The Embedding Space of Hexagonal Knots
Jorge Alberto Calvo 1
Department of Mathematics, North Dakota State University, Fargo, ND 58105
Abstract
The topology of the space of rooted oriented hexagonal knots embedded in R3 is
described, with special attention given to the number of components that make
up this space and to the topological knot types which they represent. Two cases
are considered: (i) hexagons with varying edge length, and (ii) equilateral hexagons
with unit-length edges. The structure of these spaces then gives new notions of
\hexagonal knottedness." In each case, the space consists of ve components, but
contains only three topological knot types. Therefore each type of \hexagonal equivalence" is strictly stronger than topological equivalence. In particular, unlike their
topological counterparts, hexagonal trefoils are not reversible; thus there are two
distinct components containing each type of topological trefoil. The inclusion of
equilateral hexagons into the larger class of hexagons with arbitrary edge length
maps hexagonal knot types bijectively; however the kernel of this inclusion at the
level of fundamental group is shown to be non-trivial. In addition, combinatorial
invariants are developed to distinguish between the ve dierent \hexagonal knot
types," giving a complete classication of hexagonal knots.
Key words: polygonal knots, space polygons, knot spaces, knot invariants.
AMS classication: 57M25
1 Introduction
An n-sided polygon P in R is a closed, piecewise linear loop with no selfintersections consisting of n points of R ; called vertices, joined by n straight
line segments, called edges. We think of an n-gon as the result of glueing n
sticks end to end to end. As loops in R ; n-sided polygons form a special class
of knot: more rigid than smooth knots. Added rigidity makes this class better
suited for describing macromolecules (such as DNA) in polymer chemistry and
molecular biology, but also far more resistant to investigation. The general
framework for geometric knot theory was introduced by Randell [7,8]. In this
paper we will describe the situation when n = 6 for both the general case
3
3
3
1
E-mail: jorge [email protected]
Preprint submitted to Elsevier Preprint
30 August 1999
described above, called geometric knots, and for the subset thereof consisting
of polygons with unit edge length, called equilateral knots.
1.1 Geometric knots
Consider an n-sided polygon P in R ; together with a distinguished vertex,
or root, v and a choice of orientation. We can view P as a point of R n by
listing the triple of coordinates for each of its n vertices, starting with v and
proceeding in sequence as determined by the orientation.
3
3
1
1
In the spirit of Vassiliev (see [1,10]), dene the discriminant n to be the
collection of points in R n which correspond in this way to non-embedded
polygons. A polygon fails to be embedded in R when two or more of its
edges intersect, so if n > 3 then n is the union of the closure of n (n ? 3)
cubic semi-algebraic varieties, each consisting of polygons with a given pair of
intersecting edges. For example, the collection of polygons hv ; v ; : : : ; vn? ; vni
for which v v intersects v v is the closure of the locus of the system
( )
3
3
1
2
( )
1
1 2
2
1
3 4
(v ? v ) (v ? v ) (v ? v ) = 0
(v ? v ) (v ? v ) (v ? v ) (v ? v ) < 0
(v ? v ) (v ? v ) (v ? v ) (v ? v ) < 0:
2
1
3
1
4
1
2
1
3
1
2
1
4
1
4
3
1
3
4
3
2
3
In particular, the closure of each of these semi-algebraic varieties is a codimension-1 submanifold (with boundary) of R n: Hence the subspace Geo n =
R n ? n corresponding to embedded polygons is an open 3n-manifold which
we will call the embedding space of rooted oriented n-sided geometric knots.
( )
3
3
( )
A path h : [0; 1] ! Geo n corresponds to an isotopy of polygonal simple closed
curves. If two polygons lie in the same path-component of Geo n ; we will say
they are geometrically equivalent. Also, a polygon is a geometric unknot if it is
geometrically equivalent to a standard planar polygon; since all planar n-sided
polygons are geometrically equivalent (Theorem 2.1 in [4]), the component of
geometric unknots is well-dened.
( )
( )
The geometric equivalence of two knots clearly implies their topological equivalence. However, not much is known about the converse. For instance, it is unknown whether there exist topological unknots which are geometrically knotted. In fact, there are no previously known examples of any distinct geometric
knot types which correspond to the same topological knot type (compare with
Theorem 2 below, and with Theorem 4.1 in [2]).
Since triangles are planar, the embedding space Geo of rooted oriented triangles is path-connected. A quadrilateral (tetragon) consists of two triangles hinged along a common edge; since we can change the dihedral angle at
the hinge to atten the quadrilateral out, we nd that Geo is also path(3)
(4)
2
Fig. 1. A reducing isotopy from an n-gon to an (n ? 1)-gon.
connected. To show that every pentagon can be pushed into a plane, so that
Geo is path-connected, we will need the following reduction lemma.
(5)
Lemma 1. (Reduction lemma) Let P = hv1; v2; : : : ; vn?1; vni be an n-sided
polygon embedded in R3; and let be the interior of the triangular disc with
vertices at vi?1; vi; and vi+1 : If P does not intersect ; then there is a polygonal
isotopy in Geo(n) deforming P until it coincides with the (n ? 1)-sided polygon
P 0 = hv1; v2; : : : ; vi?1; vi+1; : : : ; vn?1; vni:
Proof. Let h : [0; 1] ! Geo(n) be the isotopy which moves vi in a straight line
path across to the midpoint of the line segment vi?1vi+1 and xes the other
n ? 1 vertices of P :
h(t) = hv ; v ; : : : ; vi? ; vi(1 ? t) + (vi? + vi )t; vi ; : : : ; vn? ; vni:
1
2
1
2
1
1
+1
+1
1
See Figure 1. Since P does not intersect ; this isotopy does not introduce
any self-intersections and hence denes a path from P to P 0:
Now, suppose P = hv ; v ; v ; v ; v i is a pentagon. If P does not intersect
the interior of the triangular disc 4v v v ; then it can be deformed into a
quadrilateral by the Reduction Lemma. See Figure 2(a). On the other hand,
if P does intersect 4v v v ; it must do so in edge v v : In this case, P does not
intersect the interior of the triangular disc 4v v v ; so the Reduction Lemma
can again be applied. See Figure 2(b). In each case, P will coincide with a
quadrilateral, which can then be attened out as indicated above.
1
2
3
4
5
1 2 3
1 2 3
4 5
4 5 1
(a)
(b)
Fig. 2. All pentagons are geometric unknots.
3
Fig. 3. A hexagonal trefoil knot.
Consider the embedding space Geo of rooted oriented hexagons. It is well
known that a hexagon can be tied as a trefoil knot, as in Figure 3. Recall
that trefoils are chiral, i.e. topologically dierent than their mirror image.
This means that every hexagonal trefoil will lie in a dierent component of
Geo than its mirror image. Therefore, there must be at least three distinct
path-components in Geo ; corresponding to the unknot, the right-handed
trefoil, and the left-handed trefoil. In Sections 2.1 and 2.2, we show that there
are two distinct geometric realizations of each type of topological trefoil. In
particular, hexagonal trefoil knots are not reversible: In contrast with trefoils
in the topological setting, reversing the orientation on a hexagonal trefoil
yields a dierent geometric knot (Corollary 6). Thus geometric knottedness is
actually stronger than topological knottedness.
(6)
(6)
(6)
Theorem 2. The embedding space Geo of rooted oriented hexagons con(6)
tains ve path-components. These consist of a single component of unknots,
two components of right-handed trefoil knots, and two components of lefthanded trefoil knots.
As we shall see in Section 2.1, the distinction between the two geometric types
of right-handed trefoils is a consequence of our original choice of root and
orientation. We eliminate this choice by taking the quotient of Geo modulo
the action of the dihedral group of order 12, and nd that the spaces of nonrooted oriented hexagonal knots and of non-rooted non-oriented hexagonal
knots each consist of three components (Corollary 7).
(6)
In Section 2.3, we develop a combinatorial invariant J of hexagons, which
we call the joint chirality-curl. We show that J distinguishes between all ve
components of Geo ; taking values as follows:
(6)
8
>
>
<(0; 0)
J (H ) = >>(+1; 1)
:
i H is an unknot,
i H is a right-handed trefoil,
(?1; 1) i H is a left-handed trefoil.
In particular this gives a complete classication of hexagons embedded in R :
3
4
1.2 Equilateral knots
A path in the embedding space Geo n corresponds to a deformation which
can stretch or shrink the edges of a polygon. Let us then restrict our attention
to the class of equilateral polygons and to deformations which preserve the
lengths of their edges. Dene the embedding space Equ n of n-sided equilateral
knots as the collection of polygons hv ; v ; : : : ; vn? ; vn i in Geo n with unitlength edges. Therefore Equ n is a codimension-n quadric subvariety of Geo n
dened by the equations
( )
( )
1
( )
2
( )
1
( )
kv ? v k = kv ? v k = = kvn? ? vnk = kvn ? v k = 1:
Consider the map f : Geo n ! Rn given by the n-tuple
f (hv ; v ; : : : ; vni) = (kv ? v k; kv ? v k; : : : ; kvn? ? vnk; kvn ? v k) :
The point p = (1; 1; : : : ; 1) 2 Rn is a regular value for f (Corollary 1 in [8]), so
n
1
2
2
3
1
1
( )
1
2
1
2
2
3
1
1
that Equ = f ? (p) is a 2n-dimensional smooth submanifold which intersects
a number of the components of Geo n ; some perhaps more than once.
