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date: 2023-03-11 02:13
aliases: ["Bergman metric"]
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Last modified: `=this.file.mday`

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# Bergman metric

Let $D$ be a domain. By $\mathcal{H}^2(D)$ we denote the space of holomorphic functions that are quare integrable with respect to the Lebesgue measure. If $D$ is bounded the polynomials are in $\mathcal{H}^2(D)$, hence it is infinite dimensional. Furthermore it is a complete Hilbert space. We denote the scalar product by $(\cdot \mid \cdot)$. Consider for $w \in D$ the linear functional on $\mathcal{H}^2(D)$ given by $f \mapsto f(w)$. By the Riesz representation theorem there exists $K_w \in \mathcal{H}^2(D)$ with $f(w)=\left(f \mid K_w\right)$ for all $f \in \mathcal{H}^2(D)$. Sometimes we write $K(z, w)$ for $K_w(z)$. The function $K$ is defined on $D \times D$ and it has the property
$$
K(z, w)=K_w(z)=\left(K_w \mid K_z\right)=\overline{\left(K_z \mid K_w\right)}=\overline{K_z(w)}=\overline{K(w, z)} .
$$
Hence $K(z, z)=\mid K_z \|^2 \geq 0$. The function $K$ is called Bergman kernel.
For any complete orthogonal system $\left\{\varphi_k\right\}$ we have
$$
K(z, w)=\sum_k \varphi_k(z) \overline{\varphi_k(w)}
$$

Since $K_w \in \mathcal{H}^2(D)$, the Bergman kernel is holomorphic in the first and antiholomorphic in the second variable.

If $F: D \rightarrow D^{\prime}$ is an isomorphism ${ }^9$ between two domains with Bergman kernels $K$ respectively $K^{\prime}$, we have
$$
K(z, w)=K^{\prime}(F(z), F(w)) j_F(z) \overline{j_F(w)},
$$
where $j_F$ is the (complex) Jacobi determinant of $F$. 

Let $D$ be a bounded domain and $K$ its Bergman kernel. Let for $z \in D$
$$
g_{j k}(z):=\frac{\partial^2}{\partial z_j \partial \bar{z}_k} \log K(z, z) .
$$
**This defines an invariant Hermitian metric on $D$**. Hermitian means, that for $z \in D$ the matrix $g_{j k}(z)$ defines a Hermitian form on the tangent space $T_p D$. To see that, note that $g_{j k}(z)=\overline{g_{k j}(z)}$. It remains to show that it is positive definite on $T_z D$. This can be done by a direct calculation using a orthonormal system $\left\{\varphi_n\right\}$ of $\mathcal{H}^2(D)$. A direct calculation shows also that $F$ is an isometry between $D$ and $D^{\prime}$ equipped with their Bergman metrics.