数理解析研究所講究録1122 巻 2000 年 87-95

87

MODULI OF SEXTICS AND ITS GEOMETRY
MUTSUO OKA

岡睦雄

東京都立大学理学部

1. INTRODUCTION

be the moduli space of sextics with 6 cusps and 3 nodes. A sextic is called
type if its defining polynomial has the expression $f(x, y)=f_{2}(x, y)^{3}+$-torusoffor some polynomiak
of degree 2, 3 respectively. Hereafter we simply say
in-torus type. We denote by
the component ofof torus type the sense ofwhich consists of curves of torus type and by
the curves of non-torus type. We
denote the dual curve of by . In our previous paper [O2], we have shown that thedual curve operation
gives an involution on
and it preserves the type of the,i.e.,curve in. Let
be the moduli space ofif and only if
sextics with 3 $(3,4)$ -cusps as in [O2]. For brevity, we denote
by $N$. We have shownthat $N$ is in the closure ofand the dual curveof a generic $C\in N$ is a sextic with
6 cusps and three nodes i.e.,. Let. The quotient moduli
spaces are by definition the quotient spaces of the moduli spaces by the action of .
In \S 2, we win study the quotient moduli space
and we will show that there exists
an involution onsuch that is different from the dual curve operation andpreserves the types of the sextics (Theorem 2.3).
In \S 3, we study the quotient moduli space $N/G$ . We will show that $N/G$ is one
dimensional and consists of two components
and $N_{gen}/G$ consisting of sextics
of torus type and non-torus type respectively. Using their nornal forms, we show that
contains a unique sextic which is self dual (Theorem 3.9).
Let$\mathcal{M}$

$C$

$f$

$(\mathit{2},\mathit{3})$$f_{3}(x,y)^{2}$$f_{2},$ $f_{3}$$\mathcal{M}_{torus}$$(\mathit{2},\mathit{3})$$\mathcal{M}$$\mathcal{M}_{gen}$$C$$C^{*}$$C\mapsto C^{*}$$\mathcal{M}$$\mathcal{M}$$C^{*}\in \mathcal{M}_{t\sigma\Gamma us}$$C\in \mathcal{M}_{t\sigma rus}$$N_{3}$$N_{3}$$\overline{\mathcal{M}}$$C^{*}$$C^{*}\in \mathcal{M}([\mathrm{O}2])$$G:=\mathrm{P}\mathrm{G}\mathrm{L}(3, \mathrm{C})$$G$$\mathcal{M}/G$$\overline{\iota}$$\mathcal{M}/G$$\overline{\iota}$$\overline{\iota}$$N_{t\sigma ru}s/G$$N_{t\sigma rus}/G$

2. INVOLUTION ON THE QUOTIENT MODULI$\mathcal{M}/G$Let
andbe the moduli space of sextics with three nodes and 6 cusps and themoduli space of irreducible plane curves of degree 12 with 24 cusps and 24 nodes respectively. Note that the genus of a generic curve in(respectively in)(resp. 7).
By the class([N] or [O2]), it is easy to see that for a generic, the dual
curve
is also in. We consider the mapping$\mathcal{M}$$\overline{\mathcal{M}}$$\mathcal{M}$$\overline{\mathcal{M}}$$\mathrm{i}\mathrm{s}\underline{1}$$C\in \mathcal{M}$$\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}\mathrm{u}\mathrm{l}\underline{\mathrm{a}}$$C^{*}$$\mathcal{M}$$\pi:\mathrm{P}^{2}arrow \mathrm{P}^{2}$,

Date: November, 1999, first version.
1991 Mathematics Subject Classification.

$(X, \mathrm{Y}, Z)\vdasharrow(X^{2}, \mathrm{Y}^{2}, Z^{2})$$14\mathrm{H}\mathrm{l}\mathrm{o},14\mathrm{H}30$.

