AN INTRODUCTION TO ARITHMETIC AND RIEMANN SURFACE. We describe points on the unit circle with coordinate satisfying

Transcription

1 AN INTRODUCTION TO ARITHMETIC AND RIEMANN SURFACE 1. RATIONAL POINTS ON CIRCLE We start by asking us: How many integers x, y, z) can satisfy x 2 + y 2 = z 2? Can we describe all of them? First we can divide z and look for rational numbers instead. So the question transformed to: How many points on the unit circle of R 2 has coordinates to be integers numbers? Can we parameterized all of them? We describe points on the unit circle with coordinate satisfying 1.1) x 2 + y 2 = 1. The first step is of course set x = a c, y = b c with a, b, c being integers, 0,1) x,y) t The above line equation ) t 0 = y 1 x 0 implies x = t1 y). If x, y Q, clearly t Q. On the other hand if t Q, then implies 1 = x 2 + y 2 = t 2 y 1) 2 + y 2 t 2 + 1)y 2 2t 2 y + t 2 1 = 0. And thus using 2t 2 ) 2 4t 2 + 1)t 2 1) = 4t 4 4t 4 1) = 4, we solve Similarly x = t1 y) = The pair x, y) = 2t, t2 1 t 2 +1 t 2 +1 y = 2t2 2 2t 2 + 1) = t2 1 t t. And both x, y are rational numbers. This solves the questions. t 2 +1 ) for t Q gives all rational point in circle. This implies a = 2mn, b = m 2 n 2, c = m 2 + n 2 for m, n Z gives ALL integer solutions of a 2 + b 2 = c 2. This is formula of Euclid for Pythagorean triple a, b, c), and thus done 3rd century BC. Year -300). See [Eu] [Py]. Question How abot a 3 + b 3 = c 3? Does it have any nontrivial) integer solutions? If so, are there finite or infinite of them? Can we find them out? 1

2 2 AN INTRODUCTION TO ARITHMETIC AND RIEMANN SURFACE Question How abot a n + b n = c n, for some positive integer n? Question Given a polynomial fx, y), say with integer coefficients. Is there integer solutoin of fx, y) = 0? If so, are there finite or infinite many of them? Can we find all of them? Human beings have spent several hundred years to realize that, to approach such kind of questions, one should first think about the set fx, y) = 0 for x, y C. And then look for this set s intersection with Q 2. For example, the set {x, y, z) C 3 x 3 + y 3 = z 3.} Is a "real dimension two manifold". Q 3 C 3. And one further searches for its intersection with 2. RIEMANN SURFACES AND DEGREE-GENUS FORMULA The geometryshape) of that set, observed by Riemann, has extremely rich stories. And, the shape of that set, had deep relation with the answers to our rational point questions. Definition 2.1. A bijective map f : W 1 W 2 between two open sets in C is call biholomorphic if it is analyticnamely locally expressible by its complex Taylor expansion), and its inverse is also analytic. For example z e z, z z 3 + 5z + 1 are analytic. Definition 2.2. A complex manifold M is a topological space with an open cover charts) {U i } i I and homeomorphisms φ i : U i W i with W i open subset of C d, such that φ j φ 1 i : φ i U i U j ) φ j U i U j ) is biholomorphic. We call d = dim M the complex dimension of M. When d = 1, we call M a Riemann Surface. For example CP 1 := {x, y) C 2 {0}}/ where x, y) tx, ty) for t C. One has a homeomorphism CP 1 = S 2. Exercise 2.3. Show that CP 1 = S 2. Also CP 2 := {x, y, z) C 3 {0}}/ with x, y, z) tx, ty, tz) for t C. We use [x, y, z] to denote equivalence class of x, y, z) and call it homogeneous coordinate of point on CP 2. Exercise 2.4. Show that z 0 defines an open subset of CP 2 which is naturally homeomorphic to) C 2. Show that its complement is CP 1. CP 2 = C 2 CP 1.

