In the complex plane there exists a very large set of smooth conformal maps, (The famous Riemann Mapping Theorem states that between any two simply connected open sets in $\mathbb{C}$ there exists a biholomorphic mapping which is also conformal).
The situation is much more bleak as soon you go up to 3 dimensions and higher where the only smooth conformal maps are the mobius transformations. This is the famous Liouville's Theorem
I was thinking about trying to find a generalization of the concept of an "angle" so that the "set of all generalized-angle preserving maps" in $\mathbb{R}^3$ would have an interesting structure larger than merely mobius transformations.
There is the obvious generalization, solid angles. In 3 dimensions these are defined for collections of 4 points at a time (we select 1 anchor point and consider the 3 vectors formed by the anchor to the 3 other non collinear points. And then take the area of the spherical triangle spanned by these 3 vectors). The solid angle can be calculated explicitly using a formula given here.
So that leads to our question…. is the set of all smooth solid-angle-conformal functions from $\mathbb{R}^3 \rightarrow \mathbb{R}^3$
a larger set than the set of mobius transformations? And does there exist a generalization of the Riemann Mapping Theorem for them?
Best Answer
The basic question here is a linear algebra question: which orientation-preserving linear maps $T:\mathbb{R}^3\to\mathbb{R}^3$ preserve solid angles formed by triples of vectors? (You are then asking about smooth maps whose derivative at every point is such a linear map.) Let $G\subset GL_3(\mathbb{R})$ be the group of such maps, and let $H=\mathbb{R}^+\times SO(3)\subset GL_3(\mathbb{R})$ be the group of conformal linear maps (i.e., compositions of rotations and scalings). It is clear that $H\subseteq G$; I claim that in fact $G=H$, so your maps are just the same as conformal maps.
Suppose $T\in G$; we wish to show that $T\in H$. Let $e_1,e_2,e_3$ be the standard basis vectors for $\mathbb{R}^3$. Taking a singular value decomposition of $T$, we may compose $T$ with rotations (which does not change whether $T\in H$) to assume that $T$ is diagonal. Since $T$ is orientation-preserving, we can compose it with a further rotation to assume the diagonal entries are positive, and we can compose with a scaling to assume that $T(e_1)=e_1$. We have $T(e_2)=be_2$ and $T(e_3)=ce_3$ for positive scalars $b$ and $c$; I claim that $b=c=1$ so $T$ is just the identity.
To prove this consider the solid angle formed by the unit vectors $e_1,\frac{e_1+e_2}{\sqrt{2}},$ and $e_3$. $T$ maps these vectors to $e_1,\frac{e_1+be_2}{\sqrt{2}},$ and $ce_3$, which form the same solid angle as the unit vectors $e_1,\frac{e_1+be_2}{\sqrt{1+b^2}},$ and $e_3$. But now observe that as $b$ grows from $0$ to $\infty$, the spherical triangle formed by these vectors grows strictly (we are taking the spherical triangle formed by $e_1,e_2,$ and $e_3$ and replacing the $e_2$ vertex with a point on the edge between $e_1$ and $e_2$). So, the only way the area of this triangle would be the same as the area of the triangle formed by $e_1,\frac{e_1+e_2}{\sqrt{2}},$ and $e_3$ is if $b=1$. Swapping the roles of $e_2$ and $e_3$, we similarly conclude that $c=1$.
Similar arguments show that for any $n\geq m\geq 2$, the orientation-preserving linear maps $T:\mathbb{R}^n\to\mathbb{R}^n$ that preserve "$m$-dimensional solid angles" formed by $m$ vectors are the same as the conformal linear maps.