The first symplectic Brauer relation in dimension two

Write $v_i=(x_i,y_i)$. Then $p_{ij}=x_i{y}_j-y_i{x}_j$. Expanding the left side gives six cancelling pairs of monomials: $x_1{x}_3{y}_2{y}_4$ with $-x_1{x}_3{y}_2{y}_4$, $-x_1{x}_4{y}_2{y}_3$ with $x_1{x}_4{y}_2{y}_3$, $-x_1{x}_2{y}_3{y}_4$ with $x_1{x}_2{y}_3{y}_4$, $-x_2{x}_3{y}_1{y}_4$ with $x_2{x}_3{y}_1{y}_4$, $x_2{x}_4{y}_1{y}_3$ with $-x_2{x}_4{y}_1{y}_3$, and $-x_3{x}_4{y}_1{y}_2$ with $x_3{x}_4{y}_1{y}_2$. Thus the multilinear contraction combination vanishes.

Name the three endomorphism diagrams $I,X,E$ as in the companion convention. In the symplectic tensor representation they act as $\mathrm{id},-P,c\omega$, where $c=-e_1\otimes e_2+e_2\otimes e_1$. The map respects their multiplication table: $X^2=I$, $XE=EX=E$, and $E^2=-2E$, the last equality following from $\omega (c)=-2$ and [F2]. Moreover $I+X+E$ annihilates $e_1\otimes e_1$, $e_2\otimes e_2$, $e_1\otimes e_2$, and $e_2\otimes e_1$, so it is the zero operator.

The element $I+X+E$ is nonzero in the free Brauer module because its three coefficients on distinct basis diagrams are nonzero; [F1] gives the matching count three. At $r=0$ the action is $\mathbb{Q}\to \mathrm{End}(\mathbb{Q})$ and is an isomorphism. At $r=1$, $B_1(-2)$ is spanned by the identity diagram. For a $2\times 2$ matrix $M$, $M^{\mathsf{T}}\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)M=(\det M)\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, so both $\mathrm{diag}(2,1\mathrm{/}2)$ and $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ lie in $S{p}_2(\mathbb{Q})$. A matrix commuting with the first is diagonal; if it also commutes with the second, its diagonal entries are equal. Hence ${\mathrm{End}}_{S{p}_2}(V)=\mathbb{Q}\mathrm{id}$ and the rank-one diagram action is an isomorphism. Therefore the nonzero kernel at $r=2=n+1$ is the first failure, exactly outside the symplectic stable range.