( )
1
( )
We will say two polygons are equilaterally equivalent if they lie in the same
component of Equ n ; and that a polygon is an equilateral unknot if it is equilaterally equivalent to a standard planar polygon. Millett has shown that all
planar polygons are equilaterally equivalent, so the component of equilateral
unknots is well-dened.
( )
The embedding spaces Equ and Equ are both connected, as can be seen
by repeating the arguments given in Section 1.1 for the geometric case. In fact,
Equ consists of rotations and translations of a rigid equilateral triangle and
is thus homeomorphic to the semidirect product R o SO(3):
(3)
(4)
(3)
3
In [8], Randell showed that any equilateral pentagon can be deformed to a
planar one without changing the length of any of its edges. For suppose that
P = hv ; v ; v ; v ; v i is a generic equilateral pentagon. Then the convex hull
H (P ) spanned by the vertices of P will consist of either a tetrahedron or the
union of two tetrahedra glued along a common face. See Figure 4. In either
1
2
3
4
5
Fig. 4. There must be three consecutive vertices that span a face of H (P ).
5
Fig. 5. Flattening a pentagon without changing the length of any of its edges.
case, we can nd three consecutive vertices that span a face in the boundary
of H (P ). In particular, if H (P ) is a tetrahedron, then two of its four faces will
spanned by three non-consecutive vertices of P ; if H (P ) is the union of two
tetrahedra then four of its six faces will. Thus, after relabeling appropriately,
we can assume that v ; v ; and v span a face in the boundary of H (P ).
1
2
3
Let P be the plane determined by vertices v ; v ; and v . Then both v and
v lie to one side of P . Rotate the triangular linkage v v v about the axis
through v and v until it lies coplanar with v . We can then deform the
quadrilateral linkage v v v v in its plane until it misses the line through v
and v ; so that we can then rotate it without intersecting the linkage v v v ;
this is easy to achieve since the set of quadrilateral linkages v v v v embedded
in the plane forms a connected one-parameter family described entirely by the
angle \v v v . We can then rotate the linkage v v v about the axis through
v and v until the entire pentagon lies in a single plane. See Figure 5. Since
any equilateral pentagon can be attened out, Equ must also be connected.
1
2
3
5
4
1 2 3
1
3
4
1 2 3 4
1
4
4 5 1
1 2 3 4
4 1 2
1
4 5 1
4
(5)
Consider the case when n = 6: We have equilateral examples of each of the
ve types of hexagons in Geo : For example, the regular hexagon
(6)
H = h(1; 0; 0); (:5; :866025; 0); (?:5; :866025; 0);
(?1; 0; 0); (?:5; ?:866025; 0); (:5; ?:866025; 0)i
0
is an equilateral unknot, while the hexagon
H = h(0; 0; 0); (:886375; :276357; :371441);
(:125043; ?:363873; :473812); (:549367; :461959; :845227);
(:818041; 0; 0); (:4090205; ?:343939; :845227)i
1
is an equilateral trefoil with J (H ) = (+1; +1). Let H and rH denote the
mirror image (or obverse) and the reverse of a hexagon H ; then ; r; and r are
involutions of Geo taking H to equilateral trefoils of the other three types.
Therefore Equ intersects each of the ve components of Geo at least once.
2
1
(6)
(6)
2
1
(6)
The coordinates for H1 were kindly supplied by Kenneth Millett.
6
At rst sight it is unclear whether the intersection of Equ with any of these
components is connected, or equivalently, whether equilateral knottedness is
any dierent than geometric knottedness. Indeed, Cantarella and Johnston
show in [3] that for certain choices of edge length, there are \stuck" hexagonal
unknots. In other words, in Geo there are 12-dimensional varieties skew
to Equ which intersect the component of geometric unknots more than
once. Symmetry leads us to examine two cases: unknots and trefoils with
J () = (+1; +1): As in the case of H and H above, understanding these
two situations will completely characterize the general scenario. We shall analyze the subspace of equilateral unknots in Section 3.1 and that of equilateral
right-handed trefoils with J () = (+1; +1) in Section 3.2. In particular, we
obtain the following result.
(6)
(6)
(6)
0
1
Theorem 3. The embedding space Equ of equilateral hexagons contains ve
(6)
components. These consist of a single component of unknots, two components
of right-handed trefoil knots, and two components of left-handed trefoil knots.
As with Geo(6); the joint chirality-curl J distinguishes among these components.
In Section 3.3, we use information obtained from J to show that each component of trefoils in Equ contains essential loops which are null-homotopic
in Geo : Thus the trefoil components of Equ are not homotopy equivalent to those in Geo : In particular, this shows that, despite the fact that
two hexagons are equilaterally equivalent exactly when they are geometrically
equivalent, the two types of knottedness are quite dierent in nature.
(6)
(6)
(6)
(6)
2 Geometric hexagons
2.1 The topology of Geo(6)
The main theorem of this section provides us with the rst example of a
topological knot type realized by two distinct geometric knot types.
Theorem 2 The embedding space Geo of rooted oriented hexagons contains
ve path-components. These consist of a single component of unknots, two
components of right-handed trefoil knots, and two components of left-handed
trefoil knots.
(6)
Proof. Let H = hv1 ; v2; v3; v4; v5 ; v6i be a rooted oriented hexagon in Geo(6):
By appropriate translations and solid rotations of R3; we can arrange it so
v1 = (0; 0; 0) and v5 = (x5; 0; 0) for some x5 > 0: Recall that Geo(6) is a
manifold, so H can be slightly perturbed to a generic hexagon; hence, we can
assume that H crosses the x-axis at only these two points. Let P2; P3; and
P4 be the half-planes emerging from the x-axis and containing v2; v3; and v4;
7
respectively. Again by picking a generic H; we can assume that the Pi's are
distinct.
Consider the order in which the Pi's occur as we rotate around the x-axis in
a right-handed fashion, starting on a half-plane which does not intersect H:
This divides Geo into six dierent regions meeting along codimension-1 sets
where either:
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(i) two of the Pi's coincide, or
(ii) an edge of H crosses the x-axis.
The following structure lemma presents an analysis of each of these regions.
We delay its proof for now (see Section 2.2).
Lemma 4. (Structure lemma) Table 1 indicates the number of connected components in each of the six regions of Geo ; arranged by the topological knot
type they represent.
Table 1. Number of components in each region of Geo :
(6)
(6)
Regionof Geo 6
0
( )
2-3-4
2-4-3
3-2-4
3-4-2
4-2-3
4-3-2
31
1
1
1
1
1
1
?31
1
1
-
1
1
-
As noted above, these six regions of Geo meet along codimension-1 subsets
consisting of hexagons for which two of the Pi's coincide; for instance, regions
2-4-3 and 4-2-3 meet along a subset consisting of hexagons with P = P : See
Figure 6. The six regions also meet along codimension-1 subsets consisting of
hexagons which meet the line passing through v and v more than just twice;
for example, regions 2-4-3 and 4-3-2 meet along a set of hexagons for which
edge v v intersects this line, as in Figure 7. These connections are shown
schematically in Figure 8; solid lines represent hexagons with two coinciding
Pi's while gray lines represent hexagons for which some edge intersects line
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2
1
4
5
2 3
vv:
1 5
Consider a hexagon H in the common boundary between two regions of Geo :
Since H can be perturbed slightly to make generic hexagons of either type,
H must be of a topological knot type common to both regions. However,
the only knot type common to adjacent regions in Figure 8 is the unknot.
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8
Fig. 6. Two views of a hexagon with P2 = P4:
Fig. 7. Two views of a hexagon for which edge v2 v3 intersects line v1 v5 :
Fig. 8. Codimension-1 connections between regions.
9
Therefore hexagons in these codimension-1 subsets must be unknotted and,
in particular, the topological unknots form a single component of geometric
unknots in Geo :
(6)
Suppose that h : [0; 1] ! Geo is a path from some 2-4-3 trefoil to some 32-4 trefoil. Since Geo is an open subset of R ; there is a small open 18-ball
contained in Geo about each point in this path. Thus we can assume that
whenever h passes through a boundary of one of the six regions, it does so
through a generic point in one of the codimension-1 subsets above. But then
h must pass through either 2-3-4 or 4-3-2; see Figure 8. This is a contradiction
since only unknots live in these regions. Thus there is no path connecting
the trefoils of type 2-4-3 and those of type 3-2-4. Similarly, there is no path
between the type 4-2-3 and 3-4-2 trefoils.
(6)
(6)
18
(6)
Therefore, Geo consists of ve components: one consisting of unknots, two
of right-handed trefoils, and two of left-handed trefoils.
