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!

  • 3
    $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Aug 1, 2020 at 1:07

1 Answer 1


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.


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