Let $W_F$ denote the Weil group of a finite extension of $\mathbb{Q}_p$. Let $I$ denote the inertia subgroup and $I^{>0}$ the (pro-$p$) subgroup of wild inertia. (I hope I've got my notation right...apologies if I haven't.)

We have $W_F/ I = \mathbb{Z}$ (canonically), and $I/I^{>0} \cong \prod_{\ell \ne p} \mathbb{Z}_\ell$ (non-canonically).

Consider a Weil-Deligne representation $(V, \rho, N)$, where $V$ is a finite-dimensional complex vector space, $\rho : W_K \to GL(V)$ is a rep of $W_K$, and $N$ is a nilpotent operator satisfying a well-known condition.

Naively, I would have thought that a Weil-Deligne representation $(V,\rho,N)$ is tamely ramified if

$\rho$ vanishes on $I^{>0}$, no conditions on $N$

However, when I read literature on the Deligne-Langlands conjecture (e.g. Kazhdan-Lusztig, Chriss-Ginzburg) it seems that tamely ramified actually means

$\rho$ vanishes on $I$, no conditions on $N$

[Under this assumption an $F$-semi-simple, tamely ramified, Weil-Deligne representation becomes a (conjugacy class of a) pair $(s,N)$ consisting of a semi-simple element $s$, a nilpotent element $N$, such that $sNs^{-1} = qN$, which is what one expects from affine Hecke algebras, and what we are told the Deligne-Langlands conjecture predicts that the answer should be. [EDIT (following Ben Zvi's comments below): Actually we also need an irreducible representation of the component group of the centraliser of this pair. This extra piece of data is absent for $GL_n$ as this group is always connected.]

Passing back through Grothedieck's equivalence, the above seems to give that a tamely-ramified and continuous representation of $W_K$ in a $\mathbb{Q_\ell}$-vector space, should have each $\mathbb{Z}_{\ell'}$ factor (for $\ell' \ne \ell$) acting trivially, which seems a little strange. (I.e. the finite monodromy which could come from the other factors should magically be trivial.)

Finally, an admittedly lazy question. I can read in several places that, under the local Langlands correspondence, tamely ramified should correspond to representations admitting an Iwahori fixed vector.

Is this explained somewhere? Why should tamely ramified (whatever that actually means, c.f. the above question) translate into having an Iwahori fixed vector?

I'm sure this is easy for the pros! Thanks in advance.


3 Answers 3


$\def\R{\mathbf{R}}$ $\def\Z{\mathbf{Z}}$ $\def\Q{\mathbf{Q}}$ $\def\Qbar{\overline{\Q}}$ $\def\F{\mathbf{F}}$ $\def\GL{\mathrm{GL}}$ $\def\Gal{\mathrm{Gal}}$ Here are some thoughts on your question from a number theorist. First of all, I think the only sensible answer to your question is that your naive guess is the right one, namely, that a tamely ramified Weil-Deligne representation is one for which the wild inertia group $I_{>0}$ acts trivially. But let me take the opportunity to give a more expansive answer which touches on a few other concepts. This discussion will all be at the cartoon level, but that may still be useful to some. Let's fix ideas and always talk about the case of $G = \GL_n(\Q_p)$. Let $\rho$ be a Weil-Deligne representation and let $\pi$ be the corresponding representation associated to $\rho$ by local Langlands. The following three things are then equivalent:

  1. The Weil-Deligne representation is trivial on the entire inertia group.
  2. $\pi$ has an Iwahori fixed vector.
  3. The corresponding $l$-adic representations for $l \ne p$ are unipotent on inertia.

With that out of the way:

I would certainly only ever use "tamely-ramified" to refer to the weaker condition that the image of wild inertia is trivial, exactly as you hypothesized.

A really basic example to keep in mind is the case of characters. We have $\GL_1(\Q_p) = \Q^{\times}_p$. Let $\chi$ be a (continuous) character of $\Q^{\times}_p$. Let's specialize the conditions above in this case. We get

  1. The Weil-Deligne representation is trivial on the entire inertia group.
  2. $\pi$ has a fixed vector under $\Z^{\times}_p \subset \Q^{\times}_p$.
  3. The corresponding $l$-adic representations for $l \ne p$ are trivial (equivalently in dimension one unipotent) on inertia.

