Integral Specht Modules and Modular Simple Modules — Examples

The column stabilizer of $s_1=12\mathrm{/}34$ is $C_{s_1}=\{1,(13),(24),(13)(24)\}$, and the four tabloids of its column orbit are pairwise distinct: $1$ gives $T_1=\{\{1,2\},\{3,4\}\}$; $(13)$ gives the tableau $32\mathrm{/}14$ with row sets $\{2,3\},\{1,4\}$, so $\{(13)s_1\}=T_2=\{\{2,3\},\{1,4\}\}$; $(24)$ gives $14\mathrm{/}32$ with row sets $\{1,4\},\{2,3\}$, so $\{(24)s_1\}=T_3=\{\{1,4\},\{2,3\}\}$; and $(13)(24)$ gives $34\mathrm{/}12$, so $\{(13)(24)s_1\}=T_4=\{\{3,4\},\{1,2\}\}$. Hence $e_{s_1}=T_1-T_2-T_3+T_4$. Similarly $C_{s_2}=\{1,(12),(34),(12)(34)\}$ for $s_2=13\mathrm{/}24$: $1$ gives $U_1=\{\{1,3\},\{2,4\}\}$; $(12)$ gives $23\mathrm{/}14$, so $T_2$; $(34)$ gives $14\mathrm{/}23$, so $T_3$; and $(12)(34)$ gives $24\mathrm{/}13$, so $U_4=\{\{2,4\},\{1,3\}\}$. Hence $e_{s_2}=U_1-T_2-T_3+U_4$. The tabloids $T_1,T_2,T_3,T_4$ are distinct, as are $U_1,T_2,T_3,U_4$, by [F1] and [F2].

Because the tabloid basis is orthonormal by [F3], a polytabloid whose expansion in tabloids has all coefficients in $\{0,1,-1\}$ at pairwise distinct tabloids pairs with itself to the number of its nonzero terms; by step 1.1, $\beta (e_{s_1},e_{s_1})=4$ and $\beta (e_{s_2},e_{s_2})=4$. The supports of $e_{s_1}$ and $e_{s_2}$ meet exactly in the two tabloids $T_2$ and $T_3$, where the coefficients are $-1$ in both polytabloids, so $\beta (e_{s_1},e_{s_2})=(-1)(-1)+(-1)(-1)=2$; symmetry of $\beta$ gives $\beta (e_{s_2},e_{s_1})=2$ as well. Therefore $G_{(2,2)}=\left(\begin{array}{cc}4& 2\\ 2& 4\end{array}\right)\mathrm{.}$

Let $p$ be a prime and reduce the entries of $G_{(2,2)}$ modulo $p$. For $p=2$ all four entries vanish, so the reduced matrix is the zero matrix of rank $0$. For $p=3$ the reduction is $\left(\begin{array}{cc}1& 2\\ 2& 1\end{array}\right)$, which is nonzero while its determinant $1\cdot 1-2\cdot 2=-3$ vanishes, so its rank is $1$; its first row is nonzero, and the two columns are proportional, which confirms the rank directly. For $p>3$ the determinant $12=2^2\cdot 3$ is nonzero in $k$, so the rank is $2$. The determinant and the largest nonvanishing minor of an integer matrix depend only on the characteristic of $k$, so the same answer holds for every field of the given characteristic.

By [F5] the dimension of the modular quotient $D^{(2,2)}$ over a splitting field $k$ of characteristic $p$ is the rank computed in step 3.1, namely $0$ for $p=2$, $1$ for $p=3$ and $2$ for $p>3$. Finally, by [F4] the images of the standard polytabloids $e_{s_1},e_{s_2}$ form a $k$-basis of $S_k^{(2,2)}$ for every field $k$, so ${\dim }_k{S}_k^{(2,2)}=2$ in every characteristic; in characteristic $2$, where the reduced Gram matrix vanishes and hence $R^{(2,2)}=S_k^{(2,2)}$, this exhibits a nonzero Specht module whose invariant bilinear form is identically zero and whose form quotient is zero.

By [F2] and [F3], $M_k^{(2,1)}$ has $k$-basis $v_1,v_2,v_3$ and $S_k^{(2,1)}$ has $k$-basis $e_t=v_3-v_1$, $e_u=v_2-v_1$. Applying the orthonormal form ${\beta }_k$ of [F4], $\beta (e_t,e_t)=\beta (v_3-v_1,v_3-v_1)=1+1=2,\beta (e_u,e_u)=2,\beta (e_t,e_u)=\beta (v_3-v_1,v_2-v_1)=1,$ so $G_{(2,1)}=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$.

