Exact Temperley–Lieb truncation reduces decomposition to trivial factors

We construct $y,z:n\to r$ with $y\circ z^{\ast }=1_r$. Write $n=r+2k$. If $k=0$, take identities. If $r>0$, take $y$ with bottom cups $(1,2),(3,4),\dots ,(2k-1,2k)$ and all remaining sites propagating; take $z$ with cups $(2,3),(4,5),\dots ,(2k,2k+1)$ and all remaining sites propagating. In $y\circ z^{\ast }$, the first through strand follows the single alternating chain through sites $1,2,\dots ,2k+1$, while each remaining through strand is straight. There is no closed component, and the exterior matching is $1_r$. If $r=0$ and $\delta \mathrm{\ne }0$, choose the same adjacent perfect matching $y=z_0$ on $n=2k$ sites and put $z={\delta }^{-k}{z}_0$. Then $y\circ z^{\ast }={\delta }^{-k}{\delta }^k{1}_0=1_0$. This also handles $n=r=0$ with the empty diagram and no scalar inversion.

Set $e=z^{\ast }\circ y\in A_n$. Step 1.1 gives $e^2=e$. The map $A_r\to e{A}_{n}e$ sending $a$ to $z^{\ast }\circ a\circ y$ is multiplicative because $y\circ z^{\ast }=1_r$, and its inverse is $b\mapsto y\circ b\circ z^{\ast }$. Indeed the first composite is identity, and the second is $b\mapsto ebe=b$ on the corner. Thus it is an algebra isomorphism with identity $1_r$ mapping to $e$.

The functor $M\mapsto Me$ preserves exact sequences. If $0\to M^{\mathrm{\prime }}\to M\to M^{\mathrm{\prime }\mathrm{\prime }}\to 0$ is exact, its injection remains injective on the subspaces obtained by multiplication by $e$. If $x^{\mathrm{\prime }\mathrm{\prime }}e\in M^{\mathrm{\prime }\mathrm{\prime }}e$, lift $x^{\mathrm{\prime }\mathrm{\prime }}$ to $x\in M$ and then $xe$ lifts $x^{\mathrm{\prime }\mathrm{\prime }}e$, proving surjectivity. If $xe$ maps to zero, it belongs to $M^{\mathrm{\prime }}$ and satisfies $(xe)e=xe$, hence belongs to $M^{\mathrm{\prime }}e$, proving exactness in the middle. These are corner-module maps under step 2.1.

For every label $t$, the map $W_{n,t}e\to W_{r,t}$ sending $xe$ to $x\circ z^{\ast }$ is well-defined, since diagrams of propagation less than $t$ remain such after composition. Its inverse sends $u:r\to t$ to $u\circ y$, which lies in $W_{n,t}e$. Their composites are identity by $y\circ z^{\ast }=1_r$ and $xe=x\circ z^{\ast }\circ y$. They respect right $A_r$ action under the corner identification in step 2.1. If $t>r$, every diagram $r\to t$ has propagation less than $t$, so both spaces vanish.

We verify compatibility with radicals without requiring $e=e^{\ast }$. Swapping $y,z$ in step 3.2 identifies $W_{n,t}{e}^{\ast }$ with $W_{r,t}$ by $x\mapsto x\circ y^{\ast }$, whose inverse is $v\mapsto v\circ z$. The form pairing of $u\circ y$ and $v\circ z$ is the coefficient of $1_t$ in $u\circ y\circ z^{\ast }\circ v^{\ast }=u\circ v^{\ast }$, hence is ${\varphi }_{r,t}(u,v)$. Moreover, for $x\in W_{n,t}e$ and any $w\in W_{n,t}$, invariance gives ${\varphi }_{n,t}(x,w)={\varphi }_{n,t}(xe,w)={\varphi }_{n,t}(x,w{e}^{\ast })$. Therefore $x$ is in the full radical precisely when its pairing against $W_{n,t}{e}^{\ast }$ vanishes. Under the two identifications this is exactly the condition that $u$ belongs to $\mathrm{rad}{\varphi }_{r,t}$. Thus $(\mathrm{rad}{\varphi }_{n,t})e$ corresponds to $\mathrm{rad}{\varphi }_{r,t}$, and exactness in step 3.1 proves $D_{n,t}e\cong D_{r,t}$.

A finite-dimensional $A_n$-module has a composition series by induction on its vector-space dimension. Apply the exact functor of step 3.1 to such a series for $W_{n,m}$. By [F3] and step 4.1, each simple factor $D_{n,t}$ contributes either zero or the simple factor $D_{r,t}$. Distinct surviving labels remain distinct by [F3]. In particular, only the factor with $t=r$ contributes $D_{r,r}$, and it contributes once. Step 4.1 identifies the resulting filtered module with $W_{r,m}$, giving the claimed multiplicity equality.

The sole basis vector of $W_{r,r}$ is $1_r$; every generator lowers propagation and thus acts by zero, and its form value on that vector is $1$. Hence $D_{r,r}=W_{r,r}$ is the stated one-dimensional module. Finally, for $n>0$ even and $\delta =0$, gluing any two perfect matchings creates at least one closed component, so ${\varphi }_{n,0}$ is identically zero and $D_{n,0}=0$. This explains the precise excluded branch and completes the assertion over every field, including characteristic two.