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Let $F$ be a real quadratic field and let $E/F$ be an elliptic curve with conductor 1 (i.e. with good reduction everywhere; these things can and do exist) (perhaps also I should assume E has no CM, even over F-bar, just to avoid some counterexamples to things I'll say later on). Let me assume that $E$ is modular. Then there will be some level 1 Hilbert modular form over $F$ corresponding to $E$. But my understanding is that the cohomology of $E$ will not show up in any of the "usual suspect" Shimura varieties associated to this situation (the level 1 Hilbert modular surface, or any Shimura curve [the reason it can't show up here is that a quaternion algebra ramified at precisely one infinite place must also ramify at one finite place]).

If you want a more concrete assertion, I am saying that the Tate module of $E$, or any twist of this, shouldn't show up as a subquotient of the etale cohomology of the Shimura varieties attached to $GL(2)$ or any of its inner forms over $F$ (my knowledge of the cohomology of Hilbert modular surfaces is poor though; I hope I have this right).

But here's the question. I have it in my head that someone once told me that $E$ (or perhaps more precisely the motive attached to $E$) should not show up in the cohomology of any Shimura variety. This is kind of interesting, because here is a programme for meromorphically continuing the L-function of an arbitrary smooth projective variety over a number field to the complex plane:

1) Observe that automorphic forms for GL_n have very well-behaved L-functions; prove that they extend to the whole complex plane. [standard stuff].

2) Prove the same for automorphic forms on any connected reductive algebraic group over a number field [i.e. prove Langlands functoriality]

3) Prove that the L-functions attached to the cohomology of Shimura varieties can be interpreted in terms of automorphic forms [i.e. prove conjectures of Langlands, known in many cases]

4) Prove that the cohomology of any algebraic variety at all (over a number field) shows up in the cohomology of a Shimura variety. [huge generalisation of Taniyama-Shimura-Weil modularity conjecture]

My understanding is that this programme, nice though it looks, is expected to fail because (4) is expected not to be true. And I believe I was once assured by an expert that the kind of variety for which problems might occur is the elliptic curve over $F$ mentioned above. At the time I did not understand the reasons given to me for why this should be the case, so of course now I can't reproduce them.

Have I got this right or have I got my wires crossed?

EDIT (more precisely, "addition"): Milne's comment below seems to indicate that I did misremember, and that in fact I was probably only told what Milne mentions below. So in fact I probably need to modify the question: the question I'd like to ask now is "is (4) a reasonable statement?".

