Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 4 / 3 / i

Suppose first that $U$ is not contained in $(S^{\lambda }{)}^{\perp }$. Since polytabloids span $S^{\lambda }$, there exist $u\in U$ and a tableau $t$ with $⟨u,e_t⟩\ne 0$. The column antisymmetrizer identity from the preliminaries givesTherefore $e_t\in U$. Because $U$ is an $S_n$-submodule, it contains all translates $g{e}_t=e_{gt}$, and these span the Specht module. Hence $S^{\lambda }\subseteq U$. If the initial supposition fails, $U\subseteq (S^{\lambda }{)}^{\perp }$. This proves the James submodule theoremNow let $W$ be a submodule of $S^{\lambda }$. Apply the dichotomy to $W\subseteq M^{\lambda }$. Either $W=S^{\lambda }$, or $W\subseteq S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$. Positivity of the Hermitian form gives $S^{\lambda }\cap (S^{\lambda }{)}^{\perp }=\{0\}$: a vector in this intersection has squared norm zero. The same intersection is zero for the bilinear form by the agreement of complements explained above. Since $S^{\lambda }\ne 0$, the only submodules are $0$ and itself. ThereforeThis use of characteristic zero and nondegeneracy is why the conclusion does not automatically extend to every modular Specht module.