Are Markov traces matrix traces? When starting this question I was very hesitant - literature on the subject is vast and I thought most likely the answer is already there somewhere.
Then when the list "Questions that may already have your answer" appeared, the first on the list was Constructing Markov traces simply which is from March 2012 and has no answer, and this encouraged me to go ahead and ask:
There are famous Markov traces on (group algebras of) braid groups, providing knot/link invariants and much more, and constructed via similar traces on finite-dimensional algebras like Birman-Wenzl/Kaufmann, Iwahori-Hecke, Temperley-Lieb algebras etc.
The question is the most naïve one - can these traces be realized as plain ordinary traces of matrices in linear representations of algebras in any of these cases?
 A: You can construct Markov traces explictly in tensor product representations of braid groups. If $V$ is a finite dimensional vector space and $R\in{\rm End}(V\otimes V)$ an invertible solution on the Yang-Baxter equation, represent the generators of the $n$ strand braid group $B_n$ on $V^{\otimes n}$ by $R_k:={\rm id}_V^{\otimes(k-1)}\otimes R\otimes{\rm id}_V^{\otimes(n-k-1)}$. The idea is now to pick $A\in{\rm End}V$ such that $[A\otimes A,R]=0$ and consider the functional $X\mapsto{\rm Tr}_{V^{\otimes n}}(A^{\otimes n}X)$. This is tracial on the representation generated by the $R_i$'s, and a Markov trace if also ${\rm Tr}_2(R\,(A\otimes A))=A$, where ${\rm Tr}_2={\rm id}_{\rm End V}\otimes {\rm Tr}_V$ denotes the partial trace. The pair $(R,A)$ is then called an enhanced Yang-Baxter operator, a good reference on the subject is the 1988 article "The Yang-Baxter equation and invariants of links" by Turaev in Inventiones. In case $A=1$, the condition amounts to $R$ having trivial partial trace, and the Markov trace is given by the usual matrix trace.
