Odd degree Betti numbers are even

In characteristic 0, it follows from the Hodge decomposition and Hodge symmetry that the Betti numbers h^i(X) = \dim H^i(X^{\operatorname{an}},\mathbb C) of a smooth proper complex variety X/\mathbb C are even when i is odd. In characteristic p 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, K will be a p-adic field with ring of integers W = \O_K, residue field k of size q, and (normalised) valuation v such that v(q) = 1 (this is the q-valuation on K).

Throughout, X will be a smooth proper variety over k. We will write h^i(X) for the Betti numbers of X. It can be computed either as the dimension of H^i\et(\bar X, \Q_\ell), or that of H^i_{\operatorname{crys}}(X/W)[\frac{1}{p}].

Remark. Recall that if f is the characteristic polynomial of Frobenius acting on H^i\et(\bar X, \mathbb Q_\ell) for \ell \neq p, and \alpha \in \bar{\mathbb Q} is the reciprocal of a root of f, then for every complex embedding \sigma \colon \bar \Q \to \C we have

(1)   \begin{equation*} |\sigma(\alpha)| = q^{\frac{i}{2}}. \end{equation*}

The same holds for the eigenvalues of Frobenius on crystalline cohomology (in fact, the characteristic polynomials agree). All reciprocal roots are algebraic integers, and f \in \mathbb Z[t].

Defintion. An algebraic integer \alpha \in \bar \Q is a q^i-Weil integer if it satisfies (1) (for every embedding \sigma \colon \bar \Q \to \C).

Lemma. Let f \in \mathbb Q[t] be a polynomial, and let S be the multiset of reciprocal roots of f. Assume all \alpha \in S are q^i-Weil integers. Then v(S) = i - v(S) (counted with multiplicity).

Proof. If \alpha \in S, then \frac{q^i}{\alpha} is the complex conjugate with respect to every embedding \sigma \colon \bar \Q \to \C. Thus, it is conjugate to \alpha, hence a root of f as well (with the same multiplicity). Taking valuations gives the result. \qedsymbol

Theorem. Let X be smooth proper over k, and let i be odd. Then h^i(X) is even.

Proof. The Frobenius-eigenvalues whose valuation is not \frac{i}{2} come naturally in pairs (\alpha, \frac{q^i}{\alpha}). Now consider valuation \frac{i}{2}. Note that the p-valuation of the semilinear Frobenius F equals the q-valuation of the K-linear Frobenius F^r (which is the one used in computing the characteristic polynomial f). The sum of the p-valuations of the roots should be an integer, because f has rational coefficients. Thus, there needs to be an even number of valuation \frac{i}{2} eigenvalues, for otherwise their product would not be a rational number. \qedsymbol


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