Jones–Wenzl existence over a field is detected by one quantum-binomial row

Propagation cannot increase under stacking, and the identity is the only fully propagating diagram. Hence the nonidentity span is an ideal and the identity-coefficient map $\epsilon$ is multiplicative. By [F2] its kernel is spanned by nonempty generator words. If $d_{i}e=0$ and $\epsilon (e)=1$, [F2] gives $ae=\epsilon (a)e$ for every $a$, so $e^2=e$ follows already from these linear equations. Reflect diagrams to obtain an algebra anti-involution $\ast$ fixing the generators and preserving $\epsilon$: reflection reverses stacking and preserves each loop, so it has these properties by [F1]. Then $f=e{e}^{\ast }$ is annihilated on both sides by every $d_i$, has augmentation $1$, and is idempotent by the first observation. Since $f$ is right annihilated, $fe=\epsilon (e)f=f$; since $e$ is left annihilated, $fe=\epsilon (f)e=e$. Thus $e=f$ is itself two-sided. If $e,e^{\mathrm{\prime }}$ are two normalized two-sided annihilators, the same generator-word argument gives $e{e}^{\mathrm{\prime }}=e$ and $e{e}^{\mathrm{\prime }}=e^{\mathrm{\prime }}$, proving uniqueness.

First suppose $K$ contains a unit $q$ with $\delta =q+q^{-1}$, and put $Q=q^2$. In weight $k$, the word space of [F3] has basis the words with $k$ ones. A vector $\sum a_{w}w$ is killed by all generators exactly when $a_{u10v}=-q{a}_{u01v}$ for every adjacent mixed pair. Adjacent swaps connect every such word to $0^{n-k}{1}^k$, and each swap changes the inversion count by one. Therefore the common kernel is the free line spanned by $v_k={\sum }_w(-q{)}^{\mathrm{inv}(w)}w,$ whose coefficient at $0^{n-k}{1}^k$ is $1$. Its ordinary bilinear square is $v_k^{\mathsf{T}}{v}_k=G_{n,k}(Q)$. These calculations apply over every commutative ring containing the unit $q$, including characteristic two.

For the converse construction use first the indeterminate $q$ and $L=\mathbb{Q}(q)$. Every positive quantum integer is nonzero in $L$. There is a normalized two-sided annihilator $p_n$ over $L$, as the following algebraic induction proves. Start with $p_0=p_1=1$ and $p_2=1-{\delta }^{-1}{d}_1$; their identity coefficients are $1$, and the rank-two relations give idempotence and two-sided annihilation. For $r\ge 2$, assume $p_{r-1}$ and $p_r$ have been recursively constructed, are idempotent and annihilated on both sides by all generators available at their ranks, and satisfy $\epsilon (p_{r-1})=\epsilon (p_r)=1$. Put $P=p_r\otimes 1$, $Q_0=p_{r-1}\otimes 1\otimes 1$, $a=d_{r-1}$, $b=d_r$, and $\mu =[r-1]\mathrm{/}[r]$. The preceding recursion says $P=Q_0-\mu Q_{0}a{Q}_0$, while $Q_0^2=Q_0$, $P{Q}_0=Q_{0}P=P$, $Q_{0}b=b{Q}_0$ and $bab=b$. Consequently $bPb=(\delta -\mu )Q_{0}b=([r+1]\mathrm{/}[r])Q_{0}b$. For $X=PbP$, this gives $X^2=([r+1]\mathrm{/}[r])X$. Therefore $p_{r+1}=P-([r]\mathrm{/}[r+1])X$ is idempotent, annihilates $b$ on both sides by the displayed sandwich identity, and annihilates every earlier generator because $P$ does. Juxtaposing a vertical strand preserves the identity coefficient, so the leading term $P$ has identity coefficient $\epsilon (p_r)=1$. Every diagram term of the correction $PbP$ has fewer than $r+1$ propagating strands, since it passes through $b=d_r$ and stacking cannot increase the number of propagating strands; hence that correction has identity coefficient zero. Thus $\epsilon (p_{r+1})=1$. The initial rank-two verification uses $d_1^2=\delta d_1$; the induction step applies for $r\ge 2$. This proves the claim at every rank without using a complex-field theorem outside its hypotheses.

If $e$ exists, step 1.1 and [F3] show that every column and row of its weight-$k$ matrix is a multiple of $v_k$. Comparing first the column at $0^{n-k}{1}^k$, and then all rows, makes that block ${\alpha }_k{v}_k{v}_k^{\mathsf{T}}$ for some scalar ${\alpha }_k$. Since $e-1$ is in the augmentation ideal, [F2] and the common-kernel property give $e{v}_k=v_k$. Hence ${\alpha }_k{G}_{n,k}(Q)=1$, by comparing the coefficient of $0^{n-k}{1}^k$. Thus $G_{n,k}(Q)\mathrm{\ne }0$ and [F4] gives $B_{n,k}(\delta )\mathrm{\ne }0$ for every $k$.

The computation of steps 1.2–2.1 over $L$ now determines the entire matrix of $p_n$: its weight-$k$ block is $v_k{v}_k^{\mathsf{T}}\mathrm{/}{G}_{n,k}(q^2)$. Let $S$ be the subring of $L$ generated by $\mathbb{Z}[q,q^{-1}]$ and all $G_{n,k}(q^2{)}^{-1}$, $0\le k\le n$. That displayed matrix $P_n$ has all entries in $S$. Apply the integral linear left inverse of [F3], after base change to $S$, to obtain $z\in {\mathrm{T}\mathrm{L}}_n(S,q+q^{-1})$. Over $L$ it equals $p_n$, since $P_n={\rho }_n(p_n)$. Since both $S$ and its free diagram module inject into their scalar extensions to $L$, it follows already over $S$ that ${\rho }_n(z)=P_n$, $\epsilon (z)=1$, $z^2=z$, and $d_{i}z=z{d}_i=0$. This is the precise integrality step: no unproved centralizer or arbitrary-characteristic Schur–Weyl assertion is being used.

Return to a field containing a chosen $q_0$ with $\delta =q_0+q_0^{-1}$, and suppose all $B_{n,k}(\delta )$ are nonzero. Then [F4] makes every denominator defining $S$ nonzero at $q_0$, so evaluation of finite fractions defines a homomorphism $S\to K$, $q\mapsto q_0$. Applying it to $z$ from step 3.1 constructs the required normalized idempotent in ${\mathrm{T}\mathrm{L}}_n(K,\delta )$. All identities specialize as polynomial identities after the named nonzero denominators are inverted; division by a vanishing quantum integer is never attempted.

For a general field $K$, adjoin a root of $X^2-\delta X+1$ using [F5] if needed. This gives a field extension $K^{\mathrm{\prime }}$, with a nonzero root and the same polynomial values $B_{n,k}(\delta )$. Necessity follows in $K^{\mathrm{\prime }}$ by step 2.1. For sufficiency, step 4.1 gives a solution over $K^{\mathrm{\prime }}$ to the finite linear system $d_{i}e=0$, $\epsilon (e)=1$ in diagram coefficients. Row reduction over $K$ shows that a linear system with coefficients and constants in $K$ is consistent over $K$ whenever it is consistent over a field extension: an inconsistent row $0=1$ would persist over that extension, while a row-echelon system without such a row has a solution by setting free variables to zero. Choose such a $K$-solution. Step 1.1 then proves its idempotence, both-sided annihilation and uniqueness. Ranks zero and one give the identity, with all the displayed quantum binomials equal to $1$. This completes both implications over every field.