Specht Modules and the Irreducibles of the Symmetric Group — Examples

The three tabloids are $v_1,v_2,v_3$: the second row is a singleton, and its label determines the first row as the complementary pair. They are distinct by [F2], so they are exactly the tabloid basis of [F1].

Write $[ab;c]$ for a tableau with first row $a,b$ and second row $c$, where $\{a,b,c\}=\{1,2,3\}$. Its columns are $\{a,c\}$ and $\{b\}$, so [F4] gives $C_s=\{1,(ac)\}$. The three transpositions, in one-line notation on the labels $1,2,3$, are $213$, $321$, and $132$, with respectively $1$, $3$, and $1$ inversions, and the order-preserving relabelling $\{1,2,3\}\to \{0,1,2\}$ preserves these counts. Thus [F6] gives $\mathrm{sgn}(ac)=-1$, and [F5] yields ${\kappa }_s=1-(ac)$. The second row of $\{s\}$ is $\{c\}$, so applying $(ac)$ changes it to $\{a\}$ and $e_s=v_c-v_a$.

Applying step 1.2 to all six tableaux gives $e_{[12;3]}=v_3-v_1$, $e_{[21;3]}=v_3-v_2$, $e_{[13;2]}=v_2-v_1$, $e_{[31;2]}=v_2-v_3$, $e_{[23;1]}=v_1-v_2$, and $e_{[32;1]}=v_1-v_3$. These are precisely the three listed differences and their negatives.

The row and column inequalities in [F3] leave exactly $t=[12;3]$ and $u=[13;2]$ as standard tableaux. Their polytabloids $v_3-v_1$ and $v_2-v_1$ are linearly independent: in a relation $\alpha (v_3-v_1)+\beta (v_2-v_1)=0$, the coefficients of the distinct basis vectors $v_3$ and $v_2$ force $\alpha =\beta =0$.

By [F7], the standard polytabloids of shape $(2,1)$ form a basis of $S^{(2,1)}$; step 2.2 identifies that standard family as exactly $e_t,e_u$. Together with the six explicit calculations in step 2.1, this proves the Statement.

Let $n\ge 1$ and take the row-filled tableau $t$ of shape $(n)$. By [F4,F5], every tableau of this shape has singleton columns, so its antisymmetrizer is $1$ and every polytabloid is its tabloid. Every tableau has row set $\{1,\dots ,n\}$ by [F14], so [F6] makes all tabloids equal; [F3] then gives $S^{(n)}=\mathbb{C}{e}_t$. The group fixes the sole tabloid by [F8], so $a{e}_t\mapsto a$ identifies the action with the trivial representation [F11].

Let $n\ge 1$ and take the tableau $t$ whose single column is filled by $1,\dots ,n$ from top to bottom, which exists by [F14]. By [F4,F5], $C_t=S_n$; [F7,F8] make the tabloids $\{\sigma \cdot t\}$ distinct basis vectors, so $e_t={\sum }_{\sigma \in S_n}\mathrm{sgn}(\sigma )\{\sigma \cdot t\}$ has coefficient $1$ at $\{t\}$ by [F9] and is nonzero.

For $\tau \in S_n$, reindexing the sum of step 1.2 by $\rho =\tau \sigma$ gives $\tau \cdot e_t={\sum }_{\rho \in S_n}\mathrm{sgn}({\tau }^{-1}\rho )\{\rho \cdot t\}=\mathrm{sgn}(\tau )e_t$, since [F13] implies $\mathrm{sgn}({\tau }^{-1})=\mathrm{sgn}(\tau )$. By [F10], this orbit spans $S^{(1^n)}$, so step 1.2 gives $S^{(1^n)}=\mathbb{C}{e}_t$; the map $a{e}_t\mapsto a$ identifies its action with the sign representation [F12].

When $n=0$, [F16] gives $S^{\mathrm{\varnothing }}=\mathbb{C}{e}_{\mathrm{\varnothing }}=\mathbb{C}$, and [F15] says $S_0=\{1\}$ acts as the identity; therefore this is the one-dimensional trivial representation [F11].

