I was wondering if I could get some help understanding the following fact in an academic paper. The setup is as follows:
An open subset of $\Omega \subset \mathbb R^k$ is an open convex cone if it is closed with respect to taking linear combinations of its elements with positive coefficients. An open convex cone is homogeneous if its automorphism group
$$G(\Omega) := \left\{ A \in GL_k(\mathbb R): A \Omega = \Omega\right\}$$
acts transitively on it. Since $G(\Omega)$ is a closed subgroup of $GL_k(\mathbb R)$, it is a matrix group, and we denote its Lie algebra by $\frak g(\Omega).$ My question concerns the following lemma.
Let $\Omega \subset \mathbb R^k$ be an open convex cone not containing an entire line. Then
$$\dim {\frak g}(\Omega) \leq \frac{k^2}{2}- \frac{k}{2}+1.$$
The first two sentences of the proof are as follows: Fix a point ${\bf x} \in \Omega$ and consider its isotropy (stabiliser) subgroup $G_{\bf x}(\Omega) \subset G(\Omega).$ This subgroup is compact since it leaves invariant the bounded open set $\Omega \cap ({\bf x} – \Omega)$.
Basically, this last sentence is what I don't understand. Does anyone know the result that is being referred to here? I have tried searching for results on group actions that leave bounded open sets invariant, and have so far come up with nothing. Further, I'm not even entirely sure why $\Omega \cap ({\bf x} – \Omega)$ is a bounded open set. If anyone can shed any light, I'd be very grateful.
Best Answer
Let $A\subset R^k$ be a compact convex subset with nonempty interior invariant under a closed subgroup $H< GL(k,R)$. Since $A$ is $H$-invariant then so is the symmetrization $B=SA$ of $A$ (which is the convex hull of $A\cup -A$). Now, define the norm $||\cdot||$ on $R^k$ for which $B$ is the unit ball. This norm then is preserved by $H$. Therefore, $H$ is contained in the subset of $GL(k,R)$ consisting of matrices whose operator norm with respect to $||\cdot||$ is $\le 1$. Thus, $H$ is a closed and bounded subset of $GL(k,R)$, hence, is compact.
Edit. I just noticed that you are also asking why the intersection $\Omega\cap ({\mathbf x}-\Omega)$ is bounded. Here is an explanation. Observe that the point $p={\mathbf x}/2$ is fixed by the involution $$ \tau: {\mathbf y}\mapsto {\mathbf x}- {\mathbf y}. $$ Next, if a convex subset $C$ of $R^n$ is unbounded, then for each $c\in C$ there is a ray $\rho$ emanating from $c$ and contained in $C$. Thus, if $\Omega\cap ({\mathbf x}-\Omega)$ were to be noncompact, there would exist such a ray $\rho$ contained in both $\Omega$ and $\tau(\Omega)$. The latter means that $\tau(\rho)\subset \Omega$. But $\rho\cup \tau(\rho)$ is a straight line and you are assuming that $\Omega$ contains no straight lines.