## Characteristic zero analogues of the Weil conjectures: higher dimension April 5, 2010

Posted by David Speyer in Algebraic Geometry, characteristic p, complex analysis.

In our previous post, we proved

Theorem Let $X$ be a smooth projective curve over $\mathbb{C}$ and $F$ an endomorphism of degree $q > 0$. The eigenvalues of $F$ on $H^1(X)$ have norm $q^{1/2}$.

Today, we would like to generalize this to varieties of higher dimension. The obvious guess is

Nontheorem Let $X$ be a smooth projective variety, over $\mathbb{C}$, of dimension $d$. Let $F$ be an endomorphism of $X$ of degree $q^d$. The eigenvalues of $F$ on $H^r(X)$ have norm $q^{r/2}$.

This is not a theorem! I believe it is Serre who first figured out how to fix and prove this result. That is the topic of today’s post.

Let’s see some counter-examples to the nontheorem. Take $X = \mathbb{P}^1 \times \mathbb{P}^1$. Let $f$ and $g$ be endomorphisms of $\mathbb{P}^1$, of degrees $d$ and $e$. Then $h : (x,y) \mapsto (f(x), g(y))$ has degree $de$. But the action of $h$ on $H^2(X)$ is by the matrix $\left( \begin{smallmatrix} d & 0 \\ 0 & e \end{smallmatrix} \right)$; with eigenvalues $d$ and $e$, not the desired $\sqrt{de}$.

We can make a similar counter-example with the product of two elliptic curves. In a slight variation, we can work with an abelian surface that has complex multiplication by some rank $4$ number ring $R$ which has two nonconjugate embeddings into $\mathbb{C}$, and choose some $r \in R$ who has different norms under the two embeddings.

The fix here is to require $F$ to respect the embedding of $X$ into projective space. Specifically,
Theorem 1 Let $X$ be a smooth projective subvariety of $\mathbb{CP}^N$. Let $F$ be an endomorphism of $\mathbb{CP}^N$, of degree $q^N$, which restricts to an endomorphism of $X$. The eigenvalues of $F$ on $H^r(X)$ have norm $q^{r/2}$.

Maps to projective space are closely related to line bundles, and to Kahler classes. Here are the corresponding variants of Theorem 1.

Theorem 1′ Let $X$ be a smooth projective variety over $\mathbb{C}$; let $L$ be an ample line bundle on $X$; let $F$ be an endomorphism of $X$ such that $F^* L \cong L^{\otimes q}$. Then the eigenvalues of $F$ on $H^r(X)$ have norm $q^{r/2}$.

Theorem 1” Let $X$ be a compact Kahler manifold; let $\omega$ be the Kahler class; let $F$ be an endomorphism of $X$ such that $F^* \omega = q \omega$. Then the eigenvalues of $F$ on $H^r(X)$ have norm $q^{r/2}$.

In the previous post, the key trick was to build a positive definite Hermitian form on $H^r(X, \mathbb{C})$, for which the action of $q^{-r/2} F^*$ was unitary. We will want to do the same thing here. Guided by the theorems above, we see that the definition of our Hermitian form should involve the embedding $X \hookrightarrow \mathbb{P}^N$, or equivalently, the Kahler class $\omega$. For those who don’t know the Kahler story, you can think of $\omega$ as the image in $H^2(X)$ of the hyperplane class from $H^2(\mathbb{P}^N)$.

We will build our hermitian form in two stages. Fix $X$ smooth and projective (or Kahler) of dimension $d$, and $\omega \in H^2(X, \mathbb{C})$ a Kahler class with $F^* \omega = q \omega$.

## The naive bilinear form

For $\alpha$ and $\beta$ in $H^r(X, \mathbb{C})$, set

$\displaystyle{\langle \alpha, \beta \rangle_{\omega,\ \mathrm{naive}} = \int_X \overline{\alpha} \wedge \beta \wedge \omega^{d-r}}$.

This is either Hermitian or anti-Hermitian depending on the parity of $r$. And we have $\langle F^* \alpha, F^* \beta \rangle_{\omega, \mathrm{naive}} = q^r \langle \alpha, \beta \rangle_{\omega, \mathrm{naive}}$.

Unfortunately, this form is not positive definite. Consider $\mathbb{P}^1 \times \mathbb{P}^1$. Here $H^2$ is two dimensional. The $\omega$ form does not effect the pairing on middle cohomology (because $d-r=0$) so this is just the Hermitianization of the ordinary pairing on $H^{d/2}(X, \mathbb{R})$. In particular, it has matrix $\left( \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right)$, with signature $(1,1)$.

## The Hodge index theorem

Amazingly, there is a classical theorem which exactly addresses this issue. Consider multiplication by $\omega$ as a map from $H^{r}(X)$ to $H^{r+2}(X)$. We need:

The Hard Lefschetz Theorem: Multiplication by $\omega$ from $H^{r}(X)$ to $H^{r+2}(X)$ is an injection for $r < d$.

