The moduli space of binary quintics

Abstract

We recall the classical construction and theory of invariants for the case of binary quintics, describe the moduli space, and identify the curves in it defined by quintics having symmetry. We describe the real case, and identify the number of real roots depending on the point in moduli space. Our main interest is in five curves of binary quintics defined as linear sections of plane curves with infinite symmetry groups: these play a role in the canonical stratification of jet space, so we describe their singularities and count their intersections. All this is done in the classical case. Thereafter we analyse the changes to be made to the whole theory when we work in characteristic 2.

1 Invariant theory

We begin with classical invariant theory, for example as in [2], though we do not follow that version. We take $\mathbb {C}$ as ground field for the first two sections. The ring of ($\mathrm{SL}_{2}$-)invariants of binary quintics is generated by four invariants, which we call $I_{4}, I_{8},I_{12}$ and $I_{18}$, where $I_{d}$ has degree d (see [2, p. 131], but probably the first proof was given by Hermite [3]).

We normalise the invariants $I_{4}, I_{8}, I_{12}$ using a normal form, due to Sylvester [4] (see also [1], [2, p. 230]).

Lemma 1.1

A general binary quintic can be expressed uniquely as a sum of three fifth powers of linear forms.

Proof

We represent the quintic $a = \sum _{r=0}^{5}{5\atopwithdelims ()r}a_{r}x^{5-r}y^{r}$ by the point A with coordinates $(a_{r})$ in $\mathbb {P}^{5}$. Then a quintic that is a fifth power $(t_{1}x+t_{2}y)^{5}$ corresponds to a point T lying on the standard rational normal curve C.

If a is the sum of three fifth powers, A lies in the plane spanned by the corresponding points on C. No point may lie on two such planes, for otherwise the planes would lie in a hyperplane containing six points of C, whereas C has degree 5. But since there are three degrees of freedom for the three points on C, and hence for the plane, and two more for the point A on it, a generic point will lie on such a plane. $\square$

We may extract more from this argument. It follows by specialisation that any point A in $\mathbb {P}^{5}$ lies on a plane meeting C in three points, but the points need not be distinct. If all three coincide, so that A lies on the osculating plane corresponding to a point T, then a is divisible by the cube of $t_{1}x+t_{2}y$. If just two coincide, corresponding say to $x^{5}$ (repeated) and $y^{5}$, then we may write $a=x^{4}(px+qy) +ry^{5}$. A fuller account of the exceptional cases is in [2, p. 231].

We may thus write a general binary quintic in the form $al^{5}+bm^{5}+cn^{5}$ with $l+m+n=0$. Then the invariants are given in terms of the elementary symmetric functions $\sigma _{1}, \sigma _{2}$ and $\sigma _{3}$ of a, b and c by the formulae

$$\begin{aligned} I_{4}=4\sigma _{1}\sigma _{3}-\sigma _{2}^{2},\quad I_{8}=\sigma _{2}\sigma _{3}^{2},\quad I_{12}={-}3\sigma _{3}^{4}. \end{aligned}$$

For the above form, the discriminant vanishes if and only if, for some choice of the radicals, we have

$$\begin{aligned} a^{-1/4}+b^{-1/4}+c^{-1/4} = 0. \end{aligned}$$

Also we have, up to a constant factor,

$$\begin{aligned} I_{18} = a^{5}b^{5}c^{5}(b-c)(c-a)(a-b). \end{aligned}$$

We have just seen that a quintic which may not be written in the above form either has a cubed factor or is equivalent to $x^{4}(px+qy) +ry^{5}$. For such a form, $I_{12}= 0$.

The invariants are subject to a unique syzygy (of degree 36)

$$\begin{aligned} I_{18}^{2} = {-}144I_{12}^{3} + (I_{4}^{3}-72I_{4}I_{8})I_{12}^{2} + (24I_{8}^{3}-6I_{4}^{2}I_{8}^{2})I_{12} + 9I_{4}I_{8}^{4}. \end{aligned}$$

(1)

Taking $(I_{4},I_{8},I_{12})$ as coordinates gives an isomorphism of the moduli space $\mathcal {M}$ onto the weighted projective space $\mathbb {P}(1,2,3)$. For the values of these invariants determine $I_{18}$ up to sign, while in the weighted projective space $\mathbb {P}(4,8,12,18)$, the coordinates $(I_{4},I_{8},I_{12},I_{18})$ and $(t^{4}I_{4}, t^{8}I_{8}, t^{12}I_{12}, t^{18}I_{18})$ both give the same point; taking $t=i$ now shows that changing the sign of $I_{18}$ still gives the same point.

