Point singularities and conformal metrics on Riemann surfaces
Point singularities and conformal metrics on Riemann surfaces
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DOI：
10.1090/s0002-9939-1988-0938672-x
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发表时间：
1988
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通讯作者：
R. McOwen
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文献类型：
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作者：
R. McOwen
在给定一个闭双曲Riemann曲面和有限个点的情况下，我们证明了在给定点具有给定奇点或退化点的双曲协调度量的存在唯一性。如果M是具有负Euler特征x(M)的闭Riemann曲面，则它允许具有Gauss曲率K=-1的相容度量g.如果p e M，则我们可以要求M=M上具有Gauss曲率K=-1的相容度量g，并且在p，(1)g/g=0(R2a)asr=r(X)=dis9(x，p)^0处有一定的奇性或简并性.例如，这样的奇点产生于局部z-►zm(z EURO C，m e Z+)的映射：向前推标准度量给出对应于a=-(m-1)/m的奇点，而拉回标准度量给出对应于a=m-1的简并性。因此，我们对(1)With-1<；a<；oo特别感兴趣。更一般地，我们可以考虑有限数目的点pi，...，pn e M和ai，...，a“，并试图找到M=M\{pi，...，pn}上的相容度量g，其中(2)g/g=0(r2a‘)asrl=rl(X)=disg(x，pl)-+0。我们的主要成果如下。定理。设(M，g)是一个紧致黎曼曲面，高斯曲率K=-1，pi，…，pnEM，假设数CTi，…，an满足(I)-1<；aj，且(Ii)x(M)+et ai<；0，则M=M，…，pn}允许唯一的度规g逐点与g一致，具有Gauss曲率K=-1，且满足(2).此外，g有全曲率(3)jj-L)d=27rlx(M)+̂2Al。证据。为了方便起见，我们假定n=1，但证明的所有步骤都立即推广。我们要解(4)Au-e2u=-L关于M=Mp的问题，1986年11月14日收到，1987年2月24日以修正形式收到。1980年《数学学科分类》(1985年修订本)。初级30F10；次级35J60。
Given a closed hyperbolic Riemann surface and a finite number of points, we prove the existence and uniqueness of hyperbolic conformai metrics with prescribed singularities or degeneracies at the given points. If M is a closed Riemann surface with negative Euler characteristic x(M), then it admits a compatible metric g with Gauss curvature K = — 1. If p e M, then we can ask for a compatible metric g on M = M \ {p} with Gauss curvature K = — 1 and some prescribed singularity or degeneracy at p, (1) g/g = 0(r2a) asr = r(x) = dist9(x,p)^0. Such singularities arise, for example, from maps which are locally z —► zm (z € C, m e Z+): pushing the standard metric forward gives a singularity corresponding to a = — (m — 1)/m and pulling back the standard metric gives a degeneracy corresponding to a = m — 1. Thus we are particularly interested in (1) with — 1 < a < oo. More generally, we can consider a finite number of points pi,..., pn e M and ai,..., a„ € R and try to find a compatible metric g on M = M \ {pi,... ,pn} with (2) g/g = 0(r2a') asrl=rl(x) = distg(x,pl)-+0. Our main result is the following. THEOREM. Let (M, g) be a compact Riemann surface with Gauss curvature K = — 1 and pi,... ,pn e M. Suppose the numbers cti,..., an satisfy (i) —1 < a¿ < oo, and (ii) x(M) + Et ai < 0Then M = M \ {p\,... ,pn} admits a unique metric g which is pointwise conformai to g, has Gauss curvature K = — 1, and satisfies (2). Moreover, g has total curvature (3) Jj-l)d = 27rlx(M) + ̂ 2al\. PROOF. We shall assume for notational convenience that n = 1 but all steps of the proof generalize immediately. We want to solve (4) Au-e2u = -l on M = M\{p} Received by the editors November 14, 1986 and, in revised form, February 24, 1987. 1980 Mathematics Subject Classijication (1985 Revision). Primary 30F10; Secondary 35J60.