Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

2 Answers

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 $\chi(M) = 0$.

share|improve this answer
add comment

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 $M$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.