A Few Elementary Facts about Elliptic Curves

A FEW ELEMENTARY FACTS ABOUT ELLIPTIC CURVES 1. Introduction In our paper we shall present a number of facts regarding the doubly periodic meromorphic functions, also known as elliptic f unctions. We shall focus on the elliptic functions of order two and, in particular, on the Weierstass P-function. The doubly periodic meromorphic functions can be looked at as meromorphic functions defined on complex tori. This is because, as a topological space, a complex torus is the quotient of the complex plane over an integer lattice. 2. Periodic functions Definition 1. A meromorphic function f is said to be periodic if and only if there exists a nonzero ω ∈ C such that f (z + ω) = f (z), for all z ∈ C. The complex number ω is called period. As a first observation, we may say that if ω is a period, then any integer multiple nω is also a period. Also, if there exist two periods ω 1 and ω2 , then n1 ω1 + n2 ω2 is also a period, for all n1 , n2 ∈ Z. Given a meromorphic function f , define M to be the set of all its periods (including 0). From the above observations, we deduce that M is a Z-module. If f is a non-constant meromorphic function, the module M containing all its periods cannot have a accumulation point, since otherwise f would be a constant. Therefore each point in M is isolated. In order words, M is a discrete module. We have the following theorem regarding the module M: Theorem 1. If M is the module of periods of a meromorphic function f , it must have one of the following forms: • M = {0}. • M = {nω|n ∈ Z}, for some nonzero complex value ω. • M = {n1 ω1 + n2 ω2 |n1 , n2 ∈ Z}, for some nonnzero complex values ω1 , ω2 ∈ C, whose ratio is not real. Proof. Let us suppose that M has nonzero elements. Then take a nonzero element ω1 of smallest absolute value. This is always possible, since in any disk or radius ≤ r, there are only finitely many elements of M . Define A = {nω1 |n ∈ Z}. We have A ⊂ M . If M 6= A, choose ω2 the smallest element (in terms of absolute value) in M −A. First, note that ω 1 /ω2 cannot be real, otherwise, choose an integer m such that m ≤ ω 1 /ω2 < m + 1. It follows that |ω1 − mω2 | < |ω1 | which contradicts the minimality of ω 1 . 1 Finally, let us prove that M = {n1 ω1 + n2 ω2 |n1 , n2 ∈ Z}. We remark that since ω1 /ω2 is not real, any complex number can be written uniquely in the form tω1 +sω2 , where s and t are real numbers. In order to see this clearly, it is enough to look at ω1 and ω2 as vectors in the two-dimensional real vector space. Since ω1 and ω2 are independent as vectors, it becomes obvious why any complex number can be written uniquely as a linear combination of ω 1 and ω2 with real coefficients. Now, take an arbitrary element x of M and write it in the form sω1 + tω2 , where s and t are real numbers. Choose integers n1 and n2 such that |s − n1 | < 1/2 and |t − n2 | < 1/2. It follows easily that |x−n1 ω1 −n2 ω2 | < 1/2|ω1 |+1/2|ω2 | ≤ |ω2 | (the first inequality is strict, since ω1 /ω2 is nonreal). Because of the way ω2 was chosen, it follows that x = nω1 or x = n1 ω1 +n2 ω2 . Hence, M = {n1 ω1 +n2 ω2 |n1 , n2 ∈ Z}.  3. Elliptic Functions and Unimodular Forms Definition 2. We shall call a meromorphic function f elliptic iff its module of periods M is a linear linear combination of two periods ω 1 and ω2 , such that ω1 /ω2 is nonreal (i.e. the third case of the previous theorem). The pair (ω1 , ω2 ) mentioned above is a basis for the module M . In this section we shall discuss about the possible bases of a module of periods M . Suppose (ω10 , ω20 ) is another basis of M . Then ω10 = m1 ω1 + n1 ω2 , ω20 = m2 ω1 + n2 ω2 and ω1 = m01 ω1 + n01 ω2 , ω2 = m02 ω1 + n02 ω2 Using matrices, we can write      0 ω1 ω 1 m1 n1 ω1 ω 0 1 = m2 n2 ω2 ω 2 ω20 ω 0 2  ω1 ω 1 ω2 ω 