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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?

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Will, an interesting question! When you say that L-parameters correspond to "certain" classes of conjugation in ${}^L G(\Cmathbb C)$, can you explain how you define this correspondence, and what are, in the case of a split group say, those conjugacy classes? Also, a trivial remark: I was worried an instant that in your definition of an $L$-parameter, you forget the condition of being "relevant" (explained in Borel's paper in Corvallis), but of course, this condition is empty for a quasi-split group, and by Lang's theorem, over a finite field, all red. groups are quasi-split, so you right. – Joël Apr 16 '13 at 13:19
@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? – Will Apr 16 '13 at 13:47
@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. – Will Apr 16 '13 at 15:12
@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. – Will Apr 16 '13 at 17:11
@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. – Will Apr 17 '13 at 7:04
up vote 10 down vote accepted

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.

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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? – Will Apr 25 '13 at 16:32

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