Left invariant metric on $SL(n,\mathbb{R}$)

Question

I'm vaguely familiar with the concept of invariant Riemannian metrics on Lie groups. I am mainly interested in $SL(n,\mathbb{R})$ but I guess the following is more general.

To the best of my knowledge, for any Lie group $G$ there can always be chosen a left-invariant (or right) Riemannian metric (but usually not bi-invariant). Moreover, this Riemannian metric induces a left-invariant ("standard", non Riemannian) metric $d_G$ on $G$.

My questions are:

1. Are my statements above correct? I don't know any canonical reference for these, so I would appreciate any.
2. Is the (non Riemannian) left invariant metric $d_G$ on $G$ unique in some sense (maybe up to equivalence or something)?
3. $SL(n,\mathbb{R})$ is naturally embedded in $\mathbb{R}^{n^2}$, on which all metrics are equivalent. How is the (non Riemannian) metric $d_{SL(n,\mathbb{R})}$ on $SL(n,\mathbb{R})$ relates to the metrics on $\mathbb{R}^{n^2}$? Is there any relation at all? Are there any interesting inequalities? Even the $n=2$ case is a starting point. I vaguely remember some identities somewhat looking like $\|M\|^2\approx \cosh d(M,I)$

Answer 1

There is an old article of Milnor on invariant metrics on Lie groups, you can find it here

Answer 2

If $G$ is a Lie group with Lie algebra $\mathfrak{g}$ of left-invarant vector fields, then the left-invariant Riemannian metrics on $G$ correspond to inner products on $\mathfrak{g}$ by defining $ \langle \mathrm{d}l_g X, \mathrm{d}l_g Y \rangle_g = \langle X, Y \rangle $ for $X,Y \in T_eG \cong \mathfrak{g}$. Any connected Riemannian manifold inherits a "standard" metric by setting the distance between a pair of points $p,q$ to be the infimum of lengths of paths from $p$ to $q$.

I can't tell what uniqueness statement you're gunning for in your second question. Already the set of inner products on $\mathfrak{g} \cong \mathbb{R}^n$ is $\frac{n(n+1)}{2}$-dimensional. These are all equivalent. There can be other left-invariant "standard" metrics which are not induced by Riemannian metrics. For example, if a metric is not a path metric, it is not induced by a Riemannian metric. (E.g. the inclusion of a unit sphere into Euclidean space and just restricting the metric, so that antipodal points are at distance $2$.)

One convenient choice of left-invariant Riemannian metric on $\mathrm{SL}(n,\mathbb{R})$ corresponds to the Frobenius inner product $\langle X, Y \rangle = \mathrm{trace}(X^TY)$ at the identity. This metric is also right-invariant for the subgroup $\mathrm{SO}(n)$. Every traceless symmetric matrix $X$ defines a geodesic $c(t) = \exp(tX)$ where $\exp$ is the matrix exponential. Then $\lVert \exp(tX) \rVert^2 = \sum_{i=1}^n e^{2t\lambda_i} $ where $\{ \lambda_i \}$ are the eigenvalues of $X$, while $d(e^{tX},I) = \lVert tX \rVert$. For $\mathrm{SL}(2,\mathbb{R})$ we have $\lambda_2 = - \lambda_1$ so we get $$ \lVert \exp(tX) \rVert^2 = e^{2t\lambda} + e^{-2t\lambda} = 2 \cosh(2 t \lambda) $$ while $d(e^{tX},I)=\lVert tX \rVert = t \sqrt{\lambda^2 + \lambda^{-2}}$. This looks pretty close to what you wrote.

There's a reason I restricted to symmetric matrices: if you want to look at the distance from the identity to a general matrix $A$, then you can factor it into $A=SK$ where $S$ is symmetric and $K$ is orthogonal. The distance from $A$ to $S$ is uniformly bounded by the diameter of $\mathrm{SO}(n)$ (this bound exists because the subgroup is compact). The symmetric matrices are easier to understand while still capturing the large-scale geometry. If you want to better understand all the geodesics, I recommend these notes by Robert Bryant: https://services.math.duke.edu/~bryant/GroupGeodesicsNotes.pdf