Young's rule for M^(2,1)

The tabloids $v_1,v_2,v_3$ are distinct by [F1], since their singleton second rows $\{1\},\{2\},\{3\}$ are distinct; by [F1] they form a basis of $M^{(2,1)}$, so ${\dim }_{\mathbb{C}}{M}^{(2,1)}=3$.

The semistandard fillings with content $(2,1)$ are enumerated by shape. Shape $(3)$: the single row carries two $1$'s and one $2$ weakly increasingly, so the only filling is $112$ and $K_{(3),(2,1)}=1$. Shape $(2,1)$: the entry in the box $(2,1)$ must strictly exceed the entry in the box above it, and only the entries $1,2$ occur, so $(2,1)$ carries $2$ and $(1,1)$ carries $1$; the remaining entry $1$ fills the box $(1,2)$, whose left neighbour is $1$, and weak increase holds; hence the unique filling is $\begin{array}{cc}1& 1\\ 2& \end{array}$ and $K_{(2,1),(2,1)}=1$. Shape $(1,1,1)$: the three boxes form a column with strictly increasing entries, but the content has the entry $1$ twice, so no such filling exists and $K_{(1,1,1),(2,1)}=0$.

The shape $(2,1)$ has exactly two standard tableaux: the entry $1$ must occupy the box $(1,1)$, and the remaining boxes $(1,2)$ and $(2,1)$ receive $2$ and $3$ in either order, both fillings being standard, namely $t=\begin{array}{cc}1& 2\\ 3& \end{array}$ and $u=\begin{array}{cc}1& 3\\ 2& \end{array}$. By [F4] one has ${\kappa }_t=\mathrm{i}\mathrm{d}-(13)$ and ${\kappa }_u=\mathrm{i}\mathrm{d}-(12)$, and the action on tabloids is by relabelling, so $(13)\cdot v_3=v_1$ and $(12)\cdot v_2=v_1$. Hence $e_t=v_3-v_1$ and $e_u=v_2-v_1$, whose coordinate vectors $(-1,0,1)$ and $(-1,1,0)$ in the basis $v_1,v_2,v_3$ of [F1] are linearly independent. By [F5] the standard polytabloids of shape $(2,1)$ form a basis of $S^{(2,1)}$, whose dimension is therefore the number $2$ of standard tableaux, so $\{e_t,e_u\}$ is a basis of $S^{(2,1)}$.

Substituting step 1.2 into [F3] gives the $\mathbb{C}{S}_3$-isomorphism $M^{(2,1)}\cong (S^{(3)}{)}^{\oplus 1}\oplus (S^{(2,1)}{)}^{\oplus 1}\oplus (S^{(1,1,1)}{)}^{\oplus 0}=S^{(3)}\oplus S^{(2,1)}$, which is claim 2 and shows that the multiplicity of $S^{(2,1)}$ in $M^{(2,1)}$ is one.

Put $w:=v_1+v_2+v_3$ and let $H:=\{a_1{v}_1+a_2{v}_2+a_3{v}_3:a_1+a_2+a_3=0\}$ be the sum-zero hyperplane. Every $\sigma \in S_3$ permutes the tabloid basis, so $\sigma w=w$ and $\mathbb{C}w$ is a submodule isomorphic to the trivial representation; and $H$ is $\sigma$-stable, since $\sigma$ merely permutes the coefficients. For any $a=a_1{v}_1+a_2{v}_2+a_3{v}_3$ with $c:=(a_1+a_2+a_3)\mathrm{/}3$ one has $a=cw+(a-cw)$ with $a-cw\in H$, so $M^{(2,1)}=\mathbb{C}w+H$; and $\mathbb{C}w\cap H=0$, since $cw\in H$ forces $3c=0$, hence $c=0$. Therefore $M^{(2,1)}=\mathbb{C}w\oplus H$ with ${\dim }_{\mathbb{C}}H=2$.

By step 2.1 the standard polytabloids $e_t=v_3-v_1$ and $e_u=v_2-v_1$ form a basis of $S^{(2,1)}$, and both lie in $H$ because their coordinates sum to zero; hence $S^{(2,1)}=\mathrm{span}\{v_3-v_1,v_2-v_1\}\subseteq H$ with ${\dim }_{\mathbb{C}}{S}^{(2,1)}=2={\dim }_{\mathbb{C}}H$, so $S^{(2,1)}=H$. By [F6] $S^{(3)}$ is one-dimensional and trivial, whereas $S^{(2,1)}$ has dimension two by step 2.1; they are therefore non-isomorphic, and $\mathbb{C}w\cong S^{(3)}$ and $H=S^{(2,1)}$ are exactly the two summands found in step 2.2; combined with step 2.3 this proves claim 3 and the dimension count $3=1+2$.

Consistency and boundary audit. The third partition $(1,1,1)$ of $3$ contributes nothing, as computed directly from the strict column increase in step 1.2; the value $K_{(2,1),(2,1)}=1$ follows from the unique filling enumerated in step 1.2; the row shape $(3)$ contributes exactly one trivial summand, realized concretely as $\mathbb{C}(v_1+v_2+v_3)$; and the column shape $(1,1,1)$ would require three distinct entries, which content $(2,1)$ does not provide. The decomposition $\mathbb{C}w\oplus H$ is $\sigma$-stable for every $\sigma \in S_3$ by step 2.3, and the ambient module has dimension three over $\mathbb{C}$, so $3=1+2$ is the complete dimension count. This proves the Statement.