# Getting rid of needing the target to be a Lie Group

Something that seemed overly specific about the last approach was using elements of the translation subgroup of the isometry group on the plane. There seems to be a nice fix for this, which has the advantage of making sense even if the model space doesn't have a group action.

**The idea**

Given charts $\phi_1:U_1\to Y$ and $\phi_2:U_2\to Y$, we look at the space $\textrm{biLip}(Y,Y)$ of $k$-biLipschitz maps from $Y$ to $Y$. This set has two different functions on it, $\psi_1,\psi_2:\textrm{biLip}(Y,Y)\to \mathbb{R}^+$ defined by

$$\psi_1(f)= \min\{\epsilon>0~|~ f(U_2)\subseteq\mathcal{N}_\epsilon(f(U_1))\},$$

where $\mathcal{N}_\epsilon$ denotes the $\epsilon$-neighborhood, and

$$\psi_2(f)= \max\{x\in U_2~|~d(f(x),x)\}.$$

We then try to simultaneously minimize 3 things:

1. The biLipschitz constant of $f$, which ensures that if $\phi_2: U_2\to Y$ was a nice chart (i.e. nearly an isometric embedding), then $f\circ \phi_2$ is also a nice chart
2. The function $\psi_1$, which ensures that $f\circ\phi_1(U_1)$ is close to $\phi_2(U_2)$
3. The function $\psi_2$, which **makes f look as much like a translation as possible.**

## An example

To see this last claim, let's consider the case where the ambient space is $\mathbb{R}^2$. In this case, we can restrict our maps $f$ to global isometries of $Y=\mathbb{R}^2$, so we don't have to worry about the biLipschitz constant of $f$.

We then imagine having two sets in the plane $\phi(U_1)$ and $\phi(U_2)$. Let's think of these as somewhat densely sampled disks, say with mesh $\epsilon.$

As before, we have an $S^1$ action on the set of maps $f$ that roughly match up $\phi_1(U_1)$ and $\phi_2(U_2)$, so we can think of this set of nice $f$ as being roughly of the form $\{\rho\circ t~|~\rho\in S^1\}$ where $t$ is a fixed translation.

What is "special" about the point $t=id\circ \rho$? It is the element that moves points of $\phi_1(U_1)$ as little as possible.

## A nice property

Here's a nice property we get here, that I think was what was important about translations in the first place:

**Theorem:** Let $U_0\subset X$ be fixed, and let $\phi_i:U_i\to Y$ be a sequence of charts such that $U_i$ converges to $U_0$, and let $f_i$ be the map "matching" $\phi_i(U_i)$ to $\phi_0(U_0)$ as above. Then $f_i\circ \phi_i$ converges pointwise to $\phi_0$.

# Convex hulls and sampling density

Another thing that has been bothering me is the following. Suppose we were dealing with a metric space, but the way we sampled it has different densities at different points.

If the sampling near a $U_0\in X$ is extremely dense, then moving $\phi_1(U_1)\in X$ into the $\epsilon$-neighborhood of $\phi_0(U_0)$ will be easy to do without distortion, whereas moving $\phi_0(U_0)$ into the $\epsilon$-neighborhood of $\phi_1(U_1)$ will require ripping "holes" in $U_0$, i.e. very close points in $U_0$ will have to move apart since $U_1$ is less densly populated.

This causes a weird assymetry, and can get in the way of having nice convergence properties (e.g. having the "transport" map converge to an isometry as the sampling gets dense would require something about the sampling rates being "identical" at different parts of the manifold in the sense that they lead to the same metric density. If we think of our sensors as being sampled from a measure on a metric space, this would entail having some sort of compatibility between the measure and the metric (e.g. sufficiently small $r$-balls all have the same volume or something), which seems like an unnecessary restriction)

There's a nice way to get rid of this problem when $Y$ is a convex space. In this case, we can look at the convex hulls of $\phi(U_0)$ and $\phi(U_1)$ and try to match those up instead of the pointsets themselves.

This also has the nice property that it reduces the number of points we have to think about. The size of the boundary of the convex hull in Euclidean space has a lower order of growth than the interior points, so this should make the algorithm more efficient.