Curves of genus 2 and Desargues configurations

(de Gruyter 2002

Curves of genus 2 and Desargues configurations

Dan Avritzer and Herbert Lange*

(Communicated by K. Strambach)

1 Introduction

A Desargues configuration is the configuration of 10 points and 10 lines of the clas- sical theorem of Desargues in the complex projective plane. For a precise definition see Section 2. The Greek mathematician C. Stephanos showed in 1883 (see [10]) that one can associate to every Desargues configuration a curve of genus 2 in a canonical way. Moreover he proved that the induced map from the moduli space of Desargues configurations to the moduli space of curves of genus 2 is birational. Our main mo- tivation for writing this paper was to understand the last result. Stephanos needed about a hundred pages of classical invariant theory to prove it. We apply instead a simple argument of Schubert calculus to prove a slightly more precise version of his result.

LetM_Ddenote the (coarse) moduli space of Desargues configurations. It is a three- dimensional quasiprojective variety. On the other hand, let M₆^b denote the moduli space of stable binary sextics. There is a canonical isomorphismH2nD1GM₆^b (see [1]). Here H2 denotes the moduli space of stable curves of genus 2 in the sense of Deligne–Mumford andD₁the boundary divisor parametrizing two curves of genus 1 intersecting transversely in one point. So instead of curves of genus 2, we may speak of binary sextics. The main result of the paper consists of the following two state- ments:

(1)There is a canonical injective birational morphism F:MD,!M₆^b: (2)We determine a hypersurface HHM₆^bsuch that

M₆^bnHJimðFÞJM₆^b; where both the inclusions above are strict.

* Both authors would like to thank the GMD (Germany) and CNPq (Brasil) for support during the preparation of this paper.

A Desargues configuration is called special if one of its lines contains 4 configura- tion points. We show:

(3)A Desargues configuration D is special if and only if the binary sexticFðDÞadmits a double point,i.e. the corresponding curve of genus2is not smooth.

The moduli space M_D admits a natural compactificationM_D. We also study the configurations corresponding to boundary pointsM_DnMD, which we calldegenerate Desargues configurations. In fact, there are degenerate Desargues configurations of the first, second and third kind (see Section 2). As a rational map of normal projec- tive varieties F:M_D !M₆^b is defined in codimension 1. Hence it extends to a morphism on an open set of the divisorMDnMD. We show, however:

(4)The geometric interpretation of the morphismF:MD !M₆^bdoes not extend to an open set U: MDHUHMD, where the first inclusion is strict. In fact, the binary sextic associated to a degenerate Desargues configuration of the first kind in an analo- gous way is not semistable.

In Section 2, we construct the moduli space M_D and its compactification M_D. In Section 3, we give the definition of the map F:M_D!M₆^b. Sections 4, 5, 6 and 7 contain the proofs of statements (1), (2), (3) and (4) respectively.

The second author would like to thank W. Barth with whom he discussed the subject, already 15 years ago.

2 The moduli space of Desargues configurations

Let P₂ denote the projective plane over the field of complex numbers. The classical theorem of Desargues says: If the lines joining corresponding vertices of two triangles A1;B1;C1 and A2;B2;C2 in P2 meet in a point A then the intersections of corre- sponding sides lie on a lineaand conversely (see Figure 1).

The triangles are then said to bein perspective,Ais called thecenter of perspective and a the axis of perspective. The configuration consisting of the 10 points A;Ai; Bi;Ci,ði¼1;2;3Þand ten lines, namely the 6 sides of the triangles, the 3 lines joining Ato the vertices of the triangle and the axis of perspective, is called aDesargues con- figuration. It is a 103-configuration meaning that each of the 10 lines contains 3 of the 10 points and through each of the 10 points there pass 3 of the 10 lines.

It may happen that one of the vertices of one triangle lies on the opposite side of the other triangle, in which case Desargues’ theorem is still valid, but one line contains now 4 of the 10 points. Contrary to some authors, we consider this configuration also as a Desargues configuration and call it aspecial Desargues configuration.

As long ago as 1846, Cayley remarked (see [4], §1) that the 10 lines and 10 planes determined by 5 points e₁;e₂;e₃;e₄;e₅ in general position in P₃ meet a plane pnot containing any of the pointsei in a Desargues configuration. Conversely, it follows from the standard proof of Desargues’ theorem viaP3(see [11]) that every Desargues configuration is obtained in this way. In the sequel, we choose the coordinates ofP₃ in such a way that

e₁¼ ð1:0:0:0Þ;. . .;e₄¼ ð0:0:0:1Þ; e₅¼ ð1:1:1:1Þ:

If the planepis given by the equationP4

i¼1aixi¼0, then the assumption thateiBp fori¼1;. . .;5 amounts to

ai00 fori¼1;. . .;4 and X⁴

i¼1

ai00:

We denote byD_pthe Desargues configuration determined byp. It consists of the 10 points p_ij:¼e_ie_jVpand the 10 lineslijk :¼e_ie_je_kVpforici;j;kc5;i0j0k0i (see Figure 2). Note that the notation is meant to be symmetric, i.e. p_ij¼p_ji and lijk¼ljki, etc. From the picture it is obvious that every point of a Desargues config- uration is the center of perspective of two triangles: the point p_ij is the centre of per- spective of the trianglespikpilpimandpjkpjlpjmwherefi;j;k;l;mg ¼ f1;2;3;4;5g. In particular every point of the configuration admits an axis of perspective: for pij it is the linelklm.

By definition two Desargues configurationsD1andD2areisomorphicif there is an automorphismaAPGL2ðCÞsuch thatD2¼aðD1Þ.

