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.
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.