By [F1], the partitions of $3$ are exactly $(3),(2,1),(1^3)$.

By [F8], the Specht modules of shapes $(3)$ and $(1^3)$ are respectively one-dimensional trivial and sign modules, so each has dimension $1$.

Write $v_i$ for the tabloid whose second row is $\{i\}$; these are the three distinct tabloid basis vectors by [F4]. Give $P={\mathbb{C}}^{(\{1,2,3\})}$ the basis $w_1,w_2,w_3$ and natural action $\sigma \cdot w_i=w_{\sigma (i)}$ by [F3]. The basis map $\mathrm{\Phi }(v_i)=w_i$ is a linear isomorphism, and [F5] gives $\mathrm{\Phi }(\sigma \cdot v_i)=\mathrm{\Phi }(v_{\sigma (i)})=w_{\sigma (i)}=\sigma \cdot \mathrm{\Phi }(v_i)$, so it is $S_3$-equivariant.

In the ordered basis $(a,b)$, the left actions satisfy $(12)a=v_3-v_2=a-b$, $(12)b=v_1-v_2=-b$, $(23)a=v_2-v_1=b$, and $(23)b=v_3-v_1=a$, so their matrices (basis vectors as columns) are $\rho ((12))=\left(\begin{array}{cc}1& 0\\ -1& -1\end{array}\right)$ and $\rho ((23))=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. After the order-preserving relabelling $1,2,3\mapsto 0,1,2$, the transpositions $(12),(23),(13)$ have respectively $1,1,3$ inversions, so every transposition has sign $-1$ by [F15]; the relabelling preserves these counts. The six elements are $1,(12),(23),(13),(123),(132)$, with $(13)=(12)(23)(12)$, $(123)=(12)(23)$, and $(132)=(23)(12)$, so the two displayed actions determine the full $S_3$-action.

The image of $S^{(2,1)}$ under $\mathrm{\Phi }$ is precisely the coordinate-sum-zero subspace $P_0:=\{x_1{w}_1+x_2{w}_2+x_3{w}_3:x_1+x_2+x_3=0\}$, and it has basis $a=w_3-w_1$, $b=w_2-w_1$, hence dimension two.

The trivial and sign modules are nonisomorphic, and $S^{(2,1)}$ has dimension two rather than one.

Applying [F17] at $n=3$ shows that these three modules are all finite-dimensional complex irreducible $S_3$-representations and form a complete irredundant list.

Every tableau of shape $(3,1)$ has a first column of size two and two singleton columns; if its bottom entry is $i$ and the entry above it is $j$, then [F2] gives $C_t=\{1,(ij)\}$. Its tabloid is $v_i$, and $(ij)\cdot v_i=v_j$. By [F3] and [F4], both signs reduce to $1$ in $K$ because $1=-1$ in ${\mathbb{F}}_2$, so $e_t=v_i+v_j$. Thus all polytabloids lie in $W:=\ker ϵ$, where $ϵ({\sum }_i{a}_i{v}_i)={\sum }_i{a}_i$.

The tableaux with top rows $[4,2,3]$, $[4,1,3]$, $[4,1,2]$ and respective bottom entries $1,2,3$ give $b_1=v_1+v_4$, $b_2=v_2+v_4$, $b_3=v_3+v_4$. These vectors are independent by their first three coordinates. If $x={\sum }_i{a}_i{v}_i\in W$, then $a_4=a_1+a_2+a_3$, so $x=a_1{b}_1+a_2{b}_2+a_3{b}_3$. Therefore $W$ has basis $b_1,b_2,b_3$, and since each is a polytabloid while every polytabloid lies in $W$, $S_K^{(3,1)}=W$.

The vector $w=v_1+v_2+v_3+v_4$ is nonzero, has $ϵ(w)=4=0$ in $K$, and is fixed by every permutation in $S_4$. Hence $Kw$ is a nonzero subrepresentation of $S_K^{(3,1)}$ by [F5] and [F6]. It is proper because $S_K^{(3,1)}=W$ has the three-element basis from step 2.1, whereas $Kw$ has dimension one. Thus this Specht module is reducible, refuting the claimed field-independent irreducibility.