数理解析研究所講究録
第 764 巻 1991 年 150-160150Congruences for $M$ and $(M-1)$ -curves

with odd branches on a hyperboloid
SACHIKO MATSUOKA(松岡 幸子, 北海道教育大学函館分校)
1. Introduction.

In [2], Gudkov completed the isotopic classification of nonsingular irreducible real

algebraic curves of order 8 on a hyperboloid and conjectured two congruences concerning$M$

and $(M-1)$ -curves of order $4n$ with odd branches, where$n$

is a positive integer. In

this paper we partially prove his conjecture (see Corollary 1 and Remark 2) as a corollaryof our main theorem. Although the main theorem needs some conditions in its statement,

it treats curves of general even order. As an appendix, we give the complete isotopic
classification of curves of bidegree $(4,4)$ on a hyperboloid (see Remark 4 and the table in\S 5). Since curves of bidegree $(4,4)$ are necessarily of order 8, restrictions for curves of order
8 are also applicable to curves of bidegree $(4,4)$ . However, the existence problem should
be studied separately. Therefore the author thinks that the table is worthy of notice.

2. Formulation of the main theorem.

Let $H$ be a nonsingular quadric surface defined by a real polynomial in the complex projective 3-space p$. It is well-known that the real part $RH(=H\cap RP^{S})$ of $H$ ishomeomorphic to a 2-sphere$S^{2}$or a 2-torus$T^{2}$. In this paper we restrict ourselves to thelatter case. Such a quadric surface is caUed a hype rboloid. By an appropriate real linear

automorphism of$P^{S},$$H$is transformed into the quadric surface $\{X_{0}X_{\}-X_{1}X_{2}=0\}$ . As

is well known, there is a biholomorphic map between this surface and$P^{1}\cross P^{1}$, which is

given by some real polynomials. Hence, in what foUows, we often identify $H$ with$P^{1}\cross P^{1}$.

1980 Math ematic Subject Classifi cation $14G30,14N99$ , 57Nl$.
Research partially supported by Grant-in-Aid for Scientffic Research (No. 02952001),$s$

the Japan Ministry of Education, Science and Culture.

151
Let $A(\subset H)$ be a nonsingular irreducible algebraic curve defined by real polynomials.

Then the divisor dass $[A]$ is written as$d[\infty\cross P^{1}]+r[P^{1}\cross\infty]$

in $Pic(P^{1}xP^{1})$ for some non-negative integers$d$and ,. We $caU(d,r)$ the bidegree of$A$.

Then there exists a real bihomogeneous polynomial $F(X_{0},X_{1};Y_{0},Y_{1})$ of bidegree $(d,r)$. We may think of

$F$

in

$P^{1}\cross P^{1}$of $H^{2}(P^{1}\cross P^{1};Z)$ . The order of$A$

is the intersection number

such that$A$

is the zero locus of$Pic(P^{1}\cross P^{1})$as a subgroup.

$[A]\cdot([\infty\cross P^{1}]+[P^{1}x\infty])=d+r$

Now we set $RA=A\cap RH$ . We say $RA$ is an $(M-i)$ -curve of bidegree $(d,r)$ if thenumber of branches, i.e., connected components of that is$(d-1)(r-1)+1-i$ .
We note that Gudkov (see [2]) defines $(M-i)$-curves of a fixed even order $2m$ to be curves

with$(m-1)^{2}+1-i$

branches. For a branch$C$of $RA$ , the homology class [C] is written as$\epsilon[\infty\cross RP^{1}]+t[RP^{1}\cross\infty]$in$H_{1}$( $RP^{1}\cross RP^{1}$ ; Z), where$\ell$

and$\ell$

are some integers. Following [2], we $caU(s,t)$ thetorsion of the branch $C$. We say the branch $C$ is odd (or even) if the intersection number$[C]\cdot([\infty\cross RP^{1}]+[RP^{1}x\infty])=\ell-s$

is odd (or even). We say$C$is an oval (or non-oval) if $(s,t)=(0,0)$ (or otherwise). We

note that the torsions are equal to the same fixed value for $aU$ non-ovals of $RA$ if we givethem appropriate orientations.
Now suppose that both$d$and ’ are even. We note that if $RA$ is of even order andhas odd branches, then the number of odd branches is even, and hence, both

2

$d$

and

$r$

are

152
even (cf. Theorem below). Then we can define$B^{+}$(resp.

