First post in the cubic curves series. The goal is to understand (the two versions of) the group law.

Next post: Cubics II: Constant cross-ratios.

$\newcommand{\cC}{\mathcal C}$ $\newcommand{\cK}{\mathcal K}$ $\newcommand{\ol}{\overline}$ $\newcommand \lr[1]{\langle #1 \rangle}$ $\newcommand{\Pic}{\mathrm {Pic}}$

Introduction

What kicked this off was the following solution to IMO 2019/6 by yaron235:

… In the notation of the problem, consider the locus of all points $X$ such that the radical axis of the circles $(XEC), (XBF)$ passes through $T$, we’ll call it $\gamma$. We wish to show that $P$ is on this locus - this is just a reformulation of the problem statement.

First of all, note that this curve is an elliptic curve - indeed, one can verify that this is a cubic equation on $X$ (…).

The point $T$ is obviously on $\gamma$, and so are $D, K$. The points $E,B,C,F$ are also on $\gamma$ - one ought to choose some tangent direction $\ell$ at $E$ (or any of the other points) such that the circles $(EBF)$ and $(EEC)$ (the circle through $E$, $C$ tangent to $\ell$) will have a radical axis passing through $T$. It’s not hard to see that such a direction exists. Similarly, one shows that $I$ and $J$ lie on $\gamma$.

Consider some point $X$ on $\gamma$. Then the circles $(XEC), (XBF)$ intersect at some other point, say $Y$, which also belongs to $\gamma$. Thus, we have that the points $X, Y, E, C, I, J$ lie on a conic, so in the group of $\gamma$ we get that $X + Y + E + C+ I + J = 0$. Similarly, $X + Y + B+ F + I + J = 0$. We conclude that in particular, $E + C = B + F$, so $EC$ and $BF$ intersect on $\gamma$ - but their intersection is $A$. Thus $A \in \gamma$. Moreover, since $X, Y, T$ are collinear, we get that $X + Y + T = 0$. Thus $E + C + I + J = B + F + I + J = T$. Recall that we wish to prove that $P$ lies on $\gamma$. $P$ is defined as the intersection of the circles $(DEF), (ATD)$. These circles intersect $\gamma$ for a fourth time at $-(D + E + F + I + J)$ and $-(A + T + D + I + J)$ respectively, so it will be enough to prove that $E + F = A + T$. This is equivalent to saying that $EF$ and $AT$ intersect on the curve. But their intersection point is the infinite point in their direction, so it is also the intersection of $AT$ and the infinite line $IJ$. Thus we wish to prove that $A + T = I + J$. Recall that $A + E + C = 0$, so it remains to show that $T = I + J + E+ C$. But we already showed this!

What the hell is going on here? You might have vaguely heard that elliptic curves have a group law so it’s possible to “add points” together, but the group law looks something like (according to Wikipedia):

• First, draw the line that intersects $P$ and $Q$. This will generally intersect the cubic at a third point, $R$.
• We then take $P + Q$ to be $−R$, the point opposite $R$ (across the $x$-axis).

but this is evidently not the same “+” as what we were seeing earlier in the solution! In this article we’ll try to make sense of all of this.

Recap: Cubic curves

A cubic curve is the set of points on the plane which are the solutions to a cubic equation, e.g. $$\{(x,y): x^3 + y^3 + 3xy + 2y + 5 =0\}.$$

For what we’re about to do, we might need to add in the projective points at infinity, so we define the curve instead in a homogenized way: $$\{[x:y:z]: x^3 + y^3 + 3xyz + 2yz^2 + 5z^3 =0\}.$$

If you’ve never seen this notation before, $[x:y:z]$ refers to an equivalence class up to non-zero scalar multiplication (so $[x:y:z]$ and $[2x:2y:2z]$ refer to the same point). Using the homogenized equation instead of the original one endows the curve with points at infinity, which are points of the form $[x:y:0]$.

In some sense, the point $[x:y:0]$ is a natural answer to the question “where do two parallel lines $yx'+xy'=c_1$ and $yx' + xy' = c_2$ intersect?”

So what are elliptic curves?

An elliptic curve is essentially a cubic curve without a singular point (e.g. these guys). In those cases, we can apply some change of variables to get it into Weierstrass form: $$y^2 = x^3 + ax + b.$$

This page has some more detail, but for our purposes we’ll stick to dealing with cubic curves in general.

Recap: Bezout and “projectiveness”

Bezout’s theorem is just the fact that

A degree $d$ curve must intersect a degree $e$ at exactly $de$ points, up to multiplicity.

