arXivDaily arXiv每日学术速递 周一至周五更新
稍安勿躁,官方尚未更新
arXiv math/0402104math.CV
收藏

带边界流形上的全纯Morse不等式

Holomorphic Morse inequalities on manifolds with boundary

Robert Berman

摘要:

设X是一个有边界的紧带边流形,L^k是X上一个埃尔米特全纯线丛的高次幂.当X无边界时,Demailly全纯Morse不等式给出了L^k中的Dolbeault上同调群维数的渐近界.通过增加一个边界项,将Demailly不等式推广到X有边界的情形,该边界项表示为线丛的曲率和边界的Levi曲率的某个平均值.给出的例子表明,不等式是尖锐的,它表明,它们是兼容的洞填充。最有趣的情况是当X是一个伪凹流形,具有一个正的线丛L。若L的曲率与边界的Levi曲率共形等价,则Morse不等式成立,使得L的整体瓣空间具有极大的渐近增长,即L是大的.尖锐的例子表明,相应的上同调群的(0,1)-形式可能有最大的渐近增长,以及,除非保形等价成立。

英文摘要:

Let X be a compact manifold with boundary and let L^k be a high power of a hermitian holomorphic line bundle over X. When X has no boundary, Demailly's holomorphic Morse inequalities give asymptotic bounds on the dimensions on the Dolbeault cohomology groups with values in L^k. We extend Demailly's inequalities to the case when X has a boundary by adding a boundary term expressed as a certain average of the curvature of the line bundle and the Levi curvature of the boundary. Examples are given that show that the inequalities are sharp and it is shown that they are compatible with hole filling. The most interesting case is when X is a pseudoconcave manifold with a positive line bundle L. If the curvature of L is conformally equivalent to the Levi curvature of the boundary, the Morse inequalities are shown to be equalities, so that the space of global sections of L have maximal asymptotic growth, i.e. L is big. The sharp examples show that the corresponding cohomology group of (0,1)-forms may have maximal asymptotic growth, as well, unless the conformal equivalence holds.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

↑