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
add comment

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.