Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2007 / B / 5

A tabloid records the sets of entries in each row of a tableau. Let $C_t$ be the subgroup permuting entries independently within each column. The polytabloid isand the Specht module $S^{\lambda }$ is spanned by these vectors. A standard polytabloid is one whose tableau increases along rows and down columns.

Order tabloids by first sorting each row increasingly, then comparing the resulting row-reading words lexicographically. For a standard tableau $t$, every nonidentity column permutation produces a strictly larger word. In the first row affected by that permutation, each replacement comes from farther down the same column and is larger than the entry it replaces; all earlier rows remain unchanged. Sorting the affected row therefore increases its word. Also no nontrivial column permutation fixes its row sets, since a cell is the unique intersection of its row and column. Hence the least tabloid in $e_t$ is $\{t\}$ with coefficient one.

Distinct standard tableaux have distinct row tabloids. In a nonzero linear relation among their polytabloids, choose the least leading tabloid with a nonzero coefficient. None of the vectors with larger leading tabloids can contribute to it, so its coefficient cannot cancel. This contradiction proves linear independence of the standard polytabloids.

The hook length of a cell is one plus the number of cells to its right in the same row and below it in the same column. The hook-length formula isFor $\lambda =(n-k,k)$, with $0\le k\le n/2$, the second-row hook product is $k!$. In the first row, the hooks are $n-k-j+2$ for $1\le j\le k$ and $n-k-j+1$ for $k<j\le n-k$, so their product is $(n-k+1)!/(n-2k+1)$. ConsequentlyAt $k=0$, interpret the second binomial coefficient as zero.

For a cycle type $\mu =(m,\nu )$, the Murnaghan–Nakayama rule isThe sum is over partitions $\rho$ whose diagrams remain after removing the specified border strip; $\nu$ is the remaining cycle type, and the recursion ends with the empty character equal to one. At the identity, ${\chi }^{\rho }(1^{|\rho |})=f^{\rho }$. Incompatible total sizes contribute zero.

For shape $(2,2)$ and a transposition, a horizontal strip in the bottom row leaves $(2)$ with positive sign; a vertical strip in the right column leaves $(1,1)$ with negative sign. Both remaining dimensions are one. Thus

For shape $(n-2,2)$ with $n\ge 6$, there are exactly two removable rim hooks of size two: the bottom row, leaving $(n-2)$, and the final two cells of the first row, leaving $(n-4,2)$. Both have height zero. A vertical pair cannot be removed while leaving a Young diagram because the first row is longer than the second by at least two.

By the Murnaghan–Nakayama rule and part (b),