Given smooth vector bundles $E \to M$ and $F \to M$ over a smooth manifold $M$, a linear differential operator from $E$ to $F$ is an $\mathbb{R}$-linear sheaf homomorphism $D: \mathcal{E} \to \mathcal{F}$.

Given a smooth map $f: N \to M$, can we define a pulled back differential operator $f^*D: f^* \mathcal{E} \to f^*\mathcal{F}$? If not, under what conditions on the map $f$ or operator $D$ can we do so?

I believe that the answer is "no" for general smooth maps $f$ because $D$ is not a homomorphism of $\mathcal{O}_M$-modules, so I would not expect it to interact nicely with the pullback functor $f^*: \mathcal{O}_M\mathsf{Mod} \to \mathcal{O}_N\mathsf{Mod}$.

However, if $f$ is the inclusion of an open submanifold, then the pullback functor is just the inverse image of sheaves $f^*=f^{-1}$, so $f^*D$ makes sense and is just the restriction of $D$ to sections over $N$.

I would hope that if $f$ is the inclusion of a closed submanifold then $f^*D$ can be defined, but unfortunately in this situation $f^* \neq f^{-1}$ in general.

  • $\begingroup$ I am not sure that you can define the restriction of a differential operator to a closed submanifold. What if this submanifold is a point ? For example, take $d:\mathcal{O}_M\rightarrow\Omega^1_M$. What can be the restriction of $d$ to a point ? $\endgroup$ – Roland Jun 20 '17 at 10:20
  • $\begingroup$ Maybe I haven't thought this through thoroughly, but I would assume that the restriction of $d$ to a point should be the zero map? $\endgroup$ – ಠ_ಠ Jun 20 '17 at 10:30
  • $\begingroup$ Sure, this is only one which is canonical. But what if you start with a non constant map, take its differential and restrict, or restrict then take this zero map. Something does not commute here. Another example : take $\mathbb{R}\times\{0\}\subset\mathbb{R}^2$ and $D=\frac\partial{\partial x}+\frac{\partial}{\partial y}$. What is the restriction ? An obvious choice would be $\frac{\partial}{\partial x}$. But if you think about it, you will see that this choice depend on the basis of $\mathbb{R}^2$. $\endgroup$ – Roland Jun 20 '17 at 10:48
  • $\begingroup$ Hmm that's a good point. $\endgroup$ – ಠ_ಠ Jul 8 '17 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.