A triple braid acts identically on fusion spectators

In the grouped tree, choose an admissible local fusion-tree basis vector $f_i\in V({\tau }^3\to a_i)$ for each triple and a basis vector $g\in V(a_1\otimes \cdots \otimes a_n\to c)$ for the outer fusion. The composites obtained by inserting $f_1,\dots ,f_n$ into $g$ are exactly the grouped fusion-tree vectors. By completeness and orthonormality [F1], their orthogonal direct sum over $(a_1,\dots ,a_n)$ is the full grouped fusion space, which gives the displayed tensor/direct-sum decomposition.

A braid word supported inside triple $i$ changes only the local morphism $f_i$, by $f_i\mapsto f_i\circ b^{-1}$ under [F2]; every $f_j$ with $j\mathrm{\ne }i$ and the outer morphism $g$ are unchanged. Thus on the grouped basis its matrix is ${\rho }_{a_i}(b)$ tensored with identities. Rebracketing to the fixed tree conjugates this matrix by the unitary in [F3], proving the same operator statement in that convention. Tensoring an operator $A$ with an identity does not increase its norm: expand a vector in an orthonormal basis of the spectator factor and sum $\mathrm{\parallel }A{\xi }_j{\mathrm{\parallel }}^2\le \mathrm{\parallel }A{\mathrm{\parallel }}^2\sum \mathrm{\parallel }{\xi }_j{\mathrm{\parallel }}^2$. The norm of a finite orthogonal direct sum is at most the maximum of its block norms by the same squared-norm decomposition. These finite identities justify uniform spectator bounds.