数理解析研究所講究録
第 1540 巻 2007 年 1-12

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SURFACE KNOTS AND THEIR GENERIC PLANARPROJECTIONSYASUSHI TAKEDA (JOINT WORK WITH OSAMU SAEKI)

1. INTRODUCTION
In classical knot theory, we often use regar pltar projections of iots and links
in 3-space. Regular projections do not have triple points and the double points are
given 3-dimensional informations, that is, upper-lower informations. Therefore, we
fact,
can recover the original classical knot $hom$ the projection up to isotopy.
many knot invariants can be constructed through regular planar projections: forexample, crossing number, skein polynomials, Khovanov homology etc.
By asurface knot, we mean aclosed connected smoothly embedded surface in. In surface knot theory, we often use generic projections into 3-space. By
putting 4-dimensional informations to the singularities of ageneric projection into
recover
3-space, we get aso-called broken surface diagram [4]. Note that wethe original surface knot $hom$ its broken surface diagram. Ageneric projection into
branch points as its
3-space of asurface knot has double points, triple points,
singularities. On abroken surface diagram the projection image is cut along its
double curves.
this report, we use generic projections into the plte for studying surface
knots. Such planax projections were used, for example, in [2], [6], [7], [10]. Planax
projections of surface knots have fold points and cusps as their singularities. Cusps
appear as discrete points and fold points appear as 1-dimeoional submanifolds.
The set consisting of the cusps and the fold points in the surface is cafed thesingular set.
Given aplanar projection of asurface knot with no addItional information, we
cannotthe original surface knot. In this report, we introduce the notionof abraids with aband and show that aplanar projection together with abtded
braid attached to the image of the singular set can recover the original surface knot.
Abanded braid is abraid in the usual sense in which one of. the strings is replacedby a(possibly twisted) btd. The band part corresponds to aneighborhood of thefold curve in the surface.
Finally we prove the Whitney congruence as an application. Whitney congruenceis the congruence concerning normal Euler number which is isotopy invanrirt ofsurface knot. First, the congruence was showed by $Wl\dot{v}tney$ and he used the tool ofdifferential topology for showing the proof. However, the method was not geometric.
The latter, Caxter-Saito showed the congruence by using geometric method which
was generic projection into 3-space of surface knot. On the other hand, we $wiU$ show
the congruence by using generic pltar projection which is geometric method.
The report is organized as follows. In Section 2we introduce how to make thewe will show the
braided diagram which is analogy of broken surface diagram
diagram is determined uniquely for asurface knot. In Section 3we give two explicit$\bm{i}$$R^{4}$$C\bm{t}$$\bm{t}d$$\bm{i}$$rec^{\backslash }over$$\bm{t}d$2

FIGURE 1. A banded braid
examples. In Section 4 we explain how to calculate the normal Euler number byusing a planar projection and prove the Whitney congruence as an application.
Throughout the report, we work in the smooth category.
The author has been supported by 21st century COE program “Developmentof Dynamic Mathematics with High Functionality” at Faculty of Mathematics,
Kyushu University, Japan.

2. SURFACE KNOTS AND THEIR BRAIDED DIAGRAMSIn this section we introduce how to make braided diagram from planar projectedsurface knot and show the braided diagram is determind uniquely for a surface
knot.

2.1. Braided diagram associated with a given urface knot. Let us show
how to construct a planar projection with additional informations from a given. For this, we introduce the notion of a banded braid.
embedded surface inDefinition 2.1. We call an embedding(or) a braid with a band if satisfies following:.
.
$I_{1}=.$$=I_{r}=[0,1],$ $J_{1}=J_{2}=[0,1]$ ,(i)(ii-l): $I_{j}arrow I,j=1,2,$ $\cdots r$ , is the identity mapping, where $p:R^{2}\cross Iarrow$$s$$R^{4}$$\varphi:I_{1}\cup\cdots\cup I_{r}\cup(J_{1}\cross J_{2})arrow R^{2}\cross(I=[0;1])$$\varphi(I_{1}\cup\cdots\cup I_{r}\cup(J_{1}\cross J_{2}))$$\varphi$$p\circ\varphi|_{I_{j}}$is the projection to the second factor,(ti-2)
is the projection to the first factor,,(iii-l) for
we have $\varphi(0)\in(R\cross\{0\})\cross\{0\},$ $j=1,2,$ $\cdots r$ ,(iii-2) for $0\in J_{1}$ , we have $\varphi(0\cross J_{2})\subset(R\cross\{0\})\cross\{0\}$ ,, we have $\varphi(1)\in(R\cross\{0\})\cross\{1\},$ $j=1,2,$ $\cdots r$ ,(iii-3) for(iii-4) for $1\in J_{1}$ , we have $\varphi(1\cross J_{2})\subset(R\cross\{0\})\cross\{1\}$ .
$I$$p\circ\varphi|_{J_{1}xJ_{2}}arrow I$$0\in I_{j}$$1\in I_{j}$For example, see Fig. 1.
In the following, two banded braids are considered to be the same if they areisotopic through banded braids.