The space $\mathcal {M}$ is smooth except at two points: (0, 1, 0), where there is a singularity of type $A_{1}$, and (0, 0, 1), where there is a singularity of type $A_{2}$. It gives a moduli space for all quintics with no factor of multiplicity 3 or more; all such are stable in the sense of geometric invariant theory (for the action of $\hbox {SL}_2$ on $\mathbb {P}^{5}$). Quintics with a repeated root yield a curve $\Delta \subset \mathcal {M}$: the discriminant curve. It has equation $I_{4}^{2}=128I_{8}$.

2 Special quintics

For any set of five distinct points on $\mathbb {P}^{1}$ admitting a nontrivial symmetry group, we can choose coordinates so that the five points appear as $(0,{\pm }\,\alpha ,{\pm }\,\beta )$ for some $\alpha ,\beta$; moreover, we may take $\alpha \beta = 1$. The quintic is thus $x(x^{4} -2tx^{2}y^{2} + y^{4})$ for $t = (\alpha ^{2} +\beta ^{2})/2$. For such quintics, $I_{18} = 0$. The other invariants are

$$\begin{aligned} \begin{aligned} I_{4}&= 16t(3t^{2}+5),\\ I_{8}&= 2(t^{2}-5)^{2}(9t^{2}-5),\\ I_{12}&= 12t(t^{2}-5)^{4}.\end{aligned} \end{aligned}$$

(2)

This parametrises the curve $\Gamma _{5} \subset \mathcal {M}$ whose equation is given by the right hand side of (1). This is not a good parametrisation since t and $-t$ yield the same point of the curve.

We have the following points of particular significance on the discriminant curve $\Delta$:

• (harmonic tetrad with one point repeated) $F_{0} = (48,18,12)$, represented by $f_{0} = x^{2}y(x^{2}-y^{2})$,

• (equianharmonic tetrad with one point repeated) $F_{1} = (0,0,1)$, represented by $f_{1} = x^{2}(x^{3}-y^{3})$,

• (two repeated points, and another point) $F_{2} = (16,2,6)$, represented by $f_{2} = x^{2}y^{2}(x-y)$.

The intersection of $\Gamma _{5}$ with $\Delta$ consists of $F_{0}$, and $F_{2}$ counted twice.

The invariants of a pentad with distinct points, and with symmetry group of order greater than 2, give one of the following points in $\mathcal {M}$:

• $F_{3} = (-3,3,3)$, represented by $f_{3} = xy(x^{3}-y^{3})$,

• $F_{4} = (0,1,0)$, represented by $f_{4} = x(x^{4}-y^{4})$,

• $F_{5} = (1,0,0)$, represented by $f_{5} = x^{5}-y^{5}$,

with 3-, 4- and 5-fold symmetry respectively.

The singular points of $\mathcal {M}$ occur at $F_{4}$ and $F_{1}$, with types $A_{1}$ and $A_{2}$ respectively. The curve $\Gamma _{5}$ has singular points of type $A_{2}$ at $F_{3}$ and of type $A_{5}$ at $F_{4}$.

3 The real case

In this section we will write $\mathcal {M}(\mathbb {R})$ for the set of real points of $\mathcal {M}$.

Lemma 3.1

Each point of $\mathcal {M}(\mathbb {R})$ is represented by a quintic with real coefficients.

Proof

For one proof (which we believe is due to Hermite) we suppose that the point does not lie on $\Delta$ (other cases may be treated directly). Invariant theory provides linear covariants of a quintic, among which are $\alpha$ (of degree 5) and $\beta$ (of degree 7) in the notation of [2, p. 130]. If the quintic has no repeated root, these are linearly independent and can be taken as coordinates. Then the coefficients of the quintic are themselves invariants, hence are real.

An alternative argument may be given when $I_{12} \ne 0$. Write the quintic in the above normal form $al^{5} +bm^{5} +cn^{5}$ with $l+m+n=0$. Then the invariants are given above; since $\sigma _{3}\ne 0$, the same point on $\mathcal {M}$ is given by $(4\sigma _{1}-{\sigma _{2}^{2}}/{\sigma _{3}}, \sigma _{2}, -3\sigma _{3})$. Thus adjusting a, b and c by a common scalar factor we may suppose that these, and hence the $\sigma _{r}$ are all real.

Now either a, b, c are all real and we may take l, m, n to be real linear forms or one—say a—is real and the other two complex conjugate. We then take l real and m, n complex conjugate (e.g. $l=2x$, $m={-}x+iy$, $n={-}x-iy$) to obtain a real quintic. $\square$

If we think of the quintic as determining five points on $\mathbb {P}^{1}$, then a real form gives an anti-holomorphic involution preserving the set of points. If there are two such, their product gives a holomorphic automorphism, so the point in $\mathcal {M}$ must lie on $\Gamma _{5}$.

