Modular Specht module

Over a field of characteristic $p$, the invariant tabloid form on $S^{\lambda }$ may be degenerate. Its radical is $S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$.

The tabloid bilinear form makes the tabloid basis orthonormal and restricts to an invariant bilinear form on each Specht module. In positive characteristic of a field, its radical controls the corresponding simple quotient.

A partition is $p$-regular when no part occurs $p$ or more times.

Write $n=q(p-1)+r$, $0\le r<p-1$. The regular partition labelling the sign representation in characteristic $p>0$ is $\lambda =((q+1{)}^r,q^{p-1-r})$, with zero parts removed. Its conjugate partition is $((p-1{)}^q,r)$, which satisfies the invariant vector criterion for a Specht module. The sign-twisted duality of Specht modules turns this invariant line into a sign quotient of $S^{\lambda }$, necessarily its unique simple head. Conversely, regularity says consecutive column heights differ by less than $p$; combining this with the invariant congruences forces these heights to be $p-1$ except possibly the last. For $p=2$, the label is $(n)$ because sign equals the trivial representation.

For a $p$-regular partition,is nonzero and absolutely irreducible. Distinct $p$-regular partitions label nonisomorphic simple modules.

The James submodule theorem follows from ${\kappa }_{t}v=⟨v,e_t⟩e_t$. A nonzero restricted form therefore gives a simple radical quotient. If $m_j$ counts equal-length rows, all integral polytabloid pairings are divisible by ${\prod }_j{m}_j!$, while pairing with row reversal gives ${\prod }_j(m_j!{)}^j$. Thus the form is nonzero modulo $p$ exactly for regular partitions. The construction and the proof commute with every field extension, so the quotient is absolutely irreducible, including over the prime field.

If $\lambda =(n^{a_n},\dots ,1^{a_1})$ is $p$-regular, thenFor a tableau $t$ and its row reversal $t^{\ast }$,which is nonzero in characteristic $p$. Applying a column antisymmetrizer to an endomorphism at $e(t^{\ast })$ therefore forces its value on the cyclic generator $e(t)$ to be scalar.