# Every symplectic submanifold is J-holomorphic

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 $$J^\prime$$.

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 $$TN\subset TM|_N$$.
• We are choosing a (compatiblle) fibrewise complex structure on the $\omega$-orthogonal complement $TN^\omega\subset TM|_N$. Said another way, we are reducing the structure group of $TM|_N = TN\oplus TN^\omega$ from $Sp(2n)$ to $U(n)$ by doing it for each direct summand separately. – Mohan Swaminathan May 9 '19 at 14:32