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(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 $$ s_{\lambda/\mu}(1_N)=\det\left({N+\lambda_i-i-\mu_j+j-1\choose \lambda_i-i-\mu_j+j}\right).\tag{1}\label{1}$$

However, I was reading a paper by Chen and Stanley (A Formula for the Specialization of Skew Schur Functions) and they state that $$s_{\lambda/\mu}(1,q,q^2,\dotsc)=\frac{1}{\prod_{u\in\lambda/\mu}[N+c(u)]_q}\det\left(\genfrac[]{0pt}{}{N+\lambda_i-i}{\lambda_i-i-\mu_j+j}_q\right),\tag{2}\label{2}$$ 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 $\genfrac[]{0pt}{}a b_q=\frac{[a]_q!}{[b]_q![a-b]_q!}$.

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 \eqref{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 \eqref{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 \eqref{1} from equation \eqref{2}?

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    $\begingroup$ 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). $\endgroup$
    – Gjergji Zaimi
    Commented Jul 20, 2020 at 18:10
  • $\begingroup$ Usually $[x]_q=(1-q^x) /(1-q) $. $\endgroup$
    – Fedor Petrov
    Commented Jul 20, 2020 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
    Commented Jul 20, 2020 at 23:28

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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.

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    $\begingroup$ 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
    Commented Jul 20, 2020 at 23:29
  • $\begingroup$ Yes you combine the Jacob trudi with the finite principal specialization of the complete homogeneous symmetric functions $\endgroup$
    – Per Alexandersson
    Commented Jul 21, 2020 at 8:22

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