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In Hamilton's paper on the Nash-Moser inverse function theorem he shows that if $M$ is a smooth compact manifold and $V\to M$ a smooth vector bundle then its smooth sections $\Gamma(V)$ equipped with the norms $||\sigma||_n:= \sum_{j=1}^n|\nabla^j\sigma|$ for $\sigma\in \Gamma(V)$ is a (tame) Fréchet space. Similarly he shows that $\mathrm{Diff}(M)$ the diffeomorphism group of $M$ is a tame Fréchet Lie group.

My question is the following. We regard the pullback of differential forms as a map $$\mathrm{Diff}(M)\times \Omega^n(M)\to \Omega^n(M)$$ $$(f,w)\mapsto f^*w$$ of Fréchet manifolds and Fréchet spaces. Here $\Omega^n(M)$ denotes the Fréchet space with the topology of smooth sections of a vector bundle as illustrated above.

My question is whether this map was shown somewhere to be smooth tame.

I believe it should be similar to Hamilton's proof (in the same paper) that nonlinear differential operators are smooth tame the only difference being that $(f^*w)_p$ depends on $w_{f(p)}$ and not $w_p$. And since it suffices to look at diffeomorphisms close to the identity the proof should still go through. Any literature recommendations on this subject are greatly appreciated. Thank you!

(p.s. This is my first question on here so you are more than welcome to point out any issues with my question. Thank you!)

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    $\begingroup$ This map is a nonlinear differential operator. You can see this by writing it in local coordinates. Any nonlinear differential operator is smooth tame. $\endgroup$
    – Deane Yang
    Commented Dec 10 at 13:07
  • $\begingroup$ @DeaneYang Thank you for your answer. This was also my first intuition. However, I got stuck on a technical detail namely that for $p\in M$ $(f^∗w)_p$ depends on $w_{f(p)}$ and not $w_p$ in other words it does not map fibers to fibers which I believe it should to define a differential operator in Hamilton's sense since Hamilton defines differential operators of degree r as compositions of vector bundle operators and the map that maps a section to its $r$ -jet. $\endgroup$
    – Jan Heck
    Commented Dec 10 at 13:48
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    $\begingroup$ You must consider the pullback $f^∗\omega$ of $\omega\in\Omega^n(M)=\Gamma(\wedge^n T^*\! M)$ as a smooth section of the pullback vector bundle $f^*\wedge^n T^*\! M$ (the identification being through the graph of $f^*\omega$ as usual). Once you do that, the pullback map $\omega\mapsto f^*\omega$ becomes a proper differential operator (mapping smooth sections of $\wedge^n T^*\! M$ to smooth sections of $f^*\wedge^n T^*\! M$). $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Dec 10 at 21:13
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    $\begingroup$ Recall, though, that the notion of differential operators you're recalling from Hamilton doesn't strictly apply here because it actually defines linear differential operators. A way to see the pullback as a differential operator is to see the pullbacks of $\wedge^n T^∗\! M$ by all $f\in\text{Diff}(M)$ as sub-bundles of the (fiber) bundle $\tilde{\pi}:M\times\wedge^n T^*\! M\rightarrow M$, the bundle projection being $(p,q)\mapsto \tilde{\pi}(p,q)=p$. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Dec 10 at 21:13
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    $\begingroup$ This is consistent with seeing $f\in\text{Diff}(M)$ as a smooth section of the trivial bundle $\text{pr}_1:M\times M\ni(p_1,p_2)\mapsto p_1\in M$ once we identify $f$ with its graph as before. That's where the nonlinearity of the (first order) differential operator $(f,\omega)\mapsto f^*\omega$ comes in, for the differential operator $\omega\mapsto f^*\omega$ by itself is linear and of order zero. Likewise, one must see $(f,\omega)$ as smooth sections of the Whitney sum of $\text{pr}_1$ and $\tilde{\pi}$. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Dec 10 at 21:14

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