Ten and eleven nodes

8. Quartic surfaces with ten nodes.

We have, as before, two classes of irreducible sextics with nine nodes, viz. according as the points of the curve are or are not conjugate in pairs with regard to any seven of the nine nodes. We have also the sextic arising from two cubics or two lines and a quartic. We consider in the first place these last two cases.

In the next place, when the sextic splits up into two lines and a quartic curve, we see that through the node $x=y=z=0$ there pass two planes, each touching the surface along a conic; each is a trope. The equation of the surface is of the form

$$A^2+\rho xyB=0,$$

where $A$ and $B$ are any two quadrics.

The nodes lie on two conics: the tangent cone from each of the eight associated nodes breaks up into a plane and a quintic curve with four double points.

We now pass to irreducible sextics, first those whose points are conjugate in pairs giving syzygetic surfaces with ten nodes. Such surfaces are represented by an equation of the form

$$A^2+\rho K_1K_2=0,$$

where $K_1$ and $K_2$ are cones whose vertices lie on the quadric $A$.

9. Quartic surfaces with eleven nodes.

The three varieties of plane sextics with ten nodes (Art. 5) lead to three types of quartic surface with eleven nodes. The equation of the first variety was seen to be of the form

$$u_3^2-u_2^2K=0;$$

this sextic arises from the quartic surface

$$Kw^2+2u_3w+u_2^2=0.$$

The six nodes which lie on a conic are given by the equations $w=0$, $u_2=0$, $u_3=0$; the plane $w=0$ is a trope.

The tangent cone drawn to the surface from any one of these nodes breaks up into the plane $w=0$ and a quintic cone, the tangent cones from the remaining five nodes are irreducible.

The second kind of plane sextic with ten nodes has an infinite number of bitangent cubics through seven of its nodes (Art. 5); the quartic surface to which it corresponds must therefore be syzygetic; the equation of the ten-nodal syzygetic surface being $A^2+\rho K_1K_2=0$ (Art. 8) it may be shown that in this pencil there are twelve surfaces which have an eleventh node. It is easy to see that the equation of such a surface has the form

$$\sqrt{K_1}+\sqrt{K_2}+\sqrt{K_3}=0,$$

where $K_1=0$, $K_2=0$, $K_3=0$ are cones, and such that the vertex of $K_1$ lies upon $K_2-K_3=0$, etc. This surface is called $\mathrm{XI}_b$.

There remain two cases in which the sextic curve breaks up into simpler curves: either into two lines and a nodal quartic or into two cubic curves.

In the first case the equation of the surface is

$$A^2+\rho Kxy=0,$$

where $A$ is a quadric and $K$ a cone whose vertex lies on $A$. This is $\mathrm{XI}_d$.

In the second case we have a symmetroid, hence

$$(u_2w+u_3)^2+v_3v_3'\equiv u_2(u_2w^2+2u_3w+u_4).$$

Here either $v_3$ or $v_3'$ has a nodal line, arising from an eleventh node on the surface. The tangent cone from this eleventh node to the surface gives a plane sextic of the third variety. This case is $\mathrm{XI}_a$.

8. Ten nodes

Jessop distinguishes two kinds of irreducible nine-nodal sextic by the existence of conjugate point pairs relative to seven selected nodes. He also considers reducible sextics consisting of two cubics, or of two lines and a quartic. His treatment of the two-cubic case occupies the gap between this excerpt and the next.

Two tropes and two cones

When the branch sextic is two lines and a quartic, the corresponding planes through the projection centre are tropes. The quartic surface has equation

$$A^2+\rho xyB=0,$$

where $A,B$ are quadrics. Its nodes lie on two conics. Projection from each of the eight associated nodes gives a plane and a quintic component with four double generators; the source calls this a “quintic curve.”

The syzygetic ten-nodal surfaces associated with irreducible sextics have equation

$$A^2+\rho K_1K_2=0,$$

where each $K_i$ is a quadric cone with vertex on $A=0$.

9. Eleven nodes: the first type

The three kinds of ten-nodal plane sextic from §5 give three types of eleven-nodal quartic. The first has sextic equation $u_3^2-u_2^2K=0$ and surface equation

$$Kw^2+2u_3w+u_2^2=0.$$

Six nodes satisfy $w=u_2=u_3=0$. Their plane is a trope. Projection from one of them gives a plane and a quintic cone; projection from any of the other five gives an irreducible sextic cone.

The remaining eleven-node types

The second kind of ten-nodal sextic has a pencil of bitangent cubics through seven nodes, so the corresponding surface is syzygetic. Jessop states that the pencil $A^2+\rho K_1K_2=0$ has twelve members with an eleventh node. He writes such a member as

$$\sqrt{K_1}+\sqrt{K_2}+\sqrt{K_3}=0,$$

where the $K_i$ are quadric cones and the vertex of $K_1$ lies on $K_2-K_3=0$, cyclically. This is $\mathrm{XI}_b$. The radical notation means the polynomial equation obtained by eliminating radicals; it does not select global square-root branches.

