# Relation between the tangent bundle and the holomorphic tangent bundle of a complex manifold.

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.

• There is the obvious isomorphism between $TX_{\mathbb{C}}$ and $TM$ when $M=\Delta^m$, and for general $M$ just patch up using charts. – user10354138 Apr 3 at 15:34
• Sorry, what do you mean by $\Delta^{m}$? – Leonardo Schultz Apr 3 at 15:35
• 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, – user10354138 Apr 3 at 15:40
• Ok! Thank you for your help. – Leonardo Schultz Apr 3 at 15:42