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Let us prepare several notions from singularity theory.
theDefinition 2.2. Let be a closed connected surface. Denote by$F$
topology, endowed with the Whitney
set of all smooth mappings from to
if
to
there
exist
[5]. Let and be elements of. Then is equivalentdiffeomorphisms $p:Farrow F$ and $q:R^{2}arrow R^{2}$ such that $q\circ g=h\circ p$ .
$C^{\infty}(F, R^{2})$$F$$0\infty$$R^{2}$$C^{\infty}(F, R^{2})$

$h$

$g$

Definition 2.3. Let be an element ofofstable if there exists a neighborhood
is equivalent to .
$g$

$N_{g}$$h$$g$$C^{\infty}(F, R^{2})$$g$in. Then$C^{\infty}(F, R^{2})$is said to besuch that each in
$g$

$h$

$C^{\infty}$$N_{g}$$g$. Then $q\in F$
Definition 2.4. Let $g:Farrow R^{2}$ be a smooth mapping from $F$ tois called a fold point if we can choose local coordinates $(x, y)$ centered at and$(U, V)$ centered at $g(q)$ such that , in a neighborhood of , is of the form:$R^{2}$

$q$

$q$

$g$

$U=x,$$V=y^{2}$.

Moreover, $q\in F$ is called a cusp if we can choose local coordinates $(x, y)$ centeredat and $(U, V)$ centered at $g(q)$ such that , in a neighborhood of , is of the form:
$q$

$g$

$q$

$U=x,$ $V=xy+y^{3}$

We denote bycusps.

$S_{1}(g)$.

the set of fold points and cusps, and by$S_{1}^{2}(g)$the set ofNote that $S_{1}(g)$ is a regular l-dimensional submanifold of $F$ anddiscrete set.
stable mappings in
Recall the following well-known characterization of$S_{1}^{2}(g)$$c\infty$$R^{2})$is a,$C^{\infty}(F$.

Proposition 2.5. Let $g:Farrow R^{2}$ be a smooth mapping from a closed connected
stable if and only if has only fold points andsurface $F$ to . Then iscusps as its singu larities, its restriction to the set of fold points is an immersion
utth normal $cros8ings$, and for each cusp , we have:$R^{2}$$g$$C^{\infty}$

$g$

$q$

$g^{-1}(g(q))\cap S_{1}(g)=\{q\}$.

Let $F$ be a closed connected surface. For a smooth map $g:Farrow R^{2}$ , we set.
$S(g)=$ { $x\in F|$ rank $dg_{x}<2$ },stable, then we clearly have
which is called the singular point set of . If is$g$$S_{1}(g)=S(g)$$g$$c\infty$.

Deflnition 2.6. Let $f:Farrow R^{4}$ be an embedding of a closed connected surface.
with
Let $\pi:R^{4}arrow R^{2}$ be an orthogonal projection. Then we say that is$f(F)$
stable.
) ifisrespect to (or with respect to
$\pi$

$f$

$\pi\circ f$$gener\dot{\tau}c$$C^{\infty}$

Definition 2.7. Let $f:Farrow R^{4}$ be an embedding of a closed connected surface
and $\pi:R^{4}arrow R^{2}$ an orthogonal projection which is generic with respect to . For$x\in S(\pi\circ f)$ , the l-dimensional subspace $df_{x}(Ker(d(\pi\circ f)_{x} : T_{x}Farrow T_{\pi\circ f(x)}R^{2}))$of $df_{x}(T_{x}F)$ is called the kemel line at $f(x)\in f(F)$ .
We often regard the kernel line as included in $\pi^{-1}(\pi\circ f(x))\cong R^{2}$ .
$f$

The kernel line at a fold point and that at a cusp are depicted in Figs. 2 and 3
respectively.

