# Odd degree Betti numbers are even

In characteristic 0, it follows from the Hodge decomposition and Hodge symmetry that the Betti numbers of a smooth proper complex variety are even when is odd. In characteristic however, both Hodge-de Rham degeneration and Hodge symmetry fail (and de Rham cohomology is not a Weil cohomology theory), so we cannot use this method to obtain the result.

On the other hand, in the projective case, we can use hard Lefschetz plus the explicit description of the Poincaré pairing to conclude (we get a perfect alternating pairing, so the dimension has to be even). This leaves open the proper (non-projective) case in positive characteristic. This was settled by Junecue Suh [1]. I will explain the case for finite fields; one can easily reduce the general case to this case.

Notation. Throughout, will be a -adic field with ring of integers , residue field of size , and (normalised) valuation such that (this is the -valuation on ).

Throughout, will be a smooth proper variety over . We will write for the Betti numbers of . It can be computed either as the dimension of , or that of .

Remark. Recall that if is the characteristic polynomial of Frobenius acting on for , and is the reciprocal of a root of , then for every complex embedding we have

(1)

The same holds for the eigenvalues of Frobenius on crystalline cohomology (in fact, the characteristic polynomials agree). All reciprocal roots are algebraic integers, and .

Defintion. An algebraic integer is a -Weil integer if it satisfies (1) (for every embedding ).

Lemma. Let be a polynomial, and let be the multiset of reciprocal roots of . Assume all are -Weil integers. Then (counted with multiplicity).

Proof. If , then is the complex conjugate with respect to every embedding . Thus, it is conjugate to , hence a root of as well (with the same multiplicity). Taking valuations gives the result.

Theorem. Let be smooth proper over , and let be odd. Then is even.

Proof. The Frobenius-eigenvalues whose valuation is not come naturally in pairs . Now consider valuation . Note that the -valuation of the semilinear Frobenius equals the -valuation of the -linear Frobenius (which is the one used in computing the characteristic polynomial ). The sum of the -valuations of the roots should be an integer, because has rational coefficients. Thus, there needs to be an even number of valuation eigenvalues, for otherwise their product would not be a rational number.

References.

[1] Suh, Junecue, Symmetry and parity in Frobenius action on cohomology. Compos. Math. 148 (2012), no. 1, 295–303. MR2881317.