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 |
8
votes
Quantum group as (relative) Drinfeld double?
Yes! It is indeed, to get the quantum group $\mathcal{U}_q(\mathfrak{g})$, you have to...
start with a groupring $\mathcal{U}^0(\mathfrak{g})=\mathbb{k}[\mathbb{Z}^n]=\langle K_1,\ldots K_n\rangle_{ …
1
vote
1
answer
160
views
Example of tensor category with non-simple unit $J\to \mathbb{1} \to Q$ and suitably extensi...
Edit: Thanx very much to Neil Strickland for quickly explaining to us that the following cannot be realized over finite commutative $\mathbb{C}$-algebras, as I had originally asked.
I know that there …