By abx

2019-12-12 09:42:55 8 Comments

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?


@abx 2019-12-16 06:54:47

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).

@EBz 2019-12-14 10:09:15

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.

@Guntram 2019-12-14 10:41:17

After averaging, the singularities may not be disjoint from the fixed points anymore.

@EBz 2019-12-14 12:12:23

@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?

@abx 2019-12-14 18:31:09

$\mathbb{Z}/p$ 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 $\chi (\mathbb{S}^2)=2$.

@EBz 2019-12-14 18:34:59

@abx ofc u are correct! Sorry about that.

@Francesco Polizzi 2019-12-12 10:29:56

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 $2$.

@abx 2019-12-12 11:06:50

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.

@Francesco Polizzi 2019-12-12 12:44:07

Maybe, over $\mathbb{C}$, some simpler topological argument can work.

@abx 2019-12-12 15:13:48

That's what I was hoping for...

@Andrea Marino 2019-12-12 15:36:09

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 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 Grothendieck-Riemann-Roch to generalize to higher dims!!

@Francesco Polizzi 2019-12-12 15:38:32

@AndreaMarino: points on curves are divisors, but this is no longer true for higher dimensional varieties. In fact, the natural way to generalize the RH-formula for an involution with finite fixed locus is to blow-up the fixed points, and this leads to the original proof by the OP.

@Andrea Marino 2019-12-12 15:44:30

I see, you are right. I was thinking that the ramification divisor was the sum of branching points as for surfaces, but you are right: they are not even divisors! At this point I don't see why the OP thinks is unnatural! ^^

@Michael 2019-12-14 00:22:16

Perhaps it's possible to revive @AndreaMarino 's idea: put a smooth curve $C \subset X$ through the collection of fixed points of $\sigma$, and consider the curve $C \cup \sigma(C)$. Now $\sigma$ is an involution of the curve $C \cup \sigma(C)$ with the same fixed points as on $X$, and the fixed points are divisors on $C \cup \sigma(C)$. Then proceed with Riemann-Hurwitz as Andrea Marino described.

@Andrea Marino 2019-12-14 12:21:59

Well the problem is that now the curve is no longer irreducible, and you can't apply RH. Isn't it?

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