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I'm trying to interpret things in the following terminology:

Assume the standard conjectures, the existence of the conjectural Langlands group, and anything else you wish.

I assume the following statement: the category of isomorphism classes of irreducible continuous algebraic (I added the word `algebraic' after it was pointed out that it was needed) representations $\mathcal{L}_{\mathbb{Q}}$ (the Langlands group) $\rightarrow GL_n(\mathbb{C})$ is equivalent to the category of $\mathbb{Q}$-motives with coefficients in $\overline{\mathbb{Q}}$.

In the above terminology, there are three numbers to keep track of: $n$ (of $GL_n$), dimension of the scheme that the motive comes from, and the degree of purity of the motive within that scheme (by this I mean that if we denote it by $i$, then in any Weil cohomology it would be realize as the $i^{th}$ cohomology of this scheme).

I will give two examples:

i. Classic CFT comes from $n=1$, the dimension of the scheme that the motive comes from is $0$ (because we're dealing with fields), and the degree of purity of the motive is $0$.

I imagine that when people talk about CFT in higher dimensions they mean the $n=1$ case where you allow the other two numbers to be different.

ii. Taniyama Shimura: this is the $n=2$ case, where the dimension of the scheme is $1$ and the degree of purity of the motive is $1$. Here on the Langlands side, this must correspond to newforms of weight $2$ with rational Hecke eigenvalues.

In general, one could take $n=2$, and on the Langlands side look at a general weight $k$ newform. Let $F$ be the field generated by the Hecke eigenvalues. Then the corresponding motive is coming from a $[F:\mathbb{Q}]$-dimensional variety, and the purity degree of the motive is $k-1$. If $k=2$ one can prove that the variety from which the motive is coming is an abelian variety via an argument involving Hodge structures.

Now I wish to understand complex multiplication in this context. It seems that the goal is to classify abelian extensions of quadratic fields. This seems to imply that we are in the $n=1$ case. But what are the other two numbers? Are we really looking at motives coming from $0$-dimensional schemes? This seems weird, because in the theory of complex multiplication elliptic curves are implicated.

Can one give a similar context to complex multiplication as I did to CFT and Taniyama Shimura? How would that work? What would the three numbers I described be?

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Dear James, Higher dimensional class field theory typically means working over higher dimensional local (or global) fields, so has nothing to do with the parameters you discuss (but rather with replacing number fields or function fields in one-variable over finite field by function fields of number fields, or function fields over finite fields in several variables). See also this answer: math.stackexchange.com/questions/56222/… Regards, Matthew –  Emerton Sep 7 '11 at 18:50
    
Also, there is some confusion with weights when you discuss the $n = 2$ case. A weight $k$ modular form gives rise to a $2$-dimensional motive of weight $k - 1$. In particular, weight $2$ modular forms always give rise to motives of weight one. –  Emerton Sep 7 '11 at 18:52
    
As for your actual question, zero dimensional varieties have to do with finite order Galois characters. Elliptic curves with CM give rise to infinite order Galois characters, or if you prefer, infinite order Hecke (i.e. idele class group) characters. In some expositions of the theory, it can be hard to detect this distinction, since any $p$-adic valued character is a projective limit of characters mod $p^n$, which have finite order, so that the possibly-infinite order theory interweaves with the finite order theory. (More geometrically, the $N$-torsion of an elliptic curve is a ... –  Emerton Sep 7 '11 at 18:56
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James, you may be interested in these notes of Fargues: math.u-psud.fr/~fargues/Motifs_abeliens.ps. –  Sam Lichtenstein Sep 7 '11 at 22:10
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Regarding the question, raised in the comments to Kevin's answer, as to whether CM theory is "a corollary of Langlands": it is rather a corollary of the theory of Shimura varieties, in particular of the theory of $H^0$ of Shimura varieties. The more general question of what Galois representations arise in the higher etale cohomology of Shimura varieties plays a central role in the Langlands program (broadly understood), but it is not directly contained in your formulation. –  Emerton Sep 8 '11 at 0:00

1 Answer 1

Your assumption as written is not quite correct. The irreducible $n$-dimensional irreducible representations of the Langlands group should correspond I think to all cuspidal automorphic representations of $GL(n)$, and there are far more of these things than there are motives -- consider for example the 1-dimensional representation $|.|^s$ for $s$ a random (non-integer) complex number: this is not motivic but it is automorphic (the Langlands group will surject onto $\mathbf{A}_\mathbf{Q}^\times/\mathbf{Q}^\times$).

