On the number ofpyramids of
a
generic
spacecurve
J. J. $\mathrm{N}\mathrm{U}\tilde{\mathrm{N}}\mathrm{O}$ BALLESTEROS ($\eta_{\neg}^{\backslash }/^{\backslash }$
.
シア丸)and
OSAMU
SAEKI (佐伯修
$J$▲島大欲
)
Abstract. We prove that for a
generic
closed space curve, the number of pyramids (triple points of the tangentdevelopable) is congruent modulo2
to the sum of the indices of the torsion zero points. This index is defined as the number oftrisecant lines of the curve passing through the torsion zero point. The result is deduced from the study ofthesingularities of the tangent developablesurface of the curve.1. INTRODUCTION
In this paper we study the number of triple points of the tangent devel-opable of a space curve. The tangent develdevel-opable of a space
curve
$\alpha:S^{1}arrow \mathrm{R}^{3}$is thesurface$\chi(\alpha)$ in $\mathrm{R}^{3}$ defined bythetangent lines of$\alpha$
.
The local formofthissurface was first studied by Cleave [C], and recently the first author [N] proved that it is atopologically stablesurface when the
curve
isgeneric
(fora definition of a topologically stable surface, see $\beta \mathrm{M}$]). He obtained as a consequence thatif the curvehas no torsion zero points (topological cross caps of$\chi(\alpha)$),$.\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}$ the
number of pyramids of the curve (triple points of$\chi(\alpha)$) is even.
Here, we give one step more in this direction. The abovementioned result can be easily generalized by applying a theorem by Sz\"ucs [Sz], which gives the following
congruence:
$T( \alpha)\equiv\sum_{i=1}^{k}n(X_{i}, \alpha)$ mod
2.
The number $T(\alpha)$ is the number of pyramids of $\alpha,$ $x_{1},$ $\cdots,$$x_{k}$ are the torsion
zero points of$\alpha$ and $n(x:, \alpha)$ is the indexof each torsion zero point conveniently
defined (see also [NS]). The problem is that this index, $n(x:, \alpha)$, does not give a priori any information on the geometry ofthe curve. We will show, by using a
new
proofof the Sz\"ucs theorem given by the authors in [NS], that the index $n(x:, \alpha)$ can be interpretedgeometrically
as the number of trisecant lines of $\alpha$passing through $x:$
.
The first author is partially supported byDGICYT Grant PB91-0324. The second author is partially supported by CNPq, Brazil.
Note that thisresult can be considered dual to the congruenceobtained by
Banchoff, Gaffney and $\mathrm{M}\mathrm{c}\mathrm{C}\mathrm{r}\mathrm{o}\mathrm{r}\mathrm{y}$in [BGM], where they show that for a generic
space curve $\alpha$ : $S^{1}arrow \mathrm{R}^{3}$ we have
$\tilde{T}(\alpha)\equiv\sum_{i=1}^{k}n\sim(xi, \alpha)$ mod 2,
where now$\tilde{T}(\alpha)$ is the number oftritangent planes of$\alpha$ and the index$\sim n(x_{i}, \alpha)$ is
one half of the number of points in the intersection off the point $x_{i}$ of the curve
with the osculating plane at the torsion zero point $x_{i}$ (see also [O]). Although
their paper [BGM] is previous to the Sz\"ucs one, they implicitly use the Sz\"ucs
result for the dual surface of the curve. Remember that the dual surface of a regular space curve is the surface in $(\mathrm{R}P^{3})^{*}$ defined by the tangent planes to
$\alpha$. When the curve is generic, the dual surface is again a topologically stable
surface, the triple points corresponding to the tritangent planes to $\alpha$ and the
cross caps to the torsion zero points (see [BGM] for details and [Sch] for the
duality between the dual surface and the tangent developable of ageneric space
curve).
2. THE NUMBER OF PYRAMIDS OF A GENERIC SPACE CURVE
Suppose that $\alpha$ : $S^{1}arrow \mathrm{R}^{3}$ is a smooth space curve satisfying the general
position conditions $(\mathrm{G}\mathrm{P})$ given in [N]. We also assume the conditions (1) and (2)
in [$\mathrm{N}$, Lemma 7]. Furthermore, we assume the additional conditions as follows:
(i) If$\tau(s)=0$, then there are no quadrisecants of$\alpha$ passing through $\alpha(s)$, where $\tau$ is the torsion of $\alpha$.
(ii) If there is a trisecant to $\alpha$ at $\alpha(s\mathrm{o}),$ $\alpha(s_{1}),$$\alpha(s_{2})$ with $\tau(s\mathrm{o})=0$,
then the vectors $\alpha(s_{1})-\alpha(s\mathrm{o}),$ $\alpha(/S_{1})$ and $\alpha’(S2)$ are linearly
independent.
(iii) Thetritangentplanes of$\alpha$do not osculate at thetangencypoints.
(iv) The number of osculating planes of $\alpha$ containing a trisecant is
finite.
(v) The subset $T$ of$S^{1}\cross S^{1}\cross S^{1}-\Delta$ defined by
$\mathcal{T}=\{(S_{1}, s2, s3):(\alpha(_{\mathit{8}_{2}})-\alpha(_{S_{1}}))\cross(\alpha(s_{3})-\alpha(s1))=0\}$ ,
is a closed 1-dimensional submanifold of$S^{1}\cross S^{1}\cross S^{1}-\Delta$, where $\Delta=$
{
$(s1,$$s2,$$s3):$ $si=s_{j}$ for some $i\neq j$}.
