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Let $M$ be a complex manifold. We define the tangent bundle $TM$ as we do in the differential case. The complex structure on $M$ naturally induces a complex structure on $TM$ which makes it a complex manifold. On the other hand, if we look at the underlying differentiable manifold $X$ associated to $M$ (i.e, given $\varphi:U \rightarrow \mathbb{C}^{m}$ a holomorphic chart of M, we identify $\mathbb{C}^m$ with $\mathbb{R}^{2m}$ and then we obtain a chart of X), we can define, $$ TX_\mathbb{C}:=TX\otimes_{\mathbb{R}}(X\times\mathbb{C}). $$ the complexified tangent bundle. The manifold $X$ has a natural integrable almost complex structure $J:TX \rightarrow TX$ given by the complex structure of $M$. If we look to $J$ as a map over $TX_{\mathbb{C}}$, we have that $J$ is diagonalizable with eigenvalues $i$ and $-i$. We define $TX^{1,0}$ as the subbundle wich each fiber $T_pX^{1,0}$ is the eigenspace of $J_p:T_pX_{\mathbb{C}}\rightarrow T_pX_{\mathbb{C}}$ related to the eigenvalue $i$. Turns out that in this case, $TX^{1,0}$ is a complex manifold.

My question is: Are $TM$ and $TX^{1,0}$ is isomorphic as complex manifolds?

I'm almost convinced about this, but couldn't find this result in any book. So I decide to come here and ask to make sure that I'm understanding it well. References are very welcome.

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    $\begingroup$ There is the obvious isomorphism between $TX_{\mathbb{C}}$ and $TM$ when $M=\Delta^m$, and for general $M$ just patch up using charts. $\endgroup$ – user10354138 Apr 3 at 15:34
  • $\begingroup$ Sorry, what do you mean by $\Delta^{m}$? $\endgroup$ – Leonardo Schultz Apr 3 at 15:35
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    $\begingroup$ I mean the open polydisc $\mathbb{D}\times\mathbb{D}\times\dots\times\mathbb{D}\subset\mathbb{C}^m$, but of course the open unit ball in $\mathbb{C}^m$ would also work, $\endgroup$ – user10354138 Apr 3 at 15:40
  • $\begingroup$ Ok! Thank you for your help. $\endgroup$ – Leonardo Schultz Apr 3 at 15:42

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