni
There is a morphism from the deformation space of the datum
$(\Sigma,D,(\pi,f))$ to the deformation space of $(D,f|_D)$.  Because
both are smooth, the morphism is smooth if
and only if the induced map of Zariski tangent spaces is surjective,
i.e., if and only if
$$
H^0(\Sigma,N_{(\text{pr}_1,f)}) \rightarrow
H^0(D,N_{(\text{pr}_1,f)}|_D)
$$
is surjective.  By the long exact sequence above, the cokernel is
contained in $H^1(\Sigma,N_{(\text{pr}_1,f)}(-D))$.  This is zero by
hypothesis, therefore the map of deformation spaces is surjective.
\end{proof}

\mni
The central notion of this section is as follows.  It is closely
related to the notion of \emph{twisting family of pointed curves} from
\cite{HS2} and \cite{Sr1c}.

\begin{defn} \label{defn-q1}
\marpar{defn-q1}
For an integer $n$
an \emph{$n$-twisting surface} in $X$ is a ruled surface
$$
\pi:\Sigma \rightarrow \PP^1
$$
together with a morphism 
$$
f:\Sigma \rightarrow X
$$ 
such that 
\begin{enumerate}
\item[(i)]
$f^* T_X$ is generated by global sections,
\item[(ii)]
$(\pi_1,f)$ is finite and
$$
h^1(\Sigma,N_{(\pi,f)}(-F'-nF)) = 0.
$$
\end{enumerate}

\mni
If the pushforward of the divisor class $F$ equals
$\beta_1$ and the pushforward of $F'$ equals
$\beta_2$, then \emph{$f$ has class $(\beta_1,\beta_2)$}.  

\mni
Let $\alpha$ and $M=M_{\alpha,1}$ be given as in Lemma~\ref{lem-cic1}.
If the pushforward of every element in $|\OO_\Sigma(F)|$ is in $M_{e_1\cdot
\alpha,0}$ and the pushforward of every element in $|\OO_\Sigma(F')|$ is
in $M_{e_2\cdot \alpha,0}$, then \emph{$f$ has $M$-class $(e_1\cdot
\alpha, e_2\cdot \alpha)$}.  
\end{defn}

\mni
There are some elementary results comprising the ``sorites'' of
twisting surfaces.

\begin{lem} \label{lem-qa}
\marpar{lem-qa}
\begin{enumerate}
\item[(i)]
For $l\leq n$, every $n$-twisting surface is also $l$-twisting.

\item[(ii)]
Let $f$ be an $n$-twisting surface.  For all integers 
$l_1 \geq \text{max}(-1,-n)$ and $l_2 \geq 0$,
$$
h^1(\Sigma,N_{(\pi,f)}(l_1F'+l_2F)) = 0.
$$

\item[(iii)]
If $f$ is a $1$-twisting surface of $H$-type $0$, 
then the composition of $f$ with the
permutation morphism
$$
(\text{pr}_2,\text{pr}_1):\PP^1 \times \PP^1 \rightarrow \PP^1\times
\PP^1
$$
is also $1$-twisting of $H$-type $0$.  

\item[(iv)]
Let $h=0$ or $h=1$ and let $n=1-h$.
For $i=1,2$, let 
$$
f_i:\Sigma_i \rightarrow X
$$
be an $n$-twisting surface in $X$ of $H$-type $h$ and with class
$(\beta_i,\beta)$, respectively with $M$-class $(e_i\cdot \alpha,e\cdot
\alpha)$.  
Assume there exist irreducible divisors $D_1$ of class $E$ in
$\Sigma_1$
and $D_2$ of class
$F'$ in $\Sigma_2$
such that
$f_1|_{D_1}$ equals $f_2|_{D_2}$.  Then there exists a
$n$-twisting surface in $X$ of H-type $h$ and with class
$(\beta_1 + \beta_2,\beta)$, resp. with $M$-class $((e_1+e_2)\cdot
\alpha,e\cdot \alpha)$.  

\item[(v)]
Let $n_1$ and $n_2$ be positive integers.
For $i=1,2$, let 
$$
f_i:\Sigma_i \rightarrow X
$$ 
be an $n_i$-twisting surface in $X$ of 
class
$(\beta,\beta_i)$, respectively of $M$-class $(e\cdot \alpha,e_i\cdot
\alpha)$.  Assume the H-type of $f_1$ is $0$ and the H-type of $f_2$
is $h=0$ or $h=1$.
For $i=1,2$, assume there exists fibers $F_i$ of
$\pi_i:\Sigma_i\rightarrow \PP^1$ and an isomorphism $F_1\cong F_2$
such that
$$
f_1|_{F_1} = f_2|_{F_2}.
$$
Then there exists an $(n_1+n_2-1)$-twisting
surface $f$ in $X$ of H-type $h$ and with
class $(\beta,\beta_1+\beta_2)$, resp. with $M$-class
$(e\cdot \alpha, (e_1+e_2)\cdot \alpha)$.
Moreover, the restrictions of $f_1$, $f_2$ and $f$ to $F$-curves give
points of $\Kgnb{0,0}(X)$ that are all parametrized by the same
irreducible component.
\end{enumerate}
\end{lem}

\begin{proof}
\mni
\textbf{(i)}  
If $l=n$, this is just the hypothesis that $f$ is $n$-twisting.  Thus
assume $l<n$.  Let $C$ be a general member of the linear system
$|\OO_\Sigma((n-l)F)|$.  Thus $C$ is a disjoint union of $n-l$ fibers,
$C=C_1\cup \dots \cup C_{n-l}$.
There is a short exact sequence
$$
\begin{CD}
0 @>>> N_{(\pi,f)}(-F'-nF) @>>> N_{(\pi_1,f)}(-F'-lF) @>>> 
N_{(\pi_1,f)}(-F'-lF)|_C
@>>> 0.
\end{CD}
$$
Applying the long exact sequence in cohomology and the induction
hypothesis,
$$
h^1(\Sigma, N_{(\pi,f)}(-F'-lF)) 
\leq h^1(C,N_{(\pi,f)}(-F'-lF)|_C) =
\sum_{i=1}^{n-l} h^1(C_i,N_{(\pi,f)}(-F'-lF)|_{C_i}).
$$
There is a natural isomorphism
$$
N_{(\pi,f)}(-F'-lF)|_{C_i} \cong N_{(\pi,f)}|_{C_i}(-1).
$$
Because $f^*T_X$ is globally generated, the same holds for
$N_{(\pi,f)}|_{C_i}$.  By Grothendieck's lemma,
$$
N_{(\pi,f)}|_{C_i} \cong \text{Torsion}\oplus
\OO(a_1)\oplus \dots \oplus \OO(a_r)
$$
for integers $a_1,\dots,a_r \geq 0$.  Therefore
$$
N_{(\pi,f)}|_{C_i}(-1) \cong \text{Torsion}\oplus
\OO(a_1-1)\oplus \dots \oplus \OO(a_r-1)
$$
for integers $a_i-1\geq -1$.  Since $h^1(C_i,\OO(b)) = 0$ for $b\geq
-1$, it follows that
$$
h^1(C_i,N_{(\pi,f)}(-F'-lF)|_{C_i}) = 0.
$$
Therefore also
$$
h^1(\Sigma,N_{(\pi,f)}(-F'-lF)) = 0.
$$

\mni
\textbf{(ii)}
Let $C$ be a general member of the nonempty
linear system $|\OO((l_1+1)F'+(l_1+n)F)|$.
There is a short exact sequence
$$
\begin{CD}
0 @>>> N_{(\pi,f)}(-F'-nF) @>>> N_{(\pi,f)}(l_2F'+l_1F) 
@>>> N_{(\pi,f)}(l_2F'+l_1F)|_C @>>> 0.
\end{CD}
$$
By the long exact sequence in cohomology and the hypothesis that $f$
is $n$-twisting, it suffices to prove that
$$
h^1(C,N_{(\pi,f)}(l_2F'+l_1F)|_C)=0.
$$

\mni
Because $f^*T_X$ is globally generated, also $N_{(\pi,f)}$ is
globally generated.  Thus
$N_{(\pi,f)}(l_2F'+l_1F)|_C$
is a quotient of $\OO_\Sigma(l_2F'+l_1F)|_C^{\oplus N}$, for some integer $N$.
Since $H^1(C,-)$ is right exact, it suffices to prove that
$$
h^1(C,\OO_\Sigma(l_2F'+l_1F)|_C) = 0.
$$
Again using the long exact sequence of cohomology associated to the
short exact sequence
$$
\begin{CD}
0 @>>> \OO_\Sigma(-F'-nF) @>>> \OO_\Sigma(l_2F'+l_1F) @>>> 
\OO_\Sigma(l_2F'+l_1F)|_C @>>> 0
\end{CD}
$$
and the fact that 
$$
h^1(\Sigma,\OO(-F'-nF)) = h^2(\Sigma,\OO(-F'-nF))
= 0
$$
(using the Leray spectral sequence of $\pi$),
it is equivalent to prove that
$$
h^1(\Sigma,\OO(l_2F'+l_1F)) = 0.
$$
This holds since $l_1 \geq -1$ and $l_2 \geq 0$.  

\mni
\textbf{(iii)}
As
$$
h^1(\PP^1\times \PP^1,T_{\PP^1\times \PP^1}(-1,-1)) =
h^2(\PP^1\times \PP^1,T_{\PP^1\times \PP^1}(-1,-1)) = 0,
$$
from the long exact sequence of cohomology, 
$$
h^1(\PP^1\times \PP^1, N_{(\text{pr}_1,f)}(-1,-1)) = h^1(\PP^1\times
\PP^1,f^* T_X(-1,-1)).
$$
Therefore $f$ is $1$-twisting if and only if $f^*T_X$ is globally
generated and
$$
h^1(\PP^1\times \PP^1,f^*T_X(-1,-1)) = 0.
$$
This is clearly symmetric in the two factors.

\mni
\textbf{(iv)}
Define $\Sigma'$ to be the coproduct of $\Sigma_1$ and $\Sigma_2$
attached along $D_1\cong D_2$, the isomorphism being the unique one
compatible with $\pi_1$ and $\pi_2$.  There is a unique morphism
$$
\pi':\Sigma'\rightarrow \PP^1
$$
such that $\pi'|_{\Sigma_i} = \pi_i$ for $i=1,2$.  
There is a unique morphism
$$
f':\Sigma' \rightarrow X
$$
such that $f'|_{\Sigma_i} = f_i$ 
for $i=1,2$.  By (i) and (ii), 
$$
h^1(\Sigma_i,N_{(\pi_i,f_i)} = h^1(\Sigma_i,N_{(\pi_i,f_i)}(-E)) = 0
$$
for $i=1,2$.
Since $h=0$ or $1$, also
$$
h^1(\Sigma_i, T_{\Sigma_i}(-E)) = h^1(\Sigma_i, T_{\Sigma_i}(-E))
$$
for $i=1,2$.  Thus
$$
h^1(\Sigma_i, f_i^*T_X) = h^1(\Sigma_i, f_i^*T_X(-E)) = 0.
$$
It follows from the long exact sequence of cohomology that
$$
h^1(\Sigma',(f')^*T_X) = 0.
$$
In particular, there are no obstructions to deforming $f'$.  Since
$\Sigma'$ deforms to a ruled surface
$$
\pi:\Sigma \rightarrow \PP^1
$$
of H-type $h$, it follows that $f'$ deforms to a morphism
$$
f:\Sigma \rightarrow \PP^1.
$$
Clearly the class of $f$ is $(\beta_1+\beta_2,\beta)$, resp. with
$M$-class $((e_1+e_2)\cdot \alpha, e\cdot \alpha)$.  It only remains
to prove $f$ is $1$-twisting.

\mni
Let $\mc{L}$ be the invertible sheaf on $\PP^1\times C$ whose
restriction to $\Sigma_1$ is $\OO(-F'-nF)$ and whose
restriction to $\Sigma_2$ is $\OO(-nF)$.
There is a short exact sequence
$$
\begin{CD}
0 @>>> f_2^*T_X(-F'-nF) @>>> (f')^* T_X\otimes \mc{L} @>>>
f_2^*T_X(-F'-nF) @>>> 0. 
\end{CD}
$$
As above,
$$
h^1(\Sigma_i,T_{\Sigma_i}(-F'-nF)) = 0;
$$
for $h=1$ this is precisely the statement above, for $h=0$, this
follows from the fact that 
$$
h^1(\PP^1\times \PP^1,\OO(-1,1)) = 0.
$$
By the long exact sequence of cohomology,
$$
h^1(\Sigma_i, f_i^* T_X(-F'-nF) ) = 0.
$$
Therefore
$$
h^1(\Sigma',(f')^*T_X\otimes \mc{L}) = 0.
$$
The invertible sheaf $\mc{L}$ deforms to
$\OO(-F'-nF)$ on $\Sigma$.
Therefore, by upper semicontinuity,
$$
h^1(\Sigma, f^*T_X(-F'-nF)) = 0,
$$
i.e., $f$ is $n$-twisting.

\mni
\textbf{(v)}
Let $\Sigma'$ be the coproduct of $\Sigma_1$ and $\Sigma_2$ via the
isomorphism $F_1\cong F_2$.  Let $C'$ be the coproduct of $\PP^1$ and
$\PP^1$ via $\pi_1(F_1) \sim \pi_2(F_2)$.  There is a unique morphism
$$
\pi':\Sigma'\rightarrow C'
$$
such that $\pi'|_{\Sigma_i} = \pi_i$.  And there is a unique morphism
$$
f':\Sigma'\rightarrow X
$$
such that $f'|_{\Sigma_i} = f_i$.  
By (ii)
$$
h^1(\Sigma_1,N_{(\pi_1,f_1)}(-F)) = 0.
$$
Since also
$$
h^1(\Sigma_1, T_{\Sigma_1}(-F)) = 0,
$$
by the long exact sequence of cohomology,
$$
h^1(\Sigma_1,f_1^*T_X(-F)) = 0.
$$
By a similar argument,
$$
h^1(\Sigma_2,f_2^* T_X)=0.
$$
There is a short exact sequence
$$
\begin{CD}
0 @>>> f_1^*T_X(-F) @>>> (f')^*T_X @>>> f_2^*T_X @>>> 0.
\end{CD}
$$
Using the corresponding long exact sequence and the vanishing above,
$$
h^1(\Sigma,(f')^* T_X) = 0.
$$
In particular, there are no obstructions to deforming $f'$.  
Since $\Sigma'$ deforms to a ruled surface
$$
\pi:\Sigma \rightarrow \PP^1
$$
of H-type $h$, $f'$ deforms to a morphism
$$
f:\Sigma \rightarrow X.
$$
Clearly $f$ has class $(\beta,\beta_1+\beta_2)$, resp. $M$-class
$(e\cdot \alpha, (e_1+e_2)\cdot \alpha)$.  It remains to prove that
$f$ is $(n_1+n_2-1)$-twisting.  