LEMMA 2.1. The set ofsmooth curves $\alpha$ satisfying th$e$ above general position
conditionsisresid$\mathrm{u}\mathrm{a}l$ in the
$sp$ace $C^{\infty}(S^{1}, \mathrm{R}^{3})$, with the Whitney $C^{\infty}$-topology.
PROOF: We consider the multijet space $4J^{3}(S^{1}, \mathrm{R}^{3})$ and take the coordinates
$(s_{1},$$s_{2},$$s_{3},\mathit{8}4,$$r_{1}^{0},$$r_{2}^{0},$$r_{3}^{0},$$r_{4}^{0},$
$\ldots,$
where $s_{i}\in S^{1}$ and $r_{j}^{i}\in \mathrm{R}^{3}$
.
Then we define the subset $W_{1}\subset 4J^{3}(s^{1}, \mathrm{R}^{3})$ bythe equations:
$(r_{2}^{0}-r^{0})1\cross(r_{3^{-}}^{0}r)1=(00r-2r_{1}^{0})\cross(r_{4}^{0}-r_{1}^{0})=0$,
$\det(r_{1}^{1}, r_{1}^{2}, r^{3})1=0$.
Clearly, $W_{1}$ is an algebraicsubset$\mathrm{o}\mathrm{f}_{4}J^{3}(S1, \mathrm{R}^{3})$ of codimension
5.
By themulti-jet version of the Thomtransversality theorem, the set of curves$\alpha\in C^{\infty}(S^{1}, \mathrm{R}^{3})$
such that $4j^{3}\alpha$ : $(S^{1})^{(4})arrow 4J^{3}(s^{1}, \mathrm{R}^{3})$ is transversal to $W_{1}$ is residual in $C^{\infty}(S1, \mathrm{R}3)$. But since $(S^{1})^{(4})$ has dimension 4, it is obvious that the
transver-sality condition is equivalent to condition (i).
Analogously, we see that the rest of conditions $(\mathrm{i}\mathrm{i}),\ldots,(\mathrm{V})$ give residual
sub-sets in $C^{\infty}(S^{1}, \mathrm{R}^{3})$. $1$
LEMMA 2.2. Let $p=\alpha(S_{0})$ be a torsion zero point. If $\alpha$ satisfies the above
general position $co\mathrm{n}$dition$s$, then the$n$umber of trisecan$ts$ of$\alpha$ passing through $p$ is finite.
The proof of Lemma 2.2 will be given later (see the paragraph just after
the proof of Proposition 2.8).
DEFINITION
2.3.
For a torsion zero point $p=\alpha(S_{0})$ of $\alpha$, we define the index$n(p, \alpha)(\in \mathrm{Z})$ to be the number of trisecants of $\alpha$ passing through$p$.
Recall that if $\alpha$ satisfies the condition $(\mathrm{G}\mathrm{P})$ as in [N] (condition 3), then
the number oftorsion zero points is finite.
The main purpose of this section is to prove the following.
THEOREM 2.4. Let $\alpha:S^{1}arrow \mathrm{R}^{3}$ be asmooth space curve sa$tisfying$
,
thegeneral position $c\mathrm{o}\mathrm{n}$ditions as stated above. Then we $h$a$ve$
$T( \alpha)\equiv\sum_{i=1}^{k}n(xi, \alpha)$ mod 2,
where$T(\alpha)$ is the number of pyramids of$\alpha$
an..d
$x_{1},$$\cdots,$$x_{k}$ are the torsion zeropoints of$\alpha$.
LEMMA
2.5.
Let $f$ : $(s_{0}-\epsilon, S0+\epsilon)arrow \mathrm{R}$ be a $C^{\infty}$ function. Then there existsanother $C^{\infty}f\mathrm{u}$nction $f_{*}$ : $(s0-\epsilon, s0+\epsilon)arrow \mathrm{R}$ such that if$s\neq s_{0}$, then $f_{*}(S)= \frac{f(s)-f(s\mathrm{o})}{s-s_{0}}$,
and if$s=s_{0}$,
$f_{*}(s\mathrm{o})=f’(_{S}0),$ $\frac{df_{*}}{ds}(s\mathrm{o})=\frac{1}{2}f^{\prime J}(s_{0})$.
PROOF: Just apply a classical argument of analysis. We see easily that
$f( \mathit{8})-f(_{S_{0}})=\int_{0}^{1}\frac{d}{dt}\{f(\mathit{8}_{0}+t(s-S_{0}))\}dt$
$= \int_{0}^{1}f’(_{S_{0}+t}(S-s\mathrm{o}))(s-s\mathrm{o})dt$
$=(s-s\mathrm{o})f*(s)$,
where $f_{*}(s)= \int_{0}^{1}f’(S_{0}+t(s-S\mathrm{o}))dt$. The rest is easy to check.
1
Now let $\alpha$ : $S^{1}arrow \mathrm{R}^{3}$ be a $C^{\infty}$ curve which is regular and simple. Given
a point $p\in \mathrm{R}^{3}$ we define $t_{p}$ : $S^{1}arrow \mathrm{R}P^{2}$ by $t_{p}(s)=[\alpha(s)-p]$ if $\alpha(\mathit{8})\neq p$, and $t_{\mathrm{p}}(s)=[\alpha’(s)]$ if$\alpha(s)=p$. Note that $t_{\mathrm{p}}$ is a smooth map.
