Our goal for today is to prove the following theorem:

**Theorem 1:** Let be a projective algebraic curve of genus and an endomorphism of degree . Let be a reasonable cohomology theory. Then the action of on has eigenvalues which are algebraic integers, with norm .

For those who know the term, “reasonable cohomology theory” means “Weil cohomology theory”.

The consequence of this theorem, which does not mention cohomology, is

**Theorem 2:** Let be a projective algebraic curve of genus and an endomorphism of degree . Then there are algebraic integers , , …, with norm such that

.

In a previous post, we established this in characteristic zero, by putting a positive definite hermitian structure on such that became unitary. But, as I discussed last time, we can’t define when has characteristic . Instead, will be defined over some other field of characteristic zero, like . We will therefore need to know that the eigenvalues of are algebraic integers before we can even make sense of the statement that they have norm .

It is possible to take the proof I present here and strip it down to its bare essentials, to give a proof of Theorem 2 which doesn’t even mention cohomology. See Hartshorne Exercise V.1.10. I am going to do the opposite; I will go slowly and focus on what each step is proving about . The essential argument here is Weil’s, although I have modernized the presentation.

## The ring

We define a ring as follows: The elements of the ring are formal sums of curves in , modulo the relation that if the curves and have the same class in . We multiply as follows:

where the are the three projections .

Fans of This Week’s Finds will recognize this as a slight variant of the category of spans.

Notice that the diagonal, , of , is the multiplicative identity of .

The cohomology groups are modules for by

. In particular, if is an endomorphism of , and is the graph of , then acts on by .

Let and be the classes in of and . Then and are orthogonal idempotents. Define , so is also idempotent. Then acts by on and by on every other .

Since the are orthogonal idempotents, we have . You can compute that and , so all the interest is in . We’ll write for ; so is a module.

## The involution

We define an involution of which sends a curve to the curve obtained by switching the two factors of . You can check that sends to itself.

In our previous proof, we put a positive definite Hermitian structure on .

**Exercise:** If is a curve over , and is in , then the actions of and on are adjoint.

There is a point that confused me when I first learned this stuff. We never mention complex conjugation in the definition of , so how can we get the Hermitian conjugate? The answer is simple: the maps coming from the action of are defined over (in fact, ). So is adjoint to with respect to both the natural skew-symmetric pairing on , and with respect to its Hermitian complexification. This is just like how, for a real matrix, the transpose and the conjugate transpose are the same thing. The question of whether to think in terms of Hermitian or ordinary bilinear forms is somewhat of an aesthetic choice; perhaps at some later point I will defend my decision to use the Hermitian language.

Let be the graph of , and let . Since is not a complex vector space, we cannot say that is unitary.

But there is a statement which still makes sense:

**Exercise:** We have .

## The linear form

Finally, define a linear form by , using the intersection product in . By the Lefschetz trace formula, we have

Note that the left hand side is an integer, while the right hand side, a priori, need only lie in the coefficient ring of . It is this ability to get an integer, rather than an element of , that will resolve our difficulties with coefficients.

In our previous proof, it was crucial that the Hermitian form on was positive definite. The corresponding result here is

**Key Lemma:** The -valued inner product on is negative definite.

If is defined over , this follows from the positive definiteness of the Hermitian form. (We pick up the minus sign because is for in .) But the Key Lemma makes sense over any field because, regardless of what ring our cohomology theory takes coefficients in, the ring is still a -module, and the pairing is still -valued.

## Proof of the main theorem from the key lemma

acts by zero on and . So , where are the eigenvalues of acting on . Since is always an integer, we deduce that is an integer for every . This immediately implies that the are algebraic numbers and, working a bit harder (exercise!) that they are algebraic integers.

The Key Lemma tells us that is negative definite. So, by Cauchy-Schwartz,

.

In other words,

and

.

Similar arguments show that for all . From this is follows that . Applying similar arguments to , we can also deduce that , so .

## Proof of the key lemma

We consider the form on all of , instead of just . Then has block form

where the middle block is the pairing on . So, to prove that is negative definite on , it is enough to prove that has signature of the form on . For a proof of this fact, see Hartshorne Theorem V.1.9 and Remark V.1.9.1.

## Connection to the Hodge Index Theorem

When our ground field is , there is a nice proof that has signature on , using the Hodge index theorem. We’ll be reusing notation from my previous post, so you might want to follow that link to refresh your memory.

By definition, and, in fact, lies in . Reusing our notation from an earlier post, . But , so is one dimensional. Then the Hodge index theorem says that is positive definite on and negative definite on the orthogonal space . We see that has signature on . Its signature on the subspace must thus be of the form either or ; it is easy to check that it is the former.

The fact that has signature on can thus be thought of as a shadow of the Hodge index theorem, which remains even when our cohomology theory takes coefficients in an exotic ring. We will return to this point next time, when we tackle Grothendieck’s Standard Conjectures.

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