(6)
The main dierence between the two types of hexagonal trefoils is that trefoils
in the region 2-4-3 (or 4-2-3) in some sense \curl upward" while those in the
region 3-2-4 (or 3-4-2) \curl downward," as in Figure 9. Let the curl of a
hexagon H = hv ; v ; v ; v ; v ; v i be
1
2
3
4
5
6
curl H = sign (v3 ? v1 ) (v5 ? v1) (v2 ? v1) :
(1)
Note that this gives the sign of the z-coordinate of v if v ; v ; and v are
placed (in a counterclockwise fashion) on the xy-plane. In particular, trefoils
of type 2-4-3 (or 4-2-3) have positive curl, while those of type 3-2-4 (or 3-4-2)
have negative curl. Hence, we have the following scholium.
2
1
3
5
Corollary 5. The curl of a rooted oriented hexagonal trefoil is invariant under geometric deformations.
Fig. 9. Trefoils of type 2-4-3 curl upward.
10
Fig. 10. A hexagon of zero curl must be unknotted.
Proof. Suppose that there is some path h : [0; 1] ! Geo(6) between a hexagonal
trefoil of positive curl and one of negative curl. Then there must be some trefoil
H = hv1; v2; v3; v4; v5; v6i on this path for which the triple product in (1) is
zero. For this trefoil the vertices v1; v2; v3; and v5 lie on the same plane. If v5
lies in the exterior of the triangular disc 4v1v2v3; then the Reduction Lemma
applies; the resulting isotopy would then make H coincide with a pentagon,
and so H must be unknotted. On the other hand, if v5 lies in the interior of
4v1v2v3; as in Figure 10, then the Reduction lemma applies to the triangular
discs 4v3v4v5 and 4v5v6v1; so H is again an unknot. In either situation we
reach a contradiction.
Dene the automorphisms ; r and s on Geo by
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hv ; v ; v ; v ; v ; v i = h?v ; ?v ; ?v ; ?v ; ?v ; ?v i;
rhv ; v ; v ; v ; v ; v i = hv ; v ; v ; v ; v ; v i;
shv ; v ; v ; v ; v ; v i = hv ; v ; v ; v ; v ; v i:
Note that r and s generate the dihedral group of order twelve which acts on
1
2
3
4
5
6
1
2
3
4
1
2
3
4
5
6
1
6
5
4
3
2
1
2
3
4
5
6
2
3
4
5
6
1
5
6
Geo(6) by reversing or cyclically shifting the order of the six vertices of each
hexagon, while gives the mirror image of a hexagon. A small alteration in
the proof of Corollary 5 shows that, given any in the group generated by ; r;
and s; the function curl () remains unchanged under geometric deformations
of trefoils. Consider how the actions of r and s aect the curl of a hexagon.
Let H be the hexagonal right-handed trefoil given by the coordinates
1
H = h(0; 0; 0); (:886375; :276357; :371441);
(:125043; ?:363873; :473812); (:549367; :461959; :845227);
(:818041; 0; 0); (:4090205; ?:343939; :845227)i:
1
A simple calculation shows that each of the even-index vertices lies above the
plane determined by v ; v ; and v : Thus curl H = +1; while curl rH = ?1:
Similar calculations show that r changes the sign of the curl for each of the
1
3
5
1
11
1
hexagons H ; rH ; and rH : Since curl () is invariant under geometric
deformations, we then have
1
1
1
curl rH = ?curl H
(2)
for any hexagonal trefoil H; proving the corollary below.
Corollary 6. Rooted oriented hexagonal trefoils are not reversible.
This is surprising, since, after all, these trefoils are topologically reversible. For
instance, if the trefoil pictured in Figure 3 were built out of string, it could
be \attened" until it lay almost entirely on a plane (except for small arcs
near the three crossings). It would then be reversed by a rotation of about
an appropriate axis. The same is not true if the hexagon were made out of
rigid sticks, since we would be unable to \atten" it. This is the rst fragment
of evidence that geometric and topological knottedness dier, and one which
perhaps adds fuel to the hopes of nding a geometrically knotted topological
unknot.
However, there is something synthetic in all this. For one thing, this result
depends on the choice of the root v : For example, consider once again the
trefoil H : A series of computations like the ones above shows that for any
hexagonal trefoil H;
1
1
curl sH = ?curl H:
(3)
This means that the irreversibility of trefoils is a consequence of xing a root.
We can eliminate our original choice of root v but preserve our choice of
orientation by taking the quotient of Geo by the action of s : We refer
to this quotient as the embedding space of non-rooted oriented hexagons.
Secondly, we can remove our choice of both root and orientation by considering
the quotient of Geo by the entire dihedral group r; s : We call this
quotient the embedding space of non-rooted non-oriented hexagons.
(6)
1
(6)
By Theorem 2, Geo consists of ve components, characterized rst by topological knot type, and second (in the case of trefoils) by curl. By (2) and (3),
both r and s change the sign of curl. This proves the following corollary.
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Corollary 7. The spaces Geo = s of non-rooted oriented hexagons, and
Geo = r; s of non-rooted non-oriented hexagons, each consist of three
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(6)
path-components.
Suppose ? is an arbitrary subgroup of r; s : In the tradition of Randell
[7,8], we consider Geo =?: The following generalization of Corollary 7 follows
easily from Theorem 2 and equations (2) and (3).
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12
Theorem 8. Suppose ? is a subgroup of the dihedral group r; s : Then
Geo =? has ve components if and only if ? is contained in the index-2
subgroup s ; rs : Otherwise, Geo =? has three components.
(6)
(6)
2
2.2 The structure lemma
In this section, we prove our structure lemma, which we rst stated in Section
2.1. This will complete our proof of Theorem 2.
Lemma 4 (Structure lemma) Each of the six regions of Geo contains a
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single component of unknots. Regions 2-4-3 and 3-2-4 also each contain a
single component of right-handed trefoils, while regions 3-4-2 and 4-2-3 each
contain a single component of left-handed trefoils. Thus, Table 1 indicates the
number of path-components in each of the six regions of Geo(6):
Proof. Suppose the hexagon H = hv1 ; v2; v3; v4; v5 ; v6i is embedded in R3 with
v1 = (0; 0; 0) and v5 = (x5; 0; 0); as in Theorem 2. Suppose we rotate H
about the x-axis until v2 lies in the upper-half xy-plane. Let A be the linear
transformation xing v4 and v5; and mapping v2 to ( 12 x5; 1; 0): Then At =
tA + (1 ? t)I is a continuous family of non-degenerate linear transformations
describing a geometric deformation from H to a hexagon for which \v5v1v2 is
an acute angle. Performing a similar transformation will ensure that the angle
\v1v5v4 is also acute. Therefore, we can assume that v2 lies to the \right" of
v1 and that v4 lies to the \left" of v5:
We analyze each of the six regions of Geo separately.
(6)
2-3-4 In this case, each half-plane between P and P intersects the linkage
2
4
formed by edges v v and v v in a single point, as in Figure 11.
2 3
3 4
Fig. 11. Conguration of the linkage v1 v2 v3v4 v5 :
If v lies in one of these half-planes, the triangular linkage formed by v v
and v v loops either above or below this path. However, by shrinking or
stretching v v and v v if necessary, we can rotate the linkage v v v past
P so that v lies in a half-plane otherwise not intersecting H: See Figure 12. The Reduction Lemma can then be used to push v down to the
x-axis, showing that H is a topological unknot. Furthermore, since Geo
is connected, these hexagons are part of a single connected component of
6
5 6
6 1
5 6
2
6 1
5 6 1
6
6
Geo(6):
13
(5)
Fig. 12. Pushing the triangular linkage v5 v6 v1 past P2 :
2-4-3 In this case, each half-plane between P and P intersects edge v v
in a single point, while each half-plane between P and P intersects both
v v and v v ; each in a single point. The four-edge linkage v v v v v is
arranged in one of three ways, depending on where v v intersects P : See
Figure 13.
(i) v v intersects P to the left of v ; so v v and v v do not appear to
2
4
2 3
4
2 3
3
3 4
1 2 3 4 5
2 3
2 3
4
4
2 3
4
4 5
\cross" (see Figure 13(a)):
If v lies in a half-plane between P and P ; then the triangular linkage
v v v either loops above or below v v : As above, we can rotate this
triangle past P (perhaps by stretching or shrinking); then v can then
be pushed down to the x-axis by applying the Reduction Lemma. If
v lies between P and P ; the linkage v v v may weave over or under
v v and v v ; but can be similarly pushed past P : Thus all hexagons of
this subtype are part of a connected component of unknots in hexagon
space.
(ii) v v intersects P to the right of and above v ; so v v and v v form a
positive crossing (see Figure 13(b)):
If v lies in a half-plane between P and P ; the linkage v v v can be
pushed past P as above. If v lies between P and P ; then we can
rotate v past P if additionally
- v v v jumps over both v v and v v ;
- v v v passes under both v v and v v ; or
- v v v weaves under v v and over v v :
Under any of these conditions, the hexagon H is (topologically) unknotted. However, if v v v passes above v v and below v v ; the linkage
v v v is trapped between P and P ; then H is knotted as a right6
2
4
5 6 1
2 3
2
6
6
4
2 3
3
5 6 1
3 4
4
2 3
4
4
6
2
2
6
4 5
4
6
5 6 1
4
3
4
5 6 1
2 3
5 6 1
3 4
2 3
5 6 1
3 4
2 3
3 4
5 6 1
5 6 1
2 3
2 3
4
3 4
3
(a)
(b)
(c)
Fig. 13. Three congurations of the linkage v1 v2 v3 v4v5 :
14
handed trefoil.