LEMMA
2.6.
$S\mathrm{u}$ppose th at$p=\alpha(s_{0})$. Then$t_{p}$ isan immersion at $s_{0}$ ifand on$ly$
if$\kappa(S_{0})\neq 0$, where $\kappa$ is the $c\mathrm{u}\mathrm{r}$vature of$\alpha$.
PROOF: Suppose for instance that $\alpha_{3}’(S_{0})\neq 0$, where $\alpha(t)=(\alpha_{1}(t), \alpha_{2}(t)$,
$\alpha_{3}(t))$. We take coordinates in $\mathrm{R}P^{2}$ such that the homogeneous coordinate [X,$Y,$$Z$] corresponds to $(X/Z, Y/Z)$. Then $t_{p}$
. gives the map
$t_{p}^{\sim}$ in a
neighbour-hood of$s_{0}$ given by
$t_{p}^{\sim}(s)=( \frac{\alpha_{1}(s)-\alpha_{1}(s\mathrm{o})}{\alpha_{3}(s)-\alpha_{3}(s\mathrm{o})},$ $\frac{\alpha_{2}(s)-\alpha_{2}(s\mathrm{o})}{\alpha_{3}(s)-\alpha 3(s\mathrm{o})})$,
when $s\neq s_{0}$ and
$t_{p}(s \sim)=(\frac{\alpha_{1}’(s\mathrm{o})}{\alpha_{3}’(S_{0})},$ $\frac{\alpha_{2}’(S\mathrm{o})}{\alpha_{3}’(s\mathrm{o})})$ ,
when $s=s_{0}$. By Lemma 2.5, this map is
$t_{p}^{\sim}(s)=( \frac{\alpha_{1*}(s)}{\alpha_{3*}(s)},$$\frac{\alpha_{2*}(s)}{\alpha_{3*}(s)})$ ,
for $s$ in a neighbourhood of $s_{0}$
.
Then $t_{p}\sim$ is differentiable at$s_{0}$ and
$t_{\mathrm{p}}’(S_{0} \sim)=\frac{1}{2\alpha_{3}’(_{\mathit{8}_{0})}2}(\alpha_{1}^{\prime/}(s\mathrm{o})\alpha’(3s\mathrm{o})-\alpha_{1}/(_{S}0)\alpha’3’(\mathit{8}_{0}), \alpha_{2’}’(s\mathrm{o})\alpha’(3s\mathrm{o})-\alpha_{2}/(S\mathrm{o})\alpha(/3s\mathrm{o})/)$ .
Therefore $t_{\mathrm{p}}’(s0)\sim=0$ if and only if $\alpha’(S_{0})\cross\alpha^{\prime/}(s\mathrm{o})$ $=0$; i.e., if and only if $\kappa(s\mathrm{o})=0$.
I
LEMMA 2.7. Suppose that $p=\alpha(S_{0})$. Then $t_{p}$ is an immersion at $s\neq s_{0}$ ifand onlyif$\alpha’(s),$$\alpha(S)-\alpha(s0)$ are not collinear.
PROOF: Taking coordinates as in the proof of Lemma 2.6, we have
$t_{p}^{\sim}(s)=( \frac{\alpha_{1}(s)-\alpha_{1}(s\mathrm{o})}{\alpha_{3}(s)-\alpha_{3}(s\mathrm{o})},$$\frac{\alpha_{2}(\mathit{8})-\alpha 2(s\mathrm{o})}{\alpha_{3}(s)-\alpha_{3}(s\mathrm{o})})$ .
Hence we have that $t_{p}’(s)\sim=0$ ifand only if$\alpha’(s)\cross(\alpha(S)-\alpha(S0))=0.1$
PROPOSITION
2.8.
If$\alpha$ is a smooth curve satisfying thegeneralpositioncondi-tions and$\tau(s0)=0$, then$t_{p}$ isan immersion with $\mathrm{n}$ormalcrossingsfor$p=\alpha(S0)$
.
PROOF: Since $\kappa(s_{0})>0$ and there are no cross tangents passing through $\alpha(s_{0})$
(see the condition 7 of [N]), $t_{\mathrm{p}}$ is an immersion.
On the other hand, a self-intersection of$t_{p}$ happens when $t_{p}(s_{1})=t_{p}(s_{2})$
for $s_{1}\neq s_{2}$. Again the fact that there are no cross tangents passing through $\alpha(S_{0})$ implies that $s_{1},$ $s_{2}\neq s_{0}$; therefore $[\alpha(s_{1})-\alpha(S\mathrm{o})]=[\alpha(s_{2})-\alpha(s\mathrm{o})]$, i.e.,
thereis atrisecant to $\alpha$ passing through$\alpha(s_{0}),$$\alpha(S1),$$\alpha(S_{2})$. Byourcondition (i)
$t_{p}$ has no triple points. We prove that condition (ii) implies the normal crossing
condition at a double point, that is, that $t_{p}’(s_{1})$ and $t_{p}’(s_{2})$ are not collinear.
