Skip to main content
6 events
when toggle format what by license comment
Sep 29, 2018 at 1:57 comment added darij grinberg @FedorPetrov: $p_k^{\perp}$ is the operation of skewing by $p_k$. See, for example, §2.8 in Darij Grinberg and Victor Reiner, Hopf Algebras in Combinatorics, 11 May 2018. Actually, $p_k^{\perp}$ is also the same as the partial derivative with respect to $p_k$ when the symmetric functions are regarded as polynomials in $p_1, p_2, p_3, \ldots$; this doesn't extend to skewing by other functions.
Sep 28, 2018 at 19:23 comment added Fedor Petrov Darij, what is $p_k^\perp f$?
Sep 27, 2018 at 22:42 comment added darij grinberg Actually, here is a way to prove your identity quickly (& dirtily): It is a particular case of the more general identity $nf = \sum\limits_{k=1}^n p_k p_k^{\perp} f$, which holds for any homogeneous symmetric function $f$ of degree $n$. And this latter identity falls prey to more-or-less straightforward algebraic approaches, such as observing that both of its sides are derivations of $f$ (recall that the map sending each homogeneous symmetric function $f$ to $nf$ is a derivation), and they are equal when $f$ is one of the power sums $p_1, p_2, p_3, \ldots$.
Sep 27, 2018 at 16:43 comment added Zach H I should clarify: this identity was found by my collaborators.
Sep 27, 2018 at 15:36 comment added darij grinberg I have never seen this nice identity before.
Sep 27, 2018 at 14:09 history asked Zach H CC BY-SA 4.0