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

In many formulation of Class Field theory, the Weil group is favored as compared to the Absolute Galois group. May I asked why it is so? I know that Weil group can be generalized better to Langlands program but is there a more natural answer?

Also we know that the abelian Weil Group is the isomorphic image of the reciprocity map of the multiplicative group (in the local case) and of the idele-class group (in the global case). Is there any sense in which the "right" direction of the arrow is the inverse of the reciprocity map?

Please feel free to edit the question into a form that you think might be better.

share|improve this question
    
I have accepted the answer by Johnson but I still hope to receive the answer for the second part of the question. –  Tran Chieu Minh Jan 17 '10 at 19:28

3 Answers 3

up vote 11 down vote accepted

One reason we prefer the Weil group over the Galois group (at least in the local case) is that the Weil group is locally compact, thus it has "more" representations (over $\bf C$). In fact, all $\bf C$-valued characters of $Gal(\bar{\bf Q_p} / \bf Q_p)$ have finite image, where as that of $W_{\bf Q_p}$ can very well have infinite image. The same goes for general representations of these groups (recall that $\bf{GL}_n(\bf C)$ has no small subgroups.)

The global Weil group (which is much more complicated than the local one), on the other hand, is a rather mysterious object that is pretty much untouched in modern number theory as far as I can tell. Supposedly the global Langlands group used in the global Langlands correspondence should be the extension of the global Weil group by a compact group, but this is still largely conjectural.

The standard reference is Tate's "Number Theory Background" in the Corvallis volumes (available for free at ams.org). Also Brooks Roberts has notes on Weil representations available at his website.

share|improve this answer
1  
Thank you for the reply. Do you think it is possible to have a natural reason beside utility for which we prefer Weil group? –  Tran Chieu Minh Jan 17 '10 at 19:17
1  
Tran: Isn't utility one of the best motivators to do anything? –  Adam Hughes Mar 22 '11 at 13:21

The Weil group appears for several reasons.

Firstly: if $K$ is a non-archimedean local field with residue field $k$, the local reciprocity law induces an embedding $K^{\times} \hookrightarrow G_K^{ab}.$ The image consists of all elements in $G_k$ whose image is an integral power of Frobenius. This is the abelianized Weil group; it just appears naturally.

Secondly: suppose that $K$ is a global field of positive characteristic, i.e. the function field of a curve over a finite field $k$. Then the global reciprocity map identifies the idele class group of $K$ with a subgroup of $G_K^{ab}$ consisting of elements which act on $k$ by integral powers of Frobenius. So again, it is the abelianized Weil group that appears.

Thirdly: suppose that $E$ is an elliptic curve over a quadratic imaginary field $K$ with complex multipliction by $\mathcal O$, the ring of integers in $K$. (Thus I am implicity fixing $K$ to have class number one, but this is not so important for what I am going to say next.) If $\ell$ is a prime, then the $\ell$-adic Tate module is then free of rank one over $\mathcal O_{\ell}$ (the $\ell$-adic completion of $\mathcal O$), and the $G_K$-action on this Tate module induces a character $\psi_{\ell}:G_K^{ab} \rightarrow \mathcal O_{\ell}^{\times}$.

There is a sense in which the various $\psi_{\ell}$ are indepenent of $\ell$, but what is that sense?

Well, suppose that $\wp$ is a prime of $K$, not dividing $\ell$ and at which $E$ has good reduction. Then the value of $\psi_{\ell}$ on $Frob_{\wp}$ is indepenent of $\ell$, in the sense that its value is an element of $\mathcal O$, and this value is independent of $\ell$. More generally, provided that $\wp$ is prime to $\ell$, the restriction of $\psi_{\ell}$ to the local Weil group at $\wp$ is independent of $\ell$ (in the sense that the value at a lift of Frobenius will be an algebraic integer that is independent of $\ell$, and its restriction to inertia at $\wp$ will be a finite image representation, hence defined over algebraic integers, which again is then independent of $\ell$).

Note that independence of $\ell$ doesn't make sense for $\psi_{\ell}$ on the full local Galois group at $\wp$, since on this whole group it will certainly take values that are not algebraic, but rather just some $\ell$-adic integers, which can't be compared with one another as $\ell$ changes.

Now there is also a sense in which the $\psi_\ell$, as global Galois characters, are independent of $\ell$. Indeed, we can glue together the various local Weil group representations to get a representation $\psi$ of the global Weil group $W_K$. Since it is abelian, this will just be an idele class character $\psi$, or what is also called a Hecke character or Grossencharacter. It will take values in complex numbers. (At the finite places it even takes algebraic number values, but when we organize things properly at the infinite places, we are forced to think of it as complex valued.)

Note that $\psi$ won't factor through the connected component group, i.e. it won't be a character of $G_K^{ab}$. It is not a Galois character, but a Weil group character. It stores in one object the information contained in a whole collection of $\ell$-adic Galois characters, and gives a precise sense to the idea that these various $\ell$-adic characters are independent of $\ell$.

This is an important general role of Weil groups.

Fourthly: The Hecke character $\psi$ above will be an algebraic Hecke character, i.e. at the infinite places, it will involve raising to integral powers. But we can also raise real numbers to an arbitrary complex power $s$, and so there are Hecke characters that do not come from the preceding construction (or ones like it); in other words, there are non-algebraic, or non-motivic, Hecke characters. But they are abelian characters of the global Weil group, and they have a meaning; the variable $s$ to which we can raise real numbers is the same variable $s$ as appears in $\zeta$- or $L$-functions.

In summary: Because Weil groups are "less completed", or "less profinite", than Galois groups, they play an important role in describing how a system of $\ell$-adic representations can be independent of $\ell$. Also, they allow one to describe phenomena which are automorphic, but not motivic (i.e. which correspond to non-integral values of the $L$-function variable $s$). (They don't describe all automorphic phenomena, though --- one would need the entire Langlands group for that.)

share|improve this answer
    
Excellent answer ! By the way, the second section of Gross-Reeder paper on arithmetic invariants (Duke 2010) is also nice for Weil groups –  user4245 Jun 24 '12 at 16:56

Perhaps something worth pointing out, relating to how the Weil group appears naturally: it arises from a compatible system of group extensions at finite levels. Indeed, one of the "axioms" of class field theory, is the existence of a "fundamental class" uL/*K* in $H^2(\operatorname{Gal}(L/K),C_L)$ for each finite Galois extension $L/K$ (where $C_L$ is the class module). Each of these gives a group extension $$ 1\rightarrow C_L\rightarrow W_{L/K}\rightarrow \operatorname{Gal}(L/K)\rightarrow 1.$$ The projective limit of these gives the absolute Weil group fitting in $$ 1\rightarrow C\rightarrow W_K\rightarrow G_K$$ with the rightmost map having dense image (and $C$ is the formation module of the class formation). Thus, you can think of the Weil group as arising canonically out of the results of class field theory, thus making it a natural replacement of $G_K$ in questions related to CFT. I like section 1 of chapter III of Neukirch–Schmidt–Wingberg's Cohomology of number fields and the last chapter of Artin–Tate for this material.

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.