User david helm - MathOverflow

Answer by David Helm for Reconciling Lusztig's results with the Langlands philosophy

2013-04-23T19:21:12Z

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.

Invariants of reductive group actions and completion

2012-11-26T16:21:12Z

I'm trying to understand the extent to which taking invariants of a reductive group action "commutes" with completion. More precisely:

Let $X = \operatorname{Spec} A$ be a reduced finite type affine scheme over $\operatorname{Spec} R$, where $R$ is a discrete valuation ring. Let $G$ be a reductive algebraic group with connected fibers over $\operatorname{Spec} R$ that acts on $X$, and let $G'$ be the formal completion of $G$ at the identity in the special fiber.

Now if $x$ is a point in the special fiber of $X$, defined over the residue field of $R$, and corresponding to a maximal ideal ${\mathfrak m}$ of $A$, the action of $G$ on $X$ induces an action of $G'$ on the formal completion $X_x$ of $X$ at $x$. Let ${\mathcal O}_{X,x}^{G'}$ denote the ring of functions on $X_x$ that are invariant for this action.

If we let ${\mathcal O}_X^G$ denote the ring of $G$-invariant functions on $X$, and let ${\mathfrak m}'$ be the intersection of ${\mathfrak m}$ with ${\mathcal O}_X^G$, then we have a natural map:

$$\left({\mathcal O}_X^G\right)_{\mathfrak m'} \rightarrow {\mathcal O}_{X,x}^{G'}$$

[Here the subscript ${\mathfrak m'}$ denotes completion at ${\mathfrak m'}$.]

Under what conditions is this map an isomorphism?

It's clear that the map fails to be injective if there is an irreducible component of $X$ that meets the closure of the orbit of $x$ but does not contain $x$: the left hand side ``sees'' this component whereas the right hand side does not. Will the map always be injective if $x$ lies on every irreducible component that meets the orbit closure of $X$?

Will the map be surjective in general?

I am happy to make additional assumptions to guarantee that the map is an isomorphism. (In particular, in the applications I have in mind, $G = {\operatorname{GL}}_n$, $R$ is a $p$-adic integer ring, and $X$ is a complete intersection that is flat over $\operatorname{Spec} R$.)

Answer by David Helm for Is the $\ell$-adic cohomology of a non-proper variety unramified at good primes?

2012-06-18T22:31:05Z

I think the weaker statement should be true. Here's a sketch of an argument: by compactification theorems and resolution of singularities, there is a smooth proper scheme $Y$ over $k$ containing $X$ as an open subscheme, such that $Y \setminus X$ is a divisor $D$ with simple normal crossings. Let $D_1, \dots, D_r$ be the irreducible components of $D$. Then any $p$-fold intersection of the $D_i$'s is smooth and proper over $k$.

There should be a spectral sequence, in terms of the etale cohomology of $\overline{Y}$ and that of the intersections of the $D_i$'s, that abuts to the etale cohomology of $\overline{X}$. Thus the etale cohomology of $\overline{X}$ should be unramified at any prime of good reduction for $\overline{Y}$ and all of the intersections of the $D_i$'s. I imagine you could also use this to show that at such primes the cohomology of $\overline{X}$ was isomorphic to the cohomology of the reduction.

David

Blocks of the category of representations of $GL_n({\mathbb F}_q)$

2011-04-07T15:43:56Z

Let k be an algebraically closed field of characteristic $\ell$, and let $q = p^r$ be a prime power with $p \neq \ell$. Suppose I have a cuspidal representation $\pi$ of $GL_n({\mathbb F}_q)$, for some $n < \ell$.

The supercuspidal support of $\pi$ consists of $m$ copies of a supercuspidal representation $\sigma$ of $GL_{d}({\mathbb F}_q)$ for some integers $m$ and $d$ with $md = n$.

In this setting, there also exists a cuspidal representation $\pi'$ of $GL_m({\mathbb F}_{q^d})$ that has supercuspidal support equal to $m$ copies of the trivial character.

Let $A$ be the block of the category of $W(k)[GL_n({\mathbb F}_q)]$-modules containing $\pi$, and let $A'$ be the block of the category of $W(k)[GL_m({\mathbb F}_{q^d})]$-modules containing $\pi'$. Are $A$ and $A'$ equivalent as categories? (More precisely, is there an equivalence of $A$ and $A'$ that takes $\pi$ to $\pi'$?) I can prove this when $\pi$ is supercuspidal; i.e. $m=1$, but computations from character theory and considerations from the Langlands program make me suspect that it's true in general.

David Helm

Comment by David Helm

2012-11-26T21:05:01Z

Thanks! I'm happy to assume that $G$ has connected fibers, and that $x$ is a rational point. I've edited the question to make things clearer.

Comment by David Helm

2012-11-26T16:33:15Z

$Q$ is called the unipotent radical of $P$.

Comment by David Helm

2012-06-19T15:32:20Z

It does indeed. Thank you!