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I am looking for a left invariant metric on $SL_n(\mathbb{R})$. If this is not possible, it would be acceptable to have a metric on $SL_n(\mathbb{R})/SO_n(\mathbb{R})$ or something like that. Is there such a thing?

Note: I am not looking for a Riemannian metric, just an ordinary metric.

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What is an "ordinary metric"? –  Igor Rivin Sep 27 '12 at 19:58
I mean just a distance function. –  safsaf32 Sep 27 '12 at 20:09
The discrete metric? –  Anthony Quas Sep 27 '12 at 21:07
Every Lie group has a left-invariant metric: Start with any positive definite inner product on the Lie algebra and ntranslate it to the rest of the group using left multiplication. Note also that Riemannian metric is not the same thing as a distance function. –  Misha Sep 27 '12 at 22:50
Mischa, it is true, however, that a Riemannian metric does define a distance function, right? So I believe your answer is a full positive answer to the question, right? –  Deane Yang Sep 28 '12 at 2:34

3 Answers 3

The OP specifically asked for a(n ordinary) metric, not a Riemannian metric. While Misha and Paul have given good answers, I think that it's worth pointing out that, if one just takes an arbitrary left-invariant Riemannian metric $ds^2$ on a connected group $G$, there is no guarantee that the associated $dist$ can be computed explicitly. To do this, one would need to be able to integrate the geodesic equations explicitly enough to be able to construct the distance function; unless the Riemannian metric is quite special, this generally can't be done in any explicit way. (See for example, what one has to do to describe the free rotations of a rigid body that has 3 distinct moments of inertia, which is essentially computing the geodesics on $\text{SO}(3)$ with respect to a left-invariant metric that is not bi-invariant.)

An $\text{SL}(n,\mathbb{R})$-invariant metric on $\text{SL}(n,\mathbb{R})/\text{SO}(n)$: Since $M=\text{SL}(n,\mathbb{R})/\text{SO}(n)$ is an irreducible Riemannian symmetric space (with nonpositive sectional curvature), it has, up to constant multiples, only one $\text{SL}(n,\mathbb{R})$-invariant Riemannian metric, and the associated $dist$ in this case is not entirely trivial to write down: We can identify each element $m = A\cdot \text{SO}(n)$ with its associated positive definite, unimodular symmetric matrix $s = \sigma(m) = AA^T$, and the formula for $dist(s_1,s_2)$ is as follows: Write $$ \text{det}(ts_1-s_2) = (t-\lambda_1)(t-\lambda_2)\cdots(t-\lambda_n) $$ where each $\lambda_i$ is positive and they satisfy $\lambda_1\lambda_2\cdots\lambda_n=1$. Then, up to a constant multiple, one has $$ dist(s_1,s_2) = \left(\sum_{i=1}^n (\log\lambda_i)^2\right)^{1/2}. $$ Obviously, writing this out as a function of the entries of $s_1$ and $s_2$ would not be easy. (Of course, this is just one example of the sort of metric that Paul gave; its distinguishing characteristic is that it is the $dist$ of a Riemannian metric, which is not true for Paul's specific example.)

A left-invariant metric on $\text{SL}(n,\mathbb{R})$: Once an invariant metric on $M$ has been defined, one can use it to define a metric on $\text{SL}(n,\mathbb{R})$ itself: First, suppose that $n$ is odd, so that $\text{SL}(n,\mathbb{R})$ acts effectively on $M$. Let $\delta:M\times M\to \mathbb{R}$ be an invariant metric and let $(s_1,\ldots,s_k)\in M\times M\times \cdots \times M$ ($k$ times) be a $k$-tuple of symmetric matrices with the property that the simultaneous stabilizer of all of the $s_i$ in $\text{SL}(n,\mathbb{R})$ is the identity matrix. (I guess $k=3$ suffices to find such a $k$-tuple; $k=2$ does not.) Set $$ dist(A,B) = \delta(As_1,Bs_1) + \delta(As_2,Bs_2) + \cdots + \delta(As_k,Bs_k). $$ This defines a left-invariant metric on $\text{SL}(n,\mathbb{R})$. Note, however, that this is not derived from a Riemannian metric. When $n$ is even, $-I_n$ lies in $\text{SL}(n,\mathbb{R})$ and it acts trivially on $M$, so the above construction won't work. However, when $n$ is even, just take the metric induced on $\text{SL}(n,\mathbb{R})$ by its natural embedding into $\text{SL}(n{+}1,\mathbb{R})$, and that will do the job.

