The closure of a 1parameter subgroup in a compact Lie group is a torus. To what extent does this result generalize to compact Riemannian symmetric spaces? In other words, is the closure of a geodesic in a compact Riemannian symmetric space necessarily a flat totally geodesic submanifold?

1$\begingroup$ According to Berger in a compact symmetric space any geodesic lies in a totally geodesic flat torus, see ["A Panoramic View of Riemannian Geometry", section 4.3.5.2, page 193]. I am not sure how to prove this. I suspect an answer is somewhere is Besse's book on closed geodesics. $\endgroup$– Igor BelegradekSep 21, 2014 at 16:59

$\begingroup$ @IgorBelegradek: Thanks for the reference. $\endgroup$– Oliver JonesSep 23, 2014 at 7:27
1 Answer
I think that this is treated in Helgason's Differential Geometry, Lie Groups and Symmetric Spaces. The point is that, for any Riemannian symmetric space $G/K$, one has the notion of the rank $r$ of the symmetric space, which is the dimension $r$ of a maximal abelian subspace $\frak{a}$ in $\frak{k}^\perp\subset \frak{g}$, where $\frak{k}\subset\frak{g}$ are the respective Lie algebras of $K\subset G$. The exponential of $\frak{a}$ gives a torus that then maps down to a totally geodesic torus in $G/K$. (The images of this torus under the action of $G$ are the socalled 'flats' in $G/K$.) Then, because the $K$orbit of $\frak{a}$ in $\frak{k}^\perp$ contains everything in $\frak{k}^\perp$, it follows that every geodesic in $G/K$ is a geodesic in such a 'flat', and hence its closure is itself a closed geodesic subtorus in a flat.
The case of a compact Lie group $K$ is the special case in which $G = K\times K$ and we regard $K$ as embedded diagonally in $G$.

$\begingroup$ Thanks for the answer Robert but there's something puzzling me. The flats you describe above need not be tori, right? For example, if we have a symmetric space of noncompact type then the exponential map is a diffeomorphism and so the space can't contain a torus. Have I misunderstood something? $\endgroup$ Sep 22, 2014 at 0:35

$\begingroup$ @OliverJones: Yes, you are right. I was assuming that you wanted only to consider symmetric spaces of compact type; I should have made that clear. $\endgroup$ Sep 22, 2014 at 1:14

$\begingroup$ I guessed that's what you meant but just wanted to make sure. Thanks again for the nice answer. $\endgroup$ Sep 22, 2014 at 1:26