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.

  • $\begingroup$ 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 $n=3$? $\endgroup$ Mar 25 '15 at 11:21
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    $\begingroup$ 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. $\endgroup$
    – Suvrit
    Mar 25 '15 at 13:43
  • $\begingroup$ 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... $\endgroup$ Mar 25 '15 at 16:50
  • $\begingroup$ @VladimirDotsenko: indeed, the ideal is not attainable, but perhaps a nonnegative sum over sufficient number of suitable "parts"? $\endgroup$
    – Suvrit
    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.

Now it is clear what to do. If we add all the immanants for different types, we get \begin{equation} \sum_T d_\lambda(A_T) = \sum_T\frac{1}{m_\lambda}\text{Tr}((I_\lambda\otimes A_\lambda)\Pi_\lambda \Pi_{\nu(T)})= s_\lambda(\mu_1,\dots,\mu_n)\,. \end{equation} For the determinant, the sum over the other partitions do not matter since their projections onto the alternating irrep are zero. So we obtain that it is the Schur function in that case.

  • $\begingroup$ Thanks Hari! How about $d_\lambda$ in terms of $s_\lambda$? $\endgroup$
    – Suvrit
    Mar 25 '15 at 21:03
  • $\begingroup$ Hi Suvrit. Unfortunately, I don't know the answer to that question. This is the best answer I had to the converse direction that you mentioned in your question. $\endgroup$
    – Hari Krovi
    Mar 25 '15 at 23:33

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