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Then it is a topological sphere and z: K → P1 is a conformal equivalence between K and the projective line. Case 2. K is noncompact and p is unbounded. Then z: K → C is a conformal equivalence between K and the whole complex plane. Case 3. K is noncompact and p tends to the finite number p(∞). Then z maps K 1:1 onto a disk of radius r = exp [ p(∞)]. This is Riemann’s case. Idea of the proof. Springer [1981] explains the actual construction of p: Let ∧1 be the class of smooth 1-forms ω = ω1 d x1 + ω2 d x2 on K and let ∗ω = −ω2 d x1 + ω1 d x2 .

Punctured Spheres. The once-punctured sphere is the plane, so the twicepunctured sphere is a punctured plane, and its universal cover K may be viewed as the Riemann surface of the logarithm; in short, K is a plane. The thricepunctured sphere (= the doubly punctured plane) is different: Its universal cover is the disk. 14 Examples 47 Proof. ), so only K = C needs to be ruled out. But if K = C, then the covering group is populated by substitutions of P S L(2, C) fixing ∞; see Section 5. These are of the form z → az + b and have fixed points if a = 1, whereas covering maps do not; see ex.

Check all that by means of pictures. Think of an example in which the lifted curve is not closed. Exercise 6. Prove that every covering map arises in this way. Exercise 7. Prove that covering maps have no fixed points, the identity excepted. Fundamental Group. The loops of M, starting and ending at o, fall into deformation classes; and it is easy to see that these classes form a group: The formation of classes respects the composition of loops effected by passing first about loop 1 and then about loop 2; the identity is the class of the trivial loop = the point o itself; the inverse is the class of the loop run backward; and so on.