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 $:→M$ be a locally trivial fiber bundle with fiber $$ a contractible metrizable manifold, and base space a metrizable space. Let $N$ be a closed subspace of $M$ and $:→$ be a continuous section of $$ over $N$. There is then a continuous extension of $$ to a global section of . (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 $J^\prime$. 



Every symplectic submanifold is J-holomorphic
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 $TN\subset TM|\_N$.


