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Let $M$ be a smooth manifold of dimensions $2n$, and suppose $$TM \equiv M \times \mathbb R^{2n}. $$

Does it follow that $M$ admits a complex structure?

It seems like the almost complex structure defined by taking the standard (after choosing a global frame $\{ v_1, \dots, v_n\}$) one at each point on $TM$ $p \in M$: $$J_p: T_pM=\mathbb R^{2n} \rightarrow T_pM=\mathbb R^{2n},$$ given by $$J_p = \begin{bmatrix} 0 & I \\ -I & 0 \end{bmatrix}$$ is a good candidate for a complex structure. But how can we know if this is integrable? I am aware of the Newlander–Nirenberg theorem, but am wondering if there are other ways to know if I can't compute that.

Thanks!

user308973
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    There are example of parallelizable manifold with no complex structure, see this short paper –  Sep 26 '16 at 20:12
  • Also possibly of interest, the famous Kodaira-Thurston nilmanifold (the first published example of a closed manifold that admits both symplectic and complex structures, but does not admit a Kähler structure) is of type (iii) as listed in Theorem 2 of that paper. – Aleksandar Milivojević Nov 06 '18 at 15:33

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