A quasisymmetric surface with

no rectifiable curves

Christopher J. Bishop SUNY Stony Brook Stony Brook, NY USA 11794-3651

June 8, 1998

Abstract There is a quasiconformal mapping f of R^{3} to itself such that the

image of R^{2} × {0} contains no rectifiable curves.

1991 Mathematics Subject Classification. 30C65. Keywords: Quasisymmetric maps, quasiconformal mappings, rectifiable curves, Jacobian.

1Supported in part by NSF grant # DMS 95-00577.

1 Introduction

A mapping f : X → Y is called quasiconformal if there is an H < ∞ so that the linear dilatation

H(x) ≡ lim_{r→0} \frac{max_{|x−y|=r} |f(x) − f(y)|}{min_{|x−y|=r} |f(x) − f(y)|} ≤ H,

for every x. The purpose of this note is to prove

Theorem 1.1 There is a quasiconformal mapping f of R^{3} to itself such that

C^{−1} ≤ \frac{|f(z) − f(w)|}{φ(|z − w|)} ≤ C, ∀z, w ∈ R^{2} × \{0\},  (1)

for some C < ∞ and some increasing function φ such that

lim_{t→0} φ(t)/t = ∞.  (2)

For E ∈ R^{2}×{0}, if the image f(E) can be covered by balls {B(f(x_{j}), φ(r_{j})}, then E is covered by balls B(x_{j}, Cr_{j}). Then (??) easily implies that if f(E) has finite α-dimensional Hausdorff measure then E has zero α-measure. In particular, any curve in f(R^{2}) must have infinite length, since otherwise its preimage would be a non-degenerate curve of zero length, a contradiction. Recall that a homeomorphism f : X → Y is called quasisymmetric if there is a homeomorphism η : [0, ∞) → [0, ∞) such that

|x − y| ≤ t|x − z| implies |f(x) − f(y)| ≤ η(t)|f(x) − f(z)|.  (3)

Quasiconformal maps of R^{d}, d ≥ 2, to itself are quasisymmetric (e.g., [?]), so Theorem ?? answers a question of S. Rohde (Question 18 of [?]) by implying

Corollary 1.2 There is a quasisymmetric embedding of R^{2} into R^{3} such that the image f(R^{2}) contains no rectifiable curves.

In this case one can avoid the theorem that quasiconformal implies quasisymmetric; our function φ will be constructed so that φ(tr) ≤ φ(t)φ(r) and this combined with (??) easily implies f is quasisymmetric on R^{2}.

In [?], J. Heinonen and P. Koskela define, for a quasiconformal map f on the upper half space R_{+}^{n},

a_{f}(x) = exp{n^{−1}vol(B(x, r))^{−1} ∫_{B(x,r)} log J_{f}(y)dy},

(J_{f} denotes the Jacobian of f and r is half the distance of x = (x_{1}, …, x_{n}) to the boundary of the half space) and

E_{∞} = {x ∈ ∂R_{+}^{n} : 0 < lim inf_{r→0} a_{f}(x + re_{n}) ≤ lim sup_{r→0} a_{f}(x + re_{n}) = ∞}.

Question 6 of [?] asks if there is an example where E_{∞} has positive Lebesgue measure and whether it can be the whole boundary. In our example, the Jacobian at (x, y, t) is comparable to (φ(t)/t)^{3}, and thus E_{∞} is the whole boundary.

It is unknown whether there is an embedding of R^{2} into R^{3} which satisfies (??) with φ(t) = t^{α}, for some α < 1 (it is easy to see that α < 2/3 is impossible). Such maps are easy to find taking R into R^{2} (just take the usual parameterization of the von Koch snowflake), and are known to exist for R^{d} → R^{n} if n is large enough with respect to d (see [?]).

After seeing my first proof of Corollary ??, S. Rohde pointed out the map I constructed on R^{2} extended to to a quasiconformal map of R^{3} and hence gave Theorem ??. J. Heinonen then observed that this map solved the problem from [?]. I thank them for their comments and I thank the referees for numerous suggestions which improved the clarity of the paper.

2 The picture of the non-rectifiable surface

We will start by describing a picture of the surface f(R^{2}) which gives the main idea of the construction. We will then give a more careful construction of the map on all of R^{3}.