\mni
Let $\mc{L}$ be the invertible sheaf on $\Sigma$ whose
restriction to $\Sigma_1$ is $\OO(-F'-(n_1-1)F)$ and whose restriction
to $\Sigma_2$ is $\OO(-F'-n_2F)$.  
Because
$$
N_{(\pi',f')}|_{\Sigma_i} \cong N_{(\pi_i,f_i)}
$$
for $i=1,2$, there is a short exact sequence
$$
\begin{CD}
0 @>>> N_{(\pi_1,f_1)}(-F'-n_1F) @>>>
N_{(\pi',f')}\otimes \mc{L} @>>>
N_{(\pi_2,f_2)}(-F'-n_2F) @>>> 0.
\end{CD}
$$
By the long exact sequence of cohomology and the hypothesis that $f_i$
is $n_i$-twisting for $i=1,2$,
$$
h^1(\Sigma, N_{(\pi',f')}\otimes \mc{L}) = 0.
$$
Since $N_{(\pi',f')}$ deforms to $N_{(\pi,f)}$ and since
$\mc{L}$ deforms to $\OO(-F'-(n_1+n_2-1)F)$, by upper semicontinuity
it follows that
$$
h^1(\Sigma,N_{(\pi,f)}(-F'-(n_1+n_2-1)F)) = 0.
$$
Therefore $f$ is $(n_1+n_2-1)$-twisting.

\mni
It is clear that the $F$ curves in $\Sigma$ are deformations of the
$F$ curves in $\Sigma_1$ and deformations of the $F$-curves in $\Sigma_2$.
\end{proof}

\mni
The next result is the basic ``bootstrapping'' result producing
$1$-twisting surfaces of larger class from $1$-twisting surface of
smaller class, and producing $m$-twisting surfaces for arbitrary $m$
from a $1$-twisting surface and a $2$-twisting surface.

\begin{cor} \label{cor-qa}
\marpar{cor-qa}
\begin{enumerate}
\item[(i)]
Let $n$ be a positive integer.
Let $f:\Sigma \rightarrow X$ be an $n$-twisting surface of $H$-type
$0$.  Let $D$ be a reduced divisor in $|\OO_{\Sigma}(F'+nF)|$.  There
exists an open subset of $\Kgnb{0,0}(X)$ containing $(D,f|_D)$
parametrizing maps $(C,g_C)$ for which there exists an $n$-twisting
surface $g:\Theta\rightarrow X$ of of $H$-type $0$ and an embedding of
$C$ as a divisor in $|\OO_{\Theta}(F'+nF)|$ such that $g_C = g|_C$.  In
other words, every small deformation of $D$ is contained in a
deformation of $\Sigma$ that is $n$-twisting.

\item[(ii)]
If there exists a $1$-twisting surface $f_1:\Sigma_1 \rightarrow X$ 
of $H$-type $0$ and class
$(\beta_1,\beta_2)$, resp. $M$-class $(e_1\cdot \alpha, e_2\cdot
\alpha)$, then for every positive integer $d$ there exists a
$1$-twisting surface $f_d:\Sigma_d \rightarrow X$ of $H$-type $0$ and
class $(\beta_1,d\beta_2)$, resp. $M$-class $(e_1\cdot\alpha,de_2\cdot
\alpha)$.  Moreover, for curves $C$ in $|\OO_{\Sigma}(F)|$ and curves
$C_d$ in $|\OO_{\Sigma_d}(F)|$, the stable maps $(C,f|_C)$ and
$(C_d,f_d|_{C_d})$ are parametrized by the same irreducible component
of $\Kgnb{0,0}(X)$.

\item[(iii)]
If there exists a $1$-twisting surface of $H$-type $0$ and class
$(\beta_1,\beta_2)$, resp. $M$-class $(e_1\cdot\alpha, e_2\cdot
\alpha)$, then for every pair of positive integers $d_1,d_2$, there
exists a $1$-twisting surface of $H$-type $0$ and class
$(d_1\beta_1,d_2\beta_2)$, resp. $M$-class $(d_1e_1\cdot \alpha,
d_2e_2\cdot \alpha)$.  

\item[(iv)]
Assume there exists a $1$-twisting $f_1:\Sigma_1\rightarrow X$
surface of $H$-type $0$ and class
$(\beta,\beta_1)$, resp. $M$-class $(e\cdot \alpha, e_1\cdot \alpha)$,
and assume there exists a $2$-twisting surface $f_2:\Sigma_2\rightarrow X$
of $H$-type $0$ and class
$(\beta,\beta_2)$, resp. $M$-class $(e\cdot \alpha, e_2\cdot \alpha)$,
such that the restriction of $f_1$ to $F$-curves in $\Sigma_1$ and
the restriction of $f_2$ to $F$-curves in $\Sigma_2$ give points of
$\Kgnb{0,0}(X,\beta)$ parametrized by the same irreducible component
(this is automatic if $f_i$ has $M$-class $(e\cdot \alpha, e_i\cdot
\alpha)$ for $i=1,2$).
Then for every positive integer $m$ and every nonnegative integer $r$,
there exists an $m$-twisting surface $f_{m,r}:\Sigma_{m,r}\rightarrow
X$ 
of $H$-type $0$ and class
$(\beta, r\beta_1+(m-1)\beta_2)$, resp. $M$-class $(e\cdot \alpha,
(re_1+(m-1)e_2)\cdot \alpha)$.  Moreover, the restriction of $f_{m,r}$ to
the $F$-curves in $\Sigma_{m,r}$ give points of $\Kgnb{0,0}(X,\beta)$
parametrized by the same irreducible component as above.
\end{enumerate}
\end{cor}

\begin{proof}
\textbf{(i)}
This follows from Lemma ~\ref{lem-qmove}.

\mni
\textbf{(ii)}
This is proved by induction on $d$.  The base case $d=1$ is
tautological.  Therefore, by way of induction, assume $d>1$ and the
result is proved for $d-1$.  By (i), each of $f_1$ and $f_{d-1}$
may be chosen so that its restriction to a general $F$-curve gives a
point of $\Kgnb{0,0}(X,\beta)$ which is a general member of its
irreducible component.  By the induction hypothesis, the irreducible
component for $f_1$ equals the irreducible component for $f_{d-1}$.
Therefore, assume there exists an $F$-curve $F_1$ in $\Sigma_1$, an
$F$-curve $F_{d-1}$ in $\Sigma_{d-1}$, and an identification $F_1\cong
F_{d-1}$ such that $f_1|_{F_1} = f_{d-1}|_{F_{d-1}}$.  Then, by Lemma
~\ref{lem-qa}(v), there exists a $1$-twisting surface
$f_d:\Sigma_d\rightarrow X$ of $H$-type $0$ and class
$(\beta_1,d\beta_2)$, resp. of $M$-class $(e_1\cdot \alpha, de_2\cdot
\alpha)$.  Moreover, the restriction of $f_d$ to a general $F$-curve
gives a point of $\Kgnb{0,0}(X,\beta)$ in the same irreducible
component as for $f_1$ and $f_{d-1}$.  Thus the result is proved by
induction on $d$.

\mni
\textbf{(iii)}
Let $f:\PP^1 \times \PP^1 \rightarrow X$ be a $1$-twisting surface of
class $(\beta_1,\beta_2)$, resp. of $M$-class $(e_1\cdot \alpha,
e_2\cdot \alpha)$.  By Lemma ~\ref{lem-qa}(iii), the morphism $f\circ
(\text{pr}_2,\text{pr}_1)$ is a $1$-twisting surface of class
$(\beta_2,\beta_1)$, resp. of $M$-class $(e_2\cdot \alpha,e_1\cdot
\alpha)$.  Applying (ii) to this morphism with $d=d_1$, 
there exists a $1$-twisting morphism
$f_{d_1}$ of $H$-type $0$ and
class $(\beta_2,d_1\beta_1)$, resp. of $M$-class
$(e_2\cdot \alpha, d_1e_1\cdot \alpha)$.  Applying Lemma
~\ref{lem-qa}(iii) again, $f_{d_1}\circ (\text{pr}_2,\text{pr}_1)$ is
a $1$-twisting morphism of $H$-type $0$ and class
$(d_1\beta_1,\beta_2)$, resp. $M$-class $(d_1e_1\cdot \alpha, e_2\cdot
\alpha)$.  Applying (ii) to this morphism with $d=d_2$, there exists a
$1$-twisting morphism $f_{d_1,d_2}$ of $H$-type $0$ and class
$(d_1\beta_1,d_2\beta_2)$, resp. $M$-class $(d_1e_1\cdot \alpha,
d_2e_2\cdot \alpha)$.  

\mni
\textbf{(iv)}
This is proved by induction on $m$ and $r$.  The argument is very
similar to the argument above.  By Lemma ~\ref{lem-qa}(v), there
exists an $m$-twisting surface $f_{m,0}:\Sigma_{m,0}\rightarrow X$ 
of $H$-type $0$ and class
$(\beta,(m-1)\beta_2)$, resp. $M$-class $(e\cdot \alpha, (m-1)e_2\cdot
\alpha)$.  By (ii), there exists a $1$-twisting surface $f_r:\Sigma_r
\rightarrow X$
of $H$-type
$0$ and class $(\beta,r\beta_1)$, resp. $M$-class $(e\cdot \alpha,
re_1\cdot \alpha)$.  Finally, the restrictions of $f_{m,0}$ and $f_r$
to $F$-curves give points in $\Kgnb{0,0}(X)$ in the same irreducible
component.  Thus, by (i), $f_{m,0}$ and $f_r$ may be deformed so that
there exist $F$-curves $F_{m,0}$ in $\Sigma_{m,0}$, $F_r$ in
$\Sigma_r$ and an identification $F_{m,0}\cong F_r$ so that
$f_{m,0}|_{F_{m,0}} = f_r|_{F_r}$.  Now the result follows from Lemma
~\ref{lem-qa}(v).   
\end{proof}

\mni
By Corollary ~\ref{cor-qa}, it is clear that existence of a single
$1$-twisting surface $f_1:\PP^1\times \PP^1 \rightarrow X$ and a
single $2$-twisting surface $f_2:\PP^1\times \PP^1 \rightarrow X$ can
produce a plethora of $n$-twisting surfaces for every integer $n$.
To produce a $1$-twisting surface and a $2$-twisting surface, we
exploit a connection between $n$-twisting surfaces  
and $\lambda$-degree
$1$ free curves in a general fiber of
$$
\text{ev}:\Kgnb{0,m}(X,m\alpha) \rightarrow X^m,
$$
for $m=2n$.

\begin{lem} \label{lem-qmcp}
\marpar{lem-qmcp}
Let $n$ be a positive integer.
Assume $(X,\OO(1),m=2n)$ satisfies Hypothesis ~\ref{hyp-mpcU}.
Assume that the fiber of 
$$
\text{ev}:\Kgnb{0,m}(X,m\alpha) \rightarrow X^m
$$
is uniruled by $\lambda$-degree $1$ rational curves.  If a general
surface $\Sigma$ from Lemma~\ref{lem-mpc7} is abstractly isomorphic to
$\PP^1 \times \PP^1$ then every isomorphism 
$$
f:\PP^1\times \PP^1 \rightarrow \Sigma\subset X
$$
satisfying 
$$
f^{-1}\OO_{\Sigma}(C) \cong \OO(n,1)
$$
is $n$-twisting of type $(\alpha,n\alpha)$ 
(and there exists such an isomorphism $f$).
Here $C$ is the image in $\Sigma$ of a general fiber of the morphism
$$
\mathcal{C}_D \rightarrow D
$$
as in the proof of Lemma ~\ref{lem-mpc7}.
\end{lem}

\begin{proof}
First of all, a degree $2n$ rational surface scroll that is abstractly
isomorphic to $\PP^1\times \PP^1$ has hyperplane class $\OO(n,1)$ up
to permutation of the factors.  So an isomorphism $f$ as above
exists.  

\mni
Because $\Sigma$ is embedded in $X$, the normal bundle $N_{\Sigma/X}$
is a locally free sheaf on $\Sigma$.
The space of first order deformations of $\Sigma\subset X$, i.e., the
Zariski tangent space of the Hilbert scheme of $X$ at $[\Sigma]$, is
canonically isomorphic to 
$$
H^0(\Sigma,N_{\Sigma/X}).
$$
The space of first order deformations of $C\subset X$ is
$$
H^0(C,N_{C/X}).
$$
There is a short exact sequence
$$
\begin{CD}
0 @>>> N_{C/\Sigma} @>>> N_{C/X} @>>> N_{\Sigma/X}|_C @>>> 0.
\end{CD}
$$
This
gives rise to a diagram
$$
\begin{CD}
& & H^0(C,N_{C/X}) \\
 & & @VV u V \\
H^0(\Sigma,N_{\Sigma/X}) @> v >> H^0(C,N_{\Sigma/X}|_C)
\end{CD}
$$
Given a first order deformation $\theta_C\in H^0(C,N_{C/X})$ of $C$
and a first order deforamtion $\theta_\Sigma\in
H^0(\Sigma,N_{\Sigma/X})$, the deformation of $C$ is contained in the
deformation of $\Sigma$ if and only if
$$
u(\theta_C) = v(\theta_\Sigma).
$$
Because the $\lambda$-degree $1$ curve is free, for every first order
deformation of $C$, there exists a first order deformation of $\Sigma$
containing the first order deformation of $C$.  In other words,
$$
\text{Image}(u) \subset \text{Image}(v).
$$

\mni
On the other hand, because $C$ is a hyperplane section of $\Sigma$,
the normal sheaf
$N_{C/\Sigma}$ is globally generated and thus
$$
h^1(C,N_{C/\Sigma}) = 0.
$$
Therefore, by the long exact sequence of cohomology, $u$ is
surjective.  Thus, by the last paragraph, also $v$ is surjective.
Also
$$
h^1(\Sigma,\text{pr}_2^*T_{\PP^1}(-n,-1)) = h^2(\Sigma,\text{pr}_2^*
T_{\PP^1}(-n,-1)) = 0
$$
and
$$
h^1(C,\text{pr}_2^*T_{\PP^1}(-n,-1)|_C) = h^1(C,\OO_C) = 0.
$$
Therefore, from the long exact sequence of cohomology associated to
the short exact sequence
$$
\begin{CD}
0 @>>> \text{pr}_2^* T_{\PP^1} @>>> N_{(\text{pr}_1,f)} @>>>
N_{\Sigma/X} @>>> 0
\end{CD}
$$
it follows that
$$
H^0(\Sigma,N_{(\text{pr}_1,f)}) \rightarrow
H^0(C,N_{(\text{pr}_1,f)}|_C)
$$
is surjective.  Since $N_{\Sigma/X}|_C$ and $\text{pr}_2^*
T_{\PP^1}|_C$ are globally generated, also $N_{(\text{pr}_1,f)}|_C$ is
globally generated.  Therefore, by Lemma ~\ref{lem-qtech},
$N_{(\text{pr}_1,f)}$ is globally generated and
$$
h^1(\Sigma,N_{(\text{pr}_1,f)}(-n,-1)) = 0.
$$
In other words, $f$ is $n$-twisting.
\end{proof}

\mni
This method of producing $n$-twisting surfaces is 
compatible with the canonical irreducible components from Section
~\ref{sec-cic}.  

\begin{lem} \label{lem-qM}
\marpar{lem-qM}
For $e=1$ and $n$, assume that $M_{e\alpha,1}$ is the unique
irreducible component of $\Kgnb{0,1}(X,e)$ dominating $X$.  Then
the $n$-twisting surface from Lemma~\ref{lem-qmcp} has $M$-class
$(\alpha,n\cdot\alpha)$.  
\end{lem}

\begin{proof}
Because $N_{\Sigma/X}$ is globally generated, every rational curve in
$\Sigma$ is a free rational curve in $X$.  In particular the fibers of
the projections are free curves of $\OO(1)$-degree $e$ for $e=1,n$.
Assuming that $M_{e\alpha}$ is the unique irreducible component
parametrizing free curves of $\OO(1)$-degree $e$, 
it follows that the fibers have $M$-class
$e\cdot\alpha$.  
\end{proof}

\mni
This raises the question, when is the surface $\Sigma$ abstractly
isomorphic to $\PP^1 \times \PP^1$?

\begin{lem} \label{lem-qP1}
\marpar{lem-qP1}
Let $n$ equal $1$ or $2$, i.e., $m$ equals $2$ or $4$.  
Assume $(X,\OO(1),m=2n)$ satisfies Hypothesis ~\ref{hyp-mpcU}.
Assume that the fiber of 
$$
\text{ev}:\Kgnb{0,m}(X,m\alpha) \rightarrow X^m
$$
over a general point
is uniruled by $\lambda$-degree $1$ rational curves.
Let $\Sigma$ be a general surface as in Lemma ~\ref{lem-mpc7}.
For $n=1$, $\Sigma$ is isomorphic to $\PP^1\times
\PP^1$.  For $n=2$, $\Sigma$ is isomorphic to
$\PP^1\times \PP^1$ or a Veronese surface.
\end{lem}

\begin{proof}
Every smooth quadric
surface is abstractly isomorphic to $\PP^1\times \PP^1$.  So for
$n=1$, $\Sigma$
is isomorphic to $\PP^1\times \PP^1$.

