Smooth Cubic Curves over ${\mathbb C} $ and ${\mathbb R} $

Smooth Cubic Curves over C and R.

arXiv:1603.09018v1 [math.AG] 30 Mar 2016

Araceli Bonifant1 and John Milnor2 March 31, 2016

Abstract An expository description of smooth cubic curves in the real or complex projective plane.

1

Introduction.

This note will present elementary proofs for basic facts about smooth cubic curves C in the complex projective plane P2 (C) , or the corresponding curves CR in the real projective plane P2 (R) when the defining equation has real coefficients. The presentation will center around two different normal forms, which we refer to as the Hesse form 3 x3 + y 3 + z 3 = 3 k x y z (using homogeneous coordinates), and the Weierstrass form y 2 = (x − r1 )(x − r2 )(x − r3 )

with

r1 + r2 + r3 = 0

= x3 + s2 x − s3 (using affine coordinates (x : y : 1) with z = 1 ), where the sj are elementary symmetric functions. This last form leads to an easily computed 1 Mathematics Department, University of Rhode Island; e-mail: [email protected] 2 Institute for Mathematical Sciences, Stony Brook University; e-mail: [email protected] ∗ Key words and phrases. elliptic curve, Hesse normal form, Weierstrass normal form, projective equivalence, tangent process, flex point. ∗ 2010 Mathematics subject classification. 14-03, 01-01, 14H52, 14H37, 14H50, 14P99 3 Our notations are non-standard, but convenient. Both Hesse and Weierstrass used slightly different expressions; and our J(C) is just Felix Klein’s invariant j(C) divided by 1728. (See [H], [H2], [Wei], [Wei2], [K].) The “Weierstrass normal form” was used already by Newton (see [Ne]).

1

2 invariant 3 J(C) =

4s23 , 4s23 + 27 s32

which measures the shape of the triangle with vertices rj . (For example, J ∈ R, J < 1 if and only if this triangle is isosceles; and J = 0 if and only if it is equilateral. Compare Figure 2.) We are particularly concerned with classification up to projective equivalence (that is up to a linear change of coordinates in the projective plane). The Hesse parameter k is a complete invariant for a real projective equivalence class of curves CR . However, each real J corresponds to two quite distinct real curves, or in other words to two different values of k . (Compare Figures 3 and 8.) On the other hand, in the complex case J(C) is a complete invariant for a projective equivalence class (and also for a conformal diffeomorphism class, where we ignore the particular choice of embedding into projective space). But for a generic choice of J , there are twelve corresponding possibilities for the k -parameter. More precisely, there is a group Γ consisting of twelve M¨ obius automorphisms of the k -plane, such that two different values of k correspond to the same value of J if and only if they belong to the same orbit under Γ . Flex points (= inflection points) play an important role, both in the real case, where there are always three flex points, and in the complex case where there are always nine. In both the real and complex cases, the group of all projective automorphisms of the curve contains a maximal abelian normal subgroup consisting of automorphisms with no fixed points on the curve. In both cases, this subgroup operates simply transitively on the set of flex points; and in both cases the group of automorphisms modulo this subgroup is cyclic: of order two in the real case, and of order either two, four, or six in the complex case. The paper is organized as follows. Sections 2 through 4 concentrate on the complex case. Section 2 studies classification up to projective equivalence, and computes the automorphism group. Section 3 discusses conformal classification, and shows that a conformal diffeomorphism from C to C 0 extends to a projective automorphism of P2 if and only if it maps flex points to flex points. Section 4 describes the chord-tangent map C × C → C and the related additive group structure on the curve. Finally, Section 5 describes real cubic curves. Notation: We use the notation (x, y, z) for a non-zero point of the complex 3-space C3 , and the notation (x : y : z) for the corresponding point of P2, representing the equivalence class consisting of all multiples

3 (λ x, λ y, λ z) with λ ∈ Cr{0} . However, it is sometimes convenient to represent a point of P2 by a single bold letter such as p . Note that any linear automorphism of C3 gives rise to a projective automorphism of the projective plane P2 . Note: For historical information, see [Do]; and for real cubic curves from an older point of view, see [Wh].

2

Two Normal Forms.

Otto Hesse in 1844 (see [H], [H2]) introduced the family of cubic curves C(k) consisting of all points (x : y : z) in the projective plane P2 = P2 (C) which satisfy the homogeneous equation x3 + y 3 + z 3 − 3 k x y z = 0 .

(1)

Given a generic point (x : y : z) in P2 , we can solve equation (1) for k =

x3 + y 3 + z 3 ∈ C ∪ {∞}. 3xyz

(Thus for k = ∞ we define C(∞) to be the locus xyz = 0 .) However, there are nine exceptional points where both x3 + y 3 + z 3 = 0

and

xyz = 0 ,

so that the parameter k is not uniquely defined. All of the curves C(k) pass through all of these nine points, which have the form (0 : 1 : −γ)

or

(−γ : 0 : 1)

or

(1 : −γ : 0)

with γ 3 = 1 .

(2)

In the complement of these nine exceptional points, the space P2 (C) is the disjoint union of the Hesse curves. Remark 2.1 (Singularities). By definition, a point of the curve Φ(x, y, z) = 0 is singular if the partial derivative Φx , Φy and Φz all vanish at the point. The curve is called smooth if it has no singular points. We will show that the Hesse curve C(k) has singular points if and only if either k 3 = 1 or

4 k = ∞ . For the curve C(k) with k finite, a singular point would satisfy the equations x2 = kyz , y 2 = kxz , z 2 = kxy , which imply that x3 = y 3 = z 3 = k xyz , and hence x3 y 3 z 3 = k 3 x3 y 3 z 3 . For k 3 6= 1 with k finite, it follows easily that there are no singularities. On the other hand, if k 3 = 1 , then it is not hard to check that C(k) is the union of three straight lines of the form αx + βy + z = 0 with α3 = β 3 = 1 and αβ = k . Hence it is singular at the three points where two of these lines intersect.4 Similarly the curve C(∞) is clearly singular at the three points (0 : 0 : 1) , (1 : 0 : 0) and (0 : 1 : 0) where two of the lines x = 0 , y = 0 , and z = 0 intersect. Thus altogether there are twelve points in P2 which are singular for one of these curves. If we remove these twelve singular points and also the nine exceptional points from P2 , then we obtain a smooth foliation by Hesse curves.