The unique $(3)$-tabloid $v$ is fixed by $S_3$, so $S_k^{(3)}=kv$ is the one-dimensional trivial module, and $G_{(3)}=(1)$ has rank $1$; over a field of characteristic $2$ the sign representation is trivial by [F8], and over a field of characteristic $3$ it is nontrivial. The column stabilizer of a $(1,1,1)$-tableau $w$ is all of $S_3$, and the single standard polytabloid is $e={\sum }_{\sigma \in S_3}\mathrm{sgn}(\sigma )\{\sigma \cdot w\};$ its six tabloid coefficients are $\pm 1$ at the six distinct orderings, so $e\mathrm{\ne }0$, and for $\tau \in S_3$ the substitution ${\sigma }^{\mathrm{\prime }}=\tau \sigma$ gives $\tau \cdot e={\sum }_{\sigma }\mathrm{sgn}(\sigma )\{\tau \sigma \cdot w\}=\mathrm{sgn}(\tau ){\sum }_{{\sigma }^{\mathrm{\prime }}}\mathrm{sgn}({\sigma }^{\mathrm{\prime }})\{{\sigma }^{\mathrm{\prime }}\cdot w\}=\mathrm{sgn}(\tau )e\mathrm{.}$ Hence $S_k^{(1,1,1)}=ke$ is the one-dimensional sign representation.

Reducing $G_{(2,1)}$ modulo $p$: for $p=2$ the reduction $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ has determinant $1$, hence rank $2$; for $p=3$ the reduction is nonzero with determinant $3\equiv 0$, hence rank $1$, its columns being proportional. By the dimension formula of [F4], ${\dim }_k{D}^{(2,1)}={\mathrm{rank}}_p(G_{(2,1)}<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p)=\{\begin{array}{ll}2,& p=2,\\ 1,& p=3.\end{array}$ Also ${\dim }_k{D}^{(3)}={\mathrm{rank}}_p(1)=1$ for both primes.

Let $p=2$. Since ${\dim }_k{D}^{(2,1)}=2={\dim }_k{S}_k^{(2,1)}$ by steps 2.1 and 1.1, the radical $R^{(2,1)}$ is zero and $S_k^{(2,1)}=D^{(2,1)}$ is simple by [F6]. Since ${\dim }_k{D}^{(3)}=1={\dim }_k{S}_k^{(3)}$, also $S_k^{(3)}=D^{(3)}$, the trivial module, and by [F8] and step 1.2 the sign module satisfies $S_k^{(1,1,1)}\cong S_k^{(3)}=D^{(3)}$. Therefore $d_{(3),(3)}=1$, $d_{(3),(2,1)}=0$; $d_{(2,1),(3)}=0$, $d_{(2,1),(2,1)}=1$; $d_{(1,1,1),(3)}=1$, $d_{(1,1,1),(2,1)}=0$, which is the matrix displayed for $p=2$.

Let $p=3$. Here $D^{(3)}$ is the trivial module of dimension $1$ by step 1.2 and step 2.1, and $D^{(2,1)}$ is a simple module of dimension $1$ that is not isomorphic to $D^{(3)}$ by [F6]. The sign module $S_k^{(1,1,1)}$ of step 1.2 is one-dimensional, hence simple by [F8], so it is isomorphic to $D^{(3)}$ or to $D^{(2,1)}$ by [F6]; it is nontrivial in characteristic $3$ by step 1.2, hence it is not $D^{(3)}$ and $S_k^{(1,1,1)}\cong D^{(2,1)}$, so $d_{(1,1,1),(3)}=0$ and $d_{(1,1,1),(2,1)}=1$. For $S_k^{(2,1)}$, by [F3] $w:=e_t+e_u=(v_3-v_1)+(v_2-v_1)=v_1+v_2+v_3-3v_1=v_1+v_2+v_3$ because $3v_1=0$ in characteristic $3$; here $w\mathrm{\ne }0$ and $\sigma \cdot w=w$ for every $\sigma \in S_3$ by [F2], so $kw$ is a one-dimensional trivial submodule of $S_k^{(2,1)}$, isomorphic to $D^{(3)}$. On the quotient $S_k^{(2,1)}\mathrm{/}kw$, which is one-dimensional, the transposition $(12)$ acts by $(12)\cdot e_t=v_3-v_2\equiv -e_t$, since $v_3-v_2+e_t=2v_3-v_1-v_2=3v_3-w\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}kw)$; the quotient therefore is a nontrivial one-dimensional simple module, hence isomorphic to $D^{(2,1)}$. Its dimension $2$ equals $1+1$, so the composition factors of $S_k^{(2,1)}$ are $D^{(3)}$ and $D^{(2,1)}$, each once: $d_{(2,1),(3)}=d_{(2,1),(2,1)}=1$. With $d_{(3),(3)}=1$ and $d_{(3),(2,1)}=0$ from $S_k^{(3)}=D^{(3)}$, this is the matrix displayed for $p=3$.

