If some $a_j\geq p$, then $a_j!=0$ in $\mathbb F$. Part b(ii) makes every pairing $\langle e(t),e(u)\rangle$ zero, so the <Tabloid bilinear form> vanishes identically on $S^\lambda$ and $S^\lambda\subseteq(S^\lambda)^\perp$.

Conversely, if $\lambda$ is $p$-regular, every $a_j!$ is nonzero. Fact 1 supplies tableaux $t,u$ with
$$
\langle e(t),e(u)\rangle=\prod_{j=1}^n(a_j!)^j\ne0.
$$
Thus the restriction of the form is not identically zero and $S^\lambda\not\subseteq(S^\lambda)^\perp$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 160 / 4 / c / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-160/4/c/ii/solution