Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Linear representations of algebras and groups, Lie theory, associative algebras, multilinear algebra.
2
votes
Accepted
Classifying lisse conformal vertex algebras using singularities of associated varieties
Before I answer your question, let me begin with a brief rant about terminology. The term "lisse" is bad when applied to vertex algebras:
It is French for "smooth", but has nothing to do with smooth …
3
votes
Accepted
Lie algebras and pulled back group schemes
As the link in Erica's comment shows, you can find this in SGA3 Exp. 2, but it is not so easy to extract from the very general language. Here is a rough guide: From Definition 3.9.0, the Lie algebra …
3
votes
Existence of orbifold vertex algebras – current status?
The question of constructing $G$-orbifold vertex algebras amounts to the problem of producing a suitable multiplication operation on a sum of modules for the fixed-point vertex algebra $V^G$. The par …
2
votes
How small can a group with an n-dimensional irreducible complex representation be?
Given a finite group, the sums of squares of dimensions of irreducible representations add up to the order of the group, so the dimension of an irreducible representation is at most the square root of …
4
votes
Accepted
Examples of simple vertex operator algebras (VOAs)
I expect there will never be a classification of simple VOAs, unless perhaps one is only sorting according to very rough criteria. This is because there are too many of them - even the rational case …
17
votes
Does the 3875-dimensional rep of $E_8$ have a solution to $x\star x=0$?
Yes.
The basic representation of $E_8$ has character $j(\tau)^{1/3} = q^{-1/3}(1+248q+4124q^2 + \cdots)$, and the 4124 decomposes as $1+248+3875$. By Frenkel-Kac-Segal, the basic representation has …
4
votes
Accepted
Classification of quasi-lisse vertex algebras
I do not have a complete answer to your questions, but this is what I can say for now:
Question 1: A classification is impossible (see the response to question 3).
Question 2: Additional examples ar …
4
votes
Accepted
Relationship between irreducible representations of the Schur covering group and elements of...
The answer to your question is Yes. Consider your covering group $C$ as a central extension:
$$1 \to N \to C \to G \to 1$$
and suppose it is given by a 2-cocycle $\alpha \in H^2(G, N)$. Then for any …
7
votes
$\text{Rep}(D(G))$ as representation category of a vertex operator algebra
A lot has happened in the last four years, and we now have lots of positive results.
The current state of knowledge is given in Evans-Gannon, "Reconstruction and Local Extensions for Twisted Group Do …
11
votes
1
answer
363
views
Can we glue characteristic 0 and characteristic p representations of a finite group given eq...
Suppose I have a prime $p$ and a finite group $G$ together with representations $\sigma: G \to GL_n(\mathbb{Q}_p)$ and $\pi: G \to GL_n(\mathbb{F}_p)$. My question is:
When does there exist a rep …
5
votes
0
answers
97
views
Is there a composite-order generalization of the homomorphism on Rep(Z/p) giving total dimen...
Let $p$ be a prime, let $\mathbb{Z}_p$ be the ring of $p$-adic integers, and let $G$ be a cyclic group of order $p$. It is rather well-known that finite rank $\mathbb{Z}_p$-free representations of $G …
8
votes
Conformal blocks in genus zero
Well, the claim is bogus, so you can't expect the proof to hold much water. On the other hand, it may be instructive to try filling in details to see why it fails.
First of all, we can't define conf …
5
votes
Accepted
Irreducible representations of Heisenberg algebra
If we just consider central representations, i.e., those for which $z$ acts by a nonzero scalar, then up to a certain kind of equivalence (given by conjugation with algebra isomorphisms) there is a un …
8
votes
Accepted
Does an element in the center of universal enveloping algebra becomes a scalar in irreducibl...
No. Let $G$ be $O_2(\mathbb{R})$, so the Lie algebra is one dimensional, and the center of the universal enveloping algebra is the symmetric algebra of the Lie algebra. Then the usual 2-dimensional …
5
votes
Accepted
Any representation is a subrepresentation of a direct sum of the regular representation
This is the unique Lemma in section 3.5 of Waterhouse's "Introduction to Affine Group Schemes". It only requires that $G$ be an affine group scheme over a field.