Elliptic functions, Abelian Integrals

A modern student usually meets elliptic curves in a very polished form: $E:\quad y^2=x^3+ax+b.$ Then one defines a group law by drawing lines, introduces divisors, Abel maps, Jacobians, period lattices, and so on. This is beautiful, but historically it is backwards.

The older story began with much more concrete questions: How long is an arc of an ellipse? Can one compute integrals such as $\int \frac{dx}{\sqrt{1-x^4}}?$ Can such integrals be added, transformed, tabulated, and inverted? Can one build a new trigonometry out of them?

The answer, discovered gradually by Euler, Fagnano, Legendre, Abel, Jacobi, and others, was yes. But the route was computational. It began with integrals, not abstract curves.

The purpose of this article is to reconstruct that route. We will start with the circle, because the circle is the prototype. Then we will move to the lemniscate and elliptic integrals. Then we will see why addition formulae appear. Then we will see how Abel generalized the whole story to higher-degree algebraic integrals. Only near the end will we mention the modern language, and even then only as a summary of what the old computations were already saying.

The circle: sine as an inverse integral

Let us begin with the unit circle $x^2+y^2=1.$ On the upper semicircle, $y=\sqrt{1-x^2},$ the length element is $ds=\sqrt{dx^2+dy^2}.$ Equivalently, $ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}~dx.$ Now compute with $y=\sqrt{1-x^2}.$ Then $\frac{dy}{dx}=-\frac{x}{\sqrt{1-x^2}}.$ Therefore $\left(\frac{dy}{dx}\right)^2=\frac{x^2}{1-x^2},$ we get $ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2} ~dx. = \frac{1}{\sqrt{1-x^2}}~dx..$ Thus the arc length from $0$ to $x$ is

$\displaystyle s=\int_0^x\frac{dt}{\sqrt{1-t^2}}.$

But on the unit circle, arc length equals angle. Thus if $s=\int_0^x\frac{dt}{\sqrt{1-t^2}},$ then $x=\sin s.$ This is an important shift in viewpoint. Instead of thinking of sine first as a triangle ratio, we may think of sine as the inverse of the integral $x\mapsto \int_0^x\frac{dt}{\sqrt{1-t^2}}.$ In this view, trigonometry is the theory of the inverse function of a particular algebraic integral. This is the template for everything that follows.

Addition for Circle

Now suppose

$\displaystyle u=\int_0^x\frac{dt}{\sqrt{1-t^2}},\\ v=\int_0^X\frac{dt}{\sqrt{1-t^2}}.$

Then $x=\sin u, X=\sin v.$ Also $\sqrt{1-x^2}=\cos u, \sqrt{1-X^2}=\cos v.$

We know $\sin(u+v)=\sin u\cos v+\cos u\sin v.$ In terms of $x$ and $X$ , this becomes $Z=x\sqrt{1-X^2}+X\sqrt{1-x^2}.$ So the identity $\sin(u+v)=Z$ can be written as an integral identity:

$\displaystyle \int_0^x\frac{dt}{\sqrt{1-t^2}}+ \int_0^y\frac{dt}{\sqrt{1-t^2}}= \int_0^{x\sqrt{1-X^2}+X\sqrt{1-x^2}}\frac{dt}{\sqrt{1-t^2}}.$

This is the first model of an addition theorem. It says that the sum of two integrals of the same kind can be replaced by one integral of the same kind, with a new algebraic upper limit.

That is exactly the sort of thing eighteenth- and nineteenth-century mathematicians searched for. They wanted to know: If we replace $\sqrt{1-t^2}$ by a more complicated square root, do addition formulae still exist?

