# Defining the natural almost complex structure on a complex manifold.

The definition of an almost complex structure is as follows. If $$X$$ is a differentiable manifold and $$TX$$ is its tangent bundle, then the endomorphism $$I: TX \to TX$$ defines an almost complex structure if $$I \circ I = -1$$ on all the fibers. If $$X$$ is a complex manifold, it would be easy to define $$I$$ locally (on the holomorphic tangent bundle) on a chart $$U_i$$ with the trivialization $$\Phi_i : U_i \times \mathbb{C}^n \to \pi^{-1}(U_i)$$ by letting $$I_i(p, v) = (p, iv)$$. Then clearly, $$I_i^2 = -1$$ so every chart has an almost complex structure. This would then give a natural map $$I_i': \pi^{-1}(U_i) \to \pi^{-1}(U_i)$$ by $$I_i' = \Phi_i I_i \Phi^{-1}_i$$ which satisfies the conditions of an almost complex structure on $$TU_i$$. However, I'm having trouble seeing how this extends to a global endomorphism $$I$$. To do so, we would need $$I_i \equiv I_j$$ on $$\pi^{-1}(U_i \cap U_j).$$ If $$U_j$$ has a local trivialization $$\Phi_j$$, then the transition map $$\Phi_j^{-1} \circ \Phi_i: (U_i \cap U_j )\times \mathbb{C}^n \to (U_i \cap U_j )\times \mathbb{C}^n$$ is given by $$(p, v) \mapsto (p, \tau_p(v))$$ for some differentiable map $$\tau: U_i \cap U_j \to GL(n, \mathbb{C})$$.

Then, in order for $$I_i' = I_j'$$, we would need $$\Phi_i I_i \Phi_i^{-1} = \Phi_j I_j \Phi_j^{-1}$$ which is true iff $$\Phi_j ^{-1} \Phi_i I_i = I_j \Phi_j^{-1} \Phi_i$$ which is true iff $$(p, \tau_p(iv)) = (p, i \tau_p(v))$$ which is obviously true by the fact that $$\tau_p \in GL(n, \mathbb{C})$$. Does this prove that X has an almost complex structure? Also, is there a more clear way to construct it? The book I'm reading from dismisses this fact as obvious which makes me concerned.

Thank you!

• That's pretty much it. You wrote it pretty clearly. It's kind of a general way to do things with overlap maps, namely to use the overlap condition to prove that some structure is independent of the coordinate chart, and that's what you did. Aug 1, 2020 at 1:07

Looks good to me. Your argument does indeed prove that $$X$$ admits an almost complex structure. It can also be viewed as a proof that $$TX$$ is a complex vector bundle (which is equivalent).
Another proof I have seen is to define $$J$$ in a complex coordinate chart $$(U, (z^1, \dots, z^n))$$ by
\begin{align*} J\left(\frac{\partial}{\partial x^k}\right) &= \frac{\partial}{\partial y^k}\\ J\left(\frac{\partial}{\partial y^k}\right) &= -\frac{\partial}{\partial x^k} \end{align*}
where $$x^k = \operatorname{Re}(z^k)$$ and $$y^k = \operatorname{Im}(z^k)$$. Then the fact that this gives rise to a well-defined almost complex structure $$J$$ is equivalent to the Cauchy-Riemann equations.