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When defining the Kobayashi metric on a connected complex analytic space $X$, one makes the following auxiliary definition:

A holomorphic chain from $x\in X$ to $y\in X$ is a finite sequence of holomorphic maps $f_1,\ldots ,f_n\colon\Delta\to X$ (where $\Delta$ is the unit disk in $\mathbb{C}$) together with points $z_1,\ldots ,z_n,w_1,\ldots ,w_n\in\Delta$ such that $f_1(z_1)=x$, $f_i(z_i)=f_{i+1}(w_{i+1})$ for $1\le i< n$ and $f(w_n)=y$.

The length of a holomorphic chain is, with this notation, is $\sum_{i=1}^nd(z_i,w_i)$ (Poincaré metric on $\Delta$).

Finally the Kobayashi pseudo-distance on $X$ is obtained by setting $d(x,y)=$ infimum of lengths of all holomorphic chains from $x$ to $y$.

This "pseudo-distance" is obviously symmetric, satisfies $d(x,x)=0$ and the triangle inequality. The space is called Kobayashi hyperbolic if $d$ is in addition non-degenerate, i.e. if $d$ is a metric.

Now one could as well begin with the following much simpler construction:

Consider the function $\delta :X\times X\to [0,\infty ]$ with $\delta (x,y)=\inf d(z,w)$, the infimum running over all triples $(f,z,w)$ with $f:\Delta\to X$ holomorphic, $z,w\in\Delta$ and $f(z)=x$, $f(w)=y$.

This is still symmetric and satisfies $\delta (x,x)=0$, but now it is unclear whether

a) $\delta (x,y)$ is finite, i.e. the set of triples $(f,z,w)$ is non-empty;

b) $\delta$ satisfies the triangle inequality.

Clearly, a) and b) together are equivalent to $d=\delta$, and $d$ can be obtained from $\delta$ by an easy construction. Finally, my questions:

Under which circumstances is $d=\delta$?

Is there a simple example where $d\neq\delta$?

What is the logical relation between a) and b)?

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2 Answers

Let me give an example where $d \neq \delta$. I learned it from A Survey on Hyperbolicity of Projective Hypersurfaces, Example 1.2.1.

Consider $D$ as the following open subset of $\mathbb C^2$
$$ D = \lbrace (z,w) \in \mathbb C^2 ; |z| < 1, |zw|< 1 \rbrace \setminus \lbrace (0,w) | |w| \ge 1 \rbrace . $$ The Kobayashi distance of $p=(0,0)$ and $q=(0,1/2)$ is zero. Indeed, if $p_n = ( 1/n,0)$ and $q_n = ( 1/n, 1/2)$ then $$\lim_{n \to \infty} \delta(p_n,q_n) = 0.$$ And we can verify that this implies $d(p,q)=0$.

If $f: \Delta \to D$ is such that $f(0)=p$ and $f(a)=q$ then applying Schwarz lemma to $f_2$, the second component of $f=(f_1,f_2)$, we see that $|a|\ge 1/2$. Therefore $\delta(p,q) = 1/2$.

Notice that the Kobayashi pseudo-distance is continuous while $\delta$ is not. It seems reasonable to expect that the continuity of $\delta$ implies $\delta=d$.

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This is actually more subtle than you might think. A classification of the spaces for which $\delta = d$ is far from known, even for domains in $\mathbb{C}^n$.

However, if $\Omega \subset \mathbb{C}^n$ is convex (or biholomorphic to a lineally convex domain), then $\delta = d$, which was shown by Lempert [Lempert, László . La métrique de Kobayashi et la représentation des domaines sur la boule. Bull. Soc. Math. France 109 (1981), no. 4, 427--474.] There are also some other examples known where $\delta = d$.

One fairly simple example where $\delta$ fails to satisfy the triangle inequality is the following. Let $$\Omega_\epsilon = \lbrace z \in \mathbb{C}^2 : |z_1| < 1, |z_2| < 1, |z_1z_2| < \epsilon \rbrace.$$ Also, let $P = (1/2, 0)$ and $Q = (0, 1/2)$. You can check that $\delta(P,0)$ and $\delta(0,Q)$ (with respect to $\Omega_\epsilon$) are independent of $\epsilon$, but $\delta(P,Q) \to \infty$ as $\epsilon \to 0$. Hence, if $\epsilon$ is sufficiently small, $\delta_\Omega$ violates the triangle inequality.

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The paper by Lempert mentioned above is available at numdam.org/item?id=BSMF_1981__109__427_0 –  jvp Nov 14 '11 at 16:34
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