Weil's proof of the Riemann Hypothesis for projective curves relies upon the following positivity result: Let $\mathbb{F}q$ be the finite field with $q$ elements, $\overline{\mathbb{F}q}$ its closure, and $X$ a projective curve defined over $\overline{\mathbb{F}q}$. Let $D$ be a divisor class on $X \times X$ (where two divisors are related if they differ by a principal divisor), and let $\{p\} \times C$ and $C \times \{p\}$ denote the classes containing these elements. Define an integer valued mapping Tr on the divisor classes by

Tr$(D) = D \bullet (\{p\} \times C) + D \bullet (C \times \{p\}) - D \bullet \Delta$,

where $\bullet$ is the standard intersection product, and $\Delta$ is the diagonal divisor class. Now let $\circ$ be the multiplication induced on divisor classes by composition and let $D^t$ be the divisor given by the composition of $D$ and the permutation of the two factors of an element of $C \times C$. It can be shown that

Tr$(D \circ D^t) \geq 0$,

for all $D$. This result is usually called Castelnuovo Positivity, or Weil Positivity. My questions are as follows:

(1) Does anyone know of a good expository proof of this result available online?

(2) Does the result hold for any projective surfaces which are not curves? More specifically, does it hold for $\mathbb{C}P^2$ or for flag varieties?

(3) Is the failure of this Castelnuovo Positivity for projective varieties in general the sole reason that Weil proof does not generalise to all projective varieties?


I believe the question you meant to ask in (2) is: For $S$ a surface, is there some theorem like Castelnouvo positivity, regarding the $4$-fold $S \times S$? The answer to this question is "There is an analogous theorem, called the Hodge index theorem, but it is more complicated."

Let me explain what the Hodge index theorem says. Let $X$ be a smooth, algebraic variety over $\mathbb{C}$, with a specified projective embedding. For the purposes of the Riemann hypothesis, you would want to be working over a field of finite characteristic instead, but many of the things I want to say are much more subtle, and are conjectures rather than theorems, in finite characteristic. You should think of $X$ as $S \times S$, where $S$ is the variety for which you want to prove the Riemann hypothesis.

The cohomology $H^k(X, \mathbb{C})$ breaks up in the Hodge decomposition $H^k = \bigoplus\_{p+q=k} H^{p,q}$. For the purposes of the Riemann hypothesis, we only care about $H^{m,m}$, so I'll limit my discussion to that case. Now, our specified projective embedding $X \to \mathbb{P}^N$ gives us a map in cohomology in the other direction. $H^2(\mathbb{P}^N)$ is one dimensional and has a standard choice of generator called the hyperplane class; let $\omega$ be the image of this generator in $H^*(X)$. It turns out that $\omega$ lands in $H^{1,1}$.

Cupping with $\omega$ maps $H^{(m-1), (m-1)}$ to $H^{m,m}$. The hard Lefschetz theorem states that this map is injective when $m \leq \dim X$. Let's assume we're in this case, the other is related to this one by Poincare duality. So, $H^{m,m}$ has a filtration as $$H^{m,m} \supset \omega H^{(m-1), (m-1)} \supset \omega^2 H^{(m-2), (m-2)} \supset \cdots.$$ Abbreviate this as $$L^m \supset L^{m-1} \supset \cdots L^1 \supset L^0.$$

Define an inner product on $H^{m,m}$ by $$\langle f,g \rangle = \int \omega^{\dim X-2m} f g.$$ The Hodge index theorem says (in part) that this will be positive definite on $L^0$, negative definite on the orthogonal complement of $L^0$ within $L^1$, positive definite on the orthogonal complement of $L^1$ within $L^2$, and so forth. Let $M^i$ be the orthogonal complement of $L^{i-1}$ in $L^i$. (Not sure of the standard nomenclature here.) The case of $H^{1,1}$ of a surface is particularly easy, because $M^0$ is one-dimensional, spanned by $\omega$, and $M^1$ is the orthognonal complement of $M^0$.

