## Winding map and local injectivity

The winding map ${W}$ is a humble example that is conjectured to be extremal in a long-standing open problem. Its planar version is defined in polar coordinates ${(r,\theta)}$ by

$\displaystyle (r,\theta) \mapsto (r,2\theta)$

All this map does it stretch every circle around the origin by the factor of two — tangentially, without changing its radius. As a result, the circle winds around itself twice. The map is not injective in any neighborhood of the origin ${r=0}$.

The 3D version of the winding map has the same formula, but in cylindrical coordinates. It winds the space around the ${z}$-axis, like this:

In the tangential direction the space is stretched by the factor of ${2}$; the radial coordinate is unchanged. More precisely: the singular values of the derivative matrix ${DW}$ (which exists everywhere except when ${r=0}$) are ${2,1,1}$. Hence, the Jacobian determinant ${\det DW}$ is ${2}$, which makes sense since the map covers the space by itself, twice.

In general, when the singular values of the matrix ${A}$ are ${\sigma_1\ge \dots \ge\sigma_n}$, the ratio ${\sigma_n^{-n} \det A}$ is called the inner distortion of ${A}$. The word “inner” refers to the fact that ${\sigma_n}$ is the radius of the ball inscribed into the image of unit ball under ${A}$; so, the inner distortion compares this inner radius of the image of unit ball to its volume.

For a map, like ${W}$ above, the inner distortion is the (essential) supremum of the inner distortion of its derivative matrices over its domain. So, the inner distortion of ${W}$ is ${2}$, in every dimension. Another example: the linear map ${(x,y)\mapsto (3x,-2y)}$ has inner distortion ${3/2}$.

It is known that there is a constant ${K>1}$ such that if the inner distortion of a map ${F}$ is less than ${K}$ almost everywhere, the map is locally injective: every point has a neighborhood in which ${F}$ is injective. (Technical part: the map must be locally in the Sobolev class ${W^{1,n}}$.) This was proved by Martio, Rickman, and Väisälä in 1971. They conjectured that ${K=2}$ is optimal: that is, the winding map has the least inner distortion among all maps that are not locally injective.

But at present, there is still no explicit nontrivial lower estimate for ${K}$, for example we don’t know if inner distortion less than ${1.001}$ implies local injectivity.

## The least distorted curves and surfaces

Every subset ${A\subset \mathbb R^n}$ inherits the metric from ${\mathbb R^n}$, namely ${d(a,b)=|a-b|}$. But we can also consider the intrinsic metric on ${A}$, defined as follows: ${\rho_A(a,b)}$ is the infimum of the lengths of curves that connect ${a}$ to ${b}$ within ${A}$. Let’s assume there is always such a curve of finite length, and therefore ${\rho_A}$ is always finite. All the properties of a metric hold, and we also have ${|a-b|\le \rho_A(a,b)}$ for all ${a,b\in A}$.

If ${A}$ happens to be convex, then ${\rho_A(a,b)=|a-b|}$ because any two points are joined by a line segment. There are also some nonconvex sets for which ${\rho_A}$ coincides with the Euclidean distance: for example, the punctured plane ${\mathbb R^2\setminus \{(0,0)\}}$. Although we can’t always get from ${a}$ to ${b}$ in a straight line, the required detour can be as short as we wish.

On the other hand, for the set ${A=\{(x,y)\in \mathbb R^2 : y\le |x|\}}$ the intrinsic distance is sometimes strictly greater than Euclidean distance.

For example, the shortest curve from ${(-1,1)}$ to ${(1,1)}$ has length ${2\sqrt{2}}$, while the Euclidean distance is ${2}$. This is the worst ratio for pairs of points in this set, although proving this claim would be a bit tedious. Following Gromov (Metric structures on Riemannian and non-Riemannian spaces), define the distortion of ${A}$ as the supremum of the ratios ${\rho_A(a,b)/|a-b|}$ over all pairs of distinct points ${a,b\in A}$. (Another term in use for this concept: optimal constant of quasiconvexity.) So, the distortion of the set ${\{(x,y) : y\le |x|\}}$ is ${\sqrt{2}}$.