Kevin Buzzard
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  • The strategy of Blasius-Rogawski to construct a motive for Hilbert modular forms (base change to unitary then transfer to U(3) then to an inner form) is indeed not known to succeed in this case (because of the shape of the L-packets). I don't know if this strategy and its close cousin are known to fail (way back, Blasius and Rogawski were actually "cautiously optimistic" that it should succeed by transfer to U(4)). But you surely knew this, as you also surely know that the meromorphic continuation of the L-function of E as in your introduction is known anyway. – Olivier Sep 21 '10 at 12:31
  • I was pretty sure that one couldn't attach a motive to the level 1 eigenform. I know nothing about L-packets or U(4). I know the meromorphic continuation is known for general E/F---this is because E is potentially modular. But even proving E is potentially modular doesn't realise it in the cohomology of a Shimura variety. In fact in the situation above I assumed E was modular, so analytic continuation will be known in this setting. I wanted to emphasize that it wasn't the modularity that was the problem I was interested in, it was the Shimura variety issues. – Kevin Buzzard Sep 21 '10 at 12:46
  • [clarification: of course in the setting above one can attach a motive because I started with $E$; I mean that in general given a level 1 eigenform I agree that it might be hard to attach a geometric object, when $F$ has even degree] – Kevin Buzzard Sep 21 '10 at 12:56
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    @Kevin Buzzard: Thanks for this nice summary 1–4 of the Langlands programme! –  Sep 21 '10 at 12:59
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    Blasius has pointed out that the naive generalization of the modularity conjecture fails --- there exist elliptic curves over number fields that are not quotients of the albanese of any Shimura variety --- but I don't know of any reason why the more general version (4) can't be true. (Blasius 2004 MR2058605). – JS Milne Sep 21 '10 at 13:09
  • @JS Milne: your comment is evidence that my memory is at fault and that what I was in fact told was this Albanese result, because (and I didn't let on about this in the original post) it was Blasius who was the expert mentioned in the question. I will edit the post accordingly. – Kevin Buzzard Sep 21 '10 at 14:27
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    Kevin: I'm reasonably certain that Clozel told me (about twenty years ago) that every motive should show up in a Shimura variety, and maybe even had an argument for deducing this from some fairly 'standard' conjectures. You could ask him and then correct me if I'm remembering incorrectly. – Minhyong Kim Sep 21 '10 at 14:36
  • @Minhyong---I wonder if the standard conjectures included the one saying that the L-function of a motive should have the expected meromorphic continuation and functional equation :-) This is a pretty standard conjecture and of course it renders the plan of attack above circular. – Kevin Buzzard Sep 21 '10 at 14:50
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    That could well have been the case. As a mode of proof that would render your strategy circular. But meanwhile, knowledge of such an implication would render your strategy sound. Perhaps the idea was something like this: When asserting the automorphicity of the Galois representation associated to a motive $M$, I think the Tate conjecture, Hodge conjecture, and semisimplicity conjecture should imply that the reductive group G is the dual of the Mumford-Tate group $MT$ of $M$. Perhaps there is a local system on a Shimura variety associated to $G$ or $MT$ out of which the rep. can be recovered. – Minhyong Kim Sep 21 '10 at 15:18
  • Is it possible to make a more definite statement, perhaps conjecture, about the programme outline here, not for all motives, but in the restricted case of motives that arise from Calabi-Yau varieties? – Laie Sep 21 '10 at 16:04
  • Given $E$ as above with associated Hilbert modular form $f$, you can find $\mathrm{Asai}(f)$ in $H^2$ of the associated Hilbert modular surface. Not quite $f$ itself, but still... – David Hansen Sep 21 '10 at 19:26
  • @Minhyong: I agree with your "sound" comment! I've not heard back from Clozel yet. Regarding Mumford-Tate: consider a curve with CM over $\mathbf{Q}$. I forget conventions. If Mumford-Tate groups are defined to be connected then you're probably looking in the wrong place for your automorphic form (unless you're only "potentially" looking for it) and if they're not then there will be issues with taking "duals" because non-connectivity on the $L$ side corresponds to some sort of descent data on the other side. Maybe everything works out though! What's worrying me is that... – Kevin Buzzard Sep 22 '10 at 09:19
  • ..for a CM elliptic curve I think I'd rather be looking for it on $GL_2$ rather than on a torus, unless, as I say, I am happy to make finite base extensions. – Kevin Buzzard Sep 22 '10 at 09:19
  • @Laie: I don't even know a sensible "geometric" conjecture for elliptic curves over number fields. I want them all to be associated to automorphic forms, but I don't know an analogue of the statement "if I'm an ell curve over Q then I'm covered by a modular curve" that would be expected to hold for a general number field. @David Hansen---yes! And you can find det(f) in the $H^0$ of the associated 0-dimensional Shimura variety (modulo the fact that the 0-dimensional guy doesn't quite satisfy Deligne's axioms...). But they're not enough to give you $f$, as you well know. – Kevin Buzzard Sep 22 '10 at 09:22
  • Kevin, you are absolutely right that in my rough remarks, I ignored the 'potentiality' issue everywhere. But maybe it's sensible not to worry initially about getting the representation on the nose...

    I await the result of the Clozel query. I hope he's not mad at me for some sort of misrepresentation :=)

     – Minhyong Kim Sep 22 '10 at 11:42
  • Dear Kevin and Minhyong, did you ask Laurent Clozel in the end? What is the status of conjecture CKB as of 2020? Has the coronavirus corrupted Shimura varieties enough to make it fail? – plm Apr 06 '20 at 10:01

4 Answers4

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The question may be precised depending on what you call "show up". More precisely:

  1. Concerning the cohomology of Shimura varieties there are two points of view: intersection cohomology or ordinary cohomology (this is of course the same for compact Shimura varieties).

