The Weil Conjectures: The approach via the Standard Conjectures May 3, 2010Posted by David Speyer in Algebraic Geometry, characteristic p, Number theory.
The aim of this post is to outline a strategy for proving the Weil conjectures, proposed by Grothendieck and others. This strategy is incomplete; at various stages, we will need to assume conjectures which are still open today.
Our aim is to prove:
Let be a smooth projective variety, over a field of any characteristic. Let be a “reasonable” cohomology theory. Let be the hyperplane class for a projective embedding of . Let be an automorphism of , such that . Then the eigenvalues of are algebraic numbers and, when interpreted as elements of , have norm .
In previous posts, we gave for a proof in characteristic zero and a proof in the case that is a curve. I also explained why I need to say how I am embedding these eigenvalues into . Our proof requires all the ideas of these previous posts, plus some new ones.
Let’s be precise about what tricks we are going to use from previous posts.
- We will use the class , and the adjoint operator , to put a Hermitian bilinear form on . In characteristic zero, this form had the property that it was times a definite form on .
- We will build a ring , whose elements are linear combinations of subvarieties of , and for which is a module. There will be a trace map . It will thus make sense to say that a quantity defined in terms of the trace map is positive, leading us out of the difficulty of coefficients.
We also mention a change of perspective. In past posts, we fixed and studied . In this post, we will usually want to consider all of at once, and only focus on a particular degree towards the end of our arguments.
That said, let’s get started.
Write for the dimension of .
We write for the pairing on . Here we adopt the standard convention that for in any degree below the top degree.
It is an easy exercise that .
The ring of correspondences
Let be the sub- vector space of spanned by algebraic cycles of codimension . Note that ranges from to , and that is what was called in a previous post. In this post, we will want to work with a larger ring, so we change notation and set .
We put a ring structure on by , where are the three projections from to . Note that is a graded ring and is a graded module. (This explains our funny indexing.) The diagonal is the multiplicative identity.
We define the linear map by intersection with the diagonal . (So is zero on for .) From several uses of Kunneth’s theorem, we can prove the analogue of the Lefschetz trace formula:
Note that the left hand side is in , while the individual terms on the right are only in the coefficient ring of our cohomology theory.
Define an anti-involutation of by switching the components of . We have .
Write for the graph of ; an element of . We have . Multiplication by acts on by .
Let in be a hypersurface representing . Embed into along the diagonal and write for the resulting class in . Multiplication by acts on by cup product with . Here we see a preview of our strategy: Every important operation on will be shown (or conjectured) to be the action of some element of .
The condition that translates to the condition
It would be convenient to be able to isolate the action of on one particular cohomology group . In our previous post, we did this with the operators , and . The construction of the corresponding operators in the current setting is
There are idempotents , , …, in , so that acts by on and by on for .
There are precisely one element of which is a candidate for (exercise!), so this conjecture may be restated to say that this element in the image of the algebraic cycles of . Also, it is not hard to see that these classes in are mutual commuting and orthogonal, so the , should they exist, are likewise. We will later see that conjecture follows from other needed conjectures.
Assuming conjecture , we can put a double grading on : we have
We define . (Warning: I just made this notation up.) So .
Here is an interesting application of conjecture : Assuming conjecture , the following statement is well meaningful and true:
In particular, the right hand side is a rational number. We conclude, given Conjecture , the eigenvalues of on are algebraic numbers.
I think that conjecture C is not expected to be true if we define using integer combinations of algebraic cycles; that -combinations should be needed. If anyone can confirm this, that would be nice!
We want to define the bilinear form from our previous post. We would like to take the definition to be the following: The inner product on is given by , where is the noncommutative polynomial from the previous post.
But what is ?
Multiplication by is an isomorphism between and . There is a class in which acts by the operator introduced in the previous post.
The purpose of the first sentence is to make our previous discussion of meaningful.
In fact, the idempotents can be written as polynomials in and (exercise!). So, if we assume conjecture , we can deduce conjecture .
We prefer to work with all of at once. So we define
This is just the direct sum of our constructions on the individual ‘s. The element lives in .
From these identities, one can deduce
Proposition Assuming Conjecture , so that the statement is meaningful, we have , for and in .
Looking for positive definiteness
We have now (assuming conjecture ) been able to define a bilinear form on such that is self-adjoint, and which, in characteristic zero is the bilinear form which appeared in our earlier proof. We wish to, roughly speaking, show that this form is positive definite.
Taking a hint from our proof in the curve case, what we will actually do is create an operation on which corresponds to taking the adjoint with respect to this form. We will then show that , from to , is positive definite.
Here is the definition: It is not hard to show that is a unit in . For , define
We leave it as an exercise that
Using the relations , we have on .
We now want to establish:
Key Conjectural Lemma:
The -valued bilinear form on is times a positive definite-form on .
This sign seems to be missing from Kleiman’s article. Presumably, he has hidden it somewhere I have not noticed. Perhaps the reader should take this as a warning not to trust my signs!
Here is something that bothered me. In characteristic zero, is not positive definite on . Rather, it is times a positive definite form. So why should be positive definite on ? The point is that and will preserve each summand of , so we can compute adjoints on each summand separately. On each summand, the fact that is -definite will imply that is -definite. The key point here is that, if is the adjoint with respect to either a positive or a negative definite inner product, then is positive definite! In positive characteristic, we don’t have the Hodge decomposition, but its shadow is felt in this form.
If we have the Key Conjectural Lemma, the rest is easy. We write and . By Cauchy-Schwartz,
Using this inequality for all $k$, we see that the eigenvalues of on have norm .
Conjecturally proving the Key Conjectural Lemma
In the curve case, we deduced the conjectural lemma from the Hodge Index Theorem applied to . We wish to do something similar here.
Since we don’t have a Hodge decomposition anymore, we can’t state the Hodge index theorem as powerfully in characteristic . However, recall that algebraic cycles always live in degree . So, by stating a conjecture about algebraic cycles, we can hope to recapture just enough of the Hodge index theorem to do the job. This is:
Grothendieck’s conjecture of Hodge type:
Let be a projective algebraic variety of dimension , and the hyperplane class. Let be the -subspace of spanned by algebraic cycles. Then multiplication by is an isomorphism from , so we may define in a way analogous how we defined in the previous post. The inner product is -definite on .
The Conjecture of Hodge Type, for , implies the Key Conjectural Lemma. I don’t think there is any point in me writing up the details of this. Grothendieck attributes this to “certain arguments of Weil and Serre”, it is also explained in Kleiman’s article.