Lemma 2.1.For planespandp⁰AP₃ not containing a point eithe following conditions are equivalent:

1) D_pis isomorphic to D_p⁰.

2) There is an AAPGL₃ðCÞsuch that a) Ap¼p⁰,

b) A permutes the5points e₁;. . .;e₅.

Figure 1. Desargues’ theorem

For the proof we need the following notion: acomplete quadrangle of a Desargues configuration Dis a set of 4 points and six lines ofDwhich form the points and lines of a complete quadrangle. From Figure 2 it is obvious thatD_p(and thus anyD) admits exactly 5 complete quadrangles: anye_idetermines the complete quadrangle consisting of the points p_ij;p_ik;p_il;p_imand the lineslijk;lijl;lijm;likl;likm;lilm.

Proof of Lemma1.1. We have to show that (1) implies (2), the converse implication being obvious. So leta:p!p⁰be a linear isomorphism withaðDpÞ ¼Dp⁰. We have to show that aextends in a unique way to an AAPGL3ðCÞpermuting the 5 points

e1;. . .;e5. Certainly a maps the 5 complete quadrangles ofDp onto the 5 complete

quadrangles ofDp⁰. As outlined above a complete quadrangle ofDpis uniquely de- termined by a pointeiand similarly forDp⁰. Henceainduces a permutationsof the 5 pointse1;. . .;e5. It is easy to see that there is a one-dimensional familyfAtAPGL3ðCÞ j tAP₁g satisfying Atðe1Þ ¼esð1Þ andAtðpÞ ¼p⁰ (choose suitable coordinates for the source-P3 and the image-P3 ofA:P₃!P₃Þ. For everytAP₁the automorphismAt

maps the linee₁p₁₂onto the linee_sð1Þp_sð1Þsð2Þ. Sincee₂Ae₁p₁₂,e_sð2ÞAe_sð1Þp_sð1Þsð2Þ, and fAtðe2Þ jtAP₁g ¼e_sð1Þp_sð1Þsð2Þ, there is a uniquet₀AP₁ such that

A_t0¼e_sð2Þ:

Figure 2. Desargues’ theorem in space

We claim thatA:¼A_t0satisfies apart fromðaÞalso conditionðbÞ. But for 3ckc5 we havee_k¼e₁p_1kVe₂p_2k. Hence

AðekÞ ¼Aðe1p1kÞVAðe2p2kÞ ¼e_sð1Þp_sð1ÞsðkÞVe_sð2Þp_s2sðkÞ¼e_sðkÞ

This concludes the proof of the lemma. r

The 5 pointse1;. . .;e5form a projective basis ofP₃. Hence for every permutation sof the pointse1;. . .;e5 there is a uniqueAsAPGL3ðCÞinducing s. LetS5 denote the group of theseAs. The action ofS5 onP₃ induces an action on the dual projec- tive spaceP₃. Define

U:¼P₃n6

5 i¼1

Pei;

where P_ei denotes the plane inP₃ parametrizing the planespwithe_iAp. The action of S5 on P₃ restricts to an action on U. Since the quotient of a quasi-projective variety by a finite group is always an algebraic variety, we obtain as an immediate consequence of Lemma 2.1

Theorem 2.2. The algebraic variety MD:¼U=S5 is a moduli space for Desargues configurations.

It is clear how to define families of Desargues configurations. Doing this it is easy to see thatMDis a coarse moduli space in the sense of geometric invariant theory. In particular,MDis uniquely determined as an algebraic variety. Since a Desargues con- figuration may admit a nontrivial group of automorphisms,M_D is not a fine moduli space.

Next we work out the subspace ofM_D parametrizing special Desargues configura- tions. By definition a Desargues configurationDis calledspecialif and only ifDcon- tains a point lying on its axis. The axis of a pointp_ijis the linelklmwithfi;j;k;l;mg ¼ f1;2;3;4;5g. This implies that a configurationD_p is special if and only if the plane contains the point of intersection of a line e_ie_j with the plane e_ke_le_m. Hence a De- sargues configuration Dp is special if and only if the plane pcontains a pointQij¼ eiejVekelem, fi;j;k;l;mg ¼ f1;. . .;5g. Now let PijHU¼P₃n6⁵

i¼1Pei, denote the hyperplane parametrizing the planes p inP₃ not containing the points ei, but con- taining the point Qij. If q:U!MD denotes the natural projection map, then we conclude, since the groupS5 obviously acts transitively on the set of pointsfQijg:

Proposition 2.3. The special Desargues configurations are parametrized by the irre- ducible divisor qðP12Þin MD.

In other words: if D is a special Desargues configuration then there is a plane pHP₃ containing the pointP₁₂¼ ð1:1:0:0Þand not containing anye_i such that D_pGD.

Finally we introduce degenerate Desargues configurations. Desargues’ theorem fails if one of the two triangles is replaced by three lines passing through the center of per- spective. However, the following classical theorem (see [9], p. 99, Exercise 14) may be considered as a limiting case of Desargues’ theorem: IfABC is a proper triangle and a;b;c are 3 lines passing through a pointO(not lying on the sides of the triangle) then the pointsaVBC;bVAC;cVABare collinear if and only if there is a an involu- tion{of theP₁ of lines centered inOsuch that{ðaÞ ¼OA; {ðbÞ ¼OB; {ðCÞ ¼OC.

It is now easy to see that for any configuration of 7 points and 10 lines inP₂sat- isfying the conditions of the theorem there is a planepHP3 containing exactly 1 of the 5 pointseisuch that the configuration is isomorphic to the configurationDcut out onpby the lineseiej and the planeseiejek. A picture of a degenerate Desargues con- figurationDpis shown in Figure 3 wheree5 Apand hencep15¼p25 ¼p35¼p45¼e5

with the notation above.