88which is a 4-fold covering branched along the coordinate axes $\{X=0\}$
and let $F(X, Y, Z)$ be the defining homogeneous$\{Z=0\}$ . Take a generic curve
has three nodes, has three -tangent lines. We denotepolynomial of degree 6. As
whose three bitangent lines
which consists of curves
the subset ofby:as follows.
are $X=0,$ $Y=0$ and $Z=0$ . We define a mapping
and let $F(X, Y, Z)$ be the defining homogeneous polynomial. We defineLet$\psi(C):=\pi^{-1}(C)$ . Note that $\psi(C)$ is defined by $\overline{F}(X, Y, Z):=F(X^{2}, \mathrm{Y}^{2}, Z2)$ . Each cuspof produces 4 cusps on $\psi(C)$ . Thus $\psi(C)$ has 24 cusps. Each node of also gives 4
nodes on $\psi(C)$ , thus we get 12 nodes on $\psi(C)$ which are mapped onto the nodes of .
As the restriction of to the affine chart $\{Z\neq 0\}$ is the composition of double coverings$(x, y)rightarrow(x, .y^{2})$ and $(x, y)-\succ(x^{2}, y)$ , each simple tangent on the coordinate axis $X=0$ ,$Y=0$ gives 2 nodes on $\psi(C)$ . This is the same for the simple tangents for $Z=0$ . Thus
there are 12 nodes on $\psi(C)$ which are on the three coordinate axes and they are.
to simple tangents on coordinate axis by . Thus $\psi(C)$ has 24 nodes. Thus
$g\in
G$
.
such thatby choosing aas, we define
Now for
is
which
of
$g\in
G$
in
stabilizer
theare
The ambiguity for the choice of(scalar(the permutations of coordinates) anda direct product ofaction,is also unique up to a
. Thus the polynomialaction. Thus :is also unique up to aand therefore
is well-defined.
Recal that a polynomial $F(X, Y, Z)$ is called even in $X$ (respectively symmetric in$Z)$ ). Thus the polyno$X,$ $Y)$ if $F(-X, \mathrm{Y}, Z)=F(X, Y, Z)$ (resp. $F(Y,X,$ $Z)=F(X,$.
mial $F(X^{2}, Y^{2}, Z2)$ is even in $X,$
is defined by $F(X, Y, Z)=0$ . If is a even polynomial in the
Assume that$X$
(respectively a symmetric polynomial in $X,$ ), then 6 cusps are stable byvariable) $\vdasharrow(-X, Y, z)$ (respectively (X, $Z)\mapsto(Y,$ $X,$ $z)$ ). Then there
the involution (X,exists a homogeneous polynomial $F_{2}(X, Y, Z)$ of degree 2 which is even in $X$ (respectively
symmetric in $X,$ ) such that the conic $F_{2}(X, Y, Z)=0$ passes through the 6 cusps of .
By the criterion of Degtyarev [D], the sextic $F(X, Y, Z)=0$ is of torus type.
and letand consider the dual curve
Now we take a genericisbe a defining homogeneous polynomial of degree 12, where, so isis even in $X,$). As
the dual coordinates of (X,$\mathrm{U}\{\mathrm{Y}=0\}\cup$$C\in \mathcal{M}$$C$$C^{*}$$\mathcal{M}^{nml}$$\mathrm{b}\mathrm{i}$$C\in \mathcal{M}$$\mathcal{M}$$\mathcal{M}^{nml}arrow\overline{\mathcal{M}}$$\psi$$C\in \mathcal{M}^{nml}$

$C$

$C$

$C$

$\pi$

$\mathrm{m}\mathrm{a}\mathrm{p}\underline{\mathrm{p}\mathrm{e}}\mathrm{d}$$\psi(C)\in \mathcal{M}$$\pi$$C\in \mathcal{M}$$\overline{\psi}(C)$$C^{g}\in \mathcal{M}^{nml}$$\psi(c^{g})$$\mathcal{M}^{nml}$$G_{\lambda 4^{nm}}\iota$$\mathrm{C}^{*}\cross \mathrm{C}^{*}\cross \mathrm{C}^{*}$$\mathfrak{S}_{3}$$\overline{F}(X, Y, Z)$$G_{\Lambda 4^{nm\mathrm{t}}}$$\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{S}\underline{)}$$\overline{\psi}$$F(X, \mathrm{Y}, Z)$$\mathcal{M}/Garrow\overline{\mathcal{M}}/G$$G_{\mathrm{A}4}nm\downarrow$$Y,$$Y,$ $Z$$F$$C\in \mathcal{M}$

$Y$

$Y,$

$Y,$ $Z$$C$$Y$$C\in \mathcal{M}^{nml}$$\psi(.C)^{*}$$(X^{*}, \mathrm{Y}^{*}, z*)$$\tilde{G}(X^{*}, Y^{*,z*})$$Y,$ $Z$

in $X,$
Proposition 2.1.
$\mathrm{Y},$$Y,$ $Z$$\overline{F}(X, Y, Z)$$\tilde{G}(X^{*}, Y^{*,z*})$$Z$

has 4 nodes on each coordinate axis $X^{*}=0_{;}\mathrm{Y}^{*}=0$ or $Z^{*}=0$ .
Proof. Let $C=\{F(X, Y, Z)=0\}$ and let us consider the discriminant polynomial. We assume that the. This is a homogeneous polynomial of degree 30
axis. Assume thatthecoordinate
on
$F(X,
Y,
Z)=0$
not
are
singularities of the sextic$P:=(\alpha, \beta, \gamma)\in C$ is a singular point ofwith Milnor number and multiplicity $m$ .
with $\rho\geq\mu+m-1$ and the equality
Then $\Delta_{Y}F(X, Z)$ has a linear term(see [O2]). Thus to eachis generic with respect toholds if the line
linearterm with multiplicity 3
cusp (respectively to each node), there is an associated(resp. with multiplicity 2). The factor $X=0$ and $Z=0$ has also multiplicity 2 in
. Then theis generic in-tangent lines. Assume$\Delta_{Y}F(X, Z)=0$ , as they are$\psi(C)^{*}$$([\mathrm{O}1])$$\Delta_{\mathrm{Y}}F(X, Z)$

