I’m studying symplectic manifolds and almost complex structures. This lead to two propositions:


**Proposition 1** (from da Silva’s *Lectures on Symplectic Geometry*): If $J\_0$ and $J\_1$ are almost complex structures compatible with a symplectic manifold $(M,\omega)$, then there is a family of almost complex structures $J\_t, t \in [0,1]$ such that $J\_t$ is compatible with $(M,\omega)$.


**Proposition 2** (from Hatcher’s K-theory book): Let $E \twoheadrightarrow B \times [0,1]$ be a (topological) vector bundle. Then the restrictions of $E$ to $B \times \{0\}$ and $B \times \{1\}$ are isomorphic.


I think proposition 2 can be extended to smooth bundles, though maybe this is the source of my confusion below.


**The confusion:** I think I can prove the following absurdity using the above two propositions.


**Absurdity:** Let $(M^{2n},\omega)$ be a symplectic manifold. Then $M$ admits a holomorphic atlas.


*Proof:* Let $\phi: U \to \mathbb{R}^{2n}$ be a Darboux chart (the pullback of the standard symplectic form on $\phi(U)$ is $\omega$) of a neighborhood $U \subset M$ around an arbitrary point $p \in U$. Let $J$ be an almost complex structure on $TM$. We have two almost complex structures on $\phi(U)$. The first one, call it $j\_0$, comes from $J$:


$$ j\_0 \equiv \phi\_\* \circ j \circ \phi^{-1}\_\*.$$


The second one, call it $j\_1$, comes from identifying $\mathbb{R}^{2n}$ with $\mathbb{C}^n$. Since $\phi$ is a Darboux chart, $j\_0$ is compatible with the standard symplectic form. $j\_1$ is also compatible with the standard form. Thus, Proposition 1 tells us that we have a family $j\_t$ of almost complex structures on $\phi(U)$ compatible with the standard symplectic form.


We now construct a complex vector bundle $\pi: E \twoheadrightarrow \phi(U)$ as follows. For a point $(x,t) \in \phi(U) \times [0,1]$, the fiber $\pi^{-1}(x,t)$ is the vector space $T\_x \phi(U)$ equipped with the almost complex structure $j\_t$. From Proposition 2, we obtain an isomorphism $\psi: \left(\phi(U),j\_0\right) \to \left(\phi(U),j\_1\right)$. Notice $(\phi(U),j\_1)$ is a neighborhood in $\mathbb{C}^n$. Therefore, $\psi \circ \phi$ is a holomorphic chart around $p$.


$p$ is arbitrary, so we obtain our atlas. $\blacksquare$


I’ve been perusing this proof for a while, but I cannot clearly see where it fails. What am I missing?



(Contradiction) All symplectic manifolds are holomorphic
The comments of @JHM and @Ivan Solonenko contain an answer to OP's question.


(1) The OP is correct that $j\_0:=\phi\_\* \circ J \circ \phi^{-1}\_\*$ is an almost-complex structure on the image $V:=\phi(U) \subset \mathbb{R}^{2n}$ which is compatible with the standard symplectic form $\omega\_{std}$ satisfying $\omega=\phi^\* \omega\_{std}$, where $(U, \omega, J)$ is the initial pseudo-holomorphic structure.


(2) Since $\phi$ is a Darboux chart, then $\omega\_{std}$ is also compatible with the standard almost complex structure $j\_1$ on $\mathbb{R}^{2n}$ restricted to $V$.


So we have two $\omega\_{std}$-compatible a.c. structures $j\_0, j\_1$ on $V$, and can even find a smooth $1$-parameter family of compatible a.c structures $j\_t$ on $V$, for $0\leq t \leq 1$.


(3) The OP constructs a complex vector bundle $E \to V \times [0,1]$, where the fibre over $(p,t)$ is the $j\_t$-complex vector space over $T\_{p} V$. Now the OP correctly applies a form of Hatcher's Prop.2, obtaining a complex vector bundle isomorphism $F: E|\_{V \times \{0\}} \to E|\_{V \times \{1\}}$.


(4) As clearly stated by @Ivan Solonenko, here the OP errs in interpreting $F$ as the differential of an isomorphism between the base space $V=\phi(U)$. Indeed $F$ is the identity map on the base $V$. The existence of fibrewise complex isomorphisms does *not* imply that the base spaces $(V, j\_0)$ and $(V, j\_1)$ are holomorphic. If the isomorphism $F$ produced by Prop.2. was induced by a diffeomorphism between the bases $f: V \to V$, then we would have $F=df$ on the fibres, and would find $df\circ j\_0=j\_1\circ df$. But this is not the case.


In conclusion, fibrewise deformations like $F$ are not induced by maps on the base. $F$ is not a differential $df$ except if integrability conditions are satisfied. For a.c. structures, the tensorial form of the integrability condition is given by Nijenhuis tensor. E.g., the $\pm i$-eigenspaces of a.c. structures $j|\_p$ define distributions on $V$, and which are almost never integrable for generic a.c. structures.


This can be recurring point of frustration in studying symplectic geometry, and is basically the subject of Gromov-Eliashberg $h$-principles. You might find Eliashberg--Mishachev's AMS textbook "Introduction to $h$-principle" interesting -- it contains several such examples as above.