\mni
The case $n=2$ is more difficult.  From Lemma~\ref{lem-mpc7} it is
possible that the surface $\Sigma$ is abstractly isomorphic to the
Hirzebruch surface
$$
\Sigma \cong \PP(\OO_{\PP^1}\oplus \OO_{\PP^1}(-2)).
$$
We need to prove that in this case, a general deformation of $\Sigma$
is isomorphic to $\PP^1 \times \PP^1$.

\mni
The Hirzebruch surface contains a unique rational curve $E$ with
self-intersection $-2$.  Denote by $F$ the divisor class of the lines
of ruling of $\Sigma$.  The linear system embedding $\Sigma$ as a
quartic scroll is $\OO_{\Sigma}(E+3F)$.  Because a general deformation
of a curve $D\in |\OO_{\Sigma}(E+3F)|$ is contained in a general
deformation of $\Sigma$, the normal bundle $N_{\Sigma/X}$ satisfies
the hypotheses of Lemma ~\ref{lem-qtech} with $F'=E+2F$ and $n=1$.  In
particular, $N_{\Sigma/X}$ is globally generated and
$h^1(\Sigma,N_{\Sigma/X}(-E-3F)) = 0$.  By Lemma~\ref{lem-qmove},
every free curve in $\Sigma$ maps to a free curve in $X$ and every
deformation in $X$ of a reduced member $D \in |\OO_{\Sigma}(E+3F)|$ is
contained in a deformation of $\Sigma$.

\mni
Denote by $\alpha'$ the curve
class of $E$ in $X$.
The goal, of course, is to
argue that $\Sigma$ deforms to a quartic scroll that is isomorphic to
$\PP^1\times \PP^1$.  To prove this, we first show that if $\Sigma$ is
general then the curve $E$ is free.

\begin{claim} \label{claim-qP1a}
\marpar{claim-qP1a}
For a general surface $\Sigma$ as above, $E$ is a free curve.
\end{claim}

\mni
Let $C=C_1\cup C_2$ be a general reducible curve on $\Sigma$ with
$$
C_1 \in |\OO_{\Sigma}(F)|, \ \ C_2 \in |\OO_{\Sigma}(E+2F)|.
$$
By the argument above, $C_1$ and $C_2$ are free curves in $X$.
Moreover, because $\Sigma$ contains four general points of $X$ and
$C_1$ and $C_2$ can be chosen to contain one and three of these points
respectively, $C_2$ is $3$-dominating.

\mni
By Lemma ~\ref{lem-qmove}, 
a general deformation of $C_1\cup C_2$ is contained in a
deformation of $\Sigma$.  After replacing $C_1\cup C_2$ by a general
reducible deformation, let us assume $C_1$ and $C_2$ are general in
their deformation classes.
The curve $E$ intersects both $C_1$ and $C_2$, and the intersection
points are distinct.  Thus these intersection points
specify $E$ uniquely in $X\subset \PP^N$ since $E$ maps to a line in
$\PP^N$.  

\mni
Let $Y$ denote the union in $X$ of all non-free curves of class
$\alpha$.  To prove that $E$ is a free curve, it suffices to prove
that $E$ is not contained in $Y$.  In other words, for every
irreducible component $Y_i$, it suffices to prove that $E$ is not
contained in $Y_i$.

\mni
\textbf{(i) $\text{codim}_X(Y_i) \geq 2$.}
Because the two irreducible components of $C$ are general
free curves, they do not intersect any component of $Y$ having
codimension $\geq 2$ in $X$, cf. \cite[Proposition II.3.7]{K}.  Since
$E$ intersects $C_1$ and $C_2$, $E$ is contained in no irreducible
component $Y_i$ of $Y$ of having codimension $\geq 2$ in $X$.

\mni
\textbf{(ii) $\text{codim}_X(Y_i) = 1$, $Y_i$ not linear.}
Let
$Y_i$ be an irreducible component of $Y$ that is a codimension $1$
component of $X$.  Assume that $Y_i$ is not a linear space.
Since $C_1$ is a free curve, $C_1$ intersects $Y_i$ in general
points.  Because $C_2$ is $3$-dominating, even after specifying that
$C_2$ intersect $C_1$, $C_2$ intersects $Y_i$ in points that are
general with respect to $C_1\cap Y$.  Thus every pair of a point of
$C_1\cap Y$ and a point of $C_2\cap Y$ is a general point of $Y\times
Y$.  
Because $Y$ is not a linear space, 
there is no
line in $Y_i$ that contains a general pair of points in $Y$.  Thus
there is no line in $Y_i$ that intersects both $C_1$ and $C_2$.  
Since $E$ is a
line that intersects both $C_1$ and $C_2$, $E$ is not contained
in $Y_i$.

\mni
\textbf{(iii) $\text{codim}_X(Y_i) = 1$, $Y$ is linear.}
Finally, let $Y_i$ be an irreducible component of $Y$ that is a
codimension $1$ component of $X$ and is a linear space, i.e.,
$$
(Y_i,\OO(1)|_{Y_i}) \cong (\PP^{d-1},\OO_{\PP^{d-1}}(1))
$$
where $d$ is the dimension of $X$.  By way of contradiction, assume
$E$ lies in $Y_i$.  The intersection of $Y_i$ with $\Sigma$ contains
$E$.  Moreover, because the lines of ruling of $\Sigma$ are free
curves, they intersect $Y_i$ transversally.  Therefore the
multiplicity of $E$ in $Y_i\cap \Sigma$ is $1$, i.e.,
$$
Y_i \cap \Sigma = E + D'
$$
where $E$ is not an irreducible component of $D'$.  Therefore,
$$
(E\cdot (Y_i\cap \Sigma))_{\Sigma} = (E\cdot
E)_{\Sigma} + (E\cdot D')_{\Sigma} \geq -2.
$$
In other words,
$$
\langle c_1(N_{Y_i/X}), [E] \rangle \geq -2.
$$
Because $c_1(T_X) = \langle c_1(T_X),\alpha \rangle c_1(\OO(1))$ and
because $E$ has $\OO(1)$-degree $1$,
$$
\langle c_1(T_X),\alpha \rangle = \langle c_1(T_X),\alpha \rangle
\langle c_1(\OO(1)),[E] \rangle = \langle c_1(T_X), [E] \rangle.
$$
On the one hand, restricting the short exact sequence
$$
\begin{CD}
0 @>>> T_{Y_i} @>>> T_X|_{Y_i} @>>> N_{Y_i/X} @>>> 0
\end{CD}
$$
to $E$ gives
$$
\langle c_1(T_X), [E] \rangle = 
\langle c_1(T_{Y_i}), [E] \rangle + 
\langle c_1(N_{Y_i/X}), [E] \rangle \geq d + (-2) =d-2.
$$
On the other hand, because of the hypothesis that 
$X$ is neither a linear hypersurface
nor a quadric hypersurface,
$$
\langle c_1(T_X),\alpha \rangle \leq d-1.
$$
Together this implies that
$$
\langle c_1(T_X),\alpha \rangle = d-1 \text{ or } d-2.
$$
We will show that each of these leads to a contradiction.

\begin{claim} \label{claim-qP1b}
\marpar{claim-qP1b}
$$
\langle c_1(T_X),\alpha \rangle \neq d-1.
$$
\end{claim}

\mni
By way of contradiction, assume that $\langle c_1(T_X),\alpha \rangle$ 
does equal $d-1$.  Then the curve $D'$ is
nonempty:
$$
(E\cdot D')_{\Sigma} = -2 - (E\cdot E)_{\Sigma} = 1.
$$
Thus $D'$ must be contained in $Y_i$.  Since $E$ is a line in the
projective space $Y_i$, every curve in $Y_i$ is rationally equivalent
to a multiple of $E$, i.e.,
$$
D' \stackrel{Y_i}{\sim} a E
$$
for some positive integer $a$.  Therefore
$$
(D' \cdot Y_i)_X = \langle c_1(N_{Y_i/X}), D' \rangle = a \langle
c_1(N_{Y_i/X}), E \rangle = a(E\cdot (E+D'))_{\Sigma} = - a.
$$
On the other hand,
$$
(D'\cdot Y_i)_X = (D'\cdot(E+D'))_\Sigma = (D'\cdot D')_\Sigma + 1 >
0.
$$
This contradicts that $a$ is positive.  Therefore Claim
\ref{claim-qP1b} is valid.

\begin{claim} \label{claim-qP1c}
\marpar{claim-qP1c}
$$
\langle c_1(T_X),\alpha \rangle \neq d-2.
$$
\end{claim}

\mni
By way of contradiction, assume that $\langle c_1(T_X),\alpha \rangle$
does equal $d-2$.
Because $4\alpha$ is
$4$-dominating, the dimension of a general fiber of $\text{ev}$ is
nonnegative, i.e.,
$$
4\langle c_1(T_X),\alpha \rangle + 1\geq 3\text{dim}(X) = 3d.
$$
Substituting in $\langle c_1(T_X),\alpha \rangle = d-2$, this gives
$$
d\geq 7.
$$
But for $d\geq 7$,
$$
\langle c_1(T_X),\alpha \rangle = d-2 > \frac{1}{2}d + 1 =
\frac{1}{2}\text{dim}(X) + 1.
$$
By a theorem of Wi\'sniewski, \cite{Wis90}, this implies $\text{Pic}(X)
\cong \ZZ$.  Therefore $Y_i$ is either ample, trivial, or antiample.
Because $Y_i$ is effective, it is neither trivial nor antiample.
But since
$$
\langle [Y_i],[E] \rangle = \langle c_1(N_{Y_i/X}), [E] \rangle = -2,
$$
$Y_i$ is not ample.  This contradiction proves Claim
~\ref{claim-qP1c}.  This proves that $E$ is not contained in $Y_i$.
Since $E$ is contained in no irreducible component $Y_i$ of $Y$, $E$
is a free curve,
i.e., Claim ~\ref{claim-qP1a} is valid.

\mni
Let $D$ be a union of $E$ and three general lines of ruling $F_1,F_2,
F_3$.  Because $E$ and $F$ are free lines in $X$, $E\cup F_1$ deforms to a
conic in $X$.  Therefore $D$ deforms to a union of a conic and two
lines.  By Lemma ~\ref{lem-qmove}, there is a deformation of $\Sigma$
containing this deformation of $D$.  A quartic surface scroll contains
a conic if and only if it is isomorphic to $\PP^1 \times \PP^1$.
Therefore a general deformation of $\Sigma$ is isomorphic to
$\PP^1\times \PP^1$.
\end{proof}

\mni
As proved, the constructions above are compatible with the canonical
irreducible components of Section ~\ref{sec-cic}.  In fact, existence
of free $\lambda$-degree $1$ rational curves often implies the basic
hypothesis of Section ~\ref{sec-cic}.

\begin{lem} \label{lem-qM0.5}
\marpar{lem-qM0.5}
Let $n=1$, i.e., $m=2$.  Assume $(X,\OO(1),2)$ satisfies Hypothesis
~\ref{hyp-mpcU2}.  Assume that the fiber of
$$
\text{ev}:\Kgnb{0,2}(X,2\alpha) \rightarrow X^2
$$
over a general point is uniruled by $\lambda$-degree $1$ rational
curves. 
\begin{enumerate}
\item[(i)]
If a general fiber has dimension $\geq 2$, then there is a
unique component $M_{\alpha,1}$ of $\Kgnb{0,1}(X,\alpha)$ dominating
$X$ and a general fiber of $M_{\alpha,1}$ over $X$ is geometrically
connected.  
In particular this holds if 
$$
\langle 2\text{ch}_2(T_X),\Pi \rangle > 0.
$$
\item[(ii)]
Assume there exists a unique component $M_{\alpha,1}$ of
$\Kgnb{0,1}(X,\alpha)$ dominating $X$.  
With respect to $M_{\alpha,1}$, the surface $\Sigma$
from Lemma ~\ref{lem-qP1} has $M$-class
$(\alpha,\alpha)$.  
\end{enumerate}
\end{lem}

\begin{proof}
\textbf{(i)}
By Lemmas ~\ref{lem-mpc1}, \ref{lem-mpc3}, \ref{lem-mpcCH} and
Hypothesis ~\ref{hyp-mpcU}, the dimension of a general fiber of
$$
\text{ev}:\Kgnb{0,2}(X,2\alpha) \rightarrow X^2
$$
equals
$$
\text{dim}(X) - 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -1)
- 3.
$$
If this dimension is $\geq 2$, then
$$
2\langle c_1(T_X),\alpha \rangle \geq \text{dim}(X) + 3.
$$
Then, by Lemma ~\ref{lem-pl1.5}, a general fiber of
$$
\text{ev}:\Kgnb{0,1}(X,\alpha) \rightarrow X
$$
is irreducible, provided it is nonempty.  By hypothesis, a general
pair of points of $X$ is contained in a surface $\Sigma$.  By Lemma
~\ref{lem-qP1}, $\Sigma$ is a smooth quadric surface, which is ruled
by lines.  Thus every point is contained in a line, i.e., a general
fiber of $\text{ev}$ is nonempty, thus irreducible.  

\mni
Notice by Lemma ~\ref{lem-mpc8}, the dimension of the space of
$\lambda$-degree $1$ curves containing a general point of a general
fiber of
$$
\text{ev}:\Kgnb{0,2}(X,2\alpha) \rightarrow X^2
$$
equals
$$
\langle 2\text{ch}_2(T_X), \Pi \rangle.
$$
If this is positive, then there is at least a $1$-parameter family of
$\lambda$-degree $1$ curves in a general fiber.  Thus the fiber must
have dimension $\geq 2$.   