The nice thing about the case $n = 1$ is that the local Langlands correspondence is very transparent, namely, it is given by local class field theory. To recall (briefly) what class field theory says (in one formulation), it gives a canonical map

$$\Q^{\times}_p \hookrightarrow \Gal(\Qbar_p/\Q_p)^{\mathrm{ab}}$$

with dense image. Indeed, the image is precisely the image of the Weil group, and the image of $\Z^{\times}_p$ is the image of the inertial part of the Weil group. It follows that the wild inertia group maps to the pro-$p$ part of $\Z^{\times}_p$ which is $1 + p \Z_p$. A (Galois) character of $\Gal(\Qbar_p/\Q_p)$ canonically gives (by restriction) a character of $\Q^{\times}_p$. Certainly one would want to allow a "tamely-ramified" Galois character to be tamely ramified, so (in dimension one) we get the following equivalences:

  1. The Weil-Deligne representation is trivial on $1 + p \Z_p \subset \Q^{\times}_p$,
  2. $\pi$ has a fixed vector under $1 + p \Z_p$.
  3. The corresponding $l$-adic representations are tamely ramified on inertia.

I honestly only found very few papers in the literature in which "tamely ramified" was implied to have the meaning (1), (2), or (3). I think they were just in error. Here are a few speculations on how they could get confused. First of all, the maximal tamely ramified extension of a local field has a particularly simple Galois group. Namely, it is given by (the profinite completion of) the group

$$\Gamma: \langle \tau, \sigma | \sigma \tau \sigma^{-1} = \tau^q \rangle,$$

where $\sigma$ is Frobenius, $\tau$ is a (pro-finite) generator of tame inertia, and $q$ is the order of the residue field (So $q = p$ for $\Q_p$). In characteristics $l \ne p$, unipotent implies pro-$l$ implies tame, and thus unipotent implies tamely ramified. Let us now compare condition 3 above with a new condition 4.

  1. The corresponding $l$-adic representations for $l \ne p$ are unipotent on inertia, that is, they are representations of $\Gal(\Qbar_p/\Q_p)$ which factor through $\Gamma$ and which therefore give rise to matrices $T$ and $F$ satisfying $FTF^{-1} = T^q$, and where $T$ is unipotent.

  2. The corresponding $l$-adic representations for $l \ne p$ are tamely ramified, that is, they are representations of $\Gal(\Qbar_p/\Q_p)$ which factor through $\Gamma$ and which give rise to matrices $T$ and $F$ satisfying $FTF^{-1} = T^q$.

One reason these look quite similar is that the condition that $T$ is conjugate to $T^q$ almost looks as though it should imply that $T$ is unipotent, or equivalently that the (generalized) eigenvalues of $T$ are all $1$. But this is not quite true, it doesn't force the eigenvalues to satisfy $\lambda = 1$ but only $\lambda^{q-1}= 1$ and so only $q-1$th roots of unity. (Correction: as Will Sawin points out, this should be $\lambda^{q^k - 1} = 1$ for some $k \le n$. Indeed, in the supercuspidal case you can get primitive $(q^n-1)$th roots of unity.) So 3 and 4 are quite similar, but 4 gives a more relaxed condition and is the "correct" definition of tamely ramified.

So now, let us return to your next question (slightly modified):

Why should [...] translate into having an Iwahori fixed vector? What do tamely ramified representations correspond do?

