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Let $g$ be an element in the permutation group (symmetric group) $S_N$. Define the Markov trace of $g$ (denoted $\text{tr}_k g$) as $$\text{tr}_kg = k^\text{number of cycles in $g$} ,$$ which depends on $k$. 1-cycles (fixed points) are also counted in the cycle number. For example, in $S_3$ group, $$\text{tr}_k(1)(2)(3)=k^3,\\\text{tr}_k(3)(12)=k^2,\\\text{tr}_k(132)=k.$$ This trace natually arise as the character of the permutation representation on the $N$-fold tensor power of a $k$-dimensional vector space if $k$ is a positive integer. But I would like to consider a more general case where $k$ is a positive real number.

The qustion is: whether there is a matrix representation $D_k(g)$ of $g$ such that the matrix trace is equal to the Markov trace? $$\text{Tr}D_k(g)=\text{tr}_k g.$$ If there exist such matrix representation, is there an explicit construction of the representation (an explicit method to calculate each matrix element $D_k(g)_{ij}$)?

Everett You
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  • This fails even for $N=2$. You want a character $\chi$ such that $\chi$((12)) = 1$, however it must be a two dimensional representation and thus a sum of two trivial reps, two sign reps or the trivial and the sign. None of these have the required property. – Noah White May 15 '16 at 12:36
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    See http://mathoverflow.net/questions/228445/characters-of-permutation-groups for the real character (defined using $k$, not $x$), meaning the thing is a real linear combination of actual characters. The question there was whether it was a nonnegative real linear combination of irreducible characters (this depends on the value of $k$). – David Handelman May 15 '16 at 13:27
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    When $k$ (aka $x$) is a positive integer, this is the character of the permutation representation on the $N$-fold tensor power of a $k$-dimensional vector space (acting solely on the indices of the vector spaces). – David Handelman May 15 '16 at 13:30
  • Oops, sorry I completely ignored the $x$! Obviously this means my comment above is wrong. An easy way to see that the tensor product above gives the correct character is that the restrict of the $S_N$ action gives a permutation representation on the tensor product basis, which is the same as the set of words of length $N$ in the letters $1,2,\ldots,k$. The value of the character on a permutation is the number of fixed points of this permutation. – Noah White May 15 '16 at 13:58
  • @DavidHandelman Thanks for pointing out the other post. I am aware of the representation on $N$-fold tensor, that is actually how I come to this trace. But knowing this fact does not seem to help answer the question, in particular, what are the basis vectors and how the calculate the matrix elements. I suspect that such matrix representation does not exist. – Everett You May 16 '16 at 01:03
  • @EverettYou Let ${ e_1, \ldots, e_k}$ be a basis of $V$, then a basis for $V^{\otimes N}$ is given by elements of the form $e_{i_1} \otimes e_{i_2} \otimes \cdots \otimes e_{i_N}$. The $S_N$-action permutes this basis so the matrices are just the corresponding permutation matrices. – Noah White May 16 '16 at 09:42

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If by "matrix representation" you mean finite-dimensional complex representation, the answer is no. The characters of $S_N$ are integer-valued (see here).

Community
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Paul Gustafson
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