Math 13br HW 3 solutions February 6, 014 Problem 1 Show that for each d 1, there exists a complex torus X = C/Λ and an analytic map f : X X of degree d. Let Λ be the lattice Z Z d. It is stable under multiplication by d, and hence the map z d (fix one of the square-root of d, say) descends to a holomorphic map C/Λ C/Λ. It is clear that this is an unbranched covering map (because z dz has no critical points). The fiber of this map over 0 C/Λ are a + b = d with db Z and a Z, taken modulo Λ. Thus the fiber has exactly d element, and the map we gave has degree d. Problem Show that there is no proper holomorphic map f : C. If such an f exists, it has finitely many zeroes. Let B(z) be the (finite) Blashke product that has the same zeroes as f. Then g = f/b(z) is holomorphic and has no zeroes in. Furthermore it is still proper, because B(z) is bounded above and f as z 1 by properness of f. Thus g as z 1, so it is proper. But then g is a proper map to C with no zeroes, a contradiction. Problem 3 Give an explicit example of a polynomial p(x) of degree 6 such that the compact Riemann surface Y of genus defined by the equation y = p(x) admits a nonconstant holomorphic map to a Riemann surface of genus one.
Problem 5 There are many ways to do this, one example is that p(x) = x (x 1)(x 3). The compact Riemann surface Y corresponding to y = x (x 1)(x ) has a map to the compact Riemann surface X corresponding to y = x(x 1)(x ), the map on coordinates being given by (x, y) (x, y). (To be more precise, the map we have written is a holomorphic map between the complement of a finite set of points of Y and X. It extends uniquely to a map Y X by the Riemann extension theorem). Problem 4 Let f : X Y be a nonconstant map of degree 3 between compact Riemann surfaces. Show that if Deck(X/Y ) = Z/3, then there is no x X such that mult(f, x) =. Does the converse hold? Suppose there is a point x X such that mult(f, x) =. Then f 1 (f(x)) = {x, y}. Let c be a non-trivial deck transformation of order 3. The inverse image of of a small punctured disk around f(x) is the disjoint union of 3 punctured disk, and c permutes them non-trivially. But c(y) = y because x and c(x) have the same multiplicity for f, so c fixes the punctured disk centered around y. But then c must be a transposition and hence has order, a contradiction. The converse is false. For any compact Riemann surface Y, any non-normal subgroup H of π 1 (Y ) of index 3 determines an unbranched cover Y Y of degree 3 with trivial deck group. As an example, for Y of genus g with the usual generators a i, b i for π 1 (Y ), take the pull back of a non-normal subgroup index 3 of S 3 under the surjection π 1 (Y ) S 3 given by b i 1, a 1 a transposition and a a 3-cycle. Problem 5 Find a Belyi polynomial p(z) of degree 5 such that p 1 ([0, 1]) is homeomorphic to the letter Y, with the fork at z = 0. (The Belyi condition means p(0) = 0, p(1) = 1 and the critical values are contained in {0,1}) There are several possible solutions, depending on how exactly you draw your graph. Two possibilities are drawn in the figure below. The solid dots correspond to preimages of 0 and the open circles correspond to preimages of 1. If the graph has k edges coming out of a vertex, then p must vanish to order k 1 at that point. Math 13br HW 3
Problem 7 3 Let us find p for the first graph. We can see from the branching of the graph that p must vanish to order at 0, to order 1 at 1, and to order 1 at some other point, which we call b. Therefore p (z) = az (z 1)(z b) Since p(0) = 0, p determines p. Furthermore the conditions p(1) = 1 and p(a) = 0 imply that b = 5/3 and a = 45/4, whence p(z) = 9z5 30z 4 + 5z 3 4 For the other graph, similar considerations lead to p(z) = 9 64 z5 + 15 64 z4 + 40 64 z3 Problem 6 Find a Belyi map f : X Ĉ where X = C/Z Zi is the square torus. That is, find a holomorphic map from X to the Riemann sphere branched over just {0, 1, }. One such map is given by c (z) for a suitable constant c. The solution to a previous problem set discussed why this is branched exactly over 0, and another point, which we can normalize to be 1 by choosing c appropriately. Problem 7 Show that for any finite group G, there exists a Galois covering map f : X Y between compact Riemann surfaces such that Deck(X/Y ) = G. If G = (Z/) n, what is the smallest possible genus for Y. Math 13br HW 3
Problem 8 4 The fundamental group π 1 (Y ) of a compact Riemann surface of genus g has presentation a 1,..a g, b 1,...b g [a 1, b 1 ]...[a g, b g ] = 1. This admits a surjection onto the free group on g generators by killing b i, and hence given any finitely generated group G there is a compact Riemann surface Y such that π 1 (Y ) G, and this gives rise to a normal covering of Riemann surfaces X Y with deck group G. If G is finite, X is also compact. If G = (Z/) n, such a cover exists only when π 1 (Y ) admits G as a quotient. This happens iff its abeliazation π1 ab (Y ) = Z g admits G as a quotient, which happens iff g n. Problem 8 Let L be the union of two distinct lines through the origin in C. What is π 1 (C L)? Give a suitable topological definition of the two-sheeted covering space Y of C branched over L. Show that Y is not a manifold, and that Y is homeomorphic to the (singular quadric) surface in C 3 given by z = xy. Sketch the real points of this quadric. C L = C C, so π 1 (C L) = Z. A definition of two-sheeted cover Y πc branched over L is as follows: π 1 (L) L is a homeomorphism. Y \ π 1 (L) C L is a covering map of degree, with non-trivial local monodromy around the two lines in L. This is equivalent to this covering corresponding to the unique surjective map Za Zb Z/ sending both a and b to the non-trivial element. Here a, b denote small loops around the lines x = 0 and y = 0. Y = Y \π 1 (L) C L π 1 (L) is topologized with the weakest topology to make π continuous, where each part is topologized by the previous requirement. With this definition it is clear that Y is unique up to homeomorphism. Consider the quadric z = xy, this admits a map to C given by (x, y, z) (x, y). It is easy to see that this satisfies the above 3 items, so Y is homeomorphic to this quadric. We now show that the quadric Y is not a topological manifold. Let y Y denote the singular point (0, 0, 0). We will show that Y is not a manifold around y by computing the local cohomology group H i y(y ) = H i (Y, Y \ {y}) (all cohomology are taken with Z coefficient). By excision, this local cohomology group is the same when we replace Y by a small neighborhood of y, and thus if Y is a manifold near y, these local cohomology must agree with the local cohomology of a 4-ball at the center of the Math 13br HW 3
Problem 8 5 - - - - - - Figure 1: The cone x = y + z. ball. In particular, the cohomology is Z[ 4] (the only non-vanshing cohomology is in degree 4 and is isomorphic to Z). To compute local cohoology we use the long exact sequence Hy0 (Y ) H 0 (Y ) H 0 (Y \ {y}) Hy1 (Y ) Since Y is contractible this shows Hyi+1 (Y ) = H i (Y \ {y}) for i 1. Topologically Y \ {y} is homotopy equivalent to an S 1 -bundle over P1 = S with Euler class Z = H (S ) (Y \ {y} is the total space of the C -torsor corresponding to the line bundle O() on P1 ). The Gysin sequence thus gives H n (Y \ {y}) H n 1 (S ) H n+1 (S ) and thus Hy3 (Y ) = H (Y \ {y} = Z/Z), showing that the local cohomology of Y at y is not that of a manifold. The real points of Y is as pictured (up to affine transformation it is the cone x = y + z ). Math 13br HW 3