For the reducible sextic consisting of two lines and a nodal quartic, the equation is $A^2+\rho Kxy=0$, with the vertex of $K$ on $A$. This is $\mathrm{XI}_d$.

For two cubics the surface is a symmetroid and

$$(u_2w+u_3)^2+v_3v_3'=u_2(u_2w^2+2u_3w+u_4).$$

An eleventh node produces a double generator on one cubic cone. Projecting from that new node gives the third kind of irreducible sextic in §5. Jessop calls the surface $\mathrm{XI}_a$.

A symmetric matrix explains a special ten-node family

A quartic symmetroid is a surface

$$S=\{\det M(x)=0\},\qquad M(x)=x_0M_0+x_1M_1+x_2M_2+x_3M_3,$$

where the $M_i$ are symmetric $4\times4$ complex matrices. The matrices describe a three-dimensional projective family, or web, of quadrics in an auxiliary $\mathbf P^3$. A point of $S$ corresponds to a singular quadric in that web.

If $M$ has rank at most two, all its $3\times3$ cofactors vanish. The identity $d\det(M)=\operatorname{tr}(\operatorname{adj}(M)\,dM)$ therefore makes the point singular on $S$.

Near a rank-two matrix, choose an invertible $2\times2$ block and use its Schur complement. Up to an invertible factor, the determinant becomes the determinant of a symmetric $2\times2$ block:

$$\det\begin{pmatrix}a&b\\b&c\end{pmatrix}=ac-b^2.$$

If the web meets the rank-two locus transversely, $a,b,c$ are local coordinates, so the singularity is an ordinary node. The rank-at-most-two locus in the projective space $\mathbf P^9$ of symmetric matrices has codimension three and degree ten. A general web therefore has ten such nodes. The degree-ten assertion is an external determinantal-degree result; the local node calculation above is explicit. See Helsø, Rational Quartic Symmetroids, §2.

Rank three can also give a singularity of a special web if the web is tangent to the determinant hypersurface. Thus “ten rank-two points” is a general-web statement, not a bound on all symmetroids. This distinction matters for Jessop’s eleven- and twelve-nodal examples.

Jessop introduces the symmetroid through two cubic components of the projection sextic and develops its determinant in Chapter IX. His argument on p. 14 transports the two-cubic property from one node to another using residual sextics on cubic surfaces. The rank calculation explains the generic ten nodes; it does not by itself prove the equivalence with his projection characterization.

Why two tropes give ten nodes

Consider $F=A^2+\rho xyB$, with $A,B$ general quadrics and $\rho\ne0$. On $x=0$ the section is $A^2$, and likewise on $y=0$. Along the first conic the only potentially nonzero first derivative is proportional to $yB$; singularities occur at $A=x=y=0$ or $A=x=B=0$. The second trope contributes the analogous points.

The line $x=y=0$ meets $A=0$ twice. The complete intersections $A=x=B=0$ and $A=y=B=0$ contribute four points each. Under transversality and disjointness assumptions this gives $2+4+4=10$ ordinary nodes. Genericity is essential: coincident points, tangent intersections or extra singularities change the conclusion.

For $A^2+\rho K_1K_2$, the eight points $A=K_1=K_2=0$ are forced nodes when transverse. If the two cone vertices lie on $A=0$, each contributes another singular point. The local quadratic terms must again be nondegenerate. This explains the two additional nodes of Jessop’s syzygetic ten-node form.

Four eleven-node types

Projection from a suitable node can give an irreducible ten-nodal sextic. Its rational lift to a quadric has one of the three bidegrees already derived. Jessop’s incidence analysis then gives:

These correspondences use Jessop’s arguments in §§5 and 9. The bidegree classification alone does not establish them. In $\mathrm{XI}_c$, restricting to $w=0$ immediately explains the trope and its six singular points $u_2=u_3=0$.

In the three-cone notation, eliminate radicals to obtain

$$K_1^2+K_2^2+K_3^2-2K_1K_2-2K_2K_3-2K_3K_1=0.$$

At the vertex of $K_1$, both $K_1$ and $dK_1$ vanish. If $K_2=K_3$ there, the polynomial and its differential vanish as well. This verifies the singularity mechanism behind Jessop’s vertex condition. It does not prove ordinaryness or count all eleven points.

What the pencil counts still require

Jessop’s general nine-node pencil has thirteen members acquiring a tenth node. He subtracts base and extraneous intersections to get

$$6\cdot18-6\cdot9-3\cdot3-15\cdot2-2=13.$$

One member is a symmetroid. His eleven-node discussion also gives ten special members in the trope pencil and twelve in the syzygetic pencil. A modern description calls these intersections with a discriminant in the parameter space of surfaces. That name does not prove the numbers: one must control multiplicities, fixed singularities and the boundary members. Here these are source enumerative results; the arithmetic and local mechanisms are checked, but the complete multiplicity calculation is not independently reconstructed.