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Suppose a polynomial of the form $$\prod_i^d \sum_j^p x_i^{f_j}$$ clearly symmetric, where $f_j\in \mathbb{N}$. There is a way to find the set of $f$ numbers such that this polynomial is Schur positive? I have been trying to build a Kashiwara crystal and impose conditions on the weight function but without success beyond the fact that $\lbrace f_0,...,f_p\rbrace$ must be a sequence of consecutive numbers so $f_{i+1}= f_{i}+1 \text{ or } f_{i+1}= f_{i}, i<p$.

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Given $f_1,\dots,f_p$ and $d\geq \max f_i$, a necessary and sufficient condition is that all zeros of the polynomial $\sum x^{f_j}$ are real. See Enumerative Combinatorics, vol. 2, Exercise 7.91. Note. Your necessary condition need not hold for small $d$. If $d=1$, then $\sum x_1^{f_j}= \sum s_{f_j}(x_1)$.

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  • $\begingroup$ Thank you very much! I've been struggling with this problem for a while and this answer is just what I was looking for. Regarding the condition $d\geq max f_i$, where this come from? $\endgroup$
    – Nicolas Medina Sanchez
    Commented May 2, 2021 at 7:58
  • $\begingroup$ You need the Schur functions $s_\lambda(x_1,\dots,x_d)$ to be linearly independent when $\lambda$ is a partition of $n\leq \max f_j$. This is equivalent to $d\geq \max f_j$. $\endgroup$
    – Richard Stanley
    Commented May 3, 2021 at 15:10
  • $\begingroup$ @RichardStanley Can you please take a look at my question here, based on a remark you made in one of your papers? mathoverflow.net/q/392279/5017 $\endgroup$
    – Keshav Srinivasan
    Commented May 9, 2021 at 5:35
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A similar case which I know about are the Boolean symmetric functions, introduced by L. Billera, S. Billey, and V. Tewari. They prove Schur positivity by using Chern plethysm, which I do not know much about. It is still open to to explain combinatorially the coefficients of the Schur expansion of the boolean symmetric functions.

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    $\begingroup$ I don't see why Boolean symmetric functions are a special case. $\endgroup$
    – Richard Stanley
    Commented Apr 29, 2021 at 22:49
  • $\begingroup$ Ah, now I see - each inner sum only depends on one variable. Yes, its different. It was the unusual $\prod \sum$-form that triggered the memory, and perhaps still one can find some other connection... $\endgroup$
    – Per Alexandersson
    Commented Apr 30, 2021 at 5:53

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