Every boundary tensor-ideal component is the fixed embedded two-sided ideal

For $n\ge 1$ write $E_n={\mathrm{ptr}}_n$. Closing matching graphs gives $B_{n-1}$-bilinearity and $E_n(x\otimes 1)=\delta x$. For $n\ge 2$, it also gives $E_n(U_{n-1})=1$. In this latter range, for $x\in B_{n-1}$, $U_{n-1}(x\otimes 1)U_{n-1}=(E_{n-1}(x)\otimes 1\otimes 1)U_{n-1}\mathrm{.}$ Indeed the two neighboring cups/caps close precisely the last input/output strand of $x$, leaving its other strands and the final cup/cap unchanged; every closed component contributes the same factor on both sides. For $x=1$ this gives $U_{n-1}^2=\delta U_{n-1}$, fixing the unnormalized factor. Since $B_n$ is generated by $B_{n-1}$ and $U_{n-1}$, this identity reduces every alternating word with two final generators to one with fewer such occurrences. Hence $B_n$ is spanned by $B_{n-1}+B_{n-1}{U}_{n-1}{B}_{n-1}$.

To handle arbitrary-size morphisms we give an auxiliary diagram factorization. It does not identify a positive-loop pivotal category with the quantum ribbon category. Choose $A=i{e}^{\pi i\mathrm{/}(2\mathrm{\ell })}$, so $\delta =-A^2-A^{-2}$, and put $C=A1+A^{-1}U$. Its inverse is $A^{-1}1+AU$ because $A^2+A^{-2}+\delta =0$. The adjacent triple expansion gives Yang–Baxter, with remaining coefficient $A+A^{-1}\delta +A^{-3}=0$. Let $W_1=\cup \otimes 1$ and $W_2=1\otimes \cup$. The snakes give $U_1{W}_1=\delta W_1$, $U_2{W}_1=W_2$ and $U_1{W}_2=W_1$, whence $C_1{C}_2{W}_1=(A^2+\delta +A^{-2})W_1+W_2=W_2,C_2{C}_1{W}_2=W_1\mathrm{.}$ Reflection gives the two corresponding cap slides. Thus ordered block crossings $C_{r,t}:r+t\to t+r$ are natural with every matching, since cups, caps and identities generate those morphisms; linearity gives naturality with arbitrary diagram elements. Partitioning a block proves both block-crossing identities, and the triple relation identifies the ordered crossing words. All crossings used here are finite linear combinations of ordinary diagrams.

Every nonempty generator word kills $f$ on its appropriate side, so $J_m=\mathbb{C}f$. The boundary partial trace is $E_m(f)=\frac{[\mathrm{\ell }]}{[\mathrm{\ell }-1]}{p}_{\mathrm{\ell }-2}=0.$ Its denominator is nonzero by [F4]. Thus $E_m(J_m)=0=J_{m-1}$. This covers a seam contraction even when no two adjacent input endpoints of the projector are joined by a cap.

Denote by ${\vartheta }_t$ the planar curl of this auxiliary block crossing. Closing one strand gives ${\vartheta }_1=A\delta +A^{-1}=-A^3$. The cup/cap slides of step 1.2 give ${\vartheta }_{r+t}=C_{t,r}{C}_{r,t}({\vartheta }_r\otimes {\vartheta }_t),{\mathrm{ptr}}_t(C_{t,t})={\vartheta }_t,$ where ${\mathrm{ptr}}_t$ closes the last $t$ strands. To verify the second identity, close those strands from right to left, slide each cap through its block crossing by the displayed cap equations, then remove its matching cup by a snake. One remaining strand acquires the scalar curl $-A^3$; the unclosed strands retain the two block-crossing words in the first identity. Induction, beginning with the computed one-strand value, proves both formulas. In particular ${\vartheta }_t$ is $(-A^3{)}^t$ times a product of invertible crossings, so its inverse exists in $B_t$. This is a calculation inside the positive-loop diagram category, independent of any universal quantum $R$.

Inclusion gives $J_n\otimes 1\subseteq J_{n+1}$. We prove $E_n(J_n)\subseteq J_{n-1}$ by induction; below $m$ it is trivial and at $m$ it is step 2.1. Let $n>m$, so the embedded generator is in $B_{n-1}$. Expand each multiplier of a generator $afb$ by the spanning formula of step 1.1. With all $a_i,b_i\in B_{n-1}$ and $U=U_{n-1}$, the four types are $a_{0}f{b}_0$, $a_{1}U{a}_{2}f{b}_0$, $a_{0}f{b}_{1}U{b}_2$, and $a_{1}U{a}_{2}f{b}_{1}U{b}_2$. Their partial traces are, respectively, $\delta a_{0}f{b}_0,a_1{a}_{2}f{b}_0,a_{0}f{b}_1{b}_2,a_1{E}_{n-1}(a_{2}f{b}_1)b_2\mathrm{.}$ The first three belong to $J_{n-1}$. In the fourth, the induction hypothesis puts the middle factor in $J_{n-2}$, whose inclusion belongs to $J_{n-1}$, or is zero at the boundary. This proves trace closure. Iterated partial traces consequently carry $J_N$ into $J_n$ for every $n\le N$.

Fix arbitrary $a:t\to n$, $b:n\to t$ and $F=f\otimes 1_{t-m}$, with $t\ge m$. Set $N=n+t$ and form the actual element $X=(a\otimes b)(F{\vartheta }_t^{-1}\otimes 1_n)C_{n,t}\in B_N\mathrm{.}$ The tensor $a\otimes b$ is an endomorphism of object $N$: its source has $t+n$ strands and its target $n+t$ strands. Its middle factor equals $(f\otimes 1_{N-m})({\vartheta }_t^{-1}\otimes 1_n)$, so $X\in J_N$. Naturality gives $(1_t\otimes b)C_{n,t}=C_{t,t}(b\otimes 1_t)$, so also $X=(aF{\vartheta }_t^{-1}\otimes 1_t)C_{t,t}(b\otimes 1_t)$. Pulling maps on the untraced wires outside the closure and using step 2.2 now gives ${\mathrm{ptr}}_t(X)=aF{\vartheta }_t^{-1}{\mathrm{ptr}}_t(C_{t,t})b=aFb\mathrm{.}$ By step 3.1 this belongs to $J_n$, even when $n<m$, where it is zero. This is an explicit factorization for all external strands and all seam contractions, not a claim that every low-rank diagram has an internal cap against $f$.

A tensor-ideal word containing $f$ is a finite sum of composites $a(1_r\otimes f\otimes 1_u)b$. If it contains further copies of $f$, expand those in their finite matching basis and absorb them into the multipliers, leaving one distinguished copy. Naturality with the auxiliary crossings moves $1_r\otimes f\otimes 1_u$ to $f\otimes 1_{r+u}$ by an invertible conjugation, absorbed in $a,b$. Step 4.1 therefore puts every such consequence in $J_n$, proving $\mathcal{I}(n,n)\subseteq J_n$. Conversely tensoring $f$ with identities and composing on either side puts the generators and all their two-sided multiples in $\mathcal{I}$. Hence $J_n\subseteq \mathcal{I}(n,n)$. Their equality and all partial-trace assertions follow, including the zero components below $m$.