%% alf@mpce.mq.edu.au %% Copyright 1993 %% Alf van der Poorten %% ceNTRe for Number Theory Research %% Macquarie University %% NSW 2109 AUSTRALIA %%If you don't have a Mac with Textures comment out the %%blatantly false claim that \MacTexturestrue below. %%Now DVIPS will produce the pictures, if you have the file %%logo.art in your directory, that is. %%If you don't have either TeXtures or DVIPS, despair. %%Tell TeX that it can do without the file epsf %%when it asks for it, %%and remove the comment just above \Part, %%so as to undefine the picture. %%Also deny \MacTexturestrue of course. \newif\ifMacTextures \MacTexturestrue %%Remember, you've got to have logo.art \magnification=\magstep1 %\input amstex.tex %Make sure that it is v2.1 %Update from e-math, if not. %\documentstyle{amsppt} \vsize 9.4truein \input AlfPreamble %%I do hope you downloaded this vital file %\def\centrelogo{\relax} \def\smallcentrelogo{\relax} \Part=VII \input FermatTopmatter %%and this one \def\cosec{\roman{cosec}} \rightline{\font\dict=cmssbx10 at 9 pt{\dict ellipsis\ } \it the omission from a sentence of a word} \rightline{\it or words which would complete or clarify the construction.}\bigskip Today's story is a child's introduction to elliptic functions. Since I'll be covering a few years' coursework in a few pages --- my remarks will be elliptical --- we had best fasten our seatbelts. When there is talk of periodic functions one thinks of $\sin 2\pi z$ and $\cos 2\pi z$; I've put in the $2\pi$ as a normalisation so that these functions have {\it primitive\/} period $1$, rather than some random $\omega$, or whatever. The adjective `primitive' is there to acknowledge that they actually have periods $0$, $\pm1$, $\pm2$, $\dots$, but fundamentally the period is $1$ as said. Of course we say that $f$ has period $\omega$ if $f(z+\omega)=f(z)$ for all $z$. It is not a very sophisticated remark to observe that the {\it circular functions\/} are periodic because $e^{2i\pi }$ is periodic; after all, they are just its imaginary and real part respectively. But it is not quite trivial to add that in fact {\it every\/} periodic function is periodic because it is itself a function of $e^{2i\pi z}$. This manifests itself in practice by reasonable periodic functions having a Fourier \ex\ $$\sum_{-\infty}^{\infty}c_ne^{2ni\pi z}=c_0+ \sum_{n=1}^{\infty}\((c_n+c_{-n})\cos 2n\pi z+ i(c_n- c_{-n})\sin 2n\pi z\).$$ Mind you, it is worthwhile to pause to ask just how we might have known in the first place that $e^{2i\pi z}$ is periodic. Surely this is not obvious from its power series definition. Let me suggest two different `explanations'. In the first we treat sin as original by computing the length of an arc of the circle $x^2+y^2=1$ between the ordinates $x=0$ and $x=\sin z$, and define sin by $$z=\int^{\sin z}_0\frac{dx}{\sqrt{1-x^2}}\,.$$ Then the circularity of the circle entails the periodicity of sin. I prefer the following illustration. Here we admit that $\sin \pi z$ has simple zeros exactly at $0$, $\pm1$, $\pm2$, $\dots$, and --- rather wildly thinking of it as just a \poly\ of infinite degree --- we factorise it and write $$\sin\pi z=\pi z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)\,.$$ Of course that multiplier $\pi$ (which, after all, might have been any decent function that never vanishes) needs rather calmer justification\footnote"$^*$"{The trick is to notice that De Moivre's theorem, and then replacing $\cos^2\theta$ by $1-\sin^2\theta$, allows us to write $$\sin(2m+1)\theta=(-1)^m\sin\theta P_{2m}(\sin\theta)\,,$$ with $P_{2m}$ a monic polynomial of degree $2m$ and constant term $(-1)^m(2m+1)$. Of course its $2m$ zeros are $\pm\sin \pi k/(2m+1)$ for $k=1$, $2$, $\dots$, $m$ so we have $$\sin(2m+1)\theta=(2m+1)\sin\theta\prod_{k=1}^m \(1-\frac{\sin^2\theta}{\sin^2 \pi k/(2m+1)}\)\,.