Riemann Surfaces Mock Exam

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1 Riemann Surfaces Mock Exam Notes: 1. Write your name and student number **clearly** on each page of written solutions you hand in. 2. You are expected to give solutions in English or Dutch. 3. You are expected to explain your answers. 4. You are allowed to consult text books but not your own notes. 5. Advice: read all questions first, then start solving the ones you already know how to solve or have good idea on the steps to find a solution. After you have finished the ones you found easier, tackle the harder ones. Exercise 1. Let M be a complex manifold, i.e., M is endowed with a set of coordinates which take values in C n and such that the change of coordinates is holomorphic. Show that a M has a natural almost complex structure I. Show that this natural almost complex structure is (algebraically) integrable, i.e., the +i-eigenspace of I is an involutive subbundle of T C M. Solution. We define an almost complex structure on M as follows. For each p M, we take a holomorphic chart centered at p, ϕ : U M C n, ϕ(p) = 0. We define an automorphism I : T p M and the automorphism of T 0 C n given by multiplication by i. We must check that this is a well defined automorphism. If ψ : U C n is another holomorphic chart centered at p, then we must that that iψ ϕ 1 = ψ ϕ 1 i, but this follows from the fact that ϕ and ψ are holomorphic coordinates, i.e., ψ ϕ 1 is a holoomorphic map, a condition which is defined by the relation above. Since I in coordinate corresponds to multiplication by I, we see that the space Tp 1,0 M is the space generated by the vectors / z i p. Sine this is true for all p, we conclude that the +i-eigenbundle of I is the bundle spanned by / z i is local coordinates and to check integrability it is enough to show that the bracket of generators still lies in the space they generate. In this case [ ], = 1 [ i, i ] = 0 z i z j 4 x i y i x j y j where the last equality follows from the fact that the Lie bracket of coordinate vector fields vanish and the Lie bracket is C-linear. Exercise 2. Let Σ be a Riemann surface of genus g, p Σ and L p be the holomorphic line bundle associated to the divisor p, i.e., L p has a section with a single simple zero at p. Compute the dimensions of H 0 (Σ, L p ) and H 1 (Σ, L p );

2 H 0 (Σ, L p) and H 1 (Σ, L p). Solution. We start with the cohomology with coefficients in L p. Since L p has a holomorphic section s with a simple zero at p, its Euler characteristic is 1. To determine h 0 (Σ, L p ), we observe that any other holomorphic section of L p is given by fs where f is a meromorphic function on Σ which is either holormorphic of has at worse a single simple pole at p (and holomorphic on Σ\{p}). In the first case f is holomorphic in a compact surface hence it is constant. In the second case, f : Σ CP 1 and infinity in CP 1 is a regular value which is taken only once, that is, the degree of f is one and f is and isomorphism between Σ and CP 1. That is, if Σ is not CP 1, any other holomorphic section of L p is a constant multiple of s and hence h 0 (Σ, L p ) = 1 For Σ = CP 1, the space of meromorphic functions with a simple pole, say at z = 0, is given by the set of bi-holomorphisms of CP 1 which map zero to infinity. Since the biholomorphisms of CP 1 are given by Möbius transformations, the condition that zero is a pole means that we are considering Möbius transformations of the form z az + b. z with b 0. Notice however that b = 0 corresponds to the constant holomorphic map z a, which gives rise to the sections of L p which are constant multiples of s. Hence we conclude that H 0 (CP 1, L p ) is 2-dimensional. Finally, Riemann Roch tells us h 0 (Σ, L p ) h 1 (Σ, L p ) = d g + 1 h 1 (Σ, L p ) = g 2 + h 0 (Σ, L p ). So, if g > 0, we get h 1 (Σ, L p ) = g 1 and for g = 0, h 1 (Σ, L p ) = 0. Next we deal with L p. Since L p has a nowhere vanishing meromorphic section with a simple pole at p, we have that the degree of L p is 1. Hence L p has no holomorphic sections, as the formula χ E = #(zeros of s) #(poles of s) shows that a holomorphic line bundle can only have a holomorphic section s (i.e., s has no poles) if its Euler characteristic is positive. Therefore h 0 (Σ, L p) = 0 and Riemann Roch gives h 1 (Σ, L p) = g. Exercise 3. Let λ C, λ 0, 1 and let Σ CP 2 be the Riemann surface determined by the polynomial. and consider the map P (Z 0, Z 1, Z 2 ) = Z 2 0Z 2 2 Z 2 1(Z 1 Z 2 )(Z 1 λz 2 ) π : CP 2 \{[1, 0, 0]} CP 1 π([z 0, Z 1, Z 2 ]) = [Z 1, Z 2 ]. a) Find the singular points of Σ and resolve them to obtain a smooth Riemann surface Σ in an appropriate blow-up of CP 2, b) Show that the map π extends as a holomorphic map to the whole of Σ, c) Find the degree and the branch points of π and determine the Euler characteristic of Σ. Solution.

