I'm attempting to understand the following example from Weinstein's "Lectures on Symplectic Manifolds":
If $P\to X$ is a symplectic vector bundle, it may admit half dimensional subbundles, but no Lagrangian subbundles. Concretely, consider $E_1 \oplus E_2 \to S^2$, where $E_1$ is the tangent bundle of $S^2$ and $E_2$ is the trivial bundle $\mathbb R^2 \to S^2$.
The idea is that $c_1(E_1 \oplus E_2)= 2$, but if $E_1 \oplus E_2$ admits a Lagrangian subbundle $L$, then $E_1 \oplus E_2$ is isomorphic to $L \oplus L^*$ and hence $L_{\mathbb C}$. But this requires $c_1(E_1\oplus E_2) = 0$. Apparently "$c_1(L_{\mathbb C}) = 0$" is a very well known result since I haven't found any proofs but I've seen it stated in various forms while trying to find a proof online!
Unfortunately I know nothing about Chern classes or Chern numbers (as I'm still just learning introductory symplectic geometry), but I would like to understand this to some extent.
In McDuff-Salamon, the first Chern number of a complex vector bundle over an oriented closed surface is defined to satisfy the following four properties. (It is proved that there exists a unique functor that satisfies these.) Currently this is exactly the extent of my understanding of the first Chern number.
Two complex vector bundles over $\Sigma$ are isomorphic if and only if they have the same rank and first Chern number.
Whenever $\varphi: \Sigma_2 \to \Sigma_1$ is a smooth map between surfaces and $E \to \Sigma_1$ is a complex vector bundle, $$c_1(\varphi^* E) = \deg(\varphi) c_1(E).$$
If $E_1, E_2$ are complex vector bundles over $\Sigma$, $$c_1(E_1\oplus E_2) = c_1(E_1) + c_1(E_2.)$$
The first Chern number of the tangent bundle of $\Sigma$ is the Euler characteristic of $\Sigma$.
Is there a proof that whenever $E \to \Sigma$ is a real vector bundle over a closed oriented surface, then the complexification $E_{\mathbb C} \to \Sigma$ has vanishing first Chern number using the above axioms?
$c_k(\bar F)=(-1)^k c_k(F)$ for a complex bundle $F$ and its complex conjugate $\bar F$. Now $\bar E_{\mathbb{C}}=E_{\mathbb{C}}$ for the complexification of a real bundle $E$.