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FIGURE 2. The kernel line at a fold point

FIGURE 3. The kernel line at a cusp
like follows.
We construct abraided diagram $hom$ agiven embedded surface in$\pi:R^{4}arrow R^{2}$
Let $f:Farrow R^{4}$ be an embedding of aclosed connected surface andstable map
T.hean orthogonal projection which is generic with respect to$\pi of$ has fold points and cusps as its singularities. Let $q\in R^{2}$ be afold crossing
to fold points set. Then we canwhich is normal crossing of restriction ofcontains exactly two fold points.
that
aplane.
NoteasregardMoreover, by isotoping if necessary, we may aesume that the points $\pi^{-1}(q)\cap f(F)$the kernel lines at the fold points all lie in $R\cross\{0\}\subset R^{2}=.\pi^{-1}(q)$ as depicted
in Fig. 4for all fold crossings . Furthermore, for the image $q\in R^{2}$ of each cusp,
we arrange $f(F)$ by isotopy so that the points $\pi^{-1}(q)\cap f(F)_{\bm{t}}d$ the kernel line at
the cusp point all lie in $R\cross\{0\}\subset R^{2}=\pi^{-1}(q)$ . See Fig. 5.
. If is an open disk, then weLet be abounded region of
do $noth\dot{i}g$ for R. If is not an open disk, then we take disjointly embedded arcs
in $\tilde{R}=R\cup(\pi\circ f(S(\pi\circ f)))$ such that $(a_{1}\cup a_{2}\cup\cdots\cup a_{k})\cap(\pi of(S(\pi 0$$f)))=\partial(a_{1}\cup a_{2}\cup\cdots\cup a_{k})$ and each component of
open
isso that their end
disk (see Fig. 6). Furthermore, we take the arcs
points are not fold crossing nor.the image of cusps. We call the arcs
we take additional
additional arcs. For eai non-disk region of$R^{4}$$C^{\infty}$$f$$\pi\circ f$$\pi^{-1}(q)\cong R^{2}$$\pi^{-1}(q)$$f$$\cdot$$\bm{t}d$$q$$R^{2}\backslash \pi\circ f(S(\pi\circ f))$

$R$

$R$

$R$

$a_{1},a_{2},$ $\cdots a_{k}$$R\backslash (a_{1}\cup\cdots\cup a_{k})$$a_{1},$ $a_{2},$$\cdots$$\bm{t}$$a_{k}$$a_{1},$ $a_{2},$$R^{2}\backslash \pi\circ f(S(\pi of))$$\cdots a_{k}$5

FIGURE 4. Arranging $\pi^{-1}(q)\cap f(F)$FIGURE 5. Arranging $\pi^{-1}(q)\cap f(F)$arcs so that each bounded region of$R^{2}\backslash ((\pi\circ f(S(\pi\circ f)))\cup A$

is an open disk,

where is the union of all the additional arcs.
By above similar argument, we can fix terminals of additional arcs like follow.
Let be one of intersection between $\pi\circ f(S(\pi\circ f))$ and a additional arc. Then we
$A$

$y$

6

FIGURE 6. Additional arcs in a non-disk region

FIGURE 7. Arranging $\pi^{-1}(y)\cap f(F)$arrange $f(F)$ by isotopy so that the points $\pi^{-1}(y)\cap f(F)$ and the kernel line at thefold point all lie in $R\cross\{0\}\subset R^{2}=\pi^{-1}(y)$ as depicted in Fig. 7.
Therefore we can regard $(\pi\circ f(S(\pi\circ f)))\cup A$ as a oriented graph which is oriented
by method that the the number of pre-image of left of fold curve is larger than oneof right. The vertices are fold crossings, the images of cusps, and the end pointsof additional arcs. Then, to every edge of $\pi\circ f(S(\pi\circ f))$ we associate a banded
braid as follows. Let be edge which is a little smaller than along the . The
pre-image of consist of a braid of odd strings. The odd strings contai strings for
fold curve. Then we make the width the string by using the kernel line segment.
Therefore, we associate the banded braid to . Furthermore, to each additional arc
$e$

