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1. The original goal is to bound the character ratio $r_\lambda(\vec{x}) := \chi_\lambda(x_1, \ldots, x_n) / d_\lambda$ from above, so that we get $$ d_\lambda r_\lambda(\vec{x})^k = o(1),$$ for $k = O(1)$, uniformly for all $\lambda \neq 0^n$. Here the condition on $\vec{x} \in \mathbb{T}^n $ is that $\mathbb{P}(\mid\sum_i x_i \mid > y) < e^{-c y^2}$, that is, sub-Gaussian.

2. More specifically, I can actually assume that $\mathbb{P}(\mid \sum_i x_i^j \mid > y) < e^{-c_j y^2}$ for all $j < J$, a fixed constant that can be as large as I need. By Fourier analysis on the circle, this should imply that $\vec{x}$ viewed as a point mass distribution on $\mathbb{T}$. is close to a mixture of clumped uniform distributions $\nu_{j, \theta}$ on $\mathbb{T}$ with high probability, where $\nu_{j, \theta} := \frac{1}{j} \sum_\ell \delta_{e^{i(\theta + 2\pi \ell / j)}}$.

3. More conveniently, let $x_i = e^{\sqrt{-1} t_i}$. The above conditions should imply that $\mathbb{P}(\mid \#\{t_i \in [a, b]\} - \frac{b-a}{2\pi}\mid > \epsilon)$ is small. Then I can divide $[0, 2\pi)$ into $J$ equal contiguous segments, and let $\nu_J$ be the distribution with point mass of $1/J$ at each midpoint of the $J$ segments, and let $\vec{\xi_J}$ be the corresponding $n$-vector on $\mathbb{T}$ that has $\nu_J$ as its empirical distribution. The bigger $J$ is, the smaller $\chi_\lambda(\vec{\xi_J})$ and can be estimated by assuming the contours of $z_j$ to be centered circles of the same radius, at least for type A.

4. To estimate the contour integral in the case of $s$-picket fence distribution of $x_k$'s, we stipulate that the contour for each $z_j$ is a centered circle with radius $r_j$. Then the following formula gives an upper bound on the maximum of the integrand: $$\mid \prod_j \prod_{k=1}^s (1 - z_j e^{2\pi i k / s})^{-n/s} \prod_{j < k} (z_j - z_k) \prod_{j=1}^n z_j^{-\lambda_j + j -n -1} \mid \ll \prod_j (1 - r_j^s)^{-n/s} n^{n/2} \prod_{j=1}^n r_j^{-\lambda_j - 1}.$$ Here we use the result from this MO thread.

5. Optimizing the expression in 4 above, we find that the optimal $r_j^s = \frac{\lambda_j + 1}{n + \lambda_j + 1}$.

6. If we let $s=2$, then for $\lambda =\delta = (n-1, n-2, \ldots, 0)$, then bound obtained for $\Phi_\lambda$ above is up to lower order term exactly $2^{n^2/2}$, which agrees with $d_\lambda$, but is insufficient to prove $d_\lambda (\Phi_\lambda / d_\lambda)^{O(1)} = o(1)$.

7. Macdonald p. 47 has an expression for $s_{(a+1, 1^b)}$ in terms of an alternating sum of product of $h_s$ and $e_t$: $$s_{(a+1, 1^b)} = h_{a+1} e_b - h_{a+2} e_{b-1} +\ldots + (-1)^b h_{a+b}.$$

1. The original goal is to bound the character ratio $r_\lambda(\vec{x}) := \chi_\lambda(x_1, \ldots, x_n) / d_\lambda$ from above, so that we get $$ d_\lambda r_\lambda(\vec{x})^k = o(1),$$ for $k = O(1)$, uniformly for all $\lambda \neq 0^n$. Here the condition on $\vec{x} \in \mathbb{T}^n $ is that $\mathbb{P}(\mid\sum_i x_i \mid > y) < e^{-c y^2}$, that is, sub-Gaussian.