Ellitptic Functions

For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ , write $x=a\sin\theta$ and $y=b\cos\theta$ . Then $dx=a\cos\theta~d\theta$ and $dy=-b\sin\theta~d\theta$ , so the arc-length element is $ds=\sqrt{dx^2+dy^2}=\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}~d\theta$ . If $a>b$ and $k^2=1-\frac{b^2}{a^2}$ , this becomes $ds=a\sqrt{1-k^2\sin^2\phi}~d\phi$ . Now factor out $a^2$ : Hence the arc length from $0$ to $\phi$ is

$\displaystyle a\int_0^\phi \sqrt{1-k^2\sin^2\theta}~d\theta.$

This integral is called an elliptic integral of the second kind $E(\phi, k)$ :

Now put $t=\sin\theta$ to get

$\displaystyle \int_0^{\sin\phi} \frac{\sqrt{1-k^2t^2}}{\sqrt{1-t^2}}~dt.$

This is already an integral involving the square root of a quartic expression, because it equals

$\displaystyle \int_0^{\sin\phi} \frac{1-k^2t^2} {\sqrt{(1-t^2)(1-k^2t^2)}}~dt.$

This is the important structural point. The ellipse has led us to integrals of the form

$\displaystyle \int \frac{R(t)~dt} {\sqrt{(1-t^2)(1-k^2t^2)}}.$

The polynomial under the square root is $(1-t^2)(1-k^2t^2),$ which has degree four. This is why elliptic integrals are naturally attached to cubic and quartic polynomials.

The integral of the first kind $F(\phi, k)$ is

$\displaystyle \int_0^\phi \frac{d\theta} {\sqrt{1-k^2\sin^2\theta}}.$

Using the same substitution $t=\sin\theta$ , it becomes

$\displaystyle \int_0^{\sin\phi} \frac{dt} {\sqrt{(1-t^2)(1-k^2t^2)}}.$

This one is not exactly the arc length of the ellipse. The arc length gives $E$ , the second kind. But the first kind appears just as naturally once one studies the whole family of algebraic integrals built from the same square root.

Legendre’s achievement was to organize this entire world. Before him, many integrals appeared in different disguises: arc lengths of ellipses, pendulum motion, attraction problems, transformations of radicals, and integrals involving square roots of cubic or quartic polynomials. Legendre showed that a vast collection of such expressions could be reduced to a small number of standard forms. His three basic types were the first kind,

$\displaystyle E(\phi, k)= \int_0^\phi\frac{d\theta} {\sqrt{1-k^2\sin^2\theta}}$

the second kind,

$\displaystyle F(\phi, k)= \int_0^\phi \sqrt{1-k^2\sin^2\theta}~d\theta,$

and the third kind,

$\displaystyle \Pi(n; \phi, k) = \int_0^\phi \frac{d\theta} {(1-n\sin^2\theta)\sqrt{1-k^2\sin^2\theta}}.$

These are called incomplete elliptic integrals because the upper limit $\phi$ is variable. If $\phi=\frac{\pi}{2}$ , one gets the complete elliptic integrals

$K(k)=F\left(\frac{\pi}{2},k\right), E(k)=E\left(\frac{\pi}{2},k\right), \Pi(n, k)=\Pi(\frac{\pi}{2}, k)$

Legendre’s work was not just naming these integrals. It was a program of reduction. Given an integral of the rough form $\int \frac{R(x),dx} {\sqrt{P(x)}},$ where $P(x)$ is cubic or quartic and $R(x)$ is rational, one tries to transform the roots of $P(x)$ into a standard arrangement, usually leading to a radical like $\sqrt{(1-t^2)(1-k^2t^2)}.$ Then one decomposes the rational part into pieces corresponding to $F$ , $E$ , $\Pi$ , plus elementary terms. In this sense Legendre did for elliptic integrals something like what earlier algebraists had done for rational functions: he made a normal form theory. Abel and Jacobi later changed the emphasis. Instead of only reducing and tabulating the integrals, they inverted them. That is, from $u=F(\phi,k)$ they studied $\phi$ , or $\sin\phi$ , as a function of $u$ . This inversion is what produced elliptic functions.
.