$C$

$\mu$

$(\gamma X-\alpha Z)^{\rho}$$C$$\gamma \mathrm{Y}-\beta Z=0$$\mathrm{b}\mathrm{i}$$C$$\mathcal{M}$89

sum of degrees is $18+6+4=28$ by the above consideration. Thus there exists two simple
tangent lines of the form $X-\eta_{1}Z=0$ and $X-\eta_{2}Z=0$ for some
. Thenfour lines$i=1,2$ are bitangent lines for the curve $\psi(C)$ . This implies that
$i=1,2$ are nodes of the dual curve. Thus the coordinate axis
contains 4 nodes of. By the same argument, $X^{*}=0$ and $Z^{*}=0$ contains also 4
nodes respectively.
Definition 2.2. For, we define a polynomial of degree 6 by $G(X^{*}, Y*, Z^{*})$ $:=$
and we defineby the sextics defined by $G(X^{*},$ $Y^{*,z^{*})=0}$ . For$g\inG$, takeso thatand we define an involution
by.
Claim 1.
for a genericand is an involution which preserves the typeof sextics, that is we have the commutative diagram:$\eta_{1},$$\eta_{2}\neq 0$$X=\pm\sqrt{\eta_{i}}Z,$$(1, 0, \pm\sqrt{\eta_{i}}),$$\psi(C)^{*}$$\mathrm{Y}^{*}=0$$\psi(C)^{*}$$C\in \mathcal{M}^{nml}$$\tilde{G}(\sqrt{x*}, \sqrt{\mathrm{Y}^{*}}, \sqrt{z*})$$\iota(C)$$C^{g}\in \mathcal{M}^{nml}$$C\in \mathcal{M}$$\overline{\iota}:\mathcal{M}/Garrow \mathcal{M}/G$$\overline{\iota}(c)=b(c^{g})$$C\in \mathcal{M}$$\overline{\iota}(C)\in \mathcal{M}$$\mathcal{M}/G$$arrow\overline{\iota}\mathcal{M}/G$$\lrcorner\overline{\psi}$$\mathcal{M}/G$$\cdot\lrcorner\overline{\psi}$$arrow dud\mathcal{M}/G$

Proof. We may assume that$\overline{\iota}$$\mathcal{M}_{t\sigma\Gamma\tau lS}/G$$arrow\overline{\iota}\mathcal{M}_{t\sigma ru}s/G$$–\downarrow\overline{\psi}$$–\downarrow\overline{\psi}$$\mathcal{M}_{t\sigma\Gamma \mathfrak{U}s}/G$$arrow dud\mathcal{M}_{t\sigma rus}/c$. By the above consideration, we have seenthat the dual curveof $\psi(C)$ is defined by a polynomial $G(X^{*}, Y^{*}, z^{*})$ of degree 12
which is even in each of the three variables and it has 24 cusps and 12 nodes outside ofcoordinate axis and 4 nodes on each coordinate axis. Thus
has 6 cusps and 3 nodes.
Note that nodes ofon the coordinate axes are mapped on simple tangents on thecorresponding coordinate axes of. Thus the curve, defined by,. Finally we will show that keeps the type of the curve. As thebelongs tocurvesare topologically equivalent, the image is contained in aconnected component. Thus it is enough to show that there exists asuch. To see this, it is enough to takethatwhose defining
polynomial $F(X, Y, Z)$ is symmetric in each of $X,$ . Thenis also symnetric
in $X,$ . This implies also thatandsymmetric in.
By the Degtyarev’s criterion, this implies thatis a sextic of torus type. The following
example shows thatin general.
Thus we have proved the following:
Theorem 2.3. There exists an involution on the quotient moduli
such thatis different from the dual curue operation andthe type8 of the sextics, thatis.
Example 2.4. Let
be the sextic defined by the symmetric polynomial:$C\in \mathcal{M}^{nml}$$\psi(C)^{*}$$\iota(C)$$\psi(C)^{*}$$\iota(C)$$g(\sqrt{x^{*}},$ $\sqrt y\neg^{*}=0$$\iota(C)$$\mathcal{M}^{nml}$$\iota$$\{\overline{\iota}(C);C\in \mathcal{M}_{torus}/G\}$$C\in \mathcal{M}_{t\circ\Gamma us}/G$$\overline{\iota}(C)\in \mathcal{M}_{t\sigma\Gamma us}/G$$C\in \mathcal{M}_{tuS}^{nm_{\Gamma}l}\sigma$

$Y$

$Y$

$\overline{G}(X^{*}, Y^{*}, z*)$$\tilde{F}(X, \mathrm{Y}, Z)$$G(X^{*}, \mathrm{Y}^{*,z*})$$X^{*},$ $Y^{*}$$\iota(C)$$\overline{b}(c)\neq C^{*}$$\square$$\mathit{8}pace\mathcal{M}/G$$\overline{\iota}$$\overline{\iota}$$\overline{\iota}pre\mathit{8}erve\mathit{8}$$\overline{\iota}(C)\in \mathcal{M}_{ts}\sigma ru/G\Leftrightarrow C\in \mathcal{M}_{t\sigma\Gamma us}/G$$C\in \mathcal{M}_{t\eta lS}^{nm}\sigma rl$$\frac{1767f}{16}(_{X^{42}}.\cdot y+x^{2}y)=-684(x_{4}^{3}y+3)+\frac{881xy}{8}yx+3\frac{4055}{16}3^{-}(x^{6}\dagger y)105(_{X}3+y^{3})6^{+25(y}-\frac{87323}{8}(_{X^{5}}+y^{5})x^{2}+)2-78(_{X}+\frac{2\mathrm{t}\mathrm{K}\int 121}{4}(x^{4}+y)4-(Xy+y)+\frac{819}{\frac{97116}{8}}(x_{4}^{5}y+yx+xy)4)\mathrm{s}+-$$\frac{6947}{2}y^{2}x^{2}+2268+1038(x^{2}y+xy^{2})-4883yx-\frac{375}{2}(X^{2}y^{3}+x^{3}y^{2})$Then $\psi(C)$. is defined byandthe sextic defined by the symmetric polynomial$f(x^{2}, y^{2})$$\psi(C)^{*}\mathrm{i}_{\mathrm{S}}$.