Since $\alpha(s_{1})-\alpha(S0),$$\alpha(s2)-\alpha(s\mathrm{o})$ are collinear, we can choose a coordinate
which is not zero for both vectors, for instance, $\alpha_{3}(s_{i})-\alpha_{3}(s\mathrm{o})\neq 0,$ $i=1,2$
.
Then taking coordinates as in the proof of Lemma 2.6, $t_{p}$ gives the map
$t_{p}^{\sim}( \mathit{8})=(\frac{\alpha_{1}(s)-\alpha_{1}(s\mathrm{o})}{\alpha_{3}(s)-\alpha_{3}(s\mathrm{o})},$$\frac{\alpha_{2}(s)-\alpha_{2}(s\mathrm{o})}{\alpha_{3}(s)-\alpha_{3}(s\mathrm{o})})$ ,
for $s\neq s_{0}$ in a neighbourhood of $s_{1},$$s_{2}$ and hence
$t_{p}’( \mathit{8}_{i})=\sim\frac{1}{(\alpha_{3}(\mathit{8}_{i})-\alpha_{3}(\mathit{8}0))^{2}}(\alpha_{1}’(Si)(\alpha_{3}(S_{i})-\alpha 3(S\mathrm{o}))-(\alpha 1(_{S_{i}})-\alpha_{1}(S_{0}))(\alpha_{3}/(si))$, $\alpha_{2}’(S_{i})(\alpha_{3}(_{S_{i}})-\alpha_{3}(s\mathrm{o}))-(\alpha 2(s_{i})-\alpha_{2}(_{S}0))(\alpha 3/(S_{i})))$.
Now, ifwe set
$\alpha(S_{1})-\alpha(_{S}0)=(a_{1}1, a12, a_{13})_{)}$
$\alpha(s_{2})-\alpha(_{S}0)=\lambda(\alpha(_{S_{1}})-\alpha(s\mathrm{o}).)=\lambda(a11, a12, a13)$,
$\alpha’(_{S_{1})=()}a_{21,22,23}aa$,
$\alpha’(s_{2})=$ ($a31,$ $a_{3}2,$a33),
then an easy (but tedious) computation gives
This completes the proof.
1
Note that Lemma 2.2 is a direct consequence of Proposition 2.8, since
$n(\alpha(s_{0}), \alpha)$ is equalto the number of double points of theimmersion with normal
crossings $t_{p}$
:
$S^{1}arrow \mathrm{R}P^{2}$.Let $\alpha$ : $S^{1}arrow \mathrm{R}^{3}$ be a smooth space curve which satisfies our general
position conditions as above. We define $\chi_{\alpha}$ :
$S^{1}\cross \mathrm{R}arrow \mathrm{R}^{3}$ by $\chi_{\alpha}(s, t)=\alpha(s)+$ $t\alpha’(S)$. We denoteby$\chi(\alpha)$ thetangent developable of$\alpha$; i.e., $\chi(\alpha)=x_{\alpha}(S1\cross \mathrm{R})$
.
Furthermore, we define the smooth map $\Phi$ : $S^{1}\mathrm{x}S^{1}\cross \mathrm{R}arrow \mathrm{R}^{3}$ by
$\Phi(_{S_{1},S_{2}}, t)=\frac{\alpha(s_{1})+\alpha(s_{2})}{2}+t\int_{0}^{1}\alpha’(_{S+\tau}1(s2^{-S_{1})})d\mathcal{T}$.
Note that, if$s_{1}\neq s_{2}$, we have
$\Phi(_{S_{1},S_{2}}, t)=\frac{\alpha(s_{1})+\alpha(_{S}2)}{2}+t\frac{\alpha(S_{2})-\alpha(s1)}{s_{2}-s_{1}}$
and that, if$s_{1}=s_{2}$, we have
$\Phi(s_{1}, s_{1},t)=\alpha(s1)+t\alpha(/S_{1})$.
Thus, for a point $q\in \mathrm{R}^{3}-\chi(\alpha),$ $t_{q}$ has a double point at $s_{1}$ and $s_{2}$ if and only
if $\Phi(s_{1}, s_{2},t)=q$ for some $t$.
Set $B(\subset S^{1}\cross S^{1})$ to be the bitangency set of $\alpha$ defined in [NR]. We
know that $B$ is a closed 1-dimensional submanifold of $S^{1}\cross S^{1}$
.
Define $\Omega$ to be $\Phi(B\cross \mathrm{R})$. Furthermore we define $T(\subset \mathrm{R}^{3})$ to be the union of the trisecants of$\alpha$. We see that $T$ is of dimension 2 by our assumption (v). Note that
$\Omega$ is also
2-dimensional and that the complements $\mathrm{R}^{3}-\Omega$ and $\mathrm{R}^{3}-T$ are open and dense
in $\mathrm{R}^{3}$.
The following lemma can be proved by the same argument as in the proof of Proposition
2.8.
LEMMA
2.9.
For$p\in \mathrm{R}^{3}-(x(\alpha)\cup\Omega\cup\tau)$ the map$t_{p}$ : $S^{1}arrow \mathrm{R}P^{2}$ isan$im$mersionwith norm$\mathrm{a}l$ crossings.
LEMMA 2.10. Suppose th at$p,p’\in \mathrm{R}^{3}-(\chi(\alpha)\cup\Omega\cup T)$ and that they are in the
$s$am$e$ connected component of$\mathrm{R}^{3}-\chi(\alpha)$
.