\mni
\textbf{(ii)}
Because a general deformation of $\Sigma$ contains a general point of
$X$, the lines in $\Sigma$ are in $M_{\alpha,0}$, i.e., $\Sigma$ has
$M$-class $(\alpha,\alpha)$.
\end{proof}

\mni
To summarize, positivity hypotheses on the Chern character of $X$ and
conditions on the Chow ring of $X$ imply the basic hypothesis of
Section ~\ref{sec-cic} as well as the existence of $1$-twisting
surfaces in $X$.

\begin{cor} \label{cor-qM0.5}
\marpar{cor-qM0.5}
Let $(X,\OO(1))$ be a smooth, projective variety.  Let $\alpha$ be an
$\OO(1)$-degree $1$ curve class.  Assume $c_1(T_X) = \langle c_1(T_X),
\alpha \rangle c_1(\OO(1))$, i.e., $c_1(T_X)$ is a multiple of
$c_1(\OO(1))$.  Assume $X$ is a Fano manifold, i.e., $\langle
c_1(T_X), \alpha \rangle > 0$.  
Assume $(X,\OO(1))$ is neither a linear variety nor a
quadric hypersurface.  Assume that
$$
\text{ch}_2(T_X) = \frac{\langle 2\text{ch}_2(T_X),\Pi \rangle}{2}
c_1(\OO(1))^2 
$$
for some integer $\langle 2\text{ch}_2(T_X),\Pi \rangle$.
Assume that
$$
\langle 2\text{ch}_2(T_X),\Pi \rangle \geq 0
$$
and
$$
\text{dim}(X) \geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
-1) + 4.
$$

\mni
Assume there exists a smooth, projective variety $Y$ such that $X$ is
a codimension $c$ linear section of $Y$.  Assume
$$
c \geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle - 1) -
\text{dim}(X) + 5,
$$
$$
c > \text{dim}(X) - 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
- 1) - 2\langle 2\text{ch}_2(T_X),\Pi \rangle + 2,
$$
and 
$$
\text{CH}^p(Y) = \ZZ\{c_1(\OO(1))^p\}, \text{ for } 0\leq p \leq
(\text{dim}(X) - \langle c_1(T_X),\alpha - 1) + 2.
$$
By Barth's theorems, \cite{Barth70}, this last condition holds if
$$
c \geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -1) -
\text{dim}(X) + (h^0(X,\OO(1)) - \text{dim}(X) -2) + 5.
$$
 
\begin{enumerate}
\item[(i)]
Then there exists a unique irreducible component $M_{\alpha,1}$ of
$\Kgnb{0,1}(X,\alpha)$ dominating $X$ and a general fiber of
$\Kgnb{0,1}(X,\alpha)$ over $X$ is geometrically connected.
\item[(ii)]
And
for every pair of positive integers $d_1,d_2$, there exists a
$1$-twisting surface of $H$-type $0$ and $M$-class
$(d_1\cdot\alpha,d_2\cdot \alpha)$ (with respect to $M_{\alpha,1}$).
\end{enumerate}
The same conclusion holds for every linear variety of dimension $\geq
2$
and every smooth quadric variety of
dimension $\geq 3$.
\end{cor}

\begin{proof}
\textbf{(i)}
The proof uses Lemma ~\ref{lem-pl2a}.  The hypotheses of that lemma
are that
$$
\langle c_1(T_X),\alpha \rangle \geq 3
$$
and that $X$ is a codimension $c$ linear section of a smooth,
projective variety with
$$
c\geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -1)
-\text{dim}(X) + 5.
$$
This second condition is one of the hypotheses above.  For the first
condition, observe that since
$(X,\OO(1))$ is neither linear nor quadric,
$$
\text{dim}(X) - \langle c_1(T_X),\alpha \rangle - 1 \geq 0.
$$
Together with the inequality
$$
\text{dim}(X) \geq 2(\text{dim}(X) - \langle c_1(T_X), \alpha \rangle
-1) + 4
$$
this 
implies both that $\text{dim}(X) \geq 4$ and $\langle c_1(T_X),\alpha
\rangle \geq 3$.  

\mni
\textbf{(ii)}
The next claim is that $2\alpha$ is $2$-dominating.  This is proved
using Corollary ~\ref{cor-mpcdom4}.  The first hypothesis of that
lemma,
$$
\text{dim}(X) \geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
-1) + 3,
$$
is implied by one of the hypotheses above.  The second hypothesis
that
$$
c\geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -1)
-\text{dim}(X)
$$
follows from the first hypothesis: $c\geq -3$ holds for $c=0$.  
The final condition,
$$
\text{CH}^p(Y) = \ZZ\{ c_1(\OO(1))^p\}, \text{ for } 0 \leq p \leq
(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle - 1) + 2
$$
is precisely one of the hypotheses above.  Therefore $2\alpha$ is
$2$-dominating by Corollary ~\ref{cor-mpcdom4}.

\mni
Because $2\alpha$ is $2$-dominating, $(X,\OO(1),2)$ satisfies
Hypothesis ~\ref{hyp-mpcU}.  Together with the hypotheses above, this
implies $(X,\OO(1),2)$ satisfies Hypothesis ~\ref{hyp-mpcU2}.  The
next claim is that a general fiber of
$$
\text{ev}:\Kgnb{0,2}(X,2\alpha) \rightarrow X^2
$$
is uniruled by rational curves of $\lambda$-degree $1$.  This is
proved using Lemma ~\ref{lem-mpc9}.  For $m=2$, the first hypothesis
of Lemma ~\ref{lem-mpc9} is precisely that
$$
\langle 2\text{ch}_2(T_X),\Pi \rangle \geq 0,
$$
which is one of the hypotheses above.  The second hypothesis is that
$$ 
c > \text{dim}(X) - 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
-1) - 2\langle 2\text{ch}_2(T_X),\Pi \rangle + 2,
$$
which is precisely one of the hypotheses above.  Therefore a general
fiber of $\text{ev}$ is uniruled by  rational curves of
$\lambda$-degree $1$ by Lemma ~\ref{lem-mpc9}.  

\mni
By Lemma ~\ref{lem-qP1}, the surface $\Sigma$ swept out by a general
$\lambda$-degree $1$ curve in a fiber is a quadric surface.  In
particular, $\Sigma \cong \PP^1 \times \PP^1$.  By Lemma
~\ref{lem-qmcp}, this surface $\Sigma$ is a $1$-twisting surface of
$H$-type $0$ and class $(\alpha,\alpha)$.  Moreover, by Lemma
~\ref{lem-qM}, the surface has $M$-class $(\alpha,\alpha)$.  Finally,
by Corollary ~\ref{cor-qa}, for every pair of positive integers
$d_1,d_2$, there exists a $1$-twisting surface of $H$-type $0$ and
$M$-class $(d_1\cdot\alpha,d_2\cdot\alpha)$.  

\mni
For a linear surface $(X,\OO(1))$
of dimension $\geq 2$, let $f:\PP^1\rightarrow \PP^1 \rightarrow X$ be
the composition of a degree $2$ morphism $\PP^1\times \PP^1
\rightarrow \PP^2$ with a linear embedding $\PP^2 \hookrightarrow X$.
Then $f$ is a $1$-twisting surface of $H$-type $0$ and $M$-class
$(\alpha,\alpha)$.  And $\Kgnb{0,1}(X,\alpha)$ is irreducible.
Similarly, if $(X,\OO(1))$ is a quadric
hypersurface of dimension $\geq 2$, then a general $2$-dimensional
linear section of $X$ is a quadric surface which is a $1$-twisting
surface of $H$-type $0$ and $M$-class $(\alpha,\alpha)$.  And if the
dimension $\geq 3$, then $\Kgnb{0,1}(X,\alpha)$ is irreducible.
Thus, by
Corollary ~\ref{cor-qa}, for every pair of positive integers
$d_1,d_2$, there exists a $1$-twisting surface of $H$-type $0$ and
$M$-class $(d_1\cdot \alpha, d_2\cdot \alpha)$.  
\end{proof}

\begin{lem} \label{lem-qM1}
\marpar{lem-qM1}
Let $n=2$, i.e., $m=4$.  Assume $(X,\OO(1),4)$ satisfies Hypothesis
~\ref{hyp-mpcU2}.  Assume that the fiber of
$$
\text{ev}:\Kgnb{0,4}(X,4\alpha) \rightarrow X^4
$$
over a general point is uniruled by $\lambda$-degree $1$ rational
curves.  
\begin{enumerate}
\item[(i)]
There is a unique component $M_{\alpha,1}$ of $\Kgnb{0,1}(X,\alpha)$
dominating $X$ and a general fiber of $M_{\alpha,1}$ over $X$ is
geometrically connected.
\item[(ii)]
Every $\OO(1)$-degree $1$ curve (assuming there are any)
in a general surface $\Sigma$ from Lemma ~\ref{lem-qP1} for $m=4$ is
parametrized by a point of $M_{\alpha,0}$, cf. Notation \ref{notat-cic2}. 
\item[(iii)]
With respect to $M_{\alpha,1}$, every irreducible
$\OO(1)$-degree $2$ curve in a general surface $\Sigma$ from Lemma
~\ref{lem-qP1} 
is parametrized by a point of
$M_{2\alpha,0}$, cf. Notation \ref{notat-cic2}.  In particular, if
$\Sigma \cong \PP^1\times \PP^1$, this surface has $M$-class
$(\alpha,2\cdot \alpha)$.  
\end{enumerate}
\end{lem}

\begin{proof}
Because there exists $\lambda$-degree $1$ curves in a general fiber of
$$
\text{ev}:\Kgnb{0,4}(X,4\alpha) \rightarrow X^4,
$$
the dimension of a general fiber is $\geq 1$.
By Lemmas
~\ref{lem-mpc1}, \ref{lem-mpc3}, \ref{lem-mpcCH} and Hypothesis
~\ref{hyp-mpcU}, the dimension is
$$
(\text{dim}(X)-4) - 4(\langle \text{dim}(X) - 1 - \langle
c_1(T_X),\alpha \rangle).
$$
Also, by Lemma ~\ref{lem-mpc8}, 
$$
2\langle 2\text{ch}_2(T_X),\Pi \rangle - \langle c_1(T_X),\alpha
\rangle \geq 0.
$$

\mni
Because $X$ is neither linear nor a quadric hypersurface, 
$$
\text{dim}(X) - 1 - \langle c_1(T_X), \alpha \rangle \geq 0.
$$
Thus $\text{dim}(X) \geq 4$.  
There is one extreme case that must be
ruled out.

\begin{claim} \label{claim-qM1a}
Either $\text{dim}(X) > 4$ or $\text{dim}(X) > \langle c_1(T_X),\alpha
\rangle + 1$.
\end{claim}

\mni
By way of contradiction, assume there exists $X$ satisfying the
hypotheses of the lemma with 
$$
\text{dim}(X) = 4 \text{ and } \langle c_1(T_X), \alpha \rangle =3.
$$
Notice it also follows that
$$
\langle 2\text{ch}_2(T_X),\Pi \rangle \geq \frac{\langle
  c_1(T_X),\alpha \rangle}{2} = \frac{3}{2}.
$$
Since $\langle 2\text{ch}_2(T_X),\Pi \rangle$ is an integer, in fact
this implies that
$$
\langle 2 \text{ch}_2(T_X), \Pi \rangle \geq 2.
$$
Now, by Lemma ~\ref{lem-pl2}, the morphism
$$
\text{ev}:\Kgnb{0,1}(X,\alpha) \rightarrow X
$$
is dominant.  By Lemma ~\ref{lem-pl1}(ii), a general fiber is
$1$-dimensional.  By Lemma ~\ref{lem-pl2}(iii), every connected
component $M$ of the fiber has
$$
2c_1(T_M) = (\langle 2\text{ch}_2(T_X),\Pi \rangle +3)
\pi_*f^*[c_1(\OO(1))^2].
$$
This is positive.  Therefore $M\cong \PP^1$ which implies $c_1(T_M) =
2$.  Substituting in gives,
$$
4 = 2c_1(T_M) = 
 (\langle 2\text{ch}_2(T_X),\Pi \rangle +3)
\pi_*f^*[c_1(\OO(1))^2] \geq 5\pi_*f^*[c_1(\OO(1))^2].
$$
Since $\pi_*f^*[c_1(\OO(1))^2]$ is a positive integer, this is
impossible.  This contradiction proves that $X$ cannot have both
$$
\text{dim}(X) = 4 \text{ and } \langle c_1(T_X),\alpha \rangle = 3.
$$
Therefore Claim ~\ref{claim-qM1a} is valid.

\mni
\textbf{(i)}
Claim ~\ref{claim-qM1a} implies that
$$
(\text{dim}(X) - 5) - 2(\text{dim}(X) - 1 - \langle c_1(T_X),\alpha
\rangle) \geq 0.
$$
Therefore, by Lemma ~\ref{lem-pl1.5}, a general fiber of
$$
\text{ev}:\Kgnb{0,1}(X,\alpha) \rightarrow X
$$
is irreducible.  In particular, there exists a unique irreducible
component $M_{\alpha,1}$ of $\Kgnb{0,1}(X,\alpha)$ dominating $X$.

\mni
\textbf{(ii)}
By Lemma ~\ref{lem-qP1}, either $\Sigma \cong \PP^1\times \PP^1$ or
$\Sigma$ is a Veronese surface.  A Veronese surface contains no lines,
so the result is vacuous.  If $\Sigma \cong \PP^1\times \PP^1$, then
the lines are the fibers of one of the two projections.  In
particular, a general point of $\Sigma$ is contained in a line.  Since
a general point of $\Sigma$ is a general point of $X$, every line in
$\Sigma$ containing a general point of $\Sigma$ is a line in $X$
containing a general point of $X$.  Since $M_{\alpha,1}$ is the unique
irreducible component dominating $X$, these lines are 
in $M_{\alpha,0}$.
Since the space of lines in $\Sigma$ is irreducible, every line in
$\Sigma$ is in $M_{\alpha,0}$.

\mni
\textbf{(iii)}
Because
$$
(\text{dim}(X) - 5) - 2(\text{dim}(X) - 1 - \langle c_1(T_X),\alpha
\rangle) \geq 0,
$$
by the same bend-and-break argument used to prove the second part of Lemma
~\ref{lem-mpcdom}, every fiber of
$$
\text{ev}:\Kgnb{0,2}(X,2\alpha) \rightarrow X\times X
$$
intersects the boundary $\Delta$.  Thus, for every curve $C$ in $X$
with class $2\alpha$, and for every pair of points $p_1,p_2$ of $C$, there
is a deformation of $C$ containing $p_1$ and $p_2$ that specializes to a
union of curves $C_1\cup C_2$
of class $\alpha$ containing $p_1$ and $p_2$.  Because $\alpha$ is the
minimal curve class, both $C_1$ and $C_2$ are smooth.  By Lemma
\ref{lem-cic1}, to prove that $[C]$ is in $M_{2\alpha,0}$, it suffices
to show that $C_1$ and $C_2$ are free curves.  

\mni
Now, if
$\Sigma$ is the surface coming from a general $\lambda$-degree $1$
curve, then a general point of $\Sigma$ is a general point of $X$.
Thus, for a general conic $C$ in $\Sigma$ and general points
$p_1,p_2$ in $C$,
every irreducible curve in $X$ containing $p_i$ is a free curve.  Thus
the curves $C_1,C_2$ are free curves, unless one of them contains both
$p_1$ and $p_2$.  Therefore, to finish the proof of the lemma, it
suffices to prove that for a general $\Sigma$, a general conic $C$ in
$\Sigma$ and a general pair $p_1,p_2$ of points in $C$, there is no
line in $X$ containing both $p_1$ and $p_2$.

