Suppose we are given a vector bundle $E\to M$ and a $E$-valued $p$-form, $$\omega\in\Gamma(E\otimes \Lambda^pT^*M)$$ For any smooth $f:\Sigma\to M$ then we have the pullback $p$-form, $$f^*\omega\in\Gamma(f^*E\otimes\Lambda^p T^*\Sigma)$$ Now we may consider an operator, $D: f\mapsto f^*\omega$. This $D$ is a first order ''differential operator''. I can see this if I work with a local trivialization of the bundle $E$.

My question is how to see this $D$ as a global operator? More precisely, I wish to write $D$ as an operator between section spaces, \begin{align*}D:C^\infty(\Sigma,M)&\to\Gamma(X)\\ f&\mapsto f^*\omega\end{align*} I am unable to figure out the space $X$ above.

Any help regarding this appreciated. Thank you!

  • $\begingroup$ You already wrote $f^*\omega \in \Gamma(f^*E\otimes\bigwedge^pT^*\Sigma)$, so why isn't $X = f^*E\otimes\bigwedge^pT^*\Sigma$? $\endgroup$ Feb 28, 2020 at 21:41
  • $\begingroup$ @MichaelAlbanese Because $X$ depends on the function $f$, no? $\endgroup$
    – ChesterX
    Feb 29, 2020 at 6:19
  • $\begingroup$ Ah, I see. In that case, $D$ is not a differential operator right (at least according to the definition I have in mind)? If you disagree, would you mind including your definition of a differential operator? $\endgroup$ Feb 29, 2020 at 12:25
  • $\begingroup$ @MichaelAlbanese Yes, I agree that $D$ cannot be a lienar operator between sections of vector bundles. But it still has a local description which gives rise to a first order PDE. In fact, if I had that $E$ is a trivial vector bundle of rank $k$, then $D:C^\infty(\Sigma,M)\to \Gamma(\Lambda^p T^*M \otimes \mathbb{R}^k)$ is indeed a differential operator of order $1$. $\endgroup$
    – ChesterX
    Feb 29, 2020 at 14:16

1 Answer 1


Note: I figured out how to do this from McDuff's J-Holomorphic Curves book. Adding some details here in order to close the question.

In order to formalize the differential operator in a global setup, consider the space $\mathcal{B}=C^\infty(\Sigma, M)$. For $f:\Sigma\to M$, denote the space $\mathcal{E}_f=\Gamma \big(f^*E\otimes \Lambda^pT^*\Sigma\big)$. Then we can think of a bundle $\mathcal{E}\to\mathcal{B}$. This is an infinite dimensional vector bundle, where the spaces have Frechet topology. The operator $$D:f\mapsto f^*\omega$$ can then be realized as a section of this bundle. This seems like the correct way to study the operator.

As an aside, one can get the linearization operator associated to $D$ at $f$, by considering some arbitrary connection $\nabla$ on $E$. One then gets the operator, $$\begin{align*}L_f : \Gamma f^*TM &\to \Gamma\hom(\Lambda^p TV, f^*E) \\ \xi &\mapsto \mathfrak{L}_\xi\omega\end{align*}$$ Here $\mathfrak{L}$ is the Lie derivative of $\omega$ along $\xi$ and is given by the Cartan formula, $\mathfrak{L}_\xi\omega = \iota_\xi d_\nabla\omega + d_\nabla\iota_\xi\omega$. The details for Lie derivative by a vector field along a map, may found in Michor's book.


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