(Asked in MSE but got no response)

The generating function $\frac{1}{(1-t)^N}=\sum_k {N+k-1\choose k}t^k=\sum_k h_k(1)t^k$ and the Jacobi-Trudi formula $s_{\lambda/\mu}=\det(h_{\lambda_i-i-\mu_j+j})$ tell me that the value of the skew Schur function at the identity is $$ s_{\lambda/\mu}(1_N)=\det\left({N+\lambda_i-i-\mu_j+j-1\choose \lambda_i-i-\mu_j+j}\right).\qquad (1)$$

However, I was reading a paper by Chen and Stanley (A Formula for the Specialization of Skew Schur Functions) and they state that $$s_{\lambda/\mu}(1,q,q^2,...)=\frac{1}{\prod_{u\in\lambda/\mu}[N+c(u)]_q}\det\left(\left\lbrack \begin{matrix} N+\lambda_i-i\\\lambda_i-i-\mu_j+j\end{matrix}\right\rbrack_q\right),\qquad (2)$$ where $c(u)$ is the content of the box $u$ in the Young diagram of shape $\lambda/\mu$ and the $q$-quantities are $[x]_q=1-q^x$, $[a]_q!=[a]_q[a-1]_q\cdots$ and $\left\lbrack \begin{matrix} a\\b\end{matrix}\right\rbrack_q=\frac{[a]_q!}{[b]_q![a-b]_q!}.$

I am not an expert in this $q$-business, and I am confused by this equation. I have a few closely related questions.

  1. since the left hand side of (2) is a polynomial in $q$, it should have a limit when $q\to 1$ and this should be the skew Schur at the identity. Is this correct? But what is the number of arguments?

  2. The determinant at the right hand side on (2) has a limit when $q\to 1$, but the prefactor does not. How to take the limit $q\to 1$ of this equation?

  3. How to obtain equation (1) from equation (2)?

  • 4
    $\begingroup$ The left hand side of (2) is not a polynomial but rather a power series. Don't confuse the full principal evaluation at infinitely many variables $1,q,q^2,\dots$ with the truncated principal evaluation $1,q,q^2,\dots,q^n,0,0,\dots $. With this in mind, it doesn't make sense to directly take a limit of (2) as $q\to 1$. If you want a framework to deal with both identities look at exercise 7.102 and its solution in Stanley's EC2. In particular the reference to the Jacobi-Trudi formula for flag Schur functions which generalizes both (1) and (2). $\endgroup$ Jul 20 '20 at 18:10
  • $\begingroup$ Usually $[x]_q=(1-q^x) /(1-q) $. $\endgroup$ Jul 20 '20 at 19:25
  • $\begingroup$ @GjergjiZaimi Is there a version of formula (2) that holds for a finite number of variables? I looked in the paper by Wachs that is mentioned in exercise 102 of Stanley, but didnt find it helpful at all. $\endgroup$
    – thedude
    Jul 20 '20 at 23:28

You might want to read up on the principal specialization.

It specializes a symmetric function into a formal power series. For example, the symmetric function $s_1(x) = x_1+x_2+ \dotsb$ has principal specialization $s_1(1,q,q^2,\dotsc) = 1+q+q^2+\dotsb = \frac{1}{1-q}$, so it does not make sense to let $q\to 1$. The expression only exist as a formal power series. However, you always compute the finite version, $s_\lambda(1,q,q^2,\dotsc,q^{n})$, which is gonna be a polynomial in $q$.

So, the difference between your Jacob-Trudi formula and the cited q-formula, is the number of variables (finite vs, infinite).

The principal specialization is in many instances nicer than the finite number of variables specialization.

  • 1
    $\begingroup$ Ok, so the limit $q\to 1$ is problematic when there are infinitely many variables. Is there a version of formula (2) that holds for a finite number of variables? $\endgroup$
    – thedude
    Jul 20 '20 at 23:29
  • $\begingroup$ Yes you combine the Jacob trudi with the finite principal specialization of the complete homogeneous symmetric functions $\endgroup$ Jul 21 '20 at 8:22

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