The Branching Rule and the Young Graph — Examples

The partitions of $0,1,2,3,4$ are listed by size directly from the definition [F3]: $\mathrm{\varnothing }$; $(1)$; $(2),(1,1)$; $(3),(2,1),(1,1,1)$; $(4),(3,1),(2,2),(2,1,1),(1,1,1,1)$, giving $1,1,2,3,5$ vertices of ranks $0,1,2,3,4$, which is claim 1.

Addable nodes by the criterion of [F2]: for $\mathrm{\varnothing }$ the node $(1,1)$ gives $(1)$; for $(1)$ the nodes $(1,2)$ and $(2,1)$ give $(2)$ and $(1,1)$; for $(2)$ the nodes $(1,3)$ and $(2,1)$ give $(3)$ and $(2,1)$; for $(1,1)$ the nodes $(1,2)$ and $(3,1)$ give $(2,1)$ and $(1,1,1)$; for $(3)$ the nodes $(1,4)$ and $(2,1)$ give $(4)$ and $(3,1)$; for $(2,1)$ the nodes $(1,3)$ (here $i=1$), $(2,2)$ (here ${\lambda }_1=2>{\lambda }_2=1$) and $(3,1)$ (a new row) give $(3,1)$, $(2,2)$ and $(2,1,1)$; and for $(1,1,1)$ the nodes $(1,2)$ (here $i=1$) and $(4,1)$ (a new row) give $(2,1,1)$ and $(1,1,1,1)$. By [F1] each addable node gives exactly one edge, so the edges out of ranks $0,1,2,3$ are exactly the $1,2,4,7$ edges displayed in claim 2; in particular $(2,1)$ has the three outgoing edges to $(3,1),(2,2),(2,1,1)$.

Standard tableaux by explicit enumeration in the sense of [F3]: rank $0$ has the empty tableau; rank $1$ has $1$; rank $2$ has $12$ and $1\mathrm{/}2$; rank $3$ has $123$, the two tableaux $\begin{array}{cc}1& 2\\ 3& \end{array}$, $\begin{array}{cc}1& 3\\ 2& \end{array}$ and $1\mathrm{/}2\mathrm{/}3$; rank $4$ has $1234$, the three tableaux $\begin{array}{ccc}1& 2& 3\\ 4& \end{array}$, $\begin{array}{ccc}1& 2& 4\\ 3& \end{array}$, $\begin{array}{ccc}1& 3& 4\\ 2& \end{array}$, the two tableaux $\begin{array}{cc}1& 2\\ 3& 4\end{array}$, $\begin{array}{cc}1& 3\\ 2& 4\end{array}$, the three tableaux $\begin{array}{cc}1& 2\\ 3& \\ 4& \end{array}$, $\begin{array}{cc}1& 3\\ 2& \\ 4& \end{array}$, $\begin{array}{cc}1& 4\\ 2& \\ 3& \end{array}$ and $1\mathrm{/}2\mathrm{/}3\mathrm{/}4$. Counting these gives the values $f^{\lambda }$ displayed in claim 4.

Removable nodes by the criterion of [F2], read in the reverse direction: $(1)$ has the removable node $(1,1)$ with $(1)-(1,1)=\mathrm{\varnothing }$; $(2)$ has $(1,2)$ giving $(1)$; $(1,1)$ has $(2,1)$ giving $(1)$; $(3)$ has $(1,3)$ giving $(2)$; $(2,1)$ has $(1,2)$ giving $(1,1)$ and $(2,1)$ giving $(2)$; $(1,1,1)$ has $(3,1)$ giving $(1,1)$; $(4)$ has $(1,4)$ giving $(3)$; $(3,1)$ has $(1,3)$ giving $(2,1)$ and $(2,1)$ giving $(3)$; $(2,2)$ has $(2,2)$ giving $(2,1)$ only; $(2,1,1)$ has $(1,2)$ giving $(1,1,1)$ and $(3,1)$ giving $(2,1)$; and $(1,1,1,1)$ has $(4,1)$ giving $(1,1,1)$. In every case the resulting partition has one box fewer, and the incoming edges so obtained are exactly the edges of step 2.1 read backwards: for example the two incoming edges of $(2,1)$ come from $(2)$ and $(1,1)$, and the only incoming edge of $(2,2)$ comes from $(2,1)$.

By [F4] each ${\dim }_{\mathbb{C}}{S}_{\mathbb{C}}^{\lambda }$ equals the corresponding value $f^{\lambda }$ of step 2.2. Substituting these dimensions into [F7] with $G=S_n$ over $k=\mathbb{C}$ gives ${\sum }_{\lambda ⊢n}(f^{\lambda }{)}^2=\mathrm{\mid }{S}_n\mathrm{\mid }=n!$ by [F7]; explicitly $1^2=1$, $1^2=1$, $1^2+1^2=2$, $1^2+2^2+1^2=6$ and $1^2+3^2+2^2+3^2+1^2=24$ for $n=0,1,2,3,4$.

