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This is a cross-post of someone else's question.

I am cross-posting this question from MSE since it hasn't received any answers.

In the paper Natural Operations on Differential Forms, the author R. Palais shows that the exterior derivative $d$ is characterized as the unique "natural" linear map from $\Phi^p$ to $\Phi^{p+1}$ (Palais' $\Phi^p$ is what is perhaps more commonly written as $\Omega^p$, and "commutes with all diffeomorphisms", I believe, means $f^*(d\omega) = d(f^*\omega)$):

the exterior derivative on $p$-forms is determined to within a scalar factor by the condition that it be a linear mapping into $p+1$ forms which commutes with all diffeomorphisms.

I've tried to read the proof in the paper, but I'm struggling to follow the details and missing a sense of the big picture of the proof.

I'm interested in Palais' claim because this characterization is the most compelling one I have seen $-$ it seems far more appropriate as an axiomatic definition of $d$ than the definitions found in many textbooks, which often define $d$ based on properties such as $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^k \alpha \wedge d\beta$ (where $\alpha$ is a $k$-form), $d^2=0$, or $d(\sum \omega_{i_1...i_k}dx_{i_1} \wedge \cdots \wedge dx_{i_k}) = \sum d\omega_{i_1...i_k} \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k}$. While these are indeed quite basic properties of $d$, they are more appropriate as theorems than as a priori assumptions. (Of course, what is more "natural" is a matter of opinion, so please don't belabor the issue.)

Since it's such an innocent-looking and natural characterization, I would like to see a clear, motivated, reasonably elementary proof of it. Why ought it to be true that there is only one natural linear map from $\Omega^p$ to $\Omega^{p+1}$, up to a constant multiple? What are the key steps of a proof?

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  • $\begingroup$ Is it all "diffeomorphisms" or "smooth maps"? $\endgroup$
    – Alex Degtyarev
    Commented Jan 10, 2015 at 23:13
  • $\begingroup$ From the paper: "An automorphism of $M$ is what is usually called a diffeomorphism, that is a nonsingular differentiable homeomorphism of $M$". Here $M$ is an $n$-dimensional $C^\infty$ manifold, so I think the author means smooth maps. $\endgroup$
    – Exterior
    Commented Jan 10, 2015 at 23:20
  • $\begingroup$ Do you know how to do the case $p = 0$? It should be easier than for higher $p$. My guess is that once you figure out $p = 1$, then it should be not so hard to extend the proof to higher $p$. For $p = 1$, the exterior derivative is effectively the dual to the Lie bracket of vector fields. I suspect that there is a proof using the Lie derivative of forms. $\endgroup$
    – Deane Yang
    Commented Jan 11, 2015 at 1:04
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    $\begingroup$ I couldn't find the actual claim that you're referencing in the paper, though certainly what I did see looks like what you'd need. The main point should be that the diffeomorphism group of a manifold is huge, and that calculations with differential forms can be localized within an arbitrary neighborhood of a point. For instance, to prove that the exterior derivative of a local coordinate function $x_i$ is the differential form $dx_i$, construct a diffeomorphism which at the center of the coordinate system flows in the direction $\partial x_i$. $\endgroup$
    – Paul Siegel
    Commented Jan 11, 2015 at 1:32
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    $\begingroup$ While I agree that the perception of 'natural' varies with the person, I'd like to point out another approach that also strikes me as a natural characterization of the exterior derivative: It is the unique operation (no worries about 'up to scalar multiples') that makes Stokes' Theorem hold: $$\int_{\partial N}\omega = \int_N\mathrm{d}\omega,$$ where $N\subset M$ ranges over all compact, oriented submanifolds of dimension $p{+}1$. Of course, one must define integration of forms to state this, but it also motivates 'why exterior forms' in the first place. $\endgroup$
    – Robert Bryant
    Commented Jan 11, 2015 at 1:42

2 Answers 2

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I will try to answer this question in more detail after looking back at my paper of long ago. If anyone would like a copy of the paper they can download the pdf version at the following link:

http://vmm.math.uci.edu/PalaisPapers/NaturalOperationsOnDifferentialForms.pdf

It would help me if I had a better idea of what steps in my argument are not clear.

Dick Palais
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    $\begingroup$ I can't speak for the author of the original question, but what I would really like is: 1) A conceptual overview of why one should think of the exterior derivative as the linear map which commutes with smooth maps. 2) A conceptual summary of the proof in the sense of "what makes it tick". Ideally, both in simplest terms :) $\endgroup$
    – Exterior
    Commented Jan 11, 2015 at 12:20
  • $\begingroup$ Also, I somewhere heard the phrase even linearity [of $d$] comes from naturality. Does this somehow mean linearity is not even needed apriori? $\endgroup$
    – Exterior
    Commented Jan 11, 2015 at 12:22
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    $\begingroup$ @Exterior: Well, strictly speaking, linearity does not follow from naturality, for it does not follow when $p=0$. For example, the nonlinear operation $\hat d:\Omega^0\to\Omega^1$ defined by ${\hat d} a = \mathrm{d}(a^2)$ is natural in the sense that it commutes with pullback under diffeomorphism. $\endgroup$
    – Robert Bryant
    Commented Jan 12, 2015 at 11:08
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Here is a sketch of one possibility how to prove this:

The first key step is to see that the operator you are looking at has to be a differential operator. One usual way to ensure this is to require that it commutes with local diffeomorphisms (which include open embeddings). This implies that the operator is local and hence a differential operator by the Peetre theorem.

Having this at hand, the operator has to be induced by a vector bundle map $J^k\Lambda^pT^*M\to\Lambda^{p+1}T^*M$, where $J^k$ is the $k$th jet prolongation, and this bundle map has to be compatible with the natural action of all local diffeomorphisms. Such bundle maps are then determined by linear maps between the standard fibers of these bundles, which are equviariant for the actions of the so-called jet groups. (Formally, this is proved via higher order frame bundles.) In this case, the relevant group will be $G^k_n$, where $n$ is the dimension of $M$. This is an extension of $GL(n,\mathbb R)$, formally defined as the group of $k$-jets at $0\in\mathbb R^n$ of local diffeomorphisms of $\mathbb R^n$ fixing $0$.

Now on the target space $\Lambda^{p+1}\mathbb R^{n*}$, the jet group acts just via the usual action of $GL(n,\mathbb R)$ and this is an irreducible representation. On the other hand, restricted to the subgroup $GL(n,\mathbb R)$ the representation inducing $J^k\Lambda^pT^*M$ is the direct sum of the representations $S^{\ell}R^{n*}\otimes\Lambda^p\mathbb R^{n*}$ for $\ell=0,\dots,k$. Now there is only one summand which contains a copy of $\Lambda^{p+1}\mathbb R^{n*}$, namely the one for $\ell=1$. Hence your operator must be of first order, and its principal symbol must be the one of the exterior derivative (up to a constant multiple).

Hence you have a multiple of the exteiror derivative up to adding an operator of order zero, i.e. a bundle map $\Lambda^pT^*M\to\Lambda^{p+1}T^*M$ which commutes with the action of all diffeomorphisms. Representation theory of $GL(n,\mathbb R)$ immediately implies that such a bundle map does not exist.

There are lots of examples like this one discussed in the book Natural operations in differential geometry by Kolar, Michor and Slovak (MR 94a:58004)

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