# a variational problem related to weighted logarithmic capacity

Consider the following multiple contour integral:

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

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 $z = \frac{i + y}{i-y}$.

I am interested in the case where $x_i \in \mathbb{T}:= \{z \in \mathbb{C}: |z| = 1\}$ for all $i$. Such quantities arise naturally as irreducible characters of $U(n)$ for instance.

Even more restrictively, I am interested in the case where the centroid of the $x_i$'s is zero, that is, $\sum_{j=1}^n x_j = 0$. Thus define $V := \{(x_1, \ldots, x_n) \in \mathbb{T}^n: \sum_j x_j = 0\}$.

One way of generating such $x_i$'s is as a mixture of extremal configurations of the form $\{e^{\alpha_k + \frac{2\pi i j}{m_k}}: 1 \le j \le n_k\}$, such that $m_k \mid n_k$ and $\sum_k n_k = n$. As a special case, if $n$ is even, we can match any $n/2$ points on $\mathbb{T}$ with their antipodal points to satisfy the constraint. But there are many other ways to do it, as long as $n > 4$.

My question is whether this leads to an effective estimate of the quantity $$s^*_{\lambda, V} := \max_{(x_1, \ldots, x_n) \in V} |s_\lambda(x_1, \ldots, x_n)|.$$

In particular, my conjecture is that

$$s^*_{\lambda, V} \le \sqrt{s_\lambda(1, \ldots, 1)},$$ provided $\lambda \neq (k,\ldots, k)$ for some $k \ge 0$.

Fully credit will be given if one can show for some positive $\epsilon$,

$$s^*_{\lambda, V} \le s_\lambda(1,\ldots, 1)^{1 - \epsilon + o(1)},$$ under the same condition on $\lambda$.

The reason the conjecture should fail for such fully rectangular $\lambda$ is that the random walk on $U(n)$ generated by a conjugacy class from $V$ always lives on a single left-coset under the quotient by $SU(n)$. The fully rectangular representations correspond to the other irreducible components of the walk, whose traces are nonzero. I have actually managed to prove the conjecture with an asymptotically negligible error term when $\lambda$ consists of one row or one column only, that is, when $s_\lambda$ is either elementary or homogeneous, based on a convexity argument. In those cases the Vandermonde factor $\prod_{1 \le j < k \le n}(z_j - z_k)$ drops out, so the analysis is much easier.

I would be happy if someone can present a proof of the weaker conjecture in the case $\lambda = (kn, k(n-1), \ldots, k)$. For $k = 2$, and $n$ even, we can compute $s_\lambda(-1,1,\ldots, (-1)^j, \ldots, 1) = \sqrt{s_\lambda(1, \ldots, 1)}$, because (by Macdonald section 1.4 example 1) $$s_\lambda(x_1, \ldots, x_n) = \prod_{1 \le j < k \le n} (x_j^2 + x_j x_k + x_k^2),$$ and about half of the time $x_j, x_k$ have opposite signs, which contributes $1$ to the product. Same argument can be made for any even $k > 0$. For odd $k$, this yields $0$. This does not prove the conjecture for such $\lambda$'s, but the appearance of square root was encouraging.

Going back to the original multiple integral, in the special case $\lambda = (n,n-1, \ldots, 1)$. One way to estimate its norm is by putting absolute value around all factors, restricting all $z_j$'s to a circle of radius $r$, and estimating the maximal value the integrand can attain: $$\Psi(\vec{x},r) := \max_{\vec{\theta}} \prod_{1 \le j < k \le n} (r | e^{i \theta_j} - e^{i \theta_k}|) \prod_{j=1}^n \prod_{k=1}^n |1 - e^{i \theta_j} x_k|^{-1} r^{-n^2}.$$

For fixed $\vec{x}$ and $r$, $\Psi(\vec{x}, r)$ is related to the weighted logarithmic capacity on the unit circle, with the weight function given by $w(z) = -\sum_{k=1}^n \log| 1 - r z x_k|$.

Does the configuration $x_j = (-1)^j$ always maximize $\Phi(., r)$? It seems that for $r \approx 1$, $x_j = e^{2\pi i j/ 3}$ gives a higher value, which reminds me of the optimality competition between binary and ternary number systems. So a related question is whether $\Phi(., r)$ is maximized at an $\vec{x}$ of the form $x_j = e^{2 \pi i j / k}$ for some $k \mid n$.

Any partial result/insight/reference/comment is very welcome.

Edit (04/10/2016): For my personal note, I also derived, with the help of mathematica, the following multiple contour integral formula in the case of irreducible characters of SO(2n+1):

Recall the representations can also be indexed by weakly decreasing integer sequences $\lambda_1 \ge \lambda_2 \ge \ldots \ge \lambda_n = 0$. The dimension of $\rho_\lambda$ is given by $$\prod_{j < k} \frac{(\lambda_j - \lambda_k + k - j)( \lambda_j + \lambda_k + 2n + 1 - j -k)}{(2k - 2j + 1)(2k - 2j)} \prod_{j=1}^n (2\lambda_j + 2n + 1 - 2j).$$ This can be derived from the dimension formula derived in this paper, where the irreducible characters are indexed by weakly increasing integers, $0 \le a_1 \le a_2 \le \ldots \le a_n$ instead. I tried here to write it in similar form as the widely quoted dimension formula for dimensions of $U(n)$: $$d_\lambda = s_\lambda(1, \ldots, 1) = \prod_{j < k} \frac{\lambda_j - \lambda_k + k - j}{k - j}.$$