The lemniscate integral

The simplest non-circular example is the lemniscate integral

$\displaystyle u=\int_0^x\frac{dt}{\sqrt{1-t^4}}.$

This should be compared directly with the circular integral $u=\int_0^x\frac{dt}{\sqrt{1-t^2}}$ , whose inverse is $x=\sin u$ . By analogy, we define the inverse of the lemniscate integral by $x=\text{sl}(u)$ , called the lemniscatic sine. This means that $u=\int_0^{\text{sl}(u)}\frac{dt}{\sqrt{1-t^4}}$ . If we write $x=\text{sl}(u)$ , then the defining relation gives $\frac{du}{dx}=\frac{1}{\sqrt{1-x^4}}$ , and hence, by inverting the derivative, $\frac{dx}{du}=\sqrt{1-x^4}$ . Therefore

$\text{sl}'(u)=\sqrt{1-\text{sl}(u)^4},\quad \left(\text{sl}'(u)\right)^2=1-\text{sl}(u)^4.$

This is the exact analogue of the ordinary sine calculation. If $x=\sin u$ , then $x'=\cos u$ , so $(x')^2=1-x^2$ . Thus ordinary sine is governed by $x'^2=1-x^2$ , while the lemniscatic sine is governed by $x'^2=1-x^4$ . The change from $x^2$ to $x^4$ is small in notation but enormous in consequence: it is the first step from trigonometric functions to elliptic functions.

Now define the lemniscatic cosine by $\text{cl}(u)=\text{sl}'(u)$ . Then the previous equation becomes $\text{cl}(u)^2=1-\text{sl}(u)^4$ , or equivalently

$\text{sl}(u)^4+\text{cl}(u)^2=1.$

This is not the circular identity $\sin^2u+\cos^2u=1$ , but it plays the same role. It says that the moving point $(\text{sl}(u),\text{cl}(u))$ lies on the algebraic curve $x^4+y^2=1$ . Differentiating $\text{cl}(u)^2=1-\text{sl}(u)^4$ gives $2\text{cl}(u)\text{cl}'(u)=-4\text{sl}(u)^3\text{sl}'(u)$ . Since $\text{sl}'(u)=\text{cl}(u)$ , this becomes $2\text{cl}(u)\text{cl}'(u)=-4\text{sl}(u)^3\text{cl}(u)$ , and canceling gives $\text{cl}'(u)=-2\text{sl}(u)^3$ . Thus the pair $\text{sl},\text{cl}$ satisfies the differential system

$\text{sl}'(u)=\text{cl}(u), \quad \text{cl}'(u)=-2\text{sl}(u)^3.$

Compare this with ordinary trigonometry: $(\sin u)'=\cos u$ and $(\cos u)'=-\sin u$ . The first derivative relation is formally the same; the second has become nonlinear. That nonlinearity is the analytic signature that we have left the circular world and entered the elliptic one.

The lemniscate addition formula

Now comes the real test: does this new sine have an addition formula? For ordinary trigonometry, the inverse of $\int_0^x\frac{dt}{\sqrt{1-t^2}}$ satisfies $\sin(u+v)=\sin u\cos v+\cos u\sin v.$

For the lemniscatic sine, write

$x=\text{sl}(u),\quad y=\text{cl}(u), \\ X=\text{sl}(v),\quad Y=\text{cl}(v).$

Then $y^2=1-x^4$ and $Y^2=1-X^4$ . The addition formula is

$\displaystyle Z= \frac{xY+Xy}{1+x^2X^2}.$

The numerator $xY+Xy$ looks exactly like the sine addition formula; the denominator $1+x^2X^2$ is the new elliptic correction. It is the new feature obtained after replacing the circular equation $y^2=1-x^2$ by the quartic equation $y^2=1-x^4$ .

To verify the formula, define $Z=\frac{xY+Xy}{1+x^2X^2}.$ We want to prove that $Z=\text{sl}(u+v)$ . First check the initial value. When $u=0$ , we have $x=\text{sl}(0)=0$ and $y=\text{cl}(0)=1$ , so $Z=\frac{0\cdot Y+X\cdot1}{1+0}=X=\text{sl}(v).$ Thus $Z$ has the correct starting value: $Z|_{u=0}=\text{sl}(v)=\text{sl}(0+v)$ .

