Using the homogeneous coordinate on $CP^n$, we consider the open set $U_0:=\{[1, \ldots, z_n]\}$. Then the standard Kähler form of $CP^n$ can be written as

$$ \omega_0=\frac{\sqrt{-1}}{2}\partial\bar{\partial}\log(1+|z_1|^2+\cdots+|z_n|^2) $$

The compatability of this form can be easily checked for other chart $U_i$.

My question is, if I want to deform this Kähler form, an easy way to do this is introducing a function say $\rho: \mathbb R\to \mathbb R$ and write the new Kähler form on $U_0$ as

$$ \omega_\rho=\frac{\sqrt{-1}}{2}\partial\bar{\partial}\log(1+\rho(|z_1|^2+\cdots+|z_n|^2)) $$

Then what are the restrictions on $\rho$ and how to write the form $\omega_\rho$ in other coordinate charts, say $U_1$?

  • $\begingroup$ Either Kähler or Kaehler is a correct spelling; Kahler is not. If it were, it would be pronounced differently. I changed it to Kähler in the posting. $\endgroup$ – Michael Hardy Nov 7 '11 at 15:04

The way you've decided to deform the Kähler form you'll need to check that $\rho$ glues together on coordinate charts, then that the form you've defined is positive-definite (which should be automatic for "small" $\rho$ by compactness). I try to avoid gluing charts when I can, so allow me to address a small deformation of your question.

If you want to deform the Kähler form in such a way that you still get a Kähler form (i.e. the associated metric is Kähler, not merely hermitian$^*$), then there are slightly easier ways to go about this. Denote by $p : \mathbb C^{n+1} \setminus \{0\} \longrightarrow \mathbb P^n$ the canonical projection, then it is known that the pullback of the Fubini-Study metric by $p$ is $p^* \omega = i \partial \bar \partial \log |z|^2$, where $|z^2| = |z_0|^2 + \ldots + |z_n|^2$.

Now, the Hodge number $\dim_{\mathbb C}H^{1,1}(\mathbb P^n, \mathbb C)$ is equal to $1$, so if $\alpha$ is any other Kähler metric on $\mathbb P^n$, then by the $\partial \bar \partial$ lemma there exists a smooth real function $\phi$ on $\mathbb P^n$ such that $\alpha = \omega + i \partial \bar \partial \phi$. Pulling this back via $p$, we have that $p^* \alpha = i \partial \bar \partial (\log |z|^2 + p^*\phi)$. The function $p^* \phi$ comes from $\mathbb P^n$, so it is constant on any line in $\mathbb C^{n+1} \setminus \{0\}$ (i.e. $p^*\phi(\lambda z) = p^*\phi(z)$ for all scalars $\lambda \not= 0$).

You now get any Kähler metric on $\mathbb P^n$ by picking a smooth function $\phi$ on $\mathbb C^{n+1} \setminus \{0\}$ as above and considering the form $p^*\alpha$. As it is constant on any line, then it descends to the projective space. You do need to check that $p^* \alpha$ is positive-definite, which will end up being a condition on $\phi$ (it must be $\omega$-plurisubharmonic), and you should be able to explicit what this condition is by some violent calculations.

$^*$ If you want a non-Kähler metric on $\mathbb P^n$, multiply $\omega$ by any positive non-constant function.

  • $\begingroup$ will the curvatures, say bi-sectional curvature can be calculated explicitly? Any reference to this calculation if available? Thanks $\endgroup$ – user17150 Nov 7 '11 at 17:12
  • $\begingroup$ I suppose one would be able to say something about the different curvatures since we have a "global" potential (on the pullback). For a reference I can point you to Zheng's "Complex differential geometry", Ballmann's "Lectures on Kahler manifolds" or Moroianu's "Lectures on Kahler geometry", but you'd have to figure out how to do these calculations yourself with the techniques presented in those books. $\endgroup$ – Gunnar Þór Magnússon Nov 8 '11 at 7:21

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