Is geodesic distance equivalent to "norm distance" in $SL_n(\mathbb{R})$?

Take any norm, $\|\cdot\|$on $\mathbb{R}^n,$ and consider the resulting norm on $SL_n(\mathbb{R})$:

$$\|A\|:= sup\{\|Av\|: \|v\|=1\}.$$

Now take any left-invariant Riemannian metric, $g$, on $SL_n$. How do the geodesic balls, $B_g(I, r)$ around the identity matrix, $I$, compare with the metric balls, $B_{\|\cdot\|}(I,r)$ coming from $\|\cdot\|$? In particular do there exist $c, C$ such that $$B_{\|\cdot\|}(I,cr)\subset B_g(I, r) \subset B_{\|\cdot\|}(I,Cr)$$ for all sufficiently small $r$? Or anything of the sort?

Solution 1:

Here's what I see in Einsiedler-Ward:

Your norm on $SL_n(\mathbb{R})$ is induced by the operator norm on the vector space $M_{n}(\mathbb{R})$. Being a norm on a finite-dimensional vector space, it is equivalent to the euclidean norm $\|\cdot\|$ on $M_{n}(\mathbb{R})$.

Now any left-invariant Riemannian metric $\langle \ ,\rangle$ on $TG=G\times \mathfrak{g}$ is determined by its restriction to $TG_{I}=\{I\}\times \mathfrak{g} \cong \mathfrak{g} \subset M_n(\mathbb{R})$. Without loss of generality, we can assume that the Riemannian metric restricted to $\mathfrak{g}$ is induced by the euclidean norm on $M_n(\mathbb{R}).$ Let $d$ be the distance on $G$ induced by the path integral formula of $\langle \ , \rangle$.

Let $B$ be a pre-compact neighbourhood of $I$ in $G$ where the local inverse ($\log$) of the exponential map is defined. Assume $\log(B)$ is a convex ball in $\mathfrak{g}$. Let $B'$ be another pre-compact neighbourhood containing the closure of $B$.

Say $\phi:[0,1] \to B'$ joins $g_0, g_1 \in B$. Then since the norm of $\phi(t)$ is bounded, we get $c>0$ (independent of $g_0,g_1$) such that

$$L(\phi):= \int\langle D\phi(t), D\phi(t) \rangle^{1/2}dt = \int \left\langle DL^{-1}_{\phi(t)}\circ D\phi(t), DL^{-1}_{\phi(t)}\circ D\phi(t)\right\rangle^{1/2} dt = \int \|\phi(t)^{-1}\phi'(t)\| dt \\ \geq c\int\|\phi'(t)\|dt \geq c\| g_1-g_0\|.$$

This shows that $c\|g_1-g_0\| \leq d(g_0,g_1)$ if the infimum of path integrals is taken over paths which remain in $B'$. But since $d\left(B,(B')^c\right)>0,$ we can assume that this estimate holds in general. Hence for all $g_0,g_1 \in B$, we have

\begin{equation} c\|g_1-g_0\| \leq d(g_0,g_1) \qquad (1) \end{equation}

and it remains to show a reverse inequality.

Consider the path $\phi:[0,1] \to B$ given by $t \mapsto \exp\left(\log g_0 + t(\log g_1-\log g_0)\right)$. This is well defined since we assumed $\log(B)$ was a convex ball in $\mathfrak{g}$. Then, since the norm of $\phi(t)^{-1}$ is bounded, and since $d(\exp)$ is bounded in $\log(B)$ and since $\log$ is Lipschitz (by the mean value theorem) in a neighbourhood of $I$,

$$d(g_0,g_1) \leq \int\langle D\phi(t), D\phi(t) \rangle^{1/2}dt = \int \|\phi(t)^{-1}\phi'(t)\|dt \leq \int C_1\|\phi'(t)\|dt \leq C_1C_2\|g_1-g_0\|$$

for some $C_1, C_2>0$ (independent of $g_0, g_1$). Hence for all $g_0,g_1\in B$, we have

$$ d(g_0,g_1) \leq C \|g_0-g_1\|. \qquad (2)$$

Solution 2:

I can prove the following. If $h$ denotes the metric on $SL(n,\mathbb{R})$ coming from the operator norm, and $h_1$ denotes a left-invariant metric on $SL(n,\mathbb{R})$, then, at $g \in SL(n,\mathbb{R})$, and for any vector $x$ tangent to $SL(n,\mathbb{R})$ at $g$, we have:

$h(x,x) \leq \|g\|^2 h_1(x,x)$

I just used left translations, and things like that. If interested, I can write some more details. Also, one can prove (using the equivalence of any 2 norms on a finite-dimensional vector space) that there is a $C>0$ such that:

$h_1(x,x) <= C \|g^{-1}\|^2 h(x,x)$.

Hence, if $\|g\|$ and $\|g^{-1}\|$ are bounded above by some constants, then the two metrics induce uniformly equivalent norms on the tangent spaces of that region. In particular, there exists a neighborhood of the identity in $SL(n,\mathbb{R})$ for which the two geodesic distances are equivalent.