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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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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 …