Flip-category colored unlink and braid evaluations

If $n=1$, the braid group is trivial and $\pi$ is its unique homomorphism. If $n\ge 2$, put $s_i=(ii+1)$. For $\mathrm{\mid }i-j\mathrm{\mid }>1$, $s_i$ and $s_j$ commute because they have disjoint supports; for adjacent indices, both $s_i{s}_{i+1}{s}_i$ and $s_{i+1}{s}_i{s}_{i+1}$ equal $(ii+2)$. These are the Artin relators, so [F5] gives the homomorphism $\pi$. By [F4], each positive or negative braid generator swaps its two adjacent tensor factors, so the action on $W=V^{\otimes n}$ is the factor permutation given by $\pi (\beta )$. The twist is identity. Let $w_1,\dots ,w_{d^n}$ be the tensor basis of $W$ and $w_1^{\ast },\dots ,w_{d^n}^{\ast }$ its ordinary dual basis. Substitution in [F2]'s formula gives $u_W(w_r)(w_s^{\ast })=1$ when $r=s$ and $0$ otherwise, so the pivotal comparison is the canonical double-dual identification. The categorical trace formula [F1] is therefore the ordinary matrix trace of that permutation operator.

Fix a basis $e_1,\dots ,e_d$ of $V$. A basis tensor $e_{i_1}\otimes \cdots \otimes e_{i_n}$ is fixed by the factor permutation exactly when its indices are constant on each orbit of $\pi (\beta )$; by [F3], these orbits are exactly the cycles counted by $c(\pi (\beta ))$, including fixed points. There are $d$ independent choices for each cycle, hence exactly $d^{c(\pi (\beta ))}$ fixed basis tensors. A permutation matrix has trace equal to its number of fixed basis vectors, proving the formula.

The identity in $B_1$ has one permutation cycle, the identity in $B_2$ has two, and ${\sigma }_1$ has the transposition permutation with one cycle. Substitution into the formula gives $d$, $d^2$, and $d$, respectively; the latter closure is a single component because its endpoint permutation has one cycle. These calculations include $d=0$ because every listed cycle count is positive.