Fusion-tree basis and dimension recursion

For the empty input list, the tensor is $\mathbf{1}$. If a nonzero map $\mathbf{1}\to d$ exists, its kernel is zero and its image is the nonzero subobject $d$, so it is an isomorphism by simplicity [F6]. Since all charge labels are fixed representatives, this occurs exactly when $d=\mathbf{1}$; then $\mathrm{Hom}(\mathbf{1},d)=\mathbb{C}{\mathrm{id}}_{\mathbf{1}}$ by [F5], and the identity is the empty tree vector. Thus the empty tree has one basis vector precisely for $d=\mathbf{1}$, giving $m_0(d)$. For one input $a_1$, the same kernel/image argument shows $\mathrm{Hom}(a_1,d)=0$ unless $a_1=d$. In that case $\mathrm{End}(a_1)=\mathbb{C}{\mathrm{id}}_{a_1}$ by [F5], and the identity is the unique tree vector. This gives $m_1(d)$.

Proceed by induction on the number of leaves. Write a nontrivial tree as $T=T_L\otimes T_R$, with left and right input sublists. By induction, for each simple representative output $e$ the left-tree vectors $t_{e,\lambda }$ form an orthonormal basis of $\mathrm{Hom}(T_L,e)$. Put $i_{e,\lambda }:=t_{e,\lambda }^{†}:e\to T_L$. If $e=e^{\mathrm{\prime }}$, then $i_{e,\lambda }^{†}{i}_{e,{\lambda }^{\mathrm{\prime }}}={\delta }_{\lambda ,{\lambda }^{\mathrm{\prime }}}{\mathrm{id}}_e$ by [F4]; if $e\mathrm{\ne }{e}^{\mathrm{\prime }}$, the same composite is zero because a nonzero map between nonisomorphic simple objects is an isomorphism [F1, F6]. Thus these inclusions are orthogonal. Choose a finite semisimple decomposition of $T_L$ into simple representatives by [F1], and choose this decomposition orthogonally by [F2], so its inclusions are orthogonal and its adjoint projections sum to ${\mathrm{id}}_{T_L}$. For each simple representative $e$, its adjoint projections form an orthonormal family in $\mathrm{Hom}(T_L,e)$. Semisimplicity decomposes this Hom space as the direct sum of $\mathrm{Hom}(f,e)$ over the simple representative summands $f$ of $T_L$; only $f=e$ contributes, and that space has dimension one by [F5]. Thus the projections form a basis. Changing that basis to $t_{e,\lambda }$ multiplies the inclusion columns by a unitary matrix, which preserves orthogonality and the sum-of-projections identity. The same argument applies to $T_R$. Tensoring the child inclusions preserves these relations by [F2]. For each pair of child outputs $e,f$, choose an orthogonal finite semisimple decomposition of $e\otimes f$ into simple representatives, as permitted by [F1] and [F2]; the number of copies of $d$ is $N_{ef}^d$ by [F1] and [F7]. Composing the child inclusions with an orthonormal basis of each $V_{ef}^d$ gives orthogonal isometric inclusions of $d$ into $T$. Summing these range projections over all simple outputs gives ${\mathrm{id}}_T$, because the child decompositions and each decomposition of $e\otimes f$ do. The fixed-$d$ subfamily sums to the $d$-isotypic projection $P_d$. Every map $f:T\to d$ vanishes on all other simple-output summands by [F5] and [F6], so $f=f{P}_d$. Taking adjoints of the fixed-$d$ inclusions therefore gives a complete orthonormal basis of $\mathrm{Hom}(T,d)$, exactly the fusion-tree vectors of [F3].

For each pair of representative child outputs $e,f$, the construction in step 1.2 contributes $m_{T_L}(e)m_{T_R}(f)N_{ef}^d$ tree vectors. Summing over all representative $e,f$ yields the recurrence $m_T(d)={\sum }_{e,f}{m}_{T_L}(e)m_{T_R}(f)N_{ef}^d$. Repeated expansion of this finite recurrence chooses one representative label at each internal edge and one basis index at each vertex, so it equals the displayed sum of products over all admissible labelings.

For the left-associated tree, the right subtree at the last vertex is the single input $a_n$. Its only nonzero simple output is $a_n$, with multiplicity one by [F5] and [F6]. Substituting this into the recurrence of step 2.1 gives $m_n(d)={\sum }_e{m}_{n-1}(e)N_{e{a}_n}^d$ for $n\ge 2$, with the empty and one-input values already established in step 1.1. Every sum is finite because the set of simple charges and each multiplicity space are finite.