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Feb 26, 2011 at 13:59 comment added ARupinski Incidentally, since the highest weight of the restriction only grows quadratically (it is $\frac{n^2+n+2}{2}\beta$), it follows that for $n\geq 9$ the restriction is not multiplicity free. In fact, for $n=9$, there are 5 distinct irreducible pieces of multiplicity 2
Feb 26, 2011 at 7:43 comment added Leonid Petrov Thank you very much for this accurate computation. Actually, my question started from the combinatorial description of the decomposition of restrictions of $\Sigma_n$. This goes as follows. Take the sum $f(\xi)=\xi_1+2\xi_2+\dots+n\xi_n$, where each $\xi_i=\pm1$. Then for $n=5$ you get 16 = the number of possible values this sum can take (from 0 to $n(n+1)/2$); 10 = the number of values it can take for 2 different $\xi$'s, 6 = for three different $\xi$'s, etc. This rep of $SL_2$ appears in study of strict partitions... Then I tried to guess the construction of rep but I was not sure.
Feb 26, 2011 at 7:35 vote accept Leonid Petrov
Feb 26, 2011 at 1:03 history edited ARupinski CC BY-SA 2.5
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Feb 25, 2011 at 18:40 history answered ARupinski CC BY-SA 2.5