Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?".

share|improve this question
    
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
1  
@Kevin Buzzard: Thanks for this nice summary 1–4 of the Langlands programme! –  Timo Keller Sep 21 '10 at 12:59
4  
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

3 Answers 3

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.

share|improve this answer
    
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 $[a_1, a_2, a_3, a_4, a_6]$. –  Jamie Weigandt Sep 21 '10 at 23:59
4  
P.S.: I would upvote William, but he's currently at 389 reputation, and I have the feeling he wants to stay there. –  Jamie Weigandt Sep 22 '10 at 0:00
1  
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 9:13
    
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 9:15

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...

share|improve this answer
1  
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 4:04
    
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 4:22
    
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 5:32
    
@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 9:24
    
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 9:27

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.