OK, so answering "why" to questions in Langlands is always tricky. (What is Langlands? A vast generalization of class field theory. What is class field theory? A vast generalization of quadratic reciprocity. What is quadratic reciprocity? NOBODY KNOWS.) But let me give a little attempt to answer your question by a somewhat circuitous route. I want to start by talking about conductors. (I'm not actually going to say very much interesting about conductors, but I'm going to start out this way.) If you are a graduate student in number theory, sooner or later you are going to read about the modularity of elliptic curves. And to understand the statement, there is some mysterious invariant $N$ associated to an elliptic curve $E/\Q$. At first glance, it does something reasonable to measure the bad reduction of $E$: if $p |N$, then $E$ (or a suitably good model) has bad reduction at $p$. And there's some sense that the power of $p$ dividing $N$ measures "how bad" the reduction really is in some controllable way for $p > 3$. And then for $p = 3$ or $p = 2$ everything goes crazy, but it's OK because some guy called Tate came up with an algorithm and $\texttt{gp}$ can compute it for you. But if you try to think about it, it makes bugger all sense. Then you have to believe the same $N$ turns up as the level of the modular form, and this is even more confusing. In order to even begin to understand it, it's really key to once again step back and think about the case of $\GL(1)$. In this case, we have a character $\chi$ which we can think of as a character of $\mathbf{Q}^{\times}_p$. And what class field theory tells us is that the ramification behavior of $\chi$ is all wrapped up in the restriction of $\chi$ to $\Z^{\times}_p$. The "automorphic" and "Galois" sides are almost literally the same here, but let me still try to distinguish them. We can now define the conductor as follows:

  1. The automorphic conductor of $\chi$ is the smallest $p^n$ such that $\chi$ has an invariant under the subgroup $1 + p^n \Z_p$.

  2. The Galois conductor of $\chi$ is the smallest $p^n$ such that $\chi$ is trivial on $1 + p^n \Z_p$.

Of course these are clearly the same. But even though I haven't really said it here, what is important in class field theory is that the filtration of $\Z^{\times}_p$ by $1+p^n \Z_p$ is intimately related to the inertia filtrations on the Galois groups. The special case that $1+p \Z_p$ corresponds to $I_{>0}$ is clear, but the others are more subtle. So here the "equality" of conductors coming from "local Langlands for $\GL(1)$" is expressing something rather deep in class field theory. To give at least one concrete example, there is Hasse's Führerdiskriminantenproduktformel which expresses the discriminant of the corresponding cyclic fixed field in terms of the conductors of the non-trivial powers of $\chi$. The study of relations of this flavour were actually key to Artin's formulation of his various reciprocity laws and recognizing the correct way to define $L$-functions. I think Noah Snyder wrote an interesting undergraduate thesis about this once. Anyway, I'm drifting slightly from my main topic. To get back on point, the key thing about this example which generalizes is that the following two things are intimately related:

  1. The restriction of the Weil--Deligne representation $\rho$ to inertia.
  2. The restriction of $\pi$ as a $\GL_n(\Q_p)$ representation to $\GL_n(\Z_p)$.

There is (some sort of) converse to this, in that when $\pi$ is spherical (unramified), then it is determined by the Hecke operators at $p$, and the classical Hecke operators at $p$ come from the appropriate actions of diagonal matrices with powers of $p$, which are (in a non-technical sense) ``orthogonal'' to the copy of $\GL_n(\Z_p)$. (Warning, this is a very vague remark.) Of course, this connection is much deeper when $n > 1$ than when $n = 1$. For example, when $n > 1$, then $\pi$ is infinite dimensional, and so its restriction to $\GL_n(\Z_p)$ decomposes into infinitely many different representations. (each with finite multiplicity by admissibility). Here, I think, is one reasonable answer to your question:

Claim: $\rho$ is tamely ramified if and only if $\pi$ has an invariant vector under the $p$-congruence subgroup $K(p)$ of $\GL_n(\Z_p)$. The broader takeaway is that the image of inertia in $\rho$ is determined by the restriction of $\pi$ to $K = \GL_n(\Z_p)$.

