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Consider Schur polynomials $s_\lambda$ with $\lambda = (2m, m, m, \ldots, m, 0)$ and $\ell(\lambda) = n$ (that is, $\lambda$ has $n$ rows). Here $m \gg n$, which, for the sake of concreteness, let's assume to be $n^2$.

Let $z_1, \ldots, z_n \in \mathbb{T}$ be $n$ complex numbers of unit length, satisfying $\sum_{j=1}^n z_j^k = O(\log n)$, for all $1 \le k \le K$ and some absolute constant $K$.

Finally let $d_\lambda = \prod_{1 \le j < k \le n} (1 + \frac{\lambda_j - \lambda_k}{k - j})$ be the dimension of the corresponding $SU(n)$ irreducible representation. How can I show that $|s_\lambda(z_1, \ldots, z_n)| \le d_\lambda^{1 - \epsilon}$, for some positive $\epsilon$, for all sufficiently large $n$, provided that $K$ is sufficiently large?

If the above problem is still too hard, I will grant full credit to solutions of the problem replacing the moment conditions on $z_1, \ldots, z_n$ with $\sum_{j=1}^n z_j^k = 0$ for all $1 \le k \le 100$ (that is, $K = 100$). In fact, even result with $\epsilon$ tending to $0$ and $K$ tending to $\infty$ will be appreciated, as long as they don't converge too fast.

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  • $\begingroup$ Do you really want $K$ to be an absolute constant? And $s_\lambda$ is naturally a polynomial in $|\lambda|$ variables, not $n$ variables. $\endgroup$ Jul 10, 2016 at 23:29
  • $\begingroup$ @HughThomas yes K is an absolute constant? Why do you suspect it shouldn't? I looked at the determinant ratio definition of schur polynomials on wikipedia again and it seems they are indeed defined over n variables. Why do you think they should be defined over N variables where N is the sum of the parts of the partition? $\endgroup$
    – John Jiang
    Jul 10, 2016 at 23:34
  • $\begingroup$ On my first point, if $K$ is a constant, wouldn't you have to prove, in particular, the case $K=1$, where there is no condition on the power sums? So wouldn't that case include all cases with larger $K$ as well? (Or, if you rule out $K=1$, still, $K=2$ would include all larger cases.) $\endgroup$ Jul 10, 2016 at 23:49
  • $\begingroup$ On my second point, sorry, you're right that you can take any number of variables which is at least $n$. If we think in terms of fillings of semi-standard Young tableaux, then we are looking at semi-standard Young tableaux whose maximal entry is at most $n$. Because of the shape of $\lambda$, that's going to be very constraining. (In each of the first $m$ columns, there is exactly one number from 1 to $n$ that won't appear.) But, okay, there's nothing wrong with that. $\endgroup$ Jul 10, 2016 at 23:52
  • $\begingroup$ @ HughThomas Thanks for your first point. I meant to say K can be chosen as large as you want. This is more of a convenience. I believe even for $K=2$ the conjecture is true. I also replaced $c$ by just 1 to avoid confusion. $\endgroup$
    – John Jiang
    Jul 11, 2016 at 0:25

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