Let ψ : R → R be a smooth 1-periodic function with ψ(0) = 0 and such that the length of the graph Γ of ψ above [0, 1] is 2. Define a surface S ⊂ R^{3} by S = Γ × [0, 1]. Then S is a smooth surface of area 2 and is isometric to [0, 2] × [0, 1] (with respect to the intrinsic path metric).

Start with the unit square Q_{0} in R^{2}. Define a map f_{0} : Q_{0} → R^{3} by first stretching the square in the x direction by a factor of 2 (i.e., apply the map (x, y) → (2x, y) and then map the image rectangle isometrically onto the surface S. See Figure ??.

Figure 1: The map F_{1}: Stretching a square and folding it.

Now choose a large integer k_{1}. The dyadic subsquares of Q_{0} of size 2^{−k_{1}} are mapped by F_{1} to pieces of S which look like small rectangles of size 2^{−k_{1}+1} × 2^{−k_{1}}, and since F_{2} is smooth, the approximation is better and better, the larger k_{1} is. Replace each such approximate rectangle by a scaled copy of [0, 1] × Γ, now oriented to the “oscillations” of the surface are orthogonal to the first set. See Figure ??. The scaled copies will not quite meet along their edges, but we can make them do so by bending them with arbitrarily small angle (depending on k_{1}). Because we have stretched in the y direction this time we see that small enough squares of size r in Q_{0} are now mapped onto regions which approximate small squares of size 2r × 2r, within a small error.

We now continue by induction. After n steps, we have a smooth surface f_{n}(Q_{0}) which maps small squares of size r = 2^{−k_{n}} to pieces of the surface which approximate a 2^{n}r × 2^{n}r square with small error. Next take each dyadic square in Q_{0} of some size ≪ r and stretch by a factor of 2 in the x direction and map it onto a scaled copy of S, bending it to make the

Figure 2: Folding in the orthogonal direction

edges fit together. Then stretch in the y direction on an even smaller scale. If we choose the {k_{n}} to increase quickly enough then (??) will hold for φ(t) = t2^{α(t)} where α(t) is the minimal n so 2^{−k_{n}} < t.

3 The construction

Now we give a more careful construction of the desired map on R^{3}. It will be the limit of a sequence of compositions f_{n} = g_{1} ∘ … ∘ g_{n} where the g_{i}’s are given by the following

Lemma 3.1 There is a C < ∞, so that for any 0 < δ, ϵ, η ≤ 1, there is a ν = 2^{−m} > 0 and a quasiconformal map g : R^{3} → R^{3} so that if z = (x, y, t) ∈ R^{3}, then

1. g(z) = z if |t| > δ.

2. The linear dilatation of g is ≤ 1 + C(ϵ) in all of R^{3}, where C(ϵ) → 0 as ϵ → 0.

3. The linear dilatation of g is ≤ 1 + η if |t| < ν.

4. For z, w ∈ R^{2} × {0}, we have

\begin{matrix}(1 + ϵ)(1 − η) \\ 1 − η \\ (1 + \frac{Cδ}{|z−w|})^{−1} \end{matrix} \} ≤ \frac{|g(z) − g(w)|}{|z − w|} ≤ \{ \begin{matrix}1 + ϵ + η, & |z − w| ≤ ν. \\ 1 + ϵ + η, & ν ≤ |z − w| ≤ δ. \\ 1 + \frac{Cδ}{|z−w|}, & |z − w| ≥ δ. \end{matrix}