We wish to determine, for each real point in $\mathcal {M}(\mathbb {R})$, the number of real roots of the corresponding real quintic. The standard theory tells us that for one sign of the discriminant there are three real roots; for the other, either one or five real roots. It follows from the above that the number of real roots may only change as we cross $\Gamma _{5}$ or $\Delta$. We next investigate $\Gamma _{5}$.

As before, for five distinct points on $\mathbb {P}^{1}_{\mathbb {R}}$ admitting a nontrivial symmetry, we choose coordinates so that the points are $(0,{\pm }\alpha ,{\pm }\beta )$ for some $\alpha ,\beta$; this fixes x, y up to scalars, and the quintic is then x multiplied by a quadratic in $x^{2}$ and $y^{2}$. We adjust the scalars so that the coefficient of $x^{5}$ is 1 and that of $xy^{4}$ is $\pm 1$ (if the coefficient of $xy^{4}$ vanishes, the quintic is unstable; if that of $x^{5}$ vanishes, we have the point $F_{0}$). This fixes x, and y up to sign. However, changing the sign of the coefficient of $x^{3}y^{2}$ gives the same point of $\mathcal {M}(\mathbb {R})$ (it corresponds to substituting iy for y).

First consider $x(x^{4} -2tx^{2}y^{2} + y^{4})$. From the invariants given above, we find $I_{4}^{2}-128I_{8}=2^{8}5^{3}(t^{2}-1)^{2}>0$ except when $t=\pm 1$ corresponding to $F_{2}$, or, in the limit, $t=\infty$ corresponding to $F_{0}$. As $t^{2}$ increases from 0 to $\infty$ the point in $\mathcal {M}(\mathbb {R})$ runs from $F_{4}$ to $F_{0}$, touching $\Delta$ at $F_{2}$ when $t^{2}=1$ and passing through $F_{5}$ when $t^{2}=20$.

The quadratic for $(x/y)^{2}$ has no real roots for $t^{2}<1$ and two real roots for $t^{2}>1$, which have the same sign as t. Thus the quintic has five real roots for $t>1$ and only one for $t<1$.

For $x(x^{4} -2ux^{2}y^{2} -y^{4})$, we have $I_{4}^{2}-128I_{8}={-}2^{8}5^{3}(u^{2}+1)^{2}<0$. As $u^{2}$ increases from 0 to $\infty$, the point in $\mathcal {M}(\mathbb {R})$ runs from $F_{4}$ to $F_{0}$, passing through $F_{3}$ when $u^{2}=3$. The quadratic for $(x/y)^{2}$ has two real roots of opposite signs. Thus the quintic has three real roots.

To interpret these facts it is convenient to consider the affine chart of $\mathcal {M}(\mathbb {R})$ where $I_{4}=1$, so that we may take $(I_{8},I_{12})$ as coordinates. Taking the weights into account, we see that the compactification of this plane in $\mathcal {M}(\mathbb {R})$ may be topologically described by regarding the plane as the interior of a disc, and identifying points on the boundary via reflection in the horizontal axis. In addition, the two points at the ends of this axis are identified in $\mathcal {M}(\mathbb {R})$ (the space has a topological singularity—a quadratic cone—at $F_{4}$), but it is convenient to describe the space where these are regarded as separate: this is homeomorphic to a sphere.

In this plane picture, the curve $\Delta$ is represented by a vertical line. The curve $\Gamma _5$ runs from the point $F_4$ at infinity on the left to the corresponding point to the right, touching $\Delta$ at $F_2$, then crossing it at $F_0$, then having a simple cusp at $F_5$. It follows that the region to the right of $\Delta$, which corresponds to having three real roots, is not separated by $\Gamma _{5}$; while the region to the left is separated into two components. Both sides of that part of $\Gamma _{5}$ going off to the left (where $t^{2}<1$) give quintics with one real root, thus those with five real roots correspond to the small region bounded by $\Gamma _5$ and $\Delta$, near the centre of the figure.

4 Exceptional curves

We are interested in the five curves in $\mathcal {M}$ arising in [5]: as explained in that paper, this is part of the analysis of the partition of the $N_{16}$ stratum induced by the canonical stratification of jet space. These curves are constructed as follows (here we return to working over $\mathbb {C}$). We start with the quintic curves $H_{i}$ $(1\leqslant i\leqslant 5)$ defined in $\mathbb {P}^{2}$ by the equations $h_{i}= 0$, with $h_{i}$ given by the following table, which also gives the types of singularities of the curves $H_{i}$ at the points indicated:

Equation	(0, 0, 1)	(0, 1, 0)	(1, 0, 0)
$h_{1}$ = $xz(y^{2}z-x^{3})$	$D_{8}$	$D_{5}$
$h_{2}$ = $z(y^{3}z-x^{4})$	$E_{6}$	$A_{7}$
$h_{3}$ = $y^{3}z^{2}-x^{5}$	$E_{8}$	$A_{4}$
$h_{4}$ = $yz(y^{2}z-x^{3})$	$E_{7}$	$A_{5}$	$A_{1}$
$h_{5}$ = $xyz(yz-x^{2})$	$D_{6}$	$D_{6}$	$A_{1}$

Each curve $H_i$ admits a (semisimple) 1-parameter symmetry group $G_{i}$. Thus the lines L in the plane $\mathbb {P}^{2}$ fall into a 1-parameter family of orbits under $G_{i}$, so their intersections with $H_{i}$ give essentially a 1-parameter family of binary quintics, whose moduli trace out a curve in $\mathcal {M}$, whose closure we denote $\Gamma _{i}$.

