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.