Definition 2.2. For any curve C ⊂ P2 , the automorphism group Aut(P2 , C) is the group consisting of all projective automorphisms of P2 which map C onto itself. Curves defined by the Hesse equations (1) are clearly highly symmetric. Following is a precise statement. Lemma 2.3. For any smooth Hesse curve C(k) the automorphism group Aut(P2 , C(k)) contains a normal abelian subgroup N ∼ = Z/3 ⊕ Z/3 which acts without fixed points on C(k) , and acts simply transitively on the set of nine “exceptional points” of equation (2) . The group Aut(P2 , C(k)) also contains an element ι of order two such that conjugation by ι maps each element of N to its inverse. Thus Aut(P2 , C(k)) contains at least 18 elements. In fact, we will show in Theorem 2.9 below that any smooth cubic curve C can be put into the form (1), so that the automorphism group Aut(P2 , C) always contains a corresponding 18 element subgroup. (In most cases, this is the full automorphism group, but for some special curves there are extra symmetries, as described in Corollary 2.15.) 4 For k = 1 , one such intersection point (1 : 1 : 1) is clearly visible to the upper right in Figure 1, even though the rest of the two intersecting lines lie outside of the real projective plane.

5

Fig. 1: “Foliation” of the real projective plane by the Hesse curves C(k) ∩ P2 (R) . Here P2 (R) is represented as a unit sphere with antipodal points identified. Note the three exceptional points (−1 : 1 : 0) , (0 : −1 : 1) , and (1 : −1 : 0) where all of the C(k) intersect. Also note the three singular points where the components of C(∞) = {x = 0} ∪ {y = 0} ∪ {z = 0} intersect, and note the isolated singular point at (1 : 1 : 1) ∈ C(1) . The figure has 120◦ rotational symmetry about this point. (This figure has been borrowed from our paper [BDM], which studies rational maps preserving such cubic curves.)

Proof of Lemma 2.3. A cyclic permutation of the three coordinates x, y, z clearly acts effectively on any curve of the form (1), yielding a cyclic group of order three which acts without fixed points. (This transformation has a fixed point (1 : 1 : 1) in P2 , but it has no fixed point in C(k) since k 6= 1 .) If γ is a primitive cube root of unity, then the transformation (x : y : z)

7→

(x : γ y : γ 2 z)

is another automorphism of order three which commutes with the cyclic permutation of coordinates. It is not difficult to check that the abelian group N generated by these two transformations acts without fixed points, and that it acts simply transitively on the set of exceptional points (2).

6 Finally, the permutation (x, y, z) ↔ (y, x, z) is an element ι of order two with the required properties. (This involution does fix one of the nine exceptional points, namely (1 : −1 : 0) , and also fixes the three points of C(k) which lie on the line x = y .) Flex Points. For any smooth cubic curve C ⊂ P2 and any line L ⊂ P2 , the intersection C ∩ L consists of three points, counted with multiplicity. Here: • A transverse intersection has multiplicity one. • The intersection between C and its tangent line at a generic point has multiplicity two. • At certain special “flex points” this tangential intersection will have multiplicity three. This is the case if and only if the tangent line has no other intersection with C .

Definition 2.4. An inflection point, or briefly a flex point is a point where the tangential intersection multiplicity is three (or is ≥ 3 for a curve of higher degree). Using affine coordinates (x : y : 1) , if we can express y locally as a smooth function y = f (x) , then (x : y : 1) is a flex point if and only if the second derivative f 00 (x) is zero. Theorem 2.5. Every smooth cubic curve has precisely nine flex points. (Compare [H2].) The proof will be based on the following discussion. Let Φ(x, y, z) = 0 be the defining homogeneous equation. Working in the affine plane with z = 1 , the equation Φx dx + Φy dy = 0 shows that dy/dx = − Φx /Φy . (Since C is smooth, the partial derivatives Φx and Φy cannot both vanish at any point of C in the affine plane.5 ) After interchanging the roles of x and y if necessary, we may assume that Φy 6= 0 , so that the derivative y 0 = dy/dx is finite. It follows that y can be expressed locally as a smooth function y = f (x) satisfying the identity Φ(x, f (x), 1) = 0 . 5

This follows from the identity xΦx + yΦy + zΦz = 3 Φ = 0 .

(3)

7 Differentiating equation (3) twice, we obtain Φxx + 2Φxy y 0 + Φyy (y 0 )2 + Φy y 00 = 0 ,

(4)

where y 0 and y 00 stand for the first and second derivatives f 0 (x) and f 00 (x) . As an example, at a point where y 0 = 0 , this takes the simple form y 00 = − Φxx /Φy . Thus a point with y 0 = 0 is a flex point if and only if the second partial derivative Φxx is zero at this point. For a more general statement, we need the following. Definition 2.6. The Hessian determinant is the homogeneous polynomial function   Φxx Φxy Φxz HΦ (x, y, z) = detΦyx Φyy Φyz  . (5) Φzx Φzy Φzz Lemma 2.7. The set Flex(C) consisting of all flex points in C is equal to the set of all points in C which satisfy the homogeneous equation HΦ (x, y, z) = 0 . Proof. We will work in the affine plane with z = 1 . Fixing some point in this plane, and interchanging the coordinates x and y if necessary, we can assume that Φy 6= 0 , so that we can solve locally for y as a function y = f (x) . It will be convenient to translate the x and y coordinates so that the specified point is (0 : 0 : 1) . It follows that the coefficient of z 3 in Φ must be zero, so that Φzz (0, 0, 1) = 0 . A brief computation then shows that HΦ = 2Φxy Φyz Φxz − Φxx Φ2yz − Φyy Φ2xz (6) at the point (0, 0, 1) . On the other hand, multiplying equation (4) by Φy2 , where Φy y 0 = −Φx , we easily obtain the equation Φ3y y 00 = 2Φx Φy Φxy − Φ2y Φxx − Φ2x Φyy .

