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Think of a Young diagram as a collection of rows with numbers of elements $\mu_1 \geq \mu_2 \geq \cdots \geq \mu_d \geq \mu_{d+1}=0$ (and $\mu_k=0$ for $k>d$) and define for $s=(i,j)$ (where $i$ moves along the column, $j$ along the row; it makes sense also for $s$ outside $R$) $$ l_R(i,j) := \mu_j^t-i, \qquad a_R(i,j) := \mu_i-j $$ where $R^t$ is transpose (namely rotate by 90 degrees).

Then I would like to prove that $$ Z:= \prod_{s\in R_1} \prod_{s'\in R_2} \frac1{1 - b(l_{R_2}(s) +a_{R_1}(s)+1)} \frac1{1 + b(l_{R_1}(s') +a_{R_2}(s')+1)} $$ can be written as (notice the transpose) $$ Z=\prod_k \frac1{(1+b k)^{C_k(R_1,R_2^t)}}$$ if we define $$ f_R(q):= \sum_{i=1}^d \sum_{j=1}^{\mu_i} q^{j-i} $$ and $$ f_{R_1}(q) + f_{R_2}(q) + f_{R_1}(q)f_{R_2}(q) (q+q^{-1}-2) =:\sum_k C_k(R_1,R_2) q^k $$

Origin: the desired equality follows from equality of Nekrasov instanton partition function in supersymmetric gauge theory, as written e.g. in the Nakajima paper and by Nekrasov in the original paper.

There should also be some relation with Jack polynomials and Plancherel measure. The connection with representation theory comes from the fact that the above is computed as a virtual representation

Refs:

http://arxiv.org/abs/math/0306198 thm 2.11

http://arxiv.org/abs/hep-th/0310235 appendix B

http://arxiv.org/abs/hep-th/0206161

http://member.ipmu.jp/yuji.tachikawa/transp/masterthesis.pdf eqs 4.82,83,84

http://arxiv.org/abs/hep-th/0306238 section 3.1

The book Symmetric functions by Macdonald

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  • $\begingroup$ This is a potentially interesting question, but it would be helpful to have some context. For instance, you use the tag representation theory (and not combinatorics!) so presumably there is some representation lurking here, but you don't spell it out... $\endgroup$ Feb 7, 2015 at 15:12
  • $\begingroup$ Sorry, what's your question? I don't think "Please write the proof of this combinatorial lemma for me" is an appropriate use of the site. How the heck did you come up with this if you don't already know the proof? $\endgroup$
    – Ben Webster
    Feb 7, 2015 at 16:26
  • $\begingroup$ that's Nekrasov SU(N) instanton partition function, as computed from localisation on the ADHM moduli space (using basically a formula by Nakajima for the tangent space to a fixed point), to be proved equal to the original result obtained by Nekrasov himself. let me add some refs $\endgroup$
    – jj_p
    Feb 7, 2015 at 17:02

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