Brauer, BMW, and Partition Diagram Algebras: Examples and Counterexamples

Gluing the two vertices of $U$ to the two vertices of $A$ closes one matched component and leaves no outer vertex, so the result is one loop times the empty identity diagram, namely $\delta 1_0$.

In $d\circ d$, the two middle singleton blocks are identified with each other and with no outer vertex, so they form exactly one removed class. The outer vertices remain two singleton blocks, giving $\delta d$. Substituting $\delta =m$ or $\delta =-2n$ gives those same loop scalars in the respective classical branches.

Name the three matchings $I$ (through identity), $X$ (crossed through strands), and $E$ (top cup and bottom cap). The orthogonal tensor assignment sends them to $\mathrm{id}$, $P$, and $cB$, respectively. On $e\otimes e$, each is the identity because $P(e\otimes e)=e\otimes e$ and $cB(e\otimes e)=e\otimes e$. Also $O(V)=\{1,-1\}$ acts trivially on $V^{\otimes 2}$, whose endomorphism algebra is one-dimensional.

At $\delta =1$, stacking any two of $I,X,E$ gives one of these basis diagrams with coefficient $1$ (the possible closed loop also has coefficient $\delta =1$). Since all three images are the identity, this defines the contraction algebra map and its image is one-dimensional. Its domain has rank three by [F1] and Ranks of the Brauer and partition diagram bases, so the map has a two-dimensional kernel. For $r=0$ the map is $\mathbb{Q}\to \mathrm{End}(\mathbb{Q})$ sending identity to identity, and for $r=1$ both $B_1(1)$ and ${\mathrm{End}}_{O(V)}(V)$ are one-dimensional with identity mapping to identity. Thus $r=2=m+1$ is the first failure.

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.

The discrete partition has two blocks, while the one-block partition coarsens it. Möbius inversion on this two-element interval gives $x_d=d-c$; by [F1] it is nonzero because $d$ and $c$ are distinct basis vectors.

The one-dimensional tensor basis has only the index $1$, so its equality pattern is $c$ and it cannot realize the two-block pattern $d$. Equivalently, both $d$ and $c$ act by the $1\times 1$ matrix $(1)$, so $x_d$ acts by zero. Since $2>N=1$, this is the below-stable-range kernel predicted by [F2].

For $(g,e)=(2,0)$ and $(-1\mathrm{/}2,0)$, $g-g^{-1}=3\mathrm{/}2=z(1-e)$ and the loop and untwisting relations hold because $e=0$. For $(g,e)=(1\mathrm{/}3,25\mathrm{/}9)$, $g-g^{-1}=-8\mathrm{/}3=z(1-e)$, $ge=eg=25\mathrm{/}27={\rho }^{-1}e$, and $e^2=625\mathrm{/}81=\delta e$. Every $g$ is invertible, and BMW${}_2$ has no adjacent-index relations, so these are valid one-dimensional representations.

The three representations evaluate $g_1$ at the distinct numbers $2,-1\mathrm{/}2,1\mathrm{/}3$. Since the third sends $e_1$ to $25\mathrm{/}9\mathrm{\ne }0$, the ideal generated by $e_1$ is nonzero, so the canonical map to the quotient by that ideal is not injective. The quadratic $(g_1-2)(g_1+1\mathrm{/}2)$ evaluates to $-25\mathrm{/}18\mathrm{\ne }0$ in the third representation, whereas setting $e_1=0$ forces that quadratic by [F2].