Specht Gram rank for shape (2,2)

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.