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I have been reading about symmetric spaces and studying the correspondence between pointed symmetric Riemannian manifolds (say, $(M,o)$) and Riemannian symmetric pairs, namely a system $(G,K,\sigma)$ where $G$ is a connected Lie group, $K$ a closed subgroup s,t. $Ad(K) \le GL(\mathfrak{g})$ is compact, and $\sigma$ is an involution such that $(G^\sigma)^{\circ} \subset K \subset G^\sigma $. We can construct a $G$-invariant Riemannian structure on $M:=G/K$ and make it a Riemannian symmetric space.

It is clear that $G$ acts as isometries on $M$, but is the converse true? That is, is the identity component of the isometry group of $M$ equal to $G$?

Thanks in advance!

Wu Yaochen
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1 Answers1

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The answer to your question is yes.

It is Theorem IV.3.3 in Helgason: Differential Geometry, Lie Groups, and Symmetric Spaces; Theorem 8.3.4 and Remark 8.3.5 in Wolf: Spaces of Constant Curvature; the text before Theorem 2.3 in Takeuchi: Lie Groups II.

user538044
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    This is false and only true for semisimple Riemannian symmetric pairs (by theorem V.4.1 in the Helgason book mentioned). For a counterexample, take $G = \mathbb{R}^n, K = {0}, \sigma = -\mathrm{Id}$. This is a Riemannian symmetric pair, but $I^0(G/K) = I^0(\mathbb{R}^n) = \mathrm{E}(n)$ is larger than $G$. – Ivan Solonenko Mar 03 '21 at 20:05