(7)

Note however that Φxz (0, 0, 1) = 2Φx (0, 0, 1)

and

Φyz (0, 0, 1) = 2Φy (0, 0, 1) .

(8)

8 To see this, simply express Φ as a sum of monomials mijk = cijk xi y j z k . For most monomials of degree three, the derivatives ∂mijk /∂x and ∂ 2 mijk /∂x∂z both vanish at (0, 0, 1) . The only exception is the monomial m102 = c102 xz 2 , where taking the partial with respect to z adds an extra factor of two. Thus, after multiplying equation (7) by four, we can rewrite it using (8) as 4 Φy3 y 00 = 2Φxz Φyz Φxy − Φ2yz Φxx − Φ2xz Φyy , so that the right hand side is equal to the right hand side of equation (6). This proves that 4 Φ3y y 00 = HΦ

with

Φy 6= 0

(9)

at the given point. Thus a point of C is a flex point if and only if HΦ vanishes there, which proves Lemma 2.7. If a curve C1 of degree n1 and a curve C2 of degree n2 have only smooth transverse intersections, then it follows from B´ezout’s Theorem that the number of intersection points is precisely equal to the product n1 n2 . (See for example [BK, 6.1] or [Ha, pp. 36, 54].) We are intersecting two curves {Φ = 0} and {HΦ = 0} which both have degree three. Thus to complete the proof of Theorem 2.5. we need only show that these two curves have only smooth transverse intersections.6 Since equation (9) holds at all points of C with Φy 6= 0 , we can differentiate with respect to x to obtain 4 Φy3 y 000 =

∂HΦ ∂x

whenever y 00 = 0 . But y 00 and y 000 can never vanish simultaneously at a point of a smooth cubic curve, for this would imply that the entire tangent line at that point would have to be contained in C . Thus ∂HΦ /∂x 6= 0 . It follows easily that the two curves {Hφ = 0} and {Φ = 0} have a transverse intersection at every flex point. Thus every smooth cubic curve has exactly nine flex points, which completes the proof of Theorem 2.5. Remark 2.8. In the special case of a smooth Hesse curve C(k) , the nine flex points coincide with the nine “exceptional points” of equation (2). To see this, taking Φ(x, y, z) = x3 + y 3 + z 3 − 3 k x y z , note for example that 6

The curve {HΦ = 0} may have singular points, but is always non-singular near the intersection.

9 Φxx = 6 x and Φxy = −3 k z . A straightforward computation shows that the Hessian determinant is given by
HΦ (x, y, z) = 33 (8 − 2 k 3 )x y z − 2k 2 (x3 + y 3 + z 3 ) . If Φ(x, y, z) = 0 , then we can substitute 3kxyz for x3 + y 3 + z 3 on the right side of this equation. This yields HΦ (x, y, z) = 63 (1 − k 3 )xyz . If we are in the non-singular case, with k 3 6= 1 , then it follows that Φ = HΦ = 0 only at the nine points with x3 + y 3 + z 3 = x y z = 0 , as in (2).

Reduction to Hesse Normal Form. Recall that, two algebraic varieties in a projective space Pn are projectively equivalent if there exists a projective automorphism of Pn which carries one variety onto the other. The following result is taken from a textbook7 published by Heinrich Weber in 1898. Theorem 2.9. Every smooth cubic curve in P2 (C) is projectively equivalent to a curve C(k) in the Hesse normal form x3 + y 3 + z 3 = 3 k x y z , with k 3 6= 1 . Proof. Choose two distinct flex points for the given curve C , and choose homogeneous coordinates x, y, z so that: • the line x = 0 is tangent to C at the first flex point, • the line y = 0 is tangent at the second flex point, and • the line z = 0 passes through both flex points. 7

See [Web, v.3, p.22]. We don’t know whether this result was known earlier.

10 Working in affine coordinates with z = 1 , these conditions mean that C has no finite point on the lines x = 0 or y = 0 . In other words the polynomial function Φ(x, y, 1) must take a non-zero constant value on these two lines; say Φ(x, y, 1) = 1 whenever xy = 0 . Hence it must have the form Φ(x, y, 1) = 1 + xy(ax + by + c) . In homogeneous coordinates, this means that Φ(x, y, z) = z 3 + xy(ax + by) + cxyz . Furthermore a 6= 0 since otherwise (1 : 0 : 0) would be a singular point, and b 6= 0 since otherwise (0 : 1 : 0) would be singular. Now to put this polynomial in the required form, we must express xy(ax + by) as a sum of two cubes. In fact, consider the identity √
(γ p x + q y)3 + (−p x − γ q y)3 = 3i 3 xy − p2 q x + pq 2 y , √ where γ = (−1 + i 3)/2 . It is not difficult to choose p and q so as to satisfy the required equalities √ √ −3i 3 p2 q = a and 3i 3 pq 2 = b , since we can first solve for p/q = −a/b , and then solve for p . Remark 2.10. According to Remark 2.1, a curve in this form, with k finite, is smooth if and only if k 3 6= 1 . This Hesse form is not unique, since there are several different ways of choosing the two flex points. We will see later that generically there are twelve possible choices of the parameter k for a specified initial curve. Corollary 2.11. Every smooth cubic curve possesses an automorphism group of order at least 18 which acts transitively on its set of nine flex points, as described in Lemma 2.3 and Remark 2.8 .

Reduction to Weierstrass Normal Form. The following result was proved by Nagell in 1928. (See [Nag] or [Co]. However, the corresponding statement with conformal diffeomorphism in place of projective equivalence, was proved by Weierstrass much earlier. Compare Section 3.)