Let $k$ be between $|d-r|$ and $d$, and of the same parity as $|d-r|$. We define:

$\displaystyle{ H^{r}_k(X, \omega) :=}$

$\displaystyle{ \mathrm{Im}(H^{d-k}(X) \to H^{r}(X)) \ \cap \ \mathrm{Ker}( H^r(X) \to H^{d+k+2}(X)) }$

where both maps are multiplication by the appropriate power of $\omega$.
Set

$\displaystyle{ H^{p,q}_k(X, \omega) = H^{p+q}_k(X, \omega) \cap H^{p,q}(X)}.$

So

$\displaystyle{ H^{p,q}(X) = \bigoplus H^{p,q}_k(X, \omega)}.$

The action of $\omega$ is thus organized into strings:

$\displaystyle{ H^{d-k}_k(X, \omega) \to H^{d-k+2}_{k}(X, \omega) \to \cdots \to H^{d+k}_{k}(X, \omega) \to 0.} \quad (*)$,

where all the maps are isomorphisms. The result we need is:

Hodge Index Theorem (categorified) The form $(-1)^{k(k-1)/2} i^{p-q} \langle \ , \ \rangle_{\omega, \mathrm{naive}}$ is positive definite on $H^{p,q}_{k}(X, \omega)$.

Let’s see how this works for $\mathbb{P}^1 \times \mathbb{P}^1$. Let $H_1$ and $H_2$ be the obvious basis for $H^2(\mathbb{P}^1 \times \mathbb{P}^1)$. Let $\omega = a_1 H_1 + a_2 H_2$. The hypothesis that $\omega$ is Kahler means that $a_1$ and $a_2$ are positive. (You can think about ample line bundles instead of Kahler classes if you prefer.) So $H^{1,1}_{2}(\mathbb{P}^1 \times \mathbb{P}^1, \omega)$ is spanned by $\omega = (a_1, a_2)$. Similarly, $H^{1,1}_{0}(\mathbb{P}^1 \times \mathbb{P}^1, \omega)$ is spanned by the kernel of multiplication by $\omega$, namely $(a_1, - a_2)$. Recalling that $\langle \ , \rangle_{\omega, \mathrm{naive}}$ is given by $\left( \begin{smallmatrix} 0 & 1 \\ 1 & 0 \end{smallmatrix} \right)$, we see that this form is positive definite on $\mathrm{Span} (a_1, a_2)$ and negative definite on $\mathrm{Span} (a_1, -a_2)$ as required.

# Finishing the proof

Since $F^*$ rescales $\omega$, it preserves all of the $H^{p,q}_{k}(X, \mathbb{C})$. And they are all orthogonal for $\langle \ , \rangle_{\omega, \mathrm{naive}}$. We have already computed that $q^{-r/2} F^*$ preserves $\langle \ , \rangle_{\omega, \mathrm{naive}}$ on $H^{r}(X, \mathbb{C})$, and hence on each $H^{p,q}_{k}(X, \mathbb{C})$.

So, on each $H^{p,q}_{k}(X, \mathbb{C})$, the operator $q^{-r/2} F^*$ preserves a positive definite Hermitian form, and thus its eigenvalues have norm $1$. The eigenvalues of $F^*$ on $H^r$ are the union of the eigenvalues of $F^*$ on each individual $H^{p,r-p}_{k}(X, \omega)$.

QED

# The Lefschetz operator

We now give a reformulation of the above discussion, which will be very useful in our future presentation.

Let $\Lambda: H^r(X) \to H^{r-2}(X)$ be the operator adjoint to $\omega$. So, for $r-2 < d$, we have $\Lambda \circ \omega = \mathrm{Id}$ and, for $r > d$, we have $\omega \circ \Lambda = \mathrm{Id}$. In the string $(*)$, $\Lambda$ is the operator going the other way.

Note that we can write the projection operator $P^r_k$ from $H^r(X)$ to $H^r_k(X, \omega)$ in terms of $\omega$ and $\Lambda$. Namely,

$\displaystyle{P^r_k = \omega^{(k-d+r)/2} \Lambda^k \omega^{(k+d-r)/2} - \omega^{(k-d+r)/2+1} \Lambda^{k+2} \omega^{(k+d-r)/2+1}}$

Define an inner product on $H^r(X, \mathbb{C})$ by

$\displaystyle{ \langle \alpha, \beta \rangle_{\omega} = \sum_{k \equiv d-r \mod 2} (-1)^{k(k-1)/2} \int_X P^r_k \overline{\alpha} \wedge P^r_k \beta \wedge \omega^{d-r}}$

Hodge index theorem rephrased The restriction of $\langle \ , \ \rangle_{\omega}$ to $H^{p,q}(X, \mathbb{C})$ is $i^{q-p}$ times a positive definite Hermitian form.