Then the number of double points of$t_{p}$ has the same parity as that of$t_{p’}$
.
The above lemma is obvious, since $t_{p}$ and $t_{p’}$ : $S^{1}arrow \mathrm{R}P^{2}$ are regularly
homotopic and the parity of the number of double points is an invariant ofthe regular homotopy class.
Remark. There are exactly four regular homotopy classes of immersions $f$ : $S^{1}arrow \mathrm{R}P^{2}$, which are characterized by:
1) the $\mathrm{Z}_{2}$ class of$f$ in $\pi_{1}(\mathrm{R}P^{2})$;
2) the parity of the number of double points of$\tilde{f}$, where $\tilde{f}$ is an immersion with
normal crossings that approximates $f$. Note that, for $p\in\alpha(S^{1}),$ $t_{p}$ represents
the nontrivial class of$\pi_{1}(\mathrm{R}P^{2})$, while for$p\not\in\alpha(S^{1})$, thetrivialclass of$\pi_{1}(\mathrm{R}P^{2})$.
DEFINITION 2.11. Let $C$ be aconnected component of$\mathrm{R}^{3}-\chi(\alpha)$. Take a point
$p\in C-(\Omega\cup T)$, which is non-vacuous. We say that $C$ is a blue region (resp.
red region) if the number of double points of$t_{p}$ is odd (resp. even). Note that
this does not depend on the choice of the point $p$ by Lemma
2.10.
Furthermore,define $B$ (resp. $R$) to be the union of all blue (resp. red) regions of$\mathrm{R}^{3}-x(\alpha)$
.
LEMMA 2.12. $\overline{B}\cap\overline{R}=\partial B=\partial R=x(\alpha)$.
For the proof ofLemma 2.12, we need
th.e
following lemmas. LEMMA 2.13. $\chi(\alpha)-\Omega$ is dense in $\chi(\alpha)$.PROOF: Take a point $p\in\chi(\alpha)\cap\Omega$. We have only to show that there exists a
point $q\in\chi(\alpha)-\Omega$ arbitrarily close to $p$. We may assume that $p$ is a simple
regular point of $\chi(\alpha)$. Suppose$p\in\chi(\alpha)\cap\Omega$. Note that $p=\Phi(s, \mathit{8};t)$ for some $s\in S^{1}$ and $t\in \mathrm{R}^{3}$; in $\mathrm{o}\mathrm{t}\mathrm{h}_{J}\mathrm{e}\mathrm{r}$ words, $p=\chi_{\alpha}(s, t)$. First suppose that $(s, s)\in B$.
Then $s$ is a torsion-zero point of$\alpha$. Since $\alpha$
ha..s
only finitely many torsion-zeropoints, such a point $p$ should lie in a 1-dimensional subspace of$\chi(\alpha)$. Thus we
may assume that $p=\Phi(s_{1}, s_{2}; t’)$ for some $(s_{1}, s_{2})\in B(\mathit{8}1\neq s_{2})$ and some
$t’\in$ R. It is easily checked that the tangent space of $\Omega$ at
$p$ is spanned by
$\{\alpha(s_{2})-\alpha(S1), \alpha’(si)\}$, where $i=1$ or 2. On the other hand, the tangent space
of $\chi(\alpha)$ at $p$ is spanned by $\{\alpha(/s), \alpha’’(s)\}$. Thus, if $\Phi|(B\cross \mathrm{R})$ and $\chi_{\alpha}$ are not
transverse at $p$, then there exists a plane $P$ tangent to $\alpha$ at $s_{1},$ $s_{2}$ and $s$ which
osculates at $\alpha(s)$. Suppose that $s=s_{i}$. Then $P$ isabitangent osculating plane of $\alpha$. Since there are only finitely many pairs $(t_{1}, t_{2})\in B(t_{1}\neq t_{2})$ corresponding
to bitangent osculating planes $([\mathrm{N}\mathrm{R}])$, the point $p=\Phi(S_{1}, s_{2}; t’)$ is in a
1-dimensional set. Hence, we may assume that $s,$ $s_{1}$ and $s_{2}$ are all distinct. This
contradicts to the condition (iii). Thus $\Phi|(B\cross \mathrm{R})$ and $\chi_{\alpha}$ are transverse at $p$.
Hence the intersection of$\Omega$ and $\chi(\alpha)$ at $p$ is of1-dimension. This completes the
proof.
1
LEMMA 2.14. $\chi(\alpha)-T$ is dense in $\chi(\alpha)$.
PROOF: Take a point $p$ in $\chi(\alpha)\cap T$. We will find a point $q\in\chi(\alpha)-^{\tau}$which is
arbitrarily close to $p$. We may assume that $p$ is a simple regular point of$\chi(\alpha)$
.
Define
$\ominus:\mathcal{T}\mathrm{x}\mathrm{R}arrow \mathrm{R}^{3}$
by
Note that $\ominus$ is a smooth map of a 2-dimensional manifold (see our condition
$(\mathrm{v}))$ and $\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\ominus(\mathcal{T})=T$.
Let $V$ be the union of the tangent lines of $\alpha$ at the points where the
osculating plane contains a trisecant. By our condition (iv), it is a finite union
oflines. Thus we may assume that $p\in\chi(\alpha)-V$.