First we check that Theorem ?? follows from the lemma. Choose a sequence {ϵ_{n}} in (0, 1], so that ∑_{n} ϵ_{n} = ∞. For example, ϵ_{n} = 1 for all n will correspond to the picture described in the previous section and taking ϵ_{n} → 0, will give an asymptotically flat example. Let η_{n} = 10^{−n}. Let δ_{1} = 1. Now apply the lemma with ϵ_{1}, δ_{1}, η_{1} to obtain a quasiconformal mapping g_{1}, and a positive number ν_{1}. In general, given ν_{n−1}, choose δ_{n} < min(ν_{n−1}/8, δ_{n−1}/100), and apply the lemma to the triple ϵ_{n}, δ_{n}, η_{n}, to obtain the map g_{n} and the number ν_{n}. Let f_{n} = g_{1} ∘ … ∘ g_{n}. Then f_{n} is a finite composition of quasiconformal maps and hence is quasiconformal. It is the identity map on {z = (x, y, t) : |t| ≥ 1} and to estimate its dilatation at a point z with |t| < 1, choose k so that δ_{k+1} ≤ |t| < δ_{k}. Then g_{k+1} ∘ … ∘ g_{n} is the identity at z. The map g_{k} has linear dilatation at most 1 + C(ϵ_{k}) everywhere, and hence at z. For each 1 ≤ j < k, the image of z under g_{j} ∘ … ∘ g_{k} is in {z = (x, y, t) : |t| < δ_{j}} (since the inverse is the identity on on the complement) and hence the dilatation is at most 1 + η_{j}. Thus the linear dilatation of f_{n} at z is at most (1 + sup_{k≤n} C(ϵ_{k})) ∏_{k≤n}(1 + η_{j}) ≤ C. It is easy to check that {f_{n}} is Cauchy with respect to uniform convergence on compacta, so f = lim_{n} f_{n} exists and is quasiconformal.

To prove (??) choose z, w ∈ R^{2} and choose k so that 2δ_{k+1} ≤ |z − w| ≤ 2δ_{k}. For j = n − 1, …, 1, let z_{j} = g_{j} ∘ … ∘ g_{n}(z) and similarly for w_{j}. For k < j ≤ n − 1, induction and the fact that ∏(1 − (δ_{j}/δ_{k+1})) > 1/2 imply

|z_{j} − w_{j}| ≥ (1 − η_{j})|z_{j+1} − w_{j+1}| ≥ \frac{1}{2}|z − w| ≥ δ_{k+1} ≥ δ_{j}.

Thus case 3 of part 4 of the lemma holds and

(1 + C \frac{δ_{j}}{δ_{k+1}} )^{−1} ≤ \frac{|z_{j−1} − w_{j−1}|}{|z_{j} − w_{j}|} ≤ (1 + C \frac{δ_{j}}{δ_{k+1}} ).

For j = k we have (no matter which case holds),

(1 − η_{k})|z_{k+1} − w_{k+1}| ≤ |z_{k} − w_{k}| ≤ 2(1 + η_{k})|z_{k+1} − w_{k+1}|

For j = k − 1, |z_{k−1} − w_{k−1}| ≤ 4|z − w| ≤ 8δ_{k} ≤ ν_{k−1} so case 1 holds in part 4 of the lemma. Futhermore, for j = k − 2, …, 1, |z_{j−1} − w_{j−1}| ≤ 4|z_{j} − w_{j}| and ν_{j} < δ_{j} ≤ ν_{j}/4. Thus |z_{j} − w_{j}| ≤ ν_{j} implies |z_{j−1} − w_{j−1}| ≤ ν_{j−1}. Thus for j < k, case 1 of part 4 of Lemma ?? applies and so

1 − η_{j} ≤ \frac{|z_{j−1} − w_{j−1}|}{(1 + ϵ_{j})|z_{j} − w_{j}|} ≤ 1 + η_{j}.

Putting all these together and telescoping we get,

∏_{j=k+1}^{n}(1 + C \frac{δ_{j}}{δ_{k+1}} )^{−1} ∏_{j=1}^{k}(1 − η_{j}) ≤ \frac{|z_{1} − w_{1}|}{|z − w| ∏_{j=1}^{k}(1 + ϵ_{j})}

≤ 2 ∏_{j=k+1}^{n}(1 + C \frac{δ_{j}}{δ_{k+1}} ) ∏_{j=1}^{k}(1 + η_{j}).

Since the products converge (independent of k and n) and f(z) = z_{1}, f(w) = w_{1}, we have

C^{−1} ≤ \frac{|f(z) − f(w)|}{φ(|z − w|)} ≤ C,

where φ(t) = t ∏_{j=1}^{α(t)}(1 + ϵ_{j}), with α(t) defined by α(t) = max{k : t ≤ 2δ_{k}}. This is (??). Since ∑ ϵ_{n} = ∞, we also have (??).

