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Let $M$ be a smooth manifold and let $\Omega(M)$ be the exterior algebra of smooth differential forms over $M$.

The $\mathbb R$-linear map $D:\Omega(M)\rightarrow\Omega(M)$ is a derivation of the exterior algebra of degree $r\in\mathbb Z$ if it maps $k$-forms to $k+r$ forms and it satisfies the Leibniz-rule $$D(\omega\wedge\eta)=D\omega\wedge\eta+\omega\wedge D\eta.$$

This map is instead an antiderivation if it satisfies the anti-Leibniz rule $$D(\omega\wedge\eta)=D\omega\wedge\eta+(-1)^k\omega\wedge D\eta,$$ where $k$ is the degree of $\omega$.


It is a known fact that if $D,D^\prime$ are antiderivations of odd degrees, then the anticommutator $$\{D,D^\prime\}=DD^\prime+D^\prime D$$ is a derivation of degree $\deg D+\deg D^\prime$. For example this is often used to prove Cartan's magic formula.

I am looking for a more general version of this statement. Ideally, given any derivation or anti-derivation $D$ and derivation or antiderivation $D^\prime$, some kind of "graded commutator" $$ [D,D^\prime ]_\pm=DD^\prime \pm D^\prime D $$ should be definable, which should result in either a derivation or antiderivation, whose degree is the sum of the degrees of $D$ and $D^\prime$ .

I have not been able to find one such relation. In fact I have tried to prove that for antiderivations of arbitrary degree $DD^\prime -(-1)^{\deg D \deg D^\prime}D^\prime D$ is a derivation, but I couldn't cancel enough terms in the general case.

I have also looked up "graded commutator" and "graded super Lie algebra" on Wikipedia, however it seems to me that case is different, as a graded derivation if defined with the anti-Leibniz rule $D(\omega\eta)=(D\omega)\eta+(-1)^{k\deg D}\omega (D\eta)$, which is slightly different from our anti-Leibniz rule here.

Question:

How can I unify the treatment of derivations and antiderivations of the exterior algebra to (hopefully) turn them into a graded super Lie algebra.

In particular, given any two derivations or antiderivations (including the case of one being one, the other being the other), how do I define a (graded) commutator that will produce a derivation or antiderivation?

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  • $\begingroup$ See en.wikipedia.org/wiki/… $\endgroup$ – Travis Dec 17 '18 at 16:30
  • $\begingroup$ @Travis it seems to me that the OP already addressed the material in the source you're linking in the paragraph immediately preceding the word "Question:" $\endgroup$ – jgon Dec 17 '18 at 18:15
  • $\begingroup$ That said, OP why do you want this particular definition of antiderivation? $\endgroup$ – jgon Dec 17 '18 at 18:18

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