Unnormalized diagram closure and partial trace

Let $R$ be a commutative ring, $\delta \in R$, and use the noncrossing-pairing diagram category of Temperley–Lieb planar diagram data and endomorphism modules, with the left factor stacked above the right factor. For a basis diagram $D:n\to n$, close each top endpoint to the corresponding bottom endpoint outside the rectangle and let $c(D)$ be the number of circles in the resulting finite graph. Define

${\mathrm{Tr}}_n(D)={\delta }^{c(D)}$

and extend $R$-linearly to ${\mathrm{T}\mathrm{L}}_n(R,\delta )$. This is the unnormalized closure trace. In particular, ${\mathrm{Tr}}_0(1_0)=1$ and ${\mathrm{Tr}}_n(1_n)={\delta }^n$. No unit or nonzero condition on $\delta$ is required.

Let $\cup :0\to 2$ and $\cap :2\to 0$ be the standard cup and cap in the diagram category. For $n\ge 1$ define the right partial trace by

${\mathrm{ptr}}_n(x)=(1_{n-1}\otimes \cap )\circ (x\otimes 1_1)\circ (1_{n-1}\otimes \cup ),x\in {\mathrm{T}\mathrm{L}}_n(R,\delta )\mathrm{.}$

It is a map ${\mathrm{T}\mathrm{L}}_n\to {\mathrm{T}\mathrm{L}}_{n-1}$. If $\delta$ is a unit, define ${\tau }_n={\delta }^{-n}{\mathrm{Tr}}_n$. This agrees with the normalized trace of Normalized Temperley–Lieb trace from diagram closure over $\mathbb{C}$ when $\delta \mathrm{\ne }0$.

The partial trace closes only the rightmost strand; its cap/cup order and type are fixed by the displayed composite. The unnormalized trace remains defined at $\delta =0$, while ${\tau }_n$ does not.