The exact form of $\langle \alpha, \beta \rangle_{\omega}$ doesn’t matter. The key point is that we can write it as $\int_X \overline{\alpha} \wedge Q^r_d(\omega, \Lambda)(\beta)$, where $Q^r_d$ is a certain universal noncommutative polynomial in the operators $\omega$ and $\Lambda$. When we discuss the characteristic $p$ situation, we will take this formulation as our starting point.

Footnote: You might wonder what the standard Hodge index theorem says, and how it is related to its categorified version. We use lower case $h$‘s for the dimensions of cohomology groups. From $(*)$, we see that $h^{p, d-k-p}_k = h^{p+2,d-k+2}_k = \cdots = h^{p+k, d-p}_k$. Also, we have $h^{p,q} = \sum h^{p,q}_k$. From these linear equations, we can solve for $h^{p,q}_k$ in terms of $h^{p,q}$, and thus give a formula for the signature of $\langle \ , \rangle_{\omega, \mathrm{naive}}$ on $H^i(X)$ in terms of the $h^{p,q}(X)$. In particular, when $i=d$, this relates the signature of the Poincare pairing on $H^d(X)$ to the Hodge numbers $h^{p,q}$ of $X$. It is this last relationship which is ordinarily called the Hodge index theorem.

1. Joel Kamnitzer - April 5, 2010

Hi David,
Great post!
Some questions:
1. Is there a similar result for intersection homology of singular projective varieties?
2. Are there some hypotheses we can add, under which we can strengthen this theorem to say that F acts exactly by q^r/2 on H^r?

2. Ben Webster - April 5, 2010

Joel-

1. Almost surely. Probably exactly the same proof works (read the papers of de Cataldo and Migliorini for the relationship between Hodge theory and intersection cohomology).

2. Hard to say; a first guess would be cohomology generated by algebraic cycles, but that might not be enough in this case.

3. Sam - April 5, 2010

That’s great!

I was a bit annoyed reading through Kleiman’s treatment in “Dix Exposés…”; I’m rather unfamiliar with the material and he doesn’t give much motivation for the Standard Conjectures. It’s great to see those ideas coming out naturally (here in the characteristic 0 case).

May I ask what a recommended source for further reading might be?

Thanks again!

4. David Speyer - April 6, 2010

(EDIT: I wrote this answer assuming that F is the Frobenius. See discussion below for other automorphisms.)

Cohomology generated by algebraic cycles is certainly enough. If Y is projective and irreducible of dimension e than F always acts by q^e on H^{2e}(Y). So, if H^{2e}(X) is spanned by classes of algebraic cycles, then functoriality of cohomology for pushforward means that F acts by q^e on H^{2e}(X).

There is a conjectured converse to this. The Tate conjecture has a consequence that, if F acts identically by q^e on H^{2e}(X), then H^{2e}(X) is spanned by algebraic cycles. This is analogous to the characteristic zero situation where the Hodge conjecture implies that, if H^{2e}(X, C) = H^{e,e}(X), then H^{2e}(X,C) is spanned by algebraic cycles. I believe that even these special cases of Tate and Hodge are not known.

There is a detail here which came up in this MO question. I was speaking of the situation where H^{2e} is spanned by algebraic cycles defined over F_q. If the cycles are only defined over a finite extension of F_q, then the action will be by q^e*zeta, for zeta a root of unity.

5. David Speyer - April 6, 2010

i don’t have great suggestions for references. Milne’s expository article on Motives, about a year ago, was what inspired me to finally nail this stuff down. I also spent a lot of time reading Kleiman’s article, which you understandably found frustrating, and Grothendieck’s Standard Conjectures article.

I also remember finding a book in the Berkeley library which explained the Kahler analogue very well, back in 2005. Looking through the Berkeley library catalogue to refresh my memory, I think it might have been Zeta functions: An introduction to algebraic geometry, by Thomas? I see that MIT owns this book, so I’ll take a look later today.

6. Ben Webster - April 6, 2010

David- That argument works for the Frobenius, but F is just some endomorphism here. Why does it have to preserve the cycle?

7. David Speyer - April 6, 2010

Whoops! Good point, I lost track of that.

Right now, the best fix I can think of is H^{2e}(X) spanned by F-invariant cycles.

8. David Speyer - April 6, 2010

Thomas’s book is definitely the book I remembered: pre-LaTeX, has “zeta function” in the title, starts with number fields and moves gently through first function fields and then complex manifolds. But it only proves the case of curves, although it states the higher dimensional result. I thought the book I read had the proof which I wrote up here. I can only conclude that I was reading two books at the same time and they merged together in my memory.

I reshelved the book in the library, so it will still be there for any of our MIT readers who might be looking for it.

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