27
$\begingroup$

Let $\boldsymbol{G}$ be a reductive group over a finite field $\mathbb{F}_q$, $G = \boldsymbol{G}(\mathbb{F}_q)$, $W = \mathrm{W}(\mathbb{F}_q)$ the Witt vectors over $\mathbb{F}_q$, and $K = \mathrm{Frac}(W)$ its fraction field. I'll abuse notation by also writing $\boldsymbol{G}$ for the corresponding (unramified) reductive group over $K$.

When $\boldsymbol{G}$ has connected center, Lusztig proves the following:

Theorem. There is a canonical bijection between (isomorphism classes of) irreducible $\overline{K}$-representations of $G$ and special conjugacy classes in $\widehat{\boldsymbol{G}}(\overline{K})$ stable under $g \mapsto g^q$.

Here $\widehat{\boldsymbol{G}}$ is the dual group of $\boldsymbol{G}$. The condition of being special is a condition of Lusztig relating to special representations of Weyl groups through the Springer correspondence.

On the other hand, one could write a Langlands-type statement as follows: irreducible $\overline{K}$-representations of $G$ should correspond to $\widehat{\boldsymbol{G}}(\overline{K})$-conjugacy classes of $L$-parameters for $G$.
One possible definition of an $L$-parameter is that of a homomorphism over $\mathrm{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q)$ of the form $$\varphi \colon \langle \mathrm{Frob}_q \rangle \times \mathrm{SL}_2(\overline{K}) \to {}^L \boldsymbol{G}(\overline{K}), $$ where ${}^L \boldsymbol{G} = \widehat{\boldsymbol{G}} \rtimes \mathrm{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q)$ is the Langlands dual group of $\boldsymbol{G}$, and where we require $\mathrm{Frob}_q$ to have semisimple image in $\widehat{\boldsymbol{G}}(\overline{K})$, and the restriction of $\varphi$ to $\mathrm{SL}_2(\overline{K})$ to be algebraic.
Of course, in this situation, the data of an $L$-parameter $\varphi$ is equivalent to that of a particular kind of $\widehat{\boldsymbol{G}}(\overline{K})$-conjugacy class in ${}^L \boldsymbol{G}(\overline{K})$. However, Lusztig's conditions on the conjugacy classes do not appear anywhere. For instance, when $\boldsymbol{G}$ is split, an $L$-parameter is simply a $\widehat{\boldsymbol{G}}(\overline{K})$-conjugacy class in $\widehat{\boldsymbol{G}}(\overline{K})$.
Instead, we might want to add an extra condition which would pin down the image of $\mathrm{Frob}_q$ to lie in $\widehat{\boldsymbol{G}}(K)$, in order to parallel the following result, pertaining to the semisimple part of the correspondence:

Theorem. There is a canonical bijection between (isomorphism classes of) irreducible semisimple $\overline{K}$-representations of $G$ and semisimple conjugacy classes in $\widehat{\boldsymbol{G}}(\mathbb{F}_q)$.

However, such a modification would not account for the unipotent part, nor would it account for the condition of being special.
It seems, then, that this notion of $L$-parameter is the wrong one. What is the fix?

$\endgroup$
12
  • 5
    $\begingroup$ @Joël: I just mean that (say for $\boldsymbol{G}$ split) the $L$-parameter $\varphi$ is determined by the element $x = s u$ where $s = \varphi(\mathrm{Frob}_q)$ and $u = \varphi \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, by an abstract Jordan decomposition. @George: Yes, that's essentially my question. Do you have a reference where $\widehat{\boldsymbol{G}}(\mathbb{F}_q)$ explicitly appears? $\endgroup$
    – Will
    Apr 16, 2013 at 13:47
  • 4
    $\begingroup$ @George: It seems in that book they only use semisimple elements $s$ of $\widehat{\boldsymbol{G}}(\mathbb{F}_q)$, to explain the decomposition of representations into Lusztig series $\mathcal{E}(\mathrm{C}_{\widehat{G}}(s),1)$. There doesn't seem to be a general parametrisation of representations of $G$ by conjugacy classes in $\widehat{\boldsymbol{G}}(\mathbb{F}_q)$, unless I'm missing something. $\endgroup$
    – Will
    Apr 16, 2013 at 15:12
  • 4
    $\begingroup$ @Dror: To my understanding, these irreducible characters you mention are precisely the semisimple characters. See Proposition 8.4.6 in Carter's book for instance, which indexes the semisimple characters of $G$ by geometric conjugacy classes of pairs $(T,\theta)$ as you said. $\endgroup$
    – Will
    Apr 16, 2013 at 17:11
  • 3
    $\begingroup$ @Dror: Yes, I understand that. I only wanted to point out that the (g.c.c. of) $(T,\theta)$ naturally parametrise the semisimple representations, even though of course every representation appears inside some $\mathrm{R}_{T,\theta}$. The Jordan decomposition of characters you mention was one of my motivations for asking the question, as it parallels the semistable v.s. unipotent dichotomy for $L$-parameters. The semisimple part is then described through $\widehat{\boldsymbol{G}}(\mathbb{F}_q)$, but I don't see how the unipotent part fits in, i.e. how to relate Lusztig series to an $L$-group. $\endgroup$
    – Will
    Apr 17, 2013 at 7:04
  • 2
    $\begingroup$ Pardon for adding a comment on this very old and nice question. May I ask for a reference on the mentioned main result of Lusztig? Thanks a lot! $\endgroup$ Jul 20, 2018 at 8:26