Note that of the 5 complete quadrangles of the Desargues configuration 4 still survive in the degenerate case namely (in the case of Figure 3) fp12;p13;p14;p15g, fp12;p23;p24;p25g,fp13:p23;p34;p35g,fp14;p24;p34;p45g. This implies that the proof of Lemma 1.1 also works in this case. In other words: two degenerate Desargues configurations D_p andD_p⁰ are isomorphic if and only if there is an automorphism AAPGL₃ðCÞwithAp¼p⁰ permuting the 5 pointse₁;. . .;e₅ inP₃. Similar remarks can be made if the configuration is even more degenerate, that is if 2 triangles of a Desargues configuration collapse (equivalently if the planeppasses through 2 of the

Figure 3. A degenerate Desargues configuration of the 1st kind

pointsei (see Figure 4 where the triangle p14p24p34 of Figure 3 collapsed) or if 2 tri- angles collapse and the 2 points come together (equivalently if the plane p passes through 3 of the points ei). In this case the configuration is a complete quadrangle.

We omit the details since they are easy to work out (see also Section 6).

These remarks induce the following definitions: as above fix 5 pointse1;. . .;e5AP₃ in general position and consider any plane pHP₃. The 10 lines eiej and 10 planes eiejek inP₃cut out onpa configurationDpof points and lines. We callDpa gener- alized Desargues configuration. Ifpcontains a pointei,Dpis called adegenerate De- sargues configuration. It is calledof the i-th kindifpcontains exactlyiof the pointse_i fori¼1;2;3.

It is easy to see that Lemma 1.1 remains valid for generalized Desargues configura- tions. In fact the same proof works also in the degenerate case. One has only to remark that a degenerate Desargues configurationD_pof the first kind (respectively 2^nd kind, respectively 3^rdkind) admits 4 (respectively 3, respectively 1) complete quadrangles.

In the same way as we deduced Theorem 1.2 from Lemma 1.1 we obtain from this Theorem 2.4.The variety MD:¼P₃=S5 is a moduli space for generalized Desargues configurations.

Remark 2.5. In [6] which is the standard reference for Desargues configurations, K.

Mayer constructs the moduli spaceMD in a di¤erent way. He chooses the 4 points of a complete quadrangle as a projective basis ofP₂, sayA;A1;B1;C1 in Figure 1.

Consider the pointsA₁⁰ :¼AA1VB1C1;B₁⁰ :¼AB1VA1C1andC₁⁰¼AC1VA1B1. The three cross-ratios ðA;A₁;A₁⁰;A₂Þ;ðA;B₁;B₁⁰;B₂Þ and ðA;C₁;C₁⁰;C₂Þ determine the Desargues configuration and an open set V of P₁³ represents Desargues configura- tions. The choice of a complete quadrangle induces an action of S5 on V which Mayer worked out explicitly. Moreover he showed thatV=S₅is the moduli space of

Figure 4. A degenerate Desargues configuration of the 2nd kind

Desargues configurations. For our purposes, the construction of Theorem 1.2 is more appropriate.

3The mapF:M_D!M₆^b

Recall that a binary form fðx;yÞof degree 6 is calledstableif f admits no root of multiplicityd3 and that the moduli spaceM₆^bof stable binary sextics exists (see [7]).

In this section, we present the construction of Stephanos (see [10]—in a slightly dif- ferent set-up) associating to every Desargues configurationD_pa binary sexticJ_pin a canonical way. In Section 5 we will see thatJpis stable so that we get a holomorphic mapF:MD!M₆^b.

As in the last section we fix the coordinate system of P3 in such a way thate1 ¼ ð1:0:0:0Þ;. . .;e4¼ ð0:0:0:1Þ and e5¼ ð1:1:1:1Þ. Let p be a nondegener- ate plane inP₃, meaning that the configurationDp is nondegenerate, with equation P4

i¼1aizi¼0. For the coe‰cients ai ofpthis just means ai00 fori¼1;. . .;4 and P4

i¼1a_i00. In 1847, von Staudt proved in his fundamental book [12] that there is a unique (smooth) conics_p on psuch that the polar line of every point of the con- figurationDpis its axis. We callspthevon Staudt conicofDp.

Lemma 3.1. The von Staudt conic s_p of D_p is given by the equationP4

i¼1a_iz_i²¼0 in p:P4

i¼1a_iz_i¼0.

Proof.According to a remark of Reye, which is easy to check (see [8], p. 135), there is a unique quadric S5 in P₃ such that the tetrahedrone1;e2;e3;e4 is a polar tetrahe- dron, i.e. the polar plane ofeiwith respect toS5 is the opposite plane of the tetrahe- dron, and such that the polar plane ofe₅ is the planep. If, as usual, we denote byS₅ also the matrix of the quadric, the conditions meane_iS₅e_j¼0 for 1ci;jc4,i0j and e₅S₅¼ ða1;a₂;a₃;a₄Þ^t. But this implies S₅¼diagða1;a₂;a₃;a₄Þ. Comparing the definitions, the von Staudt conics_pis just the restriction of S₅ to p, which gives the

assertion. r

Note that the proof of Lemma 3.1 yields actually a proof of von Staudt’s theorem.

ObviouslyS5 depends on the choice ofe5as the cone vertex. One could in an analo- gous way defineS_ið1cic4Þand use it to prove Lemma 3.1.