To get a Riemannian metric on $\text{SL}(n,\mathbb{R})$ whose $dist$ is computable, one should probably take the left-invariant metric $ds^2 = \text{tr}\bigl((g^{-1}dg)^Tg^{-1}dg\bigr)$. The reason is that this metric is both left-invariant and invariant under right action by $\text{SO}(n)$, and the extra symmetries make the geodesic flow explicitly integrable. (There is, of course, no bi-invariant Riemannian metric on $\text{SL}(n,\mathbb{R})$.)

A little calculation shows that the $ds^2$-geodesic starting at $I_n$ with velocity $v\in{\frak{sl}}(n,\mathbb{R})$ is given by the formula $$ \gamma_v(t) = e^{v^Tt}e^{(v-v^T)t}, $$ where $v^T$ represents the transpose of $v$. In particular, one has the 'formula', for $A\in\mathrm{SL}(n,\mathbb{R})$, $$ dist(I_n,A) = \min\bigl\{\bigl(\text{tr}(v^Tv)\bigr)^{1/2}\ |\ e^{v^T}e^{(v-v^T)} = A\bigr\}. $$ Unfortunately, computing this $dist$ more explicitly is a challenge. By left-invariance, of course, one has $dist(A,B) = dist(I_n,A^{-1}B)$, so this determines the metric completely.

Computing $dist$ explicitly is nontrivial even in the case $n=2$. In this case, one has the fortunate circumstance that, unless $\det(v)>0$, there are no conjugate points along the geodesic $\gamma_v$, and, when $\det(v)>0$, the first conjugate point is at $t = \pi/\sqrt{\det(v)}$. However, describing the exact cut locus of $I_n$ in ${\frak{sl}}(n,\mathbb{R})$ with respect to $ds^2$ does not seem to be trivial. This does not seem to be a particularly good way to construct a left-invariant metric on $\text{SL}(2,\mathbb{R})$. Nevertheless, it's not hopeless. If I have time, I'll add a little note to this describing what one can say.

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Just to clear the air: $SL(n,R)$, as any Lie group $G$, admits a left-invariant Riemannian metric $ds^2$. This metric is not unique, but it is determined by $ds^2$ at $T_eG$, i.e., on the Lie algebra of $G$. If $G$ is connected (and $SL(n,R)$, of course, is), then $ds^2$ determines (canonically) a distance function $dist$ on $G$. Since $ds^2$ was left-invariant, so is $dist$. This answers OP's question.

If one were do deal with disconnected Lie groups, then one would have to make some further choices to define a left-invariant distance function on $G$. Here is one construction modelled on what one does in geometric group theory. Let $S\subset G$ be a smallest subset so that $G_0 \cup S$ generates $G$. Here $G_0\subset G$ is the component of identity. Then, define a "Cayley graph", where we connect elements $g, gs$ by an edge of unit length, for all $g\in G, s\in S$. Now, every two points of $G$ are connected by a path which is a concatenation of smooth paths in components of $G$ and edges of the graph. Finally, take the length-metric on $G$ defined using such paths, by minimizing lengths of paths connecting two points $x,y\in G$. By construction, this metric will be a left-invariant.

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There is a (left) $G$-invariant metric on $G/K$ for $G$ reductive, $K$ maximal compact, admitting a Cartan-type decomposition $G=KA^+K$ with $A^+$ the connected component of the identity in a maximal real-split torus (e.g., positive real diagonal matrices for $G=SL_n(\mathbb R)$. For $G=SL_n(\mathbb R)$, let $m(kak')$ be the max of the diagonal entries of $a\in A^+$ and of their inverses. Then $d(gK,hK)=\log m(h^{-1}g)$ is a metric on $G/K$.

The triangle inequality follows from the submultiplicativity of the operator norm $|\cdot|$, and noting that $m(g)=\max |g|,|g^{-1}|$.

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I'm lost here. Doesn't any Lie group have a left invariant Riemannian metric and therefore a left invariant "ordinary metric"? –  Deane Yang Sep 28 '12 at 2:35
@Deane, for the question of distance functions, probably you have to be careful when your group has $\pi_0$. But your general point is correct. –  Theo Johnson-Freyd Sep 28 '12 at 3:19
Theo, many thanks for the correction. –  Deane Yang Sep 28 '12 at 11:12
Sure, we know in advane that there is a Riemannian metric... the above is not quite it, but has tha elementary description if one wants. –  paul garrett Sep 28 '12 at 15:41

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