1 Answer 1

12
$\begingroup$

The way I like to think about this is that a Langlands parameter for the group $G({\mathbb F}_q)$ should be the "restriction to inertia" of a tame Langlands parameter for the group $G(K)$.

That is, a tame Langlands parameter (say, over ${\mathbb C}$) for $G(K)$ should be a pair $(\rho,N)$, where $\rho$ is a map $W_K \rightarrow \hat G({\mathbb C})$ that factors through the tame quotient of $W_K$ and $N$ is a nilpotent "monodromy operator", that is, a nilpotent element of the Lie algebra of $\hat G$ that satisfies a certain commutation relation with $\rho$.

The tame quotient of $W_K$ is generated by the tame inertia subgroup $I_K$ and a Frobenius element $F$; local class field theory identifies $I_K$ with the inductive limit of the groups ${\mathbb F}_{q^n}^{\times}$, and conjugation by Frobenius acts on this by raising to $q$th powers.

I don't have the details in front of me, but if I recall correctly the Deligne-Lusztig parameterization involves several choices (for instance, an identification of $\overline{\mathbb F}_q^{\times}$ with a suitable space of roots of unity in ${\mathbb C}$.) My understanding is that if one unwinds these choices, they amount to a choice of topological generator $\sigma$ for the inductive limit of the ${\mathbb F}_{q^n}^{\times}$.

Thus, if one starts with a Langlands parameter $(\rho,N)$ for $G(K)$ and restricts $\rho$ to inertia, this restriction is determined by $\rho(\sigma)$, which is a semisimple element of $\hat G({\mathbb C})$ that is conjugate to its $q$th power. The pair $(\rho(\sigma),N)$ should be the Langlands parameter for the group $G({\mathbb F}_q)$.

There should then (roughly) be a compatibility between depth zero local Langlands and the Deligne-Lusztig parameterization, as follows: let $K$ be the kernel of the reduction map $G(W) \rightarrow G({\mathbb F}_q)$, and let $\pi$ be an irrep of $G(K)$ with Langlands parameter $(\rho,N)$. Then the $K$-invariants $\pi^K$ of $\pi$ are naturally a $G({\mathbb F}_q)$-representation, and $\pi^K$ should contain the representation of $G({\mathbb F}_q)$ corresponding to $(\rho(\sigma),N)$ via Deligne-Lusztig. You should take this with a bit of a grain of salt, as I haven't thought the details through carefully. But it should be correct on a "moral" level, at least.

For $GL_n$ this falls under the rubric of the so-called "inertial local Langlands correspondence". For more general groups the picture is more conjectural, but there are ideas along these lines in the paper of DeBacker-Reeder on the depth zero local Langlands correspondence.

$\endgroup$
1
  • 2
    $\begingroup$ Thanks. I have two questions: 1) why, on a moral level, does one need to invoke the tame inertia of $K$ when working only over $\mathbb{F}_q$? (I would have thought you should only consider Galois theoretic (descent) data from $\overline{\mathbb{F}_q}$ to $\mathbb{F}_q$ to describe representations as "forms of principal series"); and 2) how come we simply forget the Frobenius; does this not clash with the semisimple case where no $g \mapsto g^q$ condition appears? $\endgroup$
    – Will
    Apr 25, 2013 at 16:32

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.