Return to Answer

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: Suppose that $\delta:M\times M\to \mathbb{R}$ is such a 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.

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.

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.)

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, determining the exact cut locus of $I_n$ with respect to $ds^2$ does not seem to be trivial. Indeed, simple calculations show that the cut locus must be quite complicated. This does not seem to be a particularly good way to construct a left-invariant metric on $\text{SL}(2,\mathbb{R})$.

NB: I thought I had an explicit construction of a left-invariant metric on $\text{SL}(2,\mathbb{R})$, but I found a mistake in the calculations, so I'm withdrawing it for more work.

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: Suppose that $\delta:M\times M\to \mathbb{R}$ is such a 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.

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.

However, an approach along the lines of just explicitly writing down a metric in the case $n=2$ does work: For any $A\in \text{SL}(2,\mathbb{R})$, define quantities $r(A)\ge0$ and $\theta(A)\in (-\pi,\pi]$ so that $$ (a_{11}{+}a_{22}) + i (a_{12}{-}a_{21}) = 2e^{i\theta(A)}\ \cosh\bigl(r(A)\bigr). $$ (The reader should check that the quantity on the left hand side really does have complex norm at least $2$; this follows from $\det(A)=1$.) Now set $$ \nu(A) = \sqrt{r(A)^2+\theta(A)^2}. $$ It is easy to check that $\nu(A) = \nu(A^{-1})\ge0$ with equality if and only if $A=I_2$. It is slightly more involved to check that $$ \nu(AB) \le \nu(A) + \nu(B) $$ for all $A,B\in \text{SL}(2,\mathbb{R})$, but this is indeed true. With these two properties checked, one finds that setting $$ d(A,B) = \nu(A^{-1}B) $$ defines a left-invariant metric on $\text{SL}(2,\mathbb{R})$ that induces the usual topology on $\text{SL}(2,\mathbb{R})$ (not the discrete topology).

In fact, this metric is a little better than that: The function $\nu^2:\text{SL}(2,\mathbb{R})\to\mathbb{R}$ is smooth everywhere except along the $2$-disk $H^-\subset \text{SL}(2,\mathbb{R})$ that consists of the negative definite symmetric matrices of determinant $1$, where it is only Lipschitz.

Presumably, there is something similar for $n>2$, but I haven't yet run across it.

NB: I thought I had an explicit construction of a left-invariant metric on $\text{SL}(2,\mathbb{R})$, but I found a mistake in the calculations, so I'm withdrawing it for more work.