Jones–Wenzl projectors: annihilation, uniqueness, partial trace, and closure

At ranks $0$ and $1$, the initial projectors are identities and there are no generators; the diagram algebras are $R$, so nonzeroness, idempotence, annihilation, identity coefficient and the scalar-corner assertion hold directly. At rank $2$, the recursion gives $p_2=1-{\delta }^{-1}{U}_1$ only when $[2]=\delta$ is a unit. Then $p_2^2=p_2$ follows from $U_1^2=\delta U_1$, $U_1{p}_2=p_2{U}_1=0$, and its identity coefficient is $1$; it is nonzero because the identity and cup-cap diagrams are distinct basis vectors. For the strong induction step, assume the stated assertions for all projectors through $p_r$ [IH]. For $r\ge 2$, put $Q=p_{r-1}\otimes 1\otimes 1$, $a=U_{r-1}$, $b=U_r$, and $\mu =[r-1]\mathrm{/}[r]$. The recursion gives $P_r=Q-\mu QaQ$. The supports of $Q$ and $b$ are disjoint, so they commute; using $b^2=\delta b$ and $bab=b$ from [F2] gives $b{P}_{r}b=\delta Qb-\mu QbabQ=(\delta -\mu )Qb=([r+1]\mathrm{/}[r])Qb$, where [F3] proves the last equality. Since $Q{P}_r=P_r$ and $Qb=bQ$, this is the absorption identity needed below.

Set $X=P_{r}b{P}_r$ and $c=[r]\mathrm{/}[r+1]$. Step 1.1 gives $X^2=P_r(b{P}_{r}b)P_r=([r+1]\mathrm{/}[r])X$, hence $p_{r+1}^2=(P_r-cX{)}^2=P_r-2cX+c^2{X}^2=P_r-cX=p_{r+1}$. Also $b{p}_{r+1}=b{P}_r-cb{P}_{r}b{P}_r=0$, because $b{P}_{r}b=([r+1]\mathrm{/}[r])Qb$ and $Qb{P}_r=b{P}_r$. By [IH], every earlier generator annihilates $P_r$ on both sides. Horizontal reflection fixes the recursion and interchanges left and right, so $p_{r+1}{U}_r=0$ too. The correction term has fewer than $r+1$ through strands, while $P_r$ has identity coefficient $1$; hence $p_{r+1}$ has identity coefficient $1$ and is nonzero.

If $D$ is any nonidentity basis diagram, [F4] writes it as a nonempty word in the generators. Its rightmost letter annihilates $p_n$ on the left and its leftmost letter annihilates $p_n$ on the right, by step 2.1. Thus $D{p}_n=p_{n}D=0$. Expanding any $x\in {\mathrm{T}\mathrm{L}}_n$ in the diagram basis gives $p_{n}x{p}_n={\lambda }_x{p}_n$ for a scalar ${\lambda }_x\in R$, hence $p_n{\mathrm{T}\mathrm{L}}_n{p}_n=R{p}_n$. If $p_n^{\mathrm{\prime }}$ is another idempotent with identity coefficient $1$ and the same annihilation, write $p_n=1+m$, $p_n^{\mathrm{\prime }}=1+m^{\mathrm{\prime }}$ with $m,m^{\mathrm{\prime }}$ linear combinations of nonidentity diagrams. The equalities $p_n^{\mathrm{\prime }}{p}_n=p_n$ and $p_n^{\mathrm{\prime }}{p}_n=p_n^{\mathrm{\prime }}$ follow respectively from $m^{\mathrm{\prime }}{p}_n=0$ and $p_n^{\mathrm{\prime }}m=0$; therefore $p_n=p_n^{\mathrm{\prime }}$.

For $n\ge 2$, write $p_n=P-([n-1]\mathrm{/}[n])P{U}_{n-1}P$ with $P=p_{n-1}\otimes 1$. Closing the last identity strand gives ${\mathrm{ptr}}_n(P)=\delta p_{n-1}$ by [F6]. Closing the last strand of $U_{n-1}$ gives the straight identity, so ${\mathrm{ptr}}_n(P{U}_{n-1}P)=p_{n-1}^2=p_{n-1}$. Therefore ${\mathrm{ptr}}_n(p_n)=(\delta -[n-1]\mathrm{/}[n])p_{n-1}=([n+1]\mathrm{/}[n])p_{n-1}$ by [F3]. For $n=1$, ${\mathrm{ptr}}_1(1_1)=\delta 1_0=[2]p_0$, the same formula.

By [F6], ${\mathrm{Tr}}_0(p_0)=1=[1]$. Inductively, the partial-trace factorization gives ${\mathrm{Tr}}_n(p_n)={\mathrm{Tr}}_{n-1}({\mathrm{ptr}}_n(p_n))=([n+1]\mathrm{/}[n]){\mathrm{Tr}}_{n-1}(p_{n-1})=[n+1]$. Together with steps 1.1–3.2, this completes the strong induction and proves all assertions.