Both matrices satisfy the constraints of [F7]. The forced zero $d_{(3),(2,1)}=0$ holds in both, because $(2,1)⋭(3)$; the $p$-regular diagonal entries $d_{(3),(3)}=d_{(2,1),(2,1)}=1$ hold in both; and with the $p$-regular rows and columns in the decreasing lexicographic order $(3),(2,1)$ the leading block is $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$ at $p=2$ and $\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ at $p=3$, lower unitriangular in both cases. The entries of the row $(1,1,1)$ lie in allowed positions since $(1,1,1)$ is dominated by every partition of $3$. The two matrices coincide with those recorded in the source reference for $S_3$, and the position $((2,1),(3))$ shows the characteristic dependence: $0$ at $p=2$ and $1$ at $p=3$.

For $\lambda =(2)$ one has $z_2(\lambda )=1<2$, so $(2)$ is $2$-regular; and ${\lambda }_1-{\lambda }_2=2-0=2$, which is not $<2$, so $(2)$ is not $2$-restricted. Thus $(2)$ is $2$-regular but not $2$-restricted.

For $\lambda =(1,1)$ one has $z_1(\lambda )=2$, which is not $<2$, so $(1,1)$ is not $2$-regular; and the padded differences are ${\lambda }_1-{\lambda }_2=1-1=0<2$ and ${\lambda }_2-{\lambda }_3=1-0=1<2$, so $(1,1)$ is $2$-restricted. Thus $(1,1)$ is $2$-restricted but not $2$-regular.

By [F2], $(2{)}^{\mathrm{\prime }}=(1,1)$ and $(1,1{)}^{\mathrm{\prime }}=(2)$: the diagram of $(2)$ has two columns of height $1$, and the diagram of $(1,1)$ has one column of height $2$. This is exactly the conjugation exchange of [F1] that matches the two partitions of steps 1.1 and 1.2.

Steps 1.1 and 1.2 exhibit partitions of the same integer $2$ that lie in exactly one of the two classes: $(2)$ is $2$-regular and not $2$-restricted, while $(1,1)$ is $2$-restricted and not $2$-regular. Hence the two families do not coincide, and labelling by one of them is not labelling by the other. Moreover the two labels describe the same simple module: by [F3] applied to the $2$-regular partition $(2)$, whose conjugate $(1,1)$ is $2$-restricted, $D^{(2)}\cong D((1,1))\otimes \mathrm{sgn}\cong D((1,1)),$ the last step because the sign representation is trivial in characteristic $2$ by [F4]. So a statement about the $p$-restricted label $D(\mu )$ cannot be applied to the James label $D^{\lambda }$ without transposing the partition (and, in odd characteristic, inserting the sign twist); the change of partition is present already at $p=2$, where the sign twist itself is invisible.

Take $k$ of characteristic $3$ and $\lambda =(2,1)$, so $n=3$. By [F1] and [F3] the module $S_k^{(2,1)}$ has $k$-basis $e_t=v_3-v_1$, $e_u=v_2-v_1$, and $e_t+e_u=(v_3-v_1)+(v_2-v_1)=v_1+v_2+v_3-3v_1=v_1+v_2+v_3=:w$ because $3v_1=0$ in characteristic $3$. In particular $w\in S_k^{(2,1)}$ and $w\mathrm{\ne }0$, since its coefficients at the basis vectors $v_1,v_2,v_3$ are all $1$. For every $\sigma \in S_3$ one has $\sigma \cdot w=v_{\sigma (1)}+v_{\sigma (2)}+v_{\sigma (3)}=w$ by [F2], so the one-dimensional subspace $kw\subseteq S_k^{(2,1)}$ is a submodule. It is proper because ${\dim }_k{S}_k^{(2,1)}=2$ by [F1]. Hence $S_k^{(2,1)}$ has a proper nonzero submodule and is not simple by [F7]; the characteristic-$3$ witness is a nonzero two-dimensional modular Specht module with a one-dimensional trivial submodule.

Take $k$ of characteristic $2$ and $\lambda =(2,2)$, so $n=4$. By [F5] the reduction of the integral Gram matrix $G_{(2,2)}$ modulo $2$ is the zero matrix, of rank $0$; by the dimension formula of [F4] this gives ${\dim }_k{D}^{(2,2)}={\mathrm{rank}}_2(G_{(2,2)}<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2)=0$, so $D^{(2,2)}=0$ while $S_k^{(2,2)}\mathrm{\ne }0$ of dimension $2$ by [F5]. Equivalently, $(2,2)$ has the part $2$ occurring twice and is $2$-singular, so [F6] also predicts $D^{(2,2)}=0$. Thus the Gram quotient of a nonzero modular Specht module can vanish.

Step 1.1 exhibits a nonzero modular Specht module that is not simple, and step 1.2 exhibits a nonzero modular Specht module whose invariant-form quotient is zero; the two failures are independent, since the first occurs for a $p$-regular label (where $D^{(2,1)}\mathrm{\ne }0$ by [F6]) and the second for a $p$-singular one. Hence the statement refuted fails in both clauses, and no field-independent appeal to the ordinary-case simplicity or to nonvanishing of the form quotient is available in prime characteristic.