Jones–Wenzl projectors over a coefficient ring and quantum-binomial units

By propagation in [F1], the span of the nonidentity diagrams is an ideal, so $\epsilon$ is a ring homomorphism; that ideal is spanned by nonempty generator words. Thus the linear equations $d_{i}e=0$, $\epsilon (e)=1$ imply $ae=\epsilon (a)e$ for every $a$, and in particular $e^2=e$. Horizontal reflection defines an anti-involution $\ast$ fixing generators and preserving $\epsilon$. The element $f=e{e}^{\ast }$ is killed by the generators on both sides and has augmentation $1$. Right annihilation gives $fe=f$, while left annihilation of $e$ gives $fe=e$; hence $e=f$ is two-sided. For two such elements $e,e^{\mathrm{\prime }}$, the same argument gives $e{e}^{\mathrm{\prime }}=e$ by right annihilation of $e$ and $e{e}^{\mathrm{\prime }}=e^{\mathrm{\prime }}$ by left annihilation of $e^{\mathrm{\prime }}$. They are equal. These arguments use no division and work over every commutative coefficient ring.

First work over any commutative ring $C$ containing a unit $q$ with $\delta =q+q^{-1}$. In the weight-$k$ word module of [F2], the equations for the common kernel are $a_{u10v}=-q{a}_{u01v}$. Successive adjacent swaps reduce each word to $0^{n-k}{1}^k$ and change its inversion number by one. Thus this kernel is exactly the free line generated by $v_k={\sum }_w(-q{)}^{\mathrm{inv}(w)}w$, whose baseword coefficient is $1$. If $e$ is normalized, step 1.1 and symmetry in [F2] put both the columns and rows of its weight block in this line. First comparing the baseword column, and then each row, expresses the block as ${\alpha }_k{v}_k{v}_k^{\mathsf{T}}$ for some ${\alpha }_k\in C$. Every nonempty generator word kills $v_k$ by its rightmost factor; consequently $(e-1)v_k=0$. Comparing the baseword coefficient in $e{v}_k=v_k$ gives ${\alpha }_k{G}_{n,k}(q^2)=1$. In particular all these Gaussian values are units. This conclusion remains valid with nilpotents or zero divisors.

For general $R$, put $R^{\mathrm{\prime }}=R[q]\mathrm{/}(q^2-\delta q+1)$. Division by this monic quadratic gives a unique representative $a+bq$: subtract successive leading terms to prove existence, while a nonzero multiple of a monic polynomial has degree at least two, proving uniqueness. Thus $R^{\mathrm{\prime }}=R\oplus Rq$ as an $R$-module, $R$ injects into $R^{\mathrm{\prime }}$, and $q$ is a unit with inverse $\delta -q$. Base-changing an existing $e$ to $R^{\mathrm{\prime }}$ and applying step 2.1 makes $G_{n,k}(q^2)$ a unit. By [F3] the factor $q^{k(n-k)}$ is a unit, so $B=B_{n,k}(\delta )$ is a unit in $R^{\mathrm{\prime }}$. Write its inverse $a+bq$. The equality $B(a+bq)=1$ in the indicated free basis gives $Ba=1$ in $R$. Hence $B$ is already a unit in $R$.

To prove sufficiency construct a universal projector with only the named denominators. Use an indeterminate $q$, $A=\mathbb{Z}[q,q^{-1}]$, and its fraction field $L=\mathbb{Q}(q)$. Facts [F3, F4] give a normalized projector $p\in {\mathrm{T}\mathrm{L}}_n(L,q+q^{-1})$. Step 2.1 determines each of its weight blocks as $v_k{v}_k^{\mathsf{T}}\mathrm{/}{G}_{n,k}(q^2)$. Let $S=A[G_{n,k}(q^2{)}^{-1}:0\le k\le n]\subset L$. That matrix has entries in $S$. Apply the base-changed linear left inverse of [F2] to obtain a diagram-coefficient vector $z\in {\mathrm{T}\mathrm{L}}_n(S,q+q^{-1})$. After extension to $L$ it equals $p$. Since $S$ injects into $L$ and the diagram module is free, the same identities already hold over $S$: $\epsilon (z)=1$, $d_{i}z=z{d}_i=0$ and $z^2=z$. Thus no unproved assertion about the image or centralizer of a quantum-group action is needed.

Suppose now every $B_{n,k}(\delta )$ is a unit in $R$, and form $R^{\mathrm{\prime }}$ as in step 3.1. Fact [F3] makes all evaluated $G_{n,k}(q^2)$ units in $R^{\mathrm{\prime }}$. Therefore evaluation defines a ring homomorphism $S\to R^{\mathrm{\prime }}$: send a fraction $a\mathrm{/}\prod G_{n,k}^{e_k}$ to the evaluated numerator times the inverses of the indicated evaluated denominators; the cross-multiplied equality defining equal fractions remains true after evaluation. Specialize $z$ from step 3.2 to get a normalized annihilator $e^{\mathrm{\prime }}\in {\mathrm{T}\mathrm{L}}_n(R^{\mathrm{\prime }},\delta )$. Let $\pi :R^{\mathrm{\prime }}\to R$ be the $R$-linear coefficient projection $\pi (a+bq)=a$ and apply $\pi$ to every diagram coefficient of $e^{\mathrm{\prime }}$, obtaining $e\in {\mathrm{T}\mathrm{L}}_n(R,\delta )$. Left multiplication by each $d_i$ has matrix entries in $R$, so it commutes with this coefficient projection. Hence $d_{i}e=0$ and $\epsilon (e)=\pi (1)=1$. Step 1.1 now makes $e$ the required unique two-sided idempotent. The projection is used only on these linear equations; it need not preserve multiplication.

Any homomorphism $R\to T$ sends the diagram coefficients and their defining linear identities to those over $T$, so the image of the projector is the unique projector there by step 1.1. For $n=0,1$ there are no generator equations, the identity is the projector, and every binomial in the row is $1$. The same argument includes the zero ring under the convention that its identity is its unique element. This establishes all stated implications and boundaries.