3 AN INTRODUCTION TO ARITHMETIC AND RIEMANN SURFACE 3 It is encouraged to check that, the open set U z = z 0), U x = x 0), U y = y 0) for an open cover of CP 2 satisfying the complex manifold definition. Another example a projective complex curve is a Riemann surface. {[x, y, z] x 3 + y 3 = z 3 } CP 2 The Fermat curve is a Riemann surface as well. Given a, b, the curve intersects with the set z 0 is equal to {[x, y, z] x n + y n = z n } CP 2 E a,b := {[x, y, z] y 2 z = x 3 + axz 2 + bz 3 } CP 2 {x, y) C 2 y 2 = x 3 + ax + b.} Under suitable conditions for a, b, the set E a,b is a smooth compact Riemann surfaces. For general a, b it is still compact, but could be singular. Definition 2.5. Let X, Y be Riemann surfaces. A map g : X Y is called holomorphic if one = = can cover X by charts {φ i : U i W i C}, Y by {ψ j : V j Y j } such that after composing with the charts, there is gw i ) Y j and g : W i Y j is an analytic function. People are particularly interested in compact Riemann surfaces. Every compact Riemann surface is by definition a compact oriented two dimensional manifold and thus topologically classified by its genus g = 0, 1, 2, 3... The geometric meaning of genus is number of holes". And if you pick any triangulation of the Riemann surface, you always have Euler formula 2 2g = v e + f where v, e, f are numbers of vertices, edges, and trianglesfaces) in your triangulation Riemann-Hurwitz formula. Definition 2.6. Let F : X Y be a holomorphic map between compact Riemann surfaces, for any p X, composed with charts of X, Y containing p, F p) as origins one can write power series near z = 0, F z) = a k z k + a k+1 z k+1 +, with a k 0. This k 1 is called the multiplicity of F at p, denoted mult p F ). If k > 1 we call p a ramification point of F, and F p) a branch point of F. Definition 2.7. For same F above, the degree of F is the number of elements in F 1 q) for any q Y which is not a branch point. From basic topology we see the size of F 1 q), even when q is branch point, will be equal to the degree of F, as long as one counts the size using weight given by the multiplicity.

4 4 AN INTRODUCTION TO ARITHMETIC AND RIEMANN SURFACE Theorem 2.8. Let F : X Y be a nonconstant holomorphic map between compact Riemann surfaces. Then 2gX) 2 = degf )2gY ) 2) + p X[mult p F ) 1]. Proof. Since X is compact, there are only finitely many ramification point. So the sum above is finite. Take a triangulation of Y, such that each branch point of F is a vertex. Say there are v vertices, e edges, and t triangles. Take preimage of it gives a triangulation on X, say of v vertices, e edges, t triangles. Then every ramification point of F is a vertex on X. Since there are no ramification points over the interior of any triangle, each triangle lifts to degf ) triangles in X. Thus t = degf )t. Similarly e = degf )e. As for each vertex q Y we have F 1 q) = degf ) + 1 multp F ) ). Thus Therefore v = vertex q Y p F 1 q) degf ) + = degf )v vertex p X 2gX) 2 = v + e f p F 1 q) = degf )v + multp F ) 1 ). vertex p X + degf )e degf )f 1 multp F ) )) multp F ) 1 ) = degf )2gY ) 2) + p X multp F ) 1 ) Bezout formula. Suppose F x, y, z) is a homogenous polynomial of degree d 1, whose zero loci X = F = 0) CP 2 is a smooth compact Riemann surface complex algebraic curve). Suppose Gx, y, z) is another such object with degree d 2 and Y = G = 0) CP 2. Then Bezout formula says the number of intersection points of X Y, counted with multiplicities, is equal to d 1 d 2. It will be very easy to convince oneself in the case d 1 = 1, namely F = 0 is a line. Our intuition tells us Bezout formula follows from Gauss s theorem saying number of root of a polynomialsingle variable) equals its degree. And Yes!, the intuition is correct degree-genus formula. Let X = F = 0) CP 2 be a smooth projective curve, for some homogeneous polynomial F = F x, y, z). Then F is also a homogeneous polynomial. Lemma 2.9. Let π : X CP 1 be defined by π[x, y, z] = [x, z]. Then multp π) 1 ) p = F = 0) X. p X