  2. Concerning the fact that the cohomology shows up in something we can add an option: it may show up potentially.

  3. Then we can add another option: to proove it shows up in the Tannakian sub category of motives generated by the motives of Shimura varieties (even weaker (?): the class in the K_0 of motives of your variety is a virtual combination of the classes of motives showing up in Shimura varieties).

Let's first say we look at intersection cohomology. As stated your hope 4) is false for trivial reasons: the only Artin motives showing up in the intersection cohomology of Shimura varieties are the abelian one...In fact by purity they show up only in the H^0 that has been computed by Deligne and is abelian.

Of course if you put option (2) in my list this counterexample disappears.

Now you may say: yes but we can twist an Artin motive by a CM character and ask the same question. This is where I come to the following point: you're saying that because the twisting operation that is a particular case of Langlands functoriality is a known Langlands functoriality. Where I want to come is that in fact if you suppose Langlands functoriality known then the fact that your variety shows up in the Tannakian category generated by motives of Shimura varieties implies its L function is automorphic (tensor product functoriality).

If you suppose Langlands functoriality and your variety shows up potentially in the motive of a Shimura variety then its L-function is automorphic (existence of automorphic induction which implies for example Artin conjecture).

About the intersection cohomology of Shimura varieties: it is now pretty well understood and I think there is no reason why any variety would show up potentially in it. More precisely the Langlands parameters of automorphic representations showing up in the intersection cohomology of Shimura varieties factor through some representation $r_\mu:\;^L G_E\rightarrow GL_n$ where $G$ is the group attached to the Shimura variety (well to be more serious I woud have to invoke cohomological Arthur's parameter but it would take 5 hours to write this in details). Thus I clearly think the class of varieties that show up potentially in the cohomology of Shimura varieties has some serious restrictions...

Now there is another thing I did not speak about: the cohomology of non-compact Shimura varieties that may not be pure. For this little is known and it may be possible some interesting Galois representations that do not show up in the intersection cohomology of Shimura varieties show up in the cohomology...I know some people are looking at this (I won't give any name, even if I'm tortured) but as I said up to now little is known.

Well, I will stop here since this is an endless story and you can speak about this during hours...

Laurent F.
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    I agree with Laurent that the question should be precised before attempting any answer. One precision is, as he said, are we talking of the intersection cohomology (which is the same as the L^2 cohomology, and which will see the motives attached to the discrete automorphic representations of the group G defining the Shimura varieties), or of the ordinary cohomology (which sees all the cuspidal automorphic representations, some of the discrete non-cuspidal, and some others, no one knows exactly which)? – Joël Sep 22 '10 at 04:04
  • That said, I'm pretty confused. What motives are supposed to appear (directly or potentially) in the cohomology of Shimura varieties? If we assume the motive regular, that is with distinct Hodge numbers, shouldn't this has a simple answer? Take F=Q, for example. Shouln't any regular motive that it is a twist of its dual appear in a Shimura variety (orthogonal, or symlectic, or unitary after restriction to quadratic imaginary field)? What about the converse?

    Now, another queston (that might be stupid): isn't any motive over Q (and with coefficients in Q) dual to a Tate twist of itself?