There are a few bells and whistles here, and I’ll provide a few examples for you to think about:

• Every two lines intersect at one point
  • …or do they? Two parallel lines don’t intersect. But in the projective world, they intersect at the point at infinity. (Hence, we work in projecive space $\mathbb P^2$.)
• A line intersects a circle at two points
  • …unless it is tangent to the circle. Then this counts as two points. (How might we distinguish “double” intersections from single oness?)
  • …unless the line doesn’t intersect the circle, which is totally possible! Fixing this requires the addition of complex points, turning $\mathbb P^2$ into $\mathbb {CP}^2$.
• A circle intersects a circle at four points.
  • Suppose the two circles intersect. Where are the other two intersections? (Hint: they are both at infinity and complex).
  • Suppose the two circles are concentric. Where are the intersection points?

Recap: Cayley-Bacharach

Given two cubic curves $\cK_1, \cK_2$ (possibly degenerate) which intersect at $P_1, \cdots, P_9$, then any cubic curve $\cK_3$ that passes through $P_1, \cdots, P_8$ must be a linear combination of $\cK_1$ and $\cK_2$ (in terms of the defining equation). In particular, $\cK_3$ must also pass through the 9th common point $P_9$.

The reasoning on Wikipedia is as follows: each cubic is uniquely determined by 9 points in general position, so having two cubics intersect at 9 points means that there is some degeneracy involved, and this degeneracy is exactly degree 1.

In more detail:

• The space of cubic curves has dimension $9 = 10 -1$, with the 10 coming from the degree 3 monomials $$x^3,y^3,z^3,x^2y,..., xyz$$ and the minus 1 coming from the projective scaling (i.e. $F(x,y,z) = 0$ and $\lambda F(x,y,z) = 0$ define the same curve for any scalar $\lambda \neq 0$).
• A curve passing through a point $(x,y,z)$ is really a linear constraint for the coefficients of the curve. Thus generically, having 9 constraints will typically determine the cubic (and hence the saying “9 points define a cubic”).
• But suppose there are two such cubics. This means that there is a linear dependence among the linear constraints (= points on the curve).
• Let’s suppose that the first 8 points were independent, and the 9th point caused a linear dependence. Then, the cubics that pass through them must exactly form a 1-dimensional subspace.
• So this implies that the third cubic is a linear combination of the first two cubics, and so we’re done.
• This is not always true, but that’s where the case checking comes in. (See the appendix for a proof.)

In the general case, Bezout’s theorem grants us that two degree $d$ curves should meet at $d^2$ points, and once we’ve had enough points to determine a curve less one, it should just pass through all the other points. For $d=4$, this means that if two quartics meet at 16 points and a third passes through 13, then it should also pass through the last three (we could call this the quartic Cayley-Bacharach).

Classical group law

Before I begin, I just want to point out that Dylan’s Elliptic Curve v2.0 post has an excellent motivation for why you might guess the group law, starting with how the (unit) circle comes with a group.

For me, where this begins is the classical group law using a cubic graph: $$y = x^3+ax+b.$$

Then, a line $y = mx+c$ intersects it at three points $(x_1, y_1), (x_2,y_2), (x_3,y_3)$. But since they are all roots of $$x^3 + (a-m)x + b-c = 0$$ by Vieta’s relations we must have $x_1+x_2+x_3=0$. Hence, for two points on the curve $X,Y$ and $O=(0,b)$, if we let $XY$ intersect the curve again at $Z$, and let $OZ$ intersect the curve again at $W$, then $x_X + x_Y = x_W$ and this operation on points defines an (abelian) group.

Remark. This is apparently the content of USAMO 2014/3.

We can repeat this on any general cubic curve to obtain a group. To be concrete:

• Pick a point $O$ on the cubic $\cK$. (For an elliptic curve, this is often the point at infinity on $x=0$.)
• $Z$ is the third intersection between $XY$ and the cubic $\cK$.
• Then we can define $X+_O Y$ is the third intersection between $OZ$ and $\cK$.

Then, $+_O$ defines an abelian group on the points of the curve. There’s a handful of things to check:

• $O$ is the identity.
• $+_O$ is commutative.
• What should $-_OX$ be? Well, if $Y=-_OX$, then $XY$ intersects $\cK$ again at $L$, the second intersection of the tangent to $\cK$ at $O$.

It remains to show associativity. We would like to show that $$(X +_O Y) +_O Z = X_O + (Y+_OZ)$$

Write $P = X+_O Y$, and let the third intersection of $XY, ZY, PY$ with $\cK$ be $A,B,C$ respectively. Then, by considering the two degenerate cubics $$\ol{XYA} \cup \ol{OB(Y+_OZ)} \cup \ol{(X+_O Y)Z}, \qquad\ol{ZYB} \cup \ol{OA(X+_O Y)} \cup\ol{X(Y+_O Z)}$$ it follows that $(X+_O Y)Z$ and $X(Y+_O Z)$ intersect on $\cK$, so the conclusion follows.

