---
tag: 5XSY1Y133X
kind: definition
name: Kobayashi metric
msc: [32F45, 32Q45]
defines: DY8QQ5GTTX
uses: [JSGWAN2RJV, 3MQTWKBBBT, 85MM2EA0TK, 0A3HJX3KTF, V1QYA9RW73]
sources:
  - url: https://en.wikipedia.org/wiki/Kobayashi_metric
---

## Definition

Let $\mathbb{D} = \set{\zeta \in \C : \abs{\zeta} < 1}$ and let $\rho$ be the Poincaré distance on $\mathbb{D}$,
$$\rho(a, b) = \tanh^{-1} \abs{\frac{a - b}{1 - \bar a b}} .$$
Let $X$ be a connected [[complex manifold|JSGWAN2RJV]] and $p, q \in X$.
A *holomorphic chain* from $p$ to $q$ consists of points $p = p_0, p_1, \dots, p_k = q$ of $X$, points $a_1, b_1, \dots, a_k, b_k$ of $\mathbb{D}$, and [[holomorphic maps|3MQTWKBBBT]] $f_1, \dots, f_k \colon \mathbb{D} \to X$ with $f_j(a_j) = p_{j-1}$ and $f_j(b_j) = p_j$.
The *Kobayashi pseudodistance* of $X$ is
$$k_X(p, q) = \inf \sum_{j=1}^k \rho(a_j, b_j),$$
the infimum over all holomorphic chains from $p$ to $q$.

## Remarks

Holomorphic chains exist because $X$ is connected.
The function $k_X$ is a [[pseudodistance|0A3HJX3KTF]]: it is symmetric, satisfies the triangle inequality and vanishes on the diagonal, but it may vanish for distinct points.
Some authors use $2\rho$, the distance of curvature $-1$, which doubles $k_X$.
The infinitesimal form is the Kobayashi–Royden pseudometric $F_X(p, v) = \inf\set{1/R : \text{there is } f \colon \mathbb{D} \to X \text{ holomorphic with } f(0) = p,\ f'(0) = R v}$ for $v$ tangent at $p$; by a theorem of Royden, $k_X$ is the integrated form of $F_X$.
Distance-decreasing property: if $g \colon X \to Y$ is holomorphic, then $k_Y(g(p), g(q)) \le k_X(p, q)$, because $g$ maps holomorphic chains to holomorphic chains.
Example: $k_{\mathbb{D}} = \rho$; the inequality $k_{\mathbb{D}} \le \rho$ comes from the chain with $f = \id$, and $k_{\mathbb{D}} \ge \rho$ comes from the [[Schwarz–Pick theorem|V1QYA9RW73]].
Non-example of a distance: $k_\C \equiv 0$; for $p \ne q$ in $\C$ and $R > 1$, the map $f(\zeta) = p + R \zeta (q - p)$ sends $0$ to $p$ and $1/R$ to $q$, so $k_\C(p, q) \le \rho(0, 1/R) \to 0$ as $R \to \infty$.