The preceding part shows that $\theta(e(t'))\ne0$ in characteristic zero, so $S^{\lambda'}\nsubseteq\ker\theta$. Apply the <James submodule theorem> to the proper submodule $\ker\theta\leq M^{\lambda'}$ to obtain
$$
\ker\theta\subseteq(S^{\lambda'})^\perp.
$$
The <Hook-length formula> gives $\dim S^\lambda=\dim S^{\lambda'}$. Surjectivity of $\theta$ therefore gives
$$
\operatorname{codim}\ker\theta=\dim S^\lambda=\dim S^{\lambda'}
=\operatorname{codim}(S^{\lambda'})^\perp,
$$
and hence
$$
\boxed{\ker\theta=(S^{\lambda'})^\perp}.
$$

Fix the original tableaux $t,u$. For a $\lambda$-tableau $w$, let $g_w$ be the unique permutation satisfying $w=g_wt$ and put $\varepsilon_w=\operatorname{sgn}(g_w)$. Since
$$
\theta(\{w'\})=\varepsilon_w e(w)\otimes e(u),
$$
the quotient pairing $M^{\lambda'}/(S^{\lambda'})^\perp\cong(S^{\lambda'})^*$ gives the explicit <conjugate Specht module as a sign-twisted dual> isomorphism
$$
e(w)\otimes e(u)longmapsto
\left[e(v')\longmapsto
\varepsilon_w\langle\{w'\},e(v')\rangle\right]
$$
for every $\lambda$-tableau $w$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 160 / 1 / c / iii / Solution

Over a field of characteristic zero,
$$
S^\lambda\otimes S^{(1^n)}\cong(S^{\lambda'})^*.
$$
The map from $M^{\lambda'}$ sending the tabloid of $g t'$ to $g e(t)\otimes g e(u)$ is a surjection whose kernel is $(S^{\lambda'})^\perp$.


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