Previous problem: Is $\operatorname{dim} H^1$ of an abelian variety the same for any Weil cohomology?

Let $X$ be an smooth projective variety over a field $k$. For any Weil cohomology theory for smooth projective varieties over $k$ with coefficient field $E$, we can define the $i$-th Betti number of $X$ to be $dim_E H^i(X)$.

The problem is whether Betti numbers are the same for any Weil cohomology theory $H$. Of course, this follows from some standard conjectures. But we don't assume standard conjectures, and it's true (see previous problem for a discussion) in the case:

  • $k$ is a finite field;
  • or $X$ is a curve or an abelian variety.

These are some old results, maybe it's also true for $K3$ surface via Kuga- Satake construction (I don't know any reference). Several years have passed, many new theories are developed, and new cohomology theory related to arithmetic geometry have been constructed, so how much do we know for independence at present (at least for comparison of two different Weil cohomology theories)?

Motivation: the Euler characteristic is independent of Weil cohomology, as it's the self intersection of the diagonal cycle inside $X \times X$.


For smooth projective varieties over $k$ separably closed field this should be known: in characteristic zero due to comparisons with Betti-de Rham cohomology, in characteristic $p>0$ due to the Weil Conjectures, esp. Deligne’s work on the Riemann Hypothesis.

For instance, over finite fields this is in Katz-Messing (see here) and for a smooth projective variety $X$ over a separably closed field $K$ of characteristic $p>0$, you can always find another smooth projective variety $X’$, this time defined over the separable closure of a finite field, such that $X’$ is the closed geometric fiber of a smooth relatively projective family $\mathcal{X}\to S$ whose geometric generic fiber is $X$.

If your cohomology is reasonable it should satisfy a version of proper base change, and so the Betti numbers of $X$ and $X’$ are the same and you apply the Katz-Messing result to $X’$.

Likewise if $X$ is defined over a separably closed field of characteristic zero, one can spread it out and specialize it to a smooth projective variety $X’$ defined over the separable closure of a number field, and if your cohomology satisfies proper base change, then the Betti numbers of $X$ are the same as the Betti numbers of (any such) $X’$.

As another instance, for $X’$ over the separable closure of a number field $K$, you can spread it out to some smooth projective family $\mathcal{X}\to V$ where $V$ is open in $\text{Spec}(R)$, $R$ a Dedekind ring, and the algebraic de Rham cohomology of $\mathcal{X}$ over $R$ interpolates both the algebraic de Rham cohomology of $X’/K$ and the crystalline cohomology of every closed fiber of $\mathcal{X}$, so all Betti numbers agree. The algebraic de Rham cohomology of $X’/K$ (base changed to the complex numbers along some complex embedding of $K$ into $\mathbf{C}$) further compares to the holomorphic de Rham cohomology of $X’(\mathbf{C})$ (a theorem of Grothendieck: see here) which further compares to Betti cohomology with $\mathbf{C}$ coefficients, which in turn compares with $\overline{\mathbf{Q}}_{\ell}$-étale cohomology (a consequence of a Theorem of Artin: see SGA 4 Exp. XI and bootstrap from the case of finite coefficients using the fact that $X’(\mathbf{C})$ has a basis of opens trivializing cohomology with constant coefficients, together with the fact that the $H^*(X’(\mathbf{C}),\mathbf{Z}/\ell^n)$ satisfy the Mittag-Leffler condition, so you can bring the inverse limit in $n$ inside cohomology).

All the respective Betti numbers agree, therefore.

If you refer to the compactly supported $\ell$-adic cohomology of arbitrary varieties (i.e. beyond the smooth case, but it seems you’re asking about smooth projective varieties), its dimensions are not known to be independent of $\ell$. This would follow from a combination of the standard conjectures (see here).

  • $\begingroup$ @zzy I mentioned most of the currently known Weil cohomologies, and argued that for smooth projective varieties over separably closed fields your question has a positive answer for all of them. If you ask about an arbitrary Weil cohomology, again for smooth projective varieties, then the answer is known (and positive) only over finite fields by Katz-Messing. If you’re asking if an analog of Katz-Messing is known over arbitrary separably closed fields (again, for smooth projective varieties), the answer is no: it should follow from a combination of the standard conjectures $\endgroup$ – John P. May 10 '19 at 16:20
  • $\begingroup$ @zzy I’m not sure why one would be interested in Weil cohomologies abstractly, unless one has a model to run calculations and arguments of geometric flavor (such as interaction with specialization maps). Also, you talk about the proper base change theorem by stressing the words “base change” as if I was talking about cohomology over non-separably closed (or non field) bases, but the proper (and smooth) base change theorem in $\ell$-adic cohomology is about geometric cohomology (i.e. over separably closed fields). $\endgroup$ – John P. May 10 '19 at 16:21
  • $\begingroup$ @zzy Obviously, étale cohomology over non-separably closed fields is not a Weil cohomology. Over arbitrary fields of characteristic zero, algebraic de Rham cohomology is. Over rings of integers, integral algebraic de Rham cohomology even puts an integral structure on it, that specializes to crystalline cohomology of every closed fiber. Algebraic de Rham cohomology and crystalline cohomology are non-geometric Weil cohomologies, in the sense that they are Weil cohomologies for (smooth proper) varieties defined over possibly non-separably closed fields. $\endgroup$ – John P. May 10 '19 at 16:25
  • $\begingroup$ Thank you! If you want to spread out, then smooth base change may also be necessay. I am interested in abstract Weil cohomology as I hope there are more of them. $\endgroup$ – sawdada May 10 '19 at 16:56

You only need Conjecture C to prove this. Let $k$ be an algebraically closed field, let $H$ be a Weil cohomology theory, and consider all smooth projective varieties over $k$ for which the Kunneth components $\pi_{i}$ of the diagonal are algebraic. Then $H$ defines a ``homological'' equivalence relation on the algebraic cycles, and hence a category of motives $Mot_{H}$, which is a rigid pseudo-abelian category with $End(1)=\mathbb{Q}$. An object $X$ of such a category has a well-defined rank (see Deligne and Milne Tannakian categories, 1.7.3) which is preserved by tensor functors. The rank of $X=(V,\pi_{i})$ is equal to the dimension of $H^{i}(V)$ (because $H$ defines a tensor functor from $Mot_{H}$ to a category of vector spaces). There is also a tensor functor $q$ from $Mot_{H}$ to the category of motives for numerical equivalence (which is even a tannakian category). Hence $rank(X)=rank(q(X))$ for $X$ in $Mot_{H}$. This equality with $X=(V,\pi_{i})$ shows that $dim(H^{i}(V))$ is independent of $H$.


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