Nodal sextics and their normalizations

4. Nodal sextics*.

For the purpose of classification of nodal quartic surfaces we discuss various properties of sextic curves with a contact-conic. In the first place it may be seen that sextic curves with six nodes lying on a conic $c_2$ can have their equation expressed as above. For if $c_3=0$ is any cubic through the six points, any other cubic through them is of the form $c_3+c_2L=0$; and any sextic through the complete intersection of $c_2$ and $c_3$ being

$$c_2c_4+c_3c_3'=0,$$

if the six points are nodes on this sextic $c_4$ and $c_3'$ must be of the form $c_3M+c_2N$, $c_3+c_2R$ respectively.

Hence the required sextic takes the form

$$c_3^2+c_2c_3A+c_2^2B=0,$$

i.e. the form $K_3^2-c_2^2V=0$, and hence has a contact-conic.

The corresponding quartic surface is $w^2V+2wK_3+c_2^2=0$; this has the plane $w=0$ as a singular tangent plane or trope, which touches the surface along a conic.

• See Rohn, l.c.

Sextics with seven nodes.

There are two different kinds of seven-nodal sextics, viz. that for which it is possible to find a pair of points $P,P'$ on the curve, such that through the seven nodes $D_1\ldots D_7$ and $P,P'$ there pass an infinite number of cubics, and the one for which it is not possible; considering the former kind, then if one such pair of points exists there is an infinite number of such pairs; for taking $c_3$ and $c_3'$ as two such cubics, then $c_6$, the given sextic, since it passes through the complete intersection of $c_3$ and $c_3'$, has an equation of the form

$$c_3\Gamma_3+c_3'\Gamma_3'=0.$$

Now $c_3$ meets $c_6$ only in $D_1\ldots D_7$, $P,P'$ and two further points $Q,Q'$, hence $\Gamma_3'$ passes through $D_1\ldots D_7$ and also through $Q$ and $Q'$; so that two and therefore an infinite number of cubics pass through $D_1\ldots D_7$, $Q$ and $Q'$. By varying the cubic through the nine points $D_1\ldots D_7$, $P,P'$ we form an involution of points $Q,Q'$ on $c_6$. If $Q$ coincides with $P$, $Q'$ will coincide with $P'$; therefore every cubic through the seven nodes which touches $c_6$ once will touch it twice.

Since $\Gamma_3$ is seen to pass through $D_1\ldots D_7$ and since only three linearly independent cubics pass through seven points, there is a linear connection between $c_3,c_3',\Gamma_3$ and $\Gamma_3'$, hence the sextic which has the property considered is represented by an equation of the form

$$(a\mathbin{\between}\phi,\psi,\chi)^2=0,$$

where $\phi,\psi,\chi$ are any three cubics through the given nodes.

Now $f$ and $c_9$ meet in $9\times6-8\times6=6$ points apart from the nodes, hence every sextic with eight nodes is touched by six cubics through these nodes.

The foregoing result as to contact-cubics is modified as follows: through any eight nodes of a sextic with nine nodes there pass four tangent cubics; through any eight nodes of a sextic with ten nodes there pass two tangent cubics.

5. Sextics with ten nodes.

The following result for ten-nodal sextics is important for our purpose: every plane sextic with ten nodes and a contact-conic is the projection of a twisted sextic on a quadric: for choosing any centre of projection $O$ and any quadric whose section by the polar plane of $O$ for the quadric projects into the given contact-conic, the sextic cone whose base is the given sextic meets the quadric in a curve $c_{12}$ which has twenty-six actual double points, since each node of the plane sextic gives rise to two nodes on $c_{12}$, and each point of contact of the contact-conic and the sextic is the projection of a point at which two branches of $c_{12}$ touch each other. Moreover $c_{12}$ has thirty apparent double points*, hence the projection of $c_{12}$ from any point has $30+26=56$ nodes, and this is one more than can be possessed by a curve of order $12$ which does not break up into simpler curves. Hence $c_{12}$ must break up into two sextic curves.

