It is known that Hodge standard conjecture is true for étale cohomology for a field $k$ of characteristic zero. It means that the following pairing $$ (x,y)\mapsto (-1)^{i}\langle L^{r-2i}(x),y\rangle $$ is positive definite, where $\langle -,-\rangle$ is the cup product, $L$ is the Lefschetz operator for this cohomology and $x,y\in C^i(X)\cap P^{2i}(X)$. In above formula $X$ is a smooth projective scheme over $k$, $C^i(X)$ is a free abelian group generated by closed irreducible subschemes of $X$ of codimension $i$ and $P^i(X)$ is a kernel of $L^{r-i-1}$.

My question is : what is the main problem with proving the similar theorem for fields of arbitrary characteristic?

  • 1
    $\begingroup$ I think the problem is that we don't have an algebraic proof: one proves it for C analytically, then applies the Lefschetz principle to extend it to other algebraically closed fields of char 0. $\endgroup$ – user00000 Jul 27 '16 at 14:42

I'm very interested myself on a better answer to this question, but let me point out the obvious: the main problem is that there is no Hodge theory on positive characteristic.

The proof in characteristic zero simply says that it is enough to consider $\mathbb{C}$ (via Lefschetz principle), and that there you can use Hodge theory, the Hodge index theorem in particular (via the comparison theorem).

Note that in positive characteristic the conjecture is known for surfaces, as it is a simple application of Riemann-Roch (see chapter V of Hartshorne for example). A proper answer should adress why no generalization of Riemann-Roch yields the result for other varieties.

| cite | improve this answer | |
  • $\begingroup$ Thanks for your answer! I'm trying find it in some articles related to problem of generalization of Riemann-Roch problem, but at the moment without success. I would write here in case I find something. $\endgroup$ – mikis Jul 28 '16 at 15:58

Another way to phrase the main obstacle in positive characteristic is the following:

Although we can formulate the Hodge standard conjecture purely cycle-theoretically, in practice the only way we get a grip on it is through a cohomological representation. For example, in characteristic $0$ we use Hodge cohomology to prove the positivity statement of the pairing on primitive cohomology.

However, in positive characteristic, there isn't even a cohomology theory taking values in a field that admits a notion of positivity! The only known cohomology theories are over fields like $\mathbb Q_\ell$ and $W(k)$, neither of which admits the structure of an ordered field (because they have too many roots of unity).

Thus, one either has to come up with a cycle-theoretic proof that doesn't make reference to any cohomology theory (as far as I can tell, this would be new even in characteristic $0$), or one has to come up with a Weil cohomology theory taking values in an ordered field (like $\mathbb Q$ or $\mathbb R$, although already over $\mathbb F_{p^2}$ cohomology theories with values in one of these two fields provably do not exist by a result of Serre).

Edit: As suggested by Will Sawin below (Serre's original argument is the case $K = \mathbb R$):

Claim. There is no Weil cohomology theory with values in any totally ordered field $K$.

Indeed, there exists a supersingular elliptic curve $E$ over $\mathbb F_{p^2}$ such that $D = \operatorname{End}(E) \otimes_\mathbb Z \mathbb Q$ is a quaternion division algebra. It is non-split at $\infty$ (e.g. because it is nontrivial but split at all $v \neq p, \infty$), so it is given by $D = (a,b)$ for $a, b \in \mathbb Q_{<0}$. Such a division algebra never splits over an ordered field (since $a$ and $b$ have to remain negative in $K$, so $b$ can never be a norm of $K(\sqrt{a})/K$). $\square$

(In fact, using the Brauer–Hasse–Noether sequence from class field theory and explicit computation of the local invariant maps, one sees that $D$ can be represented as $$D = \left\{\begin{array}{ll}(-1,-1) & p = 2,\\(-1,-p) & p \equiv 3 \pmod{4},\\(-q,-p) & p \equiv 1 \pmod{4},\end{array}\right.$$ where in the third case, $q \equiv 3 \pmod{4}$ is a prime whose residue class mod $p$ is a non-square. However, we only needed the negativity of $a$ and $b$.)

| cite | improve this answer | |
  • 5
    $\begingroup$ I believe Serre's supersingularity argument applies to an arbitrary ordered field. One just need to show that there is no nontrivial homomorphism from the quaternion algebra ramified at $p$ and $\infty$ to $M_2(k)$ for any totally ordered field $k$. Such a homomorphism would be an isomorphism upon tensoring with $\overline{k}$, hence an isomorphism upon tensoring with $k$. But all elements of the quaternion algebra tensored with $k$ have positive norm, unlike $M_2(k)$, showing they are not isomorphic. $\endgroup$ – Will Sawin Jan 7 '17 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.