Eight and nine nodes

6. Quartic surfaces with eight nodes.

Returning to the sextic curve $u_2u_4-u_3^2=0$, derived from the surface $u_2w^2+2u_3w+u_4=0$, any quadric through the node is

$$2t_1w+t_2=0;$$

if the quartic surface has any other node which also lies upon this quadric, since this node also lies on the surface

$$u_2w+u_3=0,$$

it is clear that the curve $u_2t_2-2t_1u_3=0$ will pass through the resulting node on $u_2u_4-u_3^2=0$ or $c_6$.

This quartic curve passes through the points of contact of $c_6$ with its contact-conic $u_2$, and also through the nodes of $c_6$ which result from nodes on the quartic surface. If therefore the surface has eight nodes we have seven nodes on $c_6$: to each quartic through these seven nodes and the points of contact $B_1\ldots B_6$ of $c_6$ and $u_2$, there corresponds one quadric through the eight nodes, and vice-versâ.

Now it was stated (Art. 4) that plane sextics with seven nodes form two classes; in the more general case there is a singly infinite number of quartic curves through the nodes and $B_1\ldots B_6$, and we obtain corresponding to this case a singly infinite number of quadrics through the eight nodes. For the more special case where there is a doubly infinite number of quartic curves through the thirteen points we have a doubly infinite number of quadrics through the eight nodes, which therefore form eight associated points. Such a surface is represented by an equation of the form

$$(a\mathbin{\between}A,B,C)^2=0.$$

It follows that any quadric through the eight nodes meets the quartic surface in two quadri-quartic curves which are projected from any node into two of the $\infty^2$ cubics which pass through the seven nodes of $c_6$.

These two classes of quartic surfaces will be termed asyzygetic and syzygetic respectively*.

• The general syzygetic surface is the envelope of the quadrics $\lambda^2D+\lambda E+F=0$, where $D,E,F$ are quadrics through the eight nodes.

The equation of the general seven-nodal surface being $F\equiv(a\mathbin{\between}A,B,C)^2+\rho\Sigma=0$ (Art. 2), where $A,B,C$ are quadrics through the seven nodes, if there is an eighth node we obtain, to determine it, the equations

$$A_i\frac{\partial F}{\partial A}+B_i\frac{\partial F}{\partial B}+C_i\frac{\partial F}{\partial C}+\rho\Sigma_i=0,\qquad(i=1,\ldots4);$$

hence the eighth node lies on the surface

$$J(A,B,C,\Sigma)\equiv\begin{vmatrix}A_1&B_1&C_1&\Sigma_1\\A_2&B_2&C_2&\Sigma_2\\A_3&B_3&C_3&\Sigma_3\\A_4&B_4&C_4&\Sigma_4\end{vmatrix}=0.$$

The eighth node may therefore not be taken arbitrarily, as in the case of the first seven nodes.

If $A,B$ are two quadrics through the eight nodes and $T$ any eight-nodal asyzygetic surface, the general asyzygetic surface is represented by the equation

$$\alpha A^2+\beta B^2+2\gamma AB+2\rho T=0.$$

7. Quartic surfaces with nine nodes.

From the two varieties of surfaces with eight nodes we derive two with nine nodes. Considering first syzygetic surfaces, viz.

$$\alpha A^2+\ldots+2\mu AB=0,$$

if this surface has a ninth node it must lie upon the twisted sextic

$$\left\|\begin{matrix}A_1\ldots A_4\\B_1\ldots B_4\\C_1\ldots C_4\end{matrix}\right\|=0.$$

This curve is the locus of the vertices of the cones of the system $A+\lambda B+\mu C$; hence, if a ninth node exist, there is a quadric cone $K$ whose vertex is $D_9$ which passes through the points $D_1\ldots D_8$.

Taking $D_9$ as the point from which the surface is projected by a tangent cone (giving rise to the curve $u_2u_4-u_3^2=0$), this latter curve must have eight nodes lying on a conic, and must therefore break up into this conic and a quartic curve. Therefore $K$, the quadric cone whose vertex is $D_9$, forms part of the tangent cone from $D_9$, and touches the quartic surface along a twisted quartic. The equation of the surface is therefore of the form

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

where $A=0$ is a quadric through the nine nodes and $B=0$ a quadric through the eight associated points $D_1\ldots D_8$. There is a triply infinite number of nine-nodal syzygetic quartic surfaces.

