Sites without a terminal object

Let \mathcal C be a site with a terminal object X. Then the cohomology on the site is defined as the derived functors of the global sections functor \Gamma(X,-). But what do we do if the site does not have a terminal object?

The solution is to define H^i(\mathcal C,-) as \Ext{\mathcal O}{i}(\mathcal O,-), where \mathcal O denotes the structure sheaf if \mathcal C is a ringed site. If \mathcal C is not equipped with a ring structure, we take \mathcal O to be the constant sheaf \underline{\mathbb Z}; this makes \mathcal C into a ringed site.

Lemma. Let \mathcal C be a site with a terminal object X. Then the above definitions agree, i.e.

    \[H^i(X,-) = \Ext{\mathcal O}{i}(\mathcal O,-).\]

Proof. Note that \Hom_{\mathcal O}(\mathcal O, \mathscr F) = \Gamma(X, \mathscr F), since any map \mathcal O(X) \to \mathscr F(X) can be uniquely extended to a morphism of (pre)sheaves \mathcal O \to \mathscr F, and conversely every such morphism is determined by its map on global sections. The result now follows since \Ext{\mathcal O}{i}(\mathcal O, -) and H^i(X,-) are defined as the derived functors of \Hom_{\mathcal O}(\mathcal O,-) and \Gamma(X,-) respectively. \qedsymbol

Remark. From this perspective, it seems quite magical that for a sheaf \mathscr F of \mathcal O_X-modules on a ringed space (X,\mathcal O_X), the cohomology groups \Ext{\mathcal O_X}{i}(\mathcal O_X,\mathscr F) and \Ext{\underline{\Z}}{i}(\underline{\Z},\mathscr F) agree. It turns out that this is true in the setting of ringed sites as well; see Tag 03FD.

So why is this useful? Let’s give some examples of sites that do not have a terminal object.

Example. Let G be a group scheme over k. Then we have a stack BG of G-torsors. The objects of BG are pairs (U,P), where U is a k-scheme and P is a G-torsor over U. Morphisms (U,P) \to (U',P') are pairs (f,g) \colon (U,P) \to (U',P') making the diagram

Rendered by

commutative. This forces the diagram to be a pullback, since all maps between G-torsors are isomorphisms.

The (large) Zariski site on BG is defined by declaring coverings \{(U_i, P_i) \to (U,P)\} to be families such that \{U_i \to U\} is a Zariski covering (and similarly for the étale and fppf sites).

Now does the category BG have a terminal object? This would be a G-torsor P_0 \to U_0 such that every other G-torsor P \to U admits a unique map to it, realising P as the pullback of P_0 along U \to U_0. But this object would exactly be the classifying stack U_0 = BG, which does not exist as a scheme (or algebraic space). The fact that a terminal object does not exist is the whole reason we need to define it as a stack in the first place!

Example. Let X/k be a variety in characteristic p > 0; for simplicity, let’s say k = \mathbb F_p. Then consider the crystalline site of X/\Spec(\Z/p^n\Z). Roughly speaking, its objects are triples (U,T,\delta), where U \to X is an open immersion, U \to T is a thickening with a map to \Spec{\F_p} \to \Spec{\Z/p^n\Z}, and \delta is a divided power structure on the ideal sheaf \mathcal I_U \subseteq \mathcal O_T (with a compatibility condition w.r.t. \Spec{\F_p} \to \Spec{\Z/p^n\Z}). There is a suitable notion of morphisms.

This site does not have a terminal object, basically because there are many thickenings on U = X with the respective compatibilities. (I am admittedly no expert, and it could very well be true that this is not 100% correct. However, I am certain that the crystalline site in general does not have a terminal object.)

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.