$B^{-}$) to be the set $\{F\geq 0\}(resp$ .

in $RP^{1}\cross RP^{1}$ . Moreover, we can take a double coverin $gYarrow P^{1}\cross P^{1}$ branched, and let. Let and be the two lifts of the complex conjugation of$\{F\leq 0\})$

along

$A$

$RY^{+}$

and $RY^{-}$ be their fixed point sets. Then the restrictions of the covering map make

$RY^{+}$

and $RY^{-}$ be double coverings of$T^{+}$$p\iota\cross P^{1}$$T^{-}$$B^{+}$

and$B^{-}$

branched along $RA$ respectively.

REMARK 1 (see $[5,Remark3.2]$ ). For $RA$ with non-ovals of torsion $(s,\ell),$through the covering map if and only if
regarded as the doubles of$B^{\pm}$$RY^{\pm}$$\frac{d}{2}t+\frac{\prime}{2}\iota$

can beis even.

There are several articles (for instance, [2], [7], [S], [6], and [1]) on real algebraic

curves on a hyperboloid.

Our new results are as follows.
THEOREM. Let$A$

be a nonsingular ixxed ucible real algebrai $cc$ urve of bidegree $(d,r)$ andof even order on a hyperboloid. Suppose that $RA$ has odd branch es of torsion $(s,t)$ Wi
$s$

$tA$

odd an $dt$ even, an $dr\equiv 0(mod 4)$ .
(i) ff $RA$ is an M-curve of bidegree $(d,r)$ , then we have$\chi(B^{\pm})\equiv\frac{d\prime}{2}(mod 8)$.

(i1) $HRA$ is an $(M-l)$ -curve of bidegree $(d,r)$ , then we have $\chi(B^{\pm})\equiv\frac{d}{2}\pm 1(mod 8)$ .
$CoROLLARy1$(related to Gudkov’s conjecture (see [2])). Let$A$

be a nonsingular irre-

ducible $reaI$ algebraic curve of order $8n$ on a hyperboloid. Suppose that $RAh$ as odd

branches.
(i) ff $RA$ is an M-curve of order$8\dot{n}$, then we Aave $\chi(B^{\pm})\equiv 0(mod 8)$ .

(ii) If $RA$ is an $(M-1)$ -curve of order $8n$ , then we have $\chi(B^{\pm})\equiv\pm 1(mod 8)$ .

REMARK 2. In general, $M$ and $(M-1)$-curves of order $4n$ are also that of bidegree$(2n, 2n)$ . By (4) in the proof of Lemma 1 below, for an M-curve (resp. $(M-1)$ -curve)of bidegree $(2n,2n)$ with non-ovals, we automaticaUy have $\dim H.(B^{+}; Z_{2})=(2n-1)^{2}+$(resp. $(2n-1)^{2}$ ). Hence, we do not need to assume this equdity (cf. [2]). Gudkov’s$1$conjecture is just unproved in the case$n$

is odd.

3

153
We

$wi\mathbb{I}$apply Corollary 1 to curves of bidegree $(4,4)$ in \S 5.

3. Some basic results.
In this section we only assume that bothand , are even and $RA$ has non-ovals that$d$are not necessarily odd and show some properties of the double coveringand the involutions$T^{\pm}$. In what foUows, we treat only$T^{+}$. For$T^{-}$$\pi$: $Yarrow P^{1}\cross P^{1}$, we have only to replaceby ‘–, for the reason that the hyperboloid is divided into some annffi by non-ovals.

$+$

We first note that$Y$

is a sinply connect $ed$ compact nonsingular complex algebraic

surface. Hence, in particular, we see that $H_{1}(Y;Z)=0$ and $H^{2}(Y;Z)$ is Eee. Moreover,
we can regard the divisor class group Pic(Y) as$a$subgroup of $H^{2}(Y;Z)$ . We set$E_{1}=\{ae\in H^{2}(Y;Z)|(T^{+})(ae)=\approx\}$and$E_{-1}=\{x\in H^{2}(Y;Z)|(T^{+})^{*}(x)=-x\}$

We note thatLet

$Q\pm 1$

$E_{1}$

and$E_{-1}$are orthogonal each other with respect to the intersection form.

