# 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?

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

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

• (+1) ..................... Mar 29, 2019 at 5:32
• Thank you sir. And if you get the time, could you do the same with your $11$ other accounts? Mar 29, 2019 at 17:59

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.

• Thanks for your answer. By $h(x,x)$ do you just mean the square of the operator norm of $x$, thinking of $x$ as a matrix acting on $\mathbb{R}^n$ (let's assume g=e)? Apr 23, 2016 at 0:25
• Tim kinsella: yes correct, this is what I mean. Apr 23, 2016 at 0:29
• Thanks again. I don't completely understand the last paragraph. If we were talking about two Riemannian metrics which induced uniformly equivalent norms in a neighborhood of $e$, then I think I would be able to fill in the details. But let me ponder a little more. Apr 23, 2016 at 0:43
• Tim kinsella: I must admit the last paragraph in my answer is not written in a clear way! However the 2 inequalities should give you what you were hoping for. In any case, I can provide you with more details if you want, in particular the proofs of the 2 inequalities, if you want, or how you use them. Apr 23, 2016 at 14:35