     – Joël Sep 22 '10 at 04:22
  • As I mentioned to Kevin, you two should also ask Clozel about my comment, and let us know here if my memory (or my understanding at the time) is (or was) completely stupid. I'd quite like to know myself. – Minhyong Kim Sep 22 '10 at 05:32
  • @Laurent: I agree I should make it precise. The reason I didn't make it precise initially was that I was "fishing" for a precise statement that I couldn't quite remember, so it was to my advantage to be as vague as possible! I already found the answer to that in Milne's comment, so then I had to change the question a bit and I just figured I would leave it to see if anyone could formulate a precise negative result (e.g. "the etale cohomology of a Shimura variety always has this property, hence this Galois representation can never be a subquotient"). – Kevin Buzzard Sep 22 '10 at 09:24
  • The reason I didn't mention the "potential" issue was because of the following construction: if $L/K$ is a finite Galois extension then I can consider something like $G=Res_{L/K}(GL(1))$ (or even $GL(0)$ if Deligne's axioms allow it; I forget) and get Shimura varieties whose $H^0$ is abelian over $L$ but which still give the Artin Galois representation over $K$ that I want as a subquotient. In general you're abelian over the reflex field but you can control the reflex field! – Kevin Buzzard Sep 22 '10 at 09:27
  • @Joel: I agree that for regular motives there is a chance that there is a simple conjecture! But I don't know it. As for motives being dual to Tate twists of themselves---just take a motive with Hodge-Tate weights 0,1,3 and I think this gives a counterexample. I can build reducible examples of this form easily; maybe building an irreducible one would be harder? About making the question precise: I do agree with you, but see my comments for Laurent for why I decided not to do this initially. I wonder if it really makes too much difference though? (at least conjecturally...) – Kevin Buzzard Sep 22 '10 at 09:33
  • @Kevin: You were right to ask the question (which is great) in a vague way. Making the question precise is part of answering it. About my question "is a motive with coeff in Q self-dual-up-to-a-Tate twist?", I meant "a simple motive". But I will try to clarify my mind and post again about your question. Meanwhile, sorry to pollute this discussion by a question to be erased when answered but : where on mathoevrflow can I ask the question "I have created a openId, how can I get back on this account the reputation points I earned as a non-registred user?" ? – Joël Sep 22 '10 at 13:28
  • @Joel: see the link at the top to "meta"? That's the place to ask questions about mathoverflow itself, such as yours. Or just go directly to tea.mathoverflow.net . – Kevin Buzzard Sep 22 '10 at 13:48
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    @Kevin (comment of Sep 22, 9:33): A motive with Hodge-Tate weights 0,1,3 should be "degenerate" - if the associated Galois representation were irreducible, and you believe Fontaine-Mazur, then you'd get a cuspidal GL3 automorphic form whose infinity type is not fixed by the Cartan involution, hence doesn't exist by Borel-Wallach.

    Also, how did this very interesting question peter out so quietly and without arriving at any consensus?

     – David Hansen Dec 28 '10 at 19:00
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Here is an example of an elliptic curve $E$ over $Q(\sqrt{997})$ of conductor 1: $[ 0, w, 1, -24w - 289, -144w - 2334 ]$, where $w=(1+\sqrt{997})/2$ (thanks to Lassina Dembelle; this curve even has rank 2!). Shimura's paper "Construction of class fields and zeta functions of algebraic curves" suggests (according to MathSciNet) how to construct a Shimura variety of dimension 2 that isn't a curve but is associated to the relevant quaternion algebra. Shimura lets the quaternion algebra act on the product of two copies of the upper half plane instead of 1, and is able to show the relevant variety is defined over Q by using Siegel modular forms. Perhaps the cohomology of $E$ shows up there? I don't know.

William Stein
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  • Hi William. This is a nice example to have here, but unfortunately I and probably many people don't know what the bracket notation means. I guess they're the coefficients of a Weierstrass equation, but in which order? – JBorger Sep 21 '10 at 23:23
  • @James Borger: A Weierstrass model $y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6$ for an elliptic curve is written $[a_1, a_2, a_3, a_4, a_6]$. – James Weigandt Sep 21 '10 at 23:59
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    P.S.: I would upvote William, but he's currently at 389 reputation, and I have the feeling he wants to stay there. – James Weigandt Sep 22 '10 at 00:00
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    Hey William. In fact there are lots of conductor 1 examples in the literature. The earliest one I know about is $y^2+xy+e^2y=x^3$ over $\mathbf{Q}(\sqrt{29})$ discovered by Tate, with $e=(5+\sqrt{29})/2$; this is mentioned in Serre's 1972 Inventiones article on the image of Galois in the Tate module. Richard Pinch's thesis has a bunch in, IIRC, and one of these was proved to be modular by an explicit computation, by Socrates and Whitehouse, in 2004. – Kevin Buzzard Sep 22 '10 at 09:13
  • As for the Shimura variety you mention, from what you write I think you are talking about the Hilbert modular surface attached to the data: these are built over the complexes precisely by acting on two copies of the upper half plane rather than one and have canonical models over $\mathbf{Q}$. I believe that one can check that the cohomology of $E$ does not show up in the cohomology of these surfaces. The interesting cohomology of the surface will be in the middle dimension, and the weights in the middle dimension exclude the possibility that the curve can show up there. – Kevin Buzzard Sep 22 '10 at 09:15
  • Here are several more examples: http://www.warwick.ac.uk/staff/J.E.Cremona/ecegr/ecegrqf.html – stankewicz Sep 22 '10 at 14:43
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In their 2017 preprint