There are three varieties of twisted sextics on a quadric: (1) its intersection with a cubic surface, (2) its partial intersection with a quartic surface which also contains two generators of the quadric of the same species, (3) its partial intersection with a quintic surface which also contains four generators of the quadric of the same species.

• Salmon, Geom. of three dimensions (fifth ed. 1912), vol. i. p. 356.

4. Six nodes on a conic

Let $c_2=0$ be a conic through six nodes of a sextic, and $c_3=0$ a cubic through those points. Jessop writes a sextic through their complete intersection as

$$c_2c_4+c_3c_3'=0.$$

The double-point conditions give the form

$$c_3^2+c_2c_3A+c_2^2B=0.$$

Completing the square gives $K_3^2-c_2^2V=0$, where $V$ is quadratic. Consequently the corresponding quartic

$$w^2V+2wK_3+c_2^2=0$$

has $w=0$ as a trope, a plane whose intersection with the surface is a conic counted twice. This follows Jessop’s use of the equations; the source’s implicit genericity is discussed separately.

Seven nodes and paired points

Let $D_1,\ldots,D_7$ be the nodes. Jessop distinguishes whether two further points $P,P'$ on the sextic can lie on a pencil of cubics through all seven nodes.

Suppose they can, and choose two cubics $c_3,c_3'$ in that pencil. Since the sextic contains their complete intersection, it can be written

$$c_3\Gamma_3+c_3'\Gamma_3'=0.$$

Besides the seven nodes and $P,P'$, the cubic $c_3$ meets the sextic at two residual points $Q,Q'$. The equation forces $\Gamma_3'$ through these points and the seven nodes. Thus $Q,Q'$ also lie on a pencil of cubics through the nodes. Varying $c_3$ gives the paired-point involution. When $Q=P$, its partner $Q'$ equals $P'$, so a cubic tangent once is tangent twice.

The four cubics $c_3,c_3',\Gamma_3,\Gamma_3'$ belong to the three-dimensional vector space of cubics through the nodes. Their linear dependence expresses the sextic as

$$Q(\phi,\psi,\chi)=0,$$

where $\phi,\psi,\chi$ form a basis of that vector space and $Q$ is a quadratic form.

Eight nodes

In the preceding source calculation, $f=0$ is the sextic and $c_9=0$ the degree-nine Jacobian curve associated with the pencil of cubics through the eight nodes. Their intersections away from the nodes number

$$9\cdot6-8\cdot6=6.$$

These are the tangencies, giving six tangent cubics through the eight nodes. The construction of $c_9$ is outside the selected excerpt; it is described in Reinterpreted before the alternative count.

If the sextic has nine nodes, the pencil through eight of them has four tangent cubics. For a sextic with ten nodes, the corresponding number is two.

5. Ten nodes and a lift to a quadric

Take a quadric projecting from $O$ with the given contact conic as branch conic. Intersect it with the cone over the ten-nodal sextic. Jessop counts $26$ actual double points on the degree-$12$ intersection curve: two over each of the ten nodes and one over each of the six contacts. He also counts $30$ apparent double points in a generic plane projection, invoking Salmon, Geometry of Three Dimensions, fifth edition, vol. I, p. 356.

The total $56$ exceeds $55$, the maximum genus of an irreducible plane curve of degree $12$. Jessop concludes that the intersection splits into two sextics.

He distinguishes three kinds of sextic on a quadric: the complete intersection with a cubic surface; the residual intersection with a quartic surface containing two lines of the same ruling; and the residual intersection with a quintic surface containing four lines of that ruling. The original intersection count, including its invoked input, is retained here.

