I am trying to understand the statement of the conjectures of Deligne on special values of certain $L$-functions, from his article titled, "Valuers de Fonctions L et periodes d'integrales" which appeared in Proc. of Symposia in Pure Math., 33, (1979), Part 2, 313-346.

In section 1 he makes the conjectures for a motive $M$ defined over $\mathbb{Q}$. He begins by assuming that the $\ell$-adic realizations of $M$, $H_l(M)$ form a strictly compatible system of $\ell$-adic representations. The fact that he assumes this is the reason I am asking this question, since I would have thought that this would always be the case.

If I understand correctly, the assumption that, the $\ell$-adic realizations of $M$, $H_l(M)$ form a strictly compatible system of $\ell$-adic representations, means the following: There is a finite set of primes in $\mathbb{Q}$, say $S$, such that

  1. for each $p\notin S$ and $\ell\neq p$ the representation $\rho_\ell:G_\mathbb{Q}\to GL(H_{\ell}(M))$ is unramified at $p$,
  2. Because of (1) we can consider the characteristic polynomial of the geometric Frobenius. This polynomial should have coefficients in $\mathbb{Q}$ (apriori, this is a polynomial with coefficients in $\mathbb{Q}_\ell$),
  3. For $p\notin S$, the polynomial we get in (2) should be independent of $\ell$.

Now suppose $M$ were the motive $h^i(A)=(A,\pi_i,0)$, where $A$ is an abelian variety defined over $\mathbb{Q}$, and the $\ell$-adic realization of $h^i(A)$ is $H^i_\ell(A;\mathbb{Q}_\ell)$. Let $S$ be the set of primes where $A$ does not have good reduction. Then the $\ell$-adic realizations of $M$ form a strictly compatible system of $\ell$-adic representations (condition (1) holds since good reduction implies unramified, condition (2) and (3) would hold because of the Weil conjectures, I hope I'm correct). I would have thought that for pure motives (as in Scholl's article titled, "Classical motives") the same conclusion would hold. Is this true?

Having written the above question I realize that Deligne uses the notion of absolute Hodge cycles to define motives in this article. Perhaps this is what is causing the problem. Can someone shed some light on this matter. Thanks in advance.

EDIT: From pondering over the comment by Eric and the answer by Will, I realize that it is not clear to me why the Frobenius should act on the cohomology $H_\ell(M)$. Let $M=(X,\pi,0)$ be a pure motive as in Scholl's article, and for simplicity let us also assume that it is homogeneous; I mean that the Hodge realization has only one weight. The situation I have in mind is one in which $M$ cuts out a piece from the cohomology $H^i_\ell(\bar{X},\mathbb{Q}_\ell)$. If $p$ is a prime at which $X$ has good reduction then the Galois representation $H^i_\ell(\bar{X},\mathbb{Q}_\ell)$ is unramified and so we can choose 'a' Frobenius endomorphism $$Frob_p:H^i_\ell(\bar{X},\mathbb{Q}_\ell)\to H^i_\ell(\bar{X},\mathbb{Q}_\ell)$$ If $H^i_\ell(M)$ denotes the image of the action of the projector $\pi$ on the cohomology, then I see no reason why our choice of $Frob_p$ should preserve this subspace. Is this clear? Yes it is, as pointed out by Will in the comments to his answer.

  • 2
    $\begingroup$ For the cohomology of an algebraic variety, the Galois representations will be strictly compatible. But for a general motive you're cutting out a piece of the cohomology and who knows what's going on now. How can we deduce compatibility at a finite place for a chunk of the cohomology, only knowing compatibility for all the cohomology? Letting it be 1979 only makes it harder! $\endgroup$
    – eric
    Jul 26, 2015 at 19:37
  • $\begingroup$ @eric-Thanks. In fact, I realize that it is not clear to me what the action of the Frobenius would be on the chunk cut out by the motive, there is no reason why the Frobenius should preserve that chunk, or is this something easy to see? $\endgroup$
    – Rex
    Jul 28, 2015 at 18:28

1 Answer 1


Yes, the absolute Hodge cycles cause the difficulty.

Let's consider what happens when we take a weak form of motive. Take a correspondence $C \subset X \times X$ that is an idempotent up to singular homological equivalence, and define the motive whose $\ell$-adic realization is the subspace of $H^i(X, \mathbb Q_\ell)$ on which $C$ acts with eigenvalue $1$.

Assuming enough big conjectures this will be equivalent to all the other definitions of motives, but it is a priori pretty weak.

Now let's prove independence of $\ell$ of the characteristic polynomial of $\operatorname{Frob}_p$. To compute the polynomial, it is sufficient to compute the traces of powers of $\operatorname{Frob}_p$, so we must show that those are independent of $\ell$. We can pass to characteristic $p$, where $\operatorname{Frob}_p$ becomes an endomorphism, and then the trace of $\operatorname{Frob}_p^n$ on the motive is the trace of $\operatorname{Frob}_p^n C$ on $H^i(X, \mathbb Q_\ell)$. I guess the Lefschetz operator that projects onto the $i$th cohomology exists in characteristic $p$ because of the Weil conjecture, so this is just the trace of $\operatorname{Frob}_p^n C P_i$ on $X$, which is equal to the intersection number of $\operatorname{Frob}_p^n C P_i$ with the diagonal and so is clearly independent of $\ell$.

But if $C$ is an absolute Hodge cycle, then it is not necessarily an algebraic cycle, nor does it necessarily become an algebraic cycle when we reduce mod $p$, and there is no reason to expect independence of $\ell$ for a trace of it times a power of Frobenius.

  • $\begingroup$ @Will-Thanks Will, I realized that my problem is much more basic and I've edited the question accordingly. $\endgroup$
    – Rex
    Jul 28, 2015 at 18:29
  • 1
    $\begingroup$ @Rex. $\pi$ is a Galois-invariant class on $X\times X$. Viewing cohomology classes as endomorphisms, the Galois action is by conjugation. So the conjugation action is trivial and so $\pi$ commuters with all elements of the Galois group, including Frobenius. $\endgroup$
    – Will Sawin
    Jul 28, 2015 at 19:08
  • $\begingroup$ @Will-That is a really clean way to see this. Thanks a lot. I was trying to see it directly at the level of correspondences, and it was not clear to me what was going on. $\endgroup$
    – Rex
    Jul 28, 2015 at 19:19

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.