\mni
By Lemma ~\ref{lem-qP1}, either
$\Sigma\cong \PP^2$ or
$\Sigma \cong \PP^1\times \PP^1$.  First consider the case that
$\Sigma \cong \PP^2$, i.e., $\Sigma$ is a Veronese surface.  Every
pair of points of $\Sigma$ is contained in a conic.  Moreover, a
general quadruple of points of $X$ is contained in a surface $\Sigma$,
by construction.  Therefore, a general pair of points of $X$ is a
general pair of points on a general conic in a general $\Sigma$.
Thus, if every such pair is contained in a line, then every general
pair of points of $X$ is contained in a line.  This implies $X$ is a
linear variety, contrary to hypothesis.  
Therefore, for a general
$\Sigma$, a general conic $C$ in $\Sigma$ and a general pair of points
$p_1,p_2$ of $C$, there is no line in $X$ containing $p_1,p_2$, and
thus $[C]$ is in $M_{2\cdot \alpha,0}$.

\mni
Next consider the case $\Sigma \cong \PP^1\times \PP^1$.  The conics
in $\Sigma$ are the fibers of one of the projections.  The union of
all lines connecting a pair of points of one of these fibers is a
linear projection 
of a cubic Segre variety $\PP^1\times \PP^2$, where the quartic scroll
$\PP^1\times \PP^1$ is embedded in the cubic Segre variety by
$$
\text{Id}_{\PP^1}\times (2-\text{uple Veronese}) :\PP^1 \times \PP^1
\rightarrow \PP^1 \times \PP^2.
$$
In particular, any four general points in $\PP^1\times \PP^1$ are
contained in a twisted cubic in $\PP^1\times \PP^2$.  Since four
general points of $X$ are contained in a surface $\Sigma$, this
implies that $3\alpha$ is $4$-dominating.  By Lemma ~\ref{lem-mpcCH},
this implies $X$ is either a linear variety or a quadric hypersurface,
contrary to hypothesis.  
Therefore, for a general
$\Sigma$, a general conic $C$ in $\Sigma$ and a general pair of points
$p_1,p_2$ of $C$, there is no line in $X$ containing $p_1,p_2$, and
thus $[C]$ is in $M_{2\cdot \alpha,0}$.
\end{proof}

\begin{lem} \label{lem-qV}
\marpar{lem-qV}
Assume $(X,\OO(1),m=4)$ satisfies Hypothesis ~\ref{hyp-mpcU}.  Assume
that the fiber of 
$$
\text{ev}:\Kgnb{0,4}(X,4\alpha) \rightarrow X^4
$$
over a general point
is uniruled by $\lambda$-degree $1$ rational curves.
Let $\Sigma$ be a general surface as in Lemma ~\ref{lem-mpc7}.
If $\Sigma$ is a Veronese surface, then every morphism
$$
f:\PP^1 \times \PP^1 \rightarrow \PP^2 \cong \Sigma \subset X
$$
with $f^*\OO_{\PP^2}(1) \cong \OO(1,1)$ is a $2$-twisting
surface of type $(2\alpha,2\alpha)$.

\mni
If $X$ further satisfies Hypothesis ~\ref{hyp-mpcU2}, then there
exists a unique irreducible component $M_{\alpha,1}$ of
$\Kgnb{0,1}(X,\alpha)$ dominating $X$, and $f$ has $M$-class
$(2\cdot\alpha,2\cdot \alpha)$.
\end{lem}

\begin{proof}
Let $C$ be a general conic in $\PP^2 \cong \Sigma$.  By the same
argument as in the proof of Lemma ~\ref{lem-qmcp}, the map
$$
H^0(\Sigma,N_{\Sigma/X}) \rightarrow H^0(C,N_{\Sigma/X}|_C)
$$
is surjective.  Denote by
$$
\nu:\widetilde{\Sigma} \rightarrow \Sigma
$$
the blowing up of $\Sigma$ at a point $p$ in $C$.  The surface
$\widetilde{\Sigma}$ is a rational ruled surface of Hirzebruch type
$1$.  The sheaf 
$$
N := \nu^*N_{\Sigma/X}
$$
satisfies the hypotheses of Lemma ~\ref{lem-qtech} for $n=1$: for the
divisor $D$ use the strict transform of $C$.  Thus by Lemma
~\ref{lem-qtech}, $N$ is globally generated on $\widetilde{\Sigma}$.
Therefore $N_{\Sigma/X}$ is globally generated on $\Sigma$.  Moreover
$$
h^1(\Sigma,N_{\Sigma/X}(-C)) = h^1(\Sigma,\nu_*(N(-D))) =
h^1(\widetilde{\Sigma},N(-D)) = 0.
$$ 
In other words,
$$
h^1(\Sigma,N_{\Sigma/X}(-2)) = 0.
$$
By an argument similar to the proof of Lemma ~\ref{lem-qa}(i), also
$$
h^1(\Sigma,N_{\Sigma/X}(-1)) = h^1(\Sigma,N_{\Sigma/X}) = 0.
$$
Moreover, because
$$
h^1(\Sigma,T_{\Sigma}(-2)) = h^1(\Sigma,T_{\Sigma}(-1)) =
h^1(\Sigma,T_{\Sigma}) = 0,
$$
it follows from the long exact sequence of cohomology that
$T_X|_\Sigma$ is globally generated and
$$
h^1(\Sigma,T_X|_\Sigma(-2)) = h^1(\Sigma,T_X|_\Sigma(-1)) =
h^1(\Sigma,T_X|_\Sigma) = 0.
$$

\mni
For every morphism
$$
f:\PP^1 \times \PP^1 \rightarrow \PP^2
$$
with $f^*\OO(1) \cong \OO(1,1)$,
$$
f_*\OO_{\PP^1\times \PP^1}(-1,0) \cong \OO_{\PP^2}(-1)\oplus \OO_{\PP^2}(-1)
$$
and thus
$$
f_*\OO_{\PP^1\times \PP^1})(-2,-1) \cong \OO_{\PP^2}(-2)\oplus
\OO_{\PP^2}(-2).
$$
Since $T_X|_\Sigma$ is globally generated, also $f^*T_X$ is globally
generated.  And, by the computation above
$$
h^1(\PP^1\times \PP^1,f^*T_X(-2,-1)) = h^1(\PP^2,T_X\otimes
f_*\OO(-2,-1)) = h^1(\PP^1,T_X(-2)) + h^1(\PP^1,T_X(-2)) = 0 + 0 = 0.
$$
Therefore $f$ is $2$-twisting. 

\mni
By Lemma ~\ref{lem-qM1}, if $X$ satisfies Hypothesis ~\ref{hyp-mpcU2},
then there exists a unique irreducible component $M_{\alpha,1}$ of
$\Kgnb{0,1}(X,\alpha)$ dominating $X$, and every conic in $\Sigma$
(which is the image of a line in $\PP^2$ under the Veronese morphism)
is
in $M_{2\cdot \alpha,0}$.  Since the fibers of the two projections of
$\PP^1\times \PP^1$ map to lines in $\PP^2$ under $f$, i.e., map to
conics in $\Sigma$, $f$ has $M$-class $(2\cdot\alpha,2\cdot \alpha)$.
\end{proof}

\begin{cor} \label{cor-qV}
\marpar{cor-qV}
Let $(X,\OO(1))$ be a smooth, projective variety.  Let $\alpha$ be an
$\OO(1)$-degree $1$ curve class.  Assume $c_1(T_X) = \langle c_1(T_X),
\alpha \rangle c_1(\OO(1))$, i.e., $c_1(T_X)$ is a multiple of
$c_1(\OO(1))$.  Assume $X$ is a Fano manifold, i.e., $\langle
c_1(T_X), \alpha \rangle > 0$.  
Assume $(X,\OO(1))$ is neither a linear variety nor a
quadric hypersurface.  Assume that
$$
\text{ch}_2(T_X) = \frac{\langle 2\text{ch}_2(T_X),\Pi \rangle}{2}
c_1(\OO(1))^2 
$$
for some integer $\langle 2\text{ch}_2(T_X),\Pi \rangle$.
Assume that
$$
2\langle 2\text{ch}_2(T_X), \Pi \rangle - \langle c_1(T_X),\alpha
\rangle \geq 0
$$
and 
$$
\text{dim}(X) \geq 4(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
-1) + 4.
$$ 

\mni
Assume there exists a smooth, projective variety $Y$ such that $X$ is
a codimension $c$ linear section of $Y$.  Assume
$$
c > \text{dim}(X) - 2(2\langle 2\text{ch}_2(T_X),\Pi \rangle - \langle
c_1(T_X),\alpha \rangle) - 4(\text{dim}(X) -\langle c_1(T_X),\alpha
\rangle - 1) + 6
$$
and
$$
\text{CH}^p(Y) = \ZZ\{c_1(\OO(1))^p\}, \text{ for } 0 \leq p \leq
(\text{dim}(X) - \langle c_1(T_X),\alpha\rangle -1) + 2.
$$
By Barth's theorems, \cite{Barth70}, this last condition holds if
$$
c \geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -1)
-\text{dim}(X) + (h^0(X,\OO(1)) -\text{dim}(X) - 2) +5.
$$

\begin{enumerate}
\item[(i)]
Then there exists a unique irreducible component $M_{\alpha,1}$ of
$\Kgnb{0,1}(X,\alpha)$ dominating $X$ and a general fiber of
$\Kgnb{0,1}(X,\alpha)$ over $X$ is geometrically connected.
\item[(ii)]
For every pair of positive integers $d_1,d_2$, there exists a
$1$-twisting surface of $H$-type $0$ and $M$-class
$(d_1\cdot\alpha,d_2\cdot \alpha)$ (with respect to $M_{\alpha,1}$).
\item[(iii)]
And there exists a $2$-twisting surface of $H$-type $0$ and $M$-class
either $(\alpha,2\cdot \alpha)$ or $(2\cdot\alpha,2\cdot\alpha)$.
Thus for every positive integer $m$ and nonnegative integer $r$, 
there exists an
$m$-twisting surface of $H$-type $0$ and $M$-class either
$(\alpha,(2m+r-2)\cdot\alpha)$ or $(2\cdot\alpha, (2m+r-2)\cdot \alpha)$.
\end{enumerate}
The same conclusion holds for every linear variety of dimension $\geq
2$ and every smooth quadric variety of dimension $\geq 3$.
\end{cor}

\begin{proof}
\textbf{(i) and (ii)}
First of all, the hypotheses above imply the Hypotheses of Corollary
~\ref{cor-qM0.5}.  The one hypothesis of Corollary ~\ref{cor-qM0.5}
that is a bit less obvious is,
$$
\text{dim}(X) \geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
-1) + 5.
$$
By Lemmas ~\ref{lem-mpc1}, \ref{lem-mpc3}, \ref{lem-mpcCH} and
Hypothesis ~\ref{hyp-mpcU}, this is equivalent to the hypothesis that
$$
\text{ev}:\Kgnb{0,2}(X,2\alpha) \rightarrow X^2
$$
has fiber dimension $\geq 2$.  By the hypothesis above,
$$
2\langle 2\text{ch}_2(T_X),\Pi \rangle \geq \langle c_1(T_X),\alpha
\rangle > 0,
$$
and thus
$\langle 2\text{ch}_2(T_X),\Pi \rangle >0$.  By Lemma
~\ref{lem-qM0.5}(i), this implies the fiber dimension of $\text{ev}$
is $\geq 2$.  Therefore Corollary ~\ref{cor-qM0.5} implies (i) and
(ii).

\mni
\textbf{(iii)}
The next claim is that $4\alpha$ is $4$-dominating.  This is proved
using Corollary ~\ref{cor-mpcdom4}.  The first hypothesis of that
lemma,
$$
\text{dim}(X) \geq 4(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
-1) + 3
$$ 
is implied by one of the hypotheses above.  The second hypothesis that
$$
c\geq 4(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -1)
-\text{dim}(X) + 2
$$
follows from the first hypothesis: $c\geq -1$ holds for $c=0$.  
The final condition,
$$
\text{CH}^p(Y) = \ZZ\{ c_1(\OO(1))^p\}, \text{ for } 0 \leq p \leq
(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle - 1) + 2
$$
is precisely one of the hypotheses above.  Therefore $4\alpha$ is
$4$-dominating by Corollary ~\ref{cor-mpcdom4}.

\mni
The claim is that $(X,\OO(1),4)$ satisfies Hypothesis
~\ref{hyp-mpcU}.  By the last paragraph, 
$4\alpha$ is $4$-dominating.  Because $(X,\OO(1))$ is neither linear
nor quadric,
$$
\text{dim}(X) - \langle c_1(T_X) -1 \geq 0.
$$
Therefore the hypothesis that
$$
\text{dim}(X) \geq 4(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
-1) + 4
$$
implies that $\text{dim}(X) \geq 4$.  Since $(X,\OO(1))$ is not
linear,
$$
h^0(X,\OO(1)) \geq \text{dim}(X) + 2.
$$
Therefore $4 \leq h^0(X,\OO(1))$ and
$$
h^0(X,\OO(1)) \geq \text{dim}(X) + 4-2.
$$
Therefore $(X,\OO(1))$ satisfies Hypothesis ~\ref{hyp-mpcU}.  Together
with the hypotheses above, this implies $(X,\OO(1),4)$ satisfies
Hypothesis ~\ref{hyp-mpcU2}.

\mni
The next claim is that a general fiber of
$$
\text{ev}:\Kgnb{0,4}(X,4\alpha) \rightarrow X^4
$$
is uniruled by rational curves of $\lambda$-degree $1$.  This is
proved using Lemma ~\ref{lem-mpc9}.  For $m=4$, the first hypothesis
of Lemma ~\ref{lem-mpc9} is precisely that
$$
2\langle 2\text{ch}_2(T_X),\Pi \rangle - \langle c_1(T_X),\alpha
\rangle \geq 0,
$$
which is one of the hypotheses above.  The second hypothesis is that
$$
c > \text{dim}(X) - 2(2\langle 2\text{ch}_2(T_X),\Pi \rangle - \langle
c_1(T_X),\alpha \rangle) - 4(\text{dim}(X) -\langle c_1(T_X),\alpha
\rangle - 1) + 6
$$
which is precisely one of the hypotheses above.  Therefore a general
fiber of $\text{ev}$ is uniruled by rational curves of
$\lambda$-degree $1$ by Lemma ~\ref{lem-mpc9}.

\mni
By Lemma ~\ref{lem-qP1}, the surface $\Sigma$ swept out by a general
$\lambda$-degree $1$ curve in a fiber is either a quartic scroll
abstractly isomorphic to $\PP^1\times \PP^1$ or else a Veronese
surface.  By Lemmas ~\ref{lem-qmcp} and \ref{lem-qM1}, 
in the first case $\Sigma$ is a $2$-twisting surface of $H$-type $0$
and $M$-class $(\alpha,2\cdot \alpha)$.  By Lemma ~\ref{lem-qV}, in
the second case there is a morphism $\PP^1\times \PP^1 \rightarrow
\Sigma\subset X$ which is a $2$-twisting surface of $H$-type $0$ and
$M$-class $(2\cdot \alpha, 2\cdot \alpha)$.