The other exceptional cases ($2D_6$ and $T_{2,3,10}$) in the main theorem of [5], like $h_5$, are represented by quintic curves C composed of two conics in a pencil containing a repeated line, together with that line. The intersection of C with any line thus consists of two pairs in the involution cut on the line by the pencil, together with one of its fixed points; hence is a pentad possessing a symmetry. So these all yield the curve $\Gamma _5$ in $\mathcal {M}$; we need not consider them further.

If the line L is an edge of the triangle of reference, or even if it passes through the point (0, 0, 1), the corresponding binary quintic is unstable, so does not determine a point of $\mathcal {M}$. For lines through (1, 0, 0) or (0, 1, 0) respectively, the corresponding point in $\mathcal {M}$ is given by the following table:

Line	$H_{1}$	$H_{2}$	$H_{3}$	$H_{4}$	$H_{5}$
$y=z$	$F_{3}$	$F_{4}$	$F_{5}$	$F_{1}$	$F_{0}$
$x=z$	($F_{2}$)	$F_{1}$	$F_{1}$	$F_{0}$	($F_{2}$)

Here the symbol $(F_{2})$ means that although the intersection gives an unstable binary quintic, the completed curve $\Gamma _{i}$ in $\mathcal {M}$ passes through $F_{2}$ at the corresponding limiting point.

We will see later that $F_{i}\in \Gamma _{j}$ only as indicated in this table, except that (as we already know), $F_{3}, F_{4}$ and $F_{5}$ belong to $\Gamma _{5}$.

To obtain a parametrisation of $\Gamma _{i}$ we substitute $(x,1,z{+}u)$ for (x, y, z) in $h_i$, and calculate the invariants $\{I_4,I_8,I_{12}\}$ of the resulting polynomial in x. This yields

These do not give good parametrisations of the curves in $\mathbb {P}(1,2,3)$: to obtain these, we need to substitute $u^{2}= T$ for $\Gamma _1$ and $\Gamma _4$ and $u^3=T$ for $\Gamma _2$ and $\Gamma _3$. For $\Gamma _5$ it is better to let $t^2=T$ in (2).

We determine equations for these curves by eliminating the parameter u. If $R_{i}=0$ is the equation of $\Gamma _{i}$, we have $\mathrm{deg}\,\Delta = 2$ and

i	1	2	3	4	5
deg  $\Gamma _{i}$	12	11	8	14	9

In the following equations, x, y, z stand for $I_4,I_8$ and $I_{12}$ respectively.

5 Intersections and singularities

In general in $\mathcal {M}\cong \mathbb {P}(1,2,3)$ curves of degrees $d_{1}$ and $d_{2}$ will have $d_{1}d_{2}$ points of intersection. At smooth points, intersection multiplicities are as usual. At a singular point we lift the curve-germs to the appropriate branched cover, determine the intersection number there, and divide by 2 (at the singular point of type $A_{1}$) or 3 (for that of type $A_{2}$).

First we check intersections with $\Delta$. We find

$$\begin{aligned} \Gamma _{1}\cap \Delta&= 3F_{2} + G_{1},\\ \Gamma _{2}\cap \Delta&= \frac{8}{3}\,F_{1} + G_{2},\\ \Gamma _{3}\cap \Delta&= \frac{5}{3}\,F_{1} + G_{3},\\ \Gamma _{4}\cap \Delta&= \frac{2}{3}\,F_{1} + 3F_{0} + G_{4},\\ \Gamma _{5}\cap \Delta&= F_{0} + 2F_{2}, \end{aligned}$$

where the points $G_{i}$ are distinct from each other and from the $F_{i}$. In particular, the only points $\Gamma _{i}\cap \Gamma _{j}\cap \Delta$ occur among $F_{0}, F_{1}, F_{2}$.

The mutual intersections of the $\Gamma _{i}$ can be found by substituting a parametrisation of $\Gamma _{i}$ in the equation $R_{j}$ and factorising. A first result is that (except at $F_{0}, F_{4}$), all intersection numbers with $\Gamma _{5}$ are even, so for this purpose, $\Gamma _{5}$ does behave as though its degree were $4\frac{1}{2}$. The factors are as in the following table, where $\alpha _{i}$ is of degree 1, and corresponds to $F_{i}$; $\phi _{j}$ is irreducible of degree j over $\mathbb {Q}$ (and not necessarily the same on different occasions).

