Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $ s_{\lambda} $ be the schur function associated to the partition $ \lambda $. Cauchy's identity (as in Macdonald) states that

$$ \sum_{\lambda} s_{\lambda}(X)s_{\lambda}(Y) = \prod_{i,j}(1-x_iy_j)^{-1} $$

where the sum is over all partitions. There is a generalization appearing in a paper by Ishikawa and Tagawa http://www.uec.tottori-u.ac.jp/~mi/papers/fpsac07b.pdf (Theorem 2.1(i)) which states that if $ X=(x_1,...,x_m) $ and $ Y=(y_1,...,y_m) $

$$ \sum_{\lambda} w^{\lambda_m}s_{\lambda}(X)s_{\lambda}(Y) = \frac{1-|X||Y|}{(1-w|X||Y|)\prod_{i,j=1}^m(1-x_iy_j)} $$

where the sum is over partitions $ \lambda=(\lambda_1,...,\lambda_m) $ and $ |X| = x_1x_2\cdots x_m $.

I am curious if there is a result along these lines which gives a closed form product expression for the generating function

$$ \sum_{\lambda} z_1^{\lambda_1}\cdots z_{m-1}^{\lambda_{m-1}}z_m^{\lambda_m}s_{\lambda}(X)s_{\lambda}(Y) $$

where the sum is still over partitions $ \lambda=(\lambda_1,...,\lambda_m) $. In particular, at least for now I care about such a generating function in the form

$$ \sum_{\lambda} w^{\lambda_{m-1}}z^{\lambda_m}s_{\lambda}(X)s_{\lambda}(Y) $$

It is my understanding that such a result can likely be discovered via the RSK correspondence, but this is not really my field. Thus before I delve into possibly slowly reinventing the wheel I thought I'd ask.

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.