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Consider a symmetric space G/H of compact type where rank(G) is greater than rank(H). The Euler characteristic of this space is known to be zero. What can be said about the existence of a fixed point free isometric $S^1$ action on G/H?

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    $\begingroup$ If $H$ is a simple compact connected Lie group, then it's a symmetric space of compact type, and admits the isometric action of $G=H\times H$ given by left and right multiplication. Thus it has a free action of $H$ (e.g. by left multiplication), and hence of $S^1$. $\endgroup$
    – YCor
    Sep 5, 2013 at 14:16
  • $\begingroup$ That is fine. But what about symmetric spaces of type I i.e., when it is not a Lie group? $\endgroup$
    – Atreyee
    Sep 6, 2013 at 11:44

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