defined by $g(x^{*2},y)*2=0$ and$\iota(C)$is$g(x^{*}, y^{*}):=908294_{X}*2*2-3y54000(X*y+x^{*2*}*2y)+302745(y+*4*4)X+529284(xy^{*2}+*4$$y^{*4}x^{*2})-396458(X^{*}y+y^{*}x*4*4)-722148(xy+*3*2*3x^{*}y)2+11340(y+X)*\epsilon*6-109170(X^{*\mathrm{s}_{+}}$

90$y^{*5})+86296Xy+4**82724(X^{*}y+yX)3**3*-158508(y^{*}X^{*}+y)5*5_{X}*+103096y^{*}X-2223*330(X^{*}+$$y^{*})-203920(y*3+x^{*3})+90570(y^{\mathrm{s}2}+x^{*2})+2025$

The dual curve
ily check that

$C^{*}$$\mathrm{e}\mathrm{a}s$

of

$C$

is defined by the following symmetric polynomial and we can.

$\overline{\iota}(C)\neq C^{*}$$h(X^{*},y^{*}):=3(_{X^{*4}+}y^{*}4)+14(X^{*^{3}}+y)*33(+X^{*2}+y^{*})2+4(yX+xy)**4**4+36(y^{*}X+*3$$x^{*}y^{*})3+6(yx^{*2}+*x^{*}y*2)-2y^{*}x^{*}+12(y^{*}x+*x^{*}y)242*4+84(y+Xy*2_{X}*3*2*3)+14y*2*2X+$$88y^{*3}x^{*3}+4y^{*}x4*$

3. NORMAL FORMS OF THE MODULI $N$$A:=(1,1)$of the sextics whose cusps are at
We consider the submoduli$N$
represented
by a curve$B:=(1,
-1)$
be
in
can
. Under the action of , every sextic
and. Consider the stabilizer group $G^{(1\rangle}:=\{g\in G;g(N(1))=N^{(1)}\}$ . By an easyinand and ais the semi-direct product of the group
computation, we see that
finite group where is a finite linear subgroup of , isomorphic to the permutation
is defined bygroup , and$O:=(\mathrm{O}, 0),$$N^{(1)}$$G$$N^{(1)}$$G_{0}^{(1)}$$G^{(1)}$$S_{3}$$G$$\mathcal{K}$$\mathcal{K}$$G_{0}^{(1)}$$G_{0}^{(1)}:=\{M=\in G;a3(a^{2}1-a2)2\neq 0\}$. The isomorphism
is normal inwhich fix singular points pointwise. Note thet
singularoflocus $O,$ $A,$ .
threepermutation
as theis given by identifying$N/G\congN^{(1)}/G^{(1)}$.
We will study the normal forms of the quotient moduli$G^{(1)}$$G_{0}^{(1)}$$B$$g\in \mathcal{K}$$\mathcal{K}\cong S_{3}$Lemma 3.1. For a given line $L:=\{y=bx\}$ with $b^{2}-1\neq 0$ , there exists
is given by $x=0$ .
that$M\in G_{0}^{(1)}$such$L^{M}$, where $M$ isby the action ofas above, is defined by $(a_{1}-ba_{2})y+(a_{2}-ba_{1})_{X}=0$ . Thus we take $a_{1}=ba_{2}$ . Then$a_{1}^{2}-a_{2}^{2}=a_{2}^{2}(b^{2}-1)\neq 0$ by the assumption.

Proof. By an easy computation, the image of$M^{-1}$$L$$\square$Lemma 3.2. The tangent cone at$O$

is not $y\pm x=0$ for $C\in N^{(1)}$ .

Proof. Assume for example that $y-x=0$ is the tangent cone ofatis 4 and thus
$C$

$O$

. The

,
intersection multiplicity of the line $L_{1}:=\{y-x=0\}$ and at
an obvious contradiction to Bezout theorem.
consisting of curves whose tangent cone at is given
be the subspace ofLet. By Lemma 3.1 and Lemma 3.2, we have thebe the stabilizer ofby $x=0$ . Let
isomorphism:
$C$

$O$

$L_{1}\cdot C\geq 7$$\square$$O$$N^{(1)}$$N^{(2)}$$G^{(2)}$$N^{(2)}$Corollary 3.3. $N^{(1)}/G^{(1)}\cong N(2)/G^{(2)}$ .