\mni
In the first case, since there exists a $1$-twisting surface of
$H$-type $0$ and $M$-class $(\alpha,\alpha)$ and a $2$-twisting
surface of $H$-type $0$ and $M$-class $(\alpha,2\cdot \alpha)$,
Corollary ~\ref{cor-qa}(iv) implies that for every positive integer
$m$ and nonnegative integer $r$,
there exists an $m$-twisting surface of $H$-type $0$ and $M$-class
$(\alpha,(2m+r-2)\cdot\alpha)$.  In the second case, by (ii) there exists a
$1$-twisting surface of $H$-type $0$ and $M$-class $(2\cdot
\alpha,\alpha)$ and a $2$-twisting surface of $H$-type $0$ and
$M$-class $(2\cdot \alpha, 2\cdot \alpha)$.  Thus Corollary
~\ref{cor-qa}(iv) implies that for every positive integer $m$ and
nonnegative integer $r$, there exists an $m$-twisting surface of
$H$-type $0$ and $M$-class $(2\cdot \alpha, (2m+r-2)\cdot \alpha)$.  

\mni
The conclusion for $\PP^n$, $n\geq 2$ follows by considering any
morphism $f:\PP^1 \times \PP^1 \rightarrow \PP^n$ such that $f^*\OO(1)
= \OO(1,2)$.  The conclusion for a quadric hypersurface $X$ of
dimension $\geq 3$ follows by considering any $2$-dimensional linear
section $\Sigma \cong \PP^1\times \PP^1$ of $X$ and then composing
with a morphism
$$
q\times \text{Id}_{\PP^1}:\PP^1 \times \PP^1 \rightarrow \PP^1\times
\PP^1 \cong \Sigma \subset X
$$
where $q$ is a degree $2$ morphism.   
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%
%% Section: Twisting surfaces. Rational simple connectedness
%% 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Twisting surfaces. Rational simple connectedness}
\label{sec-qrc} 
\marpar{sec-qrc}

\mni
Let $\beta_1$ and $\beta_2$ be curve classes in $X$.  Let $m$ be a
positive integer.  Denote by $\tau_m(\beta_1;\beta_2)$ the genus $0$,
stable $A$-graph with one vertex $v_0$ of class $\beta_2$ and valence $m$,
and $m$ leaves $v_i$, $i=1,\dots,m$, 
of class $\beta_1$ connected to $v_0$.  Denote by
$\tau'_m(\beta_1;\beta_2)$ the $A$-graph obtained from
$\tau_m(\beta_1;\beta_2)$ by attaching one tail to each of the leaf
vertices only, 
$v_i$, $i=1,\dots,m$.  And denote by $\tau''_m(\beta_1;\beta_2)$ the
$A$-graph obtained from $\tau_m(\beta_1;\beta_2)$ by attaching a tail
to \emph{every} vertex $v_i$, $i=0,\dots,m$. 
In other words, $\tau_m(\beta_1;\beta_2)$ is the dual graph of a
reducible, arithmetic genus $0$ curve in $X$ which is a comb with
handle class $\beta_2$ and with $m$ teeth of class $\beta_1$.  And
$\tau'_m(\beta_1;\beta_2)$ is the dual graph associated to the
$n$-pointed curve obtained by putting one marked point on each tooth of
the comb.  And $\tau''_m(\beta_1;\beta_2)$ puts one marked point on
each tooth and on the handle. [REFERENCES -- DIAGRAMS]

\mni
For a Hirzebruch surface
$$
\pi:\Sigma \rightarrow \PP^1
$$
of $H$-type $h$, an integer $m\geq h$ and a nonnegative integer $n$,
denote by 
$U_{m,n}$ the open subset of $\Kgnb{0,n}(\Sigma,F'+mF)$ parametrizing smooth
divisors
$$
C\in |\OO_{\Sigma}(F'+mF)|
$$
together with $n$ distinct points 
$$
p_1,\dots,p_n \in C.
$$
If $n=m$, denote $U_{m,m}$ by $U_m$.


\begin{lem} \label{lem-t1}
\marpar{lem-t1}
The variety $\M(\Sigma,\tau'_m(F;F'))$ is smooth and irreducible.  The
image in $\Kgnb{0,m}(\Sigma,F'+mF)$ intersects the smooth locus of the
morphism 
$$
\text{ev}:\Kgnb{0,m}(\Sigma,F'+mF) \rightarrow \Sigma^m
$$
and is
contained in $\overline{U}_m$.  
A general fiber of the restriction
$\text{ev}|_{\overline{U}_m}$ is rationally connected.  
A
general point of $\overline{U}_m$ is rationally connected to
the image of $\M(\Sigma,\tau'_m(F;F'))$ by a rational curve in a fiber
of $\text{ev}$.  
Moreover, the intersection points may be chosen general.  
\end{lem}

\begin{proof}
A point of $\M(\Sigma,\tau'_m(F;F'))$ is equivalent to the data of 
a divisor $D_0$ in
the linear system 
$|\OO_{\Sigma}(F')|$, divisors $D_i$, $i=1,\dots,m$, in the
linear system $|\OO_{\Sigma}(F)|$, and a point $p_i$ of $D_i$ for each
$i=1,\dots,m$.  The parameter space for such data is clearly smooth
and irreducible.  Moreover, for a general such datum, the divisor
$D=D_0+D_1 + \dots+D_m$ is a reduced member of the linear system
$|\OO_{\Sigma}(F'+mF)|$, and each point $p_i$ is distinct from the
intersection point $q_i$ of $D_i$ and $D_0$. 

\mni
There is a short exact sequence
$$
0 \rightarrow \oplus_{i=1}^m
N_{D/\Sigma}(-(p_1+\dots+p_m))|_{D_i/\Sigma}(-q_i) \rightarrow
N_{D/\Sigma}(-(p_1 + \dots + p_m)) \rightarrow
N_{D/\Sigma}(-(p_1+\dots+p_m))|_{D_0} \rightarrow 0.
$$
There are also isomorphisms
$$
N_{D/\Sigma}(-(p_1+\dots+p_m))|_{D_0} \cong N_{D/\Sigma}|_{D_0} \cong
N_{D_0/\Sigma}(q_1+\dots+q_m) = 
\OO_{D_0}(m-h)
$$
and
$$
N_{D/\Sigma}(-(p_1+\dots+p_m))|_{D_i}(-q_i) \cong N_{D_i/\Sigma}(-p_i)
\cong \OO_{D_i}(-1)
$$
for $i=1,\dots,m$.  
Since
$$
h^1(D_0,\OO_{D_0}(m-h)) = h^1(D_i,\OO_{D_i}(-1)) = 0,
$$
by the long exact sequence of cohomology
$$
h^1(D,N_{D/\Sigma}(-(p_1+\dots+p_m))) = 0.
$$
Thus both $\Kgnb{0,m}(\Sigma,F'+mF)$ and
the evaluation morphism
$$
\text{ev}:\Kgnb{0,m}(\Sigma,F'+mF) \rightarrow \Sigma^m
$$
are smooth at the point $(D,p_1,\dots,p_m)$.  Moreover, by a parameter
count, the divisor $D$ deforms to a smooth divisor $C$ in
$|\OO_{\Sigma}(F'+mF)|$ containing $p_1,\dots,p_m$.  Therefore the
image of $\M(X,\tau'_m(F;F'))$ is contained in $\overline{U}_m$.  

\mni
The fiber of 
$$
\text{ev}:\overline{U}_m \rightarrow \Sigma^m
$$
over a general point $(p_1,\dots,p_m)$ is birational to 
the linear system of curves in $|\OO_{\Sigma}(F'+mF)|$
with base locus $p_1,\dots,p_m$.  Therefore it is rationally connected.

\mni
Let $C$ be a smooth member of the linear system of curves in
$|\OO_{\Sigma}(F'+mF)|$ with base locus $p_1,\dots,p_m$.  
The divisors $D$ and $C$ span a pencil
$\Lambda$ of divisors in $|\OO_{\Sigma}(F'+mF)|$
containing $p_1,\dots,p_m$.  Marking each divisor in the pencil
by $p_1,\dots,p_n$, there is a $1$-morphism
$$
\zeta:\Lambda \rightarrow \Kgnb{0,m}(\Sigma,F'+mF)
$$
whose image contains $(C,p_1,\dots,p_m)$ and $(D,p_1,\dots,p_m)$ and
is contained in a fiber of $\text{ev}$.  By construction, 
$(D,p_1,\dots,p_m)$ is general in $\M(X,\tau'_m(F;F'))$ and
$(C,p_1,\dots,p_m)$ is general in $\Kgnb{0,m}(X,F'+mF)$.  
\end{proof}

\begin{lem} \label{lem-t1.5}
\marpar{lem-t1.5}
Assume the H-type of $\Sigma$ is $h=0$, i.e., $\Sigma \cong
\PP^1\times \PP^1$.  The variety $\M(\Sigma,\tau''_m(F,F'))$ is smooth
and irreducible.  The image in $\Kgnb{0,m+1}(\Sigma,F'+mF)$ intersects
the smooth locus of the morphism
$$
\text{ev}:\Kgnb{0,m+1}(\Sigma,F'+mF) \rightarrow \Sigma^{m+1}
$$
and is contained in $\overline{U}_{m,m+1}$.  A general fiber of the
restriction
$\text{ev}|_{\overline{U}_{m,m+1}}$ is rationally connected.  A
general point of $\overline{U}_{m,m+1}$ is rationally connected to the
image of $\M(\Sigma,\tau''_m(F;F'))$ by a rational curve in a fiber of
$\text{ev}$.  Moreover, the intersection points may be chosen general.
\end{lem}

\begin{proof}
The proof is very similar to the proof of Lemma ~\ref{lem-t1}.  The
one difference is that, because $\Sigma \cong \PP^1 \times \PP^1$, the
curve $D_0$ can be chosen to contain any point $p_0$ of $\Sigma$. 
\end{proof}

\begin{lem} \label{lem-t2}
\marpar{lem-t2}  
Let $h$ equal $0$ or $1$ and let $m$ be a positive integer.  
Let 
$$
\pi:\Sigma \rightarrow \PP^1, \ \ f:\Sigma \rightarrow X
$$
be an $m$-twisting surface of H-type $h$
and class $(\beta_1,\beta_2)$.  Denote $\beta_2 + m\beta_1$ by $\beta$.
By
functoriality of the Kontsevich space, for each nonnegative integer $n$
there is a $1$-morphism
$$
\Kgnb{0,n}(f): \Kgnb{0,n}(\Sigma,F'+mF) \rightarrow
\Kgnb{0,n}(X,\beta).
$$
Every point of $\Kgnb{0,n}(\Sigma,F'+nF)$ corresponding to a 
reduced divisor $D$ in $|\OO_{\Sigma}(F'+nF)|$ with $n$ distinct,
smooth marked points
is mapped to a 
smooth point of $\Kgnb{0,n}(X,\beta)$.

\mni
Denote by $M_m$ the unique irreducible component of
$\Kgnb{0,m}(X,\beta)$ containing the image of $\overline{U}_m$.  
A general point of $M_m$ is contained in the image of a $1$-morphism
$$
\zeta:\PP^1 \rightarrow M_m
$$
contained in a fiber of 
$$
\text{ev}:\Kgnb{0,m}(X,\beta) \rightarrow X^m
$$
and intersecting $\M(X,\tau'_m(\beta_1;\beta_2))$ in a general point
specializing to the image of $\M(\Sigma,\tau'_m(F;F'))$.

\mni
Finally, assume $h=0$, i.e., $\Sigma \cong \PP^1\times \PP^1$.  Denote
by $M_{m+1}$ the unique irreducible component of
$\Kgnb{0,m+1}(X,\beta)$ containing the image of
$\overline{U}_{m,m+1}$.  A general point of $M_{m+1}$ is contained in
the image of a $1$-morphism
$$
\zeta:\PP^1 \rightarrow M_{m+1}
$$
contained in a fiber of
$$
\text{ev}:\Kgnb{0,m+1}(X,\beta) \rightarrow X^{m+1}
$$
and intersecting $\M(X,\tau''_m(\beta_1;\beta_2))$ in an general point
specializing to the image of $\M(\Sigma,\tau''_m(F;F'))$.  
\end{lem}

\begin{proof}
For every reduced divisor $C$ in $|\OO_{\Sigma}(F'+mF)|$, the normal
bundle $N_{C/\Sigma}$ is globally generated.  Since also
$N_{\Sigma/X}$ is globally generated, $N_{C/X}$ is globally
generated.  Therefore $C$ is a smooth point of
$\Kgnb{0,n}(X,\beta_2+m\beta_1)$.  

\mni
By Lemma ~\ref{lem-qmove}, a general deformation in $X$ of a reduced
divisor $C$ in $|\OO_{\Sigma}(F'+mF)|$ is contained in a deformation
of $\Sigma$.  Because the H-type $h$ equals $0$ or $1$, every small
deformation of $\Sigma$ is again a Hirzebruch surface of H-type $h$.
By Lemma ~\ref{lem-t1}, a small deformation of $C$ is contained in the
image of a morphism $\zeta$ as above.  

\mni
Because the deformation of $C$
may be taken to be a general point of $M_m$, or a general deformation
of a point of $\M(\Sigma,\tau'_m(F;F'))$ to a point of
$\M(X,\tau'_m(\beta_1;\beta_2))$, the image of a general morphism
$\zeta$ as above intersects both loci in general points.

\mni
The version with $M_m$ replaced by $M_{m+1}$ and
$\overline{U}_n$ replaced by $\overline{U}_{m,m+1}$ follows in the
same way from Lemma ~\ref{lem-t1.5}.
\end{proof}

\begin{lem} \label{lem-tM}
\marpar{lem-tM}
Let $M_{\alpha,1}$ be an irreducible component of
$\Kgnb{0,1}(X,\alpha)$ dominating $X$ and whose general point is a
smooth rational curve.  If the $m$-twisting surface in Lemma
~\ref{lem-t2} has H-type $0$, i.e., $\Sigma \cong \PP^1\times
\PP^1$, and 
$M$-class $(e_1\cdot \alpha,e_2\cdot \alpha)$, then
denoting $e=e_2+me_1$, $M_m$, resp. $M_{m,m+1}$ 
equals $M_{e\cdot\alpha,m}$, resp. $M_{e\cdot \alpha,m+1}$. And the image
of $\M(\Sigma;\tau'_m(F;F'))$, resp. $\M(\Sigma;\tau''_m(F;F'))$, is
contained in the image of
$M_{\tau'_m(e_1\cdot\alpha;e_2\cdot\alpha)}$,
resp. $M_{\tau''_m(e_1\cdot\alpha; e_2\cdot \alpha)}$, cf. Notation
~\ref{notat-cic3}.    
\end{lem}

\begin{proof}
Because the H-type is $0$, every curve in $\Sigma$ is free.  Because
the $M$-class is $(e_1\cdot \alpha, e_2\cdot \alpha)$, there is a
curve in $M_m$, resp. $M_{m,m+1}$, 
whose irreducible components are free, smooth curves
parametrized by components $M_{e_i\cdot \alpha,0}$ for various
positive integer $e_i$.  Therefore the
lemma follows from Lemma ~\ref{lem-cic1.5}.
\end{proof}

\begin{lem} \label{lem-t2.5}
\marpar{lem-t2.5}
Let $Y$ be a proper algebraic space over $k$, let
$\nu:\widetilde{Y}\rightarrow Y$ be a strong resolution of singularities,
and let
$$
\phi:\widetilde{Y}\dashrightarrow Q
$$
be the MRC fibration.  