(iii) v v intersects P to the right of and below v ; so v v and v v form a
negative crossing (see Figure 13(c)):
If v falls between P and P ; or if v falls between P and P and
- v v v jumps over both v v and v v ;
- v v v passes under both v v and v v ; or
- v v v weaves over v v and under v v :
then the linkage v v v can be pushed past P ; hence H is an unknot.
Alternatively, suppose the triangular linkage v v v passes below v v
and above v v : Consider the plane P determined by v ; v ; and v :
Note that v v crosses under v v but over the x-axis; thus v lies above
and v below P : But then the linkage v v v would cross P at least
three times, between leaving v ; passing under v v ; jumping over v v ;
and arriving at v : See Figure 14. This is impossible.
2 3
4
6
4
2
4
5 6 1
6
2 3
5 6 1
4 5
4
3
3 4
2 3
5 6 1
2 3
3 4
2 3
3 4
5 6 1
2
5 6 1
3 4
2 3
4
2 3
4 5
3
2
5
1
5 6 1
5
2 3
3 4
1
Fig. 14. The triangular linkage v5 v6v1 cannot weave below v2 v3 and above v3 v4 :
If H is an unknot, then we can push v into the x-axis and then move v ;
shifting between each of these three congurations. Thus, the unknots in
these three subtypes are actually part of the same component of Geo :
Therefore, a hexagon of type 2-4-3 belongs to one of two path-components,
corresponding to the knot types O and 3 :
6
4
(6)
1
3-2-4 Just as in the 2-4-3 case, hexagons of this type are part of two components, corresponding to knot types O and 3 :
1
3-4-2 The situation here is similar to 2-4-3 case, except that left-handed tre-
foils live in this region of Geo while right-handed trefoils do not. Thus,
there are two path-components, corresponding to knot types O and ?3 :
(6)
1
4-2-3 Just as in the 3-4-2 case, hexagons of this type are part of two components, corresponding to knot types O and ?3 :
4-3-2 Just as in the 2-3-4 case, hexagons of this type are part of a single
1
path-component of topologically unknotted hexagons.
Notice that Table 1 summarizes these results, as desired.
15
2.3 An invariant of hexagonal knots
Just as curl distinguishes between the two types of right-handed trefoils, we
wish to nd a combinatorial invariant which will distinguish between the ve
dierent hexagonal knot types.
Consider the hexagon H = hv ; v ; v ; v ; v ; v i: Then v ; v ; and v determine
a triangular disc oriented by the \right hand rule;" we will call this disc the
triangle based at v : Let be the algebraic intersection number of this triangle with the rest of H: Notice that the triangle may only be pierced by edges
v v and v v ; and that if it is crossed by both of these edges, they will do
so in opposite directions, canceling out in : Thus, is equal to 0; 1; or
?1: Similarly dene and to be the intersection numbers of H with the
triangles based at v and v :
1
2
4 5
2
3
4
5
6
1
2
3
2
5 6
2
4
4
2
6
6
For example, suppose H is the right-handed trefoil shown in Figure 3. Say
the even-index vertices lie above the plane containing the odd-index vertices,
so curl H = +1: Then the triangle based at v is clearly pierced by edge v v
from top to bottom, in agreement with the orientation of that triangular disc.
Edges v v and v v intersect the triangles based at v and v (respectively)
in the same way. Therefore = = = 1: If we reverse the orientation
on H; the triangles at v ; v ; and v will be pierced from bottom to top, again
agreeing with their orientations; now curl H = ?1 but once more = =
= 1:
2
6 1
2 3
4 5
4
2
2
4
4
6
6
6
2
4
6
The following results shows that these three intersection numbers characterize
the topological and geometric knot type of H:
Lemma 9. Let H be a hexagon. Then:
(i) H is a right-handed trefoil if and only if 2 = 4 = 6 = 1;
(ii) H is a left-handed trefoil if and only if 2 = 4 = 6 = ?1;
(iii) H is an unknot if and only if i = 0 for some i 2 f2; 4; 6g:
Proof. We proceed by considering all possible values of 2; 4; and 6 :
First suppose H is a hexagon with = 0: Then either the triangle based
at v is not pierced at all, or it is pierced by both v v and v v : In the rst
case, we can use the Reduction Lemma to isotope H until it coincides with a
pentagon. See Figure 15(a). In the second, H does not intersect 4v v v ; so
again the Reduction Lemma applies. In either case, H is unknotted. Similar
arguments show that if H has = 0 or = 0; then H is an unknot.
2
2
4 5
5 6
4 5 6
4
6
Now suppose that H has = = 1: First consider the case in which the
triangle at v is pierced by v v : The orientations of the triangle and this edge
agree, so v must lie above and v below the plane P determined by v ; v ;
2
2
4
4
4 5
5
1
16
2
(a)
(b)
Fig. 15. If 2 = 0 then H must be unknotted.
and v ; as in Figure 16(a). Let w be the point where v v intersects P : Notice
that edge v v does not intersect the triangle at v : Since = 1; this triangle
must be pierced by edge v v : This edge will pass through either 4v wv or
4v wv : In the rst case, v would fall below plane P ; so = 0 and H is an
unknot. See Figure 16(b). However, if v v does pass through triangle 4v wv ;
then v lies above P with v v passing behind v v : See Figure 16(c). Then
= 1 and H is a right-handed trefoil of positive curl. Note that in either
case, 0:
3
4 5
1 2
4
4
6 1
3
4
3
6
6
6 1
6
5
3
5 6
4
2 3
6
6
(a)
(b)
Fig. 16. 2 = 4 = 1; case 1.
(c)
Next consider the case in which edge v v crosses through the triangle at v :
The orientation on the triangle forces us to have v above and v below the
plane determined by v ; v ; and v ; as in Figure 17(a). Note that v v does
not pass through the triangle based at v ; so that this triangle is pierced, if
at all, by edge v v : But since v lies behind the triangle, the intersection can
only be positive; in fact, if = 1; then H is a right-handed trefoil of negative
curl. See Figure 17(b). As before, 0 in either case.
5 6
2
5
1
2
6
3
2 3
6
3 4
3
6
6
In fact, if any two deltas are +1, the third delta must be non-negative. Similarly, if any two deltas equal to ?1; then the third delta will be non-positive,
with H knotted as a left-handed trefoil if = = = ?1:
2
4
6
Finally, suppose that H has = 1 and = ?1: If were 1, then by
the argument above would have to be non-negative; on the other hand,
2
4
4
17
6
(a)
(b)
Fig. 17. 2 = 4 = 1; case 2.
if were ?1; would have to be non-positive. Hence = 0 and H is
unknotted.
6
2
4
Corollary 10. If H is a right-handed trefoil with positive curl, then:
(i) the triangle at v2 is pierced by v4 v5;
(ii) the triangle at v4 is pierced by v6 v1; and
(iii) the triangle at v6 is pierced by v2 v3:
By reversing orientations and applying mirror images we obtain similar characterizations of the other types of hexagonal trefoils.
Proof. Since curl H = +1; v5 must lie on the \positive" side of the triangle
with vertices v1; v2; v3: Therefore if v5v6 pierced this triangle, it would do so
in a negative direction, so that 2 = ?1: This contradicts Lemma 9, so the
triangle must be pierced by v4v5 instead, as indicated by (i). Parts (ii) and
(iii) follow in the same way by considering s2H and s4H:
Consider the product (H ) = : We will refer to (H ) as the chirality
of the hexagon H; since it combines the information about topological knot
type obtained in Lemma 9 from the intersection numbers ; ; and :
The following result shows that we can use a hexagon's chirality and curl to
determine the component of Geo in which it is located. This completely
classies the embedded hexagonal knots in Geo :
2
4
6
2
4
6
(6)
(6)
Theorem 11. (i) is an invariant of hexagons under geometric deformations. In fact,
8
>
>
<0
(H ) = >1
>
:
i H is an unknot,
i H is a right-handed trefoil,
?1 i H is a left-handed trefoil.
(4)
(ii) The product 2 curl is an invariant of hexagons under geometric deformations.
18
(iii) Dene the joint chirality-curl of a hexagon H as the ordered pair J (H ) =
((H ); 2(H ) curl (H )): Then
8
>
>
<(0; 0)
J (H ) = >>(+1; c)
:
i H is an unknot,
i H is a right-handed trefoil with curl H = c; (5)
(?1; c) i H is a left-handed trefoil with curl H = c:
Therefore the geometric knot type of a hexagon H is completely determined by the value of its chirality and curl.
Proof. Note that equation (4) follows directly from Lemma 9. Thus, any geometric deformation must keep invariant, since such a deformation will preserve topological knot type. The product 2(H ) curl H is equal to curl H if H
is a trefoil and is 0 otherwise. Therefore, by Corollary 5, every geometric deformation will also keep this product invariant. Finally, according to Theorem 2,
the ve components of Geo(6) are characterized rst by the topological knot
type of their elements, and then, in the case of trefoils, by their curl. Therefore
the value of J (H ) for a hexagon H completely determines its geometric knot
type, as indicated by (5).
