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I am a physics student and am interested in the study of invariant metrics. I have searched several textbooks, including those fat books of Krantz, but the following concern seems not to be mentioned in these books.

Let $\Omega\subset\mathbb{C}^{n}$ be an open, bounded domain. In literature, the Bergman distance between two points $z_{0}$ and $z_{1}$ (with $z_{0},z_{1}\in\Omega$) is defined by $$ d\left(z_{0},z_{1}\right):=\inf\left\{ l\left(\gamma\right):\gamma\in C^{1}\left(\left[0,1\right],\Omega\right),\gamma\left(0\right)=z_{0},\gamma\left(1\right)=z_{1}\right\} . $$ Here $$ l\left(\gamma\right):=\intop_{0}^{1}\sqrt{\sum_{j,k=1}^{n}\dfrac{\partial^{2}\left(\log K\left(\gamma\left(t\right),\gamma\left(t\right)\right)\right)}{\partial z_{j}\partial\overline{z}_{k}}\gamma'_{j}\left(t\right)\overline{\gamma'_{k}}\left(t\right)}dt, $$ and $K$ denotes the Bergman kernel of $\Omega$.

My question:

Is $\left(\Omega,d\right)$ a metric space? If so, is the topology induced by $d$ the standard topology on $\Omega$?

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    $\begingroup$ Yes it's a metric space; it's true more generally for any Riemannian metric. The second should be fine as well. $\endgroup$
    – Donu Arapura
    Commented Jul 18, 2018 at 14:19
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    $\begingroup$ Yep, it's true for every distance induced by a Riemannian metric $\endgroup$
    – Qfwfq
    Commented Jul 19, 2018 at 15:49

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Let me convert my comment into an answer with references. It's easy to get distracted by the complicated formulas, but the key points are:

  1. The Bergman kernel determines a Riemannian metric on $\Omega$ for which $d$ is the Riemannian distance. See chap 1 & 8 of Helgason, Differential geometry, Lie groups and Symmetric spaces for more details.
  2. Both your questions have positive answers for Riemannian manifolds, cf. Helgason, pp 51-52.
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  • $\begingroup$ You mean that these are true for any open, bounded domain $\Omega$ in $\mathbb{C}^{n}$ without any assumption on the smoothness of its boundary. $\endgroup$
    – Luke Diahan
    Commented Jul 19, 2018 at 1:21
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    $\begingroup$ No, what happens on the boundary is not relevant for whether $d$ is a metric and for the topology it generates. $\endgroup$
    – Donu Arapura
    Commented Jul 19, 2018 at 11:31
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    $\begingroup$ Thank you for your comment. All we need further is that $\Omega$ is connected so the definition makes sense. $\endgroup$
    – Luke Diahan
    Commented Jul 19, 2018 at 11:36
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    $\begingroup$ Yes, that's right $\Omega$ should be connected. $\endgroup$
    – Donu Arapura
    Commented Jul 19, 2018 at 11:48
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For the answer see the book by Jarnicki - Pflug: Invariant distances and metrics in complex analysis, 2nd edition, Corollary 12.9.7.

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