1
$\begingroup$

Let $S$ be a symmetric space of non-compact type, specifically let $S= SL(n)/SO(n)$ for some $n\geq 2$. I know that such spaces are geodesically complete, simply connected, and have non-positive curvature. Therefore, by the Cartan-Hadamard Theorem, the Riemannian exponential map $Exp_p$ at any point $p \in S$ is a global diffeomorphism to the appropriately dimensioned Euclidean space.

Edit: Is it correct that $Exp_e(v)=\pi\circ \exp$ where $\exp$ is the matrix exponential and $\pi:SL(n)\rightarrow SL(n)/SO(n)=S$?

$\endgroup$
4
  • $\begingroup$ Do you mean $S=SL(n)/SO(n)$? $O(n)$ is not a subgroup of $SL(n)$. $\endgroup$
    – Kajelad
    Commented Dec 22, 2020 at 15:54
  • $\begingroup$ Do you mean $\pi$ has image $S$ instead of $SO(n)$? $\endgroup$ Commented Dec 22, 2020 at 16:58
  • $\begingroup$ I don't remember, but probably yes. At any rate with your example you can be explicit and check whether or not $e^{tA}$ for $A$ symmetric is a geodesic in the space of positive determinant $1$ matrices. For further reading I can recommend Helgason. $\endgroup$
    – s.harp
    Commented Dec 22, 2020 at 17:39
  • $\begingroup$ Perhaps a better description would be to ask it there if $\operatorname{Exp}_{\pi(e)}\circ d_e\pi=\pi\circ\exp$ so that both sides are maps $\mathfrak{sl}(n)\to S$. I think it would be easier to prove tht there is a subspace $A\subseteq\mathfrak{sl}(n)$ such that $d_e\pi|_{A}$ is surjective and the above equality holds when restricted to $A$. $\endgroup$
    – Kajelad
    Commented Dec 22, 2020 at 17:45

1 Answer 1

6
$\begingroup$

For a linear group like $SL(n)$, the Lie theory exponential agrees with the matrix exponential $\mathrm{exp}$. However, the Riemannian exponential map doesn't always agree with the Lie theoretic exponential map. For example, if you choose a nonzero nilpotent matrix $N$, then $c(t) = \pi \circ \exp(tN)$ is not a geodesic in the associated symmetric space. Fortunately it is easy to answer your question in terms of something called the "Cartan decomposition."

When you pick a point in a symmetric space $G/K$, you can define a Cartan decomposition of the Lie algebra $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$. Here $\mathfrak{k}$ is the Lie algebra of $K$, and this decomposition is not a Lie algebra decomposition, rather it is a vector space decomposition with the bracket satisfying $$ [\mathfrak{k},\mathfrak{k}] \subset \mathfrak{k}, \, [\mathfrak{k},\mathfrak{p}] \subset \mathfrak{p}, \, [\mathfrak{p},\mathfrak{p}] \subset \mathfrak{k} .$$

In your example, the natural basepoint to choose is the identity coset $[SO(n)]$. Then $\mathfrak{k} = \mathfrak{so}(n)$ is the Lie algebra of skew-symmetric matrices, and $\mathfrak{p}$ is the vector space of symmetric traceless matrices.

Theorem. For any $X \in \mathfrak{p}$, $c(t) = \pi \circ \exp(tX)$ is a geodesic. Morever, every geodesic through the basepoint arises this way.

So in your example, the geodesics through $[SO(n)]$ are exactly $\pi \circ \exp (tX)$ for $X$ a symmetric traceless matrix.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .