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Linear representations of algebras and groups, Lie theory, associative algebras, multilinear algebra.
45
votes
Accepted
learning Deligne-Lusztig theory
You have given no indication as to your background, so the following imagines you don’t know anything. I have purposely left interesting things out as this is designed to get you from 0 to DL theory. …
13
votes
Accepted
Analogy between product of conjugacy classes and irreps: is there analog of Thompson conject...
In the following article
Heide, Gerhard; Saxl, Jan; Tiep, Pham Huu; Zalesski, Alexandre E.
Conjugacy action, induced representations and the Steinberg square for simple groups of Lie type. Proc. …
7
votes
Accepted
The defining characteristic representations of Lie type groups
This result is false for a general connected reductive algebraic group $\mathbf{G}$ but is true if $\mathbf{G}$ is simple and simply connected. This was proved by Steinberg in Theorem 1.3 of the follo …
7
votes
How can classifying irreducible representations be a "wild" problem?
Well, to understand how this problem is wild it may be useful to contrast it with the situation of finite reductive groups where we do have a classification statement. The first part of this post cons …
7
votes
Character table of $S_7$
I am adding this because I am surprised no one has mentioned the wonderful CHEVIE package! This does precisely what the OP wants. For example.
gap> W:=CoxeterGroup("A",6);
CoxeterGroup("A",6)
gap> Dis …
6
votes
Accepted
Principal series of finite group of Lie type
So, I think the answer to your question is yes. This may not be the slickest proof but I think it works. Firstly let $\mathrm{pr}_G$ be the projection map from the space of all class functions to the …
5
votes
Accepted
Sum of skew characters over hooks and "odd" partitions
I know you were asking for a reference, and there may be better approaches, but just to offer one proof of your statement based on the Murnaghan–Nakayama formula. Assume $m>0$ then any skew tableaux $ …
5
votes
Accepted
Finite field analogue of representations in same packet have equal central character
This is quite an old question but I believe the answer to your question is given in Lemma 2.2 of Malle's paper "Height 0 characters of finite groups of Lie type" (2007) which is freely available onlin …
5
votes
Reg the motivation behind Lusztig-Vogan bijection
This isn't even vaguely an answer to your question but is more of a clarifying remark concerning the canonical quotient. Throughout I will write [Lus84] for Lusztig's orange book "Characters of reduct …
4
votes
Accepted
Equality of codimension under Lusztig-Spaltenstein induction
What you're asking for can be phrased entirely in terms of nilpotent orbits. If $\mathcal{O}$ is the nilpotent orbit of $H$ then the nilpotent orbit of $G$ you obtain via your process is the induced n …
4
votes
Character values at a cyclic permutation of a symmetric group
Here's a slightly modified version of Geoff's answer that doesn't use modular representation theory but just the representation theory of the symmetric group.
I'll assume the characters of $\mathfrak …
3
votes
Central idempotents from characters in Frobenius algebras (generalizing Lusztig arXiv:math/0...
I think you should check out Chapter 7 of Geck and Pfeiffer's book "Characters of Finite Coxeter Groups and Iwahori-Hecke Algebras". There, many properties of symmetric algebras are worked out in quit …
3
votes
Accepted
Is there any Lefschetz-like principle for representations of finite groups?
I was encouraged to make my comment an answer, so will do so.
If $G$ is a finite group and $\mathbb{K}$ is a field then many interesting results that can be proved using character theory can also be …
3
votes
Accepted
A bijection between Lusztig series induced by inflation
Your approach is correct and is proven in the book by Digne-Michel (in fact a more general statement is proven there). Indeed, by Proposition 13.22 in Digne-Michel we know that
$$R_{T\subseteq B}^G(\ …
3
votes
Accepted
Regular embeddings of reductive groups
I had cause to think about this exercise recently so I thought I’d write an answer. I think Jim’s answer is sufficient but as you seem to want more details I’ll provide them here. I am aware that you …