Now suppose that, for apoint $(s_{1}, S_{2}, S_{3};t)\in \mathcal{T}\cross \mathrm{R},$$p=\ominus(s_{1}, s_{2}, S_{3}; t)$ and
$\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}\ominus \mathrm{i}\mathrm{s}$ not transverse to $\chi(\alpha)$ at $(s_{1,23}s, s; t)$. We assume that $p=\chi_{\alpha}(S’,t’)$
$(s’\in S^{1},t’\neq 0)$. Then we see that the osculating plane of $\alpha$ at $\alpha(s’)$ contains
a trisecant. This contradicts to the fact that $\dot{p}\not\in V$. Hence, $\ominus$ is transverse to $\chi_{\alpha}$ at $p$. Thus the intersection of$\ominus(\mathcal{T})$ and $\chi(\alpha)$ is of 1-dimension at $p$. This
completes the proof.
1
PROOF OF LEMMA 2.12: Take a simple regular point $p\in\chi(\alpha)$ of the tangent
developable. We have only to show that $p\in\overline{B}\cap\overline{R}$
.
By Lemmas 2.13 and2.14, we may assume that $p\in\chi(\alpha)-(\Omega\cup T)$. Suppose $p=\alpha(s\mathrm{o})+r_{0}\alpha’$(so)
$(s_{0}\in S^{1}, r_{0}\in \mathrm{R}-\{0\})$. Note that $s_{0}$ is not a torsion-zero point, since $p\not\in\Omega$.
By the proof of Proposition 2.8, the map $t_{p}$ : $S^{1}arrow \mathrm{R}P^{2}$ is an immersion with
normalcrossingsoff$s_{0}$. If$t_{p}$ has adouble point at $s_{0}$, there exists a cross tangent
passing through $\alpha(s\mathrm{o})$. Since there are only finitely many cross tangents (see
[NR]$)$, we may assume that $t_{p}$(so) is not a double point of $t_{p}$, changing $p$ if
necessary.
We may assume that $\alpha(s\mathrm{o})=(0,0, \mathrm{o}),$ $\alpha’(s_{0})=(1,0,0)$, that the $(x_{1}, x_{2})$
-plane osculates at $\alpha(S_{0})$, that ($\alpha’(s\mathrm{o})\cross\alpha^{\prime/}$(so))$\cdot(0,0,1)>0$ and that the torsion
of $\alpha$ at
$s_{0}$ is positive. Then we have $\alpha_{2}^{\prime/}(s_{0})>0$ and $\alpha_{3}^{\prime//}(S_{0})>0$. Hence, for
some small positive number $\theta,$ $\alpha(s)\in\{x_{1}<0, x_{3}<0\}$ for $s_{0}-\theta<s<s_{0}$ and
$\alpha(s)\in\{x_{1}>0, x_{3}>0\}$ for $s_{0}<s<s_{0}+\theta$, where we identify a neighborhood of$s_{0}$ in $S^{1}$ with an interval in R.
For simplicity, we assume $r_{0}>0$. Take a point $q=p+(\mathrm{O}, 0, q3)=(r_{0},0, q_{3})$
close to$p$ with $q_{3}\neq 0$. Notethat $t_{q}$ is an immersion with normalcrossings, since
$q\not\in\chi(\alpha)\cup\Omega\cup T$. Furthermore, the number of double points of $t_{q}$ is equal to
that of $t_{p}$ off a small neighborhood of $s_{0}$. Recall that $t_{p}$ has no double points
in a neighborhood of $s_{0}$. Now we consider the number of double points of $t_{q}$
in the neighborhood of $s_{0}$. We will prove that if $q_{3}<0$, then $t_{q}$ does not have
any double points in the neighborhood of$s_{0}$, while if $q_{3}>0,$ $t_{q}$ has exactly one
double point in the neighborhood of$s_{0}$.
For asmall open disk neighborhood $D$ of$p$ with $D\cap(\Omega\cup T)=\emptyset,$ $D-\chi(\alpha)$
has exactly two connected components $D_{+}$ and $D_{-}$, where $D_{+}$ is the region
which contains a point $q$ with $(q-p)\cdot\alpha^{\prime//}(S0)>0$ (i.e., $q_{3}>0$).
LEMMA
2.15.
If $D$ is sufficiently $sm$all, $t_{q}$ has exac$tly$ one double point in aneighborhood of$s_{0}$ if$q\in D_{+;}$ and it has no $do\mathrm{u}ble$point in the neighborhood
PROOF: We identify a neighborhood of $s_{0}\in S^{1}$ with $(-\epsilon, \epsilon),$
so
being identifiedwith $0$, and we set $W=\{(s_{1}, s_{2}, t)\in(-\epsilon, \mathcal{E})\cross(-\epsilon, \epsilon)\mathrm{x}(r_{0}-\delta, r\mathit{0}+\delta):s_{1}>s_{2}\}$
$(\delta>0)$. Recall that the image of $W’=\{(s_{1}, s_{2},t) : s_{1}=s_{2}\}$ by $\Phi$ is precisely
the tangent developable of$\alpha$. We have only to show that $\Phi$ maps $W$ injectively
onto $D_{+}$ for some $\epsilon$ and
$\delta$
.