3 Equilateral hexagons.
3.1 Equilateral hexagonal unknots
The rst in-depth analysis of the space of equilateral hexagonal unknots was
done by Kenneth Millett and Rosa Orellana. We present an independent
proof of their result here.
3
Theorem 12. Any topologically unknotted equilateral hexagon can be geomet-
rically deformed into a planar one without changing the length of any of its
edges. Thus Equ(6) contains a single component of unknots.
Proof. Let H = hv1; v2; v3 ; v4; v5; v6i be a hexagonal unknot with unit-length
edges. Then by Lemma 9 there is some i = 0; without loss of generality we
can assume that 2 = 0: Let P2 be the plane determined by vertices v1; v2;
and v3: We consider the relative position of v4; v5; and v6 with respect to this
plane. There are three cases.
(i) Suppose that v ; v ; and v all lie on one side of P : Let P ; P ; and P
be the planes containing the line v v and each of the vertices v ; v ; and v :
Consider the convex hull determined by fv ; v ; v ; v ; v g; in the generic case,
one of the Pi's will cut through the interior of this hull, while the other two will
intersect it only along its boundary. In particular, at least one of the planes P
4
5
6
2
1 3
4
4
1
3
4
5
5
5
6
6
6
4
3
Their unpublished result is mentioned, for example, in Proposition 1.2 of [5].
19
Fig. 18. The vertices v4 ; v5; and v6 all lie on one side of P2:
and P is exterior to this hull. Thus we can assume without loss of generality
that both v and v lie on one side of P : See Figure 18.
6
5
6
4
Rotate the triangular linkage v v v about an axis through v and v until it
lies in P coplanar with v : Since = 0, we can deform the quadrilateral
linkage v v v v in this plane so it misses the line through v and v : This
can be done by keeping v and v xed, moving v so as to increase \v v v ;
and letting v move accordingly. Now we can rotate v v v v about the axis
through v and v until it lies coplanar with v : Neither edge outside this plane
can pierce it, so we can deform the pentagonal linkage v v v v v in its plane
until it misses the line through v and v : A nal rotation of the triangle v v v
about the axis through v and v will bring the entire hexagon into the plane.
1 2 3
4
4
1
2
1 2 3 4
1
1
4
4
3
1 4 3
2
1
3
1 2 3 4
4
5
1 2 3 4 5
1
1
5
1 6 5
5
Note that for the deformations described above, it suces to have any given
plane containing v and v separate v from v ; v ; and v : In the following,
we will make use of this observation several times.
1
3
2
4
5
6
(ii) Suppose that P separates v from v : Then without loss of generality we
can assume that v lies on the same side of P as does v : We consider 3 cases
(see Figure 19).
2
4
6
5
2
4
First suppose that edge v v crosses P \behind" the line through v and v ; as
in Figure 19(a). Let P be the plane containing the diagonal v v and parallel
to edge v v : Then v ; v ; and v all lie \underneath" P : Furthermore, we can
rotate the triangular linkage v v v about the axis through v v until v is
\above" P without introducing any self-intersections on H: In particular, the
one intersection which might occur, between v v and v v ; does not since the
dihedral angle between the triangular discs 4v v v and 4v v v increases as
v is raised. After relabeling vertices via the map
5 6
2
1
2
3 5
1 2
1
2
6
3 4 5
3 5
3 4
4
5 6
3 4 5
3 5 6
4
hv ; v ; v ; v ; v ; v i 7! hv ; v ; v ; v ; v ; v i;
1
2
3
4
5
6
5
6
1
2
3
4
we obtain a plane containing v and v which separates v from v ; v ; and v :
Therefore, the argument in (i) will deform H into a planar hexagon.
1
3
20
2
4
5
6
(a)
(b)
(c)
Fig. 19. The plane P2 separates v6 from v4 and v5 :
Now, suppose that v v crosses P \in front" of the line through v and v ;
as in Figure 19(b). We can rotate the linkage v v v \downward" until it lies
coplanar with v : Again, we avoid the only possible intersection, between edges
v v and v v ; because the dihedral angle between the discs v v v and v v v
increases with this motion. After relabeling vertices as
5 6
2
2
3
6 1 2
5
5 6
1 2
1 2 6
2 5 6
hv ; v ; v ; v ; v ; v i 7! hv ; v ; v ; v ; v ; v i;
the vertices v ; and v will lie on one side of the plane P ; while v lies on
1
5
2
3
4
5
6
2
1
6
5
4
6
3
2
4
this plane. Then following the argument in (i) will deform H into a planar
hexagon.
Thirdly, suppose that v v crosses P to the right of the diagonal v v and
between lines v v and v v : See Figure 19(c). Let P be the plane containing
v ; v ; and v : We can deform H into a planar hexagon by simply rotating
v v v \up" and both v v v and v v v \down" to P :
5 6
1 2
1
3
2
1 3
2 3
5
5 6 1
1 2 3
3 4 5
(iii) Now suppose that P separates v from v and v : Then the edges v v
and v v either both pierce the triangle based at v ; or both cross that plane
outside the triangle so that the linkage v v v pierces the triangle at v : By
relabeling vertices, we can assume the rst, so that the triangle at v is pierced
twice.
2
5
4
5 6
6
4 5
2
1 2 3
5
2
If = = 0; then at least one of the planes determined by v ; v ; and v ; or
v ; v ; and v is non-separating and the argument in (i) will deform H into a
planar hexagon. Therefore, assume that one of those two intersection numbers
is non-zero. By taking mirror images and once again relabeling vertices if
necessary, we can assume without loss of generality that = 0 and = 1:
See Figure 20(a).
4
2
3
6
1
2
6
4
4
6
Let be some very small positive number and let be the shortest distance
between edges v v and v v : Suppose we rotate the linkage v v v about the
diagonal axis through v and v to decrease : In this motion, either can be
made arbitrarily small without introducing self-intersections or else the edges
v v and v v come within a distance of each other. In the latter case, relabel
1 6
3 4
1
5 6
5 6 1
5
2 3
21
(a)
(b)
Fig. 20. If the linkage v4v5 v6 pierces the triangle at v2 twice, H can be deformed
into an \essentially singular" hexagon.
vertices as
hv ; v ; v ; v ; v ; v i 7! hv ; v ; v ; v ; v ; v i:
1
2
3
4
5
6
2
1
6
5
4
3
In either case, we will have 0 < < :
Since can be chosen as small as desired, H can be deformed until it is arbitrarily close to a singular hexagon with a self-intersection, as in Figure 20(b).
In fact, we can think of H as being \essentially singular." This is a corollary
of Theorem 1 in [8], which states that the variety of immersed equilateral
hexagons in R is a smooth manifold except at hexagons whose vertices all lie
in a straight line.
3
Let P be the plane containing v ; v ; v ; v and the singular point of selfintersection. We now show that as a singular hexagon, H can be deformed in
such a way that embedded hexagons suciently near it can be attened out.
We consider three cases.
1
3
4
6
(a) First suppose that the triangular linkage v v v can rotate about the axis
through v and v moving v past the plane containing v ; v ; and v without
introducing any self-intersections in H: Thus, the triangle based at v will be
unpierced, with v ; v ; and v all to one side of this triangle. The argument
in (i) can then be used to deform embedded hexagons near H into planar
hexagons.
4 5 6
4
6
5
1
2
3
2
4
5
6
(b) Alternatively, suppose that v v v can rotate about the axis through v
and v in the opposite direction, until v lies arbitrarily close to the plane
P : Hence, we can think of the pentagonal linkage v v v v v as \essentially
singular" in P :
4 5 6
6
4
5
3 4 5 6 1
Let a be the distance from v to v ; and let and be the measures of angles
\v v v and \v v v ; respectively. See Figure 21(a). By increasing the lesser of
the two angles, we can ensure that = : See Figure 21(b). Furthermore, we
1
3 1 6
3
1
1 3 4
1
3
22
3
(a)
(b)
(c)
Fig. 21. Deformation of an \essentially singular" pentagonal linkage.
can keep v ; v ; and v xed,pand push v and v uniformly towards each other,
keeping = ; until a < 3: Note that although this deformation will cause
v to move, we need not worry about introducing any new self-intersections
since v v and v v are the only edges outside of the plane P :
4
5
1
6
1
3
3
2
1 2
2 3
Now we can simultaneously increase = ; thus bringing v closer to the
line v v : In the limiting case, which occurs when v and v become coincident,
v will be at a distance of less than from the line v v : In the meanwhile, v
lies at a distance of at least away from the axis through v and v ; so that
the linkage v v v can easily swing past P : See Figure 21(c). Therefore, before
reaching this limiting case, we will be in the situation of case (a) above, since
we will be able to swing v v v \up and over" and past the linkage v v v :
Thus, embedded hexagons near H can also be deformed into planar hexagons.
