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Let $(X,\omega)$ be a connected symplectic manifold, possibly with boundary. Let $g_1, g_2: B(1) \to X$ be two balls in $X$. Is it true that if $\delta$ is sufficiently small, then there is an ambient isotopy from $g_1|_{B(\delta)}$ to $g_2|_{B(\delta)}$? To be clear, by this I mean a smooth family of symplectic maps $\Psi_t : X \to X$ such that $\Psi_0 = id$ and $\Psi_1(g_1) = g_0$.

If this is true, does anyone have a proof or a reference?

Thanks.

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  • $\begingroup$ This is relevant: mathoverflow.net/questions/61994/… . $\endgroup$ – Marco Golla Jul 22 '14 at 6:18
  • $\begingroup$ According to "Introduction to Symplectic Topology" by McDuff and Salamon, it is not known. See Section "Blowing up and down". I don't know what was done after the book was written. $\endgroup$ – Hwang Jul 22 '14 at 9:16
  • $\begingroup$ @Hwang Thanks for your response. I don't have a copy of Introduction to Symplectic Topology, and I couldn't find a .pdf online, but are you sure that it is stated there that this question is not known? Note that the question does not ask for a $\delta > 0$ such that any two $B(\delta)$ are isotopic, which is known to be not known; rather, the question asks about subballs of two fixed balls, which I think should be easier. $\endgroup$ – bentspoon Jul 22 '14 at 16:08
  • $\begingroup$ Now I have a problem in understanding your question. You pick two arbitrary balls of size 1 and take subballs of them. Is it different from picking arbitrary two balls of size $\delta$? $\endgroup$ – Hwang Jul 23 '14 at 0:00
  • $\begingroup$ @Hwang I think it is different, yes (it's formally easier; however, if you can convince me that the problems are equivalent that would be great) $\endgroup$ – bentspoon Jul 23 '14 at 2:25
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The answer is yes, and the construction of the isotopy follows from rather elementary techniques:

  • First, Hamiltonian isotopies act transitively on points in a connected symplectic manifold. In other words, we may assume that $g_1(0)=g_2(0)=p$.

  • Second, after another Hamiltonian isotopy, we may even suppose that $D_0g_1=D_0g_2$ (consider e.g. paths of linear symplectomorphisms in a Darboux chart appropriately cut off by using Hamiltonians).

  • Third, and finally, one can perform the "Alexander trick": a linear interpolation $g_i(tx)/t$, $t \in (0,1]$ where the (outer) scalar multiplication is defined using some Darboux chart around $p \in X$.

It can be checked that these two paths of symplectomorphisms are well-defined in some sufficiently small ball $B(\delta)$ and that both paths moreover extend to an isotopy from $g_i|_{B(\delta)}$ to the restriction of the linear symplectomorphism representing $D_0g_1=D_0g_2$ in that Darboux chart.

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