It is relatively easy to prove Castelnuovo positivity from the Hodge index theorem for surfaces; see, for example Hartshorne exercise V.1.9. I have not seen anyone write out an analogue of Castelnuovo positivity for $S \times S$ when $S$ is higher dimensional. However, it is known how to adapt Weyl's proof of the Riemann hypothesis to higher dimensional $S$, if one had an analogue of the Hodge index theorem for $S \times S$ in characteristic $p$. I've been told that a good reference for this is Kleiman's Algebraic Cycles and the Weil Conjectures but I have not read this myself.

Let me explain why it is difficult to extend the Hodge index theorem to finite characteristic. If $X$ is defined in characteristic $p$, then $H^k(X)$ must be interpreted as cohomology with coefficients in $\mathbb{Q}\_{\ell}$ (or, nowadays, $\mathbb{Q}\_p$). Since these fields aren't ordered, we can't talk about positive definiteness.

Weil dodges this obstacle by talking about the vector space of algebraic cycles. This is a the $\mathbb{Q}$-vector space spanned by algebraic cycles, which I'll denote $A^{m}$. In characteristic $0$, it is a subspace of $H^{2m}(X, \mathbb{Q}) \cap H^{m,m}(X, \mathbb{C})$. The Hodge conjecture says that this it is precisely this subspace. In any characteristic, we have a map $$A^m \to H^{2m}.$$ This map is either known or conjectured to be an injection, depending on exactly how you define $A^m$. Let's assume that it is an injection. The inner product $\langle, \rangle$ is $\mathbb{Q}$-valued on $A^m$, so it makes sense to talk about its signature restricted to subspaces of $A^m$.

I'm going to make a secret switch of notation here, and use $L^i$ and $M^i$ to now refer to constructions in $H^{2m}$ rather than $H^{m,m}$. In the end, we'll be interested in things like $A^m \cap L^i$ which, in characteristic $0$, would live in $H^{m,m}$ anyway. By making this switch, I avoid having to explain how the Hodge decomposition works (and doesn't) in characteristic $p$.

In the case where $X$ is a surface, the generator $\omega$ of $M^0$ lies in $A^m$. One can use this to show that $$A^m = (A^m \cap M^0) \oplus (A^m \cap M^1).$$ The analogue of the Hodge index theorem then says that $\langle, \rangle$ is positive definite on $A^m \cap M^0$ and negative definite on $A^m \cap M^1$.

In all higher dimensional cases, this falls apart. It is (I believe) not known that $\langle, \rangle$ is nondegenerate on $L^i$, so it is not known that we can define the $M^i$. It is certainly not known that $$A^m = \bigoplus (A^m \cap M^i).$$ And it is not known that $(-1)^i \langle, \rangle$ restricted to $A^m \cap M^i$ is positive definite. Grothendieck's standard conjectures assert that all of this works. This is a major, and challenging, field of research.

I'll close by mentioning a challenge that is more suited to a combinatorial algebraic geometer like me. Harry Tamvakis told me that he tried, and failed, to prove the hard Lefschetz and Hodge index theorems for grassmannians by brute force. Here the cohomology ring is given by well known formulas, so the difficulties are all combinatorial. I can't say this is an important problem, but it sounds fun.

| cite | improve this answer | |

I don't know what Weil's orginal proof was, but proving the Weil conjectures for P^2 is extremely easy, since the cohomology of P^2 is generated by algebraic cycles. You just note that there are 1+q+q^2 points in P^2 over F_q (which corresponds to the cohomology being one dimensional in degrees 0,2,4).

| cite | improve this answer | |
  • $\begingroup$ Yeah, I recommend writing more carefully next time. Including background information is very helpful. Also, writing a title which actually asks a question is strongly encouraged (so people actually know what they're clicking on). $\endgroup$ – Ben Webster Oct 30 '09 at 12:25
  • $\begingroup$ Yes, I know it is extremely easy to establish the result directly, however (for various reasons) I want to know if one can also do so using Weil's method. $\endgroup$ – John McCarthy Oct 30 '09 at 15:51
  • $\begingroup$ You might get more helpful answers if you explained what proof you're talking about, linked to something about it, or provided a reference. Or gave some of the reasons you think this would be an interesting thing to do. $\endgroup$ – Ben Webster Oct 30 '09 at 17:06

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.