Of course I can't really prove this because it requires (at least) the local Langlands correspondence itself. But let's talk about some examples, even for the case when $n = 2$ because you see most of the phenomena in this case. In fact, for many people, the first thing to do is just to understand the statement of local Langlands for $n = 2$ and $p > 2$ by just understanding the objects of both sides. Kevin Buzzard wrote a great note about this here: http://wwwf.imperial.ac.uk/~buzzard/maths/research/notes/old_introductory_notes_on_local_langlands.pdf and then he even gave a 20 lecture course as well: https://www.youtube.com/watch?v=Rv59aRUMfio In order to talk about it, I guess one has to say a little bit about what the representations $\pi$ really are. For $n = 2$, there are two types of basic construction: (parabolic) induction from the Borel $B$ of $\GL_2(\Q_p)$, and (compact) induction from representations of the maximal compact $K = \GL_2(\Z_p)$. (I won't even try to keep track of normalizations which I always find confusing). Let's also denote by $I$ the Iwahori subgroup of $K$ and by $P$ the pro-$p$ Iwahori (the inverse image in $K$ of the unipotent subgroup the Borel of $\GL_2(\F_p)$), and by $K(p)$ the full congruence subgroup of $K$.

Tamely Ramified Principal Series: Let $\chi_1$, $\chi_2$ be a pair of tamely ramified admissible characters of $\Q^{\times}_p$, which one can think of as a character of the torus $T$ and thus the Borel $B$, and let $\pi$ be the corresponding principal series representation, which is roughly the induction from $B$ to $G$ of the corresponding character of $B$. Suitably defined, $\pi$ has invariants under the pro-$p$-Iwahori subgroup $P$ of $K$. Note that $P$ is normalized by $I$, and thus $\pi^{P}$ has an action of $I/P$. But $I/P$ is nothing but $(\F^{\times}_p)^2$, and this acts on $\pi^{P}$. The assumption that $\chi_i$ are tame means that their restriction to $\Z^{\times}_p$ give characters on $\F^{\times}_p$, and hence the pair naturally give a character of $I/P$. If $\pi$ is irreducible (as it will be most of the time) then $\pi^{P}$ is $1$-dimensional and $I/P$ acts exactly by this character. So $\pi$ never has invariants under the Iwahori unless the $\chi_i$ are both unramified. In this case, one is either just talking about a unramified principal series, or a (possibly unramified twist of) the Steinberg representation, which have invariants under $K$ and $I$ respectively.

(tamely ramified) Supercuspidals: If we want something with $K(p)$-invariants, we can take a representation $\tau$ of $K/K(p) = \GL_2(\F_p)$ and take the compact induction from $K$ to $G$. If you take $\tau$ to be induced from a Borel then you again get a $\pi$ which has invariants under the pro-p Iwahori. But if you take $\tau$ to be one of the "exotic" representations of dimension $p - 1$, then $\tau$ turns out to have no $P$-invariant vectors. The compact inductions in these cases now turn out to be irreducible. Let's compare $\pi$ on the automorphic side with the Weil-Deligne side. On the automorphic side, I always think of $\tau$ as "almost" constructed as follows: take a character $\psi$ of the non-split Cartan (which is isomorphic to $\F^{\times}_{p^2}$), and take the induction to $\GL_2(\F_p)$. Of course, this is not quite correct, because inducing something from the non-split Cartan is not irreducible, but it at least sees $\tau$ and other similar representations. At any rate, this automorphic side is related to the induction of a character $\psi$ of $\F^{\times}_{p^2}$. But now the Weil-Deligne side is very similar --- the representation is exactly induced from a tamely ramified character ($\psi$) of the quadratic unramified extension of $\Q_p$. But, by class field theory, on inertia $\psi$ is tamely ramified and hence the same as a character of the units in the residue field $k^{\times} = \F^{\times}_{p^2}$.

These examples actually exhaust all the $\pi$ which are tamely ramified. Even for $\mathrm{GL}_n(\Q_p)$, the situation is similar. There are representations built up from smaller Levi subgroups, and there are the supercuspidals. But at least the tame supercuspidals are "easy" in that we know on the Weil-Deligne side they should be irreducible (and tamely ramified) and hence induced from an unramified cyclic degree $n$ extension of $\mathbf{Q}_p$. (The supercuspidals were constructed by Howe - the construction requires at least one understands the irreducibles of $\mathrm{GL}_n(\mathbf{F}_p)$.) And on the automorphic side they can be constructed from compact inductions from "known" representations of $\mathrm{GL}_n(\mathbf{F}_p)$.

