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Linear representations of algebras and groups, Lie theory, associative algebras, multilinear algebra.

1 vote

Is the real Jacquet module of a Harish-Chandra module still a Harish-Chandra module?

As I understand it, the Jacquet module for $(\mathfrak g, K)$-modules is defined so as to again be a $\mathfrak g$-module, and in fact it is a Harish-Chandra module, not for $({\mathfrak g},K)$, but r …
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7 votes

missing Lie algebra action on vectors in p-adic case?

If one is doing smooth representations of the $p$-adic group $G$ on complex vector spaces, then the action is locally constant (small open subgroups act trivially on vectors), and so any interpretati …
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8 votes

classification of irreducible admissible representations of GL(n)

Let me focus on the case when $K$ is non-archimedean; the archimedean case is somewhat easier. There is a coarse classification, valid for any reductive group, into supercuspidals, and all the other …
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6 votes

Can an admissible SO(n) representation contain an SO(n-1) representation with infinite multi...

Not necessarily. For example if $n = 2$, then $SO(1)$ is the trivial group, and so if $\pi$ is infinite dimensional, it is the trivial representation of $SO(1)$ with infinite multiplicity. Now suppo …
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7 votes

Prerequisites for P-adic Representations

When I learned this material, 10 or so years ago, I read Cartier's article in Corvalis (linked to by Thomas in the comments above), and then Casselman's notes (mentioned by Ramin). My experience has …
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6 votes

Schemes of Representations of Groups

This is nothing like a complete answer, but some kind of extended comment: Firstly, it is usual to quotient out by the conjugation action of $GL(n)$ (so that one is really parameterizing isomorphism …
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3 votes

Problem with the proof of a corollary of Schur's lemma

Let $v \in V$ be non-zero. Then the span of the elements $g v$ ($g \in G$) is finite-dimensional, because $G$ is finite, and is a $G$-invariant subspace of $V$. Since $V$ is irred., it is just equal …
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2 votes

geometric Langlands for GL(1)

You start with a local system $\mathcal L$ on $X$. By taking symmetric powers, you get a local system $Sym^n \mathcal L$ on $Sym^n X$ for any $n$. Now if you choose $n\geq g,$ then the natural map $S …
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6 votes

applications of Plancherel formulae

The key fact which underlies the Plancheral formula (say for real semi-simple Lie groups) is Harish--Chandra's theorem describing the discrete series (when they exist, what their infinitesimal charact …
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13 votes

Basis-free definition of Casimir element?

This is an answer to the question about the general structure of the centre of the enveloping algebra: The centre of the enveloping algebra (in particular when $\mathfrak g$ is reductive) is one of t …
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9 votes

Is there analogue of Peter–Weyl theorem for non-compact or quantum group

Marty's answer discusses the Plancherel formula, and in a comment on his answer, I mentioned Harish-Chandra's work on the Plancherel formula in the case of reductive Lie groups. Yemon Choi's answer al …
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55 votes

Why aren't representations of monoids studied so much?

One standard example of a monoid is the monoid $\mathbb N$ of natural numbers. The monoid ring $\mathbb C[\mathbb N]$ is equal to the polynomial ring $\mathbb C[T]$; the study of this ring and its mod …
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13 votes
Accepted

Weil-Deligne representations: Two monodromy operators?

This second operator $N'$ comes from Arthur's $SL_2$; the less tempered the original $(\rho,N)$ is, the more non-trivial $N'$ is. Geometrically, it comes from the Lefschetz $SL_2$ acting on the cohom …
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102 votes

Is Fourier analysis a special case of representation theory or an analogue?

I would like to elaborate slightly on my comment. First of all, Fourier analysis has a very broad meaning. Fourier introduced it as a means to study the heat equation, and it certainly remains a ma …
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20 votes
Accepted

Representations in characteristic p

Yes, you are correct. The point is that a $p$-group acting in char. $p$ always has a fixed point (and so acts trivially on an irrep.). So every irrep. of $G$ in char. $p$ factors through $G'$, as you …
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