## 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.

## Through the Looking-Glass II

Just a few illustrations that did not fit in the previous post. A brief recap: given a holomorphic map $f$ of the upper half-plane, we can extend it to the bottom half-plane by setting $f(\bar z)=f'(z)(\bar z-z)$. This extension is not holomorphic unless $f$ is linear; however, it extends $f$ to a global homeomorphism provided that $|f''(z)|\cdot |\bar z-z|<|f'(z)|$. A calculation shows that the power map $f(z)=z^p$ satisfies this inequality whenever $|p-1|<1/2$.

Here are two examples where this inequality is seriously violated: $p=1.8$ and $p=2$. The images of concentric circles $|z|=r$ are shown, the upper half (power map) in green and the lower half (extended map) in red.

When $p=2$ the extension is $f(\bar z)=2|z|^2-2z^2$, so the images of concentric circles are again circles. The plot reminded me of this:

However, here the angle is $2\arcsin (1/2)=60^{\circ}$. Oh well. Maybe I should blog about the Kelvin wake pattern instead of holomorphic maps. The Wikipedia article does not offer much in the way of explanation.

By the way, nothing forces $p$ to be a real number. The Ahlfors extension works for any complex exponent with $|p-1|<1/2$. For example, $1+i/3$:

The plot shows also images of radial segments. In the green territory, the radial segments and circles are mapped to logarithmic spirals. The red part is more complicated. Here is what happens when $|p-1|$ exceeds $1/2$:

Finally, a flower with $p=2+3i$:

## Through the Looking-Glass, and What Ahlfors Found There

There are two ways to extend a function $f\colon [0,+\infty)$ to the negative semi-axis:

• Even reflection $f(-x)=f(x)$
• Odd reflection $f(-x)=-f(x)$

As the sketch shows, the even reflection is more likely to yield a continuous function. The odd reflection has no chance of being continuous unless $f(0)=0$. On the other hand, it has the same slope to the left and to the right of $0$, so if $f$ happens to be differentiable with $f(0)=0$, then the odd reflection is differentiable too.

Most importantly for this post, the odd extension has a chance of being a bijection. For example, $f(x)=\sqrt{x}$ extends to $f(x)=x/\sqrt{|x|}$, which is a homeomorphism of $\mathbb R$ onto itself.

Let’s move one dimension up. It’s more convenient to place the looking-glass onto the horizontal axis. We have a map that $f=(u,v)$ is defined in the upper halfplane $y\ge 0$, and would like to extend its definition to $y<0$. If $f$ sends the line $y=0$ into $v=0$, we can extend it using the mixed even-u/odd-v method: $u(x,-y)=u(x,y)$, $v(x,-y)=-v(x,y)$. This is easier to write down in complex notation: $f(\bar z)=\overline{f(z)}$. If $f$ was a homeomorphism of the upper halfplane onto itself, the extension will be a homeomorphism of the entire plane. The process looks like this: But in general, the image of the horizontal axis is not a straight line. How can we find a reflection of $f$ in such a strange mirror?

Going back to the origins of geometry, we can use similar triangles to locate the image of a point:

Denoting the purple point by $z$ and the blue one by $z+h$, we see that the (complex) scaling factor of the triangle is $\displaystyle \frac{f(z+h)-f(z)}{h}$, and therefore the image of the red point is $\displaystyle f(\bar z)=f(z)+\frac{f(z+h)-f(z)}{h}(\bar z-z)$

This looks promising, but there is no obvious reason why $f(\bar z)$ should fall into the region behind the looking-glass, or why this extension should be a homeomorphism. It is also unclear how to choose $h$: should it be real, as on the picture above? should its size be fixed or depend on $z$? Or maybe… get rid of $h$ be letting $h\to 0$? Let’s do that. The extension becomes $\displaystyle f(\bar z)=f(z)+f'(z)(\bar z-z)$

Can’t get much simpler than that. The identity $f(z)=z$ extends to $f(\bar z)=z+1(\bar z-z)=\bar z$, which is encouraging but not very interesting. Let’s try other powers $f(z)=z^p$. The extension $\displaystyle f(\bar z)=z^p+pz^{p-1}(\bar z-z)$ is homogeneous of degree $p$, so concentric circles around 0 are mapped onto similar curves. The plots below show the images of concentric circles for several value of $p$: the image of upper semicircle is green, the image of the bottom (where the extension formula is used) is in red.

Let’s begin with $p>1$:

And it gets worse as $p$ increases. In the opposite direction, consider powers less than 1:
It’s surprisingly easy to get to the root of the problem. Take the complex derivatives of the extended map: $f(\bar z)_{z}=f''(z)(\bar z-z)$ and $f(\bar z)_{\bar z}=f'(z)$. To preserve orientation, the extension must satisfy $|f(\bar z)_{z}|<|f(\bar z)_{\bar z}|$, which requires the original map to satisfy $2|f''(z)|\mathrm{Im}\,z<|f'(z)|$. With $f(z)=z^p$ this boils down to $|p-1|<1/2$.
with(plots): p:=3/2: curves:={seq(complexplot((1+k/5)^p*exp(I*p*t),t=0..Pi,color=green,thickness=2), k=-4..4)} union {seq(complexplot((1+k/5)^p*((1-p)*exp(I*p*t)+p*exp((p-2)*I*t)),t=0..Pi,thickness=2) ,k=-4..4)}: display(curves);