For this, we calculate the differential of $\Phi$ at $u=$$(0,0, r_{0})$. As a basis of$T_{u}(S^{1}\cross S^{1}\cross \mathrm{R})$, we take $\{(\partial/\partial s_{1})+(\partial/\partial s_{2}),$ $(\partial/\partial s_{1})-$ $(\partial/\partial s_{2}),$ $(\partial/\partial t)\}$. Note that $\{(\partial/\partial s_{1})+(\partial/\partial s_{2}), (\partial/\partial t)\}$ constitutes a basis of
$T_{u}W’$ and that $(\partial/\partial_{\mathit{8}1})-(\partial/\partial s_{2})$ is the direction normal to $W’$ in $s^{1}\cross S^{1}\cross \mathrm{R}$
toward $W$. Then we have
$d \Phi_{u}((\partial/\partial S1)+(\partial/\partial S_{2}))=\frac{\partial}{\partial s}\Phi(_{S\mathit{8}t},,)|(s,s,t)=u$
$= \frac{\partial}{\partial s}(\alpha(_{\vee}s)+t\alpha(/s))|(s,s,\iota)=u$
$=\alpha’(s\mathrm{o})+r0\alpha(_{S}’/0)$, $d \Phi_{u}(\partial/\partial t)=\frac{\partial}{\partial t}\Phi(_{S_{1,2}}s, t)|_{(_{S_{1,2}})=}s,tu$
$=( \int_{0}^{1}\alpha’(s_{1}+\tau(_{S_{2^{-s}}}1))d\mathcal{T})|_{(s_{1}},s_{2},t)=u=\alpha’(\mathit{8}0)$,
and
$d\Phi_{u}((\partial/\partial s_{1})-(\partial/\partial s_{2}))$
$= \frac{\partial}{\partial s}\Phi(s, -s, t)|_{(s,s,t})=u$
$= \frac{\partial}{\partial s}(\frac{\alpha(s)+\alpha(-s)}{2}+t\int_{0}^{1}\alpha’(s-2s\tau)d_{\mathcal{T}}\mathrm{I}|_{(_{S,-}S},t)=u$
$=( \frac{\alpha’(s)-\alpha’(-S)}{2}+t\int^{1}0)(1-2\tau\alpha’(/-s2s\tau)dT\mathrm{I}|_{(-s,i)=}s,u--\mathrm{o}$
.
Thus the rank of $d\Phi_{u}$ is equal to 2. Note also that $d\Phi_{u}(T_{u}(S^{1}\cross S^{1}\cross \mathrm{R}))=$
$d(\Phi|W’)_{u}(\tau uW’)$ and that $\Phi|W’$ is a local diffeomorphism around $u$ onto an
open neighborhood of$p$ in $\chi(\alpha)$
.
Furthermore, we have$\frac{\partial^{2}}{\partial s^{2}}\Phi(s, -s, t)|(_{S,s},\iota)=u$
$= \frac{\partial}{\partial s}(\frac{\alpha’(s)-\alpha’(-S)}{2}+t\int_{0}^{1}(1-2\mathcal{T})\alpha^{\prime/}(S-2s\tau)d\mathcal{T})|_{(s,t)=u}s,-$
$=( \frac{\alpha^{\prime/}(S)+\alpha/\prime(-s)}{2}+t\int_{0}^{1}(1-2\mathcal{T})^{2\prime}\alpha(’/s-2S\mathcal{T})dT)|_{(s,-s,t)=u}$
Thus, ifwe take the derivative of second order along $(\partial/\partial_{\mathit{8}1})-(\partial/\partial s_{2})$, it does
not lieon the imageof$d\Phi_{u}$
.
Hence we can change the system of coordinates of$\mathrm{R}^{3}$around$p$in a homeomorphic way so that $d(\Phi|W\cup W’)_{u}$ has rank 3 with respect
to this new $C^{0}$ coordinates. Hence, $\Phi$ maps $W$ injectively onto $D_{+}$ for some
small W. (In other words, $\Phi$ has a fold singularity (or $\Sigma_{1,0}$-singularity) along
$W’$ in a neighborhood of $u$, the discriminant set being the tangent developable.
See [Mo].) This completes the proof ofLemma
2.15.
1
Thus we see that $p$ is in the closure of both $B$ and $R$. This completes the
proof of Lemma 2.12.
1
Remark. As a digression, it would be interesting to study the behavior of the
map $\Phi$. For a generic curve, is $\Phi$ stable? What are the singularities and the
critical values? Is it related to the theory of self-translation surfaces [MNR]? By the above lemma, the decomposition $\mathrm{R}^{3}-\chi(\alpha)=B\cup R$ coincides with the decomposition guaranteed by the 2-color theorem ($[\mathrm{N}\mathrm{S}$, Lemma 2.1])
applied to the topologically stable map $\chi_{\alpha}$ : $S^{1}\cross \mathrm{R}arrow \mathrm{R}^{3}$. (More precisely, we
have to compactify the map as isdone in [N] in order to apply the result of [NS]
and then we restrict to $S^{1}\cross \mathrm{R}$ and $\mathrm{R}^{3}.$) Recallthat, in [NS], we havemade the
convention that the index $n(p, x(\alpha))$ of a crosscap point $p$ of$\chi(\alpha)$ with respect
to the 2-color theorem is defined to be 1 if the outside region of $\mathrm{R}^{3}-x(\alpha)$ in
a neighborhood of$p$ is red and $0$ if it is blue. Note also that the torsion-zero
points of$\alpha$ coincides exactly to the cross cap points of$\chi(\alpha)([\mathrm{N}])$.
