Timeline for $2$-adic valuation of Schur $P$-functions in the power-sum basis
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 11, 2021 at 15:08 | vote | accept | Antoine Labelle | ||
| Sep 11, 2021 at 1:28 | answer | added | darij grinberg | timeline score: 4 | |
| Sep 5, 2021 at 14:33 | comment | added | Antoine Labelle | @darijgrinberg Thanks, your proof seems to work! You do use implicitly that $\lambda$ has distinct parts, since otherwise $Q_\lambda$ is zero, not $2^{\ell(\lambda)}P_\lambda$, so part 2 doesn't work anymlore (for example $P_{(1,1)}=\frac{1}{2}(p_1^2-p_2)$ so this hypothesis is really needed). Except for that, though, everything looks correct, so if you want to turn your comments into an answer I'll happily accept it. | |
| Sep 5, 2021 at 2:18 | comment | added | darij grinberg | Note that I have never used that $\lambda$ has distinct parts. | |
| Sep 5, 2021 at 2:16 | comment | added | darij grinberg | ... by $2$ still gives an integer). (2) Proving that $P_\lambda \in \mathbb{Z}\left[q_k / 2 \mid k \text{ is odd} \right]$. This follows from Theorem 8.1.6 in op. cit. (applied to $t = -1$), since each $q_\mu / 2^{\ell\left(\mu\right)}$ is clearly a product of $q_k / 2$s for various $k$s. Here one needs to be a bit careful, since some of the $k$s will be even, but the formula (9.1.1) in op. cit. shows (by induction) that the $q_k / 2$s for even $k$ still belong to $\mathbb{Z}\left[q_k / 2 \mid k \text{ is odd} \right]$. Here is hoping I didn't get confused. | |
| Sep 5, 2021 at 2:13 | comment | added | darij grinberg | There are two main ingredients: (1) Proving that $\mathbb{Z}_{\left(2\right)} \left[q_k/2 \mid k \text{ is odd}\right] = \mathbb{Z}_{\left(2\right)} \left[p_k \mid k \text{ is odd}\right]$, where the $q_k$ are defined by $\sum\limits_{k \in \mathbb{N}} q_k u^k = \prod\limits_j \dfrac{1+x_ju}{1-x_ju}$. This equality follows from Lemma 9.1.3 in Steven Sam's Math 740 notes (Spring 2017), because the powers of $2$ in the denominator of $z_\lambda^{-1}$ are cancelled out by the $2^{\ell\left(\lambda\right)}$ with a slack of $1$ (so dividing ... | |
| Sep 5, 2021 at 2:10 | comment | added | darij grinberg | I'm not sure if the fact is known, but I think I can prove it (modulo some results about Hall-Littlewood functions from Steven Sam's notes that I haven't properly read but have no specific reasons to distrust). Are you looking for a proof or a reference? | |
| Sep 4, 2021 at 18:43 | comment | added | Antoine Labelle | Thanks for your comment! That seems equivalent to my question actually, since unless I'm mistaken the $P_\lambda$ (for arbitrary $\lambda$) form a $\mathbb{Z}$-basis for $\Lambda$, and in particular those with $\lambda$ having distinct parts form a $\mathbb{Z}$-basis for $\Lambda \cap \mathbb{Q}[p_1,p_3,p_5,\cdots]$. | |
| Sep 4, 2021 at 18:21 | comment | added | Richard Stanley | A symmetric function $f$ is integral, denoted $f\in\Lambda$, if it is an integral linear combination of monomial symmetric functions. It is well-known that $P_\lambda\in\Lambda$ and (as stated by the proposer) $P_\lambda\in \mathbb{Q}[p_1,p_3,p_5,\dots]$. The following stronger result seems to be true: if $f\in\Lambda\cap \mathbb{Q}[p_1,p_3,p_5,\dots]$ and $f=\sum_\rho b^\lambda_\rho p_\rho$, then $v_2(b_\rho^\lambda)\geq 0$. | |
| Sep 3, 2021 at 23:08 | history | asked | Antoine Labelle | CC BY-SA 4.0 |