As remarked in the last section, each pointe_icorresponds to a complete quadrangle ofD_p, namely p_1i;. . .;pp_iiii;. . .;p_5i. Since these 4 points are in general position, there is a unique pencil of conicslq₁þmq₂passing through the 4 points. The pencillq₁þmq₂ restricts to a pencil of e¤ective divisors of degree 4 on the von Staudt conics_p. The Jacobian of this pencil is an e¤ective divisor j_pof degree 6 ons_p. It is defined as fol- lows: choose an isomorphism j:s_p!P₁ and coordinatesðx1;x₂ÞofP₁. There are binary quartics f1ðx1;x2Þandf2ðx1;x2Þwith zero divisorjðq1VspÞandjðq2VspÞ. The Jacobian of f1and f2

Jpðx1;x2Þ ¼det qfi

qx_j

is a binary sextic with zero divisorjðj_pÞ. Note the binary sextic is determined by the pencil lq₁þmq₂ up to a nonzero constant and an automorphism of P₁. We callJ_p theJacobian sexticof the pencil, for the Jacobian divisor j_pons_p.

Theorem 3.2.The Jacobian divisor j_pon the von Staudt conic s_pis uniquely determined by the Desargues configuration Dp.It is cut out on spby the cubicP4

i¼1aiz_i³¼0.

Proof. Choose firsti¼5. The pencil of conics passing through the 4 points p₁₅;p₂₅; p₃₅;p₄₅ is cut out on p by the pencil of cones in P₃ passing through the 4 lines e₁e₅;e₂e₅;e₃e₅;e₄e₅. The conditions for a quadricQHP₃to be in this pencil are:Qis singular ine5, i.e.Qe5¼0 andeiAQfori¼1;2;3;4, i.e.eiQei¼0. This yields

Q¼

0 a b ðaþbÞ

a 0 ðaþbÞ b

b ðaþbÞ 0 a

ðaþbÞ b a 0

0 BB B@

1 CC CA

Hence the pencil is generated by the cones

Q1 ¼

0 1 0 1

1 0 1 0

0 1 0 1

1 0 1 0

0 BB B@

1 CC

CA and Q2 ¼

0 0 1 1

0 0 1 1

1 1 0 0

1 1 0 0

0 BB B@

1 CC CA:

It is easy to see that the Jacobi divisor of the pencil lq₁þmq₂ on the curves_p¼ fðz1: . . . :z₄ÞAP₃jP4

i¼1a_iz_i¼P4

i¼1a_iz_i²¼0gis the intersection of the curves_pwith the hypersurface of points in P₃ where the vectors of partial derivativesdðP

a_iz_iÞ, dðP

a_iz²_iÞ,dðQ1Þ,dðQ2Þare linearly dependent. Here we considerQ_ias the quadratic form determined by the matrixQ_i. This hypersurface is given by the equation

J:¼det

a₁ a₁z₁ z₂z₄ z₃z₄ a₂ a₂z₂ z₁z₃ z₄z₃ a₃ a₃z₃ z₄z₂ z₁z₂ a₄ a₄z₄ z₃z₁ z₂z₁ 0

BB B@

1 CC CA¼0:

We claim thatJis congruent toP4

i¼1a_iz_i³ modulo the idealðP

ia_iz_i;P

ia_iz_i²Þ. But

J¼det

a₁ a₁z₁ z₂z₄ z₃z₄ a₂ a₂z₂ z₁z₃ z₄z₃ a₃ a₃z₃ z₄z₂ z₁z₂ Pa_i P

a_iz_i 0 0 0

BB B@

1 CC CA

1X

i

a_idet

a₁z₁ z₂z₄ z₃z₄ a₂z₂ z₁z₃ z₄z₃ a₃z₃ z₄z₂ z₁z₂ 0

@

1

A modX

i

a_iz_i

!

¼ X

i

a_i½2ða1z₁³þa₂z³₂þa₃z³₃Þ ða1z₁²þa₂z₂²þa₃z₃²Þz4z₄ða1z²₁þa₂z²₂þa₃z₃²Þ z₄²ða1z1þa2z2þa3z3Þ þ ðz1þz2þz3Þðz4ða1z1þa2z2þa3z3Þ

ða1z²₁þa2z²₂þa3z²₃ÞÞ 12X⁴

i¼1

ai

X⁴

j¼1

ajz³_j mod X

i

aizi;X

i

aiz_i²

!!

It remains to show that if we start with another pointeiwe obtain the same Jacobi divisor jp onsp. Without loss of generality we may start withe1, since choosing an- othereið2cic4Þamounts only to a permutation of the coordinates.

The pencil of conics inppassing through the points p12;p13;p14;p15 is cut out onp by the pencil of cones inP₃passing through the linese1e2;e1e3;e1e4;e1e5. This pencil is generated by the cones

Q₁⁰ ¼

0 0 0 0

0 0 1 0

0 1 0 1

0 0 1 0

0 BB B@

1 CC

CA and Q₂⁰ ¼

0 0 0 0

0 0 0 1

0 0 0 1

0 1 1 0

0 BB B@

1 CC CA:

A very similar computation as the one above shows that modulo the idealðP aizi; Paiz_i²Þ the hypersurface of points in P₃ where the vectors of partial derivatives dðP

aiziÞ,dðP

aiz²_iÞ,dðQ₁⁰Þ,dðQ₂⁰Þare linearly dependent is given byP4

i¼1aiz³_i, which

completes the proof of the theorem. r

Ifp⁰is another nondegenerate plane inP3such that the Desargues configurations DpandDp⁰are isomorphic, it follows immediately from Lemma 2.1 and Theorem 3.2 that the isomorphism a:p!p⁰ withaðDpÞ ¼Dp⁰ maps the Jacobi divisor jp onsp

onto the Jacobi divisor jp⁰ onsp⁰.

