Let $M$ be a contractible manifold with an almost complex structure $J:TM\to TM$. Suppose $J':TM\to TM$ is another almost complex structure. Since $M$ is contractible, so is $TM$, hence $J$ and $J'$ are homotopic, say via a homotopy $J_t:TM\to TM$. But we are not guaranteed that each $J_t$ is an almost complex structure. Can we choose a homotopy so that each $J_t$ is an almost complex structure? (actually I'm interested in the simple case $M=\Bbb C^n$)
By the way, is there a notion of "equivalence" between almost complex structures on a fixed manifold?
Thanks in advance.
Best Answer
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.
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.)