2  =  m01 n01 m02 n02  ω10 ω 0 1 ω20 ω 0 2  Consequently, we have  ω1 ω 1 ω2 ω 2  =  m01 n01 m02 n02  m1 n1 m2 n2  ω1 ω 1 ω2 ω 2  We have ω1 ω2 −ω2 ω1 6= 0, because otherwise ω1 /ω2 is real, which contradicts our assumption. It follows that  m01 n01 m02 n02  m1 n1 m2 n2 2  =  1 0 0 1  and, since all the entries are integral, m01 n01 m02 n02 = m1 n1 m2 n2 = ±1 Therefore, from (ω1 , ω2 ) one can obtain (ω10 , ω20 ) via a linear transformation of determinant 1, which is usually called unimodular transf ormation. We have thus seen that any two bases of the same module M are related to one another by a unimodular transformation. From all the possible bases of a module, one can choose a particular one, with certain characteristics which will be called cannonical basis. This fact is the object of the following Theorem 2. Given a module M , there exists a basis (ω 1 , ω2 ) such that the ratio σ = ω1 /ω2 has the following properties: • Im σ > 0 • -1/2 ≤ Re σ ≤ 1/2 • |σ| ≥ 1 • If |σ| = 1, then Re σ ≥ 0 Also, σ defined above is uniquely determined by these conditions, up to a choice of two, four or six corresponding bases. 4. General Properties of Elliptic Functions We shall use a convenient notation: z 1 ≡ z2 iff z1 − z2 belongs to M (in other words, iff z1 − z2 = n1 ω1 + n2 ω2 , for two integers n1 and n2 ). Let f be a elliptic function with (ω1 , ω2 ) as basis of the module of periods. Since f is doubly-periodic, it is entirely determined by its values on a parallelogram Pa whose vertices are a, a + ω1 , a + ω2 and a + ω1 + ω2 . The complex value a can be chosen arbitrarily. Theorem 3. If the elliptic funtion f has no poles, it is a constant. Proof. If f has no poles, it is bounded in a parallelogram P a . Since f is doubly periodic, it is bounded on the whole complex plane. By Liouville’s theorem, f must be a constant function.  As we have seen before, the set of poles of f has no accumulation point. It follows that in any parallelogram P a there are finitely many poles. When we shall refer to the poles of f , we shall mean the set of mutually incongruent poles. Theorem 4. The sum of residues of an elliptic function f is zero. Proof. Choose a ∈ C such that the parallelogram P a does not contain any pole of f . Consider the boundary ∂Pa of Pa traced in the positive sense. Then the integral Z 1 f (z) 2πi ∂Pa 3 equals the sum of residues of f. But the sum equals 0, since the integrals over the opposite sides of the parallelogram cancel each other.  A simple corollary 1 of this theorem is that an elliptic function cannot have a single simple pole, otherwise, the sum of residues would not equal 0. Theorem 5. A nonconstant elliptic function f has the same number of poles as it has zeroes. (Every pole or zero is counted according to its multiplicity) Proof. We may consider the function f 0 /f which has simple poles wherever f has a pole or a zero. The residue of a pole α of f 0 /f equals its multiplicity in f if α is a zero of f , and minus its multiplicity in f , if α is a pole of f . Applying now theorem 4 to the function f 0 /f , we get the desired result.  Since f (z) − c and f (z) have the same number of poles, we conclude that they must have the same number of zeroes. Definition 3. Given an elliptic function f , the number of mutually incongruent roots of f (z) = c is called the order of the elliptic function. Obviously, the order does not depend on the choice of c. Theorem 6. Suppose the nonconstant elliptic function f has the zeroes a1 , · · · , an and poles b1 , · · · , bn (multiple roots and poles appear multiple times). Then a1 + · · · + an ≡ b1 + · · · + bn (mod M ). Proof. Consider all ai and bi in the parallelogram Pa for some a. We consider the following integral on ∂Pa : Z zf 0 (z) 1 dz 2πi ∂Pa f (z) Given the properties of the poles