TITLE: Smooth circle action, $\chi(M^{S^1}) = \chi(M)$
QUESTION [6 upvotes]: I need a reference for the following result (which I can prove myself, but my proof is rather ugly and I would prefer to just cite the statement instead or re-proving it):
Let $M$ be a closed smooth manifold and let $S^1$ act smoothly on $M$. Then the set of fixed points of the action, $M^{S^1}$, is again a (not necessarily connected) smooth manifold and $\chi(M^{S^1}) = \chi(M)$.

REPLY [7 votes]: For the Euler characteristic there is the following.
Consider a tubular neighborhood $T$ of the fixed set (submanifold) $M^{{\mathbb S}^1}$. By Mayer-Vietoris sequence, relatively to the open covering $\{U = M-{\mathbb S}^1, T\}$ of
$M$, we get
$$\chi (M) = \chi (U) + \chi(T) - \chi(U\cap T).$$
But, $\chi (U) =\chi(U\cap T)=0$ since the circle action is free on $U$. Finally, since $T$  equivariantly retracts on $M^{{\mathbb S}^1}$ then
$$\chi (M)  = \chi (T) = \chi(M^{{\mathbb S}^1}).
$$