2. More specifically, I can actually assume that $\mathbb{P}(\mid \sum_i x_i^j \mid > y) < e^{-c_j y^2}$ for all $j < J$, a fixed constant that can be as large as I need. By Fourier analysis on the circle, this should imply that $\vec{x}$ viewed as a point mass distribution on $\mathbb{T}$. is close to a mixture of clumped uniform distributions $\nu_{j, \theta}$ on $\mathbb{T}$ with high probability, where $\nu_{j, \theta} := \frac{1}{j} \sum_\ell \delta_{e^{i(\theta + 2\pi \ell / j)}}$.

3. More conveniently, let $x_i = e^{\sqrt{-1} t_i}$. The above conditions should imply that $\mathbb{P}(\mid \#\{t_i \in [a, b]\} - \frac{b-a}{2\pi}\mid > \epsilon)$ is small. Then I can divide $[0, 2\pi)$ into $J$ equal contiguous segments, and let $\nu_J$ be the distribution with point mass of $1/J$ at each midpoint of the $J$ segments, and let $\vec{\xi_J}$ be the corresponding $n$-vector on $\mathbb{T}$ that has $\nu_J$ as its empirical distribution. The bigger $J$ is, the smaller $\chi_\lambda(\vec{\xi_J})$ and can be estimated by assuming the contours of $z_j$ to be centered circles of the same radius, at least for type A.

4. To estimate the contour integral in the case of $s$-picket fence distribution of $x_k$'s, we stipulate that the contour for each $z_j$ is a centered circle with radius $r_j$. Then the following formula gives an upper bound on the maximum of the integrand: $$\mid \prod_j \prod_{k=1}^s (1 - z_j e^{2\pi i k / s})^{-n/s} \prod_{j < k} (z_j - z_k) \prod_{j=1}^n z_j^{-\lambda_j + j -n -1} \mid \ll \prod_j (1 - r_j^s)^{-n/s} n^{n/2} \prod_{j=1}^n r_j^{-\lambda_j - 1}.$$ Here we use the result from this MO thread.

5. Optimizing the expression in 4 above, we find that the optimal $r_j^s = \frac{\lambda_j + 1}{n + \lambda_j + 1}$.

6. If we let $s=2$, then for $\lambda =\delta = (n-1, n-2, \ldots, 0)$, then bound obtained for $\Phi_\lambda$ above is up to lower order term exactly $2^{n^2/2}$, which agrees with $d_\lambda$, but is insufficient to prove $d_\lambda (\Phi_\lambda / d_\lambda)^{O(1)} = o(1)$.

7. Macdonald p. 47 has an expression for $s_{(a+1, 1^b)}$ in terms of an alternating sum of product of $h_s$ and $e_t$: $$s_{(a+1, 1^b)} = h_{a+1} e_b - h_{a+2} e_{b-1} +\ldots + (-1)^b h_{a+b}.$$

$$ \Phi_\lambda := \oint \ldots \oint \prod_{1 \le j < k \le n} (z_j - z_k) \prod_{j=1}^n \prod_{k=1}^n (1 - z_j x_k)^{-1} \prod_{j=1}^n z_j^{-\lambda_j + j - n -1} dz_1 \ldots dz_n.$$

This gives a representation of the Schur polynomial $s_\lambda(x_1, \ldots, x_n)$ according to my preliminary calculation, based on the homogenous symmetric polynomial version of the Jacobi-Trudi identity. Similar formulae seem to exist under the guise of generalized hypergeometric series, which can be defined via Jack symmetric polynomial. These can be viewed as generalization of Selberg integrals, via the conformal map $z = \frac{i + y}{i-y}$.

$$ \Phi_\lambda := \oint \ldots \oint \prod_{1 \le j < k \le n} (z_j^{-1} - z_k^{-1}) \prod_{j=1}^n \prod_{k=1}^n (1 - z_j x_k)^{-1} \prod_{j=1}^n z_j^{-\lambda_j + j - 2} dz_1 \ldots dz_n \\ =\oint \ldots \oint \prod_{1 \le j < k \le n} (z_j - z_k) \prod_{j=1}^n \prod_{k=1}^n (1 - z_j x_k)^{-1} \prod_{j=1}^n z_j^{-\lambda_j + j - n -1} dz_1 \ldots dz_n.$$

This gives a representation of the Schur polynomial $s_\lambda(x_1, \ldots, x_n)$ according to my preliminary calculation, based on the homogenous symmetric polynomial version of the Jacobi-Trudi identity; the $-2$ in the exponent of the first integral contains a $-1$ to shift $j$ to be $0$-based, and another $-1$ needed for the Cauchy integration formula. 