In the sequel we will always interpret jp as the binary sextic Jp (see above and recall thatJpis determined up to a nonzero constant and up to an automorphism of P₁). In Theorem 5.1, we will show thatJp is always stable. Hence we obtain a ca- nonical map

F:MD!M₆^b

of the moduli space of Desargues configurations into the moduli space of stable binary sextics. It is clear thatFis holomorphic, since Theorem 3.2 implies that it is given by polynomials.

Remark 3.3.The proof of Theorem 3.2 suggests another definition of the Jacobi di- visor j_p: the von Staudt conics_p is not contained in the pencil of conicslq₁þmq₂ passing through the points p₁₅;p₂₅;p₃₅;p₄₅. Hencelq₁þmq₂þns_pis a net of conics in the planep. Its discriminant locus given by

detðq1;q2;spÞ ¼0

is a cubic which intersects the conics_pin the Jacobian divisor j_p. For the proof note only that the net is general and thus its discriminant is equivalent to its Jacobian locus given by detðdq1;dq₂;ds_pÞ ¼0.

Remark 3.4. Given a nondegenerate plane p, we associated to every point ei; ð1cic5Þa pencil of quartic divisorslf₁ⁱþmf₂ⁱon the von Staudt conicsp. It is easy to check that in general these 5 pencils are di¤erent from each other.

4 Injectivity ofF:M_D !M₆^b The main object of this section is to prove the following theorem.

Theorem 4.1.The mapF:MD!M₆^b is an injective birational morphism.

For the proof we need some preliminaries. LetAandBdenote two quadrics inP_n (for us n¼2 or 3) given by the equationsx^tAx¼0 and x^tBx¼0. Recall that Ais calledapolartoBif

trðAAdjðBÞÞ ¼0:

Here AdjðBÞdenotes the adjoint matrix ofB, i.e. the matrix of the dual quadricBB^of B. Note that this definition is not symmetric (some authors say thatAis apolar toBB).^ Geometrically this means the following (see [9]): a conicAis apolar toBinP₂if and only if there is a triangle inscribed inAand self-polar with respect toB. A quadricA is apolar toBinP₃ if and only if there is a tetrahedron inscribed inAand self-polar with respect toB.

Proposition 4.2.Let D_pbe a Desargues configuration.Every conic inppassing through the4points of a complete quadrangle of Dpis apolar to the von Staudt conic sp. Proof.Since apolarity is a linear condition and by the special choice of the coordinates it su‰ces to show that the quadric conesQ1;Q2;Q₁⁰Q₂⁰ (for the notation see the proof of Theorem 3.2) are apolar to the the quadric diagða1;a2;a3;a4ÞinP₃, whose restric- tion topissp, which is an immediate computation. r In order to show that a conicqin the planeppassing through the points of a com- plete quadrangle ofD_pis the unique conic apolar tos_pand passing throughqVs_p(with multiplicities) we change the coordinates. We choose the coordinatesðx0:x₁:x₂Þof p¼P₂in such a way that the von Staudt conics_pis given by the equation

x₁²4x0x2¼0 ð1Þ

So the matrix of the dual conic is:

b sp

sp¼

0 0 2

0 4 0

2 0 0

0

@

1 A

Letðt0:t₁Þdenote homogeneous coordinates ofP₁. Then ðt0:t₁Þ ! ðx0:x₁ :x₂Þ ¼ ðt₀²:2t₀t₁ :t²₁Þ

is a parametrizationP1!^@ spofsp. Using this, any e¤ective divisor of degree 4 onsp

can be interpreted as a binary quartic fðt0;t1Þ. Hence

fðt0;t1Þ ¼a0t₀⁴þ4a1t₀³t1þ6a2t²₀t₁²þ4a3t0t₁³þa4t₁⁴ ð2Þ with ða0:a1 :a2:a3:a4ÞAP₅ representing an arbitrary divisor of degree 4 on sp. Given an e¤ective divisor d, of degree 4 on sp, i.e. a binary quartic f, consider the conic

q¼a₀x₀²þa₂x₁²þa₄x²₂þ2a₁x₀x₁þ2a₂x₀x₂þ2a₃x₁x₂:

Lemma 4.3.The conic q is the unique conic passing through the divisordand apolar to the von Staudt conic s_p.

Proof. Note first that qpasses through dby the choice of the parametrization. It is apolar tosp, since

tr

a0 a1 a2

a1 a2 a3

a2 a3 a4

0

@

1 A

0 0 2

0 4 0

2 0 0

0

@

1

A¼2a24a2þ2a2¼0:

On the other hand, apolarity is one linear condition. Hence the space of conics apolar tospis isomorphic toP₄. Now it is easy to see ([3], p. 17) that any divisor of degree 4 ons_pimposes independent conditions on conics. In other words, the space of conics passing through d is isomorphic to P₁. Hence there is exactly one conic passing throughdand apolar tos_pif we only show that not every conic passing throughdis apolar tos_p.

For this we assume d¼p₁þp₂þp₃þp₄ with p_i0p_j, for i0j, the degenerate cases being even easier to check. We may choose the coordinates in such a way that inP1we have p1¼ ð0:1Þ, p2¼ ð1:0Þ, p3¼ ð1:1Þ, p4¼ ð1:tÞwitht00;1. So in P2: p1¼ ð0:0:1Þ;p2¼ ð1:0:0Þ;p3¼ ð1:2:1Þ;p4¼ ð1:2t:t²Þand 2tðt1Þxyþ ð1t²Þy²þ ð2t2Þyz¼0 is a conic passing through pi, for i¼1;. . .;4 and not apolar tosp, ift01. Ift¼ 1 then 2xyþ4xzy²2yzsatisfies these conditions.

r We call q the conic associated to the binary quartic (2). If we associate to every divisor of a pencil ld1þmd2 of quartic divisors the unique conic of Lemma 4.3, we obtain a pencil of conics. Hence we obtain as an immediate consequence of Propo- sition 4.2 and Lemma 4.3:

Corollary 4.4. Let D_p be a Desargues configuration. The pencil of conicslq₁þmq₂ passing through the 4 points of a complete quadrangle of D_p is the unique pencil of

conics cutting out the pencil of quartic divisorsðlq1þmq₂ÞVs_pwhich is apolar to the von Staudt conic s_p.

