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Your original approach for extending the geodesic is in fact possible, after a slight modification. (Credint: I heard this Idea from Eran Assaf). (You do not pushforward the original geodesic, but i …

$\newcommand{\al}{\alpha}$ The answer is no, there are many more such metric which are not induced by a Riemannian metric. (This answer is based on the comments above, made by user89334). Examples: …

Consider the group $GL_n(\mathbb{R})$ with its standard topology. It is not hard to show that there exists Riemannian metrics on it which are left-$GL_n$ and right-$O_n$ invariant. (In fact it's pos …

Sorry to join so late to the party, but I couldn't help noticing there is a missing class of rich matrix norms (which are not operator norms). These are the p-Schatten norms on $R^{n^2}$, which see …

This paper might give you some ideas on how to calculate the geodesics. It is about left invariant metrics on $GL_n(\mathbb{R})$. The geodesics are calculated using their characterization as critical …

Let $G$ be a Lie group with a left invariant metric $g$. Let $H$ be a (closed) Lie subgroup of $G$, and assume $g$ is right-$H$-invariant. (That is $d(R_h)_e:T_eG \to T_hG$ is an isometry for every $ …

Let $M$ be a smooth manifold of dimension $\ge 3$, equipped with a conformal structure (or a Riemannian metric). Then, the group of conformal diffeomorphisms is a finite dimensional Lie group. A proo …

$\newcommand{\GL}{\operatorname{GL}}$ Let $H_{>k}$ be the space of real $d \times d$ matrices of rank bigger than $k$, for some fixed $k$. $H_{>k}$ is an open connected submanifold of $ \mathbb{R}^{d …

This is a cross-post. Let $g:(0,\infty)^n \to [0,\infty)$ be a symmetric function -i.e. $g(\sigma_1,\dots,\sigma_n)$ does not depend on the order of the $\sigma_i$, with $g(1,\dots,1)=0$. We identif …

$\newcommand{\til}{\tilde}$ Lately, I have become interested in comparing intrinsic and extrinsic metrics on Riemannian manifolds. Consider $GL_n^+$ (invertible matrices , $\det >0$) as an open Riem …