Now differentiate $Z$ with respect to $u$ , keeping $X,Y$ fixed. We use $x'=y, \quad y'=-2x^3.$ Let $N=xY+Xy, \quad D=1+x^2X^2.$ Then $Z=N/D$ , and $N'=yY-2Xx^3, \quad D'=2xyX^2.$

We need $\displaystyle \frac{N'D-ND'}{D^2}.$ Substituting $N,D,N',D'$ gives an expression in $x,y,X,Y$ . Then using only $y^2=1-x^4, Y^2=1-X^4,$ one simplifies it to

$(Z')^2=1-Z^4.$

Thus $Z$ satisfies the same differential equation as $\text{sl}(u+v)$ , namely $w'^2=1-w^4$ , and it also has the same value at $u=0$ . Therefore $Z=\text{sl}(u+v).$

So the lemniscatic addition law is

$\displaystyle\text{sl}(u+v)=\frac{\text{sl}(u)\text{cl}(v)+\text{sl}(v)\text{cl}(u)}{1+\text{sl}(u)^2\text{sl}(v)^2}.$

In integral form, the same statement says the following. Let

$\displaystyle\ I(x)=\int_0^x\frac{dt}{\sqrt{1-t^4}}, \quad y=\sqrt{1-x^4},\\ I(X)=\int_0^X \frac{dt}{\sqrt{1-t^4}}, \quad Y=\sqrt{1-X^4}.$

then

$\displaystyle I(x)+I(X)=\int_0^x\frac{dt}{\sqrt{1-t^4}} + \int_0^X \frac{dt}{\sqrt{1-t^4}} = \int_0^{\frac{xY+Xy}{1+x^2X^2}}\frac{dt}{\sqrt{1-t^4}}=I(Z). \displaystyle$

Algebraic Curve

The differential equation $x'^2=1-x^4$ can be rewritten by setting $y=x'$ . Then $y^2=1-x^4.$ So the inverse lemniscate function naturally moves on the algebraic curve

$C:\quad y^2=1-x^4.$

Indeed, if $x=\text{sl}(u)$ and $y=\text{cl}(u)=\text{sl}'(u)$ , then the point $(x,y)$ satisfies $y^2=1-x^4$ . The integral being inverted is $u=\int_0^x\frac{dt}{\sqrt{1-t^4}},$ which, on the curve $y^2=1-x^4$ , is simply

$\displaystyle du=\frac{dx}{y}.$

Thus $u$ is the special parameter measured by the differential $dx/y$ . It is not ordinary Euclidean arc length along the curve $y^2=1-x^4$ . Rather, it is the parameter for which motion along the curve satisfies $\frac{dx}{du}=y.$ This is exactly analogous to the circle. On the unit circle, if $x=\sin u$ and $y=\cos u$ , then $y^2=1-x^2$ and $du=dx/y$ . In that special case, $u$ is also the ordinary angle and arc length. But that coincidence is special to the unit circle.

Thus $u$ is the parameter measured by the differential $dx/y$ . It is the analogue of angle, not because it is literally Euclidean angle or ordinary arc length, but because it is the coordinate in which addition becomes simple. For the unit circle this $u$ is both the angle and the arc length. But that is a special accident of the circle. In the lemniscatic case, $u$ is not ordinary length along the curve $y^2=1-x^4$ ; it is the algebraic angle obtained by integrating $dx/y$ .

Now the addition formula has a precise meaning. If $P(u)=(\text{sl}(u),\text{cl}(u)), \quad P(v)=(\text{sl}(v),\text{cl}(v)),$ then adding the hidden parameters gives the new point $P(u+v)=(\text{sl}(u+v),\text{cl}(u+v)).$ The formula $\frac{\text{sl}(u)\text{cl}(v)+\text{sl}(v)\text{cl}(u)} {1+\text{sl}(u)^2\text{sl}(v)^2}$ is the rule for recovering the visible $x$ -coordinate of $P(u+v)$ from the visible coordinates of $P(u)$ and $P(v)$ . So the addition is not happening directly in the coordinate $x$ ; it is happening in the hidden angle coordinate $u$ , and the displayed formula translates that simple addition back into algebraic coordinates.

Thus the integral $u=\int\frac{dx}{y}$ is not merely a way of measuring something. It is the coordinate in which the elliptic addition law becomes ordinary addition. This is why $u$ deserves to be called a generalized angle.