A conic of nodes and a doubled plane section

Work over $\mathbf C$. Suppose six distinct points are the transverse complete intersection of a smooth conic $c_2=0$ and a cubic $c_3=0$. A sextic with double zeros at these points has, in this complete-intersection situation, an expression

$$a c_3^2+c_2c_3L+c_2^2M,$$

where $L,M$ have degrees one and two. If $a\ne0$, rescale and complete the square to write it as $K_3^2-c_2^2V$. The assumption $a\ne0$ excludes a reducible exceptional expression with a $c_2$ factor.

The corresponding surface $Vw^2+2K_3w+c_2^2=0$ restricts on $w=0$ to $c_2^2=0$. This is the precise meaning of a trope: its plane section is twice a conic. Along that conic the derivatives in $x,y,z$ vanish, while the derivative in $w$ is $2K_3$. Thus the six points $c_2=K_3=0$ are singular on the surface; ordinary nodes require a nondegenerate quadratic term. Being a doubled plane section does not make the whole conic a singular curve of the surface.

Adjoint cubics are canonical sections

Let $C\subset\mathbf P^2$ be an irreducible sextic with exactly $\delta$ ordinary nodes and no other singularities. Let $\nu:\widetilde C\to C$ be its normalization, and write $p_i^+,p_i^-$ for the two points over its $i$th node. Put $H=\nu^*\mathcal O_C(1)$, a divisor class of degree six.

A divisor records points with integer multiplicities. Its line bundle has sections that can be viewed as rational functions with the prescribed allowed zeros and poles. We write $h^0(D)$ for the dimension of the section space. The canonical divisor $K$ is the divisor class of differential forms on the smooth curve.

Adjunction for a nodal plane curve gives

$$K_{\widetilde C}=3H-\sum_{i=1}^{\delta}(p_i^++p_i^-),\qquad g(\widetilde C)=10-\delta.$$

The restrictions of plane cubics through every node are exactly these canonical sections: passing through a node removes one zero on each branch. Thus their residual intersection degree is $18-2\delta=2g-2$. Adjunction is an external theorem here; the residual count makes its content visible.

For $\delta=7$, the curve has genus three, and the three-dimensional space of adjoint cubics defines its canonical map to $\mathbf P^2$.

Jessop’s paired points are hyperelliptic fibres

Take distinct smooth points $P,Q$ of $C$, also denoting their lifts to $\widetilde C$ by $P,Q$. A pencil of cubics through the seven nodes and $P,Q$ means

$$h^0(K-P-Q)\ge2.$$

Riemann–Roch in genus three gives

$$h^0(K-P-Q)=h^0(P+Q).$$

Consequently Jessop’s condition holds exactly when a divisor of degree two moves in a pencil. Such a pencil has no base point: otherwise a degree-one moving divisor would identify the curve with $\mathbf P^1$. It defines a map $\widetilde C\to\mathbf P^1$ of degree two. A smooth curve of genus at least two admitting such a map is called hyperelliptic.

Conversely, a general fibre $P+Q$ of a hyperelliptic map has $h^0(P+Q)=2$, so gives Jessop’s pencil. The partner assignment is the involution exchanging the two sheets. One suitable pair therefore produces infinitely many pairs because the fibres vary over $\mathbf P^1$.

For genus three the canonical map of a hyperelliptic curve is the double cover followed by the conic embedding $[s:t]\mapsto[s^2:st:t^2]$. Hence the canonical coordinates $\phi,\psi,\chi$ satisfy a quadratic relation. Substituting their plane cubic representatives gives a degree-six polynomial vanishing on $C$, hence a scalar multiple of its defining equation:

$$C:\quad Q(\phi,\psi,\chi)=0.$$

If the normalization is not hyperelliptic, the canonical map is an embedding as a plane quartic, and no such paired fibres occur. The canonical-map theorem is an invoked input; the equivalence with Jessop’s pencil condition is the Riemann–Roch argument just given. See Buium, Algebraic Curves, Lecture 14.

