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Suppose that $S^1$ acts freely on a manifold M. What is the Euler characteristic of $M$?

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up vote 14 down vote accepted

In fact, the Euler characteristic of $M$ will be $0$ even if $M$ is neither orientable nor closed.

To see this, let $\mathbb{Z}/n\mathbb{Z}\subseteq S^1$ be the $n$th roots of unity. The action of $\mathbb{Z}/n\mathbb{Z}$ is still free and so $M\rightarrow M/(\mathbb{Z}/n\mathbb{Z})$ is an $n-$fold cover.

But Euler characteristic is multiplicative with respect to covers, so $\chi(M) = n\cdot\chi(M(\mathbb{Z}/n\mathbb{Z})$, so $n\mid \chi(M)$ for all $n$. This implies $\chi(M) = 0$.

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The Euler characteristic of $M$ will be $0$ (at least for a closed orientable manifold, but probably more generally). One way to see this is to note that the $S^1$ action induces a nowhere vanishing vector field on $M$.

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Doesn't this require that $M$ and the action be smooth? – Qiaochu Yuan Jun 7 '12 at 14:51
Yes, $S^1$ acts on $M$ smoothly – Peter Hu Jun 7 '12 at 14:56

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