A very famous geometer recently commented to me that "hyperbolic geometries are the only geometries invariant under affine transformations". It is unclear to me what this comment even means. Can someone clarify this for me. Some type of reference would also be nice.
2026-04-19 17:23:37.1776619417
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hyperbolic geometry and affine transformations
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One can define hyperbolic geometry as the study of points and chords of a disk $D$ invariant under affine transformations of $D$ to itself; see Klein's disk model of non-Euclidean geometry. However, there are also others models of hyperbolic space, e.g., the Poincare model.
The remark could be related to Klein's disk model of hyperbolic geometry, but I am not sure.
The "very famous geometer" might not agree.
One would need a bit more context, I think. A good reference here are excellent articles and lecture notes of W. Goldman on projective, affine and hyperbolic geometry on manifolds.
(link:http://www.math.umd.edu/~wmg/ and more in particular http://www.math.umd.edu/~wmg/pgom.pdf )
Here is a guess:
First, we need to define "geometry": I will assume that it means a complete simply-connected Riemannian manifold of constant curvature $0$ or $\pm 1$. (This, of course, excludes many other geometries.)
Next, given a Riemannian manifold $(M,g)$ with the Levi-Civita connection $\nabla$, a diffeomorphism $f: M\to M$ is called affine if it preserves $\nabla$. Similarly, $f$ is projective if it sends unparameterized geodesics to unparameterized geodesics. In particular, each affine transformation is projective.
Now, among all n-dimensional geometries, both the Euclidean space and the unit sphere admit projective transformations which are not isometric, but the hyperbolic n-space, does not admit such. On the other hand, each affine transformation of $S^n$ is an isometry. Thus, my conjecture is that "famous geometer" meant projective instead of affine and what he had in mind is:
If $(M,g)$ is a complete simply-connected Riemannian manifold of constant curvature, for which the group of projective transformations equals the group of isometric transformations, then $(M,g)$ has negative curvature; in other words, it is a hyperbolic space.
Edit. As requested, here is a sketch of a proof that every affine transformation of the round sphere $S^n$ is an isometry ($n\ge 2$). First of all, let $P(S^n)$ denote the group of projective transformations of $S^n$, i.e. maps sending (unparameterized) geodesics to (unparameterized) geodesics. Let $PGL(n+1,R)$ denote the projectivization of the general linear group $GL(n+1,R)$; this group is acting on the projective space $R P^{n}$ by projective transformations (where the Riemannian manifold $R P^{n}$ is the quotient of $S^n$ by the antipodal involution). According to the Fundamental Theorem of Projective Geometry (FTPG), $PGL(n+1,R)$ is exactly the group of projective transformations of Riemannian manifold $R P^{n}$. Each element of $PGL(n+1,R)$ lifts to a self-diffeomorphism of $S^n$; I will use the notation $G$ to denote the group consisting of such lifts: Algebraically speaking it is a central extension of $PGL(n+1,R)$: $$ 1\to {\mathbb Z}_2\to G\to PGL(n+1,R)\to 1. $$ With a bit of work, using FTPG, one proves that $G$ is exactly the group of projective transformations of $S^n$. Thus, the problem reduces to proving that the only affine transformations of $RP^n$ are the isometries of $RP^n$, i.e. the subgroup $PO(n+1)<PGL(n+1,R)$. We will use the SVD, the singular value decomposition: Every element of $PGL(n+1,R)$ can be written as a product $kak'$, with $k, k'\in PO(n+1)$ and $a$ the projectivization of a diagonal matrix with positive diagonal entries. Thus, the problem reduces to proving that if such $a$ is an affine transformation of $RP^n$ then $a=id$. Every such $a$ preserves some projective lines in $RP^n$, projectivizations of coordinate 2-planes in $R^{n+1}$. If $a\ne id$ then it acts by a non-identity map on one of these projective lines $L$, fixing precisely two points $p, q$ on $L$. But, regarding $L$ as a parameterized geodesic, it is clear that $a: L\to L$ cannot send such a parameterized geodesic to a parameterized geodesic (the speed of $a(L)$ at $p$ will be different from the speed of $a(L)$ at $q$).