Bounded graded bimodule complexes and signed tensor totalization

Fix a commutative ring $k$ and unital associative graded $k$-algebras $B,A,C$ (Associative graded algebras, bimodules, and internal shifts). A bounded cochain complex of graded $(B,A)$-bimodules is a cochain complex in the sense of Cochain complex in an abelian category that is bounded in the sense of Bounded, bounded below, and bounded above complexes, $F=(F^p,d_F^p{)}_{p\in \mathbb{Z}}$, with each $F^p$ a graded $(B,A)$-bimodule and each differential $d_F^p:F^p⟶F^{p+1}$ is a degree-zero bimodule map. Thus each $d_F^p$ preserves internal degree and commutes with both outer actions, and $d_F^{p+1}{d}_F^p=0$.

For a bounded cochain complex $F$ of graded $(B,A)$-bimodules and a bounded cochain complex $G$ of graded $(A,C)$-bimodules, the signed tensor totalization uses the graded balanced tensor product (Graded balanced tensor product and homogeneous Hom) and has cochain degree $n$ term $\mathrm{Tot}(F{\otimes }_{A}G{)}^n:={⨁}_{p+q=n}{F}^p{\otimes }_A{G}^q$ and differential on $f\in F^p$, $g\in G^q$ given by $d(f\otimes g):=d_F(f)\otimes g+(-1{)}^{p}f\otimes d_G(g)\mathrm{.}$ The balanced tensor carries its total internal grading: if $f$ has internal degree $r$ and $g$ has internal degree $s$, then $f\otimes g$ has internal degree $r+s$. Its outer actions are $b(f\otimes g)c=(bf)\otimes (gc)$ for $b\in B$ and $c\in C$. When $G$ is instead a bounded cochain complex of graded left $A$-modules, omit the right $C$-action and retain the induced left $B$-action.

This is the existing tensor-product total complex (The tensor product of a right and a left chain complex is totalized by direct sums with the Koszul differential) after reindexing cochain degree $p$ as chain degree $-p$; its Koszul sign is therefore $(-1{)}^p$. The internal $\mathbb{Z}$-grading is independent of cochain degree and contributes no additional sign. If $F$ is supported in $[a,b]$ and $G$ in $[c,d]$, then the totalization is supported in $[a+c,b+d]$, and each diagonal has only finitely many summands. If either input is the zero complex, the totalization is zero. If one input is concentrated in cochain degree $r$, then

$$\mathrm{Tot}(F{\otimes }_{A}G{)}^n=F^r{\otimes }_A{G}^{n-r},d=(-1{)}^r(1\otimes d_G)\mathrm{.}$$

If instead $G$ is concentrated in cochain degree $s$, then

$$\mathrm{Tot}(F{\otimes }_{A}G{)}^n=F^{n-s}{\otimes }_A{G}^s,d=d_F\otimes 1.$$

In particular, when both differentials vanish the total differential is zero.