3 a) First we look for singular points of Σ, i.e., solution to the equations P (Z 0, Z 1, Z 2 ) = 0 and P Z i (Z 0, Z 1, Z 2 ) = 0 for i = 0, 1, 2. Z0Z Z1(Z 2 1 Z 2 )(Z 1 λz 2 ) = 0 2Z 0 Z2 2 = 0 2Z 1 (Z 1 Z 2 )(Z 1 λz 2 ) Z1(Z 2 1 λz 2 ) Z1(Z 2 1 Z 2 ) = 0 2Z0Z Z1(Z 2 1 iλz 2 ) + λz1(z 2 1 Z 2 ) = 0 The second equation implies that at least one of Z 0 and Z 2 must vanish at a singular point. If Z 2 0, then Z 0 = 0 and the first equation means that Z 1 /Z 2 is a root of the polynomial p(z) = z 2 (z 1)(z λ). Similarly, the third equation means that Z 1 /Z 2 is also a solution to p (z) = 0 Since the only double root of p is z = 0, we see that the singular point must be of the form (0, 0, Z 2 ) and that such a point is indeed a solution to all four equations. Hence [0, 0, 1] is a singular point of Σ. Finally, if Z 2 = 0, then the third equation implies Z 1 = 0 and one can verify that (Z 0, 0, 0) is also a solution to the system, hence [1, 0, 0] is also a singular point of Σ. To resolve the singularities of Σ we take charts for CP 2 centered at the singular points and blow them up. We start with [0, 0, 1]. For that point we can take affine the chart (z 0, z 1 ) [z 0, z 1, 1] In this chart, the equation for Σ becomes z 2 0 z 2 1(z 1 1)(z 1 λ) = 0. A neighborhood of the exceptional divisor in the blow-up is parametrized by two charts which relate to the coordinates (z 0, z 1 ) by the following relations u 1 = z 0 ; v 1 = z 1 /z 0 ; u 2 = z 0 /z 1 ; v 2 = z 1. This means that Π, the blow-up map, obtained by writing (z 0, z 1 ) as a function of (u i, v i ) is given by Π(u 1, v 1 ) = (u 1, u 1 v 1 ) Π(u 2, v 2 ) = (u 2 v 2, v 2 ). In the (u 1, v 1 )-coordinate system, the pre-image of Σ\{[0, 0, 1]} is given by (Σ\{[0, 0, 1]}) = {(u 1, v 1 ) : u 2 1 u 2 1v 2 1(u 1 1)(u 1 λ) = 0 & u 1 0} (Σ\{[0, 0, 1]}) = {(u 1, v 1 ) : 1 v 2 1(u 1 1)(u 1 λ) = 0 & u 1 0} And the proper transform of Σ in this chart is given by (Σ\{[0, 0, 1]}) = {(u 1, v 1 ) : 1 v 2 1(u 1 1)(u 1 λ) = 0} and this is a smooth Riemann surface in this chart, as it is given as the zero of a function f(u 1, v 1 ) = 1 v 2 1(u 1 1)(u 1 λ) for which f = 0, f/ u 1 = 0 and f/ v 1 = 0 do not have a simultaneous solution. In the (u 2, v 2 )-coordinate system, the pre-image of Σ\{[0, 0, 1]} is given by (Σ\{[0, 0, 1]}) = {(u 2, v 2 ) : u 2 2v 2 2 v 2 2(u 2 v 2 1)(u 2 v 2 λ) = 0 & v 2 0} (Σ\{[0, 0, 1]}) = {(u 2, v 2 ) : u 2 2 (u 2 v 2 1)(u 2 v 2 λ) = 0 & v 2 0} And the proper transform of Σ in this chart is given by (Σ\{[0, 0, 1]}) = {(u 2, v 2 ) : u 2 2 (u 2 v 2 1)(u 2 v 2 λ) = 0} and this is a smooth Riemann surface in this chart, as it is given as the zero of a function f(u 2, v 2 ) = u 2 2 (u 2 v 2 1)(u 2 v 2 λ) = 0 for which f = 0, f/ u 2 = 0 and f/ v 2 = 0 do not have a simultaneous solution.

