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 |
Quantum groups, skein theories, operadic and diagrammatic algebra, quantum field theory
2
votes
Pictorial explanation of Dynkin index and quadratic Casimir?
It really sounds like you'd want to look at works of Pierre Vogel, starting from The universal Lie algebra, and also check recent results that use the same setup, like this one.
4
votes
Bialgebra pairing on ring polynomial $K[x]$
The bialgebra pairing condition implies $(1,b_1b_2)=(\Delta(1),b_1\otimes b_2)=(1,b_1)(1,b_2)$, so in particular $(1,1)=(1,1)^2$, and $(1,1)$ is zero or one. Also by a similar calculation, $(b_1b_2,1 …
6
votes
Yang–Baxter explanation
I personally like this note of John Baez: http://math.ucr.edu/home/baez/braids/node4.html
2
votes
What is a Homotopy between $L_\infty$-algebra morphisms II
First of all, you might want to look at the MO question How to define the equivalence of Maurer-Cartan elements in an $L_\infty$-algebra?, since of course this is effectively what you need (morphisms …
1
vote
Why does the type-A subdivision algebra look like the Rota-Baxter algebra axiom?
This is too long for a comment.
The reason for my question in comments was as follows. In some cases (Gerstenhaber algebras, Poisson algebras, etc.) an operad describing a certain algebraic structure …
4
votes
Accepted
What is the relation between 2-Gerstenhaber, CohFT, and Gerstenhaber geometrically?
Well, for your question 1 you presumably may ask yourself first about a relationship between (shifted) Lie algebras and Lie 2-algebras. Lie 2-algebras of Hanlon and Wachs can be viewed as $L_\infty$-a …
1
vote
Non-associative deformation quantization
I figured out that in full generality this problem has no chance of leading to a different algebraic structure for which the given one is a quasi-classical limit (like it is for associative/Poisson): …
7
votes
Accepted
Does Manin's construction of non-commutative endomorphism algebra $\mathrm{End}(A)$ produce ...
Q1: This algebra is just the Manin black product of $A$ and $A^!$ (in other words, the Koszul dual of the Segre product of $A$ and $A^!$), and hence it is Koszul. (As requested, the Segre product of t …