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Trigonometric Functions, Elliptic Functions, Elliptic Modular Forms 1 (September 24, 2013) Trigonometric functions, elliptic functions, elliptic modular forms Paul Garrett [email protected] http:=/www.math.umn.edu/egarrett/ [This document is http://www.math.umn.edu/~garrett/m/mfms/notes 2013-14/03 elliptic.pdf] 1. Reconsideration of sin x 2. Construction of singly-periodic functions 3. Arc length of ellipses: elliptic integrals, elliptic functions 4. Construction of doubly-periodic functions 5. Complex tori, elliptic curves, the modular curve 6. Elliptic modular forms 1. Reconsideration of sin x Although we already know quite a bit about trigonometric functions and their role in calculus, their treatment can be redone to emphasize parallels for elliptic integrals and elliptic functions. [1.1] Integral defining arcsin x The length of arc of a piece of a circle is x2 + y2 = 1 is Z x Z x r p d p 2 arc length of circle fragment from 0 to x = 1 + y02 dt = 1 + 1 − t2 dt 0 0 dt x s 1 x r x Z · (−2t)2 Z t2 Z dt = 1 + 2p dt = 1 + dt = p = arcsin x 2 2 2 0 1 − t 0 1 − t 0 1 − t [1.2] Periodicity The periodicity of sin x comes from the multi-valuedness of arcsin x. The multi-valuedness is ascertainable from this integral, since the path from 0 to x can meander through the complex plane, going around the two points ±1 special for the integral. The basic unit of this multi-valuedness is Z dζ = ±2π (counter-clockwise circular path γ enclosing both ±1) p 2 γ 1 − ζ That is, a path integral from 0 to x could go from 0 to x along the real axis, but then add to the path a vertical line segment from x out to a circle of radius 2, traverse the circle an arbitrary number of times, come back along the same segment (thus cancelling the contribution from this segment). 2 − 1 There are two single-valued choices for (1 − z ) 2 on any region complementary to an arc connecting ±1. For example, it is easy to see a Laurent expansion in jzj > 1: − 1 1 3 2 − 1 ±i 1 2 ±i 1 1 (− 2 )(− 2 ) 1 2 i 1=2 3=8 (1 − z ) 2 = · 1 − = · 1 − (− ) + + ::: = ± + + + ::: z z2 z 2 z2 2! z2 z z3 z5 R dz Let γ be a path traversing counterclockwise a circle of radius more than 1 centered at 0. Integrals γ zN are 0 except for N = 1, in which case the integral is 2πi. Thus, Z dz p = ±2π 2 γ 1 − z [1.3] Polynomial relation between sin x and sin0 x Rewriting the integral as Z sin x dt p = x 2 0 1 − t 1 Paul Garrett: Trigonometric functions, elliptic functions, elliptic modular forms (September 24, 2013) and taking a derivative, by the fundamental theorem of calculus, 0 1 sin x · p = 1 1 − sin2 Thus, (sin0 x)2 = 1 − (sin x)2 and the algebraic relation between sin x and its derivative (we know it is cos x) arises: (sin x)2 + (sin0 x)2 = 1 2. Construction of singly-periodic functions A singly-periodic function f on C is a (probably holomorphic or meromorphic) function such that for some ! 6= 0 f(z + !) = f(z) (for all z 2 C) or at least for z away from poles of f. Iterating the condition, for any integer n f(z + n!) = f(z) In other words, f is invariant under translation by the group Z · ! inside C. Feigning ignorance of the trigonometric (and exponential) function, whether as inverse functions to integrals or not, as a warm-up to the construction of doubly-periodic functions we should try to construct some singly-periodic functions. [2.1] Construction and comparison to sin z For simplicity, take ! = 1, and make holomorphic or meromorphic functions f such that f(z + 1) = f(z) (for all z 2 C) That is, we want Z-periodic functions on C, hypothetically closely related to sin 2πz. A fundamental approach to manufacturing such things is averaging, also called periodicization or automorphizing, as follows. For given function ', consider X f(z) = '(z + n) n2Z If this converges nicely it is certainly periodic with period 1, since X X f(z + 1) = '(z + 1 + n) = '(z + n) n2Z n2Z by replacing n by n − 1, using the fact that we have summed over the group Z, justifying rearrangement by absolute convergence. An elementary function making the sum converge, apart from poles, is '(z) = 1=z2, so put X 1 f(z) = (z + n)2 n2Z If we are lucky, this manufactures a function related to the sine function. Indeed, f(z) has double poles at the zeros of sin πz, so a plausible preliminary guess is that f(z) is 1=(sin πz)2. 