’

$e$

$e$

$e$

$e$

$e$

7

FIGURE 8. A banded braid as sociated with an edge

we associate a braid in the usual sense as follows. The pre-image of consist ofa braid of odd strings. Therefore, we associate the braid in the usual sense to .
For example, let be an edge ofand . Thenwith end points
is as depicted in Fig. 8. In particular, if do not have end points, we see it
is circle and we can not associate banded braid to . Then we only attach a label
which is the number of pre-image of inter of the circle, for, the inter of the circle isopen disk and the banded braid as sociated to is trivial closed braid. Therefore,
we can define braided diagram as following.
$a$

$a$

$a$

$\pi\circ f(S(\pi\circ f))$$e$$\pi^{-1}(e)$$x_{0}$$x_{1}$

$e$

$e$

$e$

Deflnition 2.8. Let $(\pi of(S(\pi\circ f)))\cup A$ be riented graph defined as above andwe associate banded braid or braid in the usual sense to edge of the graph. Then wecall the oriented graph with the informations (banded braid or braid in the usual
sense etc) braided diagram of surface knot.
$0$2.2. Recovering embedded surface from braided diagram. We will show.
recover a local embedding of surface in $R^{4}hom$ braided diagram is unique.
Definition 2.9. Let$F_{0,n}R^{2}$denote the subspace$F_{0,n}R^{2}=\{(p_{1}, \ldots,p_{n})\in R^{2}\cross\ldots xR^{2}|p_{i}\neq p_{j}(i\neq j)\}$

Let $Q_{m}=\{q_{1}, \ldots, q_{m}\}$ be a set of fixed distinguished points ofthe configuration spaceof
to be the space$F_{m,n}$$R^{2}$$R^{2}$.

. Then we define.

$F_{0,n}(R^{2}\backslash Q_{m})$Given braided diagram of surface knot, fold crossings, cusps, edges can be liftin
which is unique. However, disk regions constructed the image of singular set
may not lift which is unique. To show it, we consider whether it can extend loop
in configuration space which it construct points in
to disk. In fact, following
sayproposition [1]it can extend to disk which is unique.
$R^{4}$$n$$R^{2}$

8

FIGURE 9. The image of singular set with three cusps
Proposition 2.10. If $\pi_{2}(R^{2}\backslash Q_{m})=\pi_{3}(R^{2}\backslash Q_{m})=0$ for each $m\geq 0$ , then$\pi_{2}F_{0,n}R^{2}=0$.

3. EXAMPLES OF RECOVERINGIn this section we introduce an example of recovering of surface knot via planar
projection. We give a local orientation to edge by method which local width (see[7]) of left of the edge is larger than the one of right. First, we consider embedding
with the image of singular set as in Fig. 9. The existence of thisinembedding have been showed as in [10]. $m$ is an additional arc which is not imageof singular set. Then we can reconstruct original surface knot as in Fig. 10, 11. In$RP^{2}$$R^{4}$

etc correspond direction from top to bottom
these figure, the direction of foldin this report. Moreover the normal Euler number of this embedding is 2 or $-2$
since the total of twist of band is 2 half twist (see next section).
$l_{1},$$l_{2}$

4. NORMAL EULER NUMBER AND WHITNEY CONGRUENCE
In this section we introduce how to calculate the normal Euler number and as
an application we prove the Whitney congruence.

be surface knot
4.1. The calculation of normal Euler number. Letwith planar diagram. We consider edges with banded braids. Then the normal
Euler number is calculated like follows. Perturb $F$ (we will call $F’$ ) by isotopy as
depicted in Fig. 12. Then strings (braids) of the banded braid in $F\cap F’$ does notparticular, additional
intersect each other since strings in $F\cap F’$ are parallel.
arcs does not also intersect each other. However, band part of banded braid onlyintersect each other. Furthermore, the number of half-twist of the band correspondthe number of the intersection. Therefore, the normal Euler number is the number$F\subset R^{4}$$arrow In$

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$a$

$b_{1}$$a_{1}$$c_{1}$$l_{1}$$l_{2}$$1$$1$$a_{1}$$b_{1}$$d_{1}$$c_{1}$$a_{2}$$b_{2}$$d_{2}$$c_{2}$$l_{4}$$a_{2}|$$l_{3}$$a_{3}$$b_{3}$$a_{3}|$a$d_{3}$$c_{3}$$b|^{3}$$b$