11 Theorem 2.12. Let p be a flex point in the smooth cubic curve C . Then there exist coordinates x, y, z for P2 so that (1) The given flex point p has coordinates (0 : 1 : 0) , and (2) The defining equation in the affine plane z = 1 has the form y 2 = (x − r1 )(x − r2 )(x − r3 ) = x3 + s2 x − s3

(10)

with r1 + r2 + r3 = 0 , where s2 = r1 r2 + r1 r3 + r2 r3

and

s3 = r1 r2 r3

are elementary symmetric functions. Furthermore, this normal form is unique up to the change of affine coordinates (x, y) 7→ (t2 x, t3 y) which replaces each rj by t2 rj and replaces each sj by t2j sj . Here t can be any non-zero complex number. Note. This normal form seems to depend on the choice of a flex point; but it follows immediately from Corollary 2.11 that the particular choice does not matter. Proof of Theorem 2.12: Existence. In homogeneous coordinates, equation (10) takes the form Φ(x, y, z) = − y 2 z + x3 + s2 x z 2 − s3 z 3 = 0 .

(11)

Thus the intersection of the line z = 0 with the curve is given by the equation x3 = 0 ; which means that this intersection consists of the single point (0 : 1 : 0) counted three times. In other words, (0 : 1 : 0) must be a flex point, and the line z = 0 must be its tangent line. Choose preliminary coordinates X, Y, Z so that the chosen flex point has coordinates (0 : 1 : 0) and so that the tangent line at this flex point is the line Z = 0 . Then the defining equation Ψ(X, Y, Z) = 0 for the curve must have the form a X 3 + Z ψ(X, Y, Z) = 0 with a 6= 0 , where ψ is homogeneous quadratic. Furthermore, the coefficient ψ(0, 1, 0) of Z Y 2 in this expression must be non-zero, since otherwise the equations ΨX = ΨY = ΨZ = 0 would have a common solution at (0 : 1 : 0) . Therefore, after multiplying X and Y by suitable constants,

12 we may assume that the equation, in the affine space Z = 1 , takes the form Y 2 = X 3 + bX 2 + cXY + pX + qY + h = 0 . Now, after replacing Y by a carefully chosen sum y = Y + αX + β , we can assume that c = q = 0 , so that the equation becomes y 2 = X 3 + bX 2 + pX + h . Then, after replacing X by a suitable sum x = X + constant , we obtain the required form y 2 = X 3 + pX + h , where we can set s2 = p and s3 = −h . This proves existence. Proof of Uniqueness. There was only one step in the discussion above where any choice was involved, namely the step where we multiplied X and Y by suitable constants. Suppose in fact that we replace X by x = X/α and replace Y by y = Y /β . Then the expression Y 2 = X 3 + · · · would be replaced by β 2 y 2 = α3 x3 + · · · Dividing this equation by β 2 , it follows that we must have α3 = β 2 in order to obtain the required form. Setting t = β/α , it follows easily that t2 = α and t3 = β . This proves uniqueness and completes the proof. The curve defined by equation (10) or (11) is not always non-singular. Here is a precise statement. Lemma 2.13. The projective curve defined by y 2 = (x−r1 )(x−r2 )(x−r3 ) is non-singular if and only if the three roots rj are distinct, or if and only if the discriminant ∆ = (r1 − r2 )2 (r1 − r3 )2 (r2 − r3 )2 = − (27 s32 + 4 s23 )

(12)

is non-zero. Proof. The right hand equality is standard and it can be verified by direct computation. With Φ(x, y, z) as in equation (11), we must check for points where all three partial derivatives Φx = 3x2 + s2 z 2 ,

Φy = −2yz ,

and

Φz = −y 2 + 2s2 xz − 3s3 z 2

are zero. If z = 0 , it follows easily that x = y = 0 . Thus for a nontrivial solution we can assume that z = 1 . The three equations then imply respectively that s2 = −3 x2 ,

y = 0,

and that

2s2 x−3s3 = 0 ,

hence

s3 = −2 x3 .

13 It follows immediately that 27s32 + 4s23 = 0 if and only if there is a singular point. Furthermore, whenever this equation is satisfied, the factorization x3 + s2 x − s3 = (x − r)(x − r)(x + 2r) is easily verified, with r = (3s3 )/(2s2 ) . Hence this polynomial does have a double root.

Corollary 2.14. Two different curves C and C 0 in the Weierstrass normal form are projectively equivalent if and only if there is a change of coordinates (x, y, z) 7→ (t2 x, t3 y, z) transforming one to the other, thus sending rj 7→ t2 rj ,

s2 7→ t4 s2 ,

s3 7→ t6 s3 .

It follows that the complex number J(C) =

4s23

4s23 + 27s32

(13)

is a complete invariant, in the sense that J(C) = J(C 0 ) if and only if C is projectively equivalent to C 0 . Proof. The denominator of equation (13) is just the discriminant, up to sign, so that it can never vanish for a smooth curve. The definition is chosen so that J(C) = 0 if and only if s2 = 0 , and J(C) = 1 if and only if s3 = 0 . In geometric terms, J(C) is zero if and only if the triangle with vertices r1 , r2 , r3 ∈ C is equilateral, and is +1 if and only if this triangle degenerates8 to a line segment, with one rj precisely in the middle. If we label the edge lengths |rj − rk | as 0 < e1 ≤ e2 ≤ e3 , then for a sequence of such triangles we have |J| → ∞ if and only if e3 /e1 → ∞ (see Figure 2). We can now give a precise description of the automorphism group of a complex cubic curve. 8

Our “triangles” are simply a convenient way of visualizing an unordered triple of distinct points in the complex plane. The case of a “degenerate” triangle, with collinear vertices, does not imply any degeneracy in the associated cubic curve. We say that two triangles have the same shape if there is a linear transformation z 7→ c z which maps one to the other.

14

J in upper half-plane

J 1 represents curves with two connected components.

In particular, it follows that the graph, shown in Figure 3 , is invariant under the involution
(k, J) ←→ η(k), J , which maps the region k < 1 to itself with fixed √ point (1 − the region k > 1 to itself with fixed point (1 + 3, 1) .

