Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
$\mathbb{C}P^n$ is Kähler with respect to Fubini-Study metric en.wikipedia.org/wiki/K%C3%A4hler_manifold#Examples –  john mangual Oct 9 '12 at 13:42
add comment

2 Answers 2

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.

share|improve this answer
add comment

See these notes on Kähler geometry. In the preface it mentions

smooth complex projective varieties together with the Riemannian metric induced by the Fubini–Study metric are Kähler

share|improve this answer
    
Well my question is rather concrete and to be honest the wikipedia article or the reference doesn't help me with my specific question. I'd be happy if you could adress my actual question. –  MrLee Oct 10 '12 at 0:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.