Specializations of Schur functions at consecutive integers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T07:53:36Z http://mathoverflow.net/feeds/question/549 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/549/specializations-of-schur-functions-at-consecutive-integers Specializations of Schur functions at consecutive integers Armin Straub 2009-10-15T02:40:16Z 2009-11-04T12:00:19Z <p>Given a partition &lambda; = (&lambda;<sub>1</sub>, &lambda;<sub>2</sub>, ..., &lambda;<sub>n</sub>) denote with s<sub>&lambda;</sub> the associated Schur function. There exists a nice product formula for the principal specializations: <p>s<sub>&lambda;</sub>(1, q, q<sup>2</sup>, ..., q<sup>n-1</sup>) = &Pi;<sub>i&lt;j</sub> (q<sup>&lambda;<sub>i</sub>+n-i</sup> - q<sup>&lambda;<sub>j</sub>+n-j</sup>) / (q<sup>j-1</sup> - q<sup>i-1</sup>). <p>Is a similar evaluation known for specializations of the type s<sub>&lambda;</sub>(1, 2, ..., n)?</p> http://mathoverflow.net/questions/549/specializations-of-schur-functions-at-consecutive-integers/4061#4061 Answer by Greg Kuperberg for Specializations of Schur functions at consecutive integers Greg Kuperberg 2009-11-04T07:14:41Z 2009-11-04T07:14:41Z <p>Let's take the strictest meaning of "similar evaluation", i.e., a product formula. The by the remark, the answer is probably no, because for instance the Stirling number s(9,3) = 118124 = 2*2*29531.</p> <p>A different meaning is whether there is some explicit combination of nested sums and products for a Schur evaluation at consecutive integers. I don't know the answer to that. Unlike for a straight product formula, it is difficult to look for such formulas or rule them out.</p> <p>The loosest possible meaning is whether there is an efficient way to evaluate s<sub>λ</sub>(1, 2,...,n). The answer to that is an easy yes, because you can use the Jacobi-Trudi determinant.</p>