Hom N-complexes

Question

$ \newcommand{\Hom}{\mathrm{Hom}} $Let $N\geqslant 2$. An $N$-complex $X$ is a sequence $\cdots \stackrel{d_X^{i-1}}{\longrightarrow}X^i\stackrel{d_X^{i}}{\longrightarrow}X^{i+1}\stackrel{d_X^{i+1}}{\longrightarrow}\cdots$ satisfying $d_X^{i+N-1}\cdots d_X^{i+1}d_X^i=0$ for any $i$. In classical textbook, we can construct a Hom 2-complex $\Hom(X,Y)$ (i.e., complex) as follows: $\Hom^n(X,Y)=\prod_{p\in Z}\Hom(X^p,Y^{p+n})$ with the $n$-th differential $d(f)=d_Yf+(-1)^{n+1}fd_X$. I do not know how to define a Hom $N$-complex $\Hom(X,Y)$ for any $N$-complexes $X$ and $Y$. Does this approach work for $N$-complexes?

Answer 1

Yes, use the differential $d(f)=d_Yf+q^{n+1}fd_X$ where $q$ is a primitive $N$:th root of unity.

Assuming that we use a differential of the form $d(f)=d_Yf+q^{n+1}fd_X$ or a similar $q$-commutativity relation for the underlying bi-complex, then the total complex is a $N$-complex if an only if $q$ is a primitive $N$:th root of unity. For a poof of the "if" part see Prop. 1.8 in https://arxiv.org/abs/q-alg/9611005 and for the "only if" part, see Th. 3.1 in https://arxiv.org/abs/1807.02100.

It is also worth mentioning that if you work with $p$-complexes over prime characteristic $p$, then our $q$-commutativity translates to ordinary commutativity because $q=1$ is the only primitive $p$:th root of unity. This case was studied back in 1942 by Mayer, see https://www.jstor.org/stable/1968874.