Cubics and the line construction

The quartic model $y^2=1-x^4$ is good for integrals. But the group law is more visibly geometric in the cubic model

$\displaystyle E:\quad y^2=x^3+ax+b.$

Take a line $y=mx+n.$ Its intersections with $E$ are obtained by substituting into the cubic:

$\displaystyle (mx+n)^2=x^3+ax+b.$

Equivalently, define $F(x)=x^3+ax+b-(mx+n)^2.$ This is a cubic equation in $x$ , so it has three roots $x_1,x_2,x_3.$ The corresponding points are $P_1=(x_1,y_1), P_2=(x_2,y_2), P_3=(x_3,y_3).$

The central analytic fact is that, as the line moves,

$\displaystyle \frac{dx_1}{y_1}+\frac{dx_2}{y_2}+\frac{dx_3}{y_3}=0.$

Let us prove this, because it is the computational heart of the subject.

Here is the computation. Since $F(x_i)=0$ , differentiating with respect to the moving line parameters $m,n$ gives

$\displaystyle F_x(x_i)dx_i+F_m(x_i)dm+F_n(x_i)dn=0.$

But $F_m(x_i)=-2x_i(mx_i+n)=-2x_i y_i, \quad F_n(x_i)=-2(mx_i+n)=-2y_i.$

Therefore $F_x(x_i)dx_i=2x_i y_idm+2y_i dn,$ and hence

$\displaystyle = \frac{dx_1}{y_1} =\frac{2x_i,dm+2,dn}{F_x(x_i)}.$

Summing over $i=1,2,3$ gives

$\displaystyle = \frac{dx_1}{y_1} + \frac{dx_2}{y_2}+\frac{dx_3}{y_3} = 2dm\sum_i\frac{x_i}{F_x(x_i)} + 2dn\sum_i\frac{1}{F_x(x_i)}.$

Now $F(x)=(x-x_1)(x-x_2)(x-x_3),$ so $F_x(x_i)=\prod_{j\neq i}(x_i-x_j).$

For any cubic with roots $x_1,x_2,x_3$ , the identities $\sum_i\frac{1}{F_x(x_i)}=0, \quad \sum_i\frac{x_i}{F_x(x_i)}=0$ hold. They are just the Lagrange interpolation identities saying that a polynomial of degree at most one has no $x^2$ term when expanded in the basis $F(x)/(x-x_i)$ . Hence both sums vanish, and we obtain

$\displaystyle \frac{dx_1}{y_1}+\frac{dx_2}{y_2}+\frac{dx_3}{y_3}=0.$

Integrating this infinitesimal identity gives $u(P_1)+u(P_2)+u(P_3)=0$ Thus, if three points $P_1,P_2,P_3$ are collinear on a cubic, their elliptic parameters add to zero. This is the analytic content behind the chord law. Modernly one writes $P_1+P_2+P_3=O,$ but historically the key fact was the integral identity: a line relation among points produces a linear relation among the integrals $\int dx/y$ .

Why this is really addition?

The word addition can feel mysterious here. Why should points on a curve be addable? The answer is that the point is merely the visible representative of a hidden parameter. For the circle, the hidden parameter is the angle $\theta$ . A point is $(\cos\theta,\sin\theta).$ The group law is $P(\theta_1)+P(\theta_2)=P(\theta_1+\theta_2).$

For an elliptic curve, the hidden parameter is $u(P)=\int_O^P\frac{dx}{y}.$ The group law is designed so that $u(P+Q)=u(P)+u(Q).$ The line construction is not arbitrary. It is the algebraic machine that computes addition of these hidden parameters without explicitly calculating the integrals. This is the central idea.

Periods

The integral $u=\int_0^x\frac{dt}{\sqrt{1-t^4}}$ does not behave like an ordinary single-valued function once complex paths are allowed. Different paths can give values differing by constants called periods.

For the circle, one loop gives the period $2\pi.$ For elliptic functions, there are two independent periods.

For the lemniscate, define $K=\int_0^1\frac{dt}{\sqrt{1-t^4}}.$ Then, along the real direction, the inverse function passes through $0,\quad 1,\quad 0,\quad -1,\quad 0$ as $u$ moves through $0,\quad K,\quad 2K,\quad 3K,\quad 4K.$ So $4K$ is a real period.