Back to conductors: You might guess that the conductor of $\pi$ is given by the minimal $p^n$ such that $\pi$ has an invariant vector under $\Gamma(p^n)$. But in fact this is not quite correct --- the automorphic conductor is the minimal $p^n$ such that $\pi$ has an invariant vector under the group $\Gamma_1(p^n)$. In the case of the tamely ramified principal series representation the conductor is thus $p^2$ if $\chi_i$ are both ramified and $p$ if exactly one of $\chi$ is ramified. In particular, $\pi$ will have a twist of conductor $p$. Similarly, taking ramified principal series which may now be wildly ramified, their explicit construction shows that the conductor in the automorphic sense is the product (or sum if you take the exponent) of the conductors of $\chi_1$ and $\chi_2$. And similarly, the conductor on the "Galois" side is the same. On the other hand, the tamely ramified supercuspidals have conductor $p^2$, and all their twists have conductor $p^2$. In particular, they have no invariant vectors under the pro-$p$ Iwahori subgroup $P$ (this can be checked more directly). On the Galois side, the representation is tamely ramified but irreducible so also has conductor $p^2$.

  • $\begingroup$ Thanks enormously Lycurgus! This is very helpful. $\endgroup$ May 27, 2019 at 12:06
  • 1
    $\begingroup$ One small correction to thsi excellent answer: that a matrix is conjugate to its $q$th power does not force the eigenvalues $\lambda$ to satisfy $\lambda^{q-1}=1$, but rather to satisfy $\lambda^{q^n-1}=1$ for some $n$ at most the dimension of the vector space. $\endgroup$
    – Will Sawin
    May 27, 2019 at 23:52
  • $\begingroup$ @WillSawin: Completely correct, thanks! $\endgroup$ May 28, 2019 at 0:07
  • $\begingroup$ Should $P$ be the inverse image of nilradical of a Borel? $\endgroup$
    – Dr. Evil
    May 28, 2019 at 0:33
  • 1
    $\begingroup$ $$I = \left( \begin{matrix} * & * \\ 0 & * \end{matrix} \right) \mod p$$ is the Iwahori, $$P = \left( \begin{matrix} 1 & * \\ 0 & 1 \end{matrix} \right) \mod p$$ is the pro-$p$ Iwahori (= $p$-Sylow subgroup) $$K(p) = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) \mod p$$ is the congruence subgroup. If $\pi^{I}$ is non-trivial, inertia is unipotent, if $\pi^{P}$ is non-trivial (so there are "Iwahori monodromic vectors"), then inertia is unipotent after an abelian extension of $\mathbf{Q}_p$. If $\pi^{K(p)}$ is non-trivial, then inertia is tamely ramified. @Dr.Evil $\endgroup$ May 28, 2019 at 3:19

Lycurgus's answer is excellent. Here is some rambles for those who are still confused by the terminology.

Weil-Deligne vs. $\ell$-adic representations Let $\pi$ be a continuous $\ell$-adic representation of $W=W_F$. Then $\pi$ is said to be unramified if it is trivial on the inertia $I$. It is tamely ramified with unipotent monodromy if it is trivial on $I^+=I^{>0}$ and $I$ acts unipotently. (Note some people assume the unipotent monodromy but don't explicitly say it and just refer to these as tame. This is a bad habit but so entrenched now that I'm not sure there is anything we can do about it).

Now let us recall Grothendieck's theorem: if $\pi$ is a continuous $\ell$-adic representation of the Galois group, then some open subgroup of $I$ acts unipotently. Deligne encoded this unipotent part in the $N$ of the corresponding Weil-Deligne representation. Thus, if $\pi$ is unramified, then $N$ is trivial because $I$ is acting trivially. On the other hand, if $\pi$ is tamely ramified with unipotent monodromy, then the entire action of $I$ is encoded in $N$.

