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How can we show that $$\sum_{n = 0}^\infty q^n \operatorname{char}_T S^n(\mathbb C[x,y])= \prod_{p_1,p_2\geq 0}\frac{1}{1-t_1^{p_1}t_2^{p_2}q}$$ where $\operatorname{char}_T V$ denotes the character of a representation $V$ of the torus $T$, and $T$ acts on $x,y$ as $(t_1,t_2)$?

References: proposition 2.1 and above eq. (4.5)

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    $\begingroup$ Do you expect us to read these two papers in order to understand the notation? $\endgroup$
    – abx
    Nov 7, 2014 at 11:29
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    $\begingroup$ Can you explain what you tried already? Did you try to write a basis of $S^n(\mathbb{C}[x,y])$ consisting of eigenvectors of the torus, and see what kinds of formulas for characters you get? Did you try to expand the right hand side, and describe the coefficient of $q^n$ there? This question is rather elementary once you unwrap the definitions; Hilbert schemes have nothing to do with that, it is just simple combinatorics / linear algebra. $\endgroup$ Nov 7, 2014 at 14:48
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    $\begingroup$ Voting to reopen: I agree that, once the definitions are unwound, this is a question about elementary combinatorics, but I don't think it is an obvious one. $\endgroup$ Nov 7, 2014 at 15:39
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    $\begingroup$ Sketch of an answer: For each $(\boldsymbol{p}, \boldsymbol{q}) \in (\mathbb{Z}^n)^2$, we get a symmetric function $m_{\boldsymbol{p}, \boldsymbol{q}} = \sum_{\sigma \in S_n} \boldsymbol{x}^{\sigma(\boldsymbol{p})} \boldsymbol{y}^{\sigma(\boldsymbol{q})}$. Here bolded symbols denote vectors with $n$ entries, and $\boldsymbol{x}^{\boldsymbol{p}} = x_1^{p_1} \cdots x_n^{p_n}$. As $(\boldsymbol{p}, \boldsymbol{q})$ ranges over $(\mathbb{Z}^n)^2/S_n$, we get a basis of $S^n \mathbb{C}[x,y]$. The basis element $m_{\boldsymbol{p}, \boldsymbol{q}}$ has torus weight $t^{\sum p_i} u^{\sum q_i}$. $\endgroup$ Nov 7, 2014 at 15:45
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    $\begingroup$ What remain is to notice that $\sum_n \sum_{(\boldsymbol{p}, \boldsymbol{q}) \in (\mathbb{Z}^n)^2/S_n} z^n t^{\sum p_i} u^{\sum q_i} = \prod \frac{1}{1-z t^{p} u^{q}}$. $\endgroup$ Nov 7, 2014 at 15:47

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