Classical Invariant Theory

and

2-descent on Elliptic Curves

John Cremona University of Exeter

UK

Invariants of binary quartics

K: a field, char(K) ≠ 2, 3. Consider binary quartics in K[X, Y]: g(X, Y) = aX^{4} + bX^{3}Y + cX^{2}Y^{2} + dXY^{3} + eY^{4} which form a K-vector space of dimension 5. GL(2, K) acts on this via

\begin{matrix}( \begin{matrix}α & β \\ γ & δ \end{matrix} ): g(X, Y) & ↦ g(αX + βY , γX + δY) \\ (a, b, c, d, e) & ↦ (a^{∗}, b^{∗}, c^{∗}, d^{∗}, e^{∗}) \end{matrix}

Definition: An invariant of weight w of binary quartics is a polynomial I(a, b, c, d, e) such that

I(a^{∗}, b^{∗}, c^{∗}, d^{∗}, e^{∗}) = (αδ − βγ)^{w}I(a, b, c, d, e) for all transformations in GL(2, K).

Properties of Invariants

GL(2, K) is generated by matrices of the form

( \begin{matrix}α & 0 \\ 0 & δ \end{matrix} ) , ( \begin{matrix}1 & β \\ 0 & 1 \end{matrix} ) , and ( \begin{matrix}1 & 0 \\ γ & 1 \end{matrix} ) (or ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) ).

A) Invariance under ( \begin{matrix}α & 0 \\ 0 & δ \end{matrix} ) implies that I is

• homogeneous of degree d = \frac{2w}{4} = \frac{w}{2} in a, b, c, d, e.

• isobaric of weight w in a, b, c, d, e,

where  a b c d e
has weight 0 1 2 3 4

B) Invariance under ( \begin{matrix}1 & β \\ 0 & 1 \end{matrix} ) is then equivalent to satisfying the differential equation

ΩI = 4a\frac{∂I}{∂b} + 3b\frac{∂I}{∂c} + 2c\frac{∂I}{∂d} + d\frac{∂I}{∂e} = 0

C) Invariance under ( \begin{matrix}1 & 0 \\ γ & 1 \end{matrix} ) is leads to a similar condition.

Examples of Invariants

• for quadratics aX^{2} + bXY + cY^{2}, the “only” invariant is the discriminant: b^{2} − 4ac

• similarly for cubics

• but for quartics ... all invariants are generated by two of them, denoted I and J (which is called the “catalecticant”):

I = 12ae − 3bd + c^{2}

(degree 2, weight 4) and

J = 72ace + 9bcd − 27ad^{2} − 27eb^{2} − 2c^{3}

(degree 3, weight 6)

Every invariant of weight w is uniquely expressable as an isobaric polynomial in I and J:

∑_{4k+6l=w} c_{k,l}I^{k}J^{l}.

Seminvariants and their syzygy

A seminvariant is a polynomial in the coefficients a, b, c, d, e which is homogeneous, isobaric, and satisfies just the first differential equation ΩI = 0. So seminvariants are invariant with respect to the translation subgroup (X, Y) ↦ (X + βY, Y).

For quartics, there are 5 generating seminvariants:

• the invariants I and J;

• the leading coefficient a;

• p = 3b^{2} − 8ac, and r = b^{3} + 8a^{2}d − 4abc.

These are not algebraically independent, but are related by a single identity called a syzygy:

p^{3} − 48a^{2}pI − 64a^{3}J = 27r^{2}

[Proof: exercise using your favourite computer algebra package! cf. Cayley (1856), Hermite (?)]

This syzygy gives a link to elliptic curves.

The elliptic curve connection

Fix I, J ∈ K such that Δ = 4I^{3} − J^{2} ≠ 0. The equation

E_{I,J} : Y^{2} = X^{3} − 27IX − 27J

defines an elliptic curve over K, and we are interested in the set E_{I,J}(K) of K-rational points, which forms a group.

• If we have a quartic g(X, Y) with invariants I, J whose leading coefficient a is a non-zero square in K, then the syzygy says that

(X, Y) = (\frac{3p}{4a}, \frac{27r}{8a\sqrt{a}}) ∈ E_{I,J}(K)

• If a is not a square, we do not obtain a K-rational point. In this case we try to replace g by an equivalent quartic whose leading coefficient is a square.

• This is possible if and only if there is a K-rational point on the curve

C : Z^{2} = g(X, Y)

In this case we say that the quartic g is soluble.

• Hence we obtain a map ξ : C → E_{I,J} (of degree 4) mapping rational points to rational points.

