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Schur polynomials $s_\lambda(x)$ have a determinantal expression. Using that, I know how to write the polynomial $s_\lambda(\frac{x}{1-x})=s_\lambda(\frac{x_1}{1-x_1},\frac{x_2}{1-x_2},...)$ as an infinite linear combination of other Schur polynomials.

Sadly, zonal polynomials do not have a determinantal expression. Still, I would like to write the zonal polynomial $Z_\lambda(\frac{x}{1-x})$ as an infinite linear combination of other zonal polynomials.

Does someone know how to do this?

Let me focus on a particular case. When $\lambda=(A,1^a)$ is a hook, the Schur function expansion is very simple. It involves only hooks and binomial coefficients: $$ s_{(A,1^a)}\left(\frac{x}{1-x}\right)=\sum_{B\ge A,b\ge a} (-1)^{a+b}{b\choose a} {B-1\choose A-1}s_{(B,1^b)}(x).$$

Maybe $ Z_{(A,1^a)}\left(\frac{x}{1-x}\right)$ has a similar expansion?

Let me report the result of some experimenting. If we write $Z_\lambda\left(\frac{x}{1-x}\right)=\sum_\mu C_{\lambda\mu}Z_\mu(x)$, then I am confident that $$ C_{(n),(m)}={m-1\choose n-1}\frac{(2n-1)!!}{(2m-1)!!}$$ and $$ C_{(1^n),(1^m)}=(-1)^{m+n}{m-1\choose n-1}\frac{(n+1)!}{(m+1)!}.$$ (I don't know how to prove these, they are conjectures). However, $C_{(3),(4,1)}=-\frac{23}{120}$ and 23 is prime, so a very simple general formula is not so likely.

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  • $\begingroup$ Wait, what is meant by $x/(1-x)$? Do you mean the plethystic substitution which evaluates the Schur in $y,y^2,y^3,\dotsc$ for all variables? $\endgroup$
    – Per Alexandersson
    Commented May 21, 2021 at 13:19
  • $\begingroup$ @PerAlexandersson I mean $Z_\lambda(\frac{x_1}{1-x_1}, \frac{x_2}{1-x_2},\cdots,\frac{x_n}{1-x_n})$ $\endgroup$
    – Marcel
    Commented May 21, 2021 at 13:31
  • $\begingroup$ So, there are power-sum expansions of the Jack polynomials known, and on these, your substitution is easy. However, expanding this back into Jack (or Zonal, for the case alpha=2) seems hard $\endgroup$
    – Per Alexandersson
    Commented May 21, 2021 at 15:51
  • $\begingroup$ @PerAlexandersson Yes, that possibility has occurred to me. It is doable, but quite messy. For Schur functions there is a more direct approach, using a scalar product. I was hoping something like that could also work for zonals. $\endgroup$
    – Marcel
    Commented May 21, 2021 at 16:07
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    $\begingroup$ @PerAlexandersson : The Schur function case is handled generally in arxiv.org/pdf/1601.01581.pdf. Look at corollary 3.5. These are special cases of canonical stable Grothendieck polynomials. $\endgroup$
    – user35313
    Commented May 24, 2021 at 18:09

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