# The logarithm of Kähler metric is not globally defined

In reducing the existence of Kähler-Einstein metrics to the complex Monge Ampere equation, the logarithm $$-\log \det (\omega + \partial \overline{\partial} \phi)$$ appears, where $\omega$ is a Kähler metric and $\phi$ is a smooth function. In Tian's book, he writes that while this function is not globally defined, we may form the quotient $$\log \left( \frac{(\omega + \partial \overline{\partial} \phi)^n}{\omega^n} \right)$$ and this is globally defined.

I would just like to clarify some things:

1. The reason why the first logarithm is not necessarily globally defined is because $\omega + \partial \overline{\partial} \phi$ is possibly zero?
2. The reason why the second logarithm is globally defined is because $(\omega + \partial \overline{\partial} \phi)^n$ and $\omega^n$ are now volume forms which are positive?

Thanks.

## 1 Answer

Question 1: Note, $\omega + \partial\bar{\partial}\phi$ cannot be zero. Recall that $\phi$ is chosen so that $\omega + \partial\bar{\partial}\phi$ is another metric (in particular, a Kähler-Einstein one).

What does $\log\det(\omega + \partial\bar{\partial}\phi)$ mean? In holomorphic coordinates $(U, (z^1, \dots, z^n))$ we have

$$(\omega + \partial\bar{\partial}\phi)|_U = \omega_{ij}dz^i\wedge d\bar{z}^j + \frac{\partial^2\phi}{\partial z^i\partial\bar{z}^j}dz^i\wedge\ d\bar{z}^j = \left(\omega_{ij} + \frac{\partial^2\phi}{\partial z^i\partial\bar{z}^j}\right)dz^i\wedge d\bar{z}^j.$$

By $\log\det(\omega + \partial\bar{\partial}\phi)$ we mean the function $\log\det(a_{ij})$ where $a_{ij}$ is the coefficient of $dz^i\wedge d\bar{z}^j$ in the local expression of $\omega + \partial\bar{\partial}\phi$; this is only defined on $U$ in terms of the coordinates $z^1, \dots, z^n$.

Given another coordinate system $(V, (w^1, \dots, w^n))$ with $U\cap V \neq \emptyset$ and $(\omega + \partial\bar{\partial}\phi)|_V = b_{ij}dw^i\wedge d\bar{w}^j$, the functions $\log\det(a_{ij})$ and $\log\det(b_{ij})$ need not agree on $U\cap V$ and hence do not give rise to a well-defined function. Note however that $\partial\bar{\partial}\log\det(a_{ij}) = \partial\bar{\partial}\log\det(b_{ij})$ as they are both local expressions for $-\operatorname{Ric}(\omega + \partial\bar{\partial}\phi)$.

Question 2: Yes. Given two volume forms $\eta_1$ and $\eta_2$ on a manifold $M$ which induce the same orientation, then $\eta_2 = g\eta_1$ where $g : M \to (0, \infty)$ is smooth.

So $(\omega + \partial\bar{\partial}\phi)^n = g\omega^n$ and hence

$$\log\left(\frac{(\omega + \partial\bar{\partial}\phi)^n}{\omega^n}\right) = \log\left(\frac{g\omega^n}{\omega^n}\right) = \log g$$

which is globally defined.

• You are a treasure to this community. Thank you. – Zheng Hai Mu May 19 '18 at 23:04