# Complex structure on $TM$ is almost complex structure on $M$

I've seen it written in a few books and online that an almost complex structure on a manifold $M$ is a complex structure on $TM$.

Could someone explain this to me?

This confuses me and I'm not sure what it means but my impression is the following: All this happens on the corresponding tangent spaces, and locally for $U\subset M$ , $TU = U \times \mathbb R^n$ and so locally , $T(TU) = TU \times \mathbb R^{2n}.$

We get a complex structure on $TU$ by essentially identifying locally $$T(TU) = TU \times \mathbb R^{2n} = TU \times \mathbb C .$$

However, I don't understand what's really going on. Feel like I need to show why the complex structure on TM is integrable.

• I think a complex structure for a vector bundle usually just mean a bundle endomorphism that squares to $-Id$, so for the tangent bundle it really means the same as an almost complex structure of the manifold. This is like orientation of a vector bundle, and orientation of the tangent bundle being the same as orientation of the manifold.
– Qidi
Commented Jan 5, 2017 at 5:52
• Indeed, the (almost) "complex structure" on M and the "complex structure" on TM mean quite different things! Commented Jan 5, 2017 at 6:06
• I see, I guess that makes more sense. Commented Jan 5, 2017 at 16:42

As pointed out in the comments, a complex structure on a real vector bundle $$E$$ is a bundle endomorphism $$J$$ such that $$J\circ J = - \operatorname{id}_E$$, and when $$E = TM$$, we see this coincides with the notion of an almost complex structure on $$M$$.
Note, if the total space $$TM$$ is a complex manifold, $$M$$ need not admit an almost complex structure. For example, $$TS^1 = S^1\times\mathbb{R} = \mathbb{C}^*$$, but $$S^1$$ does not admit an almost complex structure as it is odd-dimensional.