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 possible to describe all of them explicitly). An immediate corollary is that there exists left-$GL\_n$ and right-$O\_n$ invariant metrics (in the sense of metric spaces) which generate the standard topology on $GL\_n$.


**Question:** Does every left-$GL\_n$ and right-$O\_n$ invariant metric (which generates the standard topology on $GL\_n$) come from a Riemannian metric? 


I suspect the answer is negative. A possible idea is the following: Take two different Riemannian metric $g\_1,g\_2$ which possess the above symmetries, and denote by $d\_1,d\_2$ the corresponding distances. Now define $d=d\_1+d\_2$, then it seems unlikely that $d$ will be induced from a Riemannian metric. (At least it is quite reasonable that it won't be induced by combinations like the $ag\_1+bg\_2$).


A criterion for when a given distance is induced by a Riemannian metric which seems relevant is given [here](https://math.stackexchange.com/a/673742/104576), but I am not sure if it's possible to verify it. (For instance how do we know if every such metric is a path metric).



Characterizing left invariant and right-$O_n$ invariant distances on $GL_n$
$\newcommand{\al}{\alpha}$


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:


1) Take any smooth norm on the space of $n \times n$ real matrices which is invariant under conjugation by the orthogonal group $O\_n$. (Smooth in the sense that $\|\cdot\|:\mathbb{R}^{n^2}\setminus \{0\}\to \mathbb{R}$ is a smooth function). An example for such a norm which is not induced by an inner product is the $p-$ [Schatten norm](https://en.wikipedia.org/wiki/Schatten_norm) (for $p \neq 2$).


Any such norm induces a left-$GL\_n$ invariant Finsler norm on $GL\_n$ which is also right $O\_n$-invariant. One just needs to remember that the induced distance function uniquely determines the Finsler norm (i.e a given distance can be induced by at most one Finsler norm), so distance functions obtained in this way from Finsler norms which are not Riemannian indeed cannot be induced from Riemannian metrics.


2) Let $d$ be a such an invariant metric. Take the square root $\sqrt d$. (It generates the same topology as $d$, and has the same isometries). There are two ways to see such a metric is not Riemannian:


**a)** It is not an intrinsic metric: Let $x,y \in GL\_n$ and assume $d(x,y)=\tilde d$. Let $\al:I \to GL\_n$ be a path connecting $x$ and $y$. We will show the intrinsic length $L\_I(\al)=\infty$ (when taken w.r.t $\sqrt d$), hence $(\sqrt d)\_I(x,y)=\infty \neq \sqrt d(x,y)$. Since any Riemannian metric is intrinsic we are done.


We can partition $\al \,$: $\, \al(t\_0),\dots,\al(t\_{n})$, such that $d(\al(t\_i),\al(t\_{i+1}))=s$ is constant. (This beacuae $d$ is continuous).


Note that $\tilde d=d(x,y)\le \sum\_{i=0}^{n-1} d(\al(t\_i),\al(t\_{i+1}))=ns$, so $s > \frac{\tilde d}{n}$. $L\_I(\al)\ge \sum\_{i=0}^{n-1} \sqrt{d(\al(t\_i),\al(t\_{i+1})})=n\sqrt s \ge \sqrt n \cdot \tilde d$, hence $L\_I(\al)=\infty$. Since $\al$ was arbitrary, this implies $(\sqrt d)\_I(x,y) = \infty$ as required.


**b)** This proof works when $d$ is induced by a Riemannian metric $g$. Let $\al:I \to GL\_n$ be a geodesic in $(GL\_n,g)$, parametrized by arclength. Assume by contraidction $\sqrt d$ is induced by the Riemannian metric $\tilde g$.


Since for small enough time $\al$ is length-minimizing, $d(\al(0),\al(t))=t$. Then: $$\sqrt t =\sqrt{d(\al(0),\al(t))} \le L\_{\tilde g}(\al)=\int\_0^t \| \dot \al(s)\|^{\tilde g}ds $$


So, denoting $g(t) =\int\_0^t \| \dot \al(s)\|^{\tilde g}ds -\sqrt t$, we get $g(t) \ge 0=g(0)$ for small enough $t$, and $g$ is differentiable on $(0,\epsilon)$, hence by Lagrang'es mean value theorem, there is asequence $t\_n \to 0$ such that $\| \dot \al(t\_n)\|^{\tilde g}-\frac{1}{2\sqrt{t\_n}} =g'(t\_n) \ge 0$, but this is a contradiction since $\| \dot \al(t\_n)\|^{\tilde g} $ converges to $\| \dot \al(0)\|^{\tilde g}$ which is finite.