The analogues of Schur polynomial are polynomials in the variables $x_1, \ldots, x_n, x_1^{-1}, \ldots, x_n^{-1}, 1$, hence Laurent polynomials in $x_1, \ldots, x_n$, where $x_j = e^{i \theta_j}$ are the eigenvalues of elements of $SO(2n+1)$. This awesome paper gives analogues of Jacobi/Trudi/Giambelli's formula in the latter case. Using their formula (3.26), which comes from Fulton & Harris, $$o_N(\lambda, x) = \mid h_{\lambda_j - j + i}(x) - h_{\lambda_j - j -i}(x) \mid,$$ it is not hard to derive a similar multiple contour formula $$o_N(\lambda, x) = \oint \ldots \oint \prod_{1 \le j, k \le 2n+1} \frac{1}{1 - z_j x_k} \det(z_j^{-\lambda_j + j -i -1} - z_j^{-\lambda_j +j +i -1}) d\vec{z} \\ = \oint \ldots \oint \prod_{j=1}^n \prod_{k=1}^{2n+1} \frac{1}{1 - z_j x_k} \prod_{j=1}^n z_j^{-2n-2 - \lambda_j + j}(1 - z_j^2) \prod_{1 \le j < k \le n} (1 - z_j z_k)(z_j - z_k) d\vec{z}.$$ The last formula is extrapolated from Mathematica calculation which I haven't verified rigorously, but looks clearly true. Thus the only difference between the contour integrals for type A and type B and D is the term $\prod_{1 \le j \le n} (1 - z_j^2) \prod_{1 \le j < k \le n} (1 - z_j z_k)$, which definitely looks familiar in other contexts related to root systems.

A few points of confusion:

1. Fulton Harris derived the formula above for complex orthogonal group $SO(N, \mathbb{C})$. I learned once that it is easy to carry results from the complex case to the real case, but forgot how exactly to justify that. Also does the formula work for both even and odd $N$?

2. Is it true that $z_j$ ranges over $j \in {1, \ldots, n}$ and $x_k$ ranges over $k \in {1, \ldots, 2n+1}$? This seems the only reasonable interpretation.

3. Is the result in Abramsky, Jahn, and King cited by Fulton Harris simply the Giambelli's formula?

Answers to each of the above would be highly appreciated.

Some more personal notes:

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 $e_t$: $$s_{(a+1, 1^b)} = h_{a+1} e_b - h_{a+2} e_{b-1} +\ldots + (-1)^b h_{a+b}.$$

This combined with the observation that $\det X \le \prod_i \|X_i\|_2$ gives an effective bound for $\lambda$'s whose Durfee dimension is $o(n)$.

1. The main remaining limitation of the contour integral bound is with respect to elementary symmetric polynomials $e_s$. The fact that $e_n = 1$ is not well captured. In fact, if $\lambda'_j = \lambda_j + s$, for $1 \le j \le n$, then $|e_{\lambda'}| = |e_\lambda|$. This suggests that we consider the dual of $\Phi$ in terms of elementary polynomials: $$\Phi_{\lambda^t} = \Phi'_{\lambda^t} := \oint \ldots \oint \prod_{1 \le j < k \le m} (z_j - z_k) \prod_{j=1}^m \prod_{k=1}^n (1 + z_j x_k) \prod_{j=1}^m z_j^{-\lambda_j + j -m -1} dz_1 \ldots dz_m.$$ Here $m$ denotes the number of nontrivial columns in $\lambda$. We could have used $m$ for the definition of $\Phi$ as well, but it is especially important for $\Phi'$.

2. Similarly, we have the dual of Macdonald's alternating sum formula: $$s_{(a+1, 1^b)} = e_{b+1} h_a - e_{b+2} h_{a-1} + \ldots + (-1)^a e_{b+a}.$$ For $s > n/2$, we should estimate $e_s$ using $\Phi'$ by first exploiting $e_s(x_1, \ldots, x_n) = \prod_j x_j / e_{n-s}(x_1^{-1}, \ldots, x_n^{-1})$. We convert $e_s$ to $e_{n-s}$ when $s > n/2$, to obtain $$\Phi'_{\lambda^t} = \oint \ldots \oint \\ \prod_k [\prod_{j \in J} (1 + z_j x_k)z_j^{-\lambda_j + j -m -1} \prod_{j \notin J} ( 1 + z_j x_k^{-1}) z_j^{-n + \lambda_j + j -m -1}] \prod_{1 \le j < k \le m} (z_j - z_k) dz_1 \ldots dz_m.$$

Here $J = \{j \in [m]: \lambda_j < n/2\}$.

1. 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: $$\left( \frac{n}{(n - j_0) \vee j_0 -1} \right)^{n/s} \left( \frac{j_0 \wedge (n - j_0) + 1}{(n - j_0) \vee j_0 -1}\right)^{-(j_0 \wedge (n - j_0) + 1) / s}.$$