PROPOSITION
2.16.
The two defiinitions of th$\mathrm{e}$ index of a torsion-zero point$p$
–the index with resp$\mathrm{e}ct$ to the number oftrisecants passingthro$\mathrm{u}ghp,$ $n(p, \alpha)$,
and that with respect to the 2-color theorem, $n(p, x(\alpha))$ –coincide with each other modulo 2.
PROOF: Let $p=\alpha(s\mathrm{o})(s_{0}\in S^{1})$ be a torsion-zero point. Take a point $q\in$
$\mathrm{R}^{3}-\chi(\alpha)$ close to
$p$. We may assume that $q\not\in\chi(\alpha)\cup\Omega\cup T$. Then the map $t_{q}$
and $t_{p}$ are immersions with normal crossings. Although $t_{q}$ and $t_{p}$ are not even
homotopic, yet they have the same number of double points offa neighborhood
$J$ of $s_{0}$. This is because the maps $t_{p}|(S^{1}-J)$ and $t_{q}|(S^{1}-J)$ are sufficiently
“close” to each other asimmersions into $\mathrm{R}P^{2}$ (note that
$t_{p}$ hasno double points
in $J$). By Lemma 2.10, the parity of the number of double points of $t_{q}$ in
the neighborhood of $s_{0}$ depends only on the region to which $q$ belongs; more
precisely it is odd when $q$ is in the
((
$\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{s}\mathrm{i}\mathrm{d}\mathrm{e}$region” of the cross-cap point $p$ of $\chi(\alpha)$, and it is even when $q$ is in the “inside region”. This is because we can
take an appropriate point $q$ in the “outside region” such that $t_{q}$ has exactly one
double point in the neighborhood $J$ of $s_{0}$. In fact, given a pair of bitangent
points $\alpha(s_{1}),$$\alpha(s_{2})$ in $J$, thesegmentjoining $\alpha(s_{1}),$$\alpha(s_{2})$ is included in the local
convex hull of the curve, which lies in the “outside region” of the cross-cap point. Thus, any point $q$ in this segment, $t_{q}$ has exactly one double point in $J$
with respect to trisecants is odd. Then the outside region is the red region and the inner region is the blue region. Hence the index $n(p, x(\alpha))$ of$p$ with respect
to the 2-color theorem is equal to 1 and coincides with the index with respect to trisecants. The other case is similar. Hencethe index with respect to the number of trisecants coincides with the index with respect to the 2-color theorem modulo 2. This completes the proof.
1
Now our Theorem 2.4 follows from [NS] (or [Sz]) and Proposition 2.16, by using the compactification $\tilde{\chi}_{\alpha}$ : $S^{1}\cross \mathrm{R}^{*}arrow \mathrm{R}P^{3}$ of$\chi_{\alpha}$ as in [N]. Note that $(\tilde{\chi}_{\alpha})_{*}[S^{1}\cross \mathrm{R}^{*}]\in H_{2}(\mathrm{R}P^{3};\mathrm{Z}2)$vanishes, since $S^{1}\cross \mathrm{R}^{*}\cong S^{1}\cross S^{1}$ is orientable,
where $[S^{1}\cross \mathrm{R}^{*}]\in H_{2}(S^{1}\cross \mathrm{R}^{*} ; \mathrm{z}_{2})$ is the fundamental class.
Remark, Theorem 2.4 implies that the number of pyramids of a
generic
spacecurve is congruent modulo 2 to the number of trisecants passing through the torsion zero points, where such trisecants are counted with multiplicities which are defined to be the numberoftorsion zero points they pass through. We note that we could add the condition that a trisecant passes through at most one torsion zero point in the general position conditions from the beginning. This condition is generic (i.e., even after adding this condition, we have Lemma 2.1),
and then we have that the number of pyramids is congruent modulo 2 to the number (counted without multiplicities) oftrisecants passingthrough thetorsion zero points.
COROLLARY 2.17.
,$Let\alpha$ : $S^{1}arrow \mathrm{R}^{3}$ be a smooth curve satisfying th$\mathrm{e}$general
position conditions, such that the number of pyram$idsT(\alpha)$ is odd. Then $\alpha$ has
at least two torsion zero points.
A special case of Theorem 2.4 is when the curve $\alpha$ is convex. By using a
result by Sedykh [Se], we can look at the structure of the convex envelope of a generic curve and deduce that a convex generic curve has no trisecants. Then we get the following immediate consequence.
COROLLARY
2.18.
Let $\alpha$:
$S^{1}arrow \mathrm{R}^{3}$ be aconvex
$c$urve satisfying our generalposition conditions and thegeneral position conditions given in [Se]. Then the number ofpyramids of$\alpha$ is even.
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Departament de Geometria $\mathrm{i}$Topologia, Universitat de Valencia, Campus de Burjasot, 46100 $\mathrm{B}\mathrm{u}\dot{\mathrm{Q}}\mathrm{a}\mathrm{s}\mathrm{o}\mathrm{t}$ (Valencia), SPAIN
$E$-mail address: nuno@mac.uv.es
Departmentof Mathematics, Facultyof Science, Hiroshima University, Higashi-Hiroshima 739, JAPAN
