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

This is somewhat related to my last MO post:

sum of the character of the symmetric group

Let $p_n$ be the $n$-th Newton symmetric function, and $s_{\nu}$ be the Schur function indexed by the partition $\nu$. We know that $$\sum_{\nu} s_{\nu}(x) s_{\nu}(y) = \prod_{i,j} \frac{1}{1-x_i y_j}$$ where the summation is over all partitions, including the empty one.

We also have $$\sum_{\nu} s_{\nu}(x)= \frac{1}{\prod_i(1-x_i)\prod_{i>j}(1-x_i x_j)}$$ see Macdonald's Symmetric functions and Hall polynomials, page 76. These two formulas are both of "infinite sum = infinite product" type.

In my current research, I was encountered with the sum $H(t) := \sum_{\nu} e^{\frac{\kappa_{\nu}}{2} t}s_{\nu}(x)$, where $\kappa_{\nu} = \sum_{i=1}^{l(\nu)} \nu_i(\nu_i-2i+1)$. Note that $H(t)$ satisfies the cut-and-join equation $$\frac{\partial}{\partial t} H(t) = \Delta H(t)$$ with initial value $H(0) = \sum_{\nu} s_{\nu}$. Here $$\Delta := \frac{1}{2}\sum_{i,j}((i+j)p_i p_j \frac{\partial}{\partial p_{i+j}} + ijp_{i+j}\frac{\partial}{\partial p_{i}}\frac{\partial}{\partial p_{j}}) $$ is the cut-and-join operator. This is because Schur functions are the eigenfunctions of $\Delta$, namely, $$\Delta s_{\nu} = \frac{\kappa_{\nu}}{2} s_{\nu}$$.

My question is: Is there an infinite product expression for $H(t)$?

Any comment is welcome!

share|improve this question
add comment

1 Answer 1

The cut-and-join operator can be easely represented in termes of eigenvalues (see e.g. (58) here) so that one can act by exponential of this operator on the second formula at your question to get the required expression.

share|improve this answer
    
hi, Sasha. I don't quite understand your point, exonentiating the cut-and-join operator at H(0) produces H(t), but this is still an infinite sum, not an infinite product, which I originally wished for. –  Hanxiong Zhang Jan 8 '12 at 5:08
    
I am not sure if there is any good infinite product expression for H(t). But I think one can obtain at least some integral transform of such infinite product. Namely, the operator $\exp (t \Delta)$ can be represented as integral of the shift operator (as in the reference I gave you), which acts on the infinite product H(0) preserving its structure. –  Sasha Jan 9 '12 at 8:30
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.