4 Notice that the blow-up process resolved the singularity of Σ at [0, 0, 1] and produced a suface which intersects the exceptional divisor in tow points, namely, in the (u 2, v 2 ) chart, these points are the solution to u 2 2 (u 2 v 2 1)(u 2 v 2 λ) = 0 & v 2 = 0. and hence u 2 = ±λ 1 2. Next we resolve the point [1, 0, 0]. Here we take affine coordinates so that Σ is given by the equation This time, the blow-up coordinates are given by Or, in terms of the blow-up map so In the (u 3, v 3 )-coordinate system, we have (z 1, z 2 ) [1, z 1, z 2 ] z 2 2 z 2 1(z 1 z 2 )(z 1 λz 2 ) = 0. u 3 = z 1, v 3 = z 2 /z 1 ; u 4 = z 1 /z 2, v 4 = z 2. Π 2 (u 3, v 3 ) = (u 3, u 3 v 3 ) Π 2 (u 4, v 4 ) = (u 4 v 4, v 4 ). 2 (Σ\{[1, 0, 0]}) = {(u 3, v 3 ) : u 2 3v 2 3 u 2 3(u 3 u 3 v 3 )(u 3 λu 3 v 3 ) = 0 & u 3 0} 2 (Σ\{[1, 0, 0]}) = {(u 3, v 3 ) : v 2 3 u 2 3(1 v 3 )(1 λv 3 ) = 0 & u 3 0} 2 (Σ\{[1, 0, 0]}) = {(u 3, v 3 ) : v 2 3 u 2 3(1 v 3 )(1 λv 3 ) = 0} Notice that in this case the proper transform of Σ is not yet smooth and (u 3, v 3 ) = (0, 0) is a singular point,so we have to blow his point up We have two new coordinate charts, (u 5, v 5 ) and (u 6, v 6 ) which relate to (u 3, v 3 ) by the blow-up map In the (u 5, v 5 )-coordinate system we have Π 3 (u 5, v 5 ) = (u 5, u 5 v 5 ) Π 3 (u 6, v 6 ) = (u 6 v 6, v 6 ). 3 2 (Σ\{[1, 0, 0]}) = {(u 5, v 5 ) : u 2 5v 2 5 u 2 5(1 u 5 v 5 )(1 λu 5 v 5 ) = 0 & u 5 0} 3 2 (Σ\{[1, 0, 0]}) = {(u 5, v 5 ) : v 2 5 (1 u 5 v 5 )(1 λu 5 v 5 ) = 0 & u 5 0}. And the proper transform is 3 2 (Σ\{[1, 0, 0]}) = {(u 5, v 5 ) : v 2 5 (1 u 5 v 5 )(1 λu 5 v 5 ) = 0}. which is now the equation of a smooth surface. The points on intersection of the resolution with the exceptional divisor correspond to v 5 = ±1. A similar computation in the chart (u 6, v 6 ) shows that (u 6, v 6 ) = (0, 0), the only point of the last exceptional divisor not included in the precious parametrization, is not part of the proper transform of Σ, so the proper transform of Σ is smooth in the charts considered so far. We have only missed the chart (u 4, v 4 ). In this chart, the proper transform of Σ is given by the equation and this is a smooth surface in this chart. 1 u 2 4v 2 4(u 4 1)(u 4 λ) = 0

5 So we conclude that after three blow-ups, we have resolved Σ into a smooth Riemann surface. b) Here we show that the map π extends to CP 2, the blow-up of CP 2 at the point [1, 0, 0] and hence it extends to Σ. To prove the claim, we simply write the map π in terms of the coordinates (u 3, v 3 ) and U 4, v 4 ) of the previous part of the exercise: π Π 2 (u 3, v 3 ) = π([1, u 3, u 3 v 3 ]) = [u 3, u 3 v 3 ] = [1, v 3 ]. π Π 2 (u 4, v 4 ) = π([1, u 4 v 4, v 4 ]) = [u 4 v 4, v 4 ] = [u 4, 1]. Showing that π Π 2 is well defined and holomorphic smooth on the blow up of CP 2 at the point [1, 0, 0]. Above we constructed a manifold CP 2, which is a blow-up of CP 2 at two points, so we have a blow-up map Π : CP 2 CP 2 since π Π 2 is smooth, we conclude that π Π 2 Π is smooth and Σ is a smooth submanifold of CP 2 hence the restriction of this smooth map to Σ is also smooth. c) In the (z 1, z 2 )-coordinates, we have that Σ is given by z 2 2 z 2 1(z 1 z 2 )(z 1 λz 2 ) = 0. For each point [a, b] CP 1, we have that the point (z 1, z 2 ) Σ maps