Apart from the $\alpha _{j}$, different $\phi _{j}$ all represent points distinct from the $F_{j}$ and from each other. Collectively, the $\phi$’s respresent 151 points, at each of which just two of the $\Gamma _{i}$ meet transversely (save when one curve is $\Gamma _{5}$, when we have simple contact.

We now consider singularities of the curves $\Gamma _{j}$. In general, a smooth curve of degree d in weighted projective space $\mathbb {P}(a_{0},a_{1},a_{2})$ has genus g equal to the number of monomials of degree $d-\sum _{0}^{2}a_{i}$. This is to be decreased by terms corresponding to any singularities (the usual rule $\mu +r-1$ will apply at points smooth on $\mathcal {M}$). Now our curves $\Gamma _{1},\ldots ,\Gamma _{5}$ have degrees 12, 11, 8, 14, 9; decreasing by 6 gives 6, 5, 2, 8, 3 and the corresponding numbers of monomials are 7, 5, 2, 10, 3: e.g. in the first case, we have $x^{7},x^{5}y,x^{3}y^{2},xy^{3},x^{3}z,xyz,xz^{2}$. Since our curves are parametrised, they are rational, so these numbers must be accounted for by singularities.

Computationally, we seek repeated factors of the discriminant of $R_{i}$ with respect to $z=I_{12}$. These are, with notation as above and where $\delta$ denotes $I_{4}^{2}-128I_{8}$,

Curve	Discriminant	Singularities
$\Gamma _{1}$	$\delta ^{2}\phi _{1}^{2}\phi _{2}^{2}\phi _{4}^{2}$	$\phi _{1}\phi _{2}\phi _{4}$
$\Gamma _{2}$	$\phi _{5}^{2}$	$\phi _{5}$
$\Gamma _{3}$	$\phi _{2}^{2}$	$\phi _{2}$
$\Gamma _{4}$	$\delta ^{2}\phi _{1}^{2}\phi _{3}^{2}\phi _{6}^{2}$	$\phi _{1}\phi _{3}\phi _{6}$
$\Gamma _{5}$	$j_{2}^{5}\phi _{1}^{3}$	$j_{2}\beta _{3}$

The factors $\delta$ for $\Gamma _{1}, \Gamma _{4}$ do not yield singular points: $\Gamma _{1}$ touches $\Delta$ at $F_{2}$ and $\Gamma _{4}$ touches it at $F_{0}$ (otherwise we would have transverse intersections at smooth points). For $\Gamma _{1},\ldots ,\Gamma _{4}$ the remaining factors yield (distinct) singular points of type $A_{1}$.

For $\Gamma _{5}$ the factor $j_{2}$ vanishes at $F_{5}$; $\beta _{3}$ vanishes at $F_{3}$. Calculating in local coordinates shows that $F_{5}$ is a singular point of $\Gamma _{5}$ of type $A_{4}$, and that $F_{3}$ is one of type $A_{2}$: together, these lower the genus by 3, confirming our calculations.

The $\phi$’s corresponding to singular points represent a further 24 points of $\mathcal {M}$. For our stratification, in the $\mathbb {C}$ case we need

$$\begin{aligned} 151 (\phi \text {'s before}) + 24 (\text {these }\phi \text {'s}) + 3 (F_{3},F_{4},F_{5}) =178 \end{aligned}$$

special points (as well as the seven points $F_{i}$ $(i=0,1,2)$, $G_{i}$ $(i=1,2,3,4)$ on $\Delta$).

6 Characteristic 2

6.1 Introduction

Over a field K of characteristic 2 (which, when convenient, we assume algebraically closed) there are several differences to the above theory. We now write a general quintic in the form

$$\begin{aligned} a = \sum _{r=0}^{5}a_{r}x^{5-r}y^{r}, \end{aligned}$$

(3)

omitting the binomial coefficients. We compute the above invariants for a where, for the moment, we still work in characteristic zero. We find that the expressions for $I_4$, $2^5I_8$ and $2^{10}I_{12}$ have no denominators in their expansions; reducing them modulo 2 we obtain $i_4,i_8$ and $i_6^2$, where the invariants $i_4,i_6$ and $i_8$ are given by the following expressions. First set

Then we have

$$\begin{aligned} i_4&=A_2A_3+A_2^2+B_1,\\ i_6&=a_2^4A_4+a_3^4A_1,\\ i_8&=A_2C+B_2^2+B_2A_2A_3+B_1A_3^2+A_3^3A_2+A_3^4. \end{aligned}$$

We can verify directly that $i_4,i_6$ and $i_8$ are indeed invariants of a under the action of $\hbox {SL}_2(K)$. Using the same technique of clearing denominators and reducing modulo 2, we find that the invariant $I_{18}$ yields $i_6^3$, and the discriminant gives $i_4^2$, so we obtain no further basic invariants. We will see in Theorem 6.2 that the ring of polynomial invariants is the polynomial ring $K[i_4,i_6,i_8]$. The moduli space $\mathcal {M}_2$ is thus isomorphic to the weighted projective space $\mathbb {P}(2,3,4)_K$. Its singular points are of type $A_2$ at (0, 1, 0) and of type $A_1$ at (0, 0, 1).