Similar formulae seem to exist under the guise of generalized hypergeometric series, which can be defined via Jack symmetric polynomial. These can be viewed as generalization of Selberg integrals, via the conformal map $z = \frac{i + y}{i-y}$.

10. To show $-\log d_\lambda / \log(\Phi_\lambda / d_\lambda) = O(1)$, we can re-parameterize the tableau $\lambda$ in terms of a weakly increasing sequence $0 = a_1 \le a_2 \le \ldots \le a_n$, where $\lambda_j = a_{n-j+1}$. Every $\lambda$ can be obtained from $\vec{a} = (0)^n$ by means of a sequence of right shifts, of the form $a_j \mapsto a_j + 1$ for $j \ge j_0$. Denote $\lambda[j_0]$ such a shifted tableau. The dimension under such shift can be bounded by $$ \frac{d_{\lambda[j_0]}}{d_\lambda} = \prod_{j < j_0 \le k}\frac{a_k + 1 - a_j + k -j}{a_k - a_j + k -j}.$$ If we again assume $|z_j| \equiv r_j$, and that $x_k = e^{2\pi i k / s}$, then with $\lambda_j^\circ := \lambda_j \wedge (n - \lambda_j)$, $$ \Phi_{\lambda^t} \ll \prod_j (1 + r_j^s)^{n/s} r_j^{-(\lambda_j^\circ + 1)} \\ \le \prod_j \left( \frac{n}{n - \lambda_j^\circ -1} \right)^{n/s} \left( \frac{\lambda_j^\circ + 1}{n - \lambda_j^\circ - 1}\right)^{-(\lambda_j^\circ + 1) / s}.$$

10. To show $-\log d_\lambda / \log(\Phi_\lambda / d_\lambda) = O(1)$, we can re-parameterize the tableau $\lambda$ in terms of a weakly increasing sequence $0 = a_1 \le a_2 \le \ldots \le a_n$, where $\lambda_j = a_{n-j+1}$. Every $\lambda$ can be obtained from $\vec{a} = (0)^n$ by means of a sequence of right shifts, of the form $a_j \mapsto a_j + 1$ for $j \ge j_0$. Denote $\lambda[j_0]$ such a shifted tableau. The dimension under such shift can be bounded by $$ \frac{d_{\lambda[j_0]}}{d_\lambda} = \prod_{j < j_0 \le k}\frac{a_k + 1 - a_j + k -j}{a_k - a_j + k -j}.$$ If we again assume $|z_j| \equiv r_j$, and that $x_k = e^{2\pi i k / s}$, then with $\lambda_j^\circ := \lambda_j \wedge (n - \lambda_j)$, $$ \Phi_{\lambda^t} \ll \prod_j (1 + r_j^s)^{n/s} r_j^{-(\lambda_j^\circ + 1)} \\ \le \prod_j \left( \frac{n}{n - \lambda_j^\circ -1} \right)^{n/s} \left( \frac{\lambda_j^\circ + 1}{n - \lambda_j^\circ - 1}\right)^{-(\lambda_j^\circ + 1) / s}.$$ Luckily the last expression involves $\lambda_j$ only rather than $\lambda_j - j$, so the inserting of a column in $\lambda$ due to the shift starting from $a_{j_0}$ corresponds to the following factor: $$ \left( \frac{n}{(n - j_0) \vee j_0 -1} \right)^{n/s} \left( \frac{j_0 \wedge (n - j_0) + 1}{(n - j_0) \vee j_0 -1}\right)^{-(j_0 \wedge (n - j_0) + 1) / s}.$$