Given a finite dimensional real vector space, we can view all the complex structures on it as a subspace of ${\rm GL}(V)$. I wonder if it is a submanifold of ${\rm GL}(V)$. And moreover, given a symplectic form $\omega$, will the $\omega$-compatible (for simplicity, I discard the requirement that $\omega(v,Jv)>0$ for it is an open condition on complex structures )complex structure consist a submanifold?
For the first problem, I have figured out a solution but I don't satisfy with it because it is not the usual metheod for proving submanifolds using constant rank theorem or transversality. Here is my proof:
view the general linear group as matrices. By linear algebra we know that for any complex structure $J$ there is a invertible matrix $P$ such that $PJP^{-1}=\Omega$, where
\begin{align*} \Omega= \begin{pmatrix}0 &-I_{n}\\\ I_{n}&0\end{pmatrix} \end{align*}
Let ${\rm GL}(V)$ act on it self by congugation, then the set of complex structures is a closed orbit of its action, and a theorem(I will post the theorem under this proof) in Topological transformation groups by L.Zippin and D.Montgomery 2.13 said that in this case the natural induced map from ${\rm GL}(V)/{\rm GL}(V)_{\Omega}$ to this orbit is a homeomorphsim. Thus the map is an embedding.
Let $G$ be a locally compact group, G' an open subgroup such that $G/G'$ is countable, and $G'/G_{0}$ ($G_{0}=$ identity component so $G_{0}\subset G'$ and $G_{0}=G_{0}')$ is compact. Let $G$ act transitively on the locally compact Hausdorff space $M$. Then for any $x\in M$, $G/G_{x}$ is homeomorphic to $M$.
( In my case, ${\rm GL}(V)$ act transitively on the closed orbit and because it is closed, the orbit is locally compact hausforff, Taking $G_{0}=G'=$ matrices with positive determinent)
For the second problem, the $\omega-$compatible complex structure can be identified with matrices satisfying $J^{2}=-I_{2n}$ and $J\Omega J^{\rm T}=\Omega$. If the map defined on complex strutures by $J\rightarrow J\Omega J^{\rm T}$ is constant rank, we are done, but I have no idea about how to prove it and I don't know whether the map is really constant rank or not. It can also be identified with symmetric matrices satisfying $A(-\Omega)$ is complex structure, so I wonder if the map defined on invertible symmetric matrices $A\rightarrow A(-\Omega)A(-\Omega)$ is constant rank. It is equivalent to show that the dimension of linear space generated by symmetric matrices $\\{X | A\Omega X \text{ is symmetric} \\}$ is indepedent of symmetric invertible matirx $A$, but I also have no idea about proving/disproving it. Or, maybe we can prove that the complex strutures and the symplectic group intersect transversally?
Oh I have done. Consider skew-symmetric matrices taking form \begin{pmatrix} A& B\\\ -B& A \end{pmatrix} where A is skew-symmetric and B is symmetric. It is easy to show that invertible matrices satisfying $P^{-1}\Omega P=$ skew-symmetric matrices like above is a submanifold. Donote it as $H$. Then $H^{-1}/{\rm GL}(V)_{\Omega}$(it is submanifold of complex structures by quotient manifold theorem) is the $\omega$-compatible complex structures.