Given a smooth conicq0, we choose the coordinates in such a way thatq0is given by Equation (1). As outlined above, any pencil of binary quartics can be interpreted (up to isomorphism) as a pencil of quartic divisors onq₀. There is a unique pencil of conics associated to it according to Lemma 4.3. We call a pencil of binary quartics (respectively the corresponding pencil of quartic divisors) admissibleif its associated pencil of conics is general, i.e. its base locus consists of 4 di¤erent points. The fol- lowing proposition is the first step in the proof of Theorem 4.1.

Proposition 4.5. Consider an admissible pencil of quartic divisors on a smooth conic q₀.There is a unique Desargues configuration D such that q₀is the von Staudt conic of D and the pencil of quartics is cut out on q₀ by the pencil of conics passing through the points of a complete quadrangle of D.

Proof.Let p1;p2;p3;p4 denote the 4 base points of the pencil of conics associated to the given pencil of quartic divisors. Consider the complete quadrangle consisting of the 4 points p₁;p₂;p₃;p₄ and the 6 lines lij¼p_ip_j for 1ci<jc4. Let p_ij be the poles of the lines lij andli the polars of the points p_i with respect to the conic q₀. Then the 10 pointsP_ij;p_i and the 10 lineslij;li form a Desargues configuration ac- cording to a theorem proved by von Staudt in 1831 (see [2], p. 62). Using Lemma 4.3

we have the assertion. r

Before we go on, let us note the following characterization of an admissible pencil of binary quartics, which we need later.

Proposition 4.6. The pencil of binary quartics lf1þmf2 with fi¼P4

j¼1 4

i a_jⁱt^4j₀ t₁^j is admissible if and only if the discriminant of the binary cubicdetðt0q₁þt₁q₂Þis nonzero, where q_i¼

a₀ⁱ a₁ⁱ a₂ⁱ a₁ⁱ a₂ⁱ a₃ⁱ a₂ⁱ a₃ⁱ a₄ⁱ 0

B@

1

CAis the matrix of the conic associated to f_ifor i¼1;2.

Proof.This is a consequence of the fact (see [11], Section 6.3) that the base locus of a pencil of conics consists of 4 di¤erent points if and only if its discriminant does not

vanish together with Lemma 4.3. r

In order to complete the proof of Theorem 4.1, we need to compute the degree of the Jacobian map. Note that the binary quartic (2) determines a point inP₄ namely ða0 : . . . :a₄Þ. If we considerP₄as the space of quartics, the space of pencils of binary quartics is Grð1;4Þ, the Grassmannian of lines inP₄. Considering in the same way the space of binary sexticsP6

i¼0ait₀⁶ⁱt₁ⁱ asP6, the map associating to every pencil of quartics its Jacobian defines a morphism

Jac:Grð1;4Þ !P₆:

Note that Grð1;4Þis also of dimension 6. We need the following lemma:

Lemma 4.7.The Jacobian mapJac:Grð1;4Þ !P₆is a ramified covering of degree5.

Proof. Let lf1þmf2 be a pencil of quartics with f1¼P4

i¼1 4

i ait⁴ⁱ₀ t₁ⁱ and f2¼ P4

i¼1 4

i bit₀⁴ⁱt₁ⁱ. Its Jacobian is given by:

Jacðf₁;f₂Þ ¼det qfi

qt_j

¼16½p01t⁶₀þ3p02t⁵₀t1þ3ð2p12þp03Þt₀⁴t₁²

þ ð8p13þp04Þt₀³t³₁þ3ð2p23þp14Þt²₀t₁⁴þ3p24t0t₁⁵þp34t₁⁶; where p_ij:¼a_ib_ja_jb_idenotes the Plu¨cker coordinates of the pencil. In particular up to a constant the Jacobian does not depend on the choice of f₁and f₂and defines a point Jacðlf1þmf₂ÞinP₆. The explicit form of Jacðf₁;f₂Þimplies that the map Jac factorizes via the Plu¨cker embedding p

P₉

p q

Grð1;4Þ ^Jac ! P₆

R

!

with a linear projection mapq. In fact Jac is given by

lf1þmf2 7! ðp01:3p02 :3ð2p12þp03Þ:8p13þp04 :3ð2p23þp14Þ:3p24 :p34ÞAP6: It is the linear projection of the P9 with coordinates ðp01: . . . :p34Þwith center the plane P with equations p01 ¼p02¼2p12þp03¼8p13þp04¼2p23þp14¼p24 ¼ p34¼0. On the other hand, the Plu¨cker variety pðGrð1;4ÞÞ in P₉ is given by the Plu¨cker relations among which there are

p12p34p13p24þp14p23¼0 p₀₁p₃₄p₀₃p₁₄þp₀₄p₁₃¼0 p₀₁p₂₃p₀₂p₁₃þp₀₃p₁₂¼0

Now ifðp01 : . . . :p34ÞAPVpðGrð1;4ÞÞ, this impliesp14p23¼p04p13¼p03p12¼0, so all pij¼0. Hence the center of projectionPdoes not intersect the Plu¨cker variety.

