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

8 votes
1 answer
245 views

Hecke algebra relation versus $\operatorname{SL}_2$ trace relation

The quadratic relation in the (type $A$) Hecke algebra is $(T-t)(T+t^{-1}) =0$, which can be rewritten as $$ T-T^{-1} = t-t^{-1}$$ Suppose $A \in \operatorname{SL}_2(\mathbb{Q})$ with eigenvalues $a …
4 votes

Where does the definition of "Tower of Algebras" come from?

I don't know what the original motivation was (or where the original definition was). But one reasonable conceptual explanation for the definition is this: let $\mathcal C$ be a category with objects …
Peter Samuelson's user avatar
2 votes

S-matrix for the HOMFLY/Hecke category

This isn't a complete answer, but it might make some partial progress. First I'll slightly restate the question. If we decompose $S^3$ into two solid tori and write $C$ and $C^{op}$ for the Homfly ske …
Peter Samuelson's user avatar
12 votes

Proving that the Jones polynomial is q-holonomic

One statement that would imply that the colored Jones polynomials are q-holonomic involves the Kauffman bracket skein module $S_q(K)$ of the knot complement. This is a module over the skein module of …
Peter Samuelson's user avatar
16 votes
Accepted

Jones polynomial of the concatenation of two braids

Here is one reason not to expect such a relationship (although I'm not sure if it can be completed to a proof). The Jones polynomial $J_\sigma$ (roughly) comes from taking the trace of a linear map $A …
Peter Samuelson's user avatar
5 votes

If tensor product of representations is a representation, must we have a bialgebra?

I think the answer to (2) is "no." If $A = \mathbb C \langle x,y\rangle / ([x,y] = 1) $ is the Weyl algebra, then the category of $A$-modules has a tensor product given by $M \odot N := M \otimes_{\ma …
Peter Samuelson's user avatar
5 votes

Representing SU(3) with 3 ropes in 3 dimensions

This doesn't exactly answer the question, but it seems like it should be closely related. In general, if $G$ is a complex semi-simple Lie group, there is an associated braid group $B_{\mathfrak g}$ (e …
Peter Samuelson's user avatar
2 votes
0 answers
239 views

Resolution of singularities of this cubic surface?

Let $A = \mathcal O(Y)^{SL_2(\mathbb C)}$ be the ring of invariant functions on $Y := \mathrm{Hom}(\mathbb Z^2, SL_2(\mathbb C))$. We can identify $A$ with the quotient of $\mathbb C[x,y,z]$ by the id …
6 votes
2 answers
1k views

Translation functors for category $\mathcal O$

Let $\mathfrak g = \mathfrak n^- \oplus \mathfrak h \oplus \mathfrak n$ be a semisimple Lie algebra over $\mathbb C$ and let $\mathcal U$ be its enveloping algebra. Category $\mathcal O$ is the catego …
7 votes
0 answers
325 views

Algebra A with Spec(A) reduced and Rep_n(A) non-reduced

As always, corrections to my misconceptions/misstatements are appreciated. This question is related to the following one, but in this question the algebras considered are commutative: Non-smooth algeb …
5 votes
1 answer
567 views

Non-smooth algebra with smooth representation variety

A not necessarily commutative algebra A (over C, say) is called formally smooth (or quasi-free) if, given any map $f:A \to B/I$, where $I \subset B$ is a nilpotent ideal, there is a lifting $F:A \to B …