2-descent on the elliptic curve

The curve C attached to each quartic g with invariants I, J is a curve of genus 1, which is a principal homogeneous space for E_{I,J}. The map ξ : C(K) → E_{I,J}(K) defined by the syzygy is called a 2-covering of E_{I,J}, and the study of E_{I,J}(K) by means of its 2-coverings is called 2-descent.

• The image of ξ is (either empty or) a complete coset of 2E_{I,J}(K) in E_{I,J}(K)

• Every point in E_{I,J}(K) is in the image of ξ for a unique GL(2, K)-equivalence class of quartics

• Hence we have a bijection between

{ Soluble quartics with invariants I, J, modulo GL(2, K)- equivalence } ⟷ E_{I,J}(K)/2E_{I,J}(K)

Thus by studying equivalence classes of quartics we can obtain group-theoretical information about E_{I,J}(K).

2-descent (continued): the number field case

If K is a number field, then E(K) is a finitelygenerated abelian group, called the Mordell–Weil group. So then

#(E(K)/2E(K)) = 2^{r+t}

where t ∈ {0, 1, 2} and 2^{t} is the order of the 2torsion subgroup of E.

We are interested in computing the rank r for a given curve. This can be divided into three parts:

• Find “all” quartics with invariants I, J K = ℚ: Birch and Swinnerton-Dyer 1963 K = ℚ(\sqrt{D}), D > 0: Serf 1995 K = ℚ(\sqrt{D}), D < 0: in progress [K : ℚ] > 2 ?

• Test for GL(2, K)-equivalence: see below.

• Test for solubility: check local solubility (Hensel’s lemma), then search for points on the curves C...

To understand equivalence, we must use some Galois Theory.

The Galois Connection

Recall: I, J ∈ K with 4I^{3} − J^{2} ≠ 0 define the elliptic curve E_{I,J} : Y^{2} = F (X) = X^{3} − 27IX − 27J.

Assume: F is irreducible, so E_{I,J}(K) has no 2torsion. Let φ be a root of F (−3X):

φ^{3} − 3Iφ + J = 0.

• [K(φ) : K] = 3;

• splitting field of F is L = K(φ, φ′, φ′′);

• [L : K] = 6 (or 3).

Let g be a quartic with invariants I, J and roots {x_{1}, x_{2}, x_{3}, x_{4}}, and set M = K(x_{1}, x_{2}, x_{3}, x_{4}).

M = K(x_{1}, x_{2}, x_{3}, x_{4}) = L(\sqrt{z}, \sqrt{z′}, \sqrt{z′′})

6 4 V_{4}
L = K(φ, φ′, φ′′)
K(x_{1}) 2 S_{4}
K(z) = K(φ)
4 3 S_{3}
K

( 1 → V_{4} ↪ S_{4} → S_{3} → 1 )

The Galois Connection (continued)

Idea: the fields M depend on the quartic g, but only up to GL(2, K)-equivalence. They all have the same resolvent cubic subfield L.

• M/L is biquadratic: Gal(M/L) ≅ V_{4}

• M = L(\sqrt{z}, \sqrt{z′}, \sqrt{z′′}), where

z ∈ K(φ)^{∗} \ (K(φ)^{∗})^{2}

and

Norm(z) = zz′z′′ ∈ (K^{∗})^{2}

Hence all such fields M are parametrized by (the non-trivial elements in) the group

ker (Norm : K(φ)^{∗}/ (K(φ)^{∗})^{2} → K^{∗}/ (K^{∗})^{2})

which is isomorphic to the Galois cohomology group H^{1}(G_{K}, E_{I,J}[2]).

Some explicit formulas

Construction of z from g: Given a quartic g with roots x_{i} and seminvariants a, p we have

\begin{aligned}z & = a^{2}(x_{1} + x_{2} − x_{3} − x_{4})^{2} = \frac{1}{3}(4aφ + p) \\ z′ & = a^{2}(x_{1} − x_{2} + x_{3} − x_{4})^{2} \\ z′′ & = a^{2}(x_{1} − x_{2} − x_{3} + x_{4})^{2} \end{aligned}

We call z the cubic seminvariant of g; its minimum polynomial is

h(Z) = Z^{3} − pZ^{2} + qZ − r^{2}

where p, q = \frac{1}{3}(p^{2} − 16a^{2}I) and r are all seminvariants of g.

Construction of g from z: Given z ∈ K(φ) with square norm r^{2} and linear in φ, we set a = \frac{3}{4}(z − z′)/(φ − φ′), b = 0 and obtain the coefficients c, d, e from the seminvariants p = 3z − 4aφ, r and I. Alternatively, we may define the roots x_{i} by 4ax_{i} = ±\sqrt{z} ± \sqrt{z′} ± \sqrt{z′′} and hence obtain a suitable g with invariants I, J.

