# Represent matrix immanants using Schur functions

For each irreducible character $\chi^\lambda$ of the symmetric group $S_n$, the immanant of an $n\times n$ square matrix $A$ is defined as \begin{equation*} d_\lambda(A) := \sum_{\sigma \in S_n} \chi^\lambda(\sigma) \prod_{i=1}^n a_{i,\sigma(i)}. \end{equation*} Observe that for $\lambda=(1^n)$, $\chi^\lambda(\sigma)=(-1)^{\text{sgn}(\sigma)}$, so that $d_\lambda(A)=\det(A)$, while for $\lambda=(n)$, $d_\lambda=\text{per}(A)$.

I've gone through numerous papers in multilinear algebra (including those of Minc, Marcus, Pate, Merris, Watkins), as well as a superficial skimming of a few in representation theory. However, I have been unable to find (except for the implicit form (9) in this paper) an explicit representation of $d_\lambda$ using Schur functions.

Question. Does anybody know of an explicit formula that represents $d_\lambda$ using Schur functions $s_\lambda$ (or as a sum of Schur functions). I'd also be happy to know about the converse direction.

• Schur functions in what variables would you aim to obtain? Do you have an idea of what you want to get for $n=2$? for $n=3$? Mar 25 '15 at 11:21
• In the ideal case, we'd have $d_\lambda(A)$ represented as some combination of Schur functions evaluated at eigenvalues of $A$. E.g., if $\lambda=(1^n)$, $d_\lambda(A)=\det(A)=s_\lambda(a_1,\ldots,a_n)=\prod_i a_i$ ... also, I looked at "Littlewood's correspondence principle" but could not get something simple out of it. Mar 25 '15 at 13:43
• That ideal case certainly can't be hoped for - the permanent, for instance, is not invariant under conjugation. That's the reason I commented in the first place... Mar 25 '15 at 16:50
• @VladimirDotsenko: indeed, the ideal is not attainable, but perhaps a nonnegative sum over sufficient number of suitable "parts"? Mar 25 '15 at 17:09

I don’t know if this is proved anywhere. It looks like one can obtain a formula for the Schur functions in terms of immanants of replicated matrices. Let me introduce the following notation. Define a type $T$ as an ordered set $(i_1,\dots, i_n)$ such that $\sum_k i_k=n$. This defines a partition $\nu(T)$ obtained by rearranging the $i_k$ in decreasing order. For a type $T$, define $A_T$ as the matrix $A$ with the first row and column repeated $i_1$ times, second row and column repeated $i_2$ times etc. This is still an $n\times n$ matrix. Then we can obtain \begin{equation} s_\lambda(\mu_1\dots,\mu_n)=\sum_T d_\lambda(A_T)\,, \end{equation} where $\mu_i$ are the eigenvalues of $A$, $m_\lambda$ is the dimension of the symmetric group irrep $\lambda$ and the sum is over all types $T$ such that the partitions $\nu(T)$ have a non-zero Kostka number $K_{\nu,\lambda}\neq 0$.
To obtain this, we can use Schur-Weyl duality. First, write the immanant as \begin{equation} d_\lambda(A) = \sum_\sigma \chi_\lambda(\sigma) \langle 1,\dots,n| A^{\otimes n} |\sigma(1),\dots,\sigma(n)\rangle \end{equation} This is equal to \begin{equation} d_\lambda(A) = \frac{n!}{m_\lambda}\langle 1,\dots,n| A^{\otimes n}\Pi_\lambda | 1,\dots,n\rangle\,, \end{equation} where $\Pi_\lambda$ is the projector on to the isotypic space of the symmetric group irrep $\lambda$. This can now be written as \begin{equation} d_\lambda(A) = \frac{n!}{m_\lambda} \text{Tr}(A^{\otimes n}\Pi_\lambda |1,\dots,n\rangle\langle 1,\dots,n|) \,. \end{equation} Finally, this can be written as \begin{equation} d_\lambda(A) = \frac{1}{m_\lambda}\text{Tr}(A^{\otimes n}\Pi_\lambda \Pi_{1^n}) \,, \end{equation} where $\Pi_{1^n}$ is the projector onto the induced representation from the trivial representation of the Young subgroup corresponding to the partition $1^n$. This induced representation is sometimes called the permutation module of the partition $1^n$. The multiplicity of the symmetric group irrep $\lambda$ in this module is the Kostka number $K_{1^n,\lambda}$. In the last step, we have also used the fact that \begin{equation} \frac{1}{n!}\sum_\sigma \sigma |1,\dots,n\rangle\langle 1,\dots,n |\sigma^{-1} = \Pi_{1^n}\,. \end{equation} Let us also block diagonalize $A^{\otimes n}$. We then obtain \begin{equation} d_\lambda(A) = \frac{1}{m_\lambda}\text{Tr}((I_\lambda\otimes A_\lambda)\Pi_\lambda \Pi_{1^n}) \,, \end{equation} where $I_\lambda$ is the identity on the symmetric group irrep space and $A_\lambda$ is the part in the $GL_n$ irrep space.
• Thanks Hari! How about $d_\lambda$ in terms of $s_\lambda$? Mar 25 '15 at 21:03