91

It is easy to see thatis generated by the groupof order two defined by $\tau(x, y)=(x, -y)$ . Note that
$G^{(2)}$$\tau$$G_{0}^{(2}$) $:=G^{(2}$)

and an element$\cap G_{0^{1}}^{()}$$G_{0}^{(2)}=\{M=\in G_{0}^{()};1 a_{1}a_{3}\neq 0\}$

For $C\in N^{(2)}$ , we associate complex numbers $b(C),$which are the directiections of the tangent cones ofat $A,$ respectively. This implies that the lines$y-1=b(C)(x-1)$ and $y+1=c(C)(X-1)$ are the tangent cones of at and $B$
respectively. We have shown that
if and only if $b(C)+c(C)=0$ and is notof torus type if and only if $c(C)^{2}+3c(c)-b(C)_{C}(C)+3-3b(c)+b(C)^{2}=0$ (\S 4, [O2]).
We consider the subspaces:$c(C)\in \mathrm{C}$

$C$

$B$

$C$

$A$

$C\in N_{t_{or}s}^{(2)}u$$N_{torus}^{(3})\{c:=\in Nt\sigma\Gamma us;b(2)(C)=1\}$,$C$$N_{gen}^{(s)}:=\mathrm{f}C\in N_{g}^{(}e2)n;b(c)=c(C)=\sqrt{-3}\}$and we put.
Remark. The common solution of the both equations: $b+c=c^{2}+3_{C}-bc+3-3b+b^{2}=0$
is $(b, c)=(1, -1)$ and in this case, degenerates into two non-reduced lines $(y^{2}-X^{2})2=0$
and a conic.
Lemma 3.4.
that $C\in N^{(2)}$ . Then there exists $C’\in N^{(3)}$ and an element$g\in G^{(2)}$such that $C^{g}=C’$ and such a $C’$ is unique. This implies that
$N^{(3)}:=N_{tuS}^{()}\sigma\Gamma 3\cup N_{g\mathrm{e}n}^{(\rangle}3$$C$$A_{\mathit{8}Su}me$$N_{torus}/G\cong N^{(2}tor\tau\iota s/))G^{(2}\cong N_{t\tau}(\mathrm{O}3)rls$’

Proof. Assume that

$C\in N_{ts}^{(1)}\rho_{\Gamma}u’ b+c=0$$N_{gen}/c\cong N(2\rangle/genG(2)\cong N_{gn}(e3)$. Consider an element$g\in G_{0}^{(1)}$,$g^{-1}=$
The imageis given by $y-x+xa_{3}-a_{3}-bxa_{3}+ba_{3}=0$. Thus we can solve theequation $a_{3}(1-b)-1=0$ in
uniquely as $a_{3}=1/(1-b)$ as $b\neq 1$ . Thusisunique if it fixes the singular points pointwise and thus $C’$ is also unique. It is easy to
see that the stabilizer ofis the cyclic group of order two generated by , as $C’$ is)even in (see the normal form below) andfor any
Thus we have
$L_{A}^{g}$$g\in G_{0}^{(1\rangle}$$a_{3}$$N_{tu}^{(3)}\sigma\Gamma S$$\tau$$C^{\prime \mathcal{T}}=C’$$y$$C’\in N_{t_{\mathit{0}}ru}^{(3}$$S^{\cdot}$$N_{t\circ r}^{()}2/us(G2\rangle\cong N_{tu}^{(}\mathrm{o}r3)s$.

Consider the case
. Then the images of the tangent cones at $A,$ by the actionof are given by $y-x+xa3-a_{3}-b_{Xa}3+ba_{3}=0$ and $y+x-Xa3+a_{3}-cxa_{3}+Ca_{3}$ respectively.
Assume that $b(C^{g})=C(c^{g})$ . Then we need to have $a_{3}(1-b)-1=a_{3}(-1-c)+1$ , which
has a unique solution in , if $(\star)b-c-2\neq 0$ . Assume that $c^{2}+3c-bc+3-3b+b^{2}=0$
and $b-c-2=0$ . Then we get $(b, c)=(1, -1)$ which is excluded as it corresponds to nonreduced sextic. Thus the conditionis always satisfied. Put $(b’, d):=(b(C^{g}), C(c^{g}))$ .
$d^{2}+3c’-b’d+3-3b’+b^{\prime2}=0$ and $b’=d$ . Thus we have
They satisfy the equality
either $b’=d=\sqrt{-3}$ or $b’=d=-\sqrt{-3}$. However in the second case, we can take theautomorphism $(x, y)arrow(x, -y)$ to change into the first case. Thus $b’=c’=\sqrt{-3}$ andas desired.
$C\in N_{gen}^{(2)}$

$B$

$g$

$a_{3}$$(\star)$$C^{g}\in N_{gen}^{()}3$

92
3.1. Normal forms of curves of torus type. In [02], we have shown that a curve inis defined by a polynomial $f(x,y)$ which is defined by a sum $f_{2}(x,y)^{3}+sf_{3}(x,y)^{2}$
where $f_{2}(x,y)$ is a smooth conic passing through $A,$ $B,$ $f_{3}(x, y)=(y^{2}-X^{2})(x-1)$ and$N_{t_{\mathit{0}}r}^{(1)}us$$O,$$s\in \mathrm{C}^{*}$.

