Let $G$ be a compact matrix Lie group under the Killing form metric $\langle \xi, \eta \rangle_g = -\frac{1}{2}\text{tr}((g^{-1}\xi)^T(g^{-1}\eta))$ for $g \in G$ and $\xi,\eta \in T_gG$. Let $C \subset G$ be a geodesically convex set. Pick finitely many $g_1,...,g_N \in C$ and define $\Omega$ to be the smallest closed convex set containing those points. What does the boundary of $\Omega$ look like? Is it like the smallest geodesic polygon that fits those points, like in Euclidean space? Or is it something more complicated?
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2$\begingroup$ What are $M$ and $C$? (Did you mean $M \subset G$ instead of $G \subset M$?) $\endgroup$– LSpiceCommented May 3, 2022 at 17:08
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1$\begingroup$ Maybe meant $C\subseteq G$. $\endgroup$– Andreas BlassCommented May 3, 2022 at 17:59
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1$\begingroup$ @LSpice Apologies. I meant $C \subset G$. I edited my question. $\endgroup$– Spencer KraislerCommented May 3, 2022 at 18:33
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$\begingroup$ You did not specify the metric on $G$. Are you assuming that $G$ is compact and endowing it with the Killing metric ? In that case, $G$ is not uniquely geodesic, so convex hulls are not a very robust notion (and it is not even clearly defined globally). $\endgroup$– NicolastCommented May 3, 2022 at 20:14
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$\begingroup$ Perhaps a first case to look at is $SU(2) = S^3$. How do you define convex hulls in a sphere ? $\endgroup$– NicolastCommented May 3, 2022 at 20:21
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1 Answer
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I guess you wanted to say smallest geodesic polytope (not polygon).
It is unclear what is polytope in a the matrix Lie group, but it seems to require geodesic hypersurfaces. They do not exist in most Riemannian manifold starting from dimension 3 and matrix Lie groups are not exceptional.
BTW, if you are interested in convex sets in general Riemannian manifolds, then check our paper "About every convex set...".
answered Feb 1, 2023 at 18:06
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$\begingroup$ Thank you. Would you by any chance know the definition for a "generic manifold". It does not appear in that paper. $\endgroup$ Commented Feb 1, 2023 at 18:50
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$\begingroup$ @SpencerKraisler It does: Given a positive integer $k$, we say that a property $P$ holds for $C^k$-generic Riemannian metric $g$ on a manifold $M$ if the property $P$ holds for a dense G-delta set (that is, a countable intersection of open subsets) of metric tensors in the $C^k$-topology. $\endgroup$ Commented Feb 2, 2023 at 12:50
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$\begingroup$ I'm confused by that sentence. That seems to describe a generic property, not a generic Riemannian manifold. Could you provide a non-Euclidean example of a generic manifold? Is, say, a matrix Lie group equipped with the sub-euclidean structure generic? $\endgroup$ Commented Feb 2, 2023 at 17:22
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$\begingroup$ @SpencerKraisler, You can do it, but I what for? For example, let us assume that "property" is something that can be expressed in a finite number of words; so there are only countable set of properties. Then say that manifold $M$ is generic if any generic property holds for $M$. But in this case any manifold that can be completely described in finite number of words is not generic, so it is not possible to describe it explicitly despite most of manifolds are generic. $\endgroup$ Commented Feb 2, 2023 at 19:35
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$\begingroup$ So you're saying that generic manifolds are non-constructible? $\endgroup$ Commented Feb 2, 2023 at 20:41