Not any even rank real vector bundle over a smooth manifold has a complex structure, because there are even dimensional manifolds that have no almost complex structures. I am curious about the following special case: Suppose $M$ is a real compact, connected smooth manifold and $E\to M$ is an orientable real $2$-plane bundle. Also suppose, for some connected open subset $U$, $E|_U$ has a complex structure (so that it can be seen as a complex line bundle). In this case can we extend the complex structure on $E|_U$ to $E$?
Locally given complex structure on a 2-plane bundle over a compact manifold can be extended globally
algebraic-topologycomplex-geometrysmooth-manifoldsvector-bundles
Related Solutions
Setup.
Yes. If $V$ is a real vector space, then $\text{Comp}(V)$, the set of complex structures on $V$ (matrices $J$ with $J^2 = -I$) is a manifold.
Now if $E \to M$ is a vector bundle, there is a (smooth) fiber bundle $\text{Comp}(E) \to M$, whose fiber above a point $x \in M$ is $\text{Comp}(E_x)$, the set of complex structures on the corresponding fiber of $E$.
An almost complex structure on $M$ is a (smoothly varying) complex structure on the tangent spaces $T_x M$. That is, it is a smooth section $$J: M \to \text{Comp}(TM).$$
Write $\mathfrak J(M) = \Gamma(\text{Comp}(TM))$ for the space of sections of this bundle. You are interested in when two almost complex structures on $M$ are homotopic through complex structures.
That is, you want to understand $\pi_0 \mathfrak J(M)$.
The case you are interested in.
The crucial information is that every fiber bundle over a contractible, paracompact space is trivializable. A proof can be found in any text that discusses fiber bundles, eg Hatcher's "Vector Bundles and K-theory".
Choose a trivalization of the bundle $\text{Comp}(TM) \to M$ --- that is, $\text{Comp}(TM) \cong M \times \text{Comp}(T_x M)$ for your favorite basepoint $x \in M$. Sections of this are maps of the form $m \mapsto (m, J_m)$ where $J_m$ is a smoothly varying complex structure on the vector space $T_x M$. This amounts to saying that a section of this bundle amounts to a smooth map $J: M \to \text{Comp}(T_x M)$.
It follows that $\mathfrak J(M)$ is homeomorphic to $\text{Map}(M, \text{Comp}(T_x M)).$ Because $M$ is contractible, the latter space deformation retracts to the space of constant maps, and it follows that $\mathfrak J(M)$ is homotopy equivalent to the space $\text{Comp}(T_x M)$ where $x$ is fixed.
Now all you need to know is the homotopy type of this space. Write $T_x M \cong \Bbb R^n$ (for $n = \dim M$). Then what you need to know is that $$\text{Comp}(M) \simeq \begin{cases} \varnothing & n \text{ odd} \\ GL_n(\Bbb R)/GL_{n/2}(\Bbb C) & n \text{ even} \end{cases}.$$
(To prove this, observe that $GL_{2n}(\Bbb R)$ acts transitively on the space of complex structures by conjugation, $A \cdot J = AJA^{-1}$; the fiber above the standard complex structure is $GL_n(\Bbb C)$, and so $$\text{Comp}(\Bbb R^{2n}) \cong GL_{2n}(\Bbb R)/GL_n(\Bbb C).$$
There are thus exactly two complex structures on $M$ up to homotopy through complex structures (because $GL_{2n}(\Bbb R)$ has exactly two path-components --- determined by the sign of the determinant --- and $GL_n(\Bbb C)$ has exactly one.)
You can understand this as saying that the two equivalence classes of complex structures are determined by the orientation they induce on $M$.
The comment.
Try to understand how this relates to the orientation a complex structure induces.
Your final question.
By the way, is there a notion of "equivalence" between almost complex structures on a fixed manifold?
Of course, because you haven't said anything about the desired properties of this notion of equivalence. I can set them all to be equivalent if I want to. I can set two to be equivalent if and only if there is a diffeomorphism pulling back one to another. I can set them to be equivalent if they are homotopic through complex structures. Not sure what you are looking for.
It may be that you are looking for what I outlined above, but without more specificity it is hard to say. (Someone else will have to look at your updates or you will have to post a new question: I do not check back on questions after answering.)
Yes, provided $M$ is finite dimensional and smooth. The key property is that a vector bundle over such a manifold has can be covered by a finite set of local trivializations $\{(U_i,\varphi_i)\}_{i=1}^N$ (see this answer for details). Armed with these trivializations, one can embed $L$ in $M\times\mathbb{C}^N$ using partitions of unity.
Best Answer
There are two obstructions, one of which is topological and second is geometric.
In particular, if $U$ is connected (as in the edit to your question) and $E\to M$ is orientable then, after possibly replacing this orientation with the opposite one, we can assume that the orientation on $E\to M$ agrees with that of $E|_U$.
Thus, from now on, I will assume that $E\to M$ is oriented as above.
Here is an example of a non-extendible complex structure in the case when $M={\mathbb R}$ (I will leave it to you to extend it to any dimension).
The rank $2$ vector bundle $E \to {\mathbb R}$ has to be trivial, $E={\mathbb R}\times {\mathbb R}^2$, with the standard fiberwise orientation. Now, take $U=(0,\infty)$ and a Hermitian metric $h$ on $E|_U$ given by the following family of Gram matrices. $$ h_x= \left[\begin{array}{cc} 1&0\\ 0& x^2\end{array}\right]. $$ The corresponding almost complex structure will not extend to the point $x=0\in M$.
One way to resolve the geometric issue is:
Proposition. Suppose that $E\to M$ is an oriented rank 2 vector bundle, $U\subset M$ is an open subset (connectedness of $U$ is irrelevant here) such that the restriction bundle $E|_U$ is equipped with a complex structure whose orientation is consistent with that of $E$. Then for every compact subset $K\subset U$, the complex structure on $E|_K$ extends to a complex structure on $E$, again, consistent with the orientation.
Proof. Let $h|_U$ be a Hermitian metric for the complex vector bundle $E|_U$. Let $V\subset U$ be a relatively compact open subset containing $K$. Take a locally finite open cover ${\mathcal W}$ of $M$ such that $V=W_0\in {\mathcal W}$, the rest of the elements $W_k\in {\mathcal W}, k\ge 1$, are disjoint from $K$ and $E$ is trivial over each $W_k, k\ge 1$. Let $h_k$ be a Hermitian metric on $E|_{W_k}, k\ge 1$; set $h_0:=h|_V$. Now, take a partition of unity $\rho_k, k\ge 0$, for the cover ${\mathcal W}$ and use it to extend the Hermitian metric $h$ from $K$ to the rest of $M$: $$ h= \sum_{k\ge 0} \rho_k h_k. $$ where, by default, $\rho_k h_k$ is extended by zero outside of $W_k$. qed