Partial-trace closure of negligible morphisms

For any $a:Y\to Z$ and any $h:Z\to X$, the trace of $h(af)$ is ${\mathrm{Tr}}_L(j_{X}haf)$. By the different-object cyclicity in [F2] and naturality $j_{Y}f=f^{\vee \vee }{j}_X$, this equals ${\mathrm{Tr}}_L(j_{Y}fha)$. Since $ha:Y\to X$, [F1] makes this zero. The same argument with a map $b:W\to X$ and a test map $g:Y\to W$ gives ${\mathrm{Tr}}_L(j_{X}bgf)=0$; hence post- and pre-composition preserve negligibility. Closure under sums and scalar multiples follows from linearity of trace.

Let $Z$ be any object and $h:Y\otimes Z\to X\otimes Z$. Write ${\stackrel{}{\mathrm{ev}}}_Z={\mathrm{ev}}_{Z^{\vee }}(j_Z\otimes 1_{Z^{\vee }}):Z\otimes Z^{\vee }\to \mathbf{1}$, the right evaluation determined by the pivotal structure, and define ${\mathrm{ptr}}_Z(h)=(1_X\otimes {\stackrel{}{\mathrm{ev}}}_Z)(h\otimes 1_{Z^{\vee }})(1_Y\otimes {\mathrm{coev}}_Z):Y\to X$. Expanding the trace of $j_{X\otimes Z}h(f\otimes 1_Z)$ gives ${\mathrm{ev}}_{Z^{\vee }\otimes X^{\vee }}(((j_X\otimes j_Z)h(f\otimes 1_Z))\otimes 1_{Z^{\vee }\otimes X^{\vee }})(1_X\otimes {\mathrm{coev}}_Z\otimes 1_{X^{\vee }}){\mathrm{coev}}_X$. The tensor-evaluation formula factors ${\mathrm{ev}}_{Z^{\vee }\otimes X^{\vee }}$ by first evaluating the middle $Z^{\vee \vee }\otimes Z^{\vee }$ pair and then the outer $X^{\vee \vee }\otimes X^{\vee }$ pair; substituting ${\stackrel{}{\mathrm{ev}}}_Z$ leaves exactly ${\mathrm{Tr}}_L(j_X{\mathrm{ptr}}_Z(h)f)$. This regrouping uses monoidality of $j$ and the dual zig-zags [F2,F3]. Since ${\mathrm{ptr}}_Z(h):Y\to X$ is an allowed test map in [F1], the value is zero, so $f\otimes 1_Z$ is negligible.

Suppose first that the two pivotal traces agree on all endomorphisms. For $h:Z\otimes Y\to Z\otimes X$ set ${\stackrel{}{\mathrm{coev}}}_Z=(1_{Z^{\vee }}\otimes j_Z^{-1}){\mathrm{coev}}_{Z^{\vee }}$ and $q=({\mathrm{ev}}_Z\otimes 1_X)(1_{Z^{\vee }}\otimes h)({\stackrel{}{\mathrm{coev}}}_Z\otimes 1_Y):Y\to X$. Expand the right pivotal trace $r_{Z\otimes X}(h(1_Z\otimes f))$ using the tensor evaluation and coevaluation formulas. Its inner closure is the $Z$-closure defining $q$; the remaining outer closure is $r_X(qf)$. This is the mirror of the nested calculation in step 2.1 and uses only the two snakes and monoidality of $j$. Thus $t_{Z\otimes X}(h(1_Z\otimes f))=r_{Z\otimes X}(h(1_Z\otimes f))=r_X(qf)=t_X(qf)=0$, by the trace-agreement hypothesis and negligibility of $f$. This proves left tensor closure in this alternative; the two changes of trace are essential.

Suppose instead that a braiding is given. Step 2.1 makes $f\otimes 1_Z$ negligible. Naturality gives $1_Z\otimes f=c_{Z,Y}^{-1}(f\otimes 1_Z)c_{Z,X}$. The two outside factors have the displayed compatible types, and composition closure from step 1.1 therefore proves $1_Z\otimes f$ negligible. This alternative does not require equality of the two pivotal traces.

For any $g:Z\to W$, $f\otimes g=(1_Y\otimes g)(f\otimes 1_Z)$ and $g\otimes f=(g\otimes 1_Y)(1_Z\otimes f)$. Composition closure from step 1.1 and step 2.1 give right tensor closure for every pivotal category. Under either stated additional hypothesis, step 3.1 or step 3.2 also gives left tensor closure. In the Temperley–Lieb category, [L1] also verifies the partial-trace identities by counting components of the closed finite stacked graph. For the full rank-one tilting category, [F5] supplies the prequotient braiding required in step 3.2, so its negligible morphisms also form a two-sided tensor ideal for every $\mathrm{\ell }\ge 3$. Together with linearity this proves every stated closure assertion.