6.2 Binary quartics in characteristic 2

We begin by revising this rather simpler case. For quartics $f(x,y):{=}ax^4+bx^3y+cx^2y^2+dxy^3+ey^4$, the ring of invariants is polynomial, generated by $i_2:{=}bd+c^2$ and $i_3:{=}ad^2+bcd+eb^2$. The vanishing of $i_3$ characterises the case of repeated roots.

The automorphism group of $\mathbb {P}^1_K$ is $\hbox {SL}_2(K)\cong \hbox {PGL}_2(K)$. Each involution (element of order 2) is conjugate to the map given affinely as $t\rightarrow t+1$, which has the unique fixed point (pole) $\infty$. Given four distinct points in $\mathbb {P}^1_K$, we can arrange them in three ways into two pairs. For each such arrangement, there is a unique involution interchanging the two elements of each pair. The poles of all three involutions coincide, and the involutions commute, forming a copy of the four group.

The derivatives $\partial f/\partial x = bx^2y + dy^3$, $\partial f/\partial y = bx^3 + dxy^2$ depend only on the class of f modulo the ideal $I_1=\langle x^2\!,y^2\rangle {\lhd } K[x,y]$ generated by perfect squares, and (provided $f\not \in I_1$) vanish together only at the point where $bx^2 + dy^2 = 0$. This point is the pole of the involutions permuting the roots of f: we can call it the pole of f. If, however, $f\in I_1$, so that $b=d=0$, f itself is a perfect square, so can be reduced under $\hbox {SL}_2(K)$ to $cx^2y^2$ (with $c\ne 0$) or $x^4$.

For f not a square, we take the pole of f as (1, 0), so that $b= 0$, $d\ne 0$, the involutions take the form $t({=}x/y)\rightarrow t+\alpha$, $t\rightarrow t+\beta$, so

$$\begin{aligned} f=a(x+\gamma y)(x+(\gamma +\alpha )y) (x+(\gamma +\beta )y)(x+(\gamma +\alpha +\beta )y); \end{aligned}$$

(4)

here we can also reduce $\gamma$ to 0, so $f=ax^4+cx^2y^2+dxy^3$, and the invariants reduce to $i_2=c^2$, $i_3=ad^2$. Under $\hbox {SL}_2(K)$ we can further reduce d to 1, so f has the unique normal form $i_3x^4+\surd i_2x^2y^2+xy^3$. If there is a further symmetry of the roots, we can take it to fix 0, and it must cyclically permute the others. In this case, $i_2=0$ and the set of four points is equivalent (under $\hbox {PGL}_2(K)$) to the affine line over $\mathbb {F}_4$.

6.3 Quintics with symmetry and special quintics

A set of five points admitting an involution must consist of a set of four points together with its pole, and hence we can take the equation as yf with f as in (4), with invariants $((\alpha \beta (\alpha +\beta ))^2,0,(\alpha ^2+\alpha \beta +\beta ^2)^6)$. In particular, the condition for a quintic to possess such a symmetry is $i_6=0$.

For a symmetry of order 3 or 5, as in the characteristic zero case, we may suppose the group a diagonal subgroup of $\hbox {SL}_2$, and obtain the above forms $f_3$ and $f_5$. Over K, these are equivalent to each other, and to $f_4$, which is the equation for the set of five points forming the projective line $\mathbb {P}^1_{\mathbb {F}_4}$ over the Galois field $\mathbb {F}_4$. This admits the group $\hbox {SL}_2(\mathbb {F}_4)$ of automorphisms, which has order 60 and acts as the alternating group. The corresponding invariants are (1, 0, 0).

As in characteristic zero, we also have cases with repeated roots: $f_{1} =x^{2}(x^{3}-y^{3})$ (equianharmonic tetrad with one point repeated), with invariants (0, 1, 0), and $f_{2} = x^{2}y^{2}(x-y)$ (two repeated points, and another point), with invariants (0, 0, 1) (over K, $f_0$ is equivalent to this).

