TITLE: Atlas of Complex Projective Space
QUESTION [5 upvotes]: In the context of Quantum Mechanics I'm trying to verify that the complex projective space $P(\mathbb{C}^n):=\mathbb{C}^n / \sim$ with $x \sim y :\iff x = \lambda \cdot y$ for some $\lambda \in \mathbb{C}  $ where $x,y\in \mathbb{C}^n$ is a Kähler manifold.
It's an exercise that I've done in my differential geometry class but in the quantum mechanical context the construction of the atlas is a bit different.
Basically I'm just experiencing one technial difficulty which I'm going to describe now: 
In the following ( , ) always denotes the euclidean scalar product on   $\mathbb{C}^n$.
Take $\phi \in S(\mathbb{C}^n) := \{ \psi \in \mathbb{C}^n : (\psi,\psi)=1 \}$
. Consider $[\phi] \in P(\mathbb{C}^n)$ the corresponding equivalence class in the complex projective space. 
Now consider: 
$V_\phi := \{ x \in P(\mathbb{C}^n) : (\phi,x)\ \neq 0 \}$, $\{\phi\}^{\perp}:=\{ x \in \mathbb{C}^n : (\phi,x)=0 \}$ and the map $b_{\phi} : V_\phi \to \{\phi\}^{\perp}$ given by $[x] \mapsto \frac{x}{(\phi,x)} - \phi$ which is just the renormalized orthogonal projection onto the orthogonal complement of $\phi$. 
The claim is that $\{ (V_{\phi}, b_{\phi}) \}_{\phi \in S(\mathbb{C}^n)}$ is an atlas for $P(\mathbb{C}^n)$. 
The technical difficulty that I'm having is the following. When computing the transition maps between different charts I want to find an explicit form of the set $b_{\phi}(V_{\phi} \cap V_{\psi})$ for some $\phi,\psi \in S(\mathbb{C}^n)$.
My guess is that we have $b_{\phi}(V_{\phi} \cap V_{\psi})$ = $\{\phi\}^{\perp}$ $\cap \{\psi\}^{\perp}$. However I'm unable to show that $b_{\phi}(V_{\phi} \cap V_{\psi})$ $\supset$ $\{\phi\}^{\perp}$ $\cap \{\psi\}^{\perp}$ and here I'm particularly worried about the case when we have $(\phi,\psi) = 0$. 
I'm not sure if there might be a mistake in my guess for the set $b_{\phi}(V_{\phi} \cap V_{\psi})$. In any case I'd be more than happy if someone could provide me with some help with this (rather technical) issue. 
Thanks a lot in advance.
Best regards.

REPLY [2 votes]: Your guess for $b_\phi(V_\phi\cap V_\psi)$ was incorrect, but the map $b_\phi$ does what you wanted (assuming you take the inner product to be conjugate linear in first variable). Personally, I would not subtract $\phi$ in the definition of $b_\phi$ because $$[x]\mapsto x/(\phi,x)$$  is easier to visualize as the map that sends complex lines through the origin (specifically, those in $V_\phi$) to their intersection with affine subspace $\phi+\{\phi\}^\perp$. This is obviously a bijection. 
When you restrict to $V_\phi\cap V_\psi$, you remove the complex lines orthogonal to $\psi$ from consideration. Intersecting with $\phi+\{\phi\}^\perp$, we get the set 
$$\{\phi+\xi : \xi\in \{\phi\}^\perp, \phi+\xi\notin \{\psi\}^\perp \} \tag1$$
Therefore, in your version (with subtracted $\phi$), the image of $V_\phi\cap V_\psi$ under $b_\phi$ is 
$$\{\xi : \xi\in \{\phi\}^\perp, \phi+\xi\notin \{\psi\}^\perp \} \tag2$$
The advantage of using affine subspaces for charts is that transition maps are also easy to visualize: they are simply radial projections. For example, $b_\psi\circ b_\phi^{-1}$ maps the set (1) onto 
$$\{\psi+\xi : \xi\in \{\psi\}^\perp, \psi+\xi\notin \{\phi\}^\perp \} \tag3$$ 
via 
$$\phi+\xi \mapsto \frac{\phi+\xi}{(\psi, \phi+\xi)} \tag4$$
which is a rational map with nonvanishing denominator.