# On MacMahon's conjecture and a Schur function identity

Recently I am reading Professor Bressoud's book "Proofs and confirmations"。And chapter 4 of his book is about using Schur functions to prove Macmahon's conjecture on symmetric plane partitions: compute the genetaring function for plane partitions in box $B(r,r,t)$ that are symmetric about their first two coorbinates. His method is Macdonald's, which employs Weyl's determinant on root systems. The crucial step is to prove the following equation and put the right end into a product of the $x_i$'s.

$$\sum_{\lambda\subseteq \{t^r\}} s_{\lambda}(x_1,\cdots,x_r) = \frac{\det(x_i^{j-1}-x_i^{t+2r-j})}{\det(x_i^{j-1}-x_i^{2r-j})}.$$

where $\lambda$ runs over all partions contained in a $r\times t$ rectangle.

The proof in quite "tedious",because it involves a determinant expansion and their are a bunch of transformations of the summations. But with the RSK correspondence, its not hard to prove this identity:

$$\sum_{\lambda} s_{\lambda}(x_1,\cdots,x_r)=\prod_{i=1}^r\frac{1}{1-x_i}\frac{1}{1-x_ix_j}$$

here $\lambda$ runs over all partitions whose length is $\leq r$. (I learned this in Stanley's book ) and what we really want is the following

$$\sum_{\lambda\subseteq \{t^r\}} s_{\lambda}(x_1,\cdots,x_r) =?$$

(it's good to have a product of the $x_i$'s at the right hand.) So my questions is, can we find a more direct way to proof Macmahon's conjecture with the help of RSK correspondence, especially find a nicer way to express the

$$\sum_{\lambda} s_{\lambda}(x_1,\cdots,x_r)?$$

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What is $x_j$ in your second formula? Do you mean the second product is over $i < j$ or something? –  John Jiang Mar 15 '11 at 21:33