Timeline for Is the pullback of differential forms on a compact manifold smooth tame as a map of Fréchet manifolds?

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Dec 10 at 21:14	comment	added	Pedro Lauridsen Ribeiro		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 $\tilde{\pi}$.
Dec 10 at 21:13	comment	added	Pedro Lauridsen Ribeiro		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 $(p,q)\mapsto \tilde{\pi}(p,q)=p$.
Dec 10 at 21:13	comment	added	Pedro Lauridsen Ribeiro		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 $f^\wedge^n T^\! M$).
Dec 10 at 13:48	comment	added	Jan Heck		@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.
Dec 10 at 13:07	comment	added	Deane Yang		This map is a nonlinear differential operator. You can see this by writing it in local coordinates. Any nonlinear differential operator is smooth tame.
Dec 10 at 13:02	history	edited	Jan Heck	CC BY-SA 4.0	edited title
Dec 10 at 12:46	history	edited	Jan Heck		edited tags
S Dec 10 at 12:35	review	First questions
Dec 10 at 13:29
S Dec 10 at 12:35	history	asked	Jan Heck	CC BY-SA 4.0