The conic image of the canonical map is not the contact conic in the original plane. They live in different planes. Jessop’s construction of the latter on p. 5 carries additional geometric information and is not replaced merely by naming the former.

Why six, four and two tangent cubics?

Choose eight nodes of an irreducible sextic with $\delta=8,9$ or $10$. Assume the pencil of cubics through them has no additional base point on the normalization. Its moving intersection divisor has degree $18-16=2$, so it defines a degree-two map $f:\widetilde C\to\mathbf P^1$.

A tangent cubic has its two residual intersections coincide: it is a ramified fibre. The Riemann–Hurwitz formula, an external input, is

$$2g-2=2(-2)+\deg R,\qquad\deg R=2g+2.$$

With $g=10-\delta$, this gives $6,4,2$. In characteristic zero a smooth double cover has simple ramification. To interpret every ramification point as an ordinary tangency in the original plane, also assume it lies away from the branches over the nodes. Otherwise the count is on the normalization and the geometric interpretation changes. See Riemann–Hurwitz, Stacks Project.

Jessop’s first count uses the Jacobian $J(f,\phi,\psi)$ of the sextic equation and two cubics through its eight nodes. Its degree is $5+2+2=9$. He subtracts eight contributions of six from the intersection number $9\cdot6$. Ramification explains why the residual six is intrinsic and why the two later counts fall by two.

Ten nodes: why the lift splits

An irreducible ten-nodal sextic has normalization $\mathbf P^1$. Let $Q_0\to\mathbf P^2$ be the double cover branched over its smooth contact conic, realized as a smooth quadric surface. In an affine trivialization its equation is $s^2=q_2$.

Assume the conic avoids the ten nodes and has six ordinary contacts with the sextic. On the normalization the divisor of $q_2$ is even, including the poles introduced by the trivialization. Every degree-zero divisor on $\mathbf P^1$ is principal. Thus $q_2=c h^2$ for a rational function $h$; over $\mathbf C$ the nonzero constant $c$ is also a square. The pulled-back cover splits as $s=\pm\sqrt c\,h$.

Its two components map birationally to the sextic and have degree six on $Q_0$. This gives the splitting in §5 without Jessop’s apparent-double-point calculation.

Why even contact is not always enough. On a smooth positive-genus curve, an even principal divisor $\operatorname{div}(f)=2D$ only says that the class of $D$ has order dividing two in the divisor-class group. If that class is nonzero, $f$ is not a square. For example, on

$$E:\quad y^2=x(x-1)(x-a),\qquad a\ne0,1,$$

the function $x$ has divisor $2P-2O$, with $P=(0,0)$ and $O$ the point at infinity. It is not a square in $\mathbf C(E)$: a square root would have divisor $P-O$ and give a degree-one map from a genus-one curve to $\mathbf P^1$. Thus adjoining $\sqrt x$ gives a connected unramified double cover. The missing information is a two-torsion divisor class. This explains exactly why rationality matters in Jessop’s ten-node step.

Three bidegrees replace three constructions

A smooth quadric is $\mathbf P^1\times\mathbf P^1$. Let $f_1,f_2$ denote the two ruling classes. They satisfy $f_1^2=f_2^2=0$ and $f_1f_2=1$. A curve of class $af_1+bf_2$ has degree $a+b$ because a hyperplane section has class $f_1+f_2$. Adjunction, with $K_{Q_0}=-2f_1-2f_2$, gives

$$p_a(a,b)=1+\frac{(af_1+bf_2)((a-2)f_1+(b-2)f_2)}2=(a-1)(b-1).$$

For an irreducible sextic both $a,b$ are positive. Up to exchanging rulings, $a+b=6$ leaves exactly three cases:

The last column is $10-p_a(a,b)$, and uses a projection with only ordinary, distinct nodes. Jessop’s numbers $6,7,10$ now follow from one genus calculation. If the lift has other singularities, arithmetic genus measures their total genus defect, not a literal number of ordinary nodes.