Real $2n$-plane bundle with a complex structure is a complex $n$-plane bundle I am trying to show that if $\xi=(E,B,\pi)$ is a real $2n$-plane bundle with a complex structure $J:E\to E$ then $\xi$ becomes a complex $n$-plane bundle if we define $(x+iy)v=xv+yJ(v)$ on each fiber.
I only need to show that $\xi$ is locally trivial of complex sense. Suppose $h:\pi^{-1}(U)\to U\times\Bbb R^{2n}$ is a local trivialization. Identifying $\Bbb R^{2n}=\Bbb C^n$ by $(x_1,y_1,\dots,x_n,y_n)=(x_1+iy_1,\dots,x_n+iy_n)$, we can view $h$ as a map $h:\pi^{-1}(U)\to U\times \Bbb C^n$, and it remains to show that $h$ is fiberwise $\Bbb C$-linear. To see this, it suffices to show that $h(J(v))=i\cdot h(v)$ where $v\in F_b$ and $F_b=\pi^{-1}(b)$. But I can't understand why does this hold. (I'm reading the book Characteristic Classes of Milnor) Any hints?
 A: Not every trivialisation is fiberwise complex-linear, however, there is a trivialisation which is. This is the bundle version of the statement that for an $n$-dimensional complex vector space $V$, not every real-linear isomorphism $V \to \mathbb{C}^n$ is complex-linear, however, there is a real-linear isomorphism which is (i.e. an isomorphism of complex vector spaces).
To see why the desired trivialisations exist, we first note that $(E_b, J)$ is a complex vector space*, so there is a real basis of the form $\{e_1, Je_1, e_2, Je_2, \dots, e_n, Je_n\}$. Now choose local sections $\sigma_i : U \to \pi^{-1}(U)$ for $i = 1, \dots, n$ such that $\sigma_i(b) = e_i$ and consider the collection $\{\sigma_1, J\sigma_1, \sigma_2, J\sigma_2, \dots, \sigma_n, J\sigma_n\}$. As the property of being a basis is an open condition, and these sections form a basis of $E_b$ when evaluated at $b \in U$, there is an open neighbourhood $U' \subseteq U$ of $b$ such that, for every $b' \in U'$, the vectors $\{\sigma_1(b'), J\sigma_1(b'), \sigma_2(b'), J\sigma_2(b'), \dots, \sigma_n(b'), J\sigma_n(b')\}$ form a real basis of $E_{b'}$. Then we can define the trivialisation $h' : \pi^{-1}(U') \to U'\times\mathbb{C}^n$ by $h'(e) = (\pi(e), (a_1 + ib_1, a_2 + ib_2, \dots, a_n + ib_n))$ where $a_1, b_1, a_2, b_2, \dots, a_n, b_n$ are the unique real numbers so that
$$e = a_1\sigma_1(\pi(e)) + b_1J\sigma(\pi(e)) + a_2\sigma_2(\pi(e)) + b_2J\sigma_2(\pi(e)) + \!\cdot\!\cdot\!\cdot\! + a_n\sigma_n(\pi(e)) + b_nJ\sigma_n(\pi(e)).$$
If we multiply this equation by $J$, we find that $Je$ is equal to
$$a_1J\sigma_1(\pi(e)) - b_1\sigma(\pi(e)) + a_2J\sigma_2(\pi(e)) - b_2\sigma_2(\pi(e)) + \dots + a_nJ\sigma_n(\pi(e)) - b_n\sigma_n(\pi(e))$$
which can be rewritten as
$$-b_1\sigma_1(\pi(e)) + a_1J\sigma_1(\pi(e)) - b_2\sigma_2(\pi(e)) + a_2J\sigma_2(\pi(e)) - \!\cdot\!\cdot\!\cdot\! - b_n\sigma_n(\pi(e_n)) + a_nJ\sigma_n(e)).$$
Since $\pi(Je) = \pi(e)$, this equation tells us that
\begin{align*}
h'(Je) &= (\pi(Je), (-b_1 +ia_1, -b_2 + ia_2, \dots, -b_n + ia_n))\\ 
&= (\pi(e), i(a_1 + ib_1, a_2 + ib_2, \dots, a_n + ib_n))\\
&= ih'(e).
\end{align*}
Therefore $h'$ is fiberwise complex-linear as desired.

*I should really be writing $J_b$ for the map $E_b \to E_b$ induced by $J$, but I suppressed the subscript for notational convenience.
