1983
DOI: 10.1016/0040-9383(83)90032-0
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The topology of complete minimal surfaces of finite total Gaussian curvature

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Cited by 195 publications
(165 citation statements)
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“…In this setting, the Gauss map g has a well-defined finite degree on M . A direct consequence of (3) is that the total curvature of an M as in Theorem 2.11 is −4π times the degree of its Gauss map g. It turns out that this degree can be computed in terms of the genus of the compactification M and the number of ends, by means of the Jorge-Meeks formula 12 [89]. Rather than stating here this general formula for an immersed surface M as in Theorem 2.11, we will emphasize the particular case when all the ends of M are embedded: (9) deg(g) = genus(M ) + #(ends) − 1.…”
Section: 3mentioning
confidence: 99%
“…In this setting, the Gauss map g has a well-defined finite degree on M . A direct consequence of (3) is that the total curvature of an M as in Theorem 2.11 is −4π times the degree of its Gauss map g. It turns out that this degree can be computed in terms of the genus of the compactification M and the number of ends, by means of the Jorge-Meeks formula 12 [89]. Rather than stating here this general formula for an immersed surface M as in Theorem 2.11, we will emphasize the particular case when all the ends of M are embedded: (9) deg(g) = genus(M ) + #(ends) − 1.…”
Section: 3mentioning
confidence: 99%
“…In these two cases, as in ours, equality implies the embeddedness of ends. (See [8,5] for the minimal surface case and [12] for the hyperbolic case.) To prove that equality implies the ends are embedded, a criterion for embeddedness of ends given in [4] will be applied.…”
Section: Introductionmentioning
confidence: 99%
“…Collin's Theorem reduces the question of topological obstructions for M ∈ M of finite topology and more than one end to the question of topological obstructions for complete embedded minimal surfaces of finite total curvature in R 3 . For example, if M is a complete embedded minimal surface in R 3 with finite total curvature, genus g and k ends, then M is properly embedded in R 3 and the Jorge-Meeks formula [20] calculates its total curvature to be −4π(g + k − 1). The first topological obstructions for complete embedded minimal surfaces M of finite total curvature were given by Jorge and Meeks [20], who proved that if M has genus zero, then M does not have 3, 4 or 5 ends.…”
Section: Conjecture 3 (Infinite Topology Conjecture (Meeks)) a Noncommentioning
confidence: 99%
“…For example, if M is a complete embedded minimal surface in R 3 with finite total curvature, genus g and k ends, then M is properly embedded in R 3 and the Jorge-Meeks formula [20] calculates its total curvature to be −4π(g + k − 1). The first topological obstructions for complete embedded minimal surfaces M of finite total curvature were given by Jorge and Meeks [20], who proved that if M has genus zero, then M does not have 3, 4 or 5 ends. Later this result was generalized by López and Ros [22] who proved that the plane and the catenoid are the only genus zero minimal surfaces of finite total curvature in M. About the same time, Schoen [37] proved that a complete embedded minimal surface of finite total curvature and two ends must be a catenoid.…”
Section: Conjecture 3 (Infinite Topology Conjecture (Meeks)) a Noncommentioning
confidence: 99%
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