# Is the structure of "solid angle" conformal maps more interesting in higher dimensions than just conformal maps?

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?

• Such a great question! I've vaguely wondered this myself, but wasn't sure how to articulate it. Oct 29, 2021 at 4:59

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.
• This argument makes sense in that locally all solid-angle preserving maps must look locally like conformal maps (a shift + rotation), but even in the case of $\mathbb{C}$ that is true: after all all biholomorphic functions locally look like shifts+rotations (see taylor series $f(z) = a_0 + a_1z + o(z^2)$ and similarly all mobius transformations are locally shifts+rotations. So just because the set of supported rotations is identical to mobius transformations, doesn't mean THEY ARE mobius transformations Jan 14, 2021 at 5:05