5 AN INTRODUCTION TO ARITHMETIC AND RIEMANN SURFACE 5 Proof. Clearly p X is a ramification point of F if and only if dπt p X) = 0, and if and only if T p X is span by T pp 2 since this vector generates kernel of dπ at p). This is equivalent to say F = 0. And thus the two set in the claimed are equal. The argument carries to sets weighted by multiplicity as follows. We may assume p lies in the open set z 0 other cases are similar). Then X = {x, y C 2 fx, y) = F x, y, 1) = 0, over which πx, y) = x. Suppose p = x 0, y 0 ) is a point of ramification, then f p) = 0. Also as X is smooth at p, one must have f p) 0 implies y is local coordinatechart) of X near p. x Implicit Function theorem implies X is locally the graph of an analytic function x = gy). This implies fgy), y) = 0 for all y near y 0. Differentiate it gives 2.1) f = f x g y). f x Now π on this open set of X is exactly π, and thus mult p π = ord y0 gy). By 2.1) and that p) 0, this implies multpπ And we verified the wanted identity. 1 = ordy0g f y) = ordp. For any smooth projective plane curve, namely X = F x, y, z) = 0) CP 2, we define the degree of X is the degree of F. It is clear to be the number of intersections of X x = 0), counted with multiplicities, using Gauss s theorem. The following is our degree genus formula. Proposition 2.10 Plücker s Formula). A smooth projective plane curve of degree d has genus g = d 1)d 2) 2. Proof. Let X = F x, y, z) = 0) CP 2 as before. Consider π : X CP 1 defined by π[x, y, z] = [x, z]. Then π has degree d. And the weighted set people call it divisor) 2.2) multp 1 ) p p X describing ramification points with multiplicitied 1) equals 2.3) F = 0) X = F = 0) F = 0). The size" sum of all coefficients) of 2.2) is then equal to number of intersections points in 2.3). And this number should be product of degree of F and F namely d 1)d. By Riemann Hurwitz forumla applied to π 2gX) 2 = d2gcp 1 ) 2) + dd 1). Since gcp 1 ) = 0. We have 2g 2 = d 2 3d, and g = d 1)d 2). 2

6 6 AN INTRODUCTION TO ARITHMETIC AND RIEMANN SURFACE 3. MORDELL CONJECTURE AND FALTING S THEOREM Let C CP n be zero loci of polynomials. Suppose C is smooth of complex dimension one i.e. Riemann surface). Then the set of rational points on C may be determined as follows: 1) Case g = 0: no points or infinitely many; for example C = x 2 + y 2 = z 2 ). 2) Case g = 1: no points, or C is an elliptic curve i.e. closure of y 2 = x 3 + ax + b in C 2 CP 2 ) and its rational points form a finitely generated abelian group Mordell s Theorem, later generalized to the Mordel-Weil theorem). Moreover Mazur s torsion theorem restricts the structure of the torsion subgroup. 3) Case g > 1: according to the Mordell conjecture1922), now Faltings s Theorem1983), C has only a finite number of rational points. The following five problems are homework to the lectures. Exercise 3.1. For a positive prime integer n, consider Fermat equation x n + y n = z n. Can it have infinite many integer solutions for n 4? Why or why not? Exercise 3.2. Suppose X, Y are compact Riemann surface with gy ) = g, and f : X Y is a nonconstant holomorphic map. If f has no ramification point, and f has degree 2, what is the genus of X? Exercise 3.3. How many intersection points are there x 2 +y 2 +z 2 = 0) with x 3 +y 3 +z 3 = 0) in CP 2. Exercise 3.4. Show that z 0 defines an open subset of CP 2 which is naturally homeomorphic to) C 2. Show that its complement is CP 1. CP 2 = C 2 CP 1. Exercise 3.5. Using CP 1 = S 2 to show there is a group action of S 1 = {e iθ θ R} on S 3 unit ball in R 4 ), whose quotient is S AN GENERAL EXERCISE NOT HOMEWORK) Exercise 4.1. For positive integer n the complex projective space is where the action of C on C n+1 is give by CP n := {x 0, x 1,, x n ) C n+1 {0}}/C t x 0, x 1,, x n ) = tx 0, tx 1,, tx n ). We denote equivalence class of x 0,, x n ) by [x 0,, x n ] CP n. Show x i 0 defines an open subset U i of CP n isomorphic to C n. And, for example taking i = n, the complement of U n in CP n is CP n 1 namely CP n = U n CP n 1. Also show that CP n = S 2n+1 /S 1 for some action of group S 1 on S 2n+1.

7 AN INTRODUCTION TO ARITHMETIC AND RIEMANN SURFACE 7 REFERENCES Eu. Wiki: "Euclid" Py. Wiki: "Pythagorean triple" Zi. Rick Miranda, Algebraic Curves and Riemann Surfaces Faltings. Wiki: "Moredell Conjecture and Faltings theorem".