Gromov observed (along with posing the Knot Distortion Problem) that every simple closed curve in a Euclidean space (of any dimension) has distortion at least ${\pi/2}$. That is, the least distorted closed curve is the circle, for which the half-length/diameter ratio is exactly ${\pi/2}$.

Here is the proof. Parametrize the curve by arclength: ${\gamma\colon [0,L]\rightarrow \mathbb R^n}$. For ${0\le t\le L/2}$ define ${\Gamma(t)=\gamma(t )-\gamma(t+L/2) }$ and let ${r=\min_t|\Gamma(t)|}$. The curve ${\Gamma}$ connects two antipodal points of magnitude at least ${r}$, and stays outside of the open ball of radius ${r}$ centered at the origin. Therefore, its length is at least ${\pi r}$ (projection onto a convex subset does not increase the length). On the other hand, ${\Gamma}$ is a 2-Lipschitz map, which implies ${\pi r\le 2(L/2)}$. Thus, ${r\le L/\pi}$. Take any ${t}$ that realizes the minimum of ${|\Gamma|}$. The points ${a=\gamma(t)}$ and ${b=\gamma(t+L/2)}$ satisfy ${|a-b|\le L/\pi}$ and ${\rho_A(a,b)=L/2}$. Done.

Follow-up question: what are the least distorted closed surfaces (say, in ${\mathbb R^3}$)? It’s natural to expect that a sphere, with distortion ${\pi/2}$, is the least distorted. But this is false. An exercise from Gromov’s book (which I won’t spoil): Find a closed convex surface in ${\mathbb R^3}$ with distortion less than ${ \pi/2}$. (Here, “convex” means the surface bounds a convex solid.)

## Embeddings II: searching for roundness in ugliness

The concluding observation of Part I was that it’s hard to embed things into a Hilbert space: the geometry is the same in all directions, and the length of diagonals of parallelepipeds is tightly controlled. One may think that it should be easier to embed the Hilbert space itself into other things. And this is indeed so.

Let’s prove that the space $L^p[0,1]$ contains an isomorphic copy of the Hilbert space $\ell_2$, for every $1\le p \le \infty$. The case $p=\infty$ can be immediately dismissed: this is a huge space which, by virtue of Kuratowski’s embedding, contains an isometric copy of every separable metric space. Assume $1\le p<\infty$.

Recall the Rademacher functions $r_n(t)=\mathrm{sign}\, \sin (2^{n+1}\pi t)$, $n=0,1,2,3,\dots$ They are simply square waves:

Define a linear operator $T\colon \ell_2\to L^p[0,1]$ by $T(c_1,c_2,\dots)=\sum_{n}c_nr_n$. Why is the sum $\sum_{n}c_nr_n$ in $L^p$, you ask? By the Хинчин inequality:

$\displaystyle A_p\sqrt{\sum c_n^2} \le \left\|\sum c_n r_n\right\|_{L^p} \le B_p\sqrt{\sum c_n^2}$

The inequality tells us precisely that $T$ is an isomorphism onto its image.

So, even the indescribably ugly space $L^1[0,1]$ contains a nice roundish subspace.
(Why is $L^1[0,1]$ ugly? Every point of its unit sphere is the midpoint of a line segment that lies on the sphere. Imagine that.) One might ask if the same holds for every Banach space, but that’s way too much to ask. For instance, the sequence space $\ell_p$ for $p\ne 2,\infty$ does not have any subspace isomorphic to $\ell_2$. Informally, this is because the underlying measure (the counting measure on $\mathbb N$) is not infinitely divisible; the space consists of atoms. For any given $N$ we can model the first $N$ Rademacher functions on sequences, but the process has to stop once we reach the atomic level. On the positive side, this shows that $\ell_p$ contains isomorphic copies of Euclidean spaces of arbitrarily high dimension, with uniform control on the distortion expressed by $A_p$ and $B_p$. And this property is indeed shared by all Banach spaces: see Dvoretzky’s theorem.