Elliptic Functions Michael Taylor Sections 30{34 of \Introduction to Complex Analysis" (And Appendices: x26 and xK) Contents 30. Periodic and doubly periodic functions { in¯nite series representations 31. The Weierstrass } in elliptic function theory 32. Theta functions and } 33. Elliptic integrals p 34. The Riemann surface of q(³) Appendices 26. The Riemann sphere and other Riemann surfaces K. Rapid evaluation of the Weierstrass }-function 1 2 Introduction We develop the basic theory of elliptic functions, starting in x30 with basic constructions of doubly periodic meromorphic functions on C as lattice sums. One basic case is the Weierstrass }-function. Its centrality in the subject is established in x31. Section 32 is devoted to a representation of } in terms of theta functions, which among other properties have the advantage of being de¯ned by rapidly convergent series. Section 33 expresses certain classes of integrals, known as elliptic integrals, as inverses of elliptic functions. To treat arbitrary elliptic integrals this way, one needs a result known as the solution to the Abel inversion problem. One convenient route top the solution to this problem involves the construction of the Riemann surface of q(³), when q(³) is a cubic (or quartic) polynomial, with no repeated roots. This is taken up in x34. Results here are precursors of a general theory of Riemann surfaces (treated in another set of notes of the author). These notes are excerpted from xx30{34 of our monograph \Introduction to Complex Analysis." They have an appendix, covering some elementary notions of a Riemann surface, taken from x26 of that monograph, and an appendix on the rapid evaluation of the Weierstrass }-function, taken from xK. 3 30. Periodic and doubly periodic functions - in¯nite series representations We can obtain periodic meromorphic functions by summing translates of z¡k. For example, X1 1 (30.1) f (z) = 1 (z ¡ n)2 n=¡1 is meromorphic on C, with poles in Z, and satis¯es f1(z + 1) = f1(z). In fact, we have X1 1 ¼2 (30.2) = : (z ¡ n)2 2 n=¡1 sin ¼z To see this, note that both sides have the same poles, and their di®erence g1(z) is seen to be an entire function, satisfying g1(z + 1) = g1(z). Also it is seen that, for z = x + iy, both sides of (30.2) tend to 0 as jyj ! 1. This forces g1 ´ 0. A second example is Xm 1 1 X³ 1 1 ´ f2(z) = lim = + + m!1 z ¡ n z z ¡ n n n=¡m n6=0 (30.3) 1 X1 2z = + : z z2 ¡ n2 n=1 This is also meromorphic on C, with poles in Z, and it is seen to satisfy f2(z + 1) = f2(z). We claim that 1 X³ 1 1 ´ (30.4) + + = ¼ cot ¼z: z z ¡ n n n6=0 In this case again we see that the di®erence g2(z) is entire. Furthermore, applying ¡d=dz to both sides of (30.4), we get the two sides of (30.2), so g2 is constant. Looking at the last term in (30.3), we see that the left side of (30.4) is odd in z; so is the right side; hence g2 = 0. As a third example, we consider Xm (¡1)n 1 X ³ 1 1 ´ lim = + (¡1)n + m!1 z ¡ n z z ¡ n n n=¡m n6=0 1 X1 2z (30.5) = + (¡1)n z z2 ¡ n2 n=1 1 X1 z(1 ¡ 2k) = ¡ 4 : z [z2 ¡ (2k ¡ 1)2][z2 ¡ (2k)2] k=1 4 We claim that 1 X ³ 1 1 ´ ¼ (30.6) + (¡1)n + = : z z ¡ n n sin ¼z n6=0 In this case we see that their di®erence g3(z) is entire and satis¯es g3(z+2) = g3(z). Also, for z = x + iy, both sides of (30.6) tend to 0 as jyj ! 1, so g3 ´ 0. We now use a similar device to construct doubly periodic meromorphic functions, following K. Weierstrass. These functions are also called elliptic functions. Further introductory material on this topic can be found in [Ahl] and [Hil]. Pick !1;!2 2 C, linearly independent over R, and form the lattice (30.7) ¤ = fj!1 + k!2 : j; k 2 Zg: In partial analogy with (30.4), we form the \Weierstrass }-function," 1 X ³ 1 1 ´ (30.8) }(z; ¤) = + ¡ : z2 (z ¡ !)2 !2 06=!2¤ Convergence on C n ¤ is a consequence of the estimate ¯ ¯ ¯ 1 1 ¯ jzj (30.9) ¯ ¡ ¯ · C ; for j!j ¸ 2jzj: (z ¡ !)2 !2 j!j3 To verify that (30.10) }(z + !; ¤) = }(z; ¤); 8 ! 