√

3, 1) , and


First Proof. The equation J η(k) = J(k) is an identity between two rational functions of degree twelve, which can be verified by direct computation.

17 However, this argument gives no clue as to how
to construct an actual projective equivalence between C(k) and C η(k) . That can be remedied as follows. Second Proof. Let X = x+y +z Y = x + γy + γz Z = x + γy + γz where γ = e2πi/3 . Then it is not hard to check that   X 3 + Y 3 + Z 3 = 3 x3 + y 3 + z 3 + 6 x y z , and that X Y Z = x3 + y 3 + z 3 − 3 x y z . Setting k = (x3 + y 3 + z 3 )/(3 x y z) , it follows easily that X3 + Y 3 + Z3 k+2 = , 3X Y Z k−1 and the conclusion follows. Theorem 2.17. Let Γ denote the group of M¨ obius transformations generated by the involution η and the rotation ρ(k) = γ k . Then Γ is equal to the twelve element tetrahedral group, consisting of all M¨ obius transformations from the Riemann sphere to itself which map the four point set {1, γ, γ, ∞} to itself. Furthermore: (1) Two Hesse curves C(k) and C(k 0 ) are projectively equivalent if and only if k 0 = µ(k) for some µ ∈ Γ . (2) The function k 7→ J(k) can be computed as J(k) =

1 Y µ(k) . 64 µ∈Γ

18 Proof. Clearly η : 1 ↔ ∞ under the involution η , and it is not hard to check that η : γ ↔ γ . It then follows easily that Γ can be identified with the group consisting of all even permutations of these four symbols. b To prove statement (1), note that the quotient C/Γ is a Riemann b b → C b clearly surface, necessarily isomorphic to C . Since the map J : C has the property that J ◦ µ = J for every µ ∈ Γ , it follows that J can be expressed as a composition b / C/Γ @@ @@ @ h J @@ 

b C @

b , C b → C/Γ b have degree twelve, it folSince both J and the projection C lows that the holomorphic map h has degree one, and hence is a conformal diffeomorphism. Since two curves C(k) and C(k 0 ) are projectively equivalent if and only if J(k) = J(k 0 ) , it follows that they are projectively equivalent if and only if k and k 0 have the same Q orbit under Γ . To prove (2), note that the function k 7→ µ µ(k) is also a rational map of degree twelve which is invariant under precomposition with any µ ∈ Γ . Furthermore, this function maps zero to zero and infinity to infinity, so it must be some constant multiple of J . To compute the precise constant, it is enough to understand one more example. √ It is not hard to check10 that the orbit√ of 1+ 3 consists of the following six points, each counted twice since 1 + 3 is a fixed point of η . √ √ √ √ √ √ 1 + 3, (1 + 3)γ, (1 + 3)γ, 1 − 3, (1 − 3)γ, (1 − 3)γ . √ √ √ Q Since (1 + 3)(1 − 3) = −2 and γγ = 1 , the product µ µ(1 + 3) is √ equal to (−2)6 = 64 . Comparing this with J(1 + 3) = 1 , the conclusion follows.

3

Conformal Structure.

Lemma 3.1. Every smooth cubic curve is conformally diffeomorphic to a flat torus of the form C/Ω where Ω is a lattice ( that is an additive subgroup generated by two complex numbers which are linearly independent √ √ In particular, note that ρ−1√ ◦ η ◦ ρ(1 + 3) = 1 − 3 . It is noteworthy that to pass between the two real points 1 ± 3 on the locus J = 1 we need to make use of a M¨ obius transformation with complex coefficients. 10

19 over R .) Here Ω is uniquely determined up to multiplication by a complex constant. Proof. We will use affine coordinates (x : y : 1) , and take the curve in the Weierstrass form y 2 = x3 + s2 x − s3 , so that 2 y dy = (3 x2 + s2 )dx . Following methods introduced by Abel, consider the holomorphic 1-form (or Abelian differential) dw which is defined by dw =

dx y

whenever

y 6= 0 ,

and by 2 dy whenever 3 x2 + s2 6= 0 . 3 x2 + s 2 (The two denominators cannot both be zero since the equations dw =

Φx = Φ y = 0 would imply that C is singular.) This form dw is clearly well defined and non-zero everywhere on C , except possibly at the flex point (0 : 1 : 0) , which is outside the affine plane. To study the behavior near this point, we must work with alternative affine plane in which y = 1 , setting x = X/Z

and y = 1/Z

Then dw =

so that

(x : y : 1) = (X : 1 : Z) .

dx d(X/Z) Z dX − X dZ = = . y 1/Z Z

Using the equation Φ(X, 1, Z) = − Z + X 3 + s2 XZ 2 − s3 Z 3 = 0 ,

(15)

we see that ΦX (0 : 1 : 0) = 0 and ΦZ (0 : 1 : 0) = −1 , so that X can be used as a local uniformizing parameter on C . In fact, we can express Z locally as a function of X of the form Z = aX n + (higher terms) , with n ≥ 2 . Substituting this expression for Z in the right hand side of the equation Z = X 3 + s2 XZ 2 − s3 Z 3 it follows easily that a = 1 and n = 3 , so that Z = X 3 + s2 X 7 + (higher terms) . A brief computation then shows that dw =


− 2 + O(X 4 ) dX .