$$ Now there is little more to do than to set $\pi z=(2m+1)\theta$ and to let $m$ go to $\infty$.}. With this evil deed done, we acknowledge that we are frightened of products, so we take the logarithm; and being bothered by logarithms, we differentiate. That yields $$\pi\cot\pi z=\frac1z-\sum_{n=1}^{\infty} \left(\frac1{n-z}-\frac1{n+z}\right)\,.$$ Unfortunately, as we catch our breath, we see that this is a mildly nasty partial fraction expansion\footnote"$^\dag$"{But now I can tell about Euler's evaluation of $\zeta(2k)$. We obtain $$i\pi z\cot i\pi z=\pi z +\frac{2\pi z}{e^{2\pi z}-1}=\pi z + \sum_{n=0}^\infty \frac{B_n}{n!}(2\pi z)^n= 1+2\sum_{k=0}^\infty(-1)^k\sum_{n=0}^\infty \frac1{n^{2(k+1)}}z^{2(k+1)}\,,$$ and comparing coefficients of $z^{2k}$ we have the claim ending the Notes III.} in that it only converges conditionally --- that is, on condition that we don't muck about with those parentheses. So we differentiate again and contemplate $$\pi^2\cosec^2\pi z=\sum_{-\infty}^{\infty}\frac1{(n-z)^2}\,,$$ and see that it shouts its periodicity. If we now backtrack, carefully, we are done. Of course I've told this story to motivate its generalisation. I'd better also announce a principle. Loosely speaking, a function is `good' if it is a convergent sum of good functions. That's why the sums above are good if we stay away from their {\it poles\/} --- the points at which a term blows up, the integers $\Z$ in our examples. All this is given that `good' --- {\it analytic\/} --- sort of means `differentiable'; or better said, that the function may be expanded as a convergent power series. At the poles our functions --- examples of {\it meromorphic\/} functions --- are quite bad, but not very bad. They just take the value $\infty$; in other words, their reciprocal is $0$. That's not too bad. Incidentally, after taking a logarithm, infinite products are just infinite sums. I'll try only to deal with functions that are good everywhere, except possibly for the odd pole; such pretty good functions are called {\it meromorphic\/} functions. These functions may well be very bad at $z=\infty$, that's permitted. A function without any singularity in the finite complex plane is called an entire function. I will use the fact that the constants comprise the only bounded entire functions. If a good function $f$, not just a constant, has essentially different periods $1$ and $\tau$, then $\tau$ must be a dinkum\footnote"$^*$"{{\it Macquarie Dictionary\/}: \font\dict=cmssbx10 at 7pt {\dict dinkum\ } true, honest, genuine: as in {\it dinkum Aussie\/}.} complex number. The point is that $\tau$ cannot be rational, either because then $1$ is not a primitive period or, anyhow, then the periods are not essentially different; and if $\tau$ were real irrational then, because there are now $\Z$-linear combinations of $1$ and $\tau$ that are arbitrarily small, $f$ would have to be just a miserable constant. Thus $\tau$ must be $\R$-linearly independent of $1$ and we may choose it in the upper half plane $\Cal H$, that is, with positive imaginary part. Just as $\Z$ is all the periods of $e^{2i\pi z}$, so all the periods of a doubly-periodic function with primitive periods $1$ and $\tau$ is the {\it lattice\/} $\Omega=\{\omega=n\tau+m: n,m\in\Z\}$. Then $$\wp'(z)=2\sum_{\omega\in\Omega}\frac1{(\omega-z)^3}$$ defines a meromorphic doubly-periodic function with period lattice $\Omega$. One confirms that, just as $\sum n^{-k}$ converges provided that $k\gt1$, so $\sum' \omega^{-k}$ converges absolutely if $k\gt2$. The ${}'$ tells one not to be silly. Integrating, we obtain $$ \wp(z)=\frac1{z^2}+\sum_{\omega\in\Omega}{}'\left( \frac1{(\omega-z)^2}-\frac1{\omega^2}\right)\,.