The upshot is that a Weil-Deligne representation $\rho$ of $W$ is unramified if it is trivial on $I$ and in addition $N$ acts trivially. On the other hand, $\rho$ is tamely ramified with unipotent monodromy if it is trivial on $I$ (but no condition on $N$).

Representations with Iwahori fixed vectors Now regarding the second question: first of all why should unramified Galois representations be in bijection with smooth irreducible representations with $G(\mathcal{O})$ fixed vector? This boils down to the Satake isomorphism. In other words, Satake isomorphisms establish the unramified Langlands correspondence. After this was pointed out by Langlands, it was natural to try to push it a bit further. On the Galois side, in view of Grothendieck's theorem, it is natural to consider the tame condition mentioned above. Then the question becomes what should tame Galois representation with unipotent monodromy correspond to on the automorphic side?

Let us note that when comparing a tame + unipotent vs. unramified representation, the only difference is in addition of a certain nilpotent Jordan block. Now on the automorphic side, Borel noticed that (roughly speaking) irreducible smooth representations with Iwahori fixed vectors can be built by Borel induction of unramified characters and that there is a partition classifying what kind of representation one gets (this partition comes from the choice of character $(\chi_1,\cdots, \chi_n)$ of the torus; for instance, if all the $\chi_i$'s are equal, we get the partition (n), which corresponds to the principal unipotent element. The resulting smooth representation is called a Steinberg representation).

Given the above story, it was natural to guess that tamely ramified Galois representations with unipotent monodromy should correspond to irreducible smooth representations with Iwahori fixed vectors. I think something that was surprising was that to make things work, one actually needed to add representations of a certain finite group, aka Lusztig's correction to the Deligne-Langlands conjecture. (Perhaps you can explain to us why this correction is not surprising. I guess there is an explanation from Lusztig's work on the representation theory of finite groups). Nowadays, one knows that this finite correction is part of the story of L-packets.

Tame but non-unipotent monodromy Finally, there are variants if we relax the conditions a bit. So for instance, instead of Iwahori fixed vector, let us suppose that we have an Iwahori monodromic vector, i.e., a vector $v$ such that $x(v)=\chi(x)v$ for all $x$ in the Iwahori, where $\chi$ is a character of the Iwahori. Then there is a version of Kazhdan-Lusztig story in this setting (as noted by Bezrukavnikov in his IHES paper). I'm not sure how to describe the corresponding Galois representations. The monodromy is no longer unipotent, so what is the right condition?

Note that this Iwahori monodromic vector is actually fixed under the pro-unipotent radical of the Iwahori. Now we can relax our condition even further and require that we have a fixed vector under pro-unipotent radical of some parahoric. (Lycurgus calls this pro-p Iwahori). Then on the Galois side, we get an $\ell$-adic representation trivial on $I^+$ (and no other conditions). This is the depths zero local Langlands correspondence, stated for $\mathrm{GL}_n$ in Lycurgus's answer.

  • 1
    $\begingroup$ As explained in my answer, this definition of "tamely ramified" is in irresolvable conflict with its usual usage in number theory. In particular, for $\mathrm{GL}(1)$, you would say that a tamely ramified quadratic character ($p \ne 2$) is not tamely ramified. I explained why (for Galois representations of local fields) tamely ramified is a strictly weaker condition than being unipotent. Can you explain why you are not making this distinction? My only guess as to how this useage seems to have arisen is that these differences (or their analogues) are not present in geometric Langlands. $\endgroup$ May 27, 2019 at 21:29
  • $\begingroup$ Hi Lycurgus, I agree with you that the terminology I used was not good (though very prevalent). I modified my answer to reflect your suggestion. $\endgroup$
    – Dr. Evil
    May 27, 2019 at 23:10
  • $\begingroup$ To clarify, I was complaining about the poor terminology [due to others] rather than your answer! $\endgroup$ May 28, 2019 at 0:22
  • 1
    $\begingroup$ @Lycurguscup This difference is very much present in geometric Langlands. In the geometric setting, unipotent representations are more or less completely understood, while general tamely ramified (w/ the definition from number theory) representations haven't received that much attention at present. Furthermore, the unipotent representations seem to be more directly relevant to understanding "classical" geometric representation theory. My guess is that because there hasn't been much GRT work on general tamely ramified reps, "tamely ramified w/ unipotent monodromy" started getting abbreviated. $\endgroup$
    – dhy
    May 28, 2019 at 0:42
  • $\begingroup$ @dhy I concur that the distinction is present and important in geometric Langlands. I also can buy "tamely ramified w unip. monodromy" becoming "tamely ramified" over time. Thanks! $\endgroup$ May 28, 2019 at 1:06

