2002
DOI: 10.1090/s0002-9947-02-02989-6
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The Bergman metric on a Stein manifold with a bounded plurisubharmonic function

Abstract: Abstract. In this article, we use the pluricomplex Green function to give a sufficient condition for the existence and the completeness of the Bergman metric. As a consequence, we proved that a simply connected complete Kähler manifold possesses a complete Bergman metric provided that the Riemann sectional curvature ≤ −A/ρ 2 , which implies a conjecture of Greene and Wu. Moreover, we obtain a sharp estimate for the Bergman distance on such manifolds.

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Cited by 28 publications
(25 citation statements)
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“…A similar criterion holds true for the existence of Bergman metric. It is a consequence of Theorem 1, p. 2998, and Observation 2, p. 3002, in [7].…”
Section: Theorem 2 Let M Be a Complex Manifold Which Has A Bounded Comentioning
confidence: 75%
“…A similar criterion holds true for the existence of Bergman metric. It is a consequence of Theorem 1, p. 2998, and Observation 2, p. 3002, in [7].…”
Section: Theorem 2 Let M Be a Complex Manifold Which Has A Bounded Comentioning
confidence: 75%
“…-Proof of Theorem 1. The existence of the Bergman metric of a hyperconvex manifold was shown in [4]. Take a smooth function χ on R such that χ = 1 on (−∞, −1] and χ = 0 on [0, ∞).…”
Section: Proposition 4 (Cf[1])mentioning
confidence: 99%
“…Chen [3] and Herbort [8] independently), and this property was verified by Blocki-Pflug [2] (independently Herbort [8]). The case of hyperconvex Riemann surfaces was shown in [4].…”
mentioning
confidence: 99%
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“…Most of the results on Bergman completeness of domains is restricted to bounded ones. However, recently some papers appeared where the problem of Bergman completeness of unbounded domains was studied (see [Chen 2004], [Chen-Zhang 2002], [Juc 2004], [Chen-Kam-Ohs 2004], [Chen-Zhang 2004]). Motivated by examples from the last two papers we present a new class of unbounded domains admitting the Bergman metric, which are Bergman exhaustive (recall that the domain D is Bergman exhaustive if for any sequence (z ν ) ⊂ D without accumulation points in D the convergence lim ν→∞ K D (z ν ) = ∞ holds) and Bergman complete (see Theorem 6).…”
mentioning
confidence: 99%