8
$\begingroup$

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!

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.