Projection from a node

3. Quartic surfaces with more than seven nodes.

The equation of a quartic surface having a node at the point $x=y=z=0$ will be of the form

$$u_2w^2+2u_3w+u_4=0,$$

where $u_2=0$, $u_3=0$, $u_4=0$ are cones whose vertex is the node.

The equation of the tangent cone drawn to the surface from this node is

$$u_2u_4-u_3^2=0.$$

The section of this cone by any plane, e.g. the plane $w=0$, is a sextic curve with a “contact-conic,” i.e. a conic which touches it wherever it meets it.

If the surface has any other node, the tangent cone will have a double line passing through this new node and giving rise to a node on this sextic; we obtain the different varieties of quartic surfaces possessing nodes by consideration of all special cases of sextic curves with a contact-conic*.

• This method is due to Rohn, see Die Flächen vierter Ordnung hinsichtlich ihrer Knotenpunkte und ihrer Gestaltung, Leipzig, 1886.

It is to be noted that the existence of a contact-conic $u_2=0$ of a sextic implies also a contact-quartic $u_4=0$; if a sextic has another contact-conic $v_2=0$, and hence another contact-quartic $v_4=0$, an identity exists of the form

$$u_2u_4-u_3^2=v_2v_4-v_3^2.$$

Now by multiplying the equation of the surface by $u_2$ we derive

$$(u_2w+u_3)^2+u_2u_4-u_3^2=0,$$

hence in the present case

$$(u_2w+u_3)^2+v_2v_4-v_3^2=0.\tag{1}$$

Denoting by $c_8$ the intersection of the quartic surface with the cone $v_2=0$, it is clear that $v_2=0$ meets the surface (1) in the curve $c_8$ and in the four lines $u_2=v_2=0$; but $v_2$ meets (1) where it meets the two nodal cubic surfaces

$$u_2w+u_3-v_3=0,\qquad u_2w+u_3+v_3=0,$$

hence in general $c_8$ must break up into two quartic curves, either of which is the partial intersection of $v_2$ with a cubic surface which contains also two generators of $v_2$. These curves are therefore quadri-quartics*.

• We denote by quadri-quartic the type of twisted quartic through which an infinite number of quadrics pass.

3. Projection from a node

Use projective coordinates $[x:y:z:w]$, with the node at $[0:0:0:1]$. The quartic equation is

$$u_2(x,y,z)w^2+2u_3(x,y,z)w+u_4(x,y,z)=0,$$

where $u_i$ is homogeneous of degree $i$. Jessop’s cone of lines tangent to the surface from the node has equation

$$u_2u_4-u_3^2=0.$$

Its section by $w=0$ is a sextic with contact conic $u_2=0$: the conic is tangent to the sextic at their intersections. A further surface node produces a double generator of this cone and hence a node of the plane sextic. Jessop therefore studies the possible sextics with contact conic to classify the surfaces.

The method is attributed to Rohn, Die Flächen vierter Ordnung hinsichtlich ihrer Knotenpunkte und ihrer Gestaltung, Leipzig, 1886.

A second contact conic

The equation $u_2u_4-u_3^2=0$ gives a contact quartic $u_4=0$ as well as the contact conic. If there is another representation

$$u_2u_4-u_3^2=v_2v_4-v_3^2,$$

multiply the surface equation by $u_2$ to obtain

$$(u_2w+u_3)^2+v_2v_4-v_3^2=0.$$

Intersect with the cone $v_2=0$. The resulting equation factors into the two cubic surfaces

$$u_2w+u_3-v_3=0,\qquad u_2w+u_3+v_3=0.$$

The multiplication by $u_2$ has introduced four generators given by $u_2=v_2=0$. Removing them, Jessop obtains two quartic curves in the intersection of the original surface and $v_2=0$. Each is residual to two generators in the intersection of the cone with a cubic surface. He calls these quadri-quartics: space quartics lying on a pencil of quadrics. This is a generic decomposition, as in the source.

A quadratic equation on each ray

Let $S\subset\mathbf P^3_{\mathbf C}$ have an ordinary node $O=[0:0:0:1]$. Its homogeneous equation has no $w^4$ or $w^3$ terms, so

$$F=u_2(x,y,z)w^2+2u_3(x,y,z)w+u_4(x,y,z)=0.$$

Here $u_2$ is nondegenerate. On a line through $O$ in direction $[x:y:z]$, the double intersection at $O$ leaves this quadratic equation for the other two points. Put

$$\Delta=u_3^2-u_2u_4,\qquad t=u_2w+u_3.$$

Completing the square gives $t^2=\Delta$. On the open set $u_2\ne0$ this is an invertible change, with $w=(t-u_3)/u_2$. The involution exchanging the two points is

$$w\longmapsto-w-\frac{2u_3}{u_2}.$$

The plane sextic $B=\{\Delta=0\}$ is the branch curve: the two roots coincide there. This is a birational double-plane description. The formula dividing by $u_2$ is not a global isomorphism, and directions of lines contained in $S$ need separate treatment.

Why the conic is tangent

On $Q=\{u_2=0\}$, the discriminant restricts to $u_3^2$. Thus the intersection divisor $B\cap Q$ is twice the divisor cut out by $u_3$ on $Q$. If $Q$ is smooth, the six intersections $u_2=u_3=0$ are distinct, and $u_4$ is nonzero there, these are ordinary tangencies.

This condition also clarifies two meanings of “tangent cone.” The local tangent cone at $O$ is quadratic, $u_2=0$. Jessop’s degree-six cone drawn from $O$ is the cone of lines with a repeated residual intersection, $\Delta=0$. They are different objects.

For another surface node whose direction lies in $u_2\ne0$, the change to $(x,y,t)$ gives a local equation $t^2=\Delta(x,y)$. At the singular point $t=0$ and $d\Delta=0$. The quadratic term is nondegenerate precisely when the plane quadratic term of $\Delta$ has rank two. Under this hypothesis the projected point is an ordinary node. Nodes in exceptional directions are not covered by this local argument.

Contact curves as splitting conditions

Suppose the same branch sextic has a second presentation

$$\Delta=v_3^2-v_2v_4.$$

Over the conic $v_2=0$, the double-plane equation factors:

$$t^2-v_3^2=(t-v_3)(t+v_3)=0.$$

The inverse image of that conic splits into two curves. Exchanging sheets exchanges the components. Returning to the quartic model gives Jessop’s two cubic equations $u_2w+u_3\pm v_3=0$. His subtraction of generators is necessary because multiplication by $u_2$ introduced an extra component; ignoring it would give the wrong degrees.

More generally, a curve $D$ away from the branch components splits in a double cover precisely when the branch function becomes a square in the function field of the normalization of $D$. Tangency supplies even orders of zeros. On a rational curve, that is enough after accounting for poles; on a curve of positive genus it need not be. The next reading gives the obstruction and an explicit counterexample.