The assumed one-dimensional-image property gives
$$
b_t e(s)=c\,e(t)
$$
for some $c\in\mathbb F$. The coefficient of $\{t\}$ in $e(t)$ is one, so
$$
c=\langle b_t e(s),\{t\}\rangle.
$$
The <tabloid bilinear form> is invariant, and the involution on the <group algebra> fixes the <Column antisymmetrizer of a Young tableau> because inversion preserves sign. Therefore
$$
c=\langle e(s),b_t\{t\}\rangle
=\langle e(s),e(t)\rangle,
$$
which proves the formula.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 160 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-160/1/b/solution