In symplectic geometry, it is commonly said that ``the set of almost complex structures tamed to a symplectic form is contractible'' even on noncompact symplectic manifolds. In my understanding one should work with strong Whitney topology on the space of almost complex structures. But it is known that the function space on noncompact manifold is not even connected on noncompact manifolds. So to correctly state the well-known Gromov's lemma on noncompact symplectic manifolds, we need to specify in what topology the lemma holds. So in what topology does Gromov's lemma hold?
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$\begingroup$ Could you specify the context of your question? Two tame metrics on what? $\endgroup$– YCorCommented Mar 9, 2023 at 7:55
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$\begingroup$ Assume a symplectic form $\omega$ is given on noncompact smooth manifold. By the way, for my question about quasi-isometriity the answer is no. Then the followup question is in what topology Gromov's contractibility holds, strong or weak Whitney topology or what? $\endgroup$– user500669Commented Mar 10, 2023 at 3:07
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$\begingroup$ Please, explain what is "Gromov's lemma on noncompact symplectic manifolds" and how does it relate to the title question. (I do not see any relation between the two notions of tameness in your question.) Incidentally, differential geometers tend to use the terminology "bounded geometry" instead of "tame" when talking about Riemannian metrics. $\endgroup$– Moishe KohanCommented Mar 13, 2023 at 18:28
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1 Answer
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Compact-open smooth topology, i.e. $C^\infty_\text{loc}$-topology, also the Fréchet topology on sections of the tangent bundle (same for Riemannian metrics). See some textbooks on J-holomorphic curves, such as Wendl's "Lectures on holomorphic curves in symplectic and contact geometry.
answered Mar 16, 2023 at 5:23
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$\begingroup$ So what you are saying is that considering compact-open topology will be enough for the applications of pseudoholomorphic curves on noncompact manifolds? Could you elaborate your answer why the latter is so? $\endgroup$ Commented Mar 16, 2023 at 6:24
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1$\begingroup$ Dear Chris, I will accept your answer that Gromov's lemma holds in weak C^\infty topology which is enough for most of applications at the moment. $\endgroup$ Commented Mar 17, 2023 at 7:11