to [a, b] if and only if z 1 = (a/b)z 2, so we see that generically, the points of Σ which map to a fixed point [a, b] CP 1 are solutions to a quadratic polynomial, and the degree of π is 2. Still in this chart, we see that if a/b = α is fixed, then the elements of Σ which project to [a, b] are given by z 2 2 α 2 z 4 2(α 1)(α λ) = 0 i.e. for α 0, 1λ there are always two solutions and for α = 0, 1 or λ the equation would force z 2 = 0 and hence z 1 = 0, which is a singular point of Σ. Similarly, if we fix b/a = β we obtain that the elements of Σ which project to [a, b] are given by β 2 z 2 1 z 4 1(1 β)(1 βλ) = 0. and β = 0 forces z 1 = 0 and hence z 2 = 0 This means that α = 0, 1, λ, are the only possible singular values of π. From b) we know that π Π 2 is well defined and that at the singular point (u 3, v 3 ) = (0, 0), π Π 2 (0, 0) = [1, 0]. Since the blow-up that resolves the singularity at (u 3, v 3 ) = (0, 0) introduced two points, we see that π 1 ([1, 0]) has two point in Σ, hence [1, 0] is not a branch point for π : Σ CP 1. Next, we must check the points which were not covered by the coordinates (z 1, z 2 ). All the remaining points are in the affine chart (z 0, z 1 ) [z 0, z 1, 1]. There Σ is given by z 2 0 z 2 1(z 1 1)(z 1 λ) and the map π is simply projection onto the z 1 -plane. From the above we see that z 1 = 0 is not a singular value, since the resolution of the singularity (z 0, z 1 ) = (0, 0) introduced two points both of which are mapped to [0, 1]. On the other hand, if z 1 = 1 or λ then the only value of z 0 that solves the euation above is z 0 = 0 and hence these are ramification points. So [1, 1] and [λ, 1] are the only ramification points of π and we can compute the Euler characteristic of Σ by the Riemann Hurwiz formula: So Σ is CP 1. χ Σ = dχ CP 1 b = = 2.

6 Exercise 4. Let f : Σ 1 Σ 2 be a non constant holomorphic map between compact connected Riemann surfaces. Let α Ω 1,0 (Σ 2 ) be a meromorphic 1-form on Σ 2 whose set of zeros and poles is disjoint from the set of critical values of f. By counting the number of zeros and poles of f α, re-obtain the Riemann Hurwitz formula: χ Σ1 = deg(f) χ Σ2 b, where b is the total branch number of f. Solution. Under the hypotheses of the exercise, for each point p Σ 2 which is a zero or a pole of α, f 1 (p) has deg(f) points and at any such point q f 1 (p) f α is a zero or a pole or the same order, since once can find coordinates centered at p andq for which f is simply the identity map. Now, if q Σ 1 is a singular point for f with branch number k and p = f(q), then we can find coordinates centered at p and q such that f is given by w = f(z) = z k and since α does not vanish at p, it can be written as α = a(w)dw, with a(0) 0. Then f α = f (a(w)dw) = a(f(z))d(f w) = a(f(z))d(z k ) = kz k 1 a(f(z))dz. That is f α has a zero of order k 1 at q. Since this is true for all singular points q, we see that the number of zeros of f α is deg(f) times the number of zeros of α plus the total branch number of f, while the number of poles of f αis simply the number of poles of ]alpha times the degree of f. Since the Euler characteristic of a Riemann surface is given by the number of poles minus the number of zeros of any meromorphic 1-form, we obtain the result.