This question is somewhat inspired by Kevin Buzzard's answer to What is the interpretation of complex multiplication in terms of Langlands? and somewhat from my own curiosity about such topics.

Let $X$ be a variety over $\mathbb{Q}$. This variety induces a pure motive of weight $1$ (it induces pure motives of other weights, but I will focus on the one with weight $1$). I understand that weight $1$ motives come (conjecturally, of course) from weight $2$ newforms.

Okay, now let's trace it back. Let's start with a weight $2$ newform. Then, as James D. Taylor alluded to (and from what I know from http://staff.science.uva.nl/~bmoonen/MTGps.pdf), the corresponding newform must be the pure motive of weight 1 that is induced by an abelian variety (this is special to the weight $2$ newforms).

If so, then it seems that this proves that any pure motive of weight 1 is equal to the pure motive of weight 1 of a motive coming from some abelian variety.

Put back in words that are not conjectural: Is it true that for every variety $X$ over $\mathbb{Q}$ there is an abelian variety over $\mathbb{Q}$, $A$, such that $H^1_{et}(X,\mathbb{Q}_l) \cong H^1 _{et} (A,\mathbb{Q}_l)$ as $Gal(\mathbb{Q})$-representations?

Or perhaps is the following weaker statement true (if I somehow managed to get something wrong in the above): For every variety $X$ over $\mathbb{Q}$, $L(X,s)$ (coming from the action on the pure motive of weight 1 -- which is well defined even without motives, since one can create it using $l$-adic cohomology) = $\prod_i L(A_i,s)$ where the $A_i$'s are (finitely many) abelian varieties over $\mathbb{Q}$ and the $L$'s are coming from their pure motives of weight $1$.

I would very much like to know, if the above is wrong, where exactly the fallacy was. But if everything above is right, then is this known without assuming crazy conjectures like the standard conjectures or forms of Langlands?

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    $\begingroup$ Doesn't the Albanese work? $\endgroup$ Sep 8, 2011 at 1:08
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    $\begingroup$ If you allow me to work with complex smooth projective varieties, then here is the point: $$Alb(X)= H^0(X,\Omega_X)^*/H_1(X,\mathbb{Z})$$ up to torsion (which you'll throw away anyway), its first (co)homology is the same as for $X$. Now use Artin to get the $\ell$-adic statement. $\endgroup$ Sep 8, 2011 at 1:23
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    $\begingroup$ It seems to be implicit in the question and in the preceding comments that $X$ is smooth and projective (e.g. otherwise its $H^1$ need not be pure of weight $1$). If you consider non-smooth or non-projective varieties, then you will have to consider not just abelian varieties, but $1$-motives. $\endgroup$
    – Emerton
    Sep 8, 2011 at 2:40
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    $\begingroup$ By the way, pure weight $1$ motives over $\mathbb Q$ of dimension $2$, and with Hodge numbers $(1,0)$ and $(0,1)$ come from weight $2$ newforms; higher rank pure weight $1$ motives (even those with Hodge number just of the form $(1,0)$ and $(0,1)$, such as those coming about as $H^1$ of a smooth projective variety) will be attached to automorphic forms on other groups. $\endgroup$
    – Emerton
    Sep 8, 2011 at 2:43
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    $\begingroup$ In fact, you can find the autom representation on several groups at the same time. The group $GL_n(A_Q)$ always works, but I guess you can also find the autom representation on the symplectic group. $\endgroup$
    – mnr
    Sep 8, 2011 at 6:58

2 Answers 2


Let me slightly expand my comment from yesterday. Unfortunately, because of various time constraints, this will still be quite sketchy. Note that I'm only addressing the titular question. I have nothing to say about the automorphic aspects, since it is too far from what I know.

Let $X$ be a smooth projective variety over $\mathbb{Q}$. For simplicity assume that $X$ has rational point. I think there are various algebraic constructions of the Albanese in the 1950's literature, but I have to confess I've never gone through the details. So instead define $A=Alb(X) =Pic^0(Pic^0(X))$. After fiddling with the Poincare sheaf and the given rational point, you should get a Abel-Jacobi morphism $\alpha:X\to A$. I claim that this induces an isomorphism $$H^1(\overline{X}_{et},\mathbb{Q}_\ell)\cong H^1(\overline{A}_{et},\mathbb{Q}_\ell)$$ (where $\overline{X}= X\otimes \overline{\mathbb{Q}}$) necessarily compatible with the Galois action. To see this, base change up to $\mathbb{C}$, then by the comparison theorem, it suffices to get an isomorphism on the first singular (aka Betti) cohomology with $\mathbb{Q}$ coefficients. But the analytic construction of $Alb(X)$ gives this immediately because $$Alb(X)= H^0(X_{an},\Omega_X^1)^*/H_1(X_{an},\mathbb{Z})$$ So the rational first homologies coincide, now dualize.

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    $\begingroup$ For the direct construction of the Albanese variety see Serre: Groupes algébriques et corps de classes. If I remember correctly he reproduces a proof by Lang but as always it is an eminently readable exposition. $\endgroup$ Sep 8, 2011 at 14:31
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    $\begingroup$ By Barberi-Viale, Rosenschon and Saito, jstor.org/pss/3597296, a similar result holds for any complex algebraic variety. Of course, one needs to consider a 1-motive instead of an Abelian variety. $\endgroup$
    – ACL
    Sep 8, 2011 at 20:23
  • $\begingroup$ So without the assumption on the existence of rational points, we only know that these two Galois reps become isomorphic when restricted to some open subgroup, right? $\endgroup$
    – shenghao
    Sep 11, 2011 at 15:51
  • $\begingroup$ From "inskspot's" answer, it appears that this is true on the nose without restriction to an open subgroup. $\endgroup$ Sep 11, 2011 at 16:51

The answer to the OP's question in his para. 5 is ``yes'', I think. Even if $X/\mathbb Q$ has no $\mathbb Q$-point there is still an Albanese torsor $T$ (Lang, Abelian Varieties, p. 45, para. 3), universal with respect to morphisms from $X$ to abelian torsors, and then $H^1(X)$ is isomorphic to $H^1(T)$. Define $A=Aut_T^0$, so that $T$ is a torsor under $A$. The isomorphism $A\times T\to T\times T: (a,t)\mapsto (a(t),t)$ gives $H^1(A)\times H^1(T)\cong H^1(T)\times H^1(T)$. Then divide both sides by the copy of $H^1(T)$ coming from the second projection to get $H^1(A)\cong H^1(T)$, so $A$ is the abelian variety you want: $H^1(X)$ is isomorphic to $H^1(A)$.

  • $\begingroup$ Could you say precisely which reference "Lang" refers to? $\endgroup$ Sep 9, 2011 at 12:00
  • $\begingroup$ Sorry, Lang's Abelian Varieties. I've edited my answer accordingly. $\endgroup$
    – inkspot
    Sep 9, 2011 at 12:06

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