\mni
Let $Z$ be a
closed subset of $Y$ such that $Z\cap Y_{\text{smooth}}$ is dense in $Z$.
If every general point of $Y$ is contained in a
rational curve intersecting $Z \cap Y_{\text{smooth}}$, then the
strict transform $\widetilde{Z}$
of $Z$ intersects the domain of definition of $\phi$,
and $\phi(\widetilde{Z})$ is dense in $Q$. 
\end{lem}

\begin{proof}
By definition of the MRC
quotient, there exists an open subset $U$ of $\widetilde{Y}$ such that
the restriction of $\phi$,
$$
\phi_U:U\rightarrow Q,
$$
is regular and proper and every curve in $\widetilde{Y}$ intersecting
the geometric generic fiber of $\phi_U$ is contained in the geometric
generic fiber of $\phi_U$ (in particular, it is contained in $U$).

\mni
Because $\nu$ is an isomorphism over $Y_{\text{smooth}}$, the strict
transform of the rational curve in the statement
intersects $\widetilde{Z}$.  Therefore
every general point of $\widetilde{Y}$ is contained in a rational
curve intersecting $\widetilde{Z}$.  By the previous paragraph,
$\widetilde{Z}$ intersects the geometric generic fiber of $\phi_U$.
In other words, $\widetilde{Z}$ intersects $U$ and $\phi_U(U\cap
\widetilde{Z})$ is dense in $Q$. 
\end{proof}

\begin{lem} \label{lem-t3}
\marpar{lem-t3}
Let $M_{\alpha,1}$ be an irreducible component of
$\Kgnb{0,1}(X,\alpha)$ dominating $X$ and whose generic fiber over $X$
is geometrically irreducible.  Let $m$ be a positive integer.  Assume
there exists an $m$-twisting surface of $H$-type $0$ and $M$-class
$(e_1\cdot \alpha, e_2\cdot \alpha)$.  

\mni
Denoting $e=e_2+me_1$, a
general point of a general fiber of the restriction
$$
\text{ev}|_M:M_{e\cdot \alpha,m+1} \subset \Kgnb{0,m+1}(X,e\alpha)
\rightarrow X^{m+1}
$$
is contained in a rational curve in the fiber intersecting the image
of the boundary map
$$
M_{e_1\cdot\alpha,2}\times_{\text{pr}_2\circ \text{ev},X,\text{pr}_1\circ
  \text{ev}} M_{(e-e_1)\cdot \alpha,m+1} \rightarrow M_{e\cdot
  \alpha,m+1},
$$
in a smooth point of the fiber.
Thus the image $\Delta_{(e_1\cdot \alpha,\{0\}),((e-e_1)\cdot \alpha,
  \{1,\dots, m\})}$ of the boundary map intersects the domain of
definition of the MRC fibration of a strong resolution of the fiber,
and this intersection dominates the MRC quotient of a strong
resolution of the fiber.  

\mni
In particular, if a general fiber of
$\Delta_{(e_1\cdot
  \alpha,\{0\}), ((e-e_1)\cdot \alpha,\{1,\dots,m\})}$ over $X^{m+1}$
is geometrically connected and geometrically rationally connected,
then a general fiber of $\text{ev}|_M$ is geometrically connected and
geometrically rationally connected.
\end{lem}


\begin{proof}
By the last part of Lemma ~\ref{lem-t2} and by Lemma ~\ref{lem-tM}, a
general point of a general fiber of $\text{ev}|_M$ is contained in a
rational curve intersecting $M_{\tau''_m(e_1\cdot\alpha;e_2\cdot
  \alpha)}$ in a smooth point.  Recall that
$M_{\tau'_m(e_1\cdot\alpha;e_2\cdot \alpha)}$ parametrizes reducible
combs $C=C_0\cup C_1\cup \dots \cup C_m$ whose handle $C_0$ has
$M$-class $e_2\cdot \alpha$ and whose teeth $C_i$, $i=1,\dots,m$, have
$M$-class $e_1\cdot \alpha$.  For each $i=1,\dots,m$, there is a point
$q_{i,0}$ of $C_0$ that is attached to a point $q_i$ of $C_i$.  Also,
there is a marked point $p_0$ of $C_0$ and for each $i=1,\dots,m$,
there is a marked point $p_i$ of $C_i$.  Altogether,
$(C,p_0,\dots,p_m)$ is a marked curve parametrized by  $M_{e\cdot
  \alpha,m+1}$.  

\mni
Now let $B_0$ be one of the teeth $C_i$, $i=1,\dots,m$ and mark it by
$r_0=p_i$ and $s_0=q_i$ so that $(B_0,r_0,s_0)$ is parametrized by
$M_{e_1\cdot\alpha,2}$.  Next let $B_1$ be the union of $C_0$ and all
of the teeth $C_j$ with $j\neq i$.  Mark $B_1$ by $s_1=q_{i,0}$ and 
$$
r_j = \lt\{ \begin{array}{cc}
p_j, & j\neq i \\
p_0, & j = i
\end{array} \rt.
$$
so that $(B_1,s_1,p_1,\dots,p_m)$ is parametrized by
$M_{(e-e_1)\cdot\alpha,m+1}$.  Thus the pair
$((B_0,r_0,s_0),(B_1,s_1,r_1,\dots,r_m))$ is parametrized by a point
of 
$$
M_{e_1\cdot \alpha,2}\times_{\text{pr}_2\circ
  \text{ev},X,\text{pr}_1\circ \text{ev}} M_{(e-e_1)\cdot\alpha,m+1}.
$$
Moreover, up to a reordering of the marked points, the image of this
point under the boundary map
$$
M_{e_1\cdot \alpha,2}\times_{\text{pr}_2\circ
  \text{ev},X,\text{pr}_1\circ \text{ev}} M_{(e-e_1)\cdot\alpha,m+1}
\rightarrow M_{e\cdot\alpha,m+1}
$$ 
is precisely $(C,p_0,\dots,p_m)$.  In other words,
$M_{\tau''_m(e_1\cdot\alpha;e_2\cdot \alpha)}$ is contained in
$\Delta_{ (e_1\cdot\alpha,\{0\}),
    ((e-e_1)\cdot\alpha,\{1,\dots,m\}) }$ up to reordering of the
  marked points.
Therefore the rational curve from the previous paragraph intersects
$\Delta_{(e_1\cdot \alpha,\{0\}),((e-e_1)\cdot
  \alpha,\{1,\dots,m\})}$.  

\mni
Also, since a general point of
$M_{\tau''_m(e_1\cdot\alpha;e_2\cdot\alpha)}$ parametrizes a curve
whose irreducible components are all free, a general point is a smooth
point of the fiber.  Thus, by Lemma ~\ref{lem-t2.5}, for a general
geometric point $(x_1,x_2)$ of $X\times X$, the boundary $\Delta\cap
\text{ev}|_M^{-1}(x_1,x_2)$ intersects the domain of definition of the
MRC fibration of $\text{ev}|_M^{-1}(x_1,x_2)$, and the intersection
dominates the MRC quotient.  If $\Delta \cap
\text{ev}|_M^{-1}(x_1,x_2)$ is connected and rationally connected, the
same holds for the MRC quotient.  Then, by \cite{GHS},
$\text{ev}|_M^{-1}(x_1,x_2)$ is connected and rationally connected. 
\end{proof}

\begin{lem} \label{lem-t4}
\marpar{lem-t4}
Let $M_{\alpha,1}$ be an irreducible component of
$\Kgnb{0,1}(X,\alpha)$ dominating $X$ and whose generic fiber over $X$
is geometrically irreducible.  Let $m$, $e'_1$, $e''_1$ and $e$ be
nonnegative integers such that $e\geq e'_1+e''_1$.  

\mni
Assume that for both $(c,c_1) = (e,e'_1)$ and $(c,c_1)=(e-e'_1,e'_2)$
a 
general point of a general fiber of the restriction
$$
\text{ev}|_M:M_{c\cdot\alpha,m_1} \subset \Kgnb{0,m+1}(X,c\alpha)
\rightarrow X^{m+1}
$$
is contained in a rational curve in the fiber intersecting the image
of 
$$
M_{c_1\cdot \alpha,2} \times_{\text{pr}_2,\circ
  \text{ev},X,\text{pr}_1 \circ \text{ev}} M_{(c-c_1)\cdot \alpha,m+1}
\rightarrow M_{c\cdot \alpha,m+1}
$$
in a smooth point.
Then a general point of a general fiber of the restriction
$$
\text{ev}|_M:M_{e\cdot \alpha, m+1} \subset \Kgnb{0,m+1}(X,e\alpha)
\rightarrow X^{m+1}
$$
is contained in a rational curve in the fiber intersecting the image
of the boundary map
$$
M_{(e'_1+e''_1)\cdot \alpha,2}\times_{\text{pr}_2\circ
  \text{ev},X,\text{pr}_1\circ \text{ev}} M_{(e-e'_1-e'_2)\cdot
  \alpha,m+1} \rightarrow M_{e\cdot \alpha,m+1}
$$
in a smooth point.
\end{lem}

\begin{proof}
Consider a general fiber of the three restrictions of the evaluation
morphism
$$
\text{ev}_{I}= \text{ev}|_M:M_{e\cdot \alpha,m+1} \subset
\Kgnb{0,m+1}(X,e\alpha) \rightarrow X^{m+1},
$$
$$
\text{ev}_{II} =
(\text{pr}_1\circ \text{ev}\circ
\text{pr}_1,\text{pr}_{2,\dots,m+1}\circ \text{ev}\circ \text{pr}_2):
M_{e'_1\cdot \alpha,2} \times_{\text{pr}_2,\circ
  \text{ev},X,\text{pr}_1 \circ \text{ev}} M_{(e-e'_1)\cdot \alpha,m+1}
\rightarrow X^{m+1},
$$
and
$$
\text{ev}_{III} = 
(\text{pr}_1\circ \text{ev}\circ
\text{pr}_1,\text{pr}_{2,\dots,m+1}\circ \text{ev}\circ \text{pr}_3):
M_{e'_1\cdot \alpha,2} \times_{\text{pr}_2 \circ \text{ev},X,
  \text{pr}_1 \circ \text{ev}} M_{e''_1\cdot \alpha,2}
\times_{\text{pr}_2 \circ \text{ev},X, \text{pr}_1 \circ \text{ev}}
M_{(e-e'_1-e''_1)\cdot \alpha,m+1} \rightarrow X^{m+1}.
$$
Of course $\text{ev}_{II}$ is the restriction of $\text{ev}_{I}$ to
the domain of $\text{ev}_{II}$, and likewise $\text{ev}_{III}$ is the
restriction of $\text{ev}_{II}$ to the domain of $\text{ev}_{III}$.
Thus the fibers of the three evaluation maps form a triple of nested
subvarieties.  

\mni
Using the hypothesis for $(c,c_1) = (e,e'_1)$, Lemma ~\ref{lem-t2.5}
implies that the MRC quotient of a strong desingularization of a
general fiber of $\text{ev}_{I}$ is dominated by the MRC quotient of a
strong desingularization of the corresponding fiber of $\text{ev}_{II}$.
Using the hypothesis for $(c,c_1) = (e-e'_1,e''_1)$, Lemma
~\ref{lem-t2.5} implies that the MRC quotient of a strong
desingularization of a general fiber of $\text{ev}_{II}$ is dominated
by the MRC quotient of a strong desingularization of the corresponding fiber
of $\text{ev}_{III}$.  Thus the MRC quotient of a strong
desingularization of a general fiber of $\text{ev}_{I}$ is dominated
by the strict transform of the corresponding fiber of
$\text{ev}_{III}$.  
Also every general point of the domain of
$\text{ev}_{III}$ is a smooth point of the image of
$$
M_{(e'_1+e''_1)\cdot \alpha,2}\times_{\text{pr}_2\circ
  \text{ev},X,\text{pr}_1\circ \text{ev}} M_{(e-e'_1-e'_2)\cdot
  \alpha,m+1} \rightarrow M_{e\cdot \alpha,m+1}.
$$
Thus the MRC quotient of a general fiber of $\text{ev}_{I}$ is
dominated by the strict transform of the corresponding fiber of
$\Delta_{((e'_1+e''_1)\cdot \alpha,\{0\}),((e-e'_1-e''_1)\cdot \alpha,
  \{1,\dots, m\})}$.
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%
%% Section: Proofs of the main theorems
%% 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Proofs of the main theorems} \label{sec-pf}
\marpar{sec-pf}

\begin{proof}[Proof of Theorem ~\ref{thm-trc1}]
First of all, the hypotheses of Corollary ~\ref{cor-qM0.5} are implied
by the hypotheses of Theorem ~\ref{thm-trc1}.  
Therefore, by Corollary ~\ref{cor-qM0.5},
there is a unique irreducible component $M_{\alpha,1}$ of
$\Kgnb{0,1}(X,\alpha)$ dominating $X$, and a general fiber of
$M_{\alpha,1}$ over $X$ is geometrically connected.  Moreover, by
Lemma ~\ref{lem-pl1}(iii), a general fiber is Fano and thus rationally
connected.  Also by Corollary ~\ref{cor-qM0.5},
for every positive integer $r$ there exists a $1$-twisting surface of
$H$-type $0$ and $M$-class $(\alpha,r\cdot \alpha)$.

\mni
The proof that the fiber of
$$
\text{ev}|_M:M_{e\cdot \alpha,2} \rightarrow X^2
$$
over a general point of $X^2$ is geometrically rationally connected is
by induction on $e\geq 2$.  The base case is $e=2$.  
First of all, by Lemmas ~\ref{lem-mpc1}, \ref{lem-mpc3} and
\ref{lem-mpcCH}, the fiber is smooth (but possibly empty).
The claim
is that the fiber is nonempty and geometrically connected.  This will
be proved by Corollary ~\ref{cor-mpcdom4}.  The first hypothesis of
Corollary ~\ref{cor-mpcdom4} for $m=2$ is,
$$
\text{dim}(X) \geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
-1) + 4.
$$
This is implied by one of the hypotheses of Theorem ~\ref{thm-trc1}.  
The second hypothesis
is
$$
c \geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle -1) -
\text{dim}(X) -2,
$$
which is implied by the first hypothesis: $c\geq -2$ holds for $c=0$.
The final hypothesis,
$$
\text{CH}^p(Y) = \ZZ\{c_1(\OO(1))^p\}, \text{ for } 0\leq p \leq
2(\text{dim}(X) -\langle c_1(T_X),\alpha \rangle -1) + 4,
$$
is one of the hypotheses of Theorem ~\ref{thm-trc1}.  
Therefore, by Corollary
~\ref{cor-mpcdom4}, a general fiber of $\text{ev}$ is nonempty and
geometrically connected.  So the fiber is nonempty, smooth and
geometrically connected.

\mni
The next claim is that a general fiber of
$$
\text{ev}:M_{2\cdot\alpha,2} \rightarrow X^2
$$
is a Fano manifold, and thus geometrically rationally connected.
Since the fiber is nonempty, smooth and geometrically connected, it
only remains to prove the first Chern class is positive.
This follows
from Lemma ~\ref{lem-mpc5}.  The hypothesis of Lemma ~\ref{lem-mpc5} is
$$
\langle 2\text{ch}_2(T_X),\Pi \rangle \geq -1.
$$
Since one of the hypotheses of Theorem ~\ref{thm-trc1} is that
$$
\langle 2\text{ch}_2(T_X),\Pi \rangle \geq 0,
$$
Lemma ~\ref{lem-mpc5} implies that a general fiber of $\text{ev}$ is
Fano.  Therefore by ~\cite{KMM92c} and ~\cite{Ca}, a general fiber
is geometrically rationally connected.  This establishes
the base case of the induction.

