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

Let $b_t$ be the Column antisymmetrizer of a Young tableau. For a tabloid $\{u\}$, if two entries from one column of $t$ lie in one row of $u$, their column transposition fixes $\{u\}$ and pairs every term of $b_t\{u\}$ with its negative. Thus $b_t\{u\}=0$.

Otherwise each row of $u$ meets each column of $t$ at most once. Matching entries within columns then gives a column permutation $c\in C(t)$ for which $\{cu\}=\{t\}$. Reindexing the antisymmetrizer gives $b_t\{u\}=\mathrm{sgn}(c)e(t)$. Hence every basis tabloid maps into $\mathbb{F}e(t)$, while $b_t\{t\}=e(t)$, and therefore

The coefficient $⟨e(t),\omega ⟩$ is nonzero exactly when every column of $t$ meets every row of $\omega$ in at most one entry. Permuting equal-length rows of $\omega$ preserves this condition, and the same holds with $u$ in place of $t$. Therefore membership in $A_{tu}$ is constant on each class of the stated equivalence relation, so $A_{tu}$ is a union of equivalence classes.

Each equivalence class has size ${\prod }_{j=1}^n{a}_j!$, because its $a_j$ rows of length $j$ may be permuted freely. Swapping two length-$j$ rows multiplies each of $⟨e(t),\omega ⟩$ and $⟨e(u),\omega ⟩$ by the same sign $(-1{)}^j$, so their product is constant on the class. Sincethe contribution of every nonzero class is a multiple of ${\prod }_j{a}_j!$. The whole inner product is therefore such a multiple.

The partition is $p$-regular precisely whenthat is, when no part occurs $p$ or more times. This is the definition of a regular partition in characteristic of a field $p$.

If some $a_j\ge p$, then $a_j!=0$ in $\mathbb{F}$. Part b(ii) makes every pairing $⟨e(t),e(u)⟩$ 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 $t,u$ withThus the restriction of the form is not identically zero and $S^{\lambda }⊈(S^{\lambda }{)}^{\perp }$.

Because $\lambda$ is $p$-regular, Fact 1 gives a tableau $u$ for whichPart a, applied linearly to the tabloid expansion of $e(u)$, gives

The nonzero homomorphism $\theta$ cannot kill $e(t)$, because the Specht module is generated by the translates of this polytabloid. Equivariance and part d(i) therefore giveThus $b_t$ acts nontrivially on $M^{\mu }/V$, and hence on $M^{\mu }$. Some $\mu$-tabloid $\{v\}$ must satisfy $b_t\{v\}\ne 0$. Fact 2 now says that $\lambda$ dominates $\mu$ in the dominance order on partitions.

Assume $\lambda =\mu$. Part a says that the image of $b_t$ on $M^{\lambda }/V$ is contained in the one-dimensional space spanned by $e(t)+V$. The identity from part d(ii) then shows that $\theta (e(t))$ itself is a scalar multiple of $e(t)+V$.

Apply this with $V=S^{\lambda }\cap (S^{\lambda }{)}^{\perp }$ and with $\theta$ obtained by composing the quotient map $S^{\lambda }\to D^{\lambda }$ with any endomorphism of the Simple symmetric-group module from a regular partition $D^{\lambda }$. Every endomorphism is scalar on the generating polytabloid and hence on all its translates. Therefore

If the two displayed simple quotients $D^{\alpha }$ and $D^{\beta }$ are isomorphic, compose the quotient map $S^{\alpha }\to D^{\alpha }$ with an isomorphism to $D^{\beta }\subseteq M^{\beta }/(S^{\beta }\cap (S^{\beta }{)}^{\perp })$. Part d(ii) gives $\alpha ⪰\beta$. Applying the inverse isomorphism gives $\beta ⪰\alpha$. Antisymmetry of the dominance order on partitions forces $\alpha =\beta$. Conversely, equality of the partitions makes the quotients identical, hence isomorphic.