All Questions
7 questions
8
votes
0
answers
233
views
$U_q(\mathfrak{g})$ is to knot theory as $U_q(\hat{\mathfrak{g}})$ is to $?$
Let $\mathfrak{g}$ be a finite dimensional semisimple Lie algebra over the complex numbers, e.g. $\mathfrak{sl}_n$.
Then every representation $\DeclareMathOperator\Rep{Rep}V\in \Rep U_q(\mathfrak{g})$ ...
7
votes
0
answers
116
views
Are there attempts to numerically finding algebraic structures over finite-dimensional vector spaces?
By "algebraic structure" I mean a finite set of linear operators between tensor products of copies of one (or more) finite-dimensional (complex or real) vector spaces, fulfilling a set of ...
10
votes
1
answer
806
views
DW, state sum models, and fully extended TQFTs
I am interested in state sum models and their relations with some other of TQFTs, especially the fully extended TQFTs and the Dijkgraaf-Witten TQFTs (generalized, in the sense that finite-group-...
6
votes
1
answer
741
views
Kauffman's state model for the Alexander polynomial, via representation theory
I've been reading Oleg Viro's paper on "quantum relatives of the Alexander polynomial" (arXiv:math/0204290), which, among other more general things, derives state-sum formulas for the Alexander ...
7
votes
2
answers
654
views
What is the state in the WRT TQFT associated to a handlebody?
Let $Y^3$ be a handlebody with boundary $\Sigma$. By definition, there is some associated vector $v_{WRT}(Y^3)\in Z(\Sigma)$, the (finite dimensional) Hilbert space associated to $\Sigma$ by the ...
18
votes
3
answers
3k
views
How to calculate the Witten-Reshetikhin-Turaev invariants from a triangulation?
I'm interested in the Witten-Reshetikhin-Turaev invariants for 3-manifolds, and in particular, how to calculate them from a triangulation of the 3-manifold (recall that as they were first introduced, ...
24
votes
3
answers
3k
views
How many definitions are there of the Jones polynomial?
Even with the connection to quantum groups being made clearer (I believe it was not known when the Jones polynomial was first introduced), it seems to me that still we don't have the "right" ...