Hence, the points form a group using $+_O$. This is the classical definition of the group law on an elliptic curve.

My version

If you were not satisfied with this, you’re not alone! I don’t like this version of the group law either, because you have to pick some $O$. One can check that three points are collinear iff $$X +_O Y +_O Z = L$$ where $L$ is the third intersection of the tangent at $O$ with $\cK$. But you’d have to pick $O$ to define the group, and you’d also have to get $L$ (which is a function of $O$), whereas the fact that three points are collinear shouldn’t care about this choice. Really, we would like this to be “$0$” instead of $L$ (as it were with the $x$-coordinates).

I can get my way by doing everything in reverse - let’s define an abelian group as follows:

• the generators are points $[X]$ on the curve. (Abelian groups are not scary! It’s just a bookkeeping object for symbols you can add together, like $2[X] + [Y]$, with some expressions equalling 0.)
• the relations are the intersections of another curve with our original curves (that is, if a line intersected our curve at $X, Y, Z$, then $[X] + [Y] + [Z] = 0$).

One checks that under this definition, $$[X +_O Y]= (- [O] - [Z]) = [X]+[Y] - [O].$$

This allows us to convert from the classical group law to our “zero-less” version. One notes that $[L] = -2[O]$, so we have $$X +_O Y +_O Z = L \Leftrightarrow [X]+[Y]+[Z]-2[O] = -2[O]$$

We also have some nice consequences: suppose that $A,B,C,D,E,F$ sum to zero under this new group law. That means that $$-[A]-[B], -[C]-[D], -[E]-[F]$$ all sum to zero (where these correspond to honest points!), so they lie on the same line. Now, by applying Cayley-Bacharach (picture below), we get that the remaining 6 points must lie on the same conic!

So we have:

• $[X] + [Y] + [Z] = 0$ iff $X,Y,Z$ are collinear.
• $[A] + [B] + [C] + [D] + [E] + [F] = 0$ iff $A,B,C,D,E,F$ lie on the same conic.

You might guess the pattern here: if 9 points sum to zero, they must be on the same cubic, and so on.

While our new abelian group seems much cleaner, it loses out because not all terms are points. We can, in fact, figure out which terms must be points: suppose $$[P_1] + [P_2] + ... + [P_k] = [Q]$$ then let’s consider a map sending every point to $1 \pmod 3$. (Notice that this sends the sum of three points on a line to $0\pmod 3$.) Hence, it follows that $k\equiv 1\pmod 3$. Conversely, one can also show inductively that every $k\equiv 1 \pmod 3$ points must sum to an honest point.

Appendices

The group law on a degenerate cubic

For just this section, let’s drop the square brackets (writing $X$ instead of $[X]$).

There’s some spooky connection between the group law and projective trasformations, which we can pick at by considering what happens on a degenerate cubic $\cK = \cC\cup \ell$ which is the union of a conic $\cC$ and a line $\ell$.

We have to be extra careful with how we apply the group law: if a line intersects $\cK$, it must intersect with $\cC$ at two points and $\ell$ at one. If a conic intersects $\cK$, it must intersect $\cC$ at 4 points and $\ell$ at 2.

Consider four points $A,B,C,D$ on the conic $\cC$. Then, we know that the pencil of conics passing through $A,B,C,D$ intersects $\ell$ along involutive pairs, and this involution must also swap the intersection points $\{P,Q\} = \cC \cap \ell$. For one such involution pair $(X,Y)$, it must satisfy $X+Y = (-A-B-C-D)$.

On the other hand, if we have a point on the line $A$ and two points on the conic such that $-2X=-2Y=A$, and $B=-X-Y$, then $(A,B; P,Q)$ are harmonic (classically), so $A,B$ are conjugate with respect to $\cC$. However, note that it’s sufficient to have $2(A-B) = 0$ under the group law, because once $X$ exists, defining $Y = BX\cap \cK$ gets us $-2Y = 2(X+B) = 2B - A = A$. Hence, adding a 2-torsion point along the line corresponds to harmonic conjugation w.r.t. $P$ and $Q$.

The idea of 2-torsion elements will be revisited in Cubics II: Constant cross-ratios.