\mni
Now, by way of induction, assume $e>2$ and assume the result is known for
$e-1$.  There exists a $1$-twisting surface of $H$-type $0$
and $M$-class $(\alpha,(e-1)\cdot \alpha)$.  
Thus, by Lemma ~\ref{lem-t3}, the fiber of
$$
\text{ev}|_M:M_{e\cdot\alpha,2} \rightarrow X^2
$$
over a general point is geometrically connected and geometrically
rationally connected if the fiber of
$$
\text{ev}_\Delta: M_{\alpha,2}\times_{\text{pr}_2\circ
  \text{ev},X,\text{pr}_1\circ \text{ev}} M_{(e-1)\cdot \alpha,2}
\rightarrow X\times X
$$
over a general point is geometrically connected and geometrically
rationally connected.  

\mni
Fix a general geometric point $(x_1,x_2)$ of
$X\times X$.  Projection onto the first factor
$$
\text{pr}_1: M_{\alpha,2}\times_{\text{pr}_2\circ
  \text{ev},X,\text{pr}_1\circ \text{ev}} M_{(e-1)\cdot \alpha,2}
\rightarrow M_{\alpha,2}
$$
restricts to give a projection of the fiber
$\text{ev}_\Delta^{-1}(x_1,x_2)$ onto the fiber of 
$$
\text{pr}_1\circ \text{ev}|_M:M_{\alpha,2} \rightarrow X\times X
\rightarrow X.
$$
Of course this is the same as the composition of the forgetful
morphism (forgetting the second marked point),
$$
\Phi_1:M_{\alpha,2} \rightarrow M_{\alpha,1}
$$
with the evaluation morphism
$$
\text{ev}|_M:M_{\alpha,1} \rightarrow X.
$$
By the first paragraph of the proof, the fiber of this last map over a
general point $x_1$ is geometrically connected and geometrically
rationally connected.  Moreover, a general fiber of $\Phi_1$ is a
smooth rational curve.  Therefore, by \cite{GHS} (or simpler
arguments), the fiber $F$ of
$$
\text{pr}_1\circ \text{ev}|_M:M_{\alpha,2} \rightarrow X\times X
\rightarrow X
$$
over a general point $x_1$ is rationally connected.  

\mni
By \cite{GHS}
again, to prove that $\text{ev}_\Delta^{-1}(x_1,x_2)$ is connected and
rationally
connected, it suffices to prove that the fiber of
$$
\text{pr}_1|_{\text{pr}_1^{-1}(F)}: \text{ev}_\Delta^{1}(x_1,x_2) \rightarrow F
$$
over a general point is geometrically connected and geometrically
rationally connected.  But the geometric
generic fiber of $\text{pr}_1|_{\text{pr}_1^{-1}(F)}$ is precisely the
same as the geometric generic fiber of
$$
\text{ev}|_M:M_{(e-1)\cdot \alpha,2} \rightarrow X^2.
$$
By the induction hypothesis, this is connected and rationally
connected.  Therefore $\text{ev}_\Delta^{-1}(x_1,x_2)$ is connected
and rationally connected.  By the arguments above, this implies the
fiber of
$$
\text{ev}|_M:M_{e\cdot \alpha,2} \subset \Kgnb{0,2}(X,e\alpha)
\rightarrow X^2
$$
over a general point of $X^2$ is geometrically connected and
geometrically rationally connected.  Therefore the theorem is proved
by induction on $e$.
\end{proof}

\begin{proof}[Proof of Theorem ~\ref{thm-main}]
The case of a linear or quadric hypersurface follows immediately from
Theorem ~\ref{thm-trc1}.  Thus assume the smooth complete intersection
is neither linear nor quadric, i.e., all $d_i \geq 2$ and
$\underline{d} \neq (2)$.  To prove the theorem, it suffices to verify
the hypotheses of Theorem ~\ref{thm-trc1}.  By Lemma ~\ref{lem-Chern},
$$
\langle c_1(T_X),\alpha \rangle = n+1 - \sum_{i=1}^c d_i,
$$
and
$$
\langle 2\text{ch}_2(T_X), \Pi \rangle = n+1 - \sum_{i=1}^c d_i^2.
$$
Thus,
$$
\text{dim}(X) - \langle c_1(T_X),\alpha \rangle - 1 = \sum_{i=1}^c
(d_i-1) - 2.
$$
So the first three inequalities in Theorem ~\ref{thm-trc1} are
$$
n+1 - \sum_{i=1}^c d_i > 0,
$$
$$
n+1 - \sum_{i=1}^c d_i^2 \geq 0,
$$
and
$$
n \geq \sum_{i=1}^c (2d_i-1).
$$
Because $d_i>1$ for all $i$, $\sum_i d_i < \sum_i d_i^2$.  Also, the
difference 
$$
[\sum_{i=1}^c d_i^2] - [\sum_{i=1}^c(2d_i-1)] = \sum_{i=1}^c
(d_i-1)^2
$$
is strictly positive.  In particular, it is $\geq 1$.  Therefore the
hypothesis
$$
n +1 \geq \sum_{i=1}^c d_i^2
$$
in Theorem ~\ref{thm-main} implies all three of the hypotheses
above.

\mni
The hypotheses on $c$ all hold.  In fact, for a smooth complete
intersection $X$, for every sufficiently positive integer $c$, $X$ is a
codimension $c$ linear section of a complete intersection $Y$ by Lemma
~\ref{lem-ext3}.  
\end{proof}

\begin{proof}[Proof of Theorem ~\ref{thm-trc2}]
First of all, the hypotheses of Theorem ~\ref{thm-trc2} imply the
hypotheses of Theorem ~\ref{thm-trc1}.  The one hypothesis of Theorem
~\ref{thm-trc1} that is a bit less obvious is,
$$
\text{dim}(X) \geq 2(\text{dim}(X) - \langle c_1(T_X),\alpha \rangle
-1) + 5.
$$
This follows by the same argument as in the proof of Corollary
~\ref{cor-qV}.  In particular, there exists a unique irreducible
component $M_{\alpha,1}$ of $\Kgnb{0,1}(X,\alpha)$ dominating $X$, and
a general fiber of $M_{\alpha,1}$ over $X$ is geometrically connected
and geometrically rationally connected.  

\mni
The second part is proved by induction on $m$.  The base case, $m=2$,
follows from Theorem ~\ref{thm-trc1}.  Thus, by way of induction,
assume the result for $m$ and consider the case $m+1$.  The hypotheses
of Theorem ~\ref{thm-trc2} imply the hypotheses of Corollary
~\ref{cor-qV}.  Thus, by Corollary ~\ref{cor-qV}, for every $e'\geq
2m-2$, there exists an $m$-twisting surface of $H$-type $0$ and
$M$-class either 
\begin{enumerate}
\item[\textbf{(I)}.]
$(\alpha,e'\cdot \alpha)$ or 
\item[\textbf{(II)}.]
$(2\cdot \alpha, e'\cdot
\alpha)$. 
\end{enumerate}

\mni
Then, by Lemma ~\ref{lem-t3}, one of the following hold
\begin{enumerate}
\item[\textbf{(I)}.]
For every $e'\geq 3m-2$, every general point of a general fiber of the
restriction
$$
\text{ev}|_M:M_{e'\cdot \alpha,m+1} \subset \Kgnb{0,m+1}(X,e'\alpha)
\rightarrow X^{m+1}
$$
is contained in a rational curve in the fiber intersecting the
boundary divisor
$\Delta_{(\alpha,\{0\}),((e'-1)\cdot\alpha,\{1,\dots,m\})}$
in a smooth point.
\item[\textbf{(II)}.]
For every $e'\geq 4m-2$, every general point of a general fiber of the
restriction 
$$
\text{ev}|_M:M_{e'\cdot \alpha,m+1} \subset \Kgnb{0,m+1}(X,e'\alpha)
\rightarrow X^{m+1}
$$
is contained in a rational curve in the fiber intersecting the
boundary divisor
$\Delta_{(2\cdot\alpha,\{0\}),((e'-2)\cdot\alpha,\{1,\dots,m\})}$
in a smooth point.
\end{enumerate}

\mni
Moreover, in Case (I), using the result for both $e'$ and $e'-1$,
Lemma ~\ref{lem-t4} implies
\begin{enumerate}
\item[\textbf{(I)}.]
For every $e'\geq 3m-1$, every general point of a general fiber of the
restriction 
$$
\text{ev}|_M:M_{e'\cdot \alpha,m+1} \subset \Kgnb{0,m+1}(X,e'\alpha)
\rightarrow X^{m+1}
$$
is contained in a rational curve in the fiber intersecting the
boundary divisor
$\Delta_{(2\cdot\alpha,\{0\}),((e'-2)\cdot\alpha,\{1,\dots,m\})}$
in a smooth point.
\end{enumerate}
Thus, both in Case (I) and Case (II), for every $e\geq 4(m+1)-6$,
every general point of a general fiber of the restriction
$$
\text{ev}|_M:M_{e\cdot \alpha,m+1} \subset \Kgnb{0,m+1}(X,e\alpha)
\rightarrow X^{m+1}
$$
is contained in a rational curve in the fiber intersecting the
boundary divisor
$\Delta_{(2\cdot\alpha,\{0\}),((e-2)\cdot\alpha,\{1,\dots,m\})}$.
Thus, a general fiber of $\text{ev}|_M$ is nonempty,
geometrically connected and
geometrically rationally connected if and only if a general fiber of
$$
\text{ev}_\Delta = (\text{pr}_1\circ\text{ev}\circ\text{pr}_1,
\text{pr}_{2,\dots,m+1}\circ \text{ev},\text{pr}_2): 
M_{2\cdot\alpha,2}\times_{\text{pr}_2\circ
  \text{ev},X,\text{pr}_1\circ \text{ev}} M_{(e-2)\cdot \alpha,m+1}
\rightarrow X\times X^m
$$
is nonempty, geometrically connected and geometrically rationally connected.

\mni
Fix a general (geometric) point $(p,(q_1,\dots,q_m))$ of $X\times
X^m$.  Because $e\geq 4(m+1)-6$, $e-2 \geq 4m-6$.  Thus, by the
induction hypothesis, the fiber of
$$
\text{ev}|_M:M_{(e-2)\cdot\alpha,m} \rightarrow X^m
$$ 
over $(q_1,\dots,q_m)$ is nonempty, geometrically connected and
geometrically rationally connected.  Moreover, for the forgetful map 
$$
\Phi_1:M_{(e-2)\cdot\alpha,m+1} \rightarrow M_{(e-2)\cdot \alpha,m}
$$
forgetting the first marked point, 
the fiber of
$$
\text{ev}|_M\circ \Phi_1:M_{(e-2)\cdot \alpha,m+1} \rightarrow
M_{(e-2)\cdot \alpha,m} \rightarrow X^m
$$
over $(q_1,\dots,q_m)$ is a flat, proper family of rationally
connected curves over $\text{ev}|_M^{-1}(q_1,\dots,q_m)$ whose general
fiber is smooth.  A conic bundle over a rationally connected variety
is rationally connected, thus the fiber of $\text{ev}|_M\circ \Phi_1$
over $(q_1,\dots,q_m)$ is rationally connected.  

\mni
Projection $\pi_2$
onto the $M_{(e-2)\cdot\alpha,m+1}$ factor defines a
projection from
the fiber of
$$
\text{ev}_\Delta:
M_{2\cdot\alpha,2}\times_{\text{pr}_2\circ
  \text{ev},X,\text{pr}_1\circ \text{ev}} M_{(e-2)\cdot \alpha,m+1}
\rightarrow X\times X^m
$$
over $(p,(q_1,\dots,q_m))$ to the fiber of
$$
\text{ev}|_M\circ \Phi_1:M_{(e-2)\cdot \alpha,m+1} \rightarrow
M_{(e-2)\cdot \alpha,m} \rightarrow X^m
$$
over $(q_1,\dots,q_m)$.  Since the second fiber is nonempty,
geometrically connected and geometrically rationally connected, by
\cite{GHS}, to prove the first fiber is rationally connected, it
suffices to prove the geometric generic fiber of $\pi_2$ is nonempty,
geometrically connected and geometrically rationally connected.

\mni
Assuming that $(q_1,\dots,q_m)$ is general,
the first marked point of a general curve parametrized by a point of
the second fiber is a general point $p'$ of $X$.  Thus, for general $p$, the
pair $(p,p')$ is a general point of $X\times X$.  Thus the geometric
generic fiber of $\pi_2$ equals the geometric generic fiber of
$$
\text{ev}|_M:M_{2\cdot \alpha,2}\subset \Kgnb{0,2}(X,2) \rightarrow 
X^2.
$$
By Theorem ~\ref{thm-trc1}, this is nonempty, geometrically connected
and geometrically rationally connected.  Thus, by the argument
above, the geometric generic fiber of
$$
\text{ev}|_M:M_{e\cdot \alpha,m+1} \subset \Kgnb{0,m+1}(X,e\alpha)
\rightarrow X^{m+1}
$$
is nonempty, connected and rationally connected.  Therefore the
theorem is proved by induction on $m$.    
\end{proof}

\begin{proof}[Proof of Theorem ~\ref{thm-main2}]
The case of a linear or quadric hypersurface follows immediately from
Theorem ~\ref{thm-trc2}.  Thus assume the smooth complete intersection
is neither linear nor quadric, i.e., all $d_i \geq 2$ and
$\underline{d} \neq (2)$.  To prove the theorem, it suffices to verify
the hypotheses of Theorem ~\ref{thm-trc2}.  
By Lemma ~\ref{lem-Chern},
$$
\langle c_1(T_X),\alpha \rangle = n+1 - \sum_{i=1}^c d_i,
$$
and
$$
\langle 2\text{ch}_2(T_X), \Pi \rangle = n+1 - \sum_{i=1}^c d_i^2.
$$
Thus,
$$
\text{dim}(X) - \langle c_1(T_X),\alpha \rangle - 1 = \sum_{i=1}^c
(d_i-1) - 2.
$$
So the first three inequalities in Theorem ~\ref{thm-trc2} are
$$
n+1 - \sum_{i=1}^c d_i > 0,
$$
$$
2(n+1-\sum_{i=1}^c d_i^2) \geq n+1-\sum_{i=1}^c d_i \ \Leftrightarrow
n+1 \geq \sum_{i=1}^c (2d_i^2-d_i),
$$
and
$$
n +4 \geq \sum_{i=1}^c (4d_i-3).
$$
Because $d_i>1$ for all $i$, $\sum_i d_i < \sum_i d_i^2$, which
implies $\sum_i d_i < \sum_i(2d_i^2-d_i)$.  Also, the difference
$$
[\sum_{i=1}^c(2d_i^2-d_i)] - [\sum_{i=1}^c(4d_i-3)] = \sum_{i=1}^c
(2d_i-3)(d_i-1)
$$
is strictly positive.  In particular, it is $\geq 1$.  Therefore the
hypothesis
$$
n+1 \geq \sum_{i=1}^c (2d_i^2-d_i)
$$
in Theorem ~\ref{thm-main} implies all three of the hypotheses above.

\mni
The hypotheses on $c$ all hold.  In fact, for a smooth complete
intersection $X$, for every sufficiently positive integer $c$, $X$ is a
codimension $c$ linear section of a complete intersection $Y$ by Lemma
~\ref{lem-ext3}.
\end{proof}

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