On MacMahon's conjecture and a Schur function identity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T04:35:36Z http://mathoverflow.net/feeds/question/58390 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58390/on-macmahons-conjecture-and-a-schur-function-identity On MacMahon's conjecture and a Schur function identity zhaoliang 2011-03-14T07:10:41Z 2011-12-30T23:00:56Z <p>Recently I am reading Professor Bressoud's book "Proofs and confirmations"ă€‚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.</p> <p>$$\sum_{\lambda\subseteq {t^r}} s_{\lambda}(x_1,\cdots,x_r) = \frac{\det(x_i^{j-1}-x_i^{t+2r-j})}{\det(x_i^{j-1}-x_i^{2r-j})}.$$</p> <p>where $\lambda$ runs over all partions contained in a $r\times t$ rectangle.</p> <p>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:</p> <p>$$\sum_{\lambda} s_{\lambda}(x_1,\cdots,x_r)=\prod_{i=1}^r\frac{1}{1-x_i}\frac{1}{1-x_ix_j}$$</p> <p>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 </p> <p>$$\sum_{\lambda\subseteq {t^r}} s_{\lambda}(x_1,\cdots,x_r) =?$$</p> <p>(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 </p> <p>$$\sum_{\lambda} s_{\lambda}(x_1,\cdots,x_r)?$$</p> http://mathoverflow.net/questions/58390/on-macmahons-conjecture-and-a-schur-function-identity/84616#84616 Answer by Benjamin Young for On MacMahon's conjecture and a Schur function identity Benjamin Young 2011-12-30T23:00:56Z 2011-12-30T23:00:56Z <p>I doubt it... as far as I know, the RSK correspondence isn't very well-behaved on the set of Young tableaux that you need. </p> <p>This doesn't entirely rule out the possibility of other "nice" proofs, though with current technology, I think you'll need to do a determinant evaluation at some point for this particular problem. That's not necessarily a bad thing, as there are some rather astonishing modern ways of evaluating determinants, especially those coming from tiling problems, plane partitions and the like. </p> <p>I'd suggest you take a look at C. Krattenthaler's inspiring papers "Advanced Determinant Calculus", <a href="http://arxiv.org/pdf/math.CO/9902004" rel="nofollow">http://arxiv.org/pdf/math.CO/9902004</a>, and "Advanced Determinant Calculus: A Complement", <a href="http://arxiv.org/pdf/math/0503507" rel="nofollow">http://arxiv.org/pdf/math/0503507</a>.</p>