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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 …
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 …
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 …
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 …
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 …
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 …
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 …
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 …
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 …
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 …
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 …
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 …
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 …
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 …
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 …