denote the restrictions of the form to$E_{\pm 1}$, and$\sigma\pm\iota$

We consider the second Wu class $v_{2}(\in H^{2}(Y;Z_{2}))$ ofclass has the property that $x\cdot v_{2}=x^{2}$ for every$v_{2}=w_{2}$by Wu’s formula, where

(1)

where

$w_{i}$

is the first Chern class of(2)$E_{-1}$. Since$c_{1}$

is contained in(for instance, see [8]). This

in $H^{2}(Y;Z_{2})$ . Since $w_{1}=0$ , we haveis the -th Stiefel-Whitney class of Y. Hence we have
$i$

$Y$

and

$K_{Y}$$h_{2}=\pi^{*}$$E_{-1}$[,is the canonical divisor of Y. We note that$[K_{Y}]=( \frac{d}{2}-2)h_{1}+(\frac{r}{2}-2)h_{2}$where we set $h_{1}=\pi[\infty xP^{1}]$ and

in

$x$

their signature.

$Y$$v_{2}=(c_{1})_{mod 2}=(-[K_{Y}])_{mod 2}$$c_{1}$.

$P^{1}\cross$,oo]. We note that

, we see that

$E_{1}$

and$h_{2}$are containedis always an even lattice. We set$E=\{x\in H_{2}(Y;Z_{2})|T^{+}(x)=x\}$4$h_{1}$,

154
and let$\alpha_{2}$: $H_{2}(Y/T^{+},RY^{+};Z_{2})\oplus H_{2}(RY^{+}; Z_{2})arrow H_{2}(Y;Z_{2})$be the homomorphism in the Smith exact sequence for the involution$T^{+}$: $Yarrow Y$ Since$H_{1}(Y;Z)=0$ , as in the proof of [4, Lemma 3.7], we have($)${\rm Im}\alpha_{2}=E$.

LEMMA 1 (cf. [5, Remark3.1]). A curve $RA$ with non-ovals is an $(M-i)$ -curve if andonly if th pair $(Y,T^{+})$ is an $(M-(i+2))$ -manifold in the sense of [8], i.e.,$e$$\dim H_{*}(RY^{+};Z_{2})=\dim H_{*}(Y;Z_{2})-2(i+2)$.

PROOF: For a curve with non-ovals, it is easy (cf. [5, \S 3]) to verify that(4)$\dim H.(B^{+};Z_{2})=\#${branches of $RA$ }and$\dim H.(RY^{+};Z_{2})=2\cdot\#${branches of $RA$ }.

We note that we must add 2 to the right-hand side of (4) if the curve has only ovals and$B^{+}$

contains the exterior of all the ovals. Here, however, we assume that the curve hassome non-ovals. On the other hand, we have (see also [5, \S 3])$\dim H_{*}(Y;Z_{2})=\chi(Y)=6+2(d-1)(r-1)$

Thus we have the required result. I
By Lemma 1 and [4, Lemma 3.7], we have(5)$|\det Q_{1}|=|\det Q_{-1}|=2^{i+2}$

for an $(M-i)$ -curve with non-ovals.

5

.

155
According to (2.4) of [8], we have$\sigma_{1}-\sigma_{-1}=-\chi(RY^{+})$

Since the signature$\sigma(Y)$of $Y$ is equal to$\sigma_{1}+\sigma_{-1}$.

, we have$\sigma(Y)+\chi(RY^{+})=2\sigma_{-1}$.

On the other hand, we have $\sigma(Y)=-dr$ (see [5, \S 3]) and $\chi(RY^{+})=2\chi(B^{+})$ . Thus we
have
(6)

.

$\chi(B^{+})-\frac{dr}{2}=\sigma_{-1}$

LEMMA 2. Let$A$

be a nonsingular irred ucible $reaI$ algebraic curve of bidegree $(d, r)$ on a

hyperboloid. $Su$ppose that $RA$ A as odd branches of orsion $(s,t)$ with$t$and $r\equiv 0(mod 4)$ . Then$h_{1}\cdot x$

is even for every

$x$

in

$E_{-1}$$\iota$odd and

$t$

even,

.