20 Thus the homomorphic 1-form dw is smooth and non-zero, even at the flex point (0 : 1 : 0) . Passing to the universal covering space Ce, and choosing a base point p ∈ Ce, we can integrate to obtain a well defined holomorphic function Z w= dw : Ce → C . (16) p

Using the flat Riemannian metric |dw| , we see that this function (16) is a conformal isometry from Ce onto the complex numbers. In fact the inverse map from C to Ce sends each straight line from the origin in C to a corresponding geodesic in Ce. Now given any smooth closed loop Λ in C we can integrate to obtain R a complex number Λ dw . This yields a homomorphism from the fundamental group π1 (C) into C . Lemma 3.2. Let Λ vary over all closed differentiable oriented loops Λ : R/Z → C , and H let Ω be the additive subgroup of C consisting of all period integrals Λ dx/y . Then Ω forms a lattice in C . It follows that the map of Equation (16) induces a conformal isomorphism C → C/Ω . Proof. Using the Weierstrass normal form y 2 = (x−r1 )(x−r2 )(x−r3 ) , let L1 be the line segment from r1 to r2 in the x -plane, and let L2 be the line segment from r2 to r3 . Then over each p Lj we can choose y = y(x) as a continuous branch of the map x 7→ ± (x − r1 )(x − r2 )(x − r3 ) . We will show that Ω is the lattice generated by the two integrals Z Z ω1 = 2 dx/y and ω2 = 2 dx/y . (17) L1

L2

Think of C as a branched 2-fold covering space of the Riemann sphere, branched at the points r1 , r2 , r3 and ∞ , with projection map (x : y : 1) 7→ x , and with covering transformation ι(x : y : 1) = (x : −y : 1) . Note that the pre-image of each Lj under the projection map is a simple closed curve, |Λj | = Λj (R/Z) ⊂ C . Now express |Λj | as the union of two subarcs Aj and ι(Aj ) , where we traverse the first arc from (rj : 0 : 1) to (rj+1 : 0 : 1) , and the second arc from (rj+1 : 0 : 1) back to (rj : 0 : 1) . The integrals over these subarcs will then be equal to each other. In fact, to pass from one integral to the other one must simply change the sign of

21 both dx and y . It follows that the sum over these two subarcs will be precisely the expression ωj given by equation (17), provided that we choose the compatible orientation for Λj . Now consider an arbitrary loop in C . In fact it suffices to consider loops in the affine plane, since any loop can be pushed away from the branch point at infinity. It is then easy to deform such a loop into an arbitrarily small neighborhood of the union |Λ1 | ∪ |Λ2 | , and in fact to deform it onto this union. The resulting loop R/Z → |Λ1 | ∪ |Λ2 | may wander around |Λ1 | ∪ |Λ2 | many times, but the corresponding integral must always be equal to n1 ω1 + n2 ω2 for some integers n1 and n2 . Finally, we must check that ω1 and ω2 are R -linearly independent, so that Ω is indeed a lattice. Otherwise, if Rω1 = Rω2 , then the integral of dx/y from any fixed base point would yield a non-constant harmonic map from C to C/Rω1 ∼ = R , which is impossible since C is compact. The converse assertion, that every flat torus T = C/Ω is conformally diffeomorphic to a smooth cubic curve, is due to Weierstrass [Wei], [Wei2], and arose from his study of doubly periodic functions. Since this result is widely known (see for example [La, Sec.2]), we will give only a brief summary. For any lattice Ω ⊂ C the Weierstrass ℘ -function is the unique holomorphic Ω -periodic map from CrΩ to C which has a pole of the form ℘(w) = 1/w2 + o(1)

as

w→0.

This satisfies a differential equation of the form
2 ℘0 (w) = 4℘(w)3 − g2 ℘(w) − g3 , where the complex constants g2 and g3 can be computed from the lattice Ω . In fact, X 1 X 1 g2 = 60 and g3 = 140 , 4 ω ω6 ω6=0

ω6=0

where ω ranges over all non-zero lattice elements. (Compare [Se, p.83-84].) Setting (X : Y : Z) = (℘(w) : ℘0 (w) : 1) , this yields a conformal diffeomorphism from the torus T = C/Ω onto the cubic curve Y 2 = 4 X 3 − g2 X − g3 .

22 This can easily be transformed into our normal form (10), with coefficients √ 3 s2 = − g2 / 4 and s3 = g3 , √ by setting x = 3 4 X and y = Y . Felix Klein showed that the J -invariant can be computed as a function of the lattice Ω = Z ⊕ τ Z . See for example [Se, p.90]. Corollary 3.3. Two smooth cubic curves are projectively equivalent if and only if they are conformally diffeomorphic, and if and only if they share a common J -invariant. Proof. This follows easily from the discussion above. Corollary 3.4. The group Aut(C) ∼ = Aut(T) of conformal automorphisms of the curve C ∼ T can be described by a split exact sequence = b → Aut(T) → Aut(T, 0) → 1 , 1 → N b ∼ where the normal subgroup N = T of automorphisms without fixed point can be identified with the group of translations of T , and where the finite cyclic subgroup Aut(T, 0) ∼ = Aut(C, p) is naturally isomorphic to the group 2 Aut(P , C, p) of Corollary 2.15 . Remark 3.5. One curious invariant of the lattice Ω is the tiling of the complex plane by Voronoi cells Vω = ω + V0 , where V0 = V0 (Ω) is the compact convex polygon consisting of all z ∈ C such that |z| = min |z − ω| . ω∈Ω

This polygon V0 is a complete invariant for Ω , since Ω is generated by the reflections of zero in its edges. Similarly, the shape of V0 is a complete invariant for the conformal diffeomorphism class of C ∼ = T (where two polyhedra have the same “shape” if a complex linear transformation maps one to the other). In particular, the group Aut(T, 0) can be identified with the group of rotational symmetries of V0 . In most cases V0 is a hexagon, regular if and only if J = 0 . However V0 is a square if J = 0 , and a rectangle if J is real with J > 1 .

23

Fig. 4: Voronoi hexagon for the lattice Z ⊕ τ Z with τ = (3 + 4 i)/5 . The Voronoi polygon for any lattice has 180 ◦ rotational symmetry. In this example, since the lattice has two generators of equal length, it also has an orientation reversing symmetry, corresponding to a real J -invariant.

4

The Chord-Tangent Map and Additive Group Structure.

