Root fusion paths have nonzero pairings

The theta formula and exact vertex convention are [F4]. At a retained triangle/parity triple, its denominator indices are $r,s,t\le \mathrm{\ell }-2$. If $h=(r+s+t)\mathrm{/}2\le \mathrm{\ell }-2$, every numerator index is at most $\mathrm{\ell }-1$ and [F3] makes the ratio positive. If $h\ge \mathrm{\ell }-1$, its numerator factorial $[h+1]!$ contains $[\mathrm{\ell }]$, whereas the denominator stays nonzero; its value is zero. These conclusions specialize the generic localization identity, not a ratio of theta values at the root.

For $1\le a\le k$ set $f_a=p_a\otimes 1$, $X_a=f_a{U}_a{f}_a$ and ${\lambda }_a=[a+1]\mathrm{/}[a]$. Define ${\alpha }_a=f_a(1_{a-1}\otimes \cup )p_{a-1},{\beta }_a=p_{a-1}(1_{a-1}\otimes \cap )f_a\mathrm{.}$ Absorption of $p_{a-1}$ next to $p_a$ gives ${\alpha }_a{\beta }_a=X_a$: insert the cup/cap and remove the absorbed lower projector. Closing the other composite is exactly the right partial trace of $p_a$ in its lower corner, so [F1] gives ${\beta }_a{\alpha }_a={\lambda }_a{p}_{a-1}$. Thus $X_a^2={\lambda }_a{X}_a$. The JW recursion gives $f_a=p_{a+1}+{\lambda }_a^{-1}{\alpha }_a{\beta }_a\mathrm{.}$ The two terms are orthogonal because $p_{a+1}$ kills the joining turnback. The down maps are inclusion ${\alpha }_a$ and retraction ${\lambda }_a^{-1}{\beta }_a$; the up maps are the tautological inclusion/retraction of $p_{a+1}$. For $a=0$ there is only $p_1$. Every denominator here has index at most $\mathrm{\ell }-1$ and is nonzero.

For retained endpoints, $\mathrm{Hom}(p_a,p_b)=0$ if $a\mathrm{\ne }b$ and is $\mathbb{C}{p}_a$ if $a=b$. Indeed any matching between unequal boundaries has an innermost adjacent cap or cup at the larger boundary and is killed by its JW projector. Between equal boundaries, every nonidentity matching has such a turnback. The identity corner survives in the quotient: its trace is $[a+1]>0$, while the boundary tensor ideal is negligible. For the latter assertion, $p_{k+1}$ has one-dimensional endomorphism corner and trace $[\mathrm{\ell }]=0$, so every test trace vanishes; tensor-ideal closure in [F5] gives the assertion at all objects. Thus the quotient introduces no extra identification in the retained scalar corners.

At $a=k$, pass to the categorical quotient and delete $p_{k+1}$ from the decomposition before taking any further normalization. The identity becomes $p_k\otimes 1={\lambda }_k^{-1}{\alpha }_k{\beta }_k$ with ${\lambda }_k=[\mathrm{\ell }-1]\mathrm{/}[\mathrm{\ell }-2]\mathrm{\ne }0$. It therefore has only its down summand. In particular at $\mathrm{\ell }=3$, $k=1$, this says that the quotient tensor square is its trivial channel; no retained highest-weight-$2$ summand is asserted. The endomorphism ideal being killed is exactly [F5]'s embedded one.

Start with the single rank-zero path and tensor its identity decomposition by $1$. Inductively split each endpoint summand $p_a\otimes 1$ by steps 1.2 and 2.1 and compose its up/down maps with the preceding path maps. Cross-composites of the new up/down terms vanish by turnback annihilation; the equal down composite is ${\lambda }_a^{-1}{\beta }_a{\alpha }_a=p_{a-1}$ and the up composite is $p_{a+1}$. Different preceding paths have zero composite by induction. Their new projections sum to the old projection tensored with $1$, because the split is complete; at the boundary only the down term remains. This proves simultaneously $r_a{i}_b=0$ for distinct paths, $r_a{i}_a=p_{a_n}$ and ${\sum }_a{i}_a{r}_a=1_n$ for every rank.

For paths with equal endpoint put $E_{a,b}=i_a{r}_b$. The composites in step 3.1 give their matrix-unit multiplication. Insert the complete path identity on each side of any morphism and use the corners of step 1.3: these units span all endomorphisms, and the same argument describes unequal-degree Hom spaces. They are independent because multiplying a putative relation by $r_a$ and $i_b$ isolates its coefficient times the nonzero endpoint projector. Closing $i_a{r}_b$ has the same finite component graph as closing $r_b{i}_a$, even though the cut occurs at a different object. Thus the trace is zero off the diagonal and $[a_n+1]$ on it. Each matrix block has nonzero weighted ordinary trace pairing, detected by its transposed matrix units. This proves completeness and nondegeneracy at every rank. Together with step 1.1 it proves all assertions without a recoupling division at the boundary.