8
$\begingroup$

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...

$\endgroup$
3
$\begingroup$

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 $$P(x_1,x_2,x_3)=s_\emptyset(x_1)(s_3(x_2,x_3)+s_{21}(x_2,x_3))+2s_1(x_1)s_2(x_2,x_3)+2s_2(x_1)2s_1(x_2,x_3)+s_3(x_1)s_\emptyset(x_2,x_3).$$

$\endgroup$
  • $\begingroup$ 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 Jun 23 '16 at 1:54
  • 1
    $\begingroup$ @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 Jun 23 '16 at 15:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.