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Given a complex manifold $M$, its complexified tangent bundle is $TM \otimes \mathbb C$. It is quite confusing for me as to why we want to do this since at each point $TM$ can already be viewed as a complex vector space.

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A complex manifold is given by two alternative definitions. First you have $M$ a "nice" topological space that has an atlas $\phi_j \colon U_j \longrightarrow V_j\subset \mathbb{C}^n$ such that $\phi_j\circ\phi_i^{-1}$ is a biholomorphism. This way you'll get a complex tangent bundle.

Another way is to consider a differentiable manifold of (real) dimension $2n$ together with an automorphism $J$ of the tangent bundle $TM$ such that $J^2= -Id$. Complexifying $TM$ gives a diagonalization for $J$ in the fibers. Locally, it decomposes $$ TM\otimes \mathbb{C} = E_i \oplus E_{-i} $$ as the sum of eigenspaces. Note that $TM\otimes \mathbb{C}$ has real rank $4n$ and from $J^2=-Id$, $E_i$ and $E_{-i}$ both have real rank $2n$. This is called an almost complex structure. If $[E_{-i},E_{-i}] \subset E_{-i}$ then this decomposition is global and we call $E_i$ (which has complex rank $n$) the holomorphic tangent bundle of $M$ and $J$ a complex structure.

The equivalence of these two definitions is not trivial and is given by the Newlander-Nirenberg theorem.

See Huybrecht's book. From the begining he starts this discussion exploring the identification of $\mathbb{R}^2$ and $\mathbb{C}$.

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  • $\begingroup$ I am confused. In your first definition, apparently, the manifold is also a smooth manifold of dimension $2n$, which means its tangent space at each point can be viewed as a complex vector space of dimension $n$. This is not consistent with the second definition, since the complexification will give you a vector space of complex dimension 2n. $\endgroup$ – Keith Sep 30 '18 at 15:35
  • $\begingroup$ I edited to keep track of the ranks and included a very readable reference. $\endgroup$ – Alan Muniz Sep 30 '18 at 16:02

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