On subquotients of the étale cohomology of Shimura varieties

C.Johannsson and J.Thorne have made substantial progresses towards question 4. In particular, they show that under reasonable conjectures, the motive attached to an elliptic curve over an imaginary CM field does not appear as subquotient of the intersection cohomology of the minimal compactification of a Shimura variety, and hence should also not appear in the ordinary cohomology of open Shimura varieties.

Very roughly, they show that the conjectures of Arthur on $A$-packets and of Kottwitz on the intersection cohomology of Shimura varieties imply that any Galois representation of the absolute Galois group of a CM field which shows up as a subquotient of the intersection cohomology of Shimura varieties and which is strongly irreducible, that is to say which remains irreducible after restriction to the absolute Galois group of any finite extension, is conjugate self-dual up to a twist. As there are motivic strongly irreducible Galois representations which are not conjugate self-dual up to a twist, this gives a negative answer to statement 4.

It seems to me, by the way, that their work is a concrete implementation of the strategy suggested in the answer of Laurent F. (whoever that guy is).

Olivier
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    +1 for the final parenthetical comment. – R. van Dobben de Bruyn Dec 09 '19 at 18:54
  • Dear Olivier B. how do you reconcile their example with the conjecture of Clozel-Buzzard-Kim? What extra degree of freedom do you need to make their example motive "come from" a Shimura variety? Thank you. – plm Apr 06 '20 at 09:56
  • @plm No idea, honestly. Maybe Jack Thorne will show up (by the way, I'm not the Olivier you are apparently thinking of). – Olivier Apr 07 '20 at 13:49
  • Ah, I was smarter than I thought, thanks to my missed joke I learnt something without meaning it. I always thought you were the Olivier from Bordeaux. Thanks for the reply. – plm Apr 08 '20 at 01:12
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Let me expand and hopefully clarify my first comment about the more specific question of whether the cohomology of a modular elliptic curve with everywhere good reduction shows up in the cohomology of a Shimura variety.

My (very limited) understanding is that the first thing to check is whether or not it shows up in the cohomology of a Picard modular surface. This is not known to happen, but I don't know if this is known to not happen. Then, one could try the following strategy: base-change to a quadratic field, base-change to $U(2)$, extension to $U(2)\times U(n)$, endoscopic transfer to $U(2+n)$ (now available, I think) and then switch to an inner form. Or in other words, one could try to look for a motive on a Picard modular variety.

As far as I understand, and this is not far at all, this strategy works when one is able to 1) construct motives on Picard varieties attached to some suitable Picard automorphic representations 2) Show that the series of operations described in the previous paragraph can yield a suitable automorphic representation. When $n=1$, we know that one can obtain a suitable representation when starting with a Hilbert modular form provided it has weight greater than 2 or a finite place at which it is Steinberg, so our case of interest is excluded. The main obstruction for 2) to work is that the Galois representations arising in the cohomology of Picard varieties will have dimensions determined by (roughly) the degree of freedom available for automorphic types at infinite places. We want this dimension to be 2, so there is a restriction there. Going to higher $n$ allows more flexibility to fiddle with these degree of freedom, so it might help.

All of this is shameful stealing from the articles of Blasius-Rogawski, which are highly recommended reading.

Olivier
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