A practical guide to the group law

(David here.) We've had at least two articles on elliptic curves so far: Zhao Yu's article talks about some neat applications across various areas of math, while Dylan's article focuses on motivating the group law. Recently, I found out that the group law can be used to tackle actual Olympiad geometry problems (!), so I thought I might write a guide as to how.

Like my last article, this is also something sitting in the intersection of geometry and algebra.

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A sample

I should probably back up my claim that there are some geometry problems out there than can be tackled using the group law. Here's a sample from a 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 does any of this mean? 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).

and then it is claimed that this "+" makes the elliptic curve into an abelian group. But evidently, this is 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 $[x:y:0]$.

Put in a different way, 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. You can think of these as "variables" (in the sense that a polynomial $P(t)$ has a variable $t$ in it)
• We're allowed to add terms together, e.g. $[X] + [Y]$. These possibly do not refer to points any more, but we'll keep them around. Relatedly, you can also multiply by a positive integer in the obvious way, e.g. $2[X] := [X] + [X]$.
• 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 $[X] + [Y] + [Z] = 0$). Conveniently, this also provides an "additive inverse", so we can always talk about $-[X]$.

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.

Circling back (curving back?)

We return to try to understand the sample now.

First, this solution is actually solving a more general problem:

Given triangle $ABC$, $E, F$ are two points on $AC, AB$ respectively. Consider any point $D$ and let $K = (DEC)\cap(DFB)$. Let $T$ be the intersection point of $DI$ and a line through $A$ and parallel to $FE$. Let $P = (ATD)\cap(FED)$ and $Q= (PEC)\cap(PFB)$. Prove that $T$ lies on $PQ$.

In my head, this is saying that if we construct $P$ as given, and then set $D'=P$ and $I'=Q$, we have that it intersects the (green) line through $A$ parallel to $FE$ at a fixed point.

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.

In the diagram, the locus $\gamma$ is the pink curve.

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

This is presumably not hard to do by coordinates - but of course as a purist I'm not completely satisfied by this.

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 $\gamma$.

This is saying that we can pick $X$ approaching e.g. $E$ such that the circles $(XEC), (XBF)$ intersect at $Y$ on $\gamma$. By rotating $X$ around $E$, we realize that we can find a suitable direction along which it can approach $E$. I'm a little less convinced that the same argument would work for $I,J$.

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, $X + Y + B+ F + I + J = 0$.

Clearly, we want to use the locus property of $X$ here, which are:

• (1) $(XEC), (XBF)$ intersect at $Y$
• (2) $X,Y,T$ are collinear So this deals with (1). The addition here is referring to our "bracketed point" version, e.g.: $$[X] + [Y] + [E] + [C] + [I] + [J] = 0$$ The circle $(XEC)$ intersects $\gamma$ at 4 known points ($X,Y,E,C$), and the two additional points are in fact the mysterious "circular points at infinity" $I, J$ which every circle must pass through (projectively!). For some unknown reason, we also know that $I, J$ must pass through $\gamma$ - this is again not hard to check by coordinates.

We conclude that in particular, $E + C = B + F$, so $EC$ and $BF$ intersect on $\gamma$ - but their intersection is $A$. Thus $A \in \gamma$.

We're using the fact now that the group law allows us to set up these equalities and "cancel" things on both sides.

Moreover, since $X, Y, T$ are collinear, we get that $X + Y + T = 0$. Thus $E + C + I + J = B + F + I + J = T$.

This is (2). So, we get that $T$ are the sums of certain sets of 4 points (and recall the sums of 4 points are always actual points).

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 $E + F = A + T$.

Magically, the definition of $P$ is also amenable to the group law, and it reduces to showing that the two lines $EF$ and $AT$ intersect on $\gamma$. But, these are parallel lines...

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!

so the final connection here was that their intersection point is the infinite point in that direction, which is "collinear" with $I,J$. So we're done!

Epilogue

We have some missing pieces left, notably that:

• We don't know that the locus is a cubic curve.
• We don't know that $I,J$ are actually on the curve.

It turns out that these are all part of a bigger picture about special cubic curves which contain the circular points at infinity, and it would make sense to call these circular cubic curves. To be continued!