On the $U_{q}(osp(1|2n))$ and $U_{-q}(so(2n+1))$ uncoloured quantum link invariants

January 2, 2009

I’ve resubmitted my paper on the $U_{q}(osp(1|2n))$ AND $U_{-q}(so(2n+1))$ uncoloured link invariants to the Journal of Knot Theory and its Ramifications. Its abstract is below (in latex).

Update: (3/1/09) the paper has been accepted for publication.

Update: (7/1/09) a preprint of the paper has been published on the website of the School of Mathematics and Statistics, University of Sydney.

On the $U_{q}(osp(1|2n))$ and $U_{-q}(so(2n+1))$ uncoloured quantum link invariants

\begin{abstract}
Let $L$ be a link and $\Phi^{A}_{L}(q)$ its link invariant associated with the vector representation of the quantum (super)algebra $U_{q}(A)$. Let $F_{L}(r,s)$ be the Kauffman link invariant for $L$ associated with the Birman–Wenzl–Murakami algebra $BWM_{f}(r,s)$ for complex parameters $r$ and $s$ and a sufficiently large rank $f$.

For an arbitrary link $L$, we show that $\Phi^{osp(1|2n)}_{L}(q) = F_{L}(-q^{2n},q)$ and $\Phi^{so({2n+1})}_{L}(-q) = F_{L}(q^{2n},-q)$ for each positive integer $n$ and all sufficiently large $f$, and that $\Phi^{osp(1|2n)}_{L}(q)$ and $\Phi^{so({2n+1})}_{L}(-q)$ are identical up to a substitution of variables.

For at least one class of links $F_{L}(-r,-s) = F_{L}(r,s)$ implying $\Phi^{osp(1|2n)}_{L}(q) = \Phi^{so({2n+1})}_{L}(-q)$ for these links.
\end{abstract}


Colour algebras and colour traces

September 21, 2007

First year pure mathematics students learn about the trace of a matrix or a linear transformation, and in dealing with Lie superalgebras and quantum superalgebras, one also deals with the supertrace of a linear transformation, which is a generalisation of the usual trace.

A number of people (e.g. David McAnally and Peter Jervis) have looked at colour algebras, which are generalisations of superalgebras. Recall that a superalgebra starts with a Z2 graded vector space on which one defines a supertrace on linear transformations of the Z2-graded vector space. With a colour algebra, the idea is to start with a Zn-graded vector space and define a colour trace on linear transformations of the Zn-graded vector space.

Colour algebras didn’t go very far as they weren’t seen as very useful – however, I’ve been wondering if they are useful in obtaining new knot invariants, and I’ve been starting to look at Zn-graded vector spaces and colour traces again.

Colour traces are interesting and non-trival – even properly defining the phase factor one obtains when applying the permutation operation P to tensor products is not straightforward. If one fixes P2 = id, one might obtain the usual supertrace (I’m not certain about this) and I’ve started investigating what happens if you don’t fix P2 = id.


Final volume of the Proceedings of ICM 2006

August 29, 2007

I recently received the final volume of the Proceedings of ICM 2006. It’s on my work desk, open at the discussion of Grigory Perelman’s work on the Poincare conjecture. It’s very nice to be able to peruse it at work and very different to my work.


Question about the Kauffman knot polynomial

August 7, 2007

Let L be a link and FL(r,s) the Kauffman knot polynomial for L where r and s are parameters in the Birman-Wenzl-Murakami algebra BWMf(r,s).

Question: is FL(-r,-s) = FL(r,s) for all L?

If the answer is yes, then the link polynomials arising from the vector (fundamental) reprsentations of Uq(-q2n,q) and U-q(q2n,-q) are the same for all L. The answer to the question is yes where L is a link with corresponding braid group element (s1)m for all integers m, but I don’t know the answer for arbitrary L.


Terence Tao – the cosmic distance ladder

June 3, 2007

Terence Tao has put up a nice post about calculating astronomical distances – the cosmic distance ladder. Download his presentation and enjoy.


Binomial trees and a closed form expression for the terms of a recurrence relation

May 30, 2007

Recently, while playing with an integral of a probabilistic function possibly inspired by the binomial trees I’ve been reading about in financial mathematics books, I was attempting to find a closed form expression for the following recurrence relation: 

Let ck(n), k, n = 0, 1, 2, …, satisfy

ck(n) = ck-1(n) + ck-1(n-k)

subject to the boundary conditions

c0(0) = 1 and c0(n) = 0 for all n=1, 2, … Read the rest of this entry »


Fun problem to tackle

May 30, 2007

Here’s a little maths problem to tackle: Read the rest of this entry »