1
1 3
5
5
4
1
2
1
2
3
6
1 3
2
1
3
1 2 3
4 5 6
1 2 3
(c) Suppose, then, that moving the triangular linkage v v v in either direction
always leads to intersections with the linkage v v v : Edges v v and v v
cannot intersect, since any deformation forcing them to intersect would rst
force an intersection between v v and v v : Similarly, an intersection between
v v and v v must be preceded by an intersection between v v and v v :
Therefore the intersections caused by rotating v v v are between v v and
v v or between v v and v v : Without loss of generality, assume that the rst
happens when rotating towards P and the second when rotating away from
P : As before, let and be the measures of angles \v v v and \v v v ;
so that we have < : See Figure 22.
4 5 6
1 2 3
2 3
1 2
2 3
5 6
5 6
1 2
4 5 6
4 5
4 5
2 3
1 2
5 6
1
1
4 5
3
3 1 6
3
23
1 3 4
Fig. 22. Worst case scenario: v1 v2 v3 traps v4 v5 v6 :
Fix v ; v ; v ; and v ; and let v and v vary so as to decrease : As we do so,
v v will move away from v v ; while v v moves closer to v v : If the latter pair
comes within of each other, we can push v v v towards P ; thus separating
v v from v v : This rotation will move v v arbitrarily close to either P or
v v : If v v can be pushed into P ; then we can apply the deformation in (b)
above. Otherwise, we can continue decreasing : Since v has more room to
move as gets smaller, this process will eventually terminate with v within
of P or with = : In the rst case, the deformation in (b) will atten
out embedded hexagons near H: In the second case, the four vertices v ; v ; v ;
and v must be the corners of a planar isosceles trapezoid, as in Figure 21(b).
Then the two remaining vertices v and v will move along circular arcs in the
plane bisecting this trapezoid orthogonally. In particular, we can push v v v
arbitrarily close to P and then follow the deformation described in case (b)
above.
1
2
3
6
4
4 5
5
1 2
3
5 6
2 3
4 5 6
5 6
2 3
1 2
4 5
4 5
3
5
3
5
1
3
1
3
4
6
2
5
4 5 6
3.2 Equilateral hexagonal trefoils
Equilateral hexagonal trefoils behave no dierently than geometric ones: Any
two equilateral trefoils with the same chirality and curl will be equivalent.
Because of symmetry, we only need consider trefoils with J () = (+1; +1):
Theorem 13. Equ contains a single component of equilateral right-handed
(6)
trefoils with positive curl.
Suppose that H = hv ; v ; v ; v ; v ; v i is an equilateral hexagonal trefoil of
type J () = (+1; +1): We view H as the sum of two isosceles quadrilaterals QA = hv ; v ; v ; v i and QB = hv ; v ; v ; v i glued along the diagonal
v v : By Corollary 10, edge v v will pierce the triangular discs 4v v v and
4v v v ; as shown in Figure 23(a).
1
1 4
4
1
2
3
2
3
4
5
6
4
1
6
5
4
2 3
4 5 6
5 6 1
We shall say that a hexagon is in standard singular position if it is the sum of
Throughout, we will say an n-sided polygon hv ; v ; : : : ; vn? ; vn i is isosceles if all
4
but the last edge (v1 vn ) have unit length.
24
1
2
1
(a)
(b)
Fig. 23. A hexagonal trefoil with J () = (+1; +1) can be deformed into a hexagon
in standard singular position.
an embedded isosceles quadrilateral QA and a singular isosceles quadrilateral
QB with a single self-intersection between its rst and third edges, as shown
in Figure 23(b). Note that any hexagon in standard singular position will lie
in rather than in Equ ; however, by the next lemma, we can always
move a type J () = (+1; +1) trefoil in Equ until it lies arbitrarily close to
a standard one in : Lemma 15 examines the connectivity of the space of
hexagons in standard singular position. Together these results will enable us
to prove Theorem 13.
(6)
(6)
(6)
(6)
Lemma 14. Every equilateral hexagonal trefoil of type J () = (+1; +1) can
be moved arbitrarily close to a hexagon in standard singular position.
Proof. Let H = hv1 ; v2; v3; v4; v5; v6 i be an equilateral hexagon with J (H ) =
(+1; +1); and let be a very small positive number. Move v6 \forward" towards
edge v4v5 by rotating the triangular linkage v5v6v1 about the diagonal axis
through v1 and v5: If we can complete this motion without introducing selfintersections until the distance between edges v4v5 and v1v6 is less than ; then
we are done. However the motion of the linkage v5v6v1 may be obstructed by
the edge v2v3:
(a)
(b)
Fig. 24. Two possible obstructions to the standard singular deformation of embedded
hexagons.
25
Suppose rst that the motion of triangular linkage v v v brings edge v v
within of v v : See Figure 24(a). If is small enough, then we can move v
\back" towards edge v v without any obstruction until the distance between
edges v v and v v is less than : We can then push v \forward" so that v v
and v v are not too close together.
5 6 1
1 6
2 3
5
1 6
1 6
4 5
2
1 6
2 3
On the other hand, suppose that v v comes within of v v : See Figure 24(b).
Then we can move either v or v away from v v until v v is within of
either v v or v v : Then we will be in the situation above and can proceed
accordingly.
5 6
2
1 6
2 3
3
5 6
2 3
4 5
In any case, we succeed in moving H arbitrarily close to a standard singular
hexagon, as desired.
We now take a motivational digression. Suppose QB = hp ; p ; p ; p i is a singular isosceles quadrilateral with a single self-intersection at p between edges
p p and p p : Without loss of generality, we can assume that QB is contained
in the upper-half xy-plane, with p and p both on the x-axis. Let a = kp ?p k;
and dene as the open triangular disc determined by p ; p ; and p: As a
simplication of the general case, we now consider the collection of embedded
isosceles quadrilaterals QA = hp ; q ; q ; p i that intersect orthogonally. This
will give us a very special type of hexagon QA + QB = hp ; q ; q ; p ; p ; p i in
standard singular position. See Figure 25.
1
1 2
2
3
4
3 4
1
4
1
2
1
1
2
4
3
4
1
1
2
4
3
2
Consider the bipolar coordinate map dened by the formula
(v) = (kv ? p k; kv ? p k) :
1
4
This map takes the upper-half plane bijectively to the innite rectangular corridor R = f(x; y) 2 R : x + y a; x ? a y x + ag in the rst quadrant
of the cartesian plane, with (p ) = (0; a) and (p ) = (a; 0): Notice that (p )
lies somewhere on the line x = 1; since kp ? p k = 1: Similarly, kp ? p k = 1
2
1
4
2
1
2
3
4
Fig. 25. A very special type of hexagon in standard singular position.
26
Fig. 26. Constructing an embedded quadrilateral through q perpendicular to the
plane of QB : The dotted vertical line indicates a \fold" along L in the diagram.
so (p ) lies somewhere on the line y = 1: On the other hand, suppose that
maps the self-intersection point p to the pair (xp; yp): Then the triangle
3
inequality applied to the perimeter of implies that
(1 ? xp) + (1 ? yp) = kp ? p k + kp ? p k > kp ? p k = 1:
2
3
2
3
Hence maps p to some point in the region of R where x + y < 1:
Suppose that (x; y) is a point in R with x 1 and y 1: Set q = ? (x; y):
Let L be the line through q perpendicular to the xy-plane, q be the point on
L above q a unit distance from p ; and q be the point on L below q a unit
distance from p : See Figure 25. Then, as indicated by Figure 26, hp ; q ; q ; p i
is an isosceles quadrilateral if and only if
1
1
1
2
4
1
1
2
4
q
p
kq ? q k = 1 ? x + 1 ? y = 1:
1
2
2
2
Squaring twice shows that the locus of such (x; y) forms a curve S dened by
the equation
x + y + 2x + 2y ? 2x y = 3:
It is clear from Figure 27 that S separates (p) from (p ) and (p ): Therefore
? S crosses both of the edges pp and pp ; and consequently intersects the
interior of : In fact, ? S will intersect in two arcs: one with endpoints
on pp and p p ; the other with endpoints on pp and p p :
Lemma 15 shows that this situation is true even when QA does not intersect 4
4
2
2
2 2
2
1
1
2
2
3
3
2 3
3
2 3
at a right angle. There are two components of hexagons in standard singular
position; hexagons in one intersect near the segment pp while hexagons in
the other intersect near the segment pp : This observation will turn out to
be crucial when proving Theorem 13.
2
3
27
Fig. 27. The image R of the bipolar map ; which sends p2 ; p3; p; and q to points
on the line x = 1 (dotted), the line y = 1 (dotted), the region x + y < 1 (shaded),
and the curve S (solid), respectively. In this case, a = 0:4:
Lemma 15. The space of hexagons in standard singular position consists of
two components.
Proof. We view the space of hexagons in standard singular position as a bred
space over the space of singular isosceles quadrilaterals QB: Then we can think
of the bre over a given QB as the space of embedded isosceles quadrilaterals
QA for which the sum QA + QB is in standard singular position. We begin by
developing two parameterizations of these bres. Then, we shall make use of
these parameterizations to gain geometric insights about the space of hexagons
QA + QB and to thereby prove our result.
(i) First, consider a singular isosceles quadrilateral QB = hp ; p ; p ; p i with a
single self-intersection between its rst and third edges. See Figure 28. Let a
be the distance between p and p and be the measure of the angle \p p p :
Up to translation and solid rotation of R ; the space of such QB can then be
parameterized as the set
1
(a; ) 2 (0; 1) (0; ) : cos > 2 (a ? 1)
by the map
1
1
*
!
