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Let $X$ be a complex manifold of dimension $n$ (holomorphic charts). As a real smooth manifold, it has dimension $2n$. I am trying to think what is exactly the tangent bundle of $X$. If we think of tangent vectors as $\mathbb{C}$-linear derivation of smooth complex valued functions, then by the answer in this stackexchange post Complexified tangent space

$$Der (C^\infty_{X,p,\mathbb C}\to \mathbb C)=T_{\mathbb{R},p}(X)\otimes_\mathbb{R}\mathbb{C}= Der_\mathbb R (X^\infty_{X,p,\mathbb R}\to \mathbb R) \otimes_\mathbb R \mathbb C $$

Therefore as a complex vector space, the tangent space of $X$ has dimension $2n$ (unlike the real case,dimension of tangent space of complex manifold is 2 times the dimension of the complex manifold), and in fact it has local decomposition of $v = v^i \partial_{z_i} + v^{\bar{j}} \partial_{\bar{z_j}}$ where $v^i,v^\bar{j}$ are smooth complex valued functions on $X$.

Now as noted in the answer above, we did not use the fact that the transition maps are holomorphic at all. In fact, we could start with a real smooth manifold of dimension $2n$ to achieve the same thing. So by Newlander-Nierenberg theorem, if we can give the real manifold an integrable almost complex structure $J$, then we will able to find holomorphic charts for the manifold, thus making the real manifold into a complex one.

Please let me know if my understanding is incorrect or not.

CuriousAlpaca
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I'm not entirely sure what your question is, but you have one misconception. You confused the complex tangent space with the complexification of the real tangent space.

The complex tangent space is a subspace of $T_{\mathbb C,p} X$, defined by $$T^{1,0}_p X= \{v \in T_{\mathbb C,p}X : (I\otimes \mathbb C)(v) = iv\}.$$ In other words, it is the eigenspace of $I \otimes \mathbb C$ to the eigenvalue $i$. The set of vectors where application of $I$ is the same as multiplication by $i$.

If $z_1, \dotsc, z_n$ are local holomorphic coordinates around $p$, then $T^{1,0}_pX$ is generated by the derivations $\frac \partial {\partial z_1}, \dotsc, \frac \partial {\partial z_n}$. So it has complex dimension $n$.

The connection to holomorphic maps comes from the following observation:

Exercise. A smooth map $f: X \to Y$ between complex manifolds is holomorphic at $p \in X$ if and only if the complexified differential map $T_pf \otimes \mathbb C: T_{\mathbb C,p}X \to T_{\mathbb C,f(p)}Y$ maps $T^{1,0}_pX$ to $T^{1,0}_{f(p)}Y$.

Also note that as complex smooth bundles, $T_{\mathbb R} X$ and $T^{1,0}X$ are isomorphic: We can split $T_{\mathbb C}X = T^{1,0}X \oplus T^{0,1}X$ where $T^{0,1} = \{v : (I \otimes \mathbb C) = -iv\}, $and then the composition $T_{\mathbb R}X \hookrightarrow T_{\mathbb C}X \twoheadrightarrow T^{1,0}X$ is an isomorphism of smooth $\mathbb C$-bundles.

red_trumpet
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  • I was thinking along this line: for a real smooth manifold, we define a tangent vector as derivation on real-valued smooth functions. Analogously, a tangent vector on a complex manifold is a a derivation on complex-valued smooth functions, and to achieve that we need to complexify the real tangent space, so therefore the tangent space of complex manifold has dimension 2n. I think that $T^{1,0}X$ is called the holomorphic tangent bundle. – CuriousAlpaca May 18 '23 at 18:50
  • @CuriousAlpaca The tangent space of a complex manifold is no different from the tangent space of a smooth manifold. Every complex manifold is a smooth manifold, and so it has a tangent space in the usual sense. It doesn't have dimension $2n$, it has dimension $n$, just as it does for a smooth manifold. If you want to find the holomorphic tangent bundle in terms of derivations, you need to look at the complex linear derivations of the local ring $\mathcal{O}_{X,x}$, the stalk of the sheaf of holomorphic functions on $X$. This already requires the presence of an integrable complex structure. – Quaere Verum May 18 '23 at 19:06
  • I am confused, by having a tangent space in the usual sense, what do you use as the definition of tangent vector? From what I have learned for real smooth manifold, a tangent vector is a derivation on germs of smooth real-valued functions on X. So I just try to extend that definition as tangent vector for complex manifold, which requires us to complexify $T_{\mathbb{R},p}(X)$ – CuriousAlpaca May 18 '23 at 21:28
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    @CuriousAlpaca If you extend the definition to a complex manifold by saying "a tangent vector is a derivation on germs of smooth complex-valued functions" you get the complexification of the tangent space of the underlying smooth manifold. If you instead say "a tangent vector is a derivation on germs of holomorphic functions" you get the complex (holomorphic) tangent space, which is usually what we think of when we talk about the tangent space of a complex manifold. – Gunnar Þór Magnússon May 19 '23 at 10:20
  • I see it now. Thank you very much. – CuriousAlpaca May 19 '23 at 16:44