Timeline for Is there for every variety X an abelian variety A such that their 1st l-adic cohomologies are isomorphic?
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19 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Sep 22, 2011 at 18:36 | comment | added | mnr | @ Kevin Buzzard, yes you are right. Thank you ! | |
| Sep 9, 2011 at 10:00 | answer | added | inkspot | timeline score: 6 | |
| Sep 8, 2011 at 22:31 | comment | added | Kevin Buzzard | @Arno Kret: the Galois representation is taking values in a symplectic group, so actually the automorphic representations would be for the dual group (in the sense of reductive groups) of a symplectic group, rather than the symplectic group itself (and actually one might even need an extra $GL(1)$ in order to have enough space to twist everything so things come out nicely). | |
| Sep 8, 2011 at 14:11 | answer | added | Donu Arapura | timeline score: 8 | |
| Sep 8, 2011 at 14:04 | comment | added | Emerton | ... worlds at all. Regards, Matthew | |
| Sep 8, 2011 at 14:04 | comment | added | Emerton | The existence of these automorphic forms is not proven yet (they fall outside the scope of existing modularity, or even potential modularity, theorems because $h^{1,0} = 2 > 1$, and present technology for $n > 2$ is restricted to (at best) the cases for which the variouss $h^{p,q}$s equal $0$ or $1$). But there are people working on their existence currently, and they take $GSp_4$ as a starting point, not $GL_4$, because this group gives a Shimura variety, whereas $GL_4$ doesn't, and without a Shimura variety at hand it's hard to make any contact between the automorphic and motiviv/Galois .. | |
| Sep 8, 2011 at 14:01 | comment | added | Emerton | Dear Makhalan, For concreteness, let's consider an abelian surface, whose $H^1$ is $4$-dim'l, and for which $h^{1,0} = h^{0,1} = 2$. Then there should be an automorphic form for $GL_4$ giving rise to the $\ell$-adic reps. on its $H^1$. But if we polarize the abelian surface than $H^1$ will be equipped with a symplectic pairing, and so (since the root system $C_2$ is self-dual) one would also expect an automorphic form on $GSp_4$ giving rise to the $H^1$. (The form on $GL_4$ will be related to the form on $GSp_4$ via functoriality, applied to the standard embedding $GSp_4 \subset GL_4$.) | |
| Sep 8, 2011 at 13:40 | comment | added | Donu Arapura | Shenghao: you're right that my comment was cryptic and incomplete (and yes, by Artin, I mean the comparison theorem). Let me flesh it out a bit below. | |
| Sep 8, 2011 at 9:12 | comment | added | shenghao | @ Donu: I don't quite follow your argument going from Betti cohom to l-adic cohom, as there's no Galois action on Betti. I suppose by "Artin" you mean Artin's comparison theorem in etale cohomology. | |
| Sep 8, 2011 at 6:58 | comment | added | mnr | 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. | |
| Sep 8, 2011 at 3:43 | comment | added | Makhalan Duff | Or maybe what you meant by "other groups" is $GL_n(\mathbb{A}_{\mathbb{Q}})$ as opposed to $GL_2(\mathbb{A}_{\mathbb{Q}})$. (at first I thought you meant other reductive group schemes applied to the adeles) | |
| Sep 8, 2011 at 3:39 | comment | added | Makhalan Duff | Emerton, that's a piece of the puzzle I haven't figured out yet. You're saying that pure weight 1 motives of dimension $>2$ don't come from an algebraic cuspidal automorphic representations of $GL_n(\mathbb{A}_{\mathbb{Q}})$? Isn't the whole point that these representations correspond exactly to motives? What is the context for looking at automorphic representations of other groups? | |
| Sep 8, 2011 at 2:43 | comment | added | Emerton | 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. | |
| Sep 8, 2011 at 2:40 | comment | added | Emerton | 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. | |
| Sep 8, 2011 at 1:23 | comment | added | Donu Arapura | 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. | |
| Sep 8, 2011 at 1:10 | comment | added | Makhalan Duff | Is that really all that is? I'm puzzled now... This is extremely basic compared with the fancy words I used, but why are their H^1's the same? | |
| Sep 8, 2011 at 1:08 | comment | added | Felipe Voloch | Doesn't the Albanese work? | |
| Sep 8, 2011 at 1:03 | history | asked | Makhalan Duff | CC BY-SA 3.0 |