Considering next asyzygetic nine-nodal surfaces, from the equation of the general eight-nodal surface it is seen that a ninth node must lie on the curve

$$\left\|\begin{matrix}A_1\ldots A_4\\B_1\ldots B_4\\T_1\ldots T_4\end{matrix}\right\|=0,$$

which is of the eighteenth order, the dianodal curve; the ninth node being taken arbitrarily on this curve, there is a singly infinite number of surfaces with the nine given nodes represented by the equation

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

where $A$ is the quadric through the nine nodes and $P$ any quartic surface with these nodes.

6. Eight nodes

Let the surface be $u_2w^2+2u_3w+u_4=0$, with projection centre $[0:0:0:1]$. A quadric through that point has equation $2t_1w+t_2=0$. At another surface node, $u_2w+u_3=0$, so elimination gives

$$u_2t_2-2t_1u_3=0.$$

This plane quartic passes through the seven projected nodes and the six contact points of the sextic with $u_2=0$. Jessop uses a one-to-one correspondence between such quartics and quadrics through the eight surface nodes.

In the general case, these quadrics form a pencil. In the special case, they form a net: the eight nodes are the common intersection of three quadrics $A,B,C$, and the surface has equation

$$Q(A,B,C)=0,$$

with $Q$ quadratic. A quadric of the net meets the quartic in two space quartics, each a complete intersection of two quadrics. Their projections are cubics through the seven plane nodes.

Jessop calls the pencil case asyzygetic and the net case syzygetic. His footnote also represents the latter as the envelope of $\lambda^2D+\lambda E+F=0$, with $D,E,F$ quadrics through the eight nodes.

The dianodal surface

For the seven-nodal family $F=Q(A,B,C)+\rho\Sigma$, differentiation gives

$$\partial_iF=Q_A\partial_iA+Q_B\partial_iB+Q_C\partial_iC+\rho\partial_i\Sigma.$$

At an additional node the four derivatives vanish. Jessop concludes that

$$\det\big(\nabla A,\nabla B,\nabla C,\nabla\Sigma\big)=0.$$

Thus an eighth node is constrained to a degree-six surface, the dianodal surface. The exception at the common base points of $A,B,C$ is explained in the edition notes.

Given eight asyzygetic nodes, two quadrics $A,B$ through them and an eight-nodal quartic $T$, the family is

$$\alpha A^2+\beta B^2+2\gamma AB+2\rho T=0.$$

7. Nine nodes

For a syzygetic quartic $Q(A,B,C)=0$, an additional node must lie where

$$\operatorname{rank}\begin{pmatrix}\nabla A\\\nabla B\\\nabla C\end{pmatrix}\le2.$$

Jessop identifies this degree-six space curve with the vertices of singular quadrics in the net $A+\lambda B+\mu C$. Thus a cone $K$ through the eight associated points has its vertex at the ninth node.

Projection from this vertex gives a plane sextic with eight nodes on the conic defined by $K$. The conic must be a component, leaving a quartic. Jessop obtains

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

where $A$ passes through all nine nodes and $B$ through the eight associated points. He states that the family of nine-nodal syzygetic surfaces is three-dimensional.

For an asyzygetic eight-nodal surface $T$, the ninth node instead lies on

$$\operatorname{rank}\begin{pmatrix}\nabla A\\\nabla B\\\nabla T\end{pmatrix}\le2,$$

the degree-$18$ dianodal curve. Once it is chosen, Jessop gives a pencil $A^2+\rho P=0$, with $A$ the quadric through the nine nodes and $P$ a quartic having them. The rectangular double bars in Original mean that all maximal minors vanish.

