# Riemannian exponential map on symmetric spaces of non-compact type

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 $$\pi:SL(n)\rightarrow SL(n)/SO(n)=S$$?

• Do you mean $S=SL(n)/SO(n)$? $O(n)$ is not a subgroup of $SL(n)$. Commented Dec 22, 2020 at 15:54
• Do you mean $\pi$ has image $S$ instead of $SO(n)$? Commented Dec 22, 2020 at 16:58
• 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. Commented Dec 22, 2020 at 17:39
• 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 $A$. Commented Dec 22, 2020 at 17:45

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.