This implies that the degree of Jac equals the degree of the Plu¨cker varietypðGrð1;4ÞÞ in P₉. It is well known that this degree is 5 (see [5], p. 247). Since Jac cannot con- tract a positive-dimensional subvariety of Grð1;4Þ this completes the proof of the

lemma. r

Using this we can finally prove Theorem 4.1.

Proof of Theorem4.1. Since admissibility of a pencil of quartics is an open condition and the Jacobian map is finite, we conclude from Proposition 4.5 thatFis dominant

and there is an open dense setUHM₆^bsuch that for every sextic jAUthe preimage Jac¹ðjÞconsists of exactly 5 pencils of quartics. ShrinkingUif necessary we may as- sume according to Remark 3.4 that every Desargues configurationDAF¹ðUÞadmits 5 di¤erent pencils of quartic divisors on the von Staudt conic. Hence for every jAU the 5 pencils of quartics of any preimageDAF¹ðjÞmust coincide. Thus by Proposi- tion 4.5F¹ðjÞconsists only of one Desargues configuration, i.e.F:F¹ðUÞ !Uis bijective. This implies thatFis of degree 1. It remains to show the injectivity.

SinceFis a finite birational morphism of normal varieties, the injectivity can only fail at points pofMDfor whichFðpÞis a singular point ofM₆^b. ButM₆^badmits only 3 singular points (see [1], Remark 5.4, p. 5551), namely a point of multiplicity 5 rep- resented by the binary sextic j5¼t₀⁶t0t₁⁵, a point of multiplicity 2 represented by j₂ ¼t₀⁶t₀²t⁴₁, a point of multiplicity 3 represented by j₃¼t₀⁵t₁t₀²t₁⁴. The moduli spaceM_Dalso admits a singular point of multiplicity 5: the Desargues configuration D_p5 with p₅ given by z₁mz₂þm²z₃m³z₄ ¼0, where mis a root of the equation x⁴þx³x²þx1¼0, admits an automorphism of order 5 namely:

0 0 0 1

1 0 0 1

0 1 0 1

0 0 1 1

0 BB B@

1 CC CA:

It permutes the pointseias follows:e17!e27!e37!e47!e57!e1. It is easy to check thatDp5 is of multiplicity 5 inMDand thatF¹ðj5Þ ¼Dp5.

As in Example 5.3 below one checks that all pencils of quartics with Jacobian j₂ are not admissible. Hence j₂ is not in the image ofF. Finally as in Example 5.4 be- low one checks that only one pencil of quartics with Jacobian j₃is admissible namely lf₃þmg₃ with f₃¼2t₀⁴þ6t₀²t²₁þ4t₀t₁³ andg₃¼₁₆¹t₀²t²₁. HenceFis also injective at the corresponding point atM_D. This completes the proof of the theorem. r

5 The image of the mapF

Using the set-up of the last section we are now in a position to prove the following theorem, which was announced (but not applied) already in Section 3.

Theorem 5.1.For any Desargues configuration D the associated binary sextic J ¼FðDÞ is stable.

Proof. LetJðt0;t₁Þbe a nonstable binary sextic, i.e. admitting a root of multiplicity d3. We choose coordinates in such a way that this root isð0:1Þ. HenceJis of the form

Jðt0;t1Þ ¼A0t₀⁶þA1t₀⁵t1þA2t⁴₀t₁²þA3t₀³t₁³: Suppoself þmgwith f ¼P4

i¼1 4

i a_it⁴ⁱ₀ t₁ⁱ andg¼P4

i¼1 4

i b_it₀⁴ⁱt₁ⁱ is a pencil of bi- nary quartics whose Jacobian isJ(up to a nonzero constant). According to the results of Section 4, it su‰ces to show that lf þmg is not admissible which according to

Proposition 4.6 means that detðt0q₁þt₁q₂Þ admits a double root where q₁ and q₂ denote the conics associated to f andg. By the equation forJgiven in the proof of Lemma 4.7 the coe‰cients of f andgsatisfy the following system of equations

2ða2b3a3b2Þ þa1b4a4b1¼0 a2b4a4b2¼0 a3b4a4b3¼0

ð3Þ

Assume first thata₄orb₄00. Without loss of generality we may assume thata₄¼1.

Replacinggbygb4f we may assumeb4¼0. But then the system (3) impliesb3¼ b2¼b1¼0 and we obtain

detðt0q₁þt₁q₂Þ ¼

t₀a₀þt₁b₀ t₀a₁ t₀a₂ t₀a₁ t₀a₂ t₀a₃ t₀a₂ t₀a₃ t₀ 0

@

1 A

which has a double rootð0:1Þ. Hencea₄¼b₄ ¼0. Then (3) just says

a₂b₃a₃b₂¼0 ð4Þ

Ifa₃¼b₃¼0,

detðt0q1þt1q2Þ ¼

t0a0þt1b0 t0a1þt1b1 t0a2þt1b2

t0a1þt1b1 t0a2þt1b2 0 t0a2þt1b2 0 0 0

@

1 A

which has a triple root. Finally, again without loss of generality, we may assume that a3¼1;b3¼0. Then (4) impliesb2¼0 and

detðt0q1þt1q2Þ ¼

t0a0þt1b0 t0a1þt1b1 t0a2

t0a1þt1b1 t0a2 t0

t0a2 t0 0

0

@

1 A

which again has a double rootð0:1Þ. This completes the proof of the theorem. r Theorems 4.1 and 5.1 lead to the question whether the mapF:MD!M₆^b is sur- jective. In order to analyze this question, recall that we represented a quartic divisor on the conics:x₁²4x0x2 by the binary quartic:

fðt0;t1Þ ¼a0t₀⁴þ4a1t₀³t1þ6a2t²₀t₁²þ4a3t0t₁³þa4t₁⁴

and the conic associated to it was given by the matrix

a₀ a₁ a₂ a₁ a₂ a₃ a₂ a₃ a₄ 0

@

1

A. We are identi- fying in this way the projective spaceP₄¼P₄ða0: . . . :a₄Þof binary quartics with the

P₄of conics apolar to the conics. Similarly we consider Grð1;4Þas a space of pencils of conics. Hence we get a map

