Number of fixed points of an involution Let $X$ be a positive-dimensional, smooth, connected projective variety (say over $\Bbb{C}$), and let $\sigma $ be an involution of $X$ with a finite number of  fixed points; then this number is even. The proof I have is somewhat artificial: blow up the fixed points, take the quotient variety, observe that the image  of the exceptional divisor in the quotient is divisible by 2 in the Picard group, and compute its self-intersection. Does someone know a more natural proof?
 A: Indeed, the number of fixed points is divisible by $2^{\dim X}$. This is actually an old result of Conner and Floyd, Periodic maps which preserve a complex structure, Bull. Amer. Math. Soc. 70, no. 4 (1964), 574-579. As @SashaP mentions, Atiyah and Bott observed that it is also a consequence of the holomorphic Lefschetz formula (Notes on the Lefschetz fixed point theorem for elliptic complexes, Matematika 10, no. 4 (1966), 101-139). 
A: This seems to be a deceptively simple statement. 
A proof not based on blow-ups, and involving instead a construction originally
used by Rost in the context of the degree formula, can be found in of O. Haution's paper 
"Diagonalizable $p$-groups cannot fix exactly one point on projective varieties", arXiv:1612.07663, 
to appear in the Journal of Algebraic Geometry. See in particular the Theorem in the introduction.
It works on every algebraically closed field of characteristic different from $2$.
A: Here´s an attempt at a more topological approach, perhaps someone can tell me where it goes wrong as I´m slightly suspicious of both argument and conclusion, nonetheless I´ll post it in case something can be gained from it. 
Let $M$ be a compact smooth oriented manifold and $G$ a finite group acting in an oriented manner on $M$. I claim that if there is a finite set $M^{G}$ of fixed points then $\chi(M)$ is divisible by the order of $G$.
To see this let $\nu$ be a vector field on $M$, chosen sufficiently generically that it has finitely many singularities and none of these lie in $M^{G}$. Now by averaging $\nu$ over $G$ we may assume that $\nu$ is $G$-equivariant, and still has singularity set disjoint from $M^{G}$. Now $G$-equivariance implies that $G$ permutes the singularity set of $\nu$ and for all $g$ in $G$ we should have $Ind_{x}(\nu)=Ind_{gx}(\nu)$ because $G$ preserves orientations, from which the theorem follows from the Hopf Index theorem.
Now for the question, $\sigma$ acts complex analytically, and so preserves orientations. By above we deduce that $\chi(M)$ is even. $\chi(M\setminus{M^{\sigma}})$ is even as $M\setminus{M^{\sigma}}:=U$ has a free action of $C_{2}$. Now one observes that $\chi(U)=\chi_{c}(U)$ as $U$ is an complex variety, and further that $\chi_{c}(U)=\chi(M)-\vert M^{\sigma}\vert$ and concludes.
Edit: the claim is false as abx notes in the comments. I'll leave this in case there's something to be salvaged by the Hopf index approach to the problem, but I'm not optimistic. 