Claim 3 follows from the two branching rules: by [F5], for each $\lambda ⊢n$ with $1\le n\le 4$ the restriction of $S_{\mathbb{C}}^{\lambda }$ is the direct sum of one copy of $S_{\mathbb{C}}^{\lambda -x}$ for each removable node $x$, that is one summand per incoming edge of step 3.1; by [F6], for each $\lambda ⊢n$ with $0\le n\le 3$ the induction of $S_{\mathbb{C}}^{\lambda }$ is the direct sum of one copy of $S_{\mathbb{C}}^{\lambda +y}$ for each addable node $y$, that is one summand per outgoing edge of step 2.1. The two displayed instances are the cases $\lambda =(2,2)$ with the single removable node and $\lambda =(2,1)$ with its three addable nodes.

Boundary and consistency audit. Rank $0$ carries the single vertex $\mathrm{\varnothing }$, whose unique standard tableau is the empty one, and the empty product $n!=0!=1$ matches $f^{\mathrm{\varnothing }}=1$; every partition of $n\ge 1$ has at least one removable node and at least one addable node by [F2] (for the addable case take the node opening a new row), so both branching sums are nonempty and each of the $1,2,4,7$ edges between consecutive ranks is counted exactly once in each direction; and the edge counts agree with the two enumerations of the same edge set, since summing the number of incoming edges over the partitions of $n$ for $n=1,2,3,4$ gives $1,2,4,7$, the same numbers as in step 2.1. All sets involved are finite and explicitly listed, so no choice principle enters. This proves claims 1 to 4 and hence the Statement.

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.

Put $p_{+}:=\frac{1+\tau }{2}$ and $p_{-}:=\frac{1-\tau }{2}$ in $\mathbb{C}[S_2]$. Using ${\tau }^2=\mathrm{i}\mathrm{d}$ one computes $p_{+}+p_{-}=\mathrm{i}\mathrm{d}$, $p_{+}^2=p_{+}$, $p_{-}^2=p_{-}$ and $p_{+}{p}_{-}=p_{-}{p}_{+}=0$; hence $E=p_{+}E\oplus p_{-}E$, and $p_{+}E$ is the fixed space ${\mathrm{Sym}}^{2}V$ while $p_{-}E$ is the anti-fixed space ${\mathrm{\Lambda }}^{2}V$.

The two rank-$2$ Specht modules are the expected one-dimensional modules. For $\lambda =(2)$ every $(2)$-tableau has the same tabloid $\{t\}=\{1,2\}$ and trivial column stabilizer, so $e_s=\{t\}$ for every such tableau and $S^{(2)}=\mathbb{C}\{t\}$ with $\tau \cdot \{t\}=\{\tau \cdot t\}=\{t\}$: the action is trivial. For $\lambda =(1,1)$ the two tabloids $\{u\}$ and $\tau \cdot \{u\}$ are distinct basis vectors of $M^{(1,1)}$; with $C_u=\{\mathrm{i}\mathrm{d},\tau \}$ one has ${\kappa }_u=\mathrm{i}\mathrm{d}-\tau$, so $e_u=\{u\}-\tau \cdot \{u\}\mathrm{\ne }0$, and $\tau \cdot e_u=\tau \cdot \{u\}-\{u\}=-e_u$; by [F3] any one polytabloid generates $S^{(1,1)}$, so $S^{(1,1)}=\mathbb{C}{e}_u$ is the sign representation, of dimension one. In particular $S^{(2)}$ is trivial and $S^{(1,1)}$ is the sign representation, as claimed in the Statement.

The two spaces are spanned by explicit tensors: set $s_{ii}:=e_i\otimes e_i$ and $s_{ij}:=e_i\otimes e_j+e_j\otimes e_i$ for $i<j$, and set $a_{ij}:=e_i\otimes e_j-e_j\otimes e_i$ for $i<j$. If $x={\sum }_{i,j}{c}_{ij}{e}_i\otimes e_j$ is fixed by $\tau$, then $c_{ij}=c_{ji}$, so $x={\sum }_i{c}_{ii}{s}_{ii}+{\sum }_{i<j}{c}_{ij}{s}_{ij}$. If $x$ is anti-fixed, then $c_{ij}=-c_{ji}$ and $2c_{ii}=0$; over $\mathbb{C}$ this gives $c_{ii}=0$, so $x={\sum }_{i<j}{c}_{ij}{a}_{ij}$. Conversely each $s_{ij}$ is fixed and each $a_{ij}$ is anti-fixed. Their respective coefficients on the tensor basis [F2] show that both displayed families are linearly independent: for the symmetric family use the coefficient of $e_i\otimes e_i$ for $s_{ii}$ and of $e_i\otimes e_j$ for $s_{ij}$ with $i<j$; for the alternating family use the coefficient of $e_i\otimes e_j$ for each $i<j$. Hence $\dim {\mathrm{Sym}}^{2}V=\left(\genfrac{}{}{0px}{}{d+1}{2}\right)=\frac{d(d+1)}{2}$ and $\dim {\mathrm{\Lambda }}^{2}V=\left(\genfrac{}{}{0px}{}{d}{2}\right)=\frac{d(d-1)}{2}$, and step 1.1 gives the direct sum $E={\mathrm{Sym}}^{2}V\oplus {\mathrm{\Lambda }}^{2}V$.