PROOF: We note that is the intersection number $[C]\cdot[\infty\cross RP^{1}]$ for each odd branch $C$$t$

of $RA$ . We may think that$\infty\cross p\iota$

and

$A$

$\infty\cross P^{1}$

intersects the curve$A$

transversely in$P^{1}\cross P^{1}$

are real curves, i.e., invariant under the complex conjugation,intersects $RA$ transversely in$RP^{1}\cross RP^{1}$. The inverse image

real curve in $Y$ and represents the cohomology class$h_{1}$$\pi^{-1}(\infty\cross P^{1})$. We set. Since$\infty\cross RP^{1}$

is a nonsingular$K=RY^{+}\cap\pi^{-1}(\infty\cross P^{1})$.

Then we have
$K=\pi^{-1}(B^{+})\cap\pi^{-1}(\infty\cross P^{1})=\pi^{-1}(B^{+}\cap\infty\cross RP^{1})$By the assumption, we see that

double ofthe double of$B^{+}\cap\infty\cross RP^{1}$is even. Hence, $RY^{+}$ can be regarded as the$\frac{d}{2}t+\frac{}{2}s$through the covering map$B^{+}$

.

(

$\pi$

$reca\mathbb{I}$Remark 1), and $K$ is also regarded as.

We will show that the cycle $K$ is a$Z_{2}$-boundary in $RY^{+}$ . To prove this, it suffices

to prove that, for every connected component$B_{h}^{+}$

of$B^{+},$$\pi^{-1}(B_{h}^{+}\cap\infty\cross RP^{1})$is a$Z_{2^{-}}$

boundary in $RY^{+}$ .

We first consider the case when the boundary

denoted by$C_{1}$

and $C;$ , and some ovals.

$C_{1}$

and6$\partial B_{h}^{+}$$C_{2}$

of$B_{h}^{+}$

consists of two odd branches,

divide the hyperboloid into two annuli.

156$B_{h}^{+}$

is obtained from one of them (say$C_{2}$) by removing the interiors of some ovals. Since

meets

$R$

together with some arcs each joining$C_{1}$$[C_{1}]\cdot[\infty\cross RP^{1}]$tois even,$R$$\infty\cross RP^{1}$in an even number of intervals joining(orand arcs always bounds, dividing $R$ into regions

bounds$\pi^{-1}(B_{h}^{+}\cap R_{+})$$C_{2}$) to itself. Such a union of intervals

and$R_{+}$$C_{1}$$R_{-}$. Then$\pi^{-1}(B_{h}^{+}\cap\infty\cross RP^{1})$.

In the case when the boundary$\theta B_{h}^{+}$

consists of only some ovals,$B_{h}^{+}$

is indeed theinterior of an oval by removing the interiors of some nested ovals since we assume that $RA$
has some non-ovals (odd branches). Hence we obtain the same fact as above by a similar(but simpler) argument.

Thus we see that$[K]=0$ in $H_{1}(RY^{+};Z_{2})$Since $K$ is a disjoint union of$S^{1}$.

, the total Stiefel-Whitney class $w(K)$ is equal to 1.

Hence, $RY^{+}$ and $K$ satisfy the conditions a) and b) of [$, Remark 2.2]. By this remark,[3, Lemma 2.3] is applicable to the involution$T^{+}$

lemma, $(h_{1})_{mod 2}(\in H_{2}(Y;Z_{2}))$ is orthogonal to$E$. Hence,$h_{1}\cdot x$

is even for every

$x$

in

LEMMA 3 (cf. [$, Lemma3.1]). Let$G\cross Garrow Z$

element$u$in$E_{-1}$$G$: $Yarrow Y$ and${\rm Im}\alpha_{2}$. By ($),$\pi^{-1}(\infty\cross P^{1})$$(h_{1})_{nod2}$. By thisis orthogonal to.I

be a $kee$ abelian group of fini $te$ rank, and$Q$:

be an even symmetric bilinear form. Suppose th at there exists a primitive$G$such that $Q(u, u)\equiv 0(mod 8)$ and $Q(u, x)$ is even for every

$x$

in

$G$

.

(i) II $|\det Q|=4$ , then Sign $Q\equiv 0(mod 8)$ .

(ii) ff $|\det Q|=8$ , then Sign $Q\equiv\pm 1(mod 8)$ .
PROOF: (i) has already appeared in [3, Lemma 3.1]. As in the proof of this lemma,

we define an even integral symmetric bilinear form$|\det\tilde{Q}|=2$. Hence, SignQ$=Sign\tilde{Q}\equiv\pm 1(mod 8)$

4. Proof of theorem,

7

.