2 Paul Garrett: Trigonometric functions, elliptic functions, elliptic modular forms (September 24, 2013) To prove equality, perhaps after adjusting details, match poles, subtract, check that the difference goes to 0 as the imaginary part of z at infinity goes to 1, and invoke Liouville. [1] To do this, first determine the Laurent expansion of f(z) near its poles. By periodicity, we'll understand all the poles if we understand the pole at z = 0. This is 1 X 1 1 f(z) = + = + (holomorphic near z = 0) z2 (z + n)2 z2 n6=0 To understand the nature of each pole of 1= sin2 πz, by periodicity it suffices to look near z = 0. Since sin πz = πz + (πz)3=3! + ::: = πz · 1 + (πz)2=3! + ::: the inverse square is [2] 1 1 1 1/π2 = = · 1 − (πz)2=3! + ::: = + holomorphic at 0 sin2 πz πz · 1 + (πz)2=3! + ::: (πz)2 z2 Correcting by π2, the poles of f(z) and π2= sin2 πz cancel: π2 X 1 π2 f(z) − = − = entire 2 (z + n)2 2 sin πz n sin πz As Im(z) becomes large, f(z) goes to zero. For apparently different reasons, also π2 2πi 2 = −! 0 (as jIm(z)j ! 1) sin2 πz eπiz − e−2πiz Thus, by Liouville, X 1 π2 = (z + n)2 sin2 πz n2Z [2.2] Differential equations for singly-periodic functions The construction produced a singly-periodic function P 1=(z + n)2 identifiable in terms of already-familiar items. The analogous discussion for doubly- periodic functions does not produce familiar objects. For practice, we now take another viewpoint that will also succeed with constructed doubly-periodic functions. [1] Liouville's theorem asserts that a bounded entire function is constant. As an immediate corollary, an entire function which is bounded and goes to 0 as the imaginary part of z goes to infinity is 0. Liouville's theorem is a striking instance of rigidity, where to prove two things equal, we need not prove something with infinite precision, but only demonstrate sufficient closeness to be able to infer equality. A trivial case of rigidity is that two integers are equal if they are within distance 1. [2] 2 The first two terms of the multiplicative inverse of a convergent power series 1 + c1z + c2z + ::: are easily 1 2 determined, using 1−r = 1 + r + r + :::: 1 1 2 = 2 1 + c1z + c2z + ::: 1 − (−c1z − c2z − :::) 2 2 2 = 1 + (−c1z − c2z − :::) + (−c1z − c2z − :::) + ::: = 1 − c1z + (higher-order) That is, with leading term 1, the coefficient of z changes simply by flipping sign. 3 Paul Garrett: Trigonometric functions, elliptic functions, elliptic modular forms (September 24, 2013) The aim is to determine a differential equation satisfied by f(z) = P 1=(z + n)2. In terms of Liouville, both f and its derivative X 1 f 0(z) = −2 (z + n)3 go to 0 as Im(z) ! ±∞, so if a polynomial P (f; f 0) in f and f 0 cancels the poles, then P (f; f 0) is necessarily constant, giving a polynomial relation [3] between f 0 and f. By periodicity, it suffices to consider the poles at z = 0, as before. Noting that f(−z) = f(z) assures vanishing of odd-order terms, let 1 −2 f(z) = + a + bz2 + O(z4) so f 0(z) = + 2bz + O(z3) z2 z3 To cancel poles by a polynomial P (f; f 0), the first step is to cancel the worst pole by f 3 −(f 0=−2)2: compute (carefully!?) that 1 2a 1 3a 3a2 + 3b f(z)2 = + + O(1) and f(z)3 = + + + O(1) z4 z2 z6 z4 z2 and f 0(z)2 1 2 1 2b = − bz + O(z3) = − + O(1) −2 z3 z6 z2 Thus, f 0(z)2 3a 3a2 + 5b − f(z)3 = − − + O(1) −2 z4 z2 The 1=z4 term can be eliminated by adjusting by a multiple of f(z)2: f 0(z)2 −3a2 − 5b + 6a2 3a2 − 5b − f(z)3 + 3a · f(z)2 = + O(1) = + O(1) −2 z2 z2 Finally, subtract a multiple of f(z) to eliminate the 1=z2 term: f 0(z)2 − f(z)3 + 3a · f(z)2 − (3a2 − 5b) · f(z) = O(1) −2 In fact, since both f and f 0 go to zero as Im(z) ! ±∞, the O(1) term must be 0. Rearrangement produces a relation anticipating Weierstraß’ analogous relation for constructed doubly-periodic functions: 02 3 2 2 X 1 1 f = 4f − 12af + (12a − 20b)f (with f(z) = = + a + bz2 + :::) (z + n)2 z2 Again, for other reasons, we know f(z) = π2= sin2 πz. [2.2.1] Remark: The existence of the relation made no use of higher Laurent coefficients, and at the same time explicitly demonstrates the dependence of the algebraic relation on the coefficients a; b in 1 2 2 4 f(z) = z2 + a + bz + :::.
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