4
2
3
4
4 1 2
3
s
!
? a + 2a cos ;
a; 7! 0; 0 ; cos ; sin ; cos 2 + a ? sin2 31 +
a ?!2a cos !+
s
sin + cos ? a 3 ? a + 2a cos ; a; 0 :
2
2
1 + a ? 2a cos Therefore the base space of singular quadrilaterals QB is connected.
2
2
2
2
28
Fig. 28. Parameterization of QB by the pair (a; ):
Next, consider an isosceles quadrilateral QA = hp ; q ; q ; p i compatible with
QB : Let be the measure of the angle \p p q : In addition, let P be the
plane containing p ; q ; and p ; and let P be the plane determined by q ; q ;
and p : Dene as the dihedral angle between P and the xy-plane, and as
the dihedral angle between P and P : See Figure 29. Notice that these three
measurements completely describe QA: Therefore, the hexagon H = QA + QB
can be parameterized as the quintuple (a; ; ; ; ):
1
1
2
4
4 1 1
1
1
4
1
2
1
4
2
1
1
2
Let p be the point of intersection between edges p p and p p ; and let be
the open triangular disc determined by p ; p ; and p: Furthermore, let B a; be
the set of points in which are contained in some standard singular hexagon
derived from QB :
1 2
2
3 4
3
(
)
Pick q 2 B a; and set t 2 (0; 1): The locus of vertices q a unit distance
away from p and a distance t away from q forms a circle perpendicular to the
xy-plane and centered on the axis p q: Thus, the locus of vertices q for which
q q is a unit length edge passing through q is again a circle perpendicular to
the xy-plane and centered on the axis p q: Denote this second circle by Ct:
See Figure 30. Since p ; p ; and q are not collinear, the intersection of each Ct
with the unit radius sphere centered about p consists of at most two points.
Furthermore, in the case that there are two such intersection points, exactly
one of these will fall below the xy-plane. Hence, for each pair (q; t) there is
at most one choice of q and q for which QA = hp ; q ; q ; p i is an isosceles
(
1
)
1
1
2
1 2
1
1
4
4
1
2
1
1
2
2
Fig. 29. Parameterization of H by the quintuple (a; ; ; ; ):
29
Fig. 30. If kq2 ? q1 k = kq1 ? p1 k = 1 and kq1 ? q k = t; then q2 lies on the circle Ct:
quadrilateral with q = q +(q ? q )t: In this way, the set of embedded isosceles
quadrilaterals QA which are compatible with QB and which pass through q can
be parameterized as an open subset J (q) of the interval (0; 1): In particular, the
collection of quadrilaterals QA compatible with QB can also be parameterized
by pairs in the set
1
2
1
n
o
Y a; = (q; t) 2 B a; (0; 1) : t 2 J (q) :
(
)
(
)
(ii) Temporarily x and ; and let vary over the interval [0; 2]: This
will have the eect of rotating q around a circle, and the edge q q around
the cone from q to this circle. See Figure 31. Let C be the piece of this cone
between q and the circle locus of q : Note that the xy-plane intersects the
cone in a hyperbola; thus as changes, the point of intersection between q q
and the xy-plane will move along a hyperbolic path with endpoints on the
circle of unit radius about p :
2
1 2
1
1
2
1 2
4
Now let vary. Unless p ; p ; and q are collinear, a change in will change the
pitch at which the xy-plane intersects the cone. If = 0 then this intersection
will be a degenerate hyperbola H ; corresponding to having the entire edge
q q inside the xy-plane. On the opposite extreme, the intersection is a single
point when the xy-plane is tangent to the circle locus of q : Between these two
situations, we obtain a continuous family of hyperbolic arcs with endpoints
on the unit circle about p : The region of the xy-plane covered by these arcs
will depend on the parameter which determines the position of q : There
are three cases.
1
4
1
0
1 2
2
4
1
p
Case 1 kq ? p k 2 (see Figure 31(a)): The cone
p will be tangent to the
sphere of unit radius about p when kq ? p k = 2: Thus, in the xy-plane,
H meets the circle of unit radius centered at p only twice. As varies, the
hyperbolae cover the open delta-shaped region between this circle and the
degenerate hyperbola H p
:
Case 2 1 kq ? p k < 2 (see Figure 31(b)): In this case, the cone in1
4
4
1
0
4
4
0
1
4
tersects the unit sphere about p in two circles. Thus, in the xy-plane, the
degenerate hyperbola H intersects the unit circle about p at four points.
4
0
4
30
p
(a) kq1 ? p4 k 2
p
(b) 1 kq1 ? p4 k < 2
(c) kq1 ? p4 k < 1
Fig. 31. Conic piece C and its intersection with the xy -plane.
Fig. 32. Region Ba 0 of possible intersections between H the xy -plane.
31
As increases, the hyperbolae ll in an open region consisting of the deltashaped region between the degenerate hyperbola and the inner circular arc,
as well as a crescent-shaped region inside the circle, which includes the area
between the circle, the two chords formed by H ; and the circle centered at
p which is tangent to these chords.
Case 3 kq ? p k < 1 (see Figure 31(c)): In this case, the vertex of the cone
is inside the unit radius sphere about p : Then the hyperbolae sweep out
an open crescent-shaped region containing the area enclosed by this circle,
H ; and the circle centered at p which is tangent to this H :
0
4
1
4
4
0
4
0
Consider the union, taken as varies from 0 to 2; of the regions described
above. The union of the delta-shaped regions from cases 1 and 2, and the
union of the crescent-shaped regions from cases 2 and 3 will each form a \half
moon." See Figure 32. Taken together, these half moons form a connected
region Ba0 of the xy-plane whose intersection with is exactly the set B a;
of points of contained in some standard singular hexagon derived from QB :
Let
(
n
)
o
Ya0 = (q; t) 2 Ba0 (0; 1) : t 2 J (q) :
Then the bre of quadrilaterals QA compatible with QB is homeomorphic to
the intersection Y a; = Ya0 \ ( (0; 1)):
(
)
Recall that for a given q 2 Ba0; the boundary points of the index set J (q)
correspond to quadrilaterals completely contained in the xy-plane. A technical
computation shows that
5
(a) there are at most four such quadrilaterals, so J (q) consists of no more
than two disjoint open intervals of (0; 1);
(b) a path in Ba0 which decreases kq ? p k and kq ? p k will not destroy either
component of J (q); and
p
(c) if kq ? p k and kq ? p k are both less than ; then J (q) is a connected
interval.
1
1
4
3
2
4
p
Let q = ( a; 3 ? 2a ? a ; 0): This is the midpoint of the second edge of
the singular isosceles quadrilateral with coordinates
p
p
*
!
!
!
!+
3
?
2
a
?
a
3
?
2
a
?
a
a
+
1
a
?
1
0; 0 ; 2 ;
; 2 ;
; a; 0
2
2
0
1
2
1
2
2
2
2
corresponding to the pair (a; arccos (a + 1)): Since the edge containing q is
horizontal, q must lie on the boundary of Ba0: Furthermore, since
1
2
0
0
kq ? p k = kq ? p k =
0
5
1
0
4
p3 ? 2a p3
2
The interested reader is referred to [2] pp. 43{45.
32
< 2;
p
there will always be points in Ba0 near q at a distance of less than of both
p and p : Therefore, if J (q) is not connected, wep can move q in Ba0 towards q
until kq ? p k and kq ? p k are both less than : In doing so, the two inside
boundary points of J (q) will merge, thus joining the two components of J (q)
into a single interval. Hence Ya0 is connected.
3
2
0
1
4
0
1
3
2
4
(iii) For each pair (a; ); the corresponding QB will have p on the unit circle
about p ; and p on the unit circle about p : Thus, the edges p p and p p
intersect the interior of Ba0 in a single interval. Furthermore, p p coincides
with part of at least one degenerate hyperbola H : Therefore there is exactly
one point of p p which lies on the boundary of Ba0: Hence B a; = Ba0 \ will consist of two components, one near the vertex p and one near the vertex
2
1
3
4
1 2
3 4
2 3
0
2 3
(
)
2
p:
3
Similarly, Y a; = Ya0 \ ( (0; 1)) will have two, three, or four components,
depending on where the edge p p intersects @ Ba0: However, if we deform QB by
bringing closer to arccos (a +1); the edge p p will become more horizontal
and its intersection with @ Ba0 will move closer to the midpoint of the edge.
This will have the eect of coalescing the four possible components of Y a;
in pairs, so that there are two components when cos = (a + 1): Letting a
vary over the interval (0, 1) will not change the situation. Therefore the space
of standard singular hexagons has two components, as desired.
With this groundwork in place, we proceed to proving Theorem 13.
(
)
2 3
1
2
2 3
1
2
(
)
First suppose that H and H are embedded, right-handed equilateral hexagonal trefoils with positive curl. By Lemma 14, we can push each of these
hexagons arbitrarily close to a hexagon in standard singular position. Thus,
we can assume they are both \essentially singular." As in Section 3.1, deformations of these singular hexagons approximate deformations of embedded
hexagons by Theorem 1 in [8], which states that the variety of immersed equilateral hexagons in R is a smooth manifold except at hexagons whose vertices
all lie on a straight line.
