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This is related to the question Hall-Littlewood functions and functions on the nilpotent cone, and arises in the construction of Coulomb branches of gauge theories. The motivation is explained at the bottom.

Let us prepare some notation. Let $\lambda$ be a dominant coweight of $GL(N)$, i.e., tuples of (not necessarily positve) integers $\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_N$. We set $|\lambda| = \lambda_1+\dots+\lambda_N$, and define $b_\lambda(t)$ (denoted by $P_{U(N)}(t,\lambda)$ in 1403.0585) by $$ \prod_{i\in\mathbb Z} \varphi_{m_i(\lambda)}(t), \qquad \varphi_r(t) = (1-t)(1-t^2)\cdots (1-t^r), $$ where $m_i(\lambda)$ denotes the number of times $i$ occurs as a part of $\lambda$. This is the inverse of the Poincare polynomial of the classifying space of the stabilizer of $\lambda$ in $GL(N)$ in view of 1601.03586.

Let $\lambda^1$, $\lambda^2$, $\dots$, $\lambda^{N-1}$ be dominant coweights of $GL(N-1)$, $GL(N-2)$, $\dots$, $GL(1)$. We define $ \Delta(\lambda,\lambda^1,\dots,\lambda^{N-1}) $ by $$ \frac12\left( \sum_{j=1}^{N-1} \sum_{i,i'} |\lambda^{j-1}_i - \lambda^{j}_{i'}| \right) - \sum_{j=1}^{N-1} \sum_{i < i'} |\lambda^j_i - \lambda^j_{i'}|, $$ where $\lambda^0 = \lambda$. We fix $\lambda$ and consider $$ H[T_{(1^N)}(SU(N)](t,x_1,\dots,x_N,\lambda) := x_1^{|\lambda|} \sum_{\lambda^1,\lambda^2,\dots,\lambda^{N-1}} t^{\Delta(\lambda,\lambda^1,\dots,\lambda^{N-1})} \times \prod_{j=1}^{N-1} \left(\frac{x_{j+1}}{x_j}\right)^{|\lambda^j|} \frac1{b_{\lambda^j}(t)}. $$ This is the monopole formula in 1403.0585 for the special case $\rho=(1^N)$. Now 1403.0585 claims that $$ H[T_{(1^N)}(SU(N)](t,x_1,\dots,x_N,\lambda) = t^{\frac{(N-1)|\lambda|}2 - n(\lambda)} R_\lambda(x_1,\dots,x_N;t) \prod_{i\neq i'} \frac1{1 - x_i^{-1}x_{i'} t} $$ where $$ n(\lambda) = \sum (i-1)\lambda_i, $$ $$ R_\lambda(x_1,\dots,x_N;t) = \sum_{w\in S_N} w\left( x_1^{\lambda_1}\cdots x_N^{\lambda_N} \prod_{i<i'} \frac{x_i - tx_{i'}}{x_i-x_{i'}}\right). $$ As far as I understand, this is checked numerically in many cases, but no rigorous proof is given.

I together with Braverman, Finkelberg am trying to give a geometric proof of this result, but we are wondering whether this is known, or has a purely combinatorial proof.

Here is the motivation. In 1503.03676 (see also 1601.03586), I conjectured that the space of sections of a line bundle over the cotangent bundle of the flag variety for $SL(N)$ is realized by the Borel-Moore homology of a certain space, whose Poincare polynomial has a combinatorial expression, known as a monopole formula by Cremonesi, et al. Since the former is given by Hall-Littlewood polynomials, we arrive at a combinatorial expression of Hall-Littlewood polynomials as above.

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Here is a simple observation (which you probably already know). If you send $t\to0$, then $b_\lambda$ in your formula disappears, and it remains to minimize $\Delta$. The minimum is then achieved on the weights $\lambda^j$ which interlace, i.e. $\lambda^j_i\ge \lambda^{j-1}_i\ge \lambda^j_{i-1}$, where I identified $\lambda$ with $\lambda^N$ instead of $\lambda^0$. For the interlacing weights the factor $t^\Delta$ disappears and what is left is the standard combinatorial formula for Schur polynomials.

By the way, doing this computation carefully in $N=2$ case, I feel that there might be a misprint in powers of $t$. Namely, we have $\lambda_1\ge \lambda_2$ and we sum over all $n$ satisfying $\lambda_1\ge n \ge \lambda_2$. Then $t^\Delta=t^{(\lambda_1-\lambda_2)/2}$, and so this is a bit different from the prefactor $t^{N(N-1)/2-n(\lambda)}$, which your formula suggests. So something like additional $t^{|\lambda|/2}$ factor is missing in this case

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  • $\begingroup$ Thank you very much for a remark. The prefactor was wrong as you said. It should be $t^{\frac{(N-1)|\lambda|}2 - n(\lambda)}$. (The original paper is correct. I made a mistake.) Your condition means $\lambda$, $\lambda^1$, $\dots$, $\lambda^{N-1}$ form a Gelfand-Tsetlin pattern. (I think the condition is $\lambda^{j-1}_i \ge \lambda^j_i \ge \lambda^{j-1}_{i-1}$.) $\endgroup$ – Hiraku Nakajima May 5 '17 at 0:52
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    $\begingroup$ Maybe this suggests that there is a combinatorial proof using the Gessel-Viennot trick. $\endgroup$ – Joel Kamnitzer May 5 '17 at 14:57

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