6.4 Normal forms

For a fifth power of a linear form we have $a_2=a_3=0$; thus a general quintic is no longer a linear combination of fifth powers: indeed, a quintic is so if and only if it belongs to the ideal $I_2{\lhd } K[x,y]$ generated by 4th powers of linear forms: equivalently, if $a_2=a_3=0$. Noting that $\partial f/\partial x = a_0x^4+a_2x^2y^2+a_4y^4$, we see that, if $a_2$ and $a_3$ do not both vanish, there is a unique differential operator $D=\alpha \partial /\partial x+\beta \partial /\partial y$ such that Df has zero coefficient of $x^2y^2$ and hence is a fourth power.

Theorem 6.1

A binary quintic f can be reduced under $\mathrm{SL}_2(K)$ to a unique normal form as follows:

• (a) if $i_4\ne 0$, $i_6\ne 0$, $a_0x^5+x^3y^2+a_4xy^4+a_5y^5$ with $a_0\ne 0$, $a_5\ne 0$, with invariants $(a_0^2a_5^2,a_5, a_0a_5^2+a_4^2)$;

• (b) if $i_4\ne 0$, $i_6=0$, $a_1x^4y+a_3x^2y^3+a_4xy^4$ with $a_1\ne 0$, $a_4\ne 0$; with invariants $(a_1^2a_4^2,0,a_1^2a_3^6)$, and either $a_3=1$ or $a_3=0$, $a_1=1$;

• (c) if $i_4=0$, $(i_6,i_8)\ne (0,0)$, $x^3y^2+a_4xy^4+a_5y^5$ with $(a_4,a_5)\ne (0,0)$, with invariants $(0,a_5,a_4^2)$;

• (d) if $i_4=i_6=i_8=0$, one of $ax^3y(x+y)$ with $a\ne 0$, $x^3y^2, x^4y$ or $x^5$.

Proof

(a) Since $i_6\ne 0$, we cannot have $a_2=a_3=0$, so have a derivation D as above. We change coordinates to take Df as (a non-zero multiple of) $y^4$. Then $A_1=0$ and $A_4\ne 0$, so $i_6=a_2^4A_4\ne 0$. We can thus also take $a_2x+a_3y$ as a non-zero multiple of x, or equivalently D as $\partial /\partial y$. Then the coordinates x and y are both fixed up to scalar multiples; and $a_1=a_3=0$, $a_5\ne 0$. Now $i_4=a_0^2a_5^2$ is non-zero, so $a_0\ne 0$. Replacing x by $a_2^{-1}x$ and y by $a_2y$, we reduce $a_2$ to 1.

(b) If $a_2$ and $a_3$ do not both vanish, we proceed as before up to the point where $A_1=0$ and $A_4\ne 0$, but now since $i_6=0$, we deduce $a_2=0$, so $D =\partial /\partial x$. We now have $a_0=a_2=0$, $a_4\ne 0$, which we recognise as the same form as for quintics with symmetry. As in the preceding paragraph, we can choose x as one of the factors of f, thus reducing $a_5$ to 0, and can reduce $a_3$ to 1.

In the case $a_2=a_3=0$, the invariants are $((a_1a_4+a_0a_5)^2,0,0)$. As $i_4\ne 0$, there are at least two distinct linear factors. Choosing x and y to be two of them we have $a_0=a_5=0$, hence $a_1\ne 0$, $a_4\ne 0$. The quintic is equivalent to $\mathbb {P}^1_{\mathbb {F}_4}$ (in particular, f is indeed a sum of two fifth powers), and this case is subsumed as $a_3=0$ in the preceding normal form.

(c) If $i_4=0$, the quintic has a repeated factor, and we can take the factor as $y^2$, and thus suppose $a_0=a_1=0$. The invariants then reduce to $(0,a_2^4(a_2a_5+a_3a_4),a_2^6a_4^2+a_2^4a_3^4)$. If $a_2=0$, all invariants vanish; otherwise we normalise coordinates first so that $a_2=1$ and then so that $a_3=0$. The invariants now reduce to $(0,a_5,a_4^2)$, so our hypothesis gives $(a_4,a_5)\ne (0,0)$.

(d) We saw in the preceding paragraph that if all invariants vanish, we may reduce till either $a_0=a_1=a_2=0$ or $a_3=a_4=a_5=0$: in either case, there is a linear factor of multiplicity at least 3. Thus there are at most three distinct roots, so by inspection the quintic can be reduced to $ax^3y(x+y)$ (with $a\ne 0$), $x^3y^2,x^4y$ or $x^5$. $\square$

We observe that the results of Geometric Invariant Theory apply to this case, and imply that all invariants vanish if and only if there is a 3-fold factor; and in all other cases the quintic is stable, and is determined up to $\mathrm{SL}_2(K)$ by its invariants, as we see directly.

Theorem 6.2

The $\mathrm{SL}_2(K)$-invariants in $K[a_0,\ldots , a_5]$ form the polynomial ring

$K[i_4,i_6,i_8]$.