But the same integral also has a vertical symmetry. If $t=is$ , then $t^4=s^4, \quad dt=ids.$ Thus $i\int_0^x\frac{ds}{\sqrt{1-s^4}}.$ So there is also an imaginary period, essentially $4iK.$

Thus the inverse function repeats in two directions. This is the analytic origin of double periodicity of the ellitpic function like $\text{sl}(u)$ ..

Higher Degree Curves

Now consider a general algebraic integral

$\displaystyle \int \frac{R(x)dx}{\sqrt{P(x)}}.$

For $P(x)$ of degree $3$ or $4$ , one integral is enough. There is essentially one basic integral, $\int \frac{dx}{y},$ and one point of the curve carries one hidden parameter. This is why elliptic addition is a law on single points.This is the elliptic case.

But if $P(x)$ has degree $5$ or $6$ , something new happens.Let $C:\quad y^2=P(x),$ with $\deg P=5$ or $6$ . One hidden parameter is no longer enough. The natural integral are the integral of the Abelian differentials:

$\displaystyle \omega_1=\frac{dx}{y}, \quad \omega_2=\frac{xdx}{y}.$

So the basic Abelian quantities (“hidden angles”) are not one number but two: $u_1=\int \frac{dx}{y}, \quad u_2=\int \frac{x,dx}{y}.$ .

This already tells us that a single point cannot be the whole story. A single point $P$ gives two numbers $\int^P dx/y$ and $\int^P x,dx/y$ , but the correct inversion problem asks for two points $P_1,P_2$ satisfying

$\displaystyle u_1= \int^{P_1}\frac{dx}{y} + \int^{P_2}\frac{dx}{y}, \\ u_2= \int^{P_1}\frac{xdx}{y} + \int^{P_2}\frac{xdx}{y}.$

Thus in genus two the basic object is not one point but a pair of points. This is the first major break from ordinary elliptic curves.

For a general degree $7$ or $8$ ciurve, one needs three points and three integrals: $\int\frac{dx}{y}, \quad \int\frac{xdx}{y}, \quad \int\frac{x^2dx}{y}.$ In general, for $y^2=P(x)$ with $\deg P=2g+1 \quad\text{or}\quad \deg P=2g+2,$ one needs $g$ integrals

$\displaystyle \int\frac{dx}{y}, \quad \int\frac{x dx}{y}, \quad \dots, \quad \int\frac{x^{g-1}dx}{y}.$

The number $g$ is the genus. It measures how many independent Abelian integrals are needed, and therefore how many points must usually be used in the inversion problem.

Let us see how the addition works concretely in genus two.

Let $C:\quad y^2=P_6(x)$ . with $P_6(x)$ a polynomial of degree six. This is a genus two curve. Before trying to add arbitrary pairs of points, let us first do the simplest geometric experiment: cut the curve by a line $y=mx+n.$ Substituting this into $y^2=P_6(x)$ gives $(mx+n)^2=P_6(x).$

Equivalently, $F(x)=P_6(x)-(mx+n)^2=0.$

Since $F(x)$ has degree six, the line meets the curve in six points, say $P_1, P_2, P_3, P_4, P_5, P_6.$

Similar computation as above gives

$\displaystyle \sum_{i=1}^6\frac{dx_i}{y_i}=0. \\ \sum_{i=1}^6\frac{x_idx_i}{y_i}=0.$

This shows that the hidden angles satisfy

$\displaystyle u_1(P_1, P_2) + u_1(P_3, P_4) +u_1(P_5, P_6)=0. \\ u_2(P_1, P_2) + u_2(P_3, P_4) +u_2(P_5, P_6)=0.$

This is the genus two analogue of the elliptic fact that three collinear points on a cubic have elliptic parameters summing to zero. But now, because the curve has genus two, there are two integral coordinates, $\int dx/y$ and $\int xdx/y$ , and because a line meets the degree-six curve in six points, the relation involves six points. This already gives a kind of addition relation. Group the six points into three pairs: Then the two integral relations say that the Abelian coordinates of these three pairs add to a fixed constant.

