Timeline for Sum of Schur functions associated to self-conjugate partitions
Current License: CC BY-SA 4.0
6 events
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| Nov 27, 2022 at 23:01 | comment | added | Richard Stanley | I don't know of any symmetric function identities that would shed light on your problem. For $\sum_{\lambda=\lambda'} (-1)^{\frac 12(|\lambda|+\mathrm{rank}(\lambda))}s_\lambda$, see Exercise 7.29(a) of Enumerative Combinatorics, volume 2. For $\sum_{\lambda}s_\lambda$, where $\lambda$ ranges over certain partitions "close to self-conjugate," see Exercise 7.29(b). | |
| Nov 26, 2022 at 19:16 | comment | added | Peter Taylor | $(\dagger\dagger)$ follows from $\sum_{|\lambda| = n} \mathrm{dim}(\lambda) s_\lambda(\vec{x}) = s_{\Box}(\vec{x})^n$. The corresponding terms when grouping by $|\lambda|$ for self-conjugate partitions don't look straightforward in any of Sage's standard bases for symmetric functions. | |
| Nov 25, 2022 at 17:25 | comment | added | Jeanne Scott | Dear Mark, I don't know about specialization(s), and I would also be very happy to find out if nice closed form expression(s) exist in such case(s). Formula $(\dagger)$ can be found on page 13 in the following paper of M. Kazarian and S. Lando, see arxiv.org/abs/1512.07172 | |
| Nov 25, 2022 at 9:05 | comment | added | Mark Wildon | Please could you add a reference for $(\dagger)$? Do you know if there is a closed form under any non-trivial specializations, for example, $x_1 = 1, x_2 = q,\ldots, x_m=q^{m-1}, x_{m+1} = \ldots = 0$? | |
| Nov 25, 2022 at 0:50 | history | edited | Jeanne Scott | CC BY-SA 4.0 |
correct double dagger formula
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| Nov 24, 2022 at 23:18 | history | asked | Jeanne Scott | CC BY-SA 4.0 |