Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2008 / B / 8

Give $M^{\lambda }$ the invariant symmetric bilinear form for which its tabloid basis is orthonormal. DefineIn characteristic $p>0$, this quotient is nonzero and simple when $\lambda$ is $p$-regular, meaning that no part occurs $p$ or more times. For other partitions it is zero, so the label “simple module” includes the nonzero/$p$-regular qualification.

The James submodule theorem states that for every $F{\Sigma }_n$-submodule $U\subseteq M^{\lambda }$, either $S^{\lambda }\subseteq U$ or $U\subseteq (S^{\lambda }{)}^{\perp }$. Applied to submodules of $S^{\lambda }$, it says that every proper submodule lies in the radical of the restricted form, giving the simplicity of the displayed nonzero quotient.

Choose the two standard tableauxand use their polytabloids $e_1,e_2$, with column-vector coordinates. Writing $[ab\mid cd]$ for the tabloid with those two row sets givesThe coefficients at $[12\mid 34]$ and $[13\mid 24]$ prove independence over every field. Applying the generators directly givesand $(34)$ acts in the same way as $(12)$. Their span is invariant under the generators, contains $e_1$, and contains its whole orbit, hence is the Specht module. Thus it is a basis without needing a characteristic-zero argument. The matrices areThese integral matrices may be reduced in the chosen characteristic.

For the characteristic-three Specht module for the two-by-two partition, reduce the matrices modulo three. The vector $e_1+e_2$ is fixed by every displayed transposition: $(12)$ and $(34)$ send it to $e_1-2e_2=e_1+e_2$, and $(23)$ interchanges its two terms. Thus its span is a proper nonzero trivial submodule.

On the quotient impose $e_2=-e_1$. Each generating transposition then acts as minus one, so the quotient is the sign representation. Consequentlyand the two composition factors are trivial and sign, each once. In the $D$ notation they are $D^{(4)}$ and $D^{(2,2)}$ respectively. Indeed the restricted tabloid form has Gram matrix $\left(\begin{array}{cc}4& 2\\ 2& 4\end{array}\right)$, whose characteristic-three radical is precisely the trivial line, so the simple quotient is $D^{(2,2)}$.

The extension does not split: the minus-one eigenspace of $\rho (12)$ is the line through $e_2$, but $\rho (23)$ sends that line to the line through $e_1$. Thus there is no sign submodule complementing the trivial line, consistent with the failure of semisimplicity.