Proposition 3.5. The direction of the tangent cones at
the tangent line of the conic $f_{2}(x,y)=0$ at those points.

$O_{y}$

$A$

and

$B$

are the same vrith

at $O,$ $A,$ are 4. See also Lemma 23
This is immediate as the multiplicity ofand aretangent cones of at $O,$
of [O2]. Assume that’ that is, the$f_{2}(x,y)=0$is alsogiven by $x=0,$ $y-1=0$ and $y+1=0$ respectively. Thus the conic
1,
centered$f_{2}(x,y)=y2+x^{2}-2X$
radiuswithcircle
is
.
This
the
uniquely determined as
is one-dimensional and it has the representaion
at $(1,0)$ . Therefore$B$$f_{3}(x,y)^{2}$

$A$

$C$

$C\in N_{t\circ}^{(3)}r\mathfrak{U}s$$B$$N_{torus}^{(3)}$: $ft\sigma\Gamma us(_{X},y, S):=f2(x,y)^{3}+Sf3(x,y)2=0$
obtains a node. As is easy
is a sextic with three $(3,4)$ cusps, while
For $s\neq 0,27,$. Thus$g\inG^{(2)}$
fixes the tangent lines $y\pm 1=0$ , then $g=e$ or andto see, if(3.6)$C_{s}$$C_{27}$$C_{s}$$\tau$$C_{s}\neq C_{t}$$C_{s}^{\prime r}=c_{s}$if $s\neq t$ .

3.2. Normal form of sextics of non-torus type. For the moduli of non-torus type, we start from the expression given in \S 4.1, [O2]. We may as sume $b=c=$
sextic. Then the parametrization is given by$N_{gen}$$\sqrt{-3}$$f_{gen}(_{X}, y, s):=f_{\mathit{0}}(x, y)+Sf3(x,y)^{2}$

where
(3.7)

$s$

is equal to$a_{06}$in [O2] and$f_{0}$,$f_{3}(x, y)=(y-x^{2})2(x-1)$

is the sextic given by$f_{0}(x, y):=y^{6}+y^{5}(6\sqrt{-3}-6\sqrt{-3}X)+y^{4}(35-76X+38x^{2})$$+y^{3}(-24\sqrt{-3}x+36\sqrt{-3}x^{2}-12\wedge\sqrt{-3}x^{3})+y^{2}(-94_{X}2+200x^{3}-103x^{4})$$+y(24\sqrt{-3}x^{3}-42\sqrt{-3}x4+18\sqrt{-3}x^{5})+64x^{3}-133X^{4}+68x^{5}$

C. Observe that $D_{0}=\{f_{0}(x,y)=0\}$ is a sextic
Let $D_{s}:=\{f_{gen}(X, y, s)=0\}$ for eachwith three $(3,4)$ -cusps and of non-torus type. For the computational reason, we take theto make the equationto be defined over rational numbers: Then
substitution$f_{0}(x, y)$ and $f_{3}(x, y)$ change into:$:=-27y^{6}+(-162+162x)y^{5}+(315-684_{X}+342x^{2})y4$$f_{0}(x, y)$(3.8)$s\in$$y\mapsto y\sqrt{-3}$$+(-216_{X}+324x^{2}-108_{X^{3})y}3+(282X2-600X^{3}+309_{X^{4}})y^{2}$$+(-54x5+126x^{4}-72x^{3})y+68x^{5}+64x^{3}-133_{X^{4}}$$f_{3}(x,y):=-(x-1)(3.y+x^{2})3$

Summerizing the discussion, we haveof twoTheorem 3.9. The quotient moduli space $N/G$ is one dimensional andcomponent8.
(1) The component $N_{toru}s/G$ has the normal form8 represented by the family of sextics
and $s\neq 0,27$$C_{s}=\{f(x, y, S)=0\}$ where $f(x,y, s)=f_{2}(x,y)^{3}+sf_{3}(x,y)^{2}$ forwhere$f_{3}(x, y)=(y^{2}-x)2(x-1)$$f_{2}(x,y)=y+X^{2}-22x$ ,$consi\mathit{8}ts$$s\in \mathrm{C}^{*}$

93

The curue
is a unique curve in $N/G$ which self-dual.
(2) The $componentN_{gen}/G$ of sextics of non-torus typethe normal form: $f_{gen}(x,y, s)=$$f_{0}(x,y)+sf_{3}(x, y)^{2}$ whereisabove and theis contained in $N_{gm}$ .
This component has no self-dual cume.
$C_{54}$$i_{\mathit{8}}$$ha\mathit{8}$$f_{3}$$\mathit{8}exticf\mathrm{o}(x,y)=0$$a\mathit{8}$Proof of Theorem 3.9. We need only prove the assertion for the dual curves. The