The chord-tangent map. Again let C ⊂ P2 be a smooth cubic curve. Recall that an arbitrary line L ⊂ P2 intersects C in exactly three points, counted with multiplicity. If these points are p, q and r , where a duplication such as p = q indicates that L is tangent to C at p , then we will refer to (p, q, r) as a collinear triple. Evidently any two points of a collinear triple determine the line L , and hence determine the third point. The resulting correspondence (p, q) 7→ r will be called the chord-tangent map, to be denoted by (p, q) 7→ p ∗ q .

(18)

Note that the equation p ∗ q = r is invariant under any permutation of p, q, r , and simply means that (p, q, r) is a collinear triple. Lemma 4.1. For any smooth cubic C , this chord-tangent map (p, q) → p ∗ q is holomorphic as a map from C × C to C . Proof. It is first necessary to show that the line L determined by two points p and q in C depends holomorphically on the pair (p, q) . This is clear if p 6= q , but we must also consider the limiting case as p and q

24 tend to a common limit. Using affine coordinates (x, y, 1) , and assuming that the slope s is finite, so that L is defined by an equation y = sx + c , it clearly suffices to prove that s depends holomorphically on p and q as p and q tend to a common point. Taking this limit point to be (0, 0, 1) , the curve C is defined locally by a power series y = a1 x + a2 x2 + · · · . Taking the two nearby points to be x1 =  and x2 = δ , the slope is given by s =

y1 − y2 x1 − x2

= =

a1 ( − δ) + a2 (2 − δ 2 ) + · · · −δ a1 + a2 ( + δ) + a3 (2 + δ + δ 2 ) + · · · ,

and therefore is locally holomorphic as a function of  and δ . Let Φ(x, y, 1) = 0 be the defining equation for the affine curve. Assuming that we have chosen coordinates so that the point r = p ∗ q also belongs to the affine plane, the function Φ(x, y, 1) restricted to the line y = sx + c determined by p and q can be expressed as a cubic polynomial Φ|L = c0 x3 + c1 x2 + c2 x + c3

with

c0 6= 0 ,

where the coefficients cj depend holomorphically on p and q . Factoring this polynomial as c0 (x − p)(x − q)(x − r) , we have p + q + r = −c1 /c0 . Therefore r = −p − q − c1 /c0 also depends holomorphically on p and q . Thus the x -coordinate of the required point r = p ∗ q ∈ L varies holomorphically, so r does also.

s = p+q

p

q

r o

Fig. 5: Constructing the sum of p and q .

Additive Group Structure. Lemma 4.2. Let o be an arbitrarily chosen base point11 in the smooth cubic curve C . Then C admits one and only one additive group structure with the following two properties: 11

The term elliptic curve is often reserved for a smooth cubic curve together with a specified base point.

25 • The base point o is the zero element, so that o + p = p for any p ∈ C. • The triple (p, q, r) is collinear ( as defined above ) if and only if the sum p + q + r takes a constant value which depends only on the choice of o . Proof of uniqueness. Assume that such a group structure exists. For any p and q , let r = p ∗ q and let s = r ∗ o as in Figure 5, using the notation (18). Then we have the identity p+q+r = r+o+s . Canceling the o and the r ’s, it follows that p + q = s , or in other words p + q = (p ∗ q) ∗ o .

(19)

This proves uniqueness. Proof of existence. Define the sum operation by the formula (19), setting r = p∗q and p+q = r∗o as illustrated by the diagram (Figure 5). Note the identity (p ∗ q) ∗ q = p for all p and q . In particular, taking q = o , we have p + o = (p ∗ o) ∗ o = p for all p . Thus o is indeed a zero element for the sum operation. For any collinear triple (p, q, r) , as in the diagram, we can compute the sum (p + q) + r = s + r = (s ∗ r) ∗ o = o ∗ o , which is constant, as required. This sum operation is clearly commutative, with inverse operation p 7→

−p=p∗o .

Over a general field, the proof of associativity is somewhat tricky. (Compare [Ca].) However, in the complex case it is quite easy: First note that for fixed q 6= o the mapping p 7→ p + q from C to itself has no fixed points. In fact, with r ∗ p = q and r ∗ (p + q) = o as in the diagram, the equation p = p + q would imply that q = o . ∼ = Now choose a conformal diffeomorphism ψ : C −→ T to some torus T = C/Ω , normalized by the requirement that ψ(o) = 0 . Then translation by q 6= o on C corresponds to a fixed point free conformal diffeomorphism from T to itself which maps zero to ψ(q) . But the only such isomorphism is

26 the translation by ψ(q) . It follows that the transformation ψ is not only a conformal diffeomorphism but also preserves the sum operation. Therefore the sum is associative; and ψ is an isomorphism of additive groups.

Fig. 6: The line between two flex points always intersects C in a third flex point. Remark 4.3. This construction is particularly convenient if we choose a flex point as base point o , so that o ∗ o = o , and so that p + q + r = o for any collinear triple. As an example, with this choice the classical12 “tangent process” p 7→ p ∗ p is given by the formula p 7→ − 2 p . With this choice, the flex points are precisely the elements of order three, satisfying p + p + p = o in the additive group. Comparing the corresponding subgroup of the torus T ∼ = C , we see that the nine flex points form a group isomorphic to Z/3 ⊕ Z/3 . One important consequence is that the line joining any two flex points must contain a third flex point (counted with multiplicity). Compare Figures 6, 7. Note. This Hesse configuration of nine points joined by twelve lines, as illustrated schematically in Figure 7, can never be realized by real straight lines, even in a high dimensional real space. However, it can be realized by nine points and twelve circles in R3 . (This is a reasonable 12

Cassels (see [Ca, Pg. 24]) remarks that the tangent process was known already to Diophantus.

27 model, since three points in a complex line do not usually lie on a real straight line, but always lie either on a real circle or a real line.)

c

d

a

a

b

b

c

d

Fig. 7:

A schematic picture of the Hesse configuration consisting of nine flex points together with the twelve lines joining them, placed on a square with opposite sides identified. ( Compare [H2, Lehrsatz 12], as well as [AD].) This configuration has the nice property that any two points determine a line and any two lines determine a point. This configuration cannot be realized by straight lines in R3 , but can be more or less realized on a flat torus, as illustrated.

