The orthogonal Brauer action has a kernel at dimension one and rank two

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.