proof will be done by a direct computation of dual curves using the method of \S 2, [O2]and the above parametrizations. We use Maple V for the practical computation. Here$z*$ be the dual coordinates$Z$
is the recipe of the proof. Letofand let$(x^{*}, y^{*}):=(X^{*}/Z^{*}, Y^{*}/Z^{*})$ be the dual affine
coordinates.
(1) Compute the defining polynomials of the dual curves
andrespectively, using the method of Lemma 2.4, [O2]. Put gtorus $(X^{*}, ys)*,$ and $g_{gen}(xys)*,*$, the defining
polynomials.
(2) Letbe the homogenization of
torus or gen.
Compute the discriminant polynomials
which is a homogeneous polynomial inof degree 30 (cf. Lemma 2.8, [O1]). Recall that the multiplicity of the pencil$X^{*}-\eta z^{*}=0$ passing through a singular point is generically given by $\mu+m-1$ where$m$ are the Milnor number and the multiplicity of the singularity. Thus thecontribution from a $(2,3)$ -cusp (respectively from a (3,4)-cusp) is 3 (resp. 8). Thus if
has three $(3,4)$ cusps, it is necessary that $\triangle_{Y^{*}}(G)=0$ has three linear factors withmultiplicity at least 8.
(3-1) For the non-torus curves, it is not possible to get a degeneration into 3 $(3,4)-$
cuspidal sextic.
(3-2) For the torus curves, we can see that $s=54$ is the only possible parameter. Thus
it is enough to show that.
(4) The dual curveofis defined by the
geneous polynomlial$X^{*},$ $Y^{*},$$X,\mathrm{Y},$$C_{s}^{*}$$G_{\epsilon}(X^{*}, \mathrm{Y}*, Z^{*}, S)$$D_{s}^{*}$$g_{\epsilon}(xyS)*,*,,$$\epsilon=$$\triangle_{Y}\cdot(G)$$X^{*},$ $Z^{*}$$([\mathrm{O}2])$$\mu,$$C_{s}^{*}$$C_{54}^{*}\cong C_{5}4$$C_{54}^{*}$$C_{54}$$\mathrm{h}\mathrm{o}\mathrm{m}o$$128X^{*5}z*+1376X^{*4}z^{*}2-192x^{*}3Y*z2*+4664X*3z*3-2x*2\mathrm{Y}*4$$G(X^{*}, Y^{*}, Z^{*})$ $:=$$-1584x^{*2}Y^{*}2Z^{*2}+7090x^{*2*4}Z+58x^{*}\mathrm{Y}*4z*-3060x^{*}\mathrm{Y}*2Z*3$$+5050x*z^{*5}+\mathrm{Y}^{*6}+349Y^{*4*}Z2-1725\mathrm{Y}^{*}2Z^{*4}+1375Z^{*6}$

We can see thattoas
$C_{54}$$C_{54}^{*}$

has also 3 $(3,4)$ -cusps. Moreover we can see that
where

$C_{54^{*}}$

is isomorphic$(C_{54}^{*})^{A}=C_{54}$$A=$. For the later purpose, we change the coordinates of so that
3.3. Involution on
the three cusps ofare at $O_{Z}:=(0,0,1),$ $\mathit{0}_{Y}:=(0,1,0),$ $O\mathrm{x}:=(1,0,1)$ . New normal
form in affine spce is given by $f(X, y, S)=f2(x, y)^{3}+sf_{3}(x, y)^{2}$ where$C_{54}$$\tau$$G$$C_{s}$(3.10)$f_{2}(x,y):=xy-X+y$ ,$f_{3}(x, y):=-xy$

94

is defined by $f(x, y)=(xy-x+y)^{3}-54X^{3}y^{3}=0$. In this coordinate,anddefined by$C_{54}$$C_{54}^{*}$

is$-28y^{3}-17_{X^{4}}y^{2}-17Xy^{4}-228x^{33}y-2y5+1788x^{3}y+1788x^{2}\mathrm{y}-17y^{4}-17_{X^{4}}$$+262xy+1788x^{23}y-1788Xy-262xy^{4}2+1788xy^{3}-1788X^{32}y-8166Xy^{2}2+28x^{3}$$+262_{X^{4}}y-2_{X}5-y2xy^{5}+1-17y^{2}-17x2+2x^{5}+2x-2y+x^{6}+y^{6}=0$

It is easy to see that

where$(C_{54}^{*})^{A_{1}}=C_{54}$$A_{1}:=(_{-1^{/}}^{-1}7/3/33$$-1/37/31/3-1/3-7/31/3\mathrm{I}$Let $F(X, Y, Z)$ be the homogenization of $f(x, y)$ . Then the Gauss map induces an au, where)
which is defined by (X,tomorphism dualc :$C_{54}arrow
C_{54}$
by theare partial derivatives. We define an isomorphism ::which is defined by the mulcomposition ofand the linear map
is given by the restriction of the rational mapping:
tiplication by
from the right.
$Y,$ $Z$$C_{54}arrow C_{54}^{*}$$F_{X},$ $F_{Y},$ $F_{Z}$$\mathrm{d}\mathrm{u}\mathrm{a}1_{c_{5}}4$$\varphi_{A_{1}}$$A_{1}$$\Psi$:$\mathrm{C}^{2}arrow \mathrm{C}^{2}$.

