4
$\begingroup$

I want to know if there is a uniqueness (in any sense) theorem for the symplectic structure in a neighborhood of a symplectic surface in a four dimensional symplectic manifold. Or more generally for any configuartion of symplectic surfaces.

$\endgroup$
2
  • 1
    $\begingroup$ Yes, of course. The neighborhood is determined by the type of the normal bundle. This follows from the Darboux-Weinstein Theorem. $\endgroup$ Commented May 16, 2015 at 23:02
  • $\begingroup$ It seems to me Darboux-Weinstein is for a Lagrangian neighborhood, do you mean we can apply it to symplectic neighborhoods as well? $\endgroup$
    – nikita
    Commented May 16, 2015 at 23:40

1 Answer 1

7
$\begingroup$

Here is the version of the Darboux-Weinstein theorem that you want to use: Let $(M_1,\omega_1)$ and $(M_2,\omega_2)$ be symplectic manifolds of dimension $2n$ and let $\iota_i:P\to M_i$ be smooth embeddings with the property that there exists an isomorphism $\phi:\iota_1^*(TM_1)\to \iota_2^*(TM_2)$ of vector bundles such that $\phi^*(\omega_2) = \omega_1$ (i.e., $\phi$ is a symplectic isomorphism) and, moreover, $\phi(\iota_1'(v)) = \iota_2'(v)$ for all $v\in TP$. Then there exist open neighborhoods $U_i\subset M_i$ of $\iota_i(P)$ and a symplectomorphism $\Phi:(U_1,\omega_1)\to(U_2,\omega_2)$ such that $\Phi\circ\iota_1 = \iota_2$.

Thus, the symplectic structure near an embedded symplectic surface $P$ in a symplectic $4$-manifold $(M,\omega)$ is determined by the isomorphism class of its (oriented) normal bundle (plus, of course, the symplectic area of $P$, i.e., the integral of $\omega$ over $P$).

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .