Does there exist higher degree graded derivations on $\Omega(M) $

Question

Does there exist any other graded derivation on $\Omega(M)$ other than the one of degree one which is the exterior derivative (i.e. maps such as $d: \Omega^p(M) \rightarrow \Omega^{(p+r)}(M) $, where $ r = 3,4,5...$).

I've been reading a bit about graded derivations on graded algebras. However, I've not been able to find an answer for specifically for the algebra of differential forms on a smooth manifold. Any help would be appreciated. Thanks.

Answer 1

Yes, there are higher degree derivations and they can be nicely described in terms of vector valued differential forms (i.e.~sections of the vector bundle $\bigwedge^*T^*M\otimes TM$). In fact you can generalize both the insertion operator and the Lie derivative from vector fields (which are just vector valued forms of degree zero) to general vector valued differential forms. This can be used to define natural bilinear operations on the space of vector valued differential forms, in particular, the Froelicher-Nijenhuis bracket.

A complete description of graded derivations of the algebra of differential forms in these terms can be found in the book by Kolar, Michor and Slovak which is available here.

Answer 2

I'm not sure I know of anything too interesting. One thing that comes to mind is the coderivative which is formed by taking the exterior derivative of the Hodge dual then Hodge dual back once more. That is, take a $p$-form $\alpha$ to an $(n-p)$-form $\star \alpha$ thus gives you an $(n-p+1)$-form $d(\star \alpha)$ and finally: a $n-(n-p+1)=p-1$ form: $$ \delta \alpha = \star d \star \alpha $$ which is the so-called coderivative of $\alpha$. In particular, $d\delta+\delta d$ gives the Laplacian. Well, I suppose there are some signs to worry about depending on your context. See the Hodge Dual Wikipedia article

I don't think this fits your desired higher-degree derivation. We already know composition of $d$ just leads to $d^2=0$. I suppose that, for a form of degree less than $1/2$ of the dimension of the space the hodge dual followed by an exterior derivative might give you something interesting. For example, in $10$-dimensional space, if I take a $2$-form $\alpha$ then take the Hodge dual I obtain $\star \alpha$ a $8$-form then $d(\star \alpha)$ is a $9$-form so the mapping $d \star$ changes the grading on $2$-forms by $7$. However, the same mapping takes $6$-forms to $10-6+1=5$-forms. Thus, the mapping $d\star$ is probably not what you're looking for as it does not have a uniform grading on the entire exterior algebra.