Math 213br HW 3 solutions

February 26, 2014

Problem 1

Show that for each d ≥ 1, there exists a complex torus X ∼= C/Λ and an analyticmap f : X → X of degree d.

Solution

Let Λ be the lattice Z⊕Z√−d. It is stable under multiplication by

√−d, and hence

the map z 7→√−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 (becausez 7→

√−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, andthe map we gave has degree d.

Problem 2

Show that there is no proper holomorphic map f : ∆→ C.

Solution

If such an f exists, it has finitely many zeroes. Let B(z) be the (finite) Blashkeproduct that has the same zeroes as f . Then g = f/B(z) is holomorphic and hasno 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 compactRiemann surface Y of genus 2 defined by the equation y2 = p(x) admits a nonconstantholomorphic map to a Riemann surface of genus one.

Problem 5 2

Solution

There are many ways to do this, one example is that p(x) = x2(x2 − 1)(x2 − 3). Thecompact Riemann surface Y corresponding to y2 = x2(x2 − 1)(x2 − 2) has a map tothe compact Riemann surface X corresponding to y2 = x(x− 1)(x− 2), the map oncoordinates being given by (x, y) 7→ (x2, y). (To be more precise, the map we havewritten is a holomorphic map between the complement of a finite set of points of Yand 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) = 2.Does the converse hold?

Solution

Suppose there is a point x ∈ X such that mult(f, x) = 2. Then f−1(f(x)) = {x, y}.Let c be a non-trivial deck transformation of order 3. The inverse image of of a smallpunctured disk around f(x) is the disjoint union of 3 punctured disk, and c permutesthem 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 transpositionand hence has order 2, a contradiction.

The converse is false. For any compact Riemann surface Y , any non-normalsubgroup H of π1(Y ) of index 3 determines an unbranched cover Y ′ → Y of degree3 with trivial deck group. As an example, for Y of genus g ≥ 2 with the usualgenerators ai, bi for π1(Y ), take the pull back of a non-normal subgroup index 3 of S3

under the surjection π1(Y ) � S3 given by bi 7→ 1, a1 7→ a transposition and a2 7→ a3-cycle.

Problem 5

Find a Belyi polynomial p(z) of degree 5 such that p−1([0, 1]) is homeomorphic to theletter Y , with the fork at z = 0.(The Belyi condition means p(0) = 0, p(1) = 1 and the critical values are containedin {0,1})

Solution

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 topreimages of 1. If the graph has k edges coming out of a vertex, then p′ must vanishto order k − 1 at that point.

Math 213br HW 3

Problem 7 3

Let us find p for the first graph. We can see from the branching of the graph thatp′ must vanish to order 2 at 0, to order 1 at 1, and to order 1 at some other point,which we call b. Therefore

p′(z) = az2(z − 1)(z − b)

Since p(0) = 0, p′ determines p. Furthermore the conditions p(1) = 1 and p(a) = 0imply that b = 5/3 and a = 45/4, whence

p(z) =9z5 − 30z4 + 25z3

4

For the other graph, similar considerations lead to p(z) = 964z5 + 15

64z4 + 40

64z3

Problem 6

Find a Belyi map f : X → C where X = C/Z⊕ Zi is the square torus. That is, finda holomorphic map from X to the Riemann sphere branched over just {0, 1,∞}.

Solution

One such map is given by c℘(z)2 for a suitable constant c. The solution to a previousproblem 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 → Ybetween compact Riemann surfaces such that Deck(X/Y ) ∼= G. If G = (Z/2)n, whatis the smallest possible genus for Y .

Math 213br HW 3

Problem 8 4

Solution

The fundamental group π1(Y ) of a compact Riemann surface of genus g has presen-tation 〈a1, ..ag, b1, ...bg|[a1, b1]...[ag, bg] = 1〉. This admits a surjection onto the freegroup on g generators by killing bi, and hence given any finitely generated group Gthere is a compact Riemann surface Y such that π1(Y ) � G, and this gives rise to anormal covering of Riemann surfaces X → Y with deck group G. If G is finite, X isalso compact.

If G = (Z/2)n, such a cover exists only when π1(Y ) admits G as a quotient. Thishappens iff its abeliazation πab

1 (Y ) ∼= Z2g admits G as a quotient, which happens iff2g ≥ n.

Problem 8

Let L be the union of two distinct lines through the origin in C2. What is π1(C2−L)?Give a suitable topological definition of the ‘two-sheeted covering space Y of C2

branched over L.’ Show that Y is not a manifold, and that Y is homeomorphic tothe (singular quadric) surface in C3 given by z2 = xy. Sketch the real points of thisquadric.

Solution

C2 − L ∼= C∗ × C∗, so π1(C2 − L) ∼= Z2.A definition of two-sheeted cover Y πC2 branched over L is as follows:

• π−1(L)→ L is a homeomorphism.

• Y \ π−1(L) → C2 − L is a covering map of degree 2, with non-trivial localmonodromy around the two lines in L. This is equivalent to this coveringcorresponding to the unique surjective map Za⊕Zb� Z/2 sending both a andb to the non-trivial element. Here a, b denote small loops around the lines x = 0and y = 0.

• Y = Y \π−1(L)→ C2−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 z2 = xy, this admits a map to C2 given by (x, y, z) 7→ (x, y).It is easy to see that this satisfies the above 3 items, so Y is homeomorphic to thisquadric.We now show that the quadric Y is not a topological manifold. Let y ∈ Y denote thesingular point (0, 0, 0). We will show that Y is not a manifold around y by computingthe local cohomology group H i

y(Y ) = H i(Y, Y \ {y}) (all cohomology are taken withZ coefficient). By excision, this local cohomology group is the same when we replaceY by a small neighborhood of y, and thus if Y is a manifold near y, these localcohomology must agree with the local cohomology of a 4-ball at the center of the

Math 213br HW 3

Problem 8 5

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Figure 1: The cone x2 = y2 + z2.

ball. In particular, the cohomology is Z[−4] (the only non-vanshing cohomology is indegree 4 and is isomorphic to Z). To compute local cohoology we use the long exactsequence

H0y (Y )→ H0(Y )→ H0(Y \ {y})→ H1

y (Y )→ · · ·

Since Y is contractible this shows H i+1y (Y ) ∼= H i(Y \ {y}) for i ≥ 1. Topologically

Y \ {y} is homotopy equivalent to an S1-bundle over P1 ∼= S2 with Euler class2 ∈ Z = H2(S2) (Y \ {y} is the total space of the C∗-torsor corresponding to the linebundle O(2) on P1). The Gysin sequence thus gives

· · · → Hn(Y \ {y})→ Hn−1(S2)→ Hn+1(S2)→ · · ·

and thus H3y (Y ) = H2(Y \ {y} ∼= Z/2Z), 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

x2 = y2 + z2).

Math 213br HW 3