Thus Lemma ?? proves Theorem ??. Now we prove the lemma. We will build g as a composition g = g_{1} ∘ g_{2} ∘ g_{3}. Let ψ : R → R be a smooth 1-periodic function with ψ(0) = 0 and such that the length of the graph Γ of ψ above [0, 1] is 1 + ϵ. Define a surface S ⊂ R^{3} by S = Γ × R. Let L(x) = ∫_{0}^{x} \sqrt{1 + |ψ′(t)|^{2}}dt, be the arclength of the graph over [0, x] and let J be the inverse, L^{−1}. Define a map F : R^{2} → S ⊂ R^{3} by

F : (x, y) → (J((1 + ϵ)x), y, ψ(J((1 + ϵ)x))).  (4)

Now extend F to a function G defined on all of R^{3} by setting G(x, y, t) = (x, y, t) if |t| ≥ 1, G is quasiconformal with dilatation at most 1 + C(ϵ), and the dilatation of G tends to 1 as t → 0. For example, we can do this by making sure the ∂_{t}G(x, y, 0) has norm 1 and is perpendicular to S = F (R^{2}), e.g., set

G(x, y, t) = F (x, y) + t · N(x, y),

for small enough t < t_{0}, where

N(x, y) = \frac{(−ψ′(J((1 + ϵ)x), 0, 1)}{\sqrt{|ψ′(J((1 + ϵ)x))|^{2} + 1}},

is the normal vector to F (R^{2}) at the point F (x, y).

To define g_{1}, let g_{1}(x, y, t) = \frac{δ}{2}G(\frac{2}{δ}(x, y, t)). Thus the tangent map near R^{2} × {0} looks like an affine stretch by a factor of 1 + ϵ in the x direction, followed by a rotation in the (x, t) plane.

To define the map g_{2} we do a similar thing, but at a much smaller scale and reversing the roles of x and y. Choose ν_{1} so small that g_{1}, restricted to the set {(x, 0, t) : t < ν_{1}} has linear dilatation ≤ 1 + η/10. Now let g_{2}(x, y, t) = ν_{1}G(ν_{1}^{−1}(y, x, t)). Choose ν_{2} so that g_{2} has linear dilatation ≤ 1 + η/10 on {(0, y, t) : t < ν_{2}}. Thus near R^{2}, the tangent map looks like an affine stretch in the y direction, followed by a rotation in the (y, t) plane.

Finally define g_{3} by

g_{3}(x, y, t) = \{ \begin{matrix}(x, y, t), & if |t| ≥ ν_{2}/2, \\ (x, y, ν_{2}h(t/ν_{2})), & if ν_{2}/8 ≤ |t| ≤ ν_{2}/2, \\ (x, y, (1 + ϵ)t), & if t ≥ ν_{2}/8, \end{matrix}

where h smoothly interpolates between (1 + ϵ)/8 and 1/2 on [\frac{1}{8}, \frac{1}{2}] and satisfies |h′ − 1| ≤ C(ϵ). Thus near R^{2}, the map is an affine stretch in the t direction and g_{3} is 1 + C(ϵ) quasiconformal.

We claim that g = g_{1} ∘ g_{2} ∘ g_{3} satisfies the lemma with ν = ν_{2}/16. Most of the conditions are clear from the construction. For example, it is clear that g is quasiconformal and is the identity on {(x, y, t) : |t| > δ}. Moreover, it has dilatation bounded everywhere by 1 + C(ϵ) (since each of the three component maps does). Finally, for points in {(x, y, t) : |t| < ν}, by considering the tangent maps of the composition, it is easy to see that the map acts as a conformal dilation of size 1+ϵ with error at most η. This also proves the expansion estimates on the small and intermediate scales (i.e., |z − w| ≤ δ). For large scales (i.e., |z − w| ≥ δ), we just use the fact that |G(z) − G(w)| = |z − w| + O(1) which implies |g(z) − g(w)| = |z − w| + O(δ). This completes the proof of the lemma.

Added in proof: G. David and T. Toro have improved Corollary

References

[1] Assouad, P. (1983) Prolongements Lipschitziens dans R^{n}, Bull. Soc. Math. France 111, 429–448.

[2] Heinonen, J. and Semmes, S. (1997) Thirty-three yes or no questions about mappings, measures and metrics, Conf. Geometry and Dynamics 1, 1–12.

[3] Heinonen, J. and Koskela, P. (1994) The boundary distortion of a quasiconformal mapping, Pacific J. Math 165, 93–114.

[4] P. Tukia and J. Väisälä (1980) Quasisymmetric embeddings of metric spaces. Ann. Acad. Sci. Fenn. Ser. A I Math. 5, 97-114.