A net of quadrics and eight forced nodes

Let $A,B,C$ be independent quadrics in $\mathbf P^3_{\mathbf C}$ whose common zero scheme consists of eight distinct transverse points. Such a complete intersection is often called a Cayley octad. Let $q$ be a nondegenerate quadratic form in three variables and set

$$S:\quad q(A,B,C)=0.$$

At a common zero, $A,B,C$ are local coordinates because their differentials are independent. The local quadratic term of the surface is therefore $q$ itself. Every base point is an ordinary node. This proves that the eight nodes are forced by the construction; it does not exclude additional nodes elsewhere.

For a concrete example take

$$A=x^2-w^2,\quad B=y^2-w^2,\quad C=z^2-w^2,\qquad F=A^2+B^2+C^2.$$

There are exactly eight singularities, $[\pm1:\pm1:\pm1:1]$. Indeed, in $w=1$, the first three derivatives force each of $x,y,z$ to be $0$ or $\pm1$. The derivative in $w$ requires $A+B+C=0$, which excludes every zero coordinate. At $w=0$, the equations $x^3=y^3=z^3=0$ give no projective point. At each of the eight nodes the affine Hessian is $8I_3$.

The genus-one family already inside the equation

Away from the base points the map

$$[A:B:C]:S\dashrightarrow\{q(a,b,c)=0\}\subset\mathbf P^2$$

lands on a smooth conic. A fibre is cut out by two independent linear combinations of $A,B,C$: it is a complete intersection of two quadrics in $\mathbf P^3$. When smooth, adjunction gives trivial canonical bundle and genus one.

After resolving the indeterminacy this describes a genus-one fibration when the general fibre is smooth and connected. Calling it an elliptic fibration would additionally require a chosen section; none is being silently assumed.

A quadric in the net is the pullback of a line in the parameter plane. A general line meets the parameter conic twice, explaining why its section of $S$ is two genus-one quartics. A tangent line gives one fibre twice. This makes Jessop’s split quadric sections and envelope of quadrics parts of the same construction.

This conic is the parameter conic of a net of space quadrics. The canonical conic in the previous reading comes from plane cubics. Their roles are related by Jessop’s projection construction, but they are not identified merely because both are conics.

Extra nodes are rank conditions

For the seven-point system $F=q(A,B,C)+\rho\Sigma$, the chain rule is

$$\nabla F=q_A\nabla A+q_B\nabla B+q_C\nabla C+\rho\nabla\Sigma.$$

If $\rho\ne0$, a singular point forces linear dependence of these four gradient columns. Their degrees are $1,1,1,3$, so the determinant has degree six. This is Jessop’s dianodal surface. For $\rho=0$ and $A=B=C=0$, all four coefficients can vanish; the determinant condition does not follow. This is the associated eighth-base-point alternative described in his Introduction.

Away from the octad, a singular point of a syzygetic surface forces the $3\times4$ matrix with rows $\nabla A,\nabla B,\nabla C$ to have rank at most two. A left-kernel vector gives a singular quadric of the net with that vertex. For an asyzygetic system $q(A,B)+\rho T$, the corresponding matrix has rows $\nabla A,\nabla B,\nabla T$.

The equations determine candidate locations. Membership in such a rank locus is not by itself proof that a chosen member of the family has an ordinary node there.

The ninth node forces a component

Suppose a ninth node is the vertex of a quadric cone $K$ through the original eight nodes. Projection from it puts those eight points on a conic in the plane. If the branch sextic did not contain that conic, Bézout would give total intersection multiplicity $12$. But eight nodes on the conic contribute at least $16$. Thus the conic is a component.

The remaining factor has degree four. In the surface model this is Jessop’s form $A^2+\rho KB=0$. The cone touches the surface along the genus-one quartic $A=K=0$, while its vertex becomes the extra singularity when it lies on $A=0$. For generic transverse data the eight points $A=K=B=0$ and that vertex are ordinary nodes; degenerations require separate checks.

The dianodal surface also contains the joins of the seven prescribed points and their twisted cubics: restrictions of the relevant quadrics and quartics vanish identically once their intersection multiplicities exceed the degree. The resulting shared tangents force dependence of the gradient columns. Jessop develops the finer $21$-line, seven-cubic and $35$-plane-cubic incidence statements on pp. 10–11. Those are retained as source results, not inferred solely from the rank-locus degree.