$1$

$\tilde{Q}$by using$Q$. If $|\det Q|=8$ , then

157
Now we prove the theorem. By Lemma 2,
to be even. Ifis$E_{-1}$. If

for every$\frac{d}{2}$$x$$\frac{d}{2}$$h_{1}\cdot x$

is even for every ae in$E_{-1}$. is assumed$\frac{}{2}$is also even, then, by (1) and (2), $H^{2}(Y;Z)$ is an even latt ice, and hence, so

is odd, then, by (1) and (2), $v_{2}=(h_{1})_{mod 2}$ . Therefore, $x^{2}\equiv h_{1}\cdot x\equiv 0(mod 2)$in$E_{-1}$, that is,in $H^{2}(Y;Z)$ , hence in$E_{-1}$is en even lattice. It is easy to check that

$E_{-1}$, and$h_{1}^{2}=0$

. If

$RA$

$h_{1}$

is primitiveis an M-curve (resp. $(M-1)$ -curve), then,by (5), we have $|\det Q_{-1}|=4$ (resp. 8). Thus, by Lemma 3 and the formula (6), we havethe required results. 1

5. Isotopic classffication of curves of bidegree $(4,4)$.

We say two real algebraic curves $RA$ and $RA’$ on a hyperboloid $RH$ are isotopic ifthere exists a continuous map
I : $RHx[0,1]arrow RH$
such that$\varphi:=\Phi$( , t) are homeomorphisms,$\varphi_{0}$is the identity map, and $\varphi_{1}(RA)=RA’$ .

For a curve of bidegree $(4,4)$ , the number of non-ovals is $0,2$ , or 4. If it is 4, thenwe have $|s|\leq 1$ and$|\ell|\leq 1$, whereno more branches (Notation:$(s,\ell)$$4(\iota,t)$

is the torsion of the non-ovals, and the curve has). If we have $(|s|, |t|)=(1,2)$ or $(2, 1)$ , then the curve

has no more branches (Notation: $2(s,t)$ ). If the number of non-ovals is 2 and $|s|\leq 1$ and$|t|\leq 1$, then the non-ovals divide$RP^{1}\cross RP^{1}$

into two annuli and the interior of each oval

does not contain any other ovals (Notation: $2(s,t;m,n)$ , where $m$ and $n(m\geq n)$ denotethe numbers of ovals contained the two annuli respectively.)
For curves of bidegree $(4,4)$ with non-ovals, the notations defined above describe theisotopy classes.

CONVENTION: We regard two isotopy classes 2 $(,,t;m, n)$ and $2(,/,t’;m, n)$ as equivalentif$(|s|, |t|)=(|s|, |t’|)$or$(|t’|, |\ell’|)$.

As for $4(s,t)$ and $2(s,t)$ , we define the equivalence relation in the same way.

Let$\frac{\overline m}{n}$denote the equivalence dass which contains $2(1,0;m, n)and/m/n$ the classwhich contains $2(1,1;m,n)$ .

8

158

$\mathbb{R}om$

Corollary 1, we obtain the following.

COROLLARY 2 ([2,THEOREM Dl]). Let $RA$ be a nonsingular $reaI$ algebraic curve of bide-gree $(4, 4)$ in $RP^{1}\cross RP^{1}$ . Suppose th at $RA$ has on-ovals of torsion $(\pm 1,0)$ or $(0, \pm 1)$ .
$n$(i) If‘ $RA$ is an M-curve, then its $iso$ topy type is$\overline{\frac{8}{0}}$or(i1) If $RA$ is an $(M-1)$ -curve, th en its isotopy type is$\overline{\frac{4}{4}}$$\frac{\overline\tau}{0}$or$\overline{\frac{4}{s}}$REMARK 3. The author confesses that the existence of curves of type$\frac{\overline 6}{2}$asserted in [5]is an error.
REMARK 4. CoroUary 2 is the last restrictions for isotopy types of curves of bidegree $(4,4)$ .
In fact, we can realize $aU$ the isotopy types listed in [5, Table 1.1] that satisfy this corollary.

The existence of some isotopy types is announced in [5] and [6]. We can show the existenceof the others by checking that the corresponding curves of order 8 constructed in [2] and[7] are just of bidegree $(4,4)$ . The following table gives all the isotopy types, which actuaUy
exist, of nonsingular real algebraic curves of bidegree $(4,4)$ in $RP^{1}xRP^{1}$ , where we use
the well-known notations for the curves which have only ovals (see [2] and [5]).
(Table is inserted here.)