2 ¤; it is convenient to di®erentiate both sides of (30.8), obtaining X 1 (30.11) }0(z; ¤) = ¡2 ; (z ¡ !)3 !2¤ which clearly satis¯es (30.12) }0(z + !; ¤) = }0(z; ¤); 8 ! 2 ¤: Hence (30.13) }(z + !; ¤) ¡ }(z; ¤) = c(!);! 2 ¤: Now (30.8) implies }(z; ¤) = }(¡z; ¤). Hence, taking z = ¡!=2 in (30.13) gives c(!) = 0 for all ! 2 ¤, and we have (30.10). Another analogy with (30.4) leads us to look at the function (not to be confused with the Riemann zeta function) 1 X ³ 1 1 z ´ (30.14) ³(z; ¤) = + + + : z z ¡ ! ! !2 06=!2¤ 5 We note that the sum here is obtained from the sum in (30.8) (up to sign) by integrating from 0 to z along any path that avoids the poles. This is enough to establish convergence of (30.14) in C n ¤, and we have (30.15) ³0(z; ¤) = ¡}(z; ¤): In view of (30.10), we hence have (30.16) ³(z + !; ¤) ¡ ³(z; ¤) = ®¤(!); 8! 2 ¤: In this case ®¤(!) 6= 0, but we can take a; b 2 C and form (30.17) ³a;b(z; ¤) = ³(z ¡ a; ¤) ¡ ³(z ¡ b; ¤); obtaining a meromorphic function with poles at (a + ¤) [ (b + ¤), all simple (if a ¡ b2 = ¤). Let us compare the doubly periodic function © constructed in (24.8){(24.11), which maps the rectangle with vertices at ¡1; 1; 1 + ip; ¡1 + ip conformally onto the upper half plane U, with ©(¡1) = ¡1; ©(0) = 0; ©(1) = 1. (Here p is a given positive number.) As seen there, (30.18) ©(z + !) = ©(z);! 2 ¤ = f4k + 2i`p : k; ` 2 Zg: Furthermore, this function has simple poles at (ip + ¤) [ (ip + 2 + ¤), and the residues at ip and at ip + 2 cancel. Thus there exist constants A and B such that (30.19) ©(z) = A³ip;ip+2(z; ¤) + B: The constants A and B can be evaluated by taking z = 0; 1, though the resulting formulas give A and B in terms of special values of ³(z; ¤) rather than in elementary terms. Exercises 1. Setting z = 1=2 in (30.2), show that X1 1 ¼2 = : n2 6 n=1 P ¡k Compare (13.79). Di®erentiate (30.2) repeatedly and obtain formulas for n¸1 n for even integers k. Hint. Denoting the right side of (30.2) by f(z), show that X1 f (`)(z) = (¡1)`(` + 1)! (z ¡ n)¡(`+2): n=¡1 Deduce that, for k ¸ 1, ³1´ X f (2k¡2) = (2k ¡ 1)!22k+1 n¡2k: 2 n¸1;odd 6 Meanwhile, use X1 X X1 n¡2k = n¡2k + 2¡2k n¡2k n=1 n¸1;odd n=1 P1 ¡2k (2k¡2) to get a formula for n=1 n , in terms of f (1=2). 1A. Set F (z) = (¼ cot ¼z) ¡ 1=z, and use (30.4) to compute F (`)(0). Show that, for jzj < 1, 1 X1 X1 ¼ cot ¼z = ¡ 2 ³(2k)z2k¡1; ³(2k) = n¡2k: z k=1 n=1 1B. Recall from Exercise 6 in x12 that, for jzj su±ciently small, 1 ez + 1 1 X1 B = + (¡1)k¡1 k z2k¡1; 2 ez ¡ 1 z (2k)! k=1 with Bk (called the Bernoulli numbers) rational numbers for each k. Note that e2¼iz + 1 1 = cot ¼z: e2¼iz ¡ 1 i Deduce from this and Exercise 1A that, for k ¸ 1, B 2³(2k) = (2¼)2k k : (2k)! Relate this to results of Exercise 1. 1C. For an alternative aproach to the results of Exercise 1B, show that G(z) = ¼ cot ¼z =) G0(z) = ¡¼2 ¡ G(z)2: Using 1 X1 G(z) = + a z2n¡1; z n n=1 0 2 compute the Laurent series expansions of G (z) and G(z) and deduce that a1 = ¡¼2=3, while, for n ¸ 2, 1 nX¡1 a = ¡ a a : n 2n + 1 n¡` ` `=1 In concert with Exercise 1A, show that ³(2) = ¼2=6; ³(4) = ¼4=90, and also compute ³(6) and ³(8). 2. Set Y1 ³ z2 ´ F (z) = ¼z 1 ¡ : n2 n=1 7 Show that F 0(z) 1 X1 2z = + : F (z) z z2 ¡ n2 n=1 Using this and (30.3){(30.4), deduce that F (z) = sin ¼z; obtaining another proof of (18.21). Hint. Show that if F and G are meromorphic and F 0=F ´ G0=G, then F = cG for some constant c. To ¯nd c in this case, note that F 0(0) = ¼. 3. Show that if ¤ is a lattice of the form (30.7) then a meromorphic function satisfying (30.20) f(z + !) = f(z); 8 ! 2 ¤ yields a meromorphic function on the torus T¤, de¯ned by (26.14). Show that if such f has no poles then it must be constant. We say a parallelogram P ½ C is a period parallelogram for a lattice ¤ (of the form (30.7)) provided it has vertices of the form p; p + !1; p + !2; p + !1 + !2. Given a meromorphic function f satisfying (30.20), pick a period parallelogram P whose boundary is disjoint from the set of poles of f.