But if you replace your Langlands group with some sort of "motivic Galois group" then some version of your assumption should be true. The $n$ in $GL_n$ is the rank of the motive. The dimension of the scheme is a complete red herring though. For example if I choose a 3-fold with no $H^1$ then the motive attached to its $H^1$ is zero and you can't see the dimension of the 3-fold from this. As a less silly example if the $H^2$ of a large-dimensional variety is 1-dimensional (which can certainly happen), then you can't tell it from the $H^2$ of any smooth projective geom connected algebraic curve. So you can't read off anything about the dimension of the scheme -- if indeed there is a scheme (motives can have negative weight of course -- you can do any linear algebra you like with them like taking duals or twists).

The "degree of purity" of the motive -- I think people say that the motive is pure of weight $w$. For motives coming from $H^i(X)$ with $X$ an algebraic variety these will be pure of weight $i$. On the side of the Langlands group you can read these off by looking e.g. at the sizes of the eigenvalues of unramified Frobenii. There's no relationship between $n$ and $i$ -- they're independent pieces of data.

$n=2$: If you're allowing coefficient fields (which you are) then a 2-dimensional representation of the motivic Galois group could easily correspond to the dual of the Tate module of a high-dimensional abelian variety with lots of real multiplications (or if you like the $H^1$ of this ab var), because the cohomology of the variety inherits the action of the totally real field and hence can be regarded as 2-dimensional. But this representation is still pure of weight 1. So again the dimension of the scheme has nothing to do with anything: $n=2$, you're pure of weight 1, and the dimension can be anything you like. In general the weight of the motive attached to a weight $k$ modular form is $k-1$ (or even $1-k$ depending on how you normalise things); this is an unfortunate notational blip.

Still with $n=2$, if you use higher weight modular forms, they're motivic by Scholl. Deligne found them in the cohomology of the modular curve with a non-trivial coefficient sheaf, but this isn't "allowed" for motives -- Scholl found them in the cohomology of some fibre product of the universal elliptic curve by itself $k-2$ times.

For CM one issue is that sometimes the base field changes. For example people talk about "elliptic curves over $\mathbb{Q}$ with CM" but the CM is only defined over some larger number field. If the CM is defined over an imaginary quadratic field $K$ then the Tate module of the curve can be thought of as a 2-dimensional representation of the absolute Galois group of $\mathbf{Q}$ but also as a 1-dimensional representation of the absolute Galois group of $K$ (the one is just induced from the other). So now $n$ is changing depending on context. As we've already seen, we shouldn't get hung up over the dimension of the elliptic curve. The torsion points on the curve generate abelian extensions of $K$, so you see both schemes of dimension 1 and 0 involved.

I've written this in rather a rush because I'm supposed to be giving lasagne to children, not doing maths, but hopefully it's mostly OK. Feel free to ask more. Let me say that I'm not at all an expert though; I don't really know what a motive is :-) Eew I smell burning :-/

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Oh, excellent! I do have a couple of questions, if you don't mind: 1. For your first point, about me not being precise, if I wanted to put a restriction on the kinds of irreducible representations of $\mathcal{L}_{\mathbb{Q}}$ I'm looking at so that there would be a correspondence with motives, what would that be? I imagine this has something to do with the term "Langlands parameters" I've been hearing about. Do you have a reference? 2. Is there an easy way to see that torsion points on the elliptic curve over $\mathbb{Q}$ generate abelian extensions of $K$, where by "easy" I mean –  James D. Taylor Sep 7 '11 at 19:29
    
a way of saying why it's true that would make it easy for me to generalize to other contexts that I can consider using the terminology above. –  James D. Taylor Sep 7 '11 at 19:30
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...here's a proof. Say $E/\mathbf{Q}$ is an elliptic curve and say that it has CM. This forces the Tate module of the curve to be an induced representation, induced from a 1-d representation of the absolute Galois group of $K$, the im quad field in question. So now the $n$-torsion also has to be induced from a 1-d rep of the absolute Galois group of $K$ and there's your abelian extension of $K$. Well, that's the very bare bones of a proof, at least! –  Kevin Buzzard Sep 7 '11 at 19:40
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@James: the hard point is the other way -- that all abelian extensions arise in this way. This is harder work. See Serre's article in Cassels-Froehlich for a nice exposition. –  Kevin Buzzard Sep 7 '11 at 19:42
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I don't know what you mean by "Langlands" but somehow this is I think "explicit class field theory", which is known for the rationals (adjoin torsion points of $\mathbb{G}_m$) and for im quad fields (same but use ell curves) but which I think is unknown, even conjecturally, for a general number field -- I guess it's Hilbert's $n$th problem for some value of $n$ (perhaps $n=12$ going from memory...) –  Kevin Buzzard Sep 7 '11 at 21:51

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