5

Real Cubic Curves.

This section is concerned with smooth cubic curves CR ⊂ P2 (R) defined by equations Φ(x, y, z) = 0 with real coefficients. However, even when studying this real locus CR , we will always assume that the full complex locus C is nonsingular.13 The problem of classifying such real curves was studied already by Isaac Newton. (See [Ne] and compare [BK, p. 284].) In general, the real classification is parallel to the complex classification, but there are important 13

Thus we do not allow examples such as Φ(x, y, z) = x(x2 +y 2 +z 2 ) . (In this example, the real locus is just a non-singular line x = 0 ; but there are complex singular points at (0 : 1 : ±i) where the two irreducible components intersect.)

28 differences. In looking at pictures of real cubic curves, it is important to remember that the real projective plane is a non-orientable manifold, and that every real cubic curve has a non-orientable neighborhood, which can never be completely pictured within an affine plane. We first consider the Weierstrass normal form. Lemma 5.1. Every such curve CR is real projectively equivalent to a curve in the normal form y 2 = x3 + s 2 x − s 3 which is unique up to the transformation (x, y) 7→ (t2 x, t3 y) ,

(s2 , s3 ) 7→ (t4 s2 , t6 s3 )

with t real. In particular, the invariant J = (4s32 )/(4s32 + 27s23 ) is always real. Conversely however, each real J can be realized by two such curves CR which are distinct with respect to real projective equivalence. If J 6= 1 , a complete invariant for real projective equivalence is provided by J together with the sign of s3 . In the anomalous case where J = 1 we have s3 = 0 , but a complete invariant is then provided by J together with the sign of s2 . Proof. Since the associated complex curve C has an odd number of flex points, at least one must be invariant under complex conjugation (x : y : z) 7→ (x : y : z) within P2 (C) . It is then not difficult to choose a real representative (x, y, z) ∈ R3 ⊂ C3 . The reduction to Weierstrass normal form now goes through without difficulty in the real case, just as in the proof of Theorem 2.12. Examining Figures 2 and 8, we see easily that for real J there are exactly two ways of placing the corresponding triangle in the complex plane so that it is invariant under complex conjugation. In the isosceles case J < 1 , the vertex on the real axis can point either left or right; and in the case J > 1 , where the vertices of the “triangle” are collinear, we can put all three roots on the real line, but the middle root can be closer to either the left hand root or to the right hand root. This left-to-right flip correspond precisely to a change in sign of the product s3 = r1 r2 r3 . In the exceptional case J = 1 , the three roots can either be placed on the real axis or the imaginary axis; but in every case there are exactly two possibilities. Theorem 5.2. Every smooth real cubic curve CR is real projectively equivalent to the real Hesse curve C(k)R for one and only one real k 6= 1 . It follows that every such CR has exactly three flex points.

29

Fig. 8: Examples of pairs of distinct real curves in Weierstrass normal form which have the same J invariant. The top two have J = −.583 · · · and the bottom two J = 3.43 · · · . (If we rotate the triangle of Figure 2 with vertices r1 , r2 , r3 by 180◦ , then we must rotate the y coordinate by 90◦ . This maps the complex curve isomorphically onto itself, but has a drastic effect on the real curve.) In the figures, the curve in the real (x, y) -plane is shown in solid curves, and the triangle in the complex x -plane is shown below in dotted lines. For these four figures we have respectively: s3 < 0, k ≈ −3.91 ; s3 > 0, k ≈ −.582 ; s3 > 0 , k ≈ 1.63 ; and s3 < 0, k ≈ 5.75 .

Even if we use the Weierstrass normal form, the invariant k provides the most concise way of specifying a particular real curve. As examples, in the Weierstrass normal form we can solve for x as a smooth function of y (as in Figure 8, upper right) if and only if −2 < k < 1 ; and the curve CR is connected if and only if k < 1 . Proof of Theorem 5.2. Our proof of reduction to Hesse normal form

30

k = −20 (J = −124.5 · · · )

k = 2 (J = 1.492 · · · )

k = −2 (J = 0)

k = 4 (J = 1.492 · · · )

k = 0 (J = 0)

k = 10 (J = 16.05 · · · )

Fig. 9: Six canonical pictures of real curves in order of increasing k . ( See Remark 5.4.) The three asymptotic lines are also shown in each case. Top row: Connected examples: the case k < 1 . Bottom row: Examples with two components: k > 1 . The three asymptotic lines pass through a common point if and only if k = −2 . (In Weierstrass normal form, this corresponds to the case where the tangent lines at the two finite flex points are parallel.)

doesn’t work in the real case since it depends on the use of a primitive cube root of unity. However, we can bypass this difficulty by using the explicit computation of the map k 7→ J(k) : See equation (14) as well as Figure 3. This map is precisely two-to-one as a map from R to R . In fact, we claim that the two preimages of a real J correspond precisely to the two real CR for a given real J , as described in Lemma 5.1. It is not hard to check that the two preimages correspond to the two choices of sign for s3 , with √ √ s3 > 0 ⇐⇒ 1 − 3 < k < 1 + 3 . (Compare Figure 3.) The statement in the exceptional case s3 = 0 can also easily be checked.

31 The automorphism group of a smooth real cubic can be described as follows.
Corollary 5.3. The projective automorphism group Aut P2 (R), CR is non-abelian of order six and can be identified with the group of permutations of the three flex points of CR . That is, every permutation of these flex points
extends uniquely to a projective automorphism of the pair P2 (R), CR . Remark 5.4. Every real cubic curve can be presented by a canonical picture which makes its six symmetries evident. (See Figure 9.) Simply put the three flex points at equally spaced points on the line at infinity, and put the center of symmetry at the origin. The tangent lines at the three flex points will then appear as asymptotic lines. Proof of Corollary 5.3. Using the Hesse normal form, it follows easily that the permutations of the three coordinates yield a group of six automorphisms. But inspection of this canonical picture shows that there cannot be any other automorphisms.

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