$$ The periodicity of the Weierstra\ss\ $\wp$-function follows by observing that certainly $\wp(z+\omega)-\wp(z)$ is constant. Now take $\omega$ a primitive period and $z=-\frac12\omega$, to see that the constant is zero because $\wp(z)$ is an even function of $z$. Just as we produced the Riemann $\zeta$-function in expanding $\cosec^2 \pi z$ as a power series, so expanding $\wp'(z)$ yields the Eisenstein series $G_{2k}$, $$\wp'(z)=-2z^{-3}+6G_4z+20G_6z^3+\cdots \quad\hbox{where}\quad G_{2k}=G_{2k}(\tau)= \sum_{\omega\in\Omega}{}'\frac1{\omega^{2k}}.$$ There is now nothing for it other than brute computation to discover that $$\(\wp'(z)\)^2=4\(\wp(z)\)^3-60G_4z-140G_6\,,$$ because the difference of the sinister and dexter sides of this equation is a doubly-periodic function without poles and vanishing at $z=0$. An amusing corollary is that the $G_{2k}$ must all be \poly s in just $G_4$ and $G_6$. Thus, for example, we must have $$\left(\sum_{\omega\in\Omega}{}'\frac1{\omega^{4}}\right) \left(\sum_{\omega\in\Omega}{}'\frac1{\omega^{6}}\right) =c\sum_{\omega\in\Omega}{}'\frac1{\omega^{10}},$$ a child's dream; $c$ is a constant, independent of $\Omega$, which is easy to determine, but which I have been too lazy to compute. Put $y=\wp'(z)$ and $x=\wp(z)$. Then $$y^2=4x^3-g_2x-g_3, \quad\hbox{ with\ }\quad g_2=60G_4 \hbox{ and\ } g_3=140G_6,$$ is the equation of an {\it elliptic curve\/}. In truth, this equation has nothing much to do with ellipses. The background is that the integral that gives the length of an arc of an ellipse gives rise to a doubly-periodic function; thence the terminology whereby the Weierstra\ss\ $\wp$-function (and indeed all doubly-periodic meromorphic functions) are called {\it elliptic functions\/}. All this goes back to Fagnano's {\it rectification\/} of the {\it lemniscate\/}, the locus of a point which moves such that the product of its distances from two given points remains a positive constant, which leads to the integral $$\int^\infty_w\frac {dr}{\sqrt{1-r^4}}\,.$$ Our curve is an {\it elliptic\/} curve because it may be {\it parametrised\/} by elliptic functions. I will eventually try to explain how it became believed that these curves can also be parametrised by modular forms. My earlier remarks readily provide the factorisation $$4x^3-g_2x-g_3=4\(x-\wp(\tfrac\tau2)\)\(x-\wp(\tfrac12)\) \(x-\wp(\tfrac{\tau+1}2)\)=(x-e_1)(x-e_2)(x-e_3)\,,$$ showing that the cubic has distinct zeros. Thus its discriminant $g_2^3-27g_3^2$, the square of the difference product $(e_1-e_2)(e_2-e_3)(e_3-e_1)$, is non-zero. Now, a word about the lattice $\Omega$. Let $a$, $b$, $c$ and $d$ be integers so that $ad-bc=1$. Then the ordered pair $(a\tau+b,c\tau+d)$ generates the same lattice as did $(\tau,1)$, and $(a\tau+b)/(c\tau+d)$ is still in $\Cal H$, the upper half-plane. Set $$M=\pmatrix a&b\\c&d\endpmatrix \quad\hbox{ and define\ }\quad M\tau=(a\tau+b)/(c\tau+d)\,.$$ Then $$G_{2k}(M\tau)=(c\tau+d)^{2k}G_{2k}(\tau)\,.$$ The Eisenstein series are the basic examples of {\it modular forms\/}, of functions exhibiting an invariance under transformations by a discrete group, in this case the full modular group $\Gamma={\roman{SL}\/}_2(\Z)$ of integer matrices of determinant $1$. In the mid-eighties Gerhard Frey suggested that if $a^p+b^p+c^p=0$ then the elliptic curve $y^2=x(x-a^p)(x+b^p)$ --- note that its discriminant is essentially $(abc)^{2p}$, so the definition is symmetric --- would have trouble existing. Indeed, in 1987, Ken Ribet showed on the presumption of the conjecture of Taniyama-Weil-Shimura --- to the effect that every elliptic curve is {\it modular\/}, namely that it is parametrised by certain modular forms --- that the {\it Frey curve\/} cannot exist. Andrew Wiles has demonstrated the Taniyama-Weil-Shimura conjecture for {\it semi-stable\/} elliptic curves, a class that includes the Frey curve. \footnote""{Andr\'e Weil, {\it Elliptic functions according to Eisenstein and Kronecker\/}, Ergebnisse der Math. 88, Springer-Verlag, 1976 makes instructive reading.} \end