Explicit formulas for certain elements in $Z(U(\mathfrak{gl_n}))$

Question

Let $\lambda$ be a partition with $\leq n$ rows and let $L_{\lambda}$ be the corresponding irreducible representation of ${\rm GL}_n(\mathbb{C})$. Let $e_m(X_1,\dots,X_n)$ be the $m$th elementary symmetric polynomial. Is there a known formula for the element $c_m \in Z(U(\mathfrak{gl}_n))$ which acts on $L_{\lambda}$ by the scalar $e_m(\lambda_1,\dots,\lambda_n)$?

It is easy to check that $c_1 = E_{11} + \dots + E_{nn}$, so I suspect that $c_m$ should be related to the coefficients of the characteristic polynomial. Also, I am aware of the Harish-Chandra isomorphism, but have not been able to use it to produce a formula.

I would be particularly interested in references where this is worked out. I couldn't find it in any of the standard representation theory text books.

Answer 1

I believe the elements of $\mathcal{Z}(\mathfrak{g})$ you are looking for are (suitably renormalised) the Capelli elements; these come from the (renormalised) Capelli determinant \[C(u) = \det\left(E_{jk} + \left(u - \frac{n - 2j + 1}{2}\right) \delta_{jk}\right),\] where by the determinant of a matrix $A = (A_{jk})$ we mean the column determinant \[\det A = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) A_{\sigma(1),1} \cdots A_{\sigma(n),n}. \] For all $u$, $C(u)$ lies in $\mathcal{Z}(\mathfrak{g})$.

One can show that given an irreducible representation $\pi_{\lambda}$ of $\mathrm{GL}_n(\mathbb{C})$ of highest weight $\lambda = (\lambda_1,\ldots,\lambda_n)$, the Harish-Chandra isomorphism $\chi_{\pi_{\lambda}}$ acts on $\mathcal{Z}(\mathfrak{g})$ by sending $E_{jj}$ to $\lambda_j + \frac{n - 2j + 1}{2}$ and $E_{jk}$ to $0$ whenever $j \neq k$. So from the definition of $C(u)$, $\pi_{\lambda}(C(u))$ acts by the scalar \[\prod_{\ell = 1}^{n} (u + \lambda_{\ell}).\] Expanding $C(u)$ as a polynomial in $u$, we see that the coefficients of each power of $u$ are elements of $\mathcal{Z}(\mathfrak{g})$ that act by the scalar $e_m(\lambda_1, \lambda_2, \ldots, \lambda_n)$.