Antecedentes

Consider a plane cubicC =V(f)with equation in the Hesse canonical form (3.7). The partials of1₃f are

t2₀+ 2αt1t2, t21+ 2αt0t2, t22+ 2αt0t1. (3.22) Thus the Hessian ofChas the following equation:

He(C) = t0 αt2 αt1 αt2 t1 αt0 αt1 αt0 t2 = (1 + 2α3)t0t1t2−α2(t30+t 3 1+t 3 2). (3.23) In particular, the Hessian of the member of the Hesse pencil corresponding to the parameter(λ, µ) = (1,6α), α6= 0,is equal to

t3₀+t3₁+t3₂−1 + 2α 3

α2 t0t1t2= 0, (3.24) or, if(λ, µ) = (1,0)or(0,1), then the Hessian is equal toV(t0t1t2).

Lemma 3.2.4 LetCbe a nonsingular cubic in a Hesse’s canonical form. The following assertions are equivalent:

(i) dimSing(Pa(C))>0; (ii) a∈Sing(He(C));

(iii) He(C)is the union of three nonconcurrent lines; (iv) Cis isomorphic to the Fermat cubicV(t3

0+t31+t31); (v) He(C)is a singular cubic;

(vi) Cis an equianharmonic cubic; (vii) α(α3_{−1) = 0.}

Proof Use the Hesse equation for a cubic and for its Hessian. We see that He(C)is singular if and only if eitherα= 0or 1 + 8(−1+2α3

6α2 )

3 _{= 0. Ob-} viously,α= 1is a solution of the second equation. Other solutions are, 2. This corresponds to He(C), whereCis of the formV(t3

0+t31+t31), or is given by the equation t30+t31+t32+ 6it0t1t2= (it0+t1+t2)3+ (t0+it1+t2)3 +(t0+t1+it2)3= 0, wherei= 1,2, or t3₀+t3₁+t₂3+ 6t0t1t2= (t0+t1+t2)3+ (t0+t1+2t2)3 +(t0+2t1+t2)3= 0.

This computation proves the equivalence of (iii), (iv), (v), and (vii).

Assume (i) holds. Then the rank of the Hessian matrix is equal to 1. It is easy to see that the first two rows are proportional if and only ifα(α3_{−1) = 0.} Thus (i) is equivalent to (vii), and hence to (iii), (iv), (v) and (vii). The pointa

is one of the three intersection points of the lines such that the cubic is equal to the sum of the cubes of linear forms defining these lines. Direct computation shows that (ii) holds. Thus (i) implies (ii).

Assume (ii) holds. Again the previous computations show thatα(α3_{−1) =} 0and the Hessian curve is the union of three lines. Now (i) is directly verified. The equivalence of (iv) and (vi) follows from Theorem3.1.3since the trans- formation[t0, t1, t2]→[t1, t0, e2πi/3t2]generates a cyclic group of order 6 of automorphisms ofCleaving the point[1,−1,0]fixed.

Corollary 3.2.5 Assume that C = V(f)is not projectively isomorphic to the Fermat cubic. Then the Hessian cubic is nonsingular, and the mapa 7→ Sing(Pa(C))is an involution onHe(C)without fixed points.

Proof The only unproved assertion is that the involution does not have fixed points. A fixed pointahas the property thatDa(Da(f)) = Da2(f) = 0. It

follows from Theorem1.1.5that this implies thata∈Sing(C).

Remark3.2.6 Consider the Hesse pencil of cubics with parameters(λ, µ) = (α0,6α1) C(α0,α)=V(α0(t 3 0+t 3 1+t 3 2) + 6α1t0t1t2). Taking the Hessian of each curve from the pencil we get the pencil

H(α0,α)=V(α0t 3 0+t 3 1+t 3 2+ 6α1t0t1t2).

The mapC(α0,α)→H(α0,α)defines a regular map h:_P1→_P1_{, [α} 0, α1]7→[t0, t1] = [−α0α21, α 3 0+ 2α 3 1]. (3.25) This map is of degree3. For a general value of the inhomogeneous parameter

λ=t1/t0, the preimage consists of three points with inhomogeneous coordi- nateα=α1/α0satisfyfing the cubic equation

6λα3−2α2+ 1 = 0. (3.26)

We know that the points[α0, α1] = [0,1],[1,−₂1],[1,−₂],[1,−

2

2]correspond to singular members of theλ-pencil. These are the branch points of the maph. Over each branch point we have two points in the preimage. The points

(α0, α1) = [1,0],[1,1],[1, ],[1, 2]

are the ramification points corresponding to equianharmonic cubics. A non- ramication point in the preimage corresponds to a singular member.

