A liouville theorem on the PDE $$det (f_{ibar{jmath }})=1$$

A liouville theorem on the PDE $$det (f_{ibar{jmath }})=1$$
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偏微分方程的刘维尔定理 $$det (f_{ibar{jmath }})=1$$

DOI:
10.1007/s00209-020-02571-z
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发表时间:
2020
影响因子:
0.8
通讯作者:
Li Sheng
Li Sheng
中科院分区:
数学2区
文献类型:
--
作者:
An-Min Li;Li Sheng

文献摘要

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设f是求解(f_ i _ \jmath)= 1\;\;\;\;在\;\ \;\mathbb C^ n. det (fi <e:1>¯)= 1 on C n.假设度量ω _ f=-1 f_ i \jmath dz_ i∧d \ z_ j ω f=-1 fi <s:1>¯dzi∧dz¯j是完备的,并且f满足生长条件N_ 0^-1 (1+| z|^ 2)≤f≤\mathsf N_ 0 (1+| z|^ 2),\;\;\;作为\;\ \;| z|→∞,N 0-1 (1+| z| 2)≤f≤n0 (1+| z| 2),当| z|→∞时,对于某些数学N_ 0> 0, N 0> 0,则f是一个二次多项式。
Let f be a smooth plurisubharmonic function which solves (f_ i ̄\jmath)= 1\;\;\;\;\;\; on\;\;\;\mathbb C^ n. det (fi ȷ¯)= 1 on C n. Suppose that the metric ω _ f=-1 f_ i ̄\jmath dz_ i ∧ d ̄ z _ j ω f=-1 fi ȷ¯ dzi∧ dz¯ j is complete and f satisfies the growth condition N_ 0^-1 (1+| z|^ 2) ≤ f ≤\mathsf N_ 0 (1+| z|^ 2),\;\;\;\; as\;\;\;| z| → ∞, N 0-1 (1+| z| 2)≤ f≤ N 0 (1+| z| 2), as| z|→∞, for some\mathsf N_ 0> 0, N 0> 0, then f is a quadratic polynomial.