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?

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