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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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    $\begingroup$ For a big picture, maybe try Kolar, Michor, and Slovak's Natural Operations in Differential Geometry (emis.de/monographs/KSM/kmsbookh.pdf). $\endgroup$ Sep 4 '14 at 5:16
  • $\begingroup$ Anyway, here is roughly a simpler and more direct characterization of the exterior derivative and differential forms simultaneously. The claim is more or less that $\Omega^{\bullet}(X)$, equipped with the exterior derivative, is the free commutative differential graded algebra on $C^{\infty}(X)$. Unfortunately this is not quite true and one has to modify it a bit, and I'm not sure how annoying the modification gets. $\endgroup$ Sep 4 '14 at 5:17
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    $\begingroup$ @QiaochuYuan I'm not familiar with the terminology, but it seems to me that a "differential graded algebra" is just an abstraction of the algebraic properties of $A=(\Omega^{\bullet}(X),d)$; e.g. the Leibniz rule is one of its axioms. The interesting thing about the characterization I described is that we don't directly impose this requirement on the algebra $A$; we just require that $A$ "transform correctly" in response to transformations of the underlying $X$. "Differential graded algebra" doesn't seem to have the notion of "underlying $X$", so it's not generalizing in the right direction. $\endgroup$ Sep 4 '14 at 6:23
  • $\begingroup$ Have you checked out spivak's Calculus on Manifolds? Definitely coherent and clear and they do explain how to prove step by step exactly what you're after. No mention of linearity, but how to use d it is clear anways $\endgroup$
    – MKF
    Apr 11 '16 at 13:36
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I think this might be a fairly intuitive way of understanding this statement. You are defining the operator $d$ by the property $f^*(d\omega)=df^*(\omega)$ for all forms $\omega$ and all diffeomorphisms $f$. A more general linear map from $\Lambda^p$ to $\Lambda^{p+1}$ (I like this notation for the exterior p forms) would be a general affine connection $\nabla$. Now, using your defining identity twice, we find \begin{equation} f^*(d^2\omega)=d^2f^*(\omega)=0 \end{equation} which is clearly true of the exterior derivative. For a more general connection, we have \begin{equation} f^*(\nabla^2\omega)=f^*(\Omega \omega)=\tilde{\nabla}^2 f^*(\omega)=\tilde{\Omega}f^*(\omega), \end{equation} where $\Omega$ is the curvature of the connection, and $\tilde{\nabla}$ and $\tilde{\Omega}$ denote the induced connection and curvature under the diffeomorphism. Clearly, in general the form of the induced curvature of the connection depends on the diffeomorphism. However, in the case where the curvature is zero, the curvature remains form invariant under all diffeomorphisms.

So we have established a relationship between your definition and the more usual requirement that $d^2=0$. As $d$ is the unique affine connection that satisfies this requirement, we have established the result.

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    $\begingroup$ and how do we know $d^2f^*\omega=0$? $\endgroup$
    – peter
    Feb 20 '19 at 16:12
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I personally think Palais' proof is clear, motivated, reasonably elementary. Maybe I can add some intuition.

  • The key observation is that if $T:\Omega^p\to\Omega^{p+1}$ is linear and commutes with diffeomorphisms, then it must be local, in the sense that if $V$ is open, then $T\omega|_V$ is determined by $\omega|_V$. To see this, we observe that for any point $x$ we can cook up a diffeo $f$ that acts like a dilatation by $r\neq 1$ on the tangent space at $x$ and which is identity outside some neighborhood $U$ of $x$. Now, if $\omega$ vanishes on this neighborhood, then $f^\star\omega=\omega$. So, $$r^{q+1}(T\omega)|_x=f^\star(T\omega)|_x=T(f^\star \omega)|_x=T\omega|_x,$$ thus $T\omega|_x=0.$
  • You can further restrict possible properties of $T$ by observing that in any coordinate chart, it must commute with shifts (which makes sense due to locality). Indeed, you can cook up a diffeo that looks like a shift locally in a coordinate chart and which is an identity far away.
  • It is clear that this is already quite restrictive. For example, a continuous map on the space of test functions that commutes with shifts is clearly a convolution with a generalized function. If our map is local, then the support of the latter must be the origin. It is known that such a generalized function is a finite linear combination of Dirac's delta and its derivatives. Of course we don't know that our map is continuous, but it is plausible that commuting with all diffeos is strong enough to outrule pathological examples. Thus, we may conclude that in any coordinate chart, our map should be a differential operator with constant coefficients.
  • Using scaling again, we can moreover infer that our operator must be first-order.
  • At this point, just looking at a bunch of concrete linear maps (say, ones that dilate one coordinate) will fix $T$ to be the exterior derivative.
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    $\begingroup$ Thanks for your answer to my question of 6 years ago :) It's a great summary of key points, although I'm not sure if your 3rd point lines up with Palais's proof. It seems to be a key part of Palais's story to decompose forms into "basic" forms that look nice in some coordinates. Only then does it look at what $T$ commuting with shifts implies, for basic forms. No talk of $T$ being a differential operator with constant coeffs. Do you think introducing basic forms is not really essential (not that there's anything wrong with the technique) and your 3rd point can be made rigorous directly? $\endgroup$ Apr 15 '20 at 9:37
  • $\begingroup$ @echinodermata, I think, actually, it can. Say, if I have a sequence $\varphi_n$ of test functions converging to $\varphi$, and $d \varphi\neq 0$ at $x$, then by inverse function theorem I can cook up a diffeos $S,S_n$ a such that $S^\star \varphi \equiv S^\star \varphi_n$ is a fixed (affine) function $\psi$ in a neighborhood of $x$, and $S_n\to S$. From this, $$T\varphi_n=T(S^{-1}_n)^\star \psi=(S^{-1}_n)^\star (T\psi)\to (S^{-1})^\star T\psi=T\varphi,$$ which shows that $T$ must be continuous. Of course you need a version of this for forms not functions... $\endgroup$
    – Kostya_I
    Apr 15 '20 at 10:17

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