Proof

Any invariant is determined by its value on a generic quintic, which we can take in the normal form (a). As we have just seen, the invariants $(i_4,i_6,i_8)$ here take the values $(a_0^2a_5^2,a_5, a_0a_5^2+a_4^2)$. We can thus write $a_5=i_6$, $a_0=i_6^{-1}\surd (i_4)$ and $a_4=\surd (i_8+i_6\surd (i_4))$. The field of invariants is thus contained in $K(i_6, i_6^{-1}\surd (i_4),$

$\surd (i_8+i_6\surd (i_4))$, which has degree 4 over $K(i_4,i_6,i_8)$. We see by inspection that each strictly intermediate field contains $\surd (i_4)$. However, for a generic quintic (3), $i_4$ is not a perfect square. Thus the field of invariants coincides with $K(i_4,i_6,i_8)$.

It remains to show that an element of this field which restricts to a polynomial in $a_0,\ldots ,a_5$ is a polynomial in $i_4,i_6$ and $i_8$. We see from the preceding paragraph that we have a polynomial in $i_6,i_6^{-1}\surd (i_4)$ and $\surd (i_8+i_6\surd (i_4))$, so the only denominator that may occur is a power of $i_6$. If such a case occurs, there must be an example with denominator $i_6$, and we may suppose without loss of generality that the numerator depends on $i_4$ and $i_8$ only. But for the normal form for case (b), $i_4$ and $i_8$ are independent, while $i_6=0$. Hence no such example exists. $\square$

6.5 Exceptional curves

As before, given a plane quintic curve $H_i$ with 1-parameter symmetry group $G_{i}$, the lines L in the plane $\mathbb {P}^{2}_K$ fall into a 1-parameter family of orbits under $G_{i}$, so we expect their intersections with $H_{i}$ to give a family of binary quintics whose moduli trace out a curve $\Gamma _{i}$ in $\mathcal {M}_2$.

If we take the above calculations, remove the appropriate power of 2, and reduce mod 2, $R_1$ reduces to $i_6^8$, $R_2$ to $i_6^2i_8^4$, $R_3$ to $(i_4i_6^2+i_8^2)^2$ and $R_5$ to $i_6^6$, but $R_4$ remains irreducible. However, it is easy to calculate directly.

For each of $h_1,\ldots ,h_5$ in turn, we first substitute $z=x+uy$, then list the coefficients of $x^{5-r}y^r$ for $r=0,1,2,3,4,5$. We then calculate in turn $A_1,A_2,A_3,A_4$, $B_1,B_2,C$ and finally the invariants $(i_4,i_6,i_8)$. The results are as follows (recall that we are calculating mod 2):

For $h_1$, the quintic $x^4(x+uy)+xy^2(x+uy)^2$ has coefficients $[1,u,1,0,u^2,0]$. We obtain $A_1=u$, $A_2=u^3$, $A_3=0$, $A_4=0$, $B_1=0$, $B_2=u^6$, $C=0$, and $(i_4,i_6,i_8)=(u^6,0,u^{12})$.

For $h_2$ we find in turn: $x^4(x+uy)+y^3(x+uy)^2$, coefficients $[1,u,0,1,0,u^2]$ $A_1=1$, $A_2=u^2$, $A_3=0$, $A_4=0$, $B_1=0$, $B_2=u$, $C=1$ and $(i_4,i_6,i_8)=(u^4,1,0)$.

Next, $h_3$ gives $x^5+y^3(x+uy)^2$, with coefficients $[1,0,0,1,0,u^2]$, so $A_1=1$, $A_2=u^2$, $A_3=0$, $A_4=0$, $B_1=0$, $B_2=0$, $C=1$, and $(i_4,i_6,i_8)=(u^4,1,u^2)$.

Next, $h_4$ yields the quintic $x^3y(x+uy)+y^3(x+uy)^2$, coefficients $[0,1,u,1,0,u^2]$, giving $A_1=u$, $A_2=0$, $A_3=u$, $A_4=u^3$, $B_1=u^4$, $B_2=1$, $C=u^7$, and $(i_4,i_6,i_8)=(u^4, u+u^5, 1+u^4+u^6)$.

Finally, $h_5$ gives $xy^3(x+uy)+x^2y(x+uy)^2$, $[0,1,0,1+u^2,u,0]$, $A_1=0$, $A_2=u$, $A_3=0$, $A_4=u+u^3$, $B_1=0$, $B_2=1+u^2+u^4+u^6$, $C=0$, and $[i_4,i_6,i_8]=[u^2,0,1+u^4+u^8+u^{12}]$.

Thus the loci in $\mathcal {M}_2$ are indeed defined by $i_6=0$, $i_8=0$, $i_4i_6^2=i_8^2$, a complicated expression, and $i_6=0$.