Jac:Grð1;4Þ !P6

associating to every pencil of conics apolar tosthe Jacobian of its associated pencil of binary quartics. According to the definition of the mapFa binary sextic jis in the image ofFif and only if the preimage Jac¹ðjÞ(which consists of 5 pencils counted with multiplicities) consists not only of special pencils of conics, i.e. pencils of conics lq₁þmq₂ such that detðlq1þmq₂Þadmits a multiple root ðl:mÞ. LetDHGrð1;4Þ denote the hypersurface of Grð1;4Þgiven by the equation

discrðdetðlq1þmq₂ÞÞ ¼0

Jac being a finite morphism, the image JacðDÞis a hypersurface inP₆. IfUdenotes the open set of P₆ parametrizing stable sextics andr:U!M₆^b the natural projec- tion, it is easy to see that also

H:¼pðJacðDÞVUÞ is a hypersurface inM₆^b. We obtain

Proposition 5.2.The imageimðFÞofFsatisfies M₆^bnHHimðFÞHM₆^b:

The following two examples show that both inclusions are strict, i.e. imðFÞ0M₆^b andM₆^bnH0imðFÞ.

Example 5.3.Consider the binary sextic

j0¼ 16t₁⁶48t⁴₀t₁²þ128t₀³t₁³48t₀²t₁⁴16t₁⁶:

Jac¹ðj0Þcontains the following 3 pencils of conics pi¼lq₁ⁱþmq₂ⁱ,ði¼1;2;3) with

q¹₁¼

1 0 0

0 0 0

0 0 1

0

@

1

A q¹₂ ¼

8 1 0

1 0 1

0 1 0

0

@

1 A

q²₁¼

1 1 1

1 1 0

1 0 1

0

@

1 A q²₂ ¼

0 1 0

1 0 1

0 1 0

0

@

1 A

q³₁¼

1 1 0

1 0 0

0 0 1

0

@

1

A q₂³¼

0 1 0 1 0 1 0 1 0 0

@

1 A

Since the Jacobian matrix of the map Jac:Grð1;4Þ !P₆ at the points p₁;p₂;p₃ of Grð1;4Þ is of rank 5;6;5 the pencils p₁ and p₃ are in the ramification locus of Jac.

Hence Jac¹ðj₀Þ ¼ fp1;p₂;p₃g. Now it is easy to check that discrðdetðlq₁ⁱþmq₂ⁱÞÞ ¼0 fori¼1;2;3. Hence there is no Desargues configurationDpwithFðDpÞ ¼j0. Example 5.4.Consider the binary sextic

j₁¼48t₀⁵t₁þ48t²₀t₁⁴

Jac¹ðj1Þcontains the 2 pencils of conics pi¼lq₁ⁱþmq₂ⁱ,ði¼1;2) with

q₁¹¼

2 ffiffiffiffiffiffiffi

32

p 1 ¹₈ ffiffiffi

34 p 1 ¹₈ ffiffiffi

34 p 1

4

ffiffiffiffiffiffiffi

32 p

1 8

ffiffiffi4 p3

1 4

ffiffiffiffiffiffiffi

32

p 1

0 B@

1 CA q₂¹¼

2 ffiffiffiffiffiffiffi

32

p 1 0

1 0 0

0 0 0

0 B@

1 CA

q₁²¼

1 0 1

0 1 ¹₂ 1 ¹₂ 0 0

B@

1

CA q₂²¼

0 0 1

0 1 0

1 0 0

0

@

1 A

One easily checks discrðdetðlq¹₁þmq₂¹ÞÞ ¼0 and discrðdetðlq²₁þmq₂²ÞÞ00. So lq₁²þmq²₂ comes from a Desargues configuration, whereas lq¹₁þmq₂¹ does not. This implies j1AimðFÞ, but j1AH.

6 Special Desargues configurations

Consider the 10 points inP₃

S₁₂:¼e₁e₂Ve₃e₄e₅¼ ð1:1:0:0Þ S₃₄:¼e₃e₄Ve₁e₂e₅ ¼ ð0:0:1:1Þ S₁₃:¼e₁e₃Ve₂e₄e₅¼ ð1:0:1:0Þ S₁₅:¼e₁e₅Ve₂e₃e₄ ¼ ð0:1:1:1Þ S₁₄:¼e₁e₄Ve₂e₃e₅¼ ð1:0:0:1Þ S₂₅:¼e₂e₅Ve₁e₃e₄ ¼ ð1:0:1:1Þ S₂₃:¼e₂e₃Ve₁e₄e₅¼ ð0:1:1:0Þ S₃₅:¼e₃e₅Ve₁e₂e₄ ¼ ð1:1:0:1Þ S24:¼e2e4Ve1e3e5¼ ð0:1:0:1Þ S34:¼e3e4Ve1e2e5 ¼ ð1:1:1:0Þ Recall that a Desargues configurationDpis special if and only if the planepcontains one of the pointsSij(and no pointei). The following theorem gives a characterization of special Desargues configurations in terms of their associated binary sextics.

Theorem 6.1.For a Desargues configuration D_pthe following statements are equivalent:

(1) Dpis a special Desargues configuration.

(2) The binary sexticFðDpÞadmits a double point.