Equivalence of Quartics

Proposition: Let g, g^{∗} be quartics with invariants I, J and cubic seminvariants z, z^{∗}. Then

g ∼_{GL(2,K)} g^{∗} ⟺ zz^{∗} ∈ (K(φ)^{∗})^{2} .

Proof:

( \begin{matrix}1 & β \\ 0 & 1 \end{matrix} ) : z ↦ z; ( \begin{matrix}α & 0 \\ 0 & δ \end{matrix} ) : z ↦ (α/δ)^{2}z;

( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) : z ↦ z^{∗}, where

zz^{∗} = ((2φ^{2} − 2cφ + 9bd − 4c^{2})/9)^{2}

Conversely, if z, z^{∗} ∈ K(φ) are equal modulo squares, the fields M, M^{∗} are equal; if z and z^{∗} are both linear in φ, then the cross-ratios of the x_{i} and the x_{i}^{∗} are both equal (to \frac{φ−φ′′}{φ′−φ′′}), so there is a matrix A ∈ GL(2, M) taking the x_{i} to the x_{i}^{∗}; and a little Galois theory shows that we may take A ∈ GL(2, K).

We can turn this into a simple algebraic test for equivalence of quartics: to test if zz^{∗} is a square, we form a third quartic (using the seminvariants of the two quartics) and test if it has a root in K.

Trivial Quartics

The whole picture degenerates when g has a root x_{1} in K, or equivalently when z is a square in K(φ): then K(x_{1}) = K and M = L.

Equivalent conditions:

• z is a square in K(φ) = K(z); this occurs when h(Z^{2}) = −h_{1}(Z)h_{1}(−Z) for some h_{1}(Z) ∈ K[Z]. Writing h_{1}(Z) = Z^{3} + uZ^{2} + vZ − r, this is possible iff u is a root of the quartic (U^{2} − p)^{2} + 8rU − 4q = a^{−1}g(U − b, 4a), so is equivalent to

• g has a root in K; this in turn is equivalent (for soluble g) to:

• the point P ∈ E_{I,J}(K) attached to g is in 2E(K); for g is (essentially) the 2-division polynomial of P. [Alternatively, y(P)^{2} = ∏(x(P) + 3φ) = ∏(9z/4a); by a well-known criterion, P is a double iff each factor is a square.]

• The cocycle attached to g in H^{1}(G_{K}, E[2]) is trivial.

Hence trivial quartics (which form a single GL(2)- orbit) may be easily recognised.

Group structure on quartics

The equivalence principle means that the set of (equivalence classes of) soluble (resp. locally soluble) quartics with given invariants I, J forms an elementary abelian 2-group, isomorphic to E_{I,J}(K)/2E_{I,J}(K) (resp. the 2-Selmer group of E_{I,J}/K). We would like to make this group structure explicit in terms of the quartics.

(g_{1}, g_{2}) ⟶ (z_{1}, z_{2}) ⟶ z_{3} = z_{1}z_{2} ⟶^{?} g_{3} = g_{1} ∗ g_{2}

• Problem: since z_{3} is not linear in φ, the resulting quartic does not have invariants I, J.

• Solution: replace z_{3} = z_{1}z_{2} by z_{3} = z_{1}z_{2}w^{2}, where w ∈ K(φ) is chosen to make z_{3} linear in φ. This can be done provided that the g_{i} are both soluble. Existence of w is given by the existence of a nontrivial zero of a ternary quadratic form. Hence it is even sufficient for the g_{i} to be everywhere locally soluble. It is also sufficient for the product of the leading coefficients a_{1}a_{2} to be a square in K (possibly after replacing the g_{i} by equivalent quartics). Given three quartics g_{1}, g_{2}, g_{3} we can test whether g_{3} ∼ g_{1} ∗ g_{2}, by testing whether z_{1}z_{2}z_{3} is a square in K(φ).

Questions

• Question 1: Can this group law be extended to all quartics with invariants I, J regardless of solubility?

• Question 2: Can we use the group structure to make 2-descent algorithms more efficient? It would suffice to find quartics which generated the group of soluble (or everywhere locally soluble) quartics; can we use this to reduce the size of the search region?

• Question 3: Can we extend to other number fields the reduction theory which bounds the search region in (a, b, c, d, e)-space, and hence obtain workable algorithms for doing 2-descents over these fields?

• Question 4: Can we use invariant theory for higher descents? Yes: See Merriman–Siksek–Smart (+C) for 4descents and beyond.