# Almost complex structure on a contractible manifold

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?

• What happens for the almost complex structures $i$ and $-i$ on $\mathbb{C}$. May 9 at 2:52

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.