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Suppose we have a symmetric polynomial $P$ in $n$ variables. We can partition this alphabet into sets with one or two letters, e.g. ${ {x_1}, {x_2, x_3}}.

We can thus see $P$ as an element in $Q[x_1][x_2,x_3]$, and expand it in the Schur basis in each sub-set of variables. E.g, $P=5 s_{2}(x_1)s_{32}(x_2,x_3)$ or similar.

Suppose now that this expansion always have non-negative coefficients, for every choice of partition of letters into 1-or-2-element subsets.

Can we conclude that $P$ itself is Schur-positive in $n$ variables? Note that the Littlewood-Richardson rule tells us that the converse is true.

If not, is there some $k = k(n)$ such that Schur-positivity on subsets of size $\leq k$ implies Schur-positivity in $n$ variables?

The intuition behind why 2-element sets might be enough is as follows: If we want to show that a sum over some combinatorial objects is Schur-positive, it suffices to create a map to SSYTs, or equivalently, to reading-words which are Knuth-equivalent to the SSYTs. To completely describe a reading-word, it suffices to know how many times $i$ appear before $j$, for every pair $i$, $j$.

So, assume that I have a set of objects in bijection with SSYTs of shape $\lambda$. By expanding the sum over these it in different choices of 2-variable subsets, we can almost figure out this data above. But I cannot make this formal, at least without using type A crystals or similar...

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My intuition is that if $P$ is in three variables then your requirements only force $P$ to be unimodal in each pair of variables, while being Schur positive is much more restrictive. For example, take $$P(x_1,x_2,x_3)=s_3(x_1,x_2,x_3)+s_{21}(x_1,x_2,x_3)-s_{111}(x_1,x_2,x_3).$$ Equivalently, $$P(x_1,x_2,x_3)=m_3(x_1,x_2,x_3)+2m_{21}(x_1,x_2,x_3)+2m_{111}(x_1,x_2,x_3).$$ Now, partition the variables as $\{\{x_1\},\{x_2,x_3\}\}$, this partition is essentially unique. And then we have $$P(x_1,x_2,x_3)=s_\emptyset(x_1)(s_3(x_2,x_3)+s_{21}(x_2,x_3))+2s_1(x_1)s_2(x_2,x_3)+2s_2(x_1)2s_1(x_2,x_3)+s_3(x_1)s_\emptyset(x_2,x_3).$$

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  • $\begingroup$ Right, I had a suspicion of this as well. I wonder if it is the same for 3 variables, or if Schur positivity in parts of size less than or equal to 3 is enough... This is really close to Stembridges combinatorial crystal characterization, that basically verifies three variables at a time. $\endgroup$
    – Per Alexandersson
    Commented Jun 23, 2016 at 1:54
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    $\begingroup$ @PerAlexandersson actually, I think there is an "easy" counterexample if you even replace $2$ with any number. Let $P(x_1,\dots,x_{k+1})=s_{21^{k-1}}-s_{1^{k+1}}$. Then fixing any monomial in $m\leq k$ first variables corresponds to basically skewing both shapes by a column $1^m$, so applying the L-R rule gives the result $\endgroup$
    – Pavel Galashin
    Commented Jun 23, 2016 at 15:49

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