The Picard group

A very fancy way to phrase all of this is that an elliptic curve is isomorphic to its Picard variety. It comes from the quotient of two things:

• A formal sum of points $\sum_P n_P \lr P$, which are called divisors. The sum of coefficients $\sum_P n_P$ is also called the degree of the divisor.
• We can define a rational function of $x$ and $y$ on the elliptic curve, and record the order of vanishing at each point $P$ as $n_P$.
• Let’s consider $ax+by-c$. Suppose that the line $ax+by - c = 0$ intersects the cubic curve at three distinct points $P, Q, R$. Then,
  • $ax+by-c$ has three roots, at $P, Q, R$ respectively
  • generically, the cubic curve has three points at infinity $\infty_1, \infty_2, \infty_3$, and at each of them we have a pole (because the value of $ax+by-c$ diverges).
  • the corresponding divisor is therefore $\lr P + \lr Q + \lr R - \lr \infty_1 - \lr \infty_2 - \lr \infty_3$.
  • You could imagine that for a quadratic form, we instead have something like $$\lr A + \lr B + \lr C + \lr D + \lr E + \lr F - 2(\lr \infty_1 + \lr \infty_2 + \lr \infty_3).$$
• The divisor of a rational function $p(x,y) / q(x,y)$ is simply the divisor of $p$ minus the divisor of $q$. Analogous fact holds for products too.
• Divisors coming from a rational function will be called principal divisors.

Naturally, we could ask the following questions:

(a) Can all divisors be realized by a rational function?

Well, it doesn’t take too much work to see that there is an invariant at play: if a divisor corresponded to an actual rational function, then for every additional real root we have, we must have an additional pole at infinity. In other words, all principal divisors have degree 0.

We could refine this and instead ask

(b) Can all degree 0 divisors be realized by a rational function?

It turns out that the answer is still no, but we’re getting closer to the meat of the problem. We could ask if something like $\lr P - \lr Q$ can be realized, but we realize that the answer is no: if our rational function is $\frac{A(x,y)}{B(x,y)}$, then we’re asking for $\{A(x,y) = 0, (x,y) \in \cK\}$ and $\{B(x,y) = 0, (x,y) \in \cK\}$ to differ by only 1 point, which we know to be impossible by the group law (since they must individually “sum to 0”).

To actually measure what this difference is, we can consider the quotient of degree 0 divisors by principal divisors. (This sounds scary, but really the quotient $D/P$ just means that we zero out all the elements of $P$ and see what we get.) So, corresponding to the two questions above: (a) We’ll call divisors modded out by principal divisors the Picard group, denoted $\Pic(\cK)$. (b) We’ll call the degree 0 divisors modded out by principal divisors the degree 0 Picard group, denoted $\Pic^0(\cK)$.

The classical group law says that there is an isomorphism $\cK \cong \Pic^0(\cK)$, and it looks like $$[P]_O \mapsto \lr P - \lr O.$$

On the other hand, our group law is equivalent to setting $\lr \infty_1 + \lr \infty_2 + \lr \infty_3 = 0$ in the full Picard group.

Proof of Cayley-Bacharach

We copy the proof from Chapter 11 in the Taiwanese geometry book.

Notice that we can pick $t_1\cK_1 + t_2\cK_2 + t_3 \cK_3$ such that $\overline{P_1P_2}$ is wholly contained in the combination, where not all of $t_1,t_2,t_3$ are zero. (This is easiest seen by a change of basis that sets $\overline{P_1P_2} = \{y=0\}$, then $(x-p_1)(x-p_2)$ divides $\cK_i|_{y=0}$.)

If there’s no third point on $P_1P_2$, this means that the other six points $P_{3-8}$ must lie on the same conic. Now we’ll show that $P_1P_2\cap (P_3...P_8)$ must lie in the linear combination of any of the two cubics, which is sufficient. Consider a point on the conic $P^*$, then some combination of $\cK_1, \cK_2$ must contain $P^*$. But a cubic can only intersect a conic at 6 points (due to Bezout), so said cubic contains the entire conic. Hence, this implies that the linear combination is $P_1P_2\cap (P_3...P_8)$ unless $P_1$ and/or $P_2$ is also on the conic, but then that would mean that the conic is in both cubics and the conclusion follows anyway.

If there’s a third point $P_3$ on line $P_1P_2$ but not a 4th, then we can find a linear combination of any two curves that contains all of $\overline{P_1P_2P_3}$. The other 5 points define a conic, so this linear combination is exactly $\overline{P_1P_2P_3}\cap (P_4...P_8)$. Thus, our three curves are linearly dependent.

If there’s 4th point $P_4$ on the line, then $\ell$ is on each curve. Then, this reduces to conics passing through 4 points and they’re well known to be linearly dependent.