I am familiar with the property that expands the exterior derivative acting on the wedge product of two forms,

$$d(\alpha \wedge \beta) = d \alpha \wedge \beta + (-1)^p \alpha \wedge d \beta$$

where $\alpha$ is a p-form and $\beta$ is a q-form.

Can this be generalized in the case

$$d(\alpha \wedge \beta \wedge \gamma)$$

with $\gamma$ being a z-form, and if yes what is the relevant formula?

And can this also be generalized for more terms in the wedge product? Something like $d(\alpha \wedge \beta \wedge \gamma \wedge \cdots)$?

I have tried to find the answer online but I have not been successful.


1 Answer 1


You can use the associativity of the wedge product as follows (or the other way around): $$d(\alpha \wedge \beta \wedge \gamma) = d((\alpha \wedge \beta) \wedge \gamma) $$ and then apply the graded Leibniz rule twice. (As a matter of fact, associativity is what allows you to even write such a thing as $\alpha \wedge \beta \wedge \gamma$.)

  • 1
    $\begingroup$ oh, I see. Thanks for the clear answer! $\endgroup$
    – user494842
    Mar 27, 2020 at 12:11

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