$(P_1,P_2)+(P_3,P_4)+(P_5,P_6)=0.$

Thus a line gives a special three-pair relation. This is already the right genus two pattern: the objects being added are not single points, but pairs of points. However, a line does not give the most general addition of two arbitrary pairs. If we start with two arbitrary pairs $(P_1,P_2), (Q_1,Q_2),$ a line is too small a family to pass through all four prescribed points. So for the general pair-addition law one uses a larger auxiliary curve, for example $y= R(x)=Ax^3+Bx^2+Cx+D.$ This cubic graph has four coefficients, so it can be chosen to pass through $P_1,P_2,Q_1,Q_2$ . Substituting into $y^2=P_6(x)$ gives $R(x)^2=P_6(x).$ This is again a degree-six equation. Four intersections are the prescribed points, and two new intersections remain which give the pair satisfying the addition law $(P_1,P_2)+(Q_1,Q_2)+(R_1,R_2)=0.$ So to add arbitrary pairs, one uses a higher auxiliary curve chosen to contain the desired known points and lets the remaining intersections produce the sum.

Singularities

The degree of $P(x)$ is not enough. What matters is whether the roots of $P(x)$ are distinct. For a smooth curve $y^2=P(x),$ with $\deg P=2g+1$ or $\deg P=2g+2$ , one normally gets $g$ independent integrals: $\int\frac{dx}{y},\int\frac{xdx}{y},\dots, \int\frac{x^{g-1}dx}{y}.$

So degree $3$ or $4$ gives one integral; degree $5$ or $6$ gives two; degree $7$ or $8$ gives three, and so on. But if two roots collide, the integral becomes simpler. Let $P(x)=(x-a)^2Q(x)$ , then writing $y=(x-a)Y$ gives $Y^2=Q(x)$ , and the basic differential becomes $dx/y=dx/((x-a)Y)$ . Near $x=a$ , if $Y(a)\neq0$ , this behaves like $dx/(x-a)$ , whose integral is $\log(x-a)$ . Thus when two roots collide, one Abelian direction degenerates into a logarithmic direction: a genus two integral may partly collapse to an elliptic integral plus a logarithm, a genus three integral may collapse toward genus two plus logarithmic pieces, and so on.

In the simplest case $y^2=(1-x^2)^2$ , one gets $u=\int dx/(1-x^2)=\frac12\log\left(\frac{1+x}{1-x}\right)$ , so $x=\tanh u$ . Thus distinct roots give genuinely elliptic or Abelian integrals, while repeated roots make part of the theory degenerate into logarithms, exponentials, hyperbolic functions, or rational functions.

Modern Language

Modern language packages the old calculation by saying that a function on a curve has a divisor: its zeros counted positively and its poles counted negatively. Thus, if a line $L$ meets a cubic elliptic curve at $P,Q,R$ , then as a function on the curve it vanishes at those three points and has a triple pole at the point at infinity $O$ . So one writes $\text{div}(L)=P+Q+R-3O$ . In older language, this simply says: the algebraic function $L$ has zeros at $P,Q,R$ and its compensating pole at infinity.

Abel’s theorem says that such a zero-pole relation forces the corresponding Abelian integral sum to vanish, up to periods. In genus one, with $u(P)=\int_O^P dx/y$ , the divisor relation above gives $u(P)+u(Q)+u(R)-3u(O)=0$ . Since $u(O)=0$ , we get $u(P)+u(Q)+u(R)=0$ . This is exactly the analytic content behind the geometric statement that three collinear points on a cubic add to zero. A modern reader says “the divisor of a function is zero in the Jacobian”; the older computational statement is “the zeros and poles of an algebraic function give an identity among Abelian integrals.