LetCα=C(1,α). If we fix a group law on aHα=He(Cα), we will be able to identify the involution described in Corollary3.2.5with the translation auto- morphism by a nontrivial 2-torsion pointη. Given a nonsingular cubic curveH

together with a fixed-point-free involutionτ, there exists a unique nonsingular cubicCαsuch thatH =Hαand the involutionτ is the involution described in the corollary. Thus the three roots of Equation (3.26) can be identified with 3 nontrivial torsion points onHα. We refer the reader to Exercise 3.2 for a reconstruction ofCαfrom the pair(Hα, η).

Recall that the Cayleyan curve of a plane cubicC is the locus of linespq

in the dual plane such thata ∈ He(C)andbis the singular point ofPa(C). Each such line intersects He(C)at three pointsa, b, c. The following gives the geometric meaning of the third intersection point.

Proposition 3.2.7 Letcbe the third intersection point of a line`∈Cay(C) andHe(C). Then`is a component of the polarPd(C)whose singular point is

c. The pointdis the intersection point of the tangents ofHe(C)at the pointsa

andb.

Proof From the general theory of linear system of quadrics, applied to the net of polar conics ofC, we know that`is a Reye line, i.e. it is contained in some polar conicPd(C)(see subsection1.1.7). The pointdmust belong to He(C) and its singular pointcbelongs to`. Thuscis the third intersection point of`

withC.

It remains for us to prove the last assertion. Chose a group law on the curve He(C)by fixing an inflection point as the zero point. We know that the Steine- rian involution is defined by the translationx 7→ x⊕η, whereη is a fixed

2-torsion point. Thusb=a⊕η. It follows from the definition of the group law on a nonsingular cubic that the tangents_Ta(He(C))andTb(He(C))intersect at a pointdon He(C). We haved⊕2a = 0, henced = −2a. Sincea, b, c

lie on a line, we getc = −a−bin the group law. After subtracting, we get

d−c = b−a = η. Thus the pointsxandc are an orbit of the Steinerian involution. This shows thatc is the singular point ofPd(C). By Proposition 1.2.5,Pd(C)contains the pointsa, b. Thusabis a component ofPd(C).

It follows from the above Proposition that the Cayleyan curve of a nonsin- gular cubicC parameterizes the line components of singular polar conics of

C. It is also isomorphic to the quotient of He(C)by the Steinerian involution from Corollary3.2.5. Since this involution does not have fixed points, the quo- tient map He(C)→Cay(C)is an unramified cover of degree 2. In particular, Cay(C)is a nonsingular curve of genus 1.

Let us find the equation of the Cayleyan curve. A line`belongs to Cay(X) if and only if the restriction of the linear system of polar conics of X to`

is of dimension 1. This translates into the condition that the restriction of the partials ofX to`is a linearly dependent set of three binary forms. So, write

`in the parametric form as the image of the mapP1 →P2given by[u, v] 7→

[a0u+b0v, a1u+b1v, a2u+b2v]. The condition of the linear dependence is given by det   a2₀+ 2αa1a2 2a0b0+ 2α(a1b2+a2b1) b20+ 2αb1b2 a2 1+ 2αa0a2 2a1b1+ 2α(a0b2+a2b0) b21+ 2αb0b2 a22+ 2αa0a1 2a2b2+ 2α(a0b1+a1b0) b22+ 2αb0b1  = 0.

The coordinates of`in the dual plane are

[u0, u1, u2] = [a1b2−a2b1, a2b0−a0b2, a0b1−a1b0].

Computing the determinant, we find that the equation of Cay(X)in the coor- dinatesu0, u1, u2is

u3₀+u3₁+u3₂+ 6α0u0u1u2= 0, (3.27) whereα0 = (1−4α3_{)/6α. Note that this agrees with the degree of the Cay-} leyan curve found in Proposition1.1.26. Using formula (3.9) for the absolute invariant of the curve, this can be translated into an explicit relationship be- tween the absolute invariant of an elliptic curveC and the isogenous ellip- tic curve C/(τe), where τe is the translation automorphism by a nontrivial 2-torsion pointe.