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

The symmetric group acts transitively on the tableaux of a fixed shape. Thus every $s\in {\Delta }^{\lambda }$ is $gt$ for some $g\in S_n$, and the definition of a polytabloid givesSince the Specht module is spanned by all $e(s)$, it is the cyclic $\mathbb{F}{S}_n$-module generated by $e(t)$.

Ina fixed tabloid $\{u\}$ can occur at most once. Indeed, if $\{\pi t\}=\{\sigma t\}$ for $\pi ,\sigma \in C(t)$, then ${\sigma }^{-1}\pi$ belongs to both the row and column stabilizers of $t$. Their intersection is trivial, so $\pi =\sigma$. The coefficient $⟨e(t),\{u\}⟩$ is consequently $0$, $1$, or $-1$.

For a tabloid $U$, let $r_U(k)$ be the row containing $k$, and order tabloids lexicographically byIf $t$ is standard and $1\ne \pi \in C(t)$, let $k$ be the least number moved by $\pi$. Every smaller entry is fixed, while ${\pi }^{-1}(k)$ is a larger entry below $k$ in the same column. Hence $\{\pi t\}$ agrees with $\{t\}$ before $k$ and has $r_{\pi t}(k)>r_t(k)$. Thus $\{t\}$ is the unique least tabloid in $e(t)$ and has coefficient one.

Distinct standard tableaux have distinct tabloids, since increasing each row recovers the tableau from its tabloid. A linear relation among standard polytabloids now has a least leading tabloid, which cannot cancel. They are therefore linearly independent, as in the linear independence of standard polytabloids.

The assumed one-dimensional-image property givesfor some $c\in \mathbb{F}$. The coefficient of $\{t\}$ in $e(t)$ is one, soThe tabloid bilinear form is invariant, and the involution on the group algebra fixes the Column antisymmetrizer of a Young tableau because inversion preserves sign. Thereforewhich proves the formula.

The equality $h\{t\}=h^{\ast }\{t^{\ast }\}$ says that $h(i)$ and $h^{\ast }(i)$ lie in corresponding rows for every entry $i$. For $i$ in the leftmost column, both $h(i)$ and $h^{\ast }(i)$ remain in that column. Corresponding rows therefore force $h(i)=h^{\ast }(i)$.

Remove the leftmost entry of every row and repeat the argument on the shortened tableaux. Induction across the columns gives equality on every entry, so $h=h^{\ast }$.

The nonzero coefficients and part a(ii) give unique elements $h\in C(t)$ and $h^{\ast }\in C(t^{\ast })$ such thatPart i gives $h=h^{\ast }=:k$, so $k\in C(t)\cap C(t^{\ast })$. Both coefficients are then $\mathrm{sgn}(k)$ and hence are equal.

By part ii, the common tabloids in the supports of $e(t)$ and $e(t^{\ast })$ are precisely $k\{t\}$ with $k\in C(t)\cap C(t^{\ast })$, and each contributes $\mathrm{sgn}(k{)}^2=1$. ConsequentlyFor the $a_j$ rows of length $j$, an element of the intersection applies the same arbitrary permutation of those rows independently in each of their $j$ columns. Hence that row length contributes $(a_j!{)}^j$, and

Let $\theta \in {\mathrm{End}}_{\mathbb{F}{S}_n}(S^{\lambda })$ and putSince $\lambda$ is $p$-regular, every $a_j<p$, so $q\ne 0$ in $\mathbb{F}$. Parts b and c give $b_{t}e(t^{\ast })=qe(t)$. Since $\theta$ commutes with the group-algebra action,The left side belongs to $b_t{M}^{\lambda }=\mathbb{F}e(t)$. Division by $q$ shows that $\theta (e(t))$ is a scalar multiple of $e(t)$. Part a(i) says that $e(t)$ generates $S^{\lambda }$, so $\theta$ is that scalar multiple of the identity. This proves the endomorphism theorem for a regular Specht module.