$(x.v)\mapsto \mathrm{r}_{x_{d}.\lrcorner_{d}}\prime 1$$\mathcal{T}1\mathrm{S}$$C_{54}^{*}arrow C_{54}$$\tau$

and.
$(1/4, -1,1),$ $F_{3}.--(4, -4,1)$ andObserve thatderlned over$\mathrm{q}$$\mathrm{C}_{54}’$

has tllree tlexes 01 oraer$\angle$at $F_{\dot{1}}:=(\perp, -\perp/4,1),$$F_{2}$$:=$

exchanges flexes and cusps:$\tau$$\tau(O_{X})=F_{1},$ $\tau(O_{Y})=F_{2},$ $\tau(O_{Z})=F_{3}$(3.11)$\mapsto(F_{X}, F_{Y}, F_{Z})$$\tau$,$\{$$\tau(F_{1})=o_{x,\tau}(F_{2})=O_{Y},$ $\tau(F_{3})=Oz$

Furthermore we assert that
Proposition 3.12. The morphism$\tau$

is an involution$C_{54}$.

definedFor the proof, we prepare a lemma. Let be a given irreducible curve in. Thenis defined byby a homogeneous polynomial $F(X, Y, Z)$ and let$G(X,\mathrm{Y}, Z):=F((X, Y, Z)B-1)$ . Let dualc: $Carrow C^{*}$ be the Gauss map which is defined) $\mapsto(F_{x}(x, Y, Z), FY(X, Y, z), F_{Z}(X, Y, z))$ .
by (X,$C$$\mathrm{P}^{2}$$B\in \mathrm{G}\mathrm{L}(3, \mathrm{C})$$C^{B}$$Y,$ $Z$

Lemma 3.13. Two curve8$(C^{B})^{*}$

and$(C^{*})^{t}B^{-1}$coincide and the following diagram com-mutes.
$carrow Carrow\iota_{B}\varphi_{B}l_{C^{B}}duaduab$$(c)c*\mathrm{g}^{\varphi_{t}-}|B1*$

Proof. This is essentially the same as Lemma 2, [02]. The assertion follows from thefollowing equalities. Let $(a, b, c)\in C$ .

$\mathrm{d}\mathrm{u}\mathrm{a}1_{c}B(\varphi B(a, b, C))=(Gx(\varphi B(a, b, C)), G_{Y}(\varphi B(a, b, C)), c_{z}(\varphi B(a, b, C)))$$=(F_{X}(a, b, c), F_{Y}(a, b, c), Fy(a, b, c))tB^{-}1=\varphi \mathrm{r}B-1(\mathrm{d}\mathrm{u}\mathrm{a}1c(a, b, c)$$\square$95

Proof of Proposition 3.12. By the definition of , we have $(C:=C_{54})$ :$\tau$($\tau\circ\tau=$

as$\varphi_{t}A_{1}^{-1\circ}$

dualc)2$=(\mathrm{d}\mathrm{u}\mathrm{a}1_{c^{A}1}\circ\varphi_{A_{1}})\circ$($\varphi_{t}A_{1}^{-1\circ}$

dualc)$=\mathrm{i}\mathrm{d}$is a symmetric matrix.
Of course, the same assertion is true forin the old normal form.
has another$C_{54}arrow
C_{54}$
obvious involution :which is defined by $(x, y)\mapsto(x, -y)$ in the old normal
form. For the application to arithmetic prop.erty of cubic curves, see [O3].
$A_{1}$$C_{54}$$C_{54}$$\iota$REFERENCES
[B-K]

[D]
[Ko]

[N]
[O1]
[O2]

E. Brieskorn and H. Kn\"orrer, Ebene Algebraische Kurven, Birkh\"auser (1981), Basel-Boston
Stuttgart.
A. Degtyarev, Alexander polynomial of a curve of degree six, J. Knot Theory and its Ramification,
Vol. 3, No. 4, 439-454, 1994
K. Kodaira, On compact analytic surfaces II, Ann. of Math. 77 (1963) 563-626 and III, Ann. of
Math. 78 (1963) 1-40.
M. Namba, Geometry of projective algebraic curves, Decker, New York, 1984
M. Oka, Flex Curves and their Applications, Geometriae Dedicata, Vol. 75 (1999), 67-100
M. Oka, Geometry of cuspidal sextics and their dual curves, to appear in Advanced Studies inPure Math. $2^{7},1999^{7}$ , Singularities and arrangements,
M. Oka, Moduli of Sextics and Elliptic Curves, preprint 1999.
R. Walker, Algebraic curves, Dover Publ. Inc., New York, 1949.
-$\mathrm{s}_{\mathrm{a}_{\mathrm{P}\mathrm{p}\succ}}\mathrm{o}\mathrm{r}(\mathrm{T}\mathrm{o}\mathrm{k}\mathrm{y}\mathrm{o}$

[O3]
[W]

DEPARTMENT OF MATHEMATICS, TOKYO METROPOLITAN UNIVERSITYMINAMI-OHSAWA, HACHIOJI-SHI TOKYO 192-03, JAPAN
$E$-mail address:. metro-u. . jp$\mathrm{o}\mathrm{k}\mathrm{a}\Phi \mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}$$\mathrm{a}\mathrm{c}$199.8.