0
$\begingroup$

Say $\lambda \vdash n$ and $\mu \vdash m$ and $|\rho|=|\lambda|+|\mu|$ then for all $\lambda$ and $m$ it appears to hold that

$$ \sum_\mu [s_\lambda](s_\lambda s_\mu)_{/\mu} = \sum_\rho [s_\rho](s_{\rho /\mu}) s_\mu$$ with the (abusive?) shorthand $(s_\lambda s_\mu)_{/\mu}$ standing for the expansion of the product $s_\lambda s_\mu$ and taking the skew Schur function $s_{\xi/\mu}$ of each term $s_\xi$, and analogously $(s_{\rho /\mu}) s_\mu$ standing for the expansion of $s_{\rho/\mu}$ and multiplying each term with $s_\mu$ and expanding again. The $[s_\lambda ]$ denotes the coefficient of $s_\lambda$ in the polynome.

Example: take $\lambda = \{3, 1\}$ and $m=3$

then the left hand side evaluates to $6+11+4 = 21$ (running over the 3 partitions of 3) and the right hand side gives $0+1+2+2+2+6+2+2+1+2+1+0+0+0+0 = 21$ over the 15 partitions of 7.

Does this hold generally? Is this well known? Is it in EC2? Is it easy to prove?

$\endgroup$
3
  • $\begingroup$ You should recall the meaning of "majoring" notation $\lambda \vdash n$ and/or give a reference. $\endgroup$
    – Jean Marie
    Mar 19 at 17:18
  • 1
    $\begingroup$ @JeanMarie, the condition that $\lambda$ majors $\mu$ is not needed. A propos, "majoring" is equivalent to "at least covering" when seen as Young diagrams. $\endgroup$
    – Wouter M.
    Mar 20 at 18:38
  • $\begingroup$ Thanks for your answer. $\endgroup$
    – Jean Marie
    Mar 21 at 7:53

0

Reset to default

You must log in to answer this question.