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

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

  • $\begingroup$ (+1) ..................... $\endgroup$ – Tim kinsella Mar 29 '19 at 5:32
  • $\begingroup$ Thank you sir. And if you get the time, could you do the same with your $11$ other accounts? $\endgroup$ – Insubordinate Mar 29 '19 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.

  • $\begingroup$ 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)? $\endgroup$ – Tim kinsella Apr 23 '16 at 0:25
  • $\begingroup$ Tim kinsella: yes correct, this is what I mean. $\endgroup$ – Malkoun Apr 23 '16 at 0:29
  • $\begingroup$ 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. $\endgroup$ – Tim kinsella Apr 23 '16 at 0:43
  • $\begingroup$ 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. $\endgroup$ – Malkoun Apr 23 '16 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.