Every symplectic submanifold is J-holomorphic

Question

I am trying to show that every symplectic submanifold $N$ of a 2n- dimensional symplectic manifold $(M,\omega)$ is J-holomorphic for some compatible almost complex structure $J$.

The way I am currently approaching it is as follows:

Step 1: Let $i: N \hookrightarrow M$. Consider a almost complex structure $J^\prime$ which is compatible with $i^*\omega$.

Step2: I would like to extend $J^\prime$ to act on vectors normal to $N$. (Lets call this extended almost complex structure $J^\prime$ as well) and then use the following general fact to complete the proof:

Let $𝜋:𝐸→M$ be a locally trivial fiber bundle with fiber $𝐹$ a contractible metrizable manifold, and base space 𝐵 a metrizable space. Let $N$ be a closed subspace of $M$ and $𝜎:𝐴→𝐸$ be a continuous section of $𝐸$ over $N$. There is then a continuous extension of $𝜎$ to a global section of 𝐸. (In our case we would use the fact that $F= Sp(2n)/U(n)$ is contractible).

However I am unable to carry out the extension of $J^\prime$ to normal vectors to $N$. Is this the correct approach to the problem? If so would one go about extending $J^\prime$.

Answer 1

The normal bundle is a symplectic vector bundle (the fibres are symplectic vector spaces), and so, it has a compatible almost complex structure. Further, the normal bundle to $N$ can be realized as the symplectic orthogonal complement to $TN\subset TM|_N$.