Commuting versus anticommuting double complex conventions

Suppose horizontal and vertical homological arrows $h,v$ separately square to zero and satisfy the commuting convention $hv=vh$. Define ${\tilde{v}}_{p,q}=(-1{)}^p{v}_{p,q}$, leaving $h$ unchanged. On $C_{p,q}$ the two mixed composites have sum $h_{p,q-1}{\tilde{v}}_{p,q}+{\tilde{v}}_{p-1,q}{h}_{p,q}=(-1{)}^p{h}_{p,q-1}{v}_{p,q}+(-1{)}^{p-1}{v}_{p-1,q}{h}_{p,q}=0.$ Also ${\tilde{v}}_{p,q-1}{\tilde{v}}_{p,q}=(-1{)}^{2p}{v}_{p,q-1}{v}_{p,q}=0$. Thus $(C,h,\tilde{v})$ satisfies Homological double complex.

Applying the same twist twice restores $v$. Starting with anticommuting arrows instead, the same calculation gives commuting twisted arrows. Bidegree-preserving morphisms commute with the twist because their source and target have the same $p$. Zero arrows and characteristic two cause no exception: an additive inverse and its original still sum to zero.

Consequently the expression for the total differential is $h+(-1{)}^{p}v$ in the commuting convention and $h+\tilde{v}$ after translation. The library starts with anticommuting arrows, so its total differential is simply $h+v$. An additional sign twist on those arrows would change the convention again. All signs are specified integers; no selection or infinite summation occurs.