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. $\endgroup$ 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
    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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.