20
$\begingroup$

I was reading a series of article from the Corvallis volume. There are couple of questions which came to my mind:

  1. Why do we need to consider representation of Weil-Deligne group? That is what is an example of irreducible admissible representation of $ Gl(n,F)$ which does not correspond to a representation of $W_F$ of dimension $n$ ? An example for $ n=2 $ will be of great help.

  2. In the setting of global Langlands conjecture, why extension of $W_F$ by $G_a$ or products of $W'_{F_v}$ does not work?

Thank you.

$\endgroup$
0

3 Answers 3

29
$\begingroup$

Regarding (1), from the point of view of Galois representations, the point is that continuous Weil group representations on a complex vector space, by their nature, have finite image on inertia.

On the other hand, while a continuous $\ell$-adic Galois representation of $G_{\mathbb Q_p}$ (with $\ell \neq p$ of course) must have finite image on wild inertia, it can have infinite image on tame inertia. The formalism of Weil--Deligne representations extracts out this possibly infinite image, and encodes it as a nilpotent operator (something that is algebraic, and doesn't refer to the $\ell$-adic topology, and hence has a chance to be independent of $\ell$).

As for (2): Representations of the Weil group are essentially the same thing as representations of $G_{\mathbb Q}$ which, when restricted to some open subgroup, become abelian. Thus (as one example) if $E$ is an elliptic curve over $\mathbb Q$ that is not CM, its $\ell$-adic Tate module cannot be explained by a representation of the Weil group (or any simple modification thereof). Thus neither can the weight 2 modular form to which it corresponds.

In summary: the difference between the global and local situations is that an $\ell$-adic representation of $G_{\mathbb Q_p}$ (or of $G_E$ for any $p$-adic local field) becomes, after a finite base-change to kill off the action of wild inertia, a tamely ramified representation, which can then be described by two matrices, the image of a lift of Frobenius and the image of a generator of tame inertia, satisfying a simple commutation relation.

On the other hand, global Galois representations arising from $\ell$-adic cohomology of varieties over number fields are much more profoundly non-abelian.

Added: Let me also address the question about a product of $W_{F_v}'$. Again, it is simplest to think in terms of Galois representations (which roughly correspond to motives, which, one hopes, roughly correspond to automorphic forms).

So one can reinterpret the question as asking: is giving a representation of $G_F$ (for a number field $F$) the same as giving representations of each $G_{F_v}$ (as $v$ ranges over the places of $F$). Certainly, by Cebotarev, the restriction of the global representation to the local Galois groups will determine it; but it will overdetermine it; so giving a collection of local representations, it is unlikely that they will combine into a global one. ($G_F$ is very far from being the free product of the $G_{F_v}$, as Cebotarev shows.)

To say something on the automorphic side, imagine writing down a random degree 2 Euler product. You can match this with a formal $q$-expansion, which will be a Hecke eigenform, by taking Mellin transforms, and with a representation of $GL_2(\mathbb A_F)$, by writing down a corresponding tensor product of unramified representations of the various $G_{F_v}$. But what chance is there that this object is an automorphic representation? What chance is there that your random formal Hecke eigenform is actually a modular form? What chance is there that your random Euler product is actually an automorphic $L$-function? Basically none.

You have left out some vital global glue, the same glue which describes the interrelations of all the $G_{F_v}$ inside $G_F$. Teasing out the nature of this glue is at the heart of proving the conjectured relationship between automorphic forms and motives; its mysterious nature is what makes the theories of automorphic forms, and of Galois representations, so challenging.

$\endgroup$
0
5
$\begingroup$

The answer to your first question would be a Steinberg representation (i.e. under suitable normalizations, the infinite-dimensional subquotient of the induction of $(\chi|\cdot|^{-1/2},\chi|\cdot|^{1/2})$). Kudla's article in Motives II is a nice place to see this. I don't have an answer for number two.

$\endgroup$
1
  • $\begingroup$ I see, since it corresponds to a nontrivial nilpotent operator, it can not come from a representation of Weil group. Or is there some other argument? Thanks $\endgroup$ Feb 28, 2010 at 5:07
4
$\begingroup$

(I'm putting an "answer" to clarify Rob's question, and answer Dipramit's question in the comments, because I don't yet have the reputation to comment).

Let's first recall that the L-function of the Steinberg representation $\sigma = \sigma(\chi|\cdot|^{-1/2},\, \chi|\cdot|^{1/2})$ (for $\chi$ an unramified character) is $(1 - \chi(\varpi)q^{-s-1/2})^{-1}$, where $\varpi$ is a uniformizer (Bump shows this in detail in his book). In particular, its reciprocal is a degree-one polynomial in $q^{-s}$.

By the ideas of Bernstein-Zelevinski (described in Kudla' article in "Motives"), $\sigma$ corresponds to the Weil-Deligne representation $\rho' = (\rho,\, V,\, N)$, where $\rho = \chi |\cdot|^{-1/2} \oplus \chi |\cdot|^{1/2}$, and the operator $N$ takes the first summand to the second, and the second summand to $0$.

If we only look at $\rho$, then we see that $V^I = V$ and therefore the $L$-function of $\rho$ would be the reciprocal of a degree-two polynomial. Thus the monodromy operator becomes necessary for match of $L$-function: we see that $V^I_N \cong \chi|\cdot|^{1/2}$, which has the desired $L$-function.

This is a good specific example if you like thinking about match of $L$-functions. More generally, you need to consider Weil-Deligne representations because if $\pi_1$ and $\pi_2$ have the same cuspidal support and correspond to $\rho_1' = (\rho_1,\, N_1,\, V_1)$ and $\rho_2' = (\rho_2,\, N_2,\, V_2)$, then $\rho_1\cong \rho_2$ as Weil representations; this follows from the ideas of Bernstein-Zelevinski.

$\endgroup$
2
  • 1
    $\begingroup$ Actually the Steinberg representation $\sigma$ is ramified of conductor $1$, and its $L$-function equals $(1 - \chi(\varpi)q^{-s-1/2})^{-1}$. In general, let $\pi$ be a local irreducible generic representation of $\mathrm{GL}_n$. Then $L(s,\pi)^{-1}$ is a polynomial of $q^{-s}$ of degree at most $n$, and the degree equals $n$ if and only if $\pi$ is unramified. See the Proposition on page 203 of Jacquet-PiatetskiShapiro-Shalika: Conducteur des répresentations du groupe linéaire (Math. Ann. 256 (1981), 199-214). $\endgroup$
    – GH from MO
    Jul 31, 2013 at 17:12
  • $\begingroup$ Fixed. Thank you for pointing out my error. Apparently my understanding of the Artin conductor is limited, looks like it's time to go back and read Tate's Corvallis article again. As a side question, my construction depends on the ideas of Bernstein-Zelevinski. Does anyone know of a construction of a "counterexample" that does not rely on these ideas? $\endgroup$ Jul 31, 2013 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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