The reason divisors are tied to integrals can be seen from the logarithm of a function. Let $f$ be an algebraic function on the curve, and suppose $f$ has a zero of order $n$ at a point $P$ . Choose a local coordinate $t$ near $P$ , with $t(P)=0$ . Then locally $f(t)=t^n h(t)$ , where $h(0)\neq0$ , so $\log f(t)=n\log t+\log h(t)$ . When one goes once around $P$ , $\log t$ changes by $2\pi i$ , and therefore $\log f$ changes by $2\pi i n$ . At a pole of order $n$ , the same calculation gives a jump of $-2\pi i n$ . Thus the jumps of $\log f$ remember exactly the divisor of $f$ : zeros contribute positively, poles negatively. Now multiply by an Abelian differential $\omega$ , such as $dx/y$ , and integrate $\log f,\omega$ around a cut-up version of the curve. Across a cut ending at $P$ , the two values of $\log f$ differ by $2\pi i n$ , so that cut contributes $2\pi i n\int_O^P\omega$ . Summing over all zeros and poles gives

$\displaystyle 2\pi i\sum_i n_i\int_O^{P_i}\omega,$

which is precisely the Abel sum of the divisor $\text{div}(f)=\sum_i n_iP_i$ . Since $f$ is a globally defined function, these logarithmic jumps must balance, except for possible period contributions coming from loops on the curve. Hence

$A(\text{div}(f))=0 \quad\text{mod periods}.$

This is the analytic bridge: zeros and poles give a divisor, the logarithm detects that divisor by its jumps, and integrating those jumps against $\omega$ gives the corresponding Abelian integral relation.

Abel’s theorem tells us what happens to sums of Abelian integrals when the points satisfy an algebraic relation. Jacobi inversion asks the opposite question: if the integral sums are given, can we recover the points? In genus one this means: given $u=\int_O^P dx/y$ , recover the point $P$ . Writing the coordinates of $P$ as functions of $u$ gives elliptic functions. This is exactly the same pattern as ordinary trigonometry: given $u=\int_0^x dt/\sqrt{1-t^2}$ , recover $x=\sin u$ .

In genus two the inversion problem already becomes genuinely different. Given two numbers $u_1,u_2$ , one must recover two points $P_1,P_2$ from

$\displaystyle u_1= \int^{P_1}\frac{dx}{y} + \int^{P_2}\frac{dx}{y},\\ u_2= \int^{P_1}\frac{x,dx}{y} + \int^{P_2}\frac{x,dx}{y}.$

The solutions $x(P_1),x(P_2)$ then become functions of the two variables $u_1,u_2$ ; these are Abelian functions. In genus $g$ , the same pattern uses $g$ points and $g$ integrals. Modern language says that the variables $u_1,\dots,u_g$ live on the Jacobian, modulo periods. But historically the question was more concrete: can we invert these multi-integral systems and express the algebraic coordinates of the points as functions of the new generalized angle variables?

Conclusion

The historical progression can be summarized in one line: trigonometry begins with the circle, elliptic functions begin with elliptic integrals, and Abelian functions begin when several such integrals must be inverted at once. For the circle, the integral $\int dx/\sqrt{1-x^2}$ is inverted by $\sin u$ . For elliptic integrals, one meets radicals such as $\sqrt{(1-x^2)(1-k^2x^2)}$ , and the inverse functions become elliptic functions. Their addition formulae generalize the ordinary sine and cosine addition formulae. The deeper reason is that algebraic relations among points on a curve force linear relations among the corresponding integrals.

Abel’s great step was to see that this is not limited to elliptic integrals. For higher-degree equations $y^2=P(x)$ , one needs several integrals, such as $\int dx/y,\int x,dx/y,\dots$ , and several moving points. Thus higher genus means several generalized angle variables. Jacobi inversion then asks how to recover the points from those variables. Modern language says that these variables live on a Jacobian, and that divisors package the zero-pole relations of algebraic functions. But historically the essential phenomenon was simpler and more concrete:

$\displaystyle \text{algebraic relations among points} \quad\Longrightarrow\quad \text{linear relations among integrals}.$

This is the heart of Abel’s theorem. The algebraic curve is not decoration; it is the machine that organizes the upper limits of integration and makes addition formulae possible. Ordinary trigonometric functions are the natural trigonometry of the circle. Elliptic functions are the natural trigonometry of cubic and quartic curves. Abelian functions are the natural trigonometry of higher-genus curves.

Share this:

Related

Leave a comment Cancel reply