## Effective Motives

This is a brief follow up concerning a question asked by Felipe. Suppose we assume the standard conjectures. Let $M$ be a pure motive, and consider the following problems:

1. Problem A: (“effectivity”) Suppose that $M$ has non-negative Hodge-Tate weights. Then is $M$ effective?
2. Problem B: (“ordinary primes”) Does the Hodge polygon = Newton polygon for infinitely primes $p$?
3. Problem C: (“Katz”) Suppose the characteristic polynomials of Frobenius have coefficients in $\mathbf{Z}$. Then is $M$ effective?

An affirmative answer to Problem C implies an affirmative answer to Problem A. Conversely, a positive answer to Problems A & B implies a positive one for Problem C.

The relevance of Problem A was for deducing that a weight zero regular algebraic cuspidal automorphic form for $\mathrm{GL}(2)/F$ could be associated to an abelian variety of $\mathrm{GL}_2$-type over $F$. I claimed that this was probably “Standard Conjectures hard.” It seems that this is partly right and partly wrong.

(Completely unrelated remark: wordpress seems to be a vastly inferior typesetter than LaTeX, since it happily takes LaTeX expressions followed by full stops without a space and separates them by line breaks, and doesn’t even seem to align math formulas within a sentence correctly. Is there a way to integrate the LaTeX more seamlessly into wordpress?)

As mentioned previously, if $M$ has weight zero, then Problem A already follows from Kisin-Wortmann (always assuming the standard conjectures), because then $M$ will be an Artin motive.

As was pointed out to me, the case of weight one follows from the Hodge conjecture. Namely, the Hodge realization gives a polarized Hodge structure of weight one which gives a polarized complex torus. By Riemann, such a torus is actually an an abelian variety $A$, which (using the standard conjectures) one can descend to $F$. This argument doesn’t obviously extend to the general case, because the image of the period map from (say) pure Motives with Hodge-Tate weights $[0,k]$ to polarized Hodge structures will not be surjective for Griffiths transversality reasons. As an aside, it was also pointed out that the Hodge conjecture is not one of the standard conjectures.

1. There’s no evidence for Problem A beyond the fact that it would be nice,
2. The Hodge conjecture is false, and
3. Grothendieck already mentioned that his (Grothendieck’s) modification of the generalized Hodge conjecture implies that the answer the Problem A is positive.

Here the generalized Hodge conjecture says (roughly) that a sub-Hodge structure of $H^k$ with weights in the range $[k-q,q]$ to $[q,k-q]$ arises via the Gysin map from an algebraic cohomology class on an $\ge q$-codimensional subvariety. In particular, if $M$ has non-negative Hodge-Tate weights and is of weight $w$, and $M(n)$ is effective inside some smooth proper variety $X$, then $M$ gives rise to a sub-Hodge structure of $H^{w+2n}(X)$ with weights from $[n,n+w]$ to $[n+w,n]$, and hence come from some algebraic subvariety $Y$ of codimension at least $n$. However, the Gysin map on etale cohomology involves a Tate twist by $\mathbf{Q}_p(n)$, and so (using the standard conjectures) one recovers $M$ effectively in $Y$. Grothendieck also points out that, in the case when $M$ has weight one, the generalized Hodge conjecture follows from the usual Hodge conjecture after replacing $X$ by $X \times C$ for proper smooth curves $C$, essentially by the same argument of the previous paragraph. (I guess one also has to use the easy fact that any abelian variety is a quotient of a Jacobian.)

Talking of Deligne and Grothendieck, the Farbster sent me the following link to an interview of Deligne by MacPherson:

https://www.simonsfoundation.org/science_lives_video/pierre-deligne/

which contains the following slightly terrifying exchange about Grothendieck:

MacPherson: I’ve heard people say that he [Grothendieck] was always very kind to students when they didn’t understand, but if someone was older and had pretentions he could be less …

Deligne: That’s quite possible, and I think he was completely willing to explain something once, I don’t think he would have be willing to explain it three times, even to students.

(In my original memory of this passage, “three times” was replaced by “twice.”)

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### 5 Responses to Effective Motives

1. I’ve looked at Grothendieck’s paper and I cannot find where he shows that his version of Hodge implies A.

Also, Mazur has some notes “Descending cohomology, geometrically” of a talk he gave at a conference in honor of Joe Harris, which discusses this problem. They don’t seem to available on his webpage.

2. student says:

Did Deligne mean that 2 is false rationally as well as integrally?

• I think the implication was integrally, but then who knows?

3. Thomas says:

Barry Mazur just allowed to share his preliminary notes, a revised version will appear later on his website: http://pdfcast.org/pdf/descending-cohomology-geometrically-barry-mazur