and zeroes of f 0 /f mentioned above, we deduce that f has a zero (or a pole) at t, of order k ∈ Z >0 , iff zf 0 /f has a simple pole at t with residue nt (or −nt). It follows that the integral equals a1 + · · · + an − b1 − · · · − bn . Now, we must prove that the integral is in M . For this purpose, we write the integral on the pairs of two opposite sides: 1 2πi Z a a+ω1 zf 0 (z) dz − f (z) Z a+ω1 +ω2 a+ω2 zf 0 (z) dz f (z)  −ω2 = 2πi Z a a+ω1 f 0 (z) dz f (z) Also, −ω2 2πi Z a+ω1 a f 0 (z) −ω2 dz = f (z) 2πi Z ∂D 1 dw w R 1 1 where D is the curve given by f (z) when z varies from a to a+ω 1 . 2πi ∂D w dw is an integer (=the winding number with respect to 0). Taking into account both pairs of opposite sides of Pa , we get the desired result.  1Latin corona ”garland” > diminutive corolla > corollarium ”gratuity” > English 4 5. The Weierstass P-function The simplest example of an elliptic function is of order 2. As we have seen before, there is no elliptic function of order 1. An elliptic function of degree 2 can have either one pole of degree 2 or two distinct simple poles. We shall analyse, following Weierstass, the case of an elliptic function with a double pole. We may place the pole at the origin and consider the coefficient of z −2 as being 1 (translations and multiplications by constants do not change essential properties of elliptic functions). If we consider f(z)-f(-z) we get a elliptic function that has no singular part. (The function f (z) − f (−z) could only have a single simple pole at 0, but this is impossible). Hence, f (z) − f (−z) is constant and setting z = ω 1 /2, we get that this constant is zero. Therefore, f (z) = f (−z) and we can write P(z) = z −2 + a0 + a1 z 2 + ... We may suppose that a0 = 0, because adding/substracting a constant from f is irrelevant. What we get is the so called W eierstass P-function. This elliptic function can be written as (1) P(z) = z −2 + a1 z 2 + a2 z 4 + ... The existence of a elliptic function of order 2 has not yet been proven. We shall prove that the Weierstrass P-function is uniquely determined for a basis (ω1 , ω2 ), being given by the formula:  X 1 1 1 P(z) = 2 + − (2) z (z − ω)2 ω 2 ω6=0 where the summation is over all ω = n1 ω1 + n2 ω2 , ω 6= 0. We shall prove first that this sum is convergent. For every |ω| > 2|z|, we have 1 z(2ω − z) 10|z| 1 ≤ − 2 = 2 2 2 (z − ω) ω ω (z − ω) |ω|3 This is because |2ω − z| < 5/2|ω| and |z − ω| ≥ |ω| − |z| ≥ |ω|/2. We conclude that in order for the sum (2) to converge, it is enough to prove that the sum X 1 <∞ |ω|3 ω6=0 Also, since ω1 /ω2 is not real |n1 ω1 + n2 ω2 | = |ω2 ||n1 ω1 /ω2 + n2 | ≥ c|n1 | for some positive real constant c (for any integers n 1 , n2 ). Similarly, we find a constant d such that |n1 ω1 + n2 ω2 | ≥ d|n2 |, for any integers n1 , n2 . In conclusion, |n1 ω1 + n2 ω2 | ≥ k(|n1 | + |n2 |), where k = cd/(c + d) ≥ 0. Since there are exactly 4n ordered pairs (n 1 , n2 ) of integers such that |n1 | + |n2 | = 5 n, we have ∞ X 1 1 X 1 < . |ω|3 4k 3 n2 1 ω6=0  P 1 We also need to prove that P(z) = z12 + ω6=0 (z−ω) 2 − ω1 and ω2 . We denote, for this purpose  X 1 1 1 f (z) = 2 + (3) − z (z − ω)2 ω 2 1 ω2  has periods ω6=0 Since the series is absolutely convergent, we may differentiate term by term: X X 2 2 1 f 0 (z) = − 3 − = −2 3 z (z − ω) (z − ω)3 ω ω6=0 This series is also absolutely convergent, since if |ω| > 2|z|, we have X 1 X 1 ≤ 16 (z − ω)3 |ω|3 ω ω and we deduce that (3) converges absolutely. Consequently, f (z + ω 1 ) − f (z) and f (z + ω2 ) − f (z) are constant functions. By definition, f is even and, therefore, setting z = −ω1 /2 and z = −ω2 /2 we get that the constants are 0. Therefore, f (z + ω1 ) = f (z) and f (z + ω2 ) = f (z), for all z ∈ C. Hence, we proved that f is elliptic. If P has the form in (1) and is doubly