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Suppose that $f, g$ are chain maps between chain complexes, and $P$ an operator that takes $k$-chains to $k+1$-chains. Let $\partial$ be the usual boundary operator.

Then the condition $$\partial P + P\partial = g - f$$ is that $P$ is a chain homotopy between $f$ and $g$.

This almost looks like a Leibniz rule. Is there any deeper connection here?

user801923
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  • In what way does it look like Leibniz's rule? – Mariano Suárez-Álvarez Dec 03 '13 at 05:04
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    Particularly, the $\partial P + P \partial$ part which applies to $k$-chains. de Rham theory should provide a way to translate statements for chains into statements for forms, and the boundary operator translates to the exterior derivative, which satisfies a Leibniz rule by definition. I'm not sure how to make this precise, or if other similar statements hold for other types of derivatives. – user801923 Dec 03 '13 at 05:12
  • I cannot make sense of that, really. The answer I just posted is as close as I think one can get to your idea. – Mariano Suárez-Álvarez Dec 03 '13 at 05:16
  • @Mariano The Leibniz rule for (singular) chains (here, also here or on nLab) $\partial([0,1]\times C) =$ $\partial [0,1]\times C + I\times\partial C =$ $({1}-{0})\times C + [0,1]\times\partial C$ looks relevant. – Alex Shpilkin Jan 09 '19 at 11:17

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If $P$ and $Q$ are cochain complexes, with differentials $d_P$ and $d_Q$, and if we write $\hom(P,Q)_n$ the abelian group of all morphisms of graded groups $P\to Q$ of degree $n$ (on which we do not impose any conditions involving the differentials), then there is a cochain complex, which we write $\hom(P,Q)_\bullet$, whose differential $\partial$ maps $f\in\hom(P,Q)_n$ to $d_Q\circ f+(-1)^nf\circ d_P\in\hom(P,Q)_{n+1}$.

With this notation in place, it is easy to check that a morphism of complexes $f:P\to Q$ is preciely an element of $\hom(P,Q)_0$ such that $\partial(f)=0$, that is, a $0$-cocycle in our complex $\hom(P,Q)_\bullet$. Moreover, two morphisms $f,g:P\to Q$ are homotopic precisely when the element $f-g$, which is a $0$-cocycle, is a $0$-coboundary in the compex $\hom(P,Q)_\bullet$.

Let us now consider the special case in which $Q=P$. Now $\hom(P,P)_\bullet$ is not only a cochain complex: it is also a ring, with respect to the composition of maps, and with respect to this ring structure the differential $\partial$ is a (graded) derivation, that is, it satisfies Leibniz's rule: $$\partial(f\circ g)=\partial(f)\circ g\pm f\circ\partial(g)$$ (for some choice of signs...)

Mariano Suárez-Álvarez
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