Bergman metric

Remarks

On a bounded domain $K(z, z) > 0$, so $\log K(z, z)$ is defined and smooth. The matrix $(g_{j\bar k})$ is positive definite at every point; the bounded functions $1$ and $z_j$ in $A^2(\Omega)$ give this positivity. If $F \colon \Omega_1 \to \Omega_2$ is a biholomorphism, then $K_1(z, w) = \det F'(z) \, K_2(F(z), F(w)) \, \overline{\det F'(w)}$, and the factors $\det F'$ disappear under $\partial\bar\partial \log$, so $F$ is an isometry. Example: for the unit disc, $K(z, w) = \frac{1}{\pi (1 - z \bar w)^2}$, so $\log K(z, z) = -\log \pi - 2 \log(1 - \abs{z}^2)$ and $g_{1\bar 1} = \frac{2}{(1 - \abs{z}^2)^2}$. Example: for the unit ball of $\C^n$, the Bergman metric is $n + 1$ times the metric with potential $-\log(1 - \abs{z}^2)$. Non-example: for $\Omega = \C^n$ the Bergman space is $\set{0}$, so $K \equiv 0$ and the definition does not apply. The Bergman metric is a Kähler metric, since it has a global potential.