# Cotangent bundle of symmetric space is symmetric space?

Let $G$ be a connected Lie group. Then a symmetric space for $G$ is a homogeneous space $G/H$ where the stabilizer $H$ of a typical point is an open subgroup of the fixed point set of an involution involution $σ$ in $Aut(G)$. Thus $σ$ is an automorphism of $G$ with $σ^2 = id$ and $H$ is an open subgroup of the set $G^\sigma=\{ g\in G: \sigma(g) = g\}.$

Now let $G/H$ be symmetric space then $T^*(G/H)\cong G^{\mathbb C}/H^{\mathbb C}$ is symmetric space?. Where $G^{\mathbb C}$ means complexification of lie group $G$

• $T^*(S^2)$ is not diffeomorphic to any symmetric space. – Misha Apr 8 '14 at 19:24
• thanks a lot for nice comment, but can you explain more why? – user21574 Apr 8 '14 at 19:25
• de Rham decomposition: Every simply-connected symmetric space is isometric to a product of compact symmetric spaces, Euclidean spaces and symmetric spaces of noncompact type. In this example, the manifold does not split topologically as a product of $S^2$ and $R^2$ (why?). – Misha Apr 8 '14 at 21:45
• Helgason in his book said that if $\frak g/\frak h$ is symmetric then $\frak g^{\mathbb C}/\frak h^{\mathbb C}$ is symmetric. So $G^{\mathbb C}/H^{\mathbb C}$ is symmetric , where is the point? – user21574 Apr 23 '14 at 16:32
• My guess is that you are misreading what Helgason said. What you wrote is a vector space (quotient of two Lie algebras); of course, it admits a flat metric which is symmetric. This has nothing to do with your question. – Misha Apr 23 '14 at 17:46

• if $M$ is a symmetric space, then $TM$ has a canonical structure of symmetric space (whatever the way you realize your symmetric space, apply the tangent functor to it);
Hence, $T^*M$ inherits a canonical structure of symmetric space from that of $TM$.