User mrf - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T04:37:22Z http://mathoverflow.net/feeds/user/19246 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80638/subtlety-in-the-definition-of-the-kobayashi-metric/80880#80880 Answer by mrf for Subtlety in the definition of the Kobayashi metric mrf 2011-11-14T10:02:46Z 2011-11-14T10:02:46Z <p>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 $\mathbb{C}^n$.</p> <p>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 $\delta = d$.</p> <p>One fairly simple example where $\delta$ fails to satisfy the triangle inequality is the following. Let $$\Omega_\epsilon = \lbrace z \in \mathbb{C}^2 : |z_1| &lt; 1, |z_2| &lt; 1, |z_1z_2| &lt; \epsilon \rbrace.$$ 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.</p>