periodic, with periods ω1 and ω2 , it follows that P − f has no singular part, and is therefore constant. Since both P and f have no constant term (i.e. the coefficient of z 0 is 0), it follows that P = f . Hence, we deduce that  X 1 1 1 P(z) = 2 + − z (z − ω)2 ω 2 ω6=0 and also that P 0 (z) = −2 X ω 1 (z − ω)3 6. The function ζ(z) Because P has no residues, it is the derivative of a function -ζ(z), where   1 X 1 z 1 ζ(z) = + + + z z − ω ω ω2 ω6=0 In order to prove that this series converges, we note that 1 1 z z2 + + 2 = 2 z−ω ω ω ω (z − ω) 6 Hence, if |ω| > 2|z|, X 1 ω6=0 z−ω + X 2 X z 1 z 2 ≤ |z| + 2 = ω ω ω 2 (z − ω) |ω|3 and, consequently, the series converges absolutely. Since f (z + ωi ) = f (z), for i = 1, 2 it follows that ζ(z + ω i ) = ζ(z) + ηi , for i = 1, 2, where ηi are complex constants. These complex constants have a beautiful property which comes from the fact that Z 1 (4) ζ(z)dz = 1 2πi ∂Pa We prove first (4). If we look at (2), we deduce that a1 a2 ζ(z) = z −1 − z 3 − z 5 + · · · 3 5 Now, if we evaluate the integral on two opposite sides of the parallelogram Pa , we get Z a+ω1  Z a+ω1 +ω2 1 −η1 ω2 ζ(z)dz − ζ(z)dz = 2πi 2πi a a+ω2 Integrating similarly on the other pair of opposide sides, we get that −η2 ω1 −η1 ω2 − =1 2πi 2πi and, finally, η1 ω2 − η2 ω1 = 2πi which is called Legendre0 s relation. 7. The Differential equation corresponding to P Using the definition of ζ(z), we have the following identity 1 1 z z2 z3 + + 2 = − 3 − 4 − ··· z−ω w ω ω ω Consequently, we can write ∞ 1 X Gk z 2k−1 ζ(z) = + z k=2 where by Gk we mean (5) Gk = X 1 ω 2k ω6=0 We note that ζ(z) has no terms of the form z 2n since P is an even function (and ζ is odd). Differentiating (5), we get ∞ X 1 P= 2+ (2k − 1)Gk z 2k−2 z k=2 7 Writing only the significant parts of the following functions, we have: 1 P(z) = 2 + 3G2 z 2 + 5G3 z 4 + · · · z P 0 (z) = − 2 + 6G2 z + 20G3 z 3 + · · · z3 P 0 (z)2 = 24G2 4 − 2 − 80G3 + · · · z6 z 4P(z)3 = 36G2 4 + 2 + 60G3 + · · · z6 z 60G2 P(z) = 60G2 +0 + ··· z2 It follows that P 0 (z)2 − 4P(z)3 + 60G2 P(z) = −140G3 + · · · The left side of the last identity is an elliptic function and it does not have any singular part. Therefore, it must equal a constant. Setting z = 0, we get that the constant is −140G3 . Hence P 0 (z)2 − 4P(z)3 + 60G2 P(z) + 140G3 = 0 If we set g2 = 60G2 and g3 = 140G3 , we get P 0 (z)2 = 4P(z)3 − g2 P(z) − g3 Provided that e1 , e2 and e3 are the complex roots of the polynomial 4y 3 − g2 y − g3 , we can write P 0 (z)2 = 4(P(z) − e1 )(P(z) − e2 )(P(z) − e3 ) Because P is even and periodic, we have P(ω 1 − z) = P(z). It follows that P 0 (ω1 − z) = P 0 (z). Hence P 0 (ω1 /2) = 0. Similarly, P 0 (ω2 /2) = 0. We have also, P 0 (ω1 + ω2 − z) = P 0 (z), and therefore, P 0 ((ω1 + ω2 )/2) = 0. Since P 0 has only one pole (of order 3), it must have exactly three roots (counting multiplicity). On the other hand, the complex numbers ω 1 /2, ω2 /2 and (ω1 + ω2 )/2 are mutually incongruent. Therefore, they are the three distinct roots of P 0 . For every ei there exists di such that P(di ) = ei (this equation in fact has a double solutions, as we shall see below). Also, d i are roots of P 0 . Therefore, if we apply P on the set of roots of P 0 (a set which has three elements, ω1 /2, ω2 /2 and (ω1 + ω2 )/2), we get the whole set {e1 , e2 , e3 }. Consequently, we may set P(ω1 /2) = e1 and P(ω2 /2) = e2 and P((ω1 + ω2 )/2) = e3 . Every root of P − ei is also a root of P 0 . If, for instance, e1 = e2 , then the root d1 of P − e1 has multiplicity at least 2 in P 0 and, therefore, multiplicity at least 3 in P − e1 , which is impossible for an elliptic curve of order 2. In 8 conclusion, we derive an important observation, namely that all roots e i are distinct. References [1] L.Ahlfors, Complex Analysis, (1979), 263-278 9