Short version: I wondering how to simply check if a proposed Markov trace, $\phi$ had the correct property using techniques similar to those from the Akutsu-Wadati 1987 paper `Exactly solvable models and new link polynomials. I: N-state vertex models'. The property $\phi(AB)=\phi(BA)$ appears to be simple while I am unaware of a simple and general way to check the other property.

Long version: My question is about constructing a Markov trace associated with a particular solution to the braid equation. For the braid group $B_{n}$ (for $n$ strands with generators $b_{i}$), the Markov trace, $\phi$, is required to satisfy two properties:

1) $\phi(AB) = \phi(BA)$ for $A,B \in B_{n}$,

2) $\phi(Ab_{n}) = \tau\phi(A)$ for $A \in B_{n}$ for some $\tau$ independent of $n$.

Using the Akutsu-Wadati 1987 paper, the say that given a representation of the generators, $b \rightarrow \tilde{b} \in \mbox{End}(V\otimes V)$ for some vector space $V$, then it is possible to define Markov trace as $\phi(A) = \mbox{Tr}(H\tilde{A})$ ($\tilde{A}$ denotes the representation of the braid word $A$). Tr is the usual matrix trace while $H=h\otimes h \cdots \otimes h$ with $h\in \mbox{End}(V)$. To test the first property one needs merely to check $$ \left[\tilde{b}, h\otimes h\right] = 0. $$

Does anyone know of a simple test for the second condition? I think $$ \mbox{id} \otimes \mbox{Tr} [\tilde{b} (I_{V} \otimes h)] \propto I_{V}, $$ is sufficient but from what I have read this is not true for the Jones polynomial.

If anyone knows of other simple systematic methods for constructing link invariants I would be interested to find out. Also, would the procedure generalise if the representation mapped $b \rightarrow \tilde{b} \in \mbox{End}(V\otimes V\otimes V)$

  • $\begingroup$ The paper by Turaev that I cited in the answer to this question mathoverflow.net/questions/206448/… should be relevant for this one as well. For invertible $h$, the partial trace condition that you give is exactly what you need. I am not aware of a classification of the pairs ($\tilde b, h$) satisfying both these conditions. $\endgroup$ – Gandalf Lechner Mar 14 '18 at 17:30

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