Any complex manifold of dimension $n$ is a real manifold of dimension $2n$

Question

I was thinking about how to prove that $\mathbb{CP}^n$ is a real manifold. I have never studied complex geometry, but I would cover $\mathbb{CP}^n$ with the charts given by $U_i=\{[z_0,...,z_n]|z_i\neq 0\}$ and $\phi_i: [z_0,...,z_n]\in U_i\mapsto\frac{1}{z_i}(z_0,...,z_{i-1},z_{i+1},...,z_n)\in \mathbb{C}^n$ then I would go to $\mathbb{R}^{2n}$ by $(z_o,...,z_n)\mapsto(x_o,y_o,...,x_n,y_n)$ where $z_i=x_i+iy_i$. But is the whole thing $C^\infty$? How to prove it explicitely? Does this work in general?

Answer 1

One just needs to construct the relevant transition maps and check that all partial derivatives exist.

Fix $0 \leq i < j \leq n$. As complex functions on $\phi_j(U_i \cap U_j) \subset \mathbb{C}^n$, we have the transition maps $\phi_i \circ \phi_j^{-1}$ given by $$\phi_i \circ \phi_j^{-1}((z_0,...,z_{j-1},z_{j+1},...,z_n)) = \phi_i([z_0,...,z_{j-1},1,z_{j+1},...,z_n]) = \frac{1}{z_i}(z_0,...,z_{i-1},z_{i+1},...,z_{j-1},1,z_{j+1},...,z_n) = \frac{\bar{z_i}}{|z_i|^2}( z_0,...,z_{i-1},z_{i+1},...,z_{j-1},1,z_{j+1},...,z_n)$$

Using your identification of $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ and noting $\bar{z_i}z_k = (x_i x_k + y_i y_k)+(x_i y_k - y_i x_k)i$, we can interpret this transition function as a map $V_{ij} \subset \mathbb{R}^{2n} \to \mathbb{R}^{2n}$ (for the appropriate open set $V_{ij}$-- what should it be?) with component functions of the forms

$$(x_0,y_0,...,x_{j-1},y_{j-1},x_{j+1},y_{j+1},...x_n,y_n) \mapsto \frac{x_i x_k + y_i y_k}{x_i^2+y_i^2} \text{ or } \frac{x_i y_k - y_i x_k}{x_i^2+y_i^2}. $$ Here, $k \in \{0,...,i-1,i+1,...,n\}$ and we define $x_j:=1, \, y_j:=0$ (notice there is no input $z_j$, so we are free to do this). It is simple to observe that these expressions are smooth with respect to the $2n$ inputs.

More generally, since the transition maps $\phi_i \circ \phi_j^{-1}$ are required to be holomorphic as maps $\mathbb{C}^n \to \mathbb{C}^n$ (again, for appropriate restrictions of the domain), in particular their coordinate functions are holomorphic with respect to each complex coordinate (i.e. when we fix all but one $z_i$), and this implies that each of these are smooth as maps $\mathbb{R}^2 \to \mathbb{R}^2$ (since, for example, the real and imaginary parts of a holomorphic function are harmonic). That is to say, as maps $\mathbb{R}^{2n} \to \mathbb{R}^{2n}$, all partial derivatives of each coordinate function exists, i.e. they are smooth. So yes, this works in general.