This is more of an extended comment. If I'm understanding the question correctly, I don't believe it's correct to say that tamely ramified Galois representations with unipotent monodromy correspond precisely to representations with Iwahori fixed vector, outside of say $GL_n$ -- it seems to me one can't characterize reps with an Iwahori-fixed vector in terms of only their Langlands parameter.

[Edit: More precisely, there are strictly more unipotent tamely ramified representations than reps with an Iwahori-fixed vector. This is precisely analogous to the situation with Chevalley groups: there are more representations which are unipotent in the Deligne-Lusztig sense, i.e. are attached the trivial semisimple conjugacy class in the dual group, than there are unipotent principal series, i.e., reps with a Borel-fixed vector. This is accounted for by Lusztig's generalized Springer correspondence -- geometrically, it's the statement that not all equivariant perverse sheaves on the nilpotent cone are generated by the Springer sheaf, outside of $GL_n$.]

Kazhdan-Lusztig's result identifies representations with Iwahori-fixed vector as having unipotent Langlands parameter - i.e. q-commuting pairs of a semisimple and nilpotent in the dual group. However representations aren't classified by their Langlands parameter - rather one should think of them as suitable sheaves on a stack of Langlands parameters, and one has to identify the action of the stabilizer (i.e. the L-packet). (This I think is a modern form of local Langlands correspondence, one seeks to identify categories of reps of groups over local fields as or inside suitable categories of sheaves on stacks of Langlands parameters). Kazhdan-Lusztig say we only get SOME of the objects in the corresponding L-packets: the representations of stabilizers that appear are only those that take part in a form of the Springer correspondence. This has to do with the existence of cuspidal local systems on Levis (as in the generalized Springer correspondence) and hence is invisible for $GL_n$.

Lusztig then proved [Classification of unipotent representations of simple p-adic groups. Internat. Math. Res. Notices 1995, no. 11, 517–589] an "affine generalized Springer correspondence". To paraphrase: unipotent Langlands parameters exactly correspond to unipotent representations -- a generalization of reps with an Iwahori fixed vector, in which the notion of Iwahori fixed vector (given by the trivial representation of the finite torus, hence of the finite Borel, hence of Iwahori) is replaced by a unipotent cuspidal representation of a Levi of the corresponding Chevalley group.

This distinction (unipotent vs Iwahori-fixed) does not arise in geometric Langlands -- it appears that the categorical analogs of unipotent representations have Iwahori-fixed vectors. This is e.g. well understood for reductive groups (rather than loop groups), where all unipotent representations come from decategorifying the finite Hecke category. Likewise, if you formally take the categorical trace of Frobenius on a (suitably mixed!!) version of Bezrukavnikov's Langlands duality for the Iwahori-Hecke category (cited by Dr. Evil), you get (up to singular support issues) the full category of coherent sheaves on the stack of unipotent Langlands parameters -- i.e. the home (thanks to Lusztig's theorem) of all unipotent reps of our original group.

  • $\begingroup$ There are groups other than $\mathrm{GL}_n$? $\endgroup$ Aug 14, 2019 at 1:43
  • $\begingroup$ Thanks BZ! I haven't fully digested what you're saying yet, but I certainly forgot the rep of the centralizer of the pair. I added a sentence to the question to reflect this. $\endgroup$ Aug 19, 2019 at 11:19
  • $\begingroup$ @GeordieWilliamson - sure, tried to clarify with an edit $\endgroup$ Aug 19, 2019 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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