FIGURE 10. The reconstructing of region$c_{3}|$$c$$\lceil$$d$$l_{1}l_{2}l_{3}l_{4}$of the half-twist of the band. Let be edge (without additional arc)
the terminal’s vertexes. See Fig. 12.
We give aorientation to disk regions on graph made the image of singular set
how it is induced from the orientation of. Moreover, we give aorientation
boundaries of disk regions how it is induced&om the orientation of the disk regions(it is different Rom aorientation $concer\iota ling$ the local width).
For every region which is constructed by the image of singular set, the boundaryof the region is associated by trivial braid. Every two adjacent strings of the trivial
braid ist times. Therefore the sum of twist of every two adjacent strings for
total regions is also . In this sum we consider the kind of half-twist of two adjacent$\bm{t}dx_{0},$$e$$R^{2}$$t^{(}w$$x_{1}$$t\dot{o}$

$0$

$0$

strings.

In general, fold curve is associated by banded braid. The strings of banded
braid is divided by braid part and band part. Let
be aleft region of fold curveandaright region of fold curve. Then the half-twist of braid part is canceledeach other since the orientation ofandare reverse direction each other.
Therefore non-cancel part are the half-twist of between boundaries of band partand the strings in braid part. The former contribute normal Euler number. The
latter correspond twice of the sum of twists of between core of band and the stringsin braid part.
On the other hand in neighborhood of additional arc we can cancel the strings
because there is no band. In neighborhood of cusps twist of banded braid is times.
$R_{1}$$R_{2}$$\partial R_{1}$$\partial R_{2}$

$0$

10
a

$b$

a

$b$

FIGURE 11. The constructing of region$ml_{4}l_{1}l_{2}l_{3}m^{-1}l_{5}^{-1}$

In neighborhood of fold crossing twist of banded braid also is times. Therefore
we can see the following Theorem.
Theorem 4.1. We consider generic planar projection of a surface knot. Then foldcurve is associated by banded braid. Let be normal Euler number of surface knots
and $D$ be the sum of half-twist of between the core of band in banded braid and theother strings in banded braid for all regions. Then
$0$

$e$

$e=-2D$ .

By above theorem we can see that normal Euler number is even.

4.2. Proof of Whitney congruence. Whitney [9] showed relation of between
normal Euler number and Euler characteristic (we call it Whitney congruence).
However Carter-Saito [3] gived the diagrammatic proof of Whitney congruence.
They showed it by using generic projections into 3-space.
On the other hand we will show it by using generic planar projections. Whitneycongruence is following theorem.

11

FIGURE 12. Perturbation of $F$

and $e(F)$ be no rmal Euler
Theorem 4.2. Let $F$ be embedding surface into
number of $F$ and $\chi(F)$ be Euler chamcteris tic of F. Then
$R^{4}$$(1/2)e(F)\equiv\chi(F)$$(mod 2)$.

The following theorem is also well-known [8].

Theorem 4.3. Let $F$ be a closed 2-dimensional manifold, $N$ an $07\dot{v}entable$ surface,. Then$f:Farrow N$ a stable map and $C(f)$ the set of the cusp singularities of$f$$|C(f)|\equiv\chi(M)$$(mod 2)$

Therefore by Theorem 4.1, 4.3 we may show the.

$fo\mathfrak{U}ow\dot{m}g$theorem.

and $\pi:R^{4}arrow R^{2}$ be
Theorem 4.4. Let Let be embedding surface into
orthogonal projection and $D$ be $D$ in Theorem 4.1 and $|C(\pi of)|$ be the number ofcusps. Then
$R^{4}$$f$$D\equiv|C(\pi\circ f)|$$(mod 2)$.

Any way, we have a problem, which is to prove the Whitney conjecture by using
pure geometric method. Whitney conjecture was showed by Massey but the proof
was not geometric. We want to prove the Whitney conjecture by using genericprojection. If we use the generic planar projection, we may prove the conjecture.
That is the open problem.
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$Grad$ . Texts Math.,
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$c$

FACULTY OF MATHEMATICS, KYUSHU UNIVERSITY, HAKOZAKI, FUKUOKA 812-8581, JAPANE-mail address: takedaQmath. kyushu-u. ac. jp