Past exam of the representation Theory of Symmetric Groups course of the University of Oxford / 2011 / 2 / b / Solution

Put $W=S_F^{(n-2,1,1)}$ and let $H\cong S_{n-1}$ be the subgroup fixing one. For $\tau =(r,s)\in H$, let $u=\min \{\tau (i),\tau (j)\}$ and $v=\max \{\tau (i),\tau (j)\}$. Permuting a tableau's entries and then ordering its first column givesThe underlying basis element is $e_{1ij}$ exactly when $\tau$ preserves the unordered pair $\{i,j\}$: either $\{r,s\}=\{i,j\}$, giving the minus sign, or the two pairs are disjoint, giving the plus sign.

If $p\ne 2$, invariance of $x$ under $(i,j)$ forces its $e_{1ij}$ coefficient to satisfy $a_{ij}=-a_{ij}$, because no other basis vector is sent to that basis vector. Thus every coefficient vanishes. Consequently $W^H=0$, and in particular $W^{S_n}=0$, so there is no nonzero homomorphism from the trivial module. Any homomorphism $M_F^{(n-1,1)}\to W$ is determined by the image of its tabloid with singleton row $\{1\}$. That image is fixed by $H$ and is zero, making the whole homomorphism zero.

The printed vanishing assertion is false in characteristic two. Signs then coincide, so $H$ permutes the basis indexed by unordered pairs from $\{2,\dots ,n\}$. This action is transitive, andThe vector is nonzero by standard-basis independence. It also gives a nonzero map from the permutation module: send the singleton-row basis vector indexed by $r$ to $g_{r}v$, where $g_r(1)=r$. This is well defined because two choices of $g_r$ differ by an element of $H$, and it is $S_n$-equivariant by construction. Thus the claimed vanishing of permutation-module homomorphisms fails for every $n\ge 4$ in characteristic two.

The full-group fixed space can be determined explicitly. Denote a hook tabloid by $[a,b]$, recording its second- and third-row entries. In characteristic two, $e_{1ij}$ is the sum of the six tabloids using ordered pairs from $\{1,i,j\}$. In $v$, a tabloid whose two recorded entries both differ from one appears once; a tabloid using one appears $n-2$ times. Hence for odd $n$, $v$ is the sum of all hook tabloids and is fixed by $S_n$. For even $n$, it is the sum of precisely the tabloids avoiding one and is moved by $(1,2)$. Since every full-group fixed vector already lies in $Fv$, the corrected dimensions areThus even the trivial-module vanishing fails for odd $n$ in characteristic two; for example $n=5$ gives the explicit nonzero invariant $v$. These corrected fixed spaces allow the valid later conclusions to be proved in all characteristics.

The corrected parity dependence is described by fixed vectors in a hook Specht module in characteristic two.