Diagram closure trace is cyclic and satisfies the tower rules

For basis diagrams $D,E:n\to n$, closing the stack $DE$ makes a finite graph whose vertices are glued endpoints and whose components are circles. Moving the cut between the $D$-layer and the $E$-layer changes the cyclic order in which the same two layers are read but neither the component partition nor the number of circles. Hence the closed graph of $DE$ and that of $ED$ have the same number of circles. Definition [F3] assigns both the same power of $\delta$, so ${\mathrm{Tr}}_n(DE)={\mathrm{Tr}}_n(ED)$. Bilinearity extends this equality to arbitrary $x,y$.

In the closure of $D\otimes 1_1$, the extra vertical strand closes to one circle disjoint from every component of the closure of $D$. Thus the closed graph has exactly $c(D)+1$ circles; its value is ${\delta }^{c(D)+1}=\delta {\mathrm{Tr}}_n(D)$, an identity valid also at $\delta =0$.

Let $n\ge 1$. In the closure of $(D\otimes 1_1)U_n$, the cap and cup of $U_n$ reconnect the added strand to the final old strand. Following the two arcs through the stack shows they form part of the same exterior-to-exterior path that was closed in $D$; they create no separate circle. Every other component is unchanged. The circle count is therefore $c(D)$, proving ${\mathrm{Tr}}_{n+1}((D\otimes 1)U_n)={\mathrm{Tr}}_n(D)$. Bilinearity gives the result for $x$.

Let $n\ge 1$. By [F3], ${\mathrm{ptr}}_n(h)$ is obtained by adding the rightmost cup, stacking $h$ on the enlarged graph, and then joining the rightmost pair by the cap. Closing the remaining $n-1$ strands produces exactly the same finite graph as closing $h(x\otimes 1)$ after the same $x$ has been stacked on the first $n-1$ strands. The two closures therefore have equal circle counts and equal powers of $\delta$ for each basis pair $(h,x)$. Bilinearity proves the partial-trace identity. Together with steps 1.1–1.3 this proves all claims.