REFERBNCES

1. P. Gilmer, Algebraic curves in $RP(1)\cross RP(1)$ , Proc. Amer. Math. Soc. (to appear).
2. D.A. Gudkov, On the topology of algebraic curves on a hyperboloid, Russian Math.
Surveys 34 (1979), 27-35.

3. V.M. Kharlamov, Additional congruences for the Euler characteristic of real algebraic

manifolds of even dimensions, Funktsional’nyi Analiz i Ego Prilozheniya $=Ehnctional$9

159

Anal. Appl. 9 (1975), 134-141.

4. V.M. Kharlamov,’ The topological types of nonsingular surfaces in $RP^{\}$ of degree
four, Funktsional’nyi Analiz i Ego Prilozheniya $=Functional$ Anal. Appl. 10 (1976),

295-305.
5. S. Matsuoka, Nonsingular algebraic curves in $RP^{1}\cross RP^{1}$ , Ikans. Amer. Math. Soc.
324 (1991) (to appear), 87-107.

6. S. Matsuoka, The configurations bf the M-curves of degre e(4,4) in $RP^{1}\cross RP^{1}$ andperiods of real K$ surfaces, Hokkaido Math. J. 19 (1990), 361-378.

7. O.Ya. Viro, Construction of multicomponent real algebraic sutfaces, Dokl. Akad. NaukSSSR $=Soviet$ Math. Dokl. 20 (1979), 991-995.
8. G. Wilson, Hilbert’s sixteenth pw blem, Topology 17 (1978), 53-73.

$(\star L\# 1^{1\underline{\not\in}\ovalbox{\tt\small REJECT}^{l}\hslash x\epsilon’s/\forall^{\grave{\gamma}}\ovalbox{\tt\small REJECT},?k^{\chi\ovalbox{\tt\small REJECT})}}\backslash \overline{T}^{0_{\backslash k}}.+o\not\in\ovalbox{\tt\small REJECT} 7_{/}\cdot g_{\sim_{g\nu^{/_{\grave{A}_{-}}}}^{ffl_{7}}}$,$\backslash \backslash \sigma)_{\hat{b}}^{1}\phi X_{\backslash }IXB\iota_{\lambda}//eT/^{-}n$ド$*e\S$\Re \‘‘れる予紅 7。

of de$L_{0\cap}do\eta/\sqrt{}\iota demat_{4\grave{c}\ }/s_{oc};_{e}$

10

160
Curves without non-ovals(46 types)$\frac{1}{1}8$$\frac{5}{1}4$$\frac{9}{1}$$\frac{5}{1}$$\frac{8}{1}$$\frac{s}{1}4$$\frac{s}{1}3$$\frac{4}{1}2$$\frac{5}{1}1$$\frac{6}{1}$54$\frac{1}{1}2$$\frac{2}{1}1$$\frac{s}{1}$6$\frac{1}{1}3$$\frac{2}{1}2$$\frac{s}{1}1$$\frac{4}{1}$7$\frac{1}{1}5$$\frac{1}{1}4$$\frac{2}{1}3$8$\frac{1}{1}6$$\frac{}{1}5$$\frac{}{1}4$$\frac{s}{1}2$$\frac{4}{1}1$$\frac{s}{1}$9$\frac{1}{1}7$$\frac{4}{1}4$$\frac{4}{1}3$$\frac{\}{1}2$$\frac{t}{1}1$$\frac{7}{1}$$$\frac{1}{1}\frac{1}{1}$3$\frac{1}{1}1$$\frac{}{1}$2$\frac{1}{1}$1$\phi$(20 types)

Curves vrith two non-ovals of torsion $(1,0)$$\overline{\frac{m}{\iota}}$$\frac{\overline 8}{0}$$\frac{\overline 4}{4}$$\frac{\overline 7}{0}$$\frac{\overline 4}{\}$( $m\geq n\geq 0$ andCurves with two non-ovals of torsion $(1,1)$$/m/n$

Otherwise$m+\mathfrak{n}\leq 6$)(25 types)( $m\geq n\geq 0$ and $m+n\leq 8$ )($ types)$4(1,0)$$4(1,1)$$2(2,1)$TABLE. All the isotopy types of curves of bidegree $(4,4)$

11(94 types)