Let $X$ be a positivedimensional, 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 selfintersection. Does someone know a more natural proof?

3$\begingroup$ $\mathbb{A}^1$ is not projective $\endgroup$– Francesco PolizziDec 12, 2019 at 10:07

1$\begingroup$ By the way, you must assume $X$ irreducible, otherwise one can consider an involution over three points fixing just one of them. $\endgroup$– Francesco PolizziDec 12, 2019 at 10:28

2$\begingroup$ A very partial remark. A Kahler manifold with complex dimension $2n+1$ has even topological Euler characteristic by Hodge theory. So any diffeomorphism with nondegenerate fixed points has an even number of fixed points by the LefschetzHopf formula (for an involution the Lefschetz number is equal to the Euler characteristic mod $2$). $\endgroup$– Nick LDec 12, 2019 at 16:42

17$\begingroup$ Doesn't the holomorphic Lefschetz fixed point formula imply that the number of fixed points has to be divisible by $2^{\dim X}$? On one hand, $\sum (1)^i Tr(\sigma:H^i(O))$ is an integer because eigenvalues of an involution are $\pm 1$. On the other hand, for an isolated fixed point all the eigenvalues of the involution on the tangent space have to be equal to $1$, so the local contribution of each fixed point is $1/\det(1d\sigma)=1/2^{\dim X}$. $\endgroup$– SashaPDec 12, 2019 at 19:33

3$\begingroup$ @EBz: $X=(\mathbb{P}^1)^n$, $\sigma =(\sigma _1,\ldots ,\sigma _n)$, where $\sigma _i$ is a nontrivial involution of $\mathbb{P}^1$. $\endgroup$– abxDec 14, 2019 at 18:34
3 Answers
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), 574579. 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), 101139).
This seems to be a deceptively simple statement.
A proof not based on blowups, 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 $2$.

$\begingroup$ Many thanks, but I must say I was expecting something less sophisticated! I'll wait a few days before accepting your answer, in the hope that someone comes with some simpler idea. $\endgroup$– abxDec 12, 2019 at 11:06

$\begingroup$ Maybe, over $\mathbb{C}$, some simpler topological argument can work. $\endgroup$ Dec 12, 2019 at 12:44


1$\begingroup$ Well, I am not an algebraic geometer, so I get easily puzzled by hard papers... But I guess what are you looking for, at least for X/\sigma smooth, is the generalization of Riemann hurwitz formula. See math.leidenuniv.nl/scripties/MasterJavanpeykar.pdf. Indeed, suppose X is a riemann surface, and X/sigma smooth. Then the riemann hurwitz formula says that 2 g_X 2 = deg(sigma) 2 g_X/\sigma 2 + #number of ramified points, which yields that the nnumber of fixed points is even. Maybe you experts can try to use GrothendieckRiemannRoch to generalize to higher dims!! $\endgroup$ Dec 12, 2019 at 15:36

1$\begingroup$ @AndreaMarino: points on curves are divisors, but this is no longer true for higher dimensional varieties. In fact, the natural way to generalize the RHformula for an involution with finite fixed locus is to blowup the fixed points, and this leads to the original proof by the OP. $\endgroup$ Dec 12, 2019 at 15:38
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.

$\begingroup$ After averaging, the singularities may not be disjoint from the fixed points anymore. $\endgroup$– GuntramDec 14, 2019 at 10:41

$\begingroup$ @guntram I'm not sure I follow, if P is fixed by \sigma then surely \nu +sigma^{*} \nu singular at P implies \sigma is. What am I missing? $\endgroup$– user108998Dec 14, 2019 at 12:12

2$\begingroup$ $\mathbb{Z}/p$ acts on $ \mathbb{S}^2 \cong\mathbb{P}^1_{\mathbb{C}}$ by $z\mapsto e^{\frac{2\pi i}{p} }z$, the fixed points are $0$ and $\infty$. But $\chi (\mathbb{S}^2)=2$. $\endgroup$– abxDec 14, 2019 at 18:31

$\begingroup$ @abx ofc u are correct! Sorry about that. $\endgroup$– user108998Dec 14, 2019 at 18:34