0
1
3
That there is a deformation from H to H is clear when these lie in the same
component of the space of standard singular hexagons. Suppose, then, that H
and H lie in dierent components of this space, and consider the equilateral
hexagonal trefoil with coordinates
0
1
0
1
H = h(?:45; 0; 0); (:1673333; :63; :4711683);
(?:1673333; :63; ?:4711683); (:45; 0; 0);
(?:0638889; :6999999; :4959014);
(:0638889; :6999999; ?:4959014)i
2
shown in Figure 33. Notice that this right-handed positive curl trefoil is symmetric about the y-axis.
33
Fig. 33. A right-handed trefoil which is symmetrical about the y -axis.
Following the recipe given in the proof of Lemma 14, we can deform H by
pushing v and v until they both lie on the xy-plane. Suppose we rst move
v up into the xy-plane by rotating the linkage v v v about the axis through
v and v ; and then move v down by rotating the linkage v v v about the
axis through v and v : Then we will arrive at the standard singular hexagon
2
5
6
6
5 6 1
1
5
5
4
4 5 6
6
H a = h(?:45; 0; 0); (:1673333; :63; :4711683);
(?:1673333; :63; ?:4711683); (:45; 0; 0);
(?:1495275; :8003540; 0); (:5465967; :0824328; 0)i:
However, if we move v down to the xy-plane rst, and then move v up, we
( )
2
5
6
will arrive at the hexagon
H b = h(?:45; 0; 0); (:1673333; :63; :4711683);
(?:1673333; :63; ?:4711683); (:45; 0; 0);
(?:5465967; :0824328; 0); (:1495275; :8003540; 0)i:
( )
2
For both of these hexagons, v v intersects the xy-plane at (0; :63; 0): However,
in H a this point lies very close to edge v v ; while in H b it lies very close
to edge v v : Thus H a and H b lie in dierent components of the space of
standard singular hexagons. The deformations of H to these two singular
hexagons provide a bridge between the two components, and hence a path for
deforming H into H : This proves the theorem.
2 3
( )
2
( )
2
1 6
( )
2
4 5
( )
2
2
0
1
3.3 The fundamental group of the subspace of trefoils
By Theorem 3, two hexagons are equilaterally equivalent exactly when they
are geometrically equivalent. Thus Equ intersects each component of Geo
exactly once. In this section, we show that the inclusion i : Equ ,! Geo
has a nontrivial kernel in fundamental group. In particular, the restriction of
i to each of the components of trefoil knots is not a homotopy equivalence.
(6)
(6)
(6)
34
(6)
We begin with the following observation.
Lemma 16. The odd-index vertices of an equilateral hexagonal trefoil cannot
determine an equilateral triangle.
Proof. Suppose that H = hv1 ; v2; v3; v4; v5 ; v6i is an equilateral hexagonal trefoil. By reversing orientation and taking mirror images we can assume that
J (H ) = (+1; +1): Translate H through R3 so that v1 lies at the origin. A
solid rotation of R3 can then place v3 on the positive x-axis and v5 on the
upper-half xy-plane. Since curl H = +1; the vertices v2; v4; and v6 have will
have positive z-coordinates.
Let a; b; and c be the distances between v and v ; v and v ; and v and
v ; respectively. Furthermore, let ; ; and be the dihedral angles between
the triangle 4v v v and
p the triangles based at v ; v ; and v ; respectively.
Finally, set d(a; b; c) = 2a b + 2a c + 2b c ? a ? b ? c : Then H can be
parameterized by the map (a; b; c; ; ; ) 7! hv ; v ; v ; v ; v ; v i for which
1
3
3
5
5
1
1 3 5
2
2 2
2 2
2 2
4
4
4
1
2
3
6
4
4
5
6
!
v = 0; 0; 0
!
p
p
1
a
1
v = 2 ; 2 4 ? a cos ; 2 4 ? a sin !
v = a; 0; 0
1
2
2
2
3
p
v = 3a ?4ba + c ? 4dab 4 ? b cos ;
!
d ? a + b ? c p4 ? b cos ; 1 p4 ? b sin 4a ! 4ab
2
v = a ? 2ba + c ; 2da ; 0
p
v = a ? 4ba + c + 4dac 4 ? c cos ;
!
d ? a ? b + c p4 ? c cos ; 1 p4 ? c sin :
4a
4ac
2
2
2
2
2
4
2
2
2
2
2
2
2
2
2
2
2
2
2
5
2
6
2
2
2
Note that curl H = +1; so the angles ; ; and take values in the interval
(0; ):
By Corollary 10, the triangle based at v and edge v v intersect in such a way
that their orientations agree. Thus v lies on the positive side of the triangle
determined by v ; v ; and v ; as in Figure 34. Similarly, the triangles based
at v and v are pierced by edges v v and v v ; so that v and v lie on the
2
4 5
2
1
4
6
4
5
1 6
35
2 3
4
6
Fig. 34. The triangle based at v2 is pierced by v5 v6 :
positive side of triangles 4v v v and 4v v v : Hence
3 6 1
5 2 3
f = (v ? v ) (v ? v ) (v ? v ) > 0;
f = (v ? v ) (v ? v ) (v ? v ) > 0;
f = (v ? v ) (v ? v ) (v ? v ) > 0:
1
1
5
4
5
2
5
2
3
1
6
1
4
1
3
5
3
2
3
6
3
Consider the function
p
p
p
c f sin p+ 4 ?pa f sin + 4 ? b f sin :
(H ) = 4 ? p
4 ? a 4 ? b 4 ? c sin sin sin 2
1
2
2
2
2
3
2
2
Since ; ; and lie in the open interval (0; ) and f ; f and f are all
positive, we must have (H ) > 0: However, simple algebraic manipulation
shows that
1
2
3
4(H ) = c ?a b cot + a ?b c cot + b ?c a cot :
Therefore, if the odd-index vertices of H determined an equilateral triangle,
a = b = c; making (H ) equal to zero and giving a contradiction.
2
2
2
2
2
2
Lemma 16 reects the empirical observation that the \most symmetric" hexagonal trefoils seem to be not equilateral but dihedral, with alternating short
and long edges, and whose even and odd vertices form equilateral triangles in
parallel planes (see Section 4.2.3 in [9]).
Let : Equ ! [0; 2] [0; 2] [0; 2] be the map taking an equilateral hexagon
hv ; v ; v ; v ; v ; v i to the triple (kv ? v k; kv ? v k; kv ? v k): In addition, let
D = f(x; x; x) : x 2 [0; 2]g be the diagonal across the cube [0; 2] [0; 2] [0; 2]:
Then ? D consists of those equilateral hexagons whose odd vertices form an
equilateral triangle. By Lemma 16, these will all be unknotted, so that the
image of the subset of equilateral trefoil knots under is contained in the
open solid torus (0; 2) (0; 2) (0; 2) ? D:
(6)
1
2
3
4
5
6
3
1
1
36
5
3
1
5
Theorem 13 states that the subset of equilateral right-handed hexagonal trefoils with positive curl is connected. Let h be a path in Equ from
(6)
H = h(:4090205; ?:343939; :845227); (0; 0; 0);
(:886375; :276357; :371441); (:125043; ?:363873; :473812);
(:549367; :461959; :845227); (:818041; 0; 0)i
1
to s H : Notice that
2
1
H = (0:914936; 0:610324; 0:818027);
s H = (0:610324; 0:818027; 0:914936);
2
1
1
so h is a path in the solid torus wrapping one third of the way around
D; from the region f(a; b; c) : b < c < ag to the region with the region
f(a; b; c) : a < b < cg: Hence the image under of the concatenation h s h s h
is a loop wrapping entirely around the \hole" of the solid torus. Therefore
h s h s h is an essential loop of innite order in the component of equilateral
hexagonal trefoils. In particular, each component of Equ corresponding to
trefoils must have innite fundamental group.
2
2
4
4
(6)
However, the component T of Geo corresponding to right-handed hexagonal
trefoils with positive curl has fundamental group isomorphic to Z : For suppose
Ht : [0; 1] ! Geo is a loop of trefoils with J (Ht ) = (+1; +1): The R oSO(3)
action obtained by translating and rotating R is free on this component of
Geo ; so that T is a principal bundle with bre R o SO(3): Since the vectors
v ? v ; v ? v ; and v ? v form a basis for R for each t 2 [0; 1]; we can choose
a unique representative for each equivalence class in T =R o SO(3) by placing
v (t) on the origin, v (t) on the positive x-axis, and v (t) on the upper-half xyplane. This gives us a section : T =R o SO(3) ! T into the bundle, so that
T is homeomorphic to the cartesian product (T =R o SO(3)) (R o SO(3)):
Therefore
(6)
2
(6)
3
3
(6)
3
1
3
5
1
2
1
3
1
3
3
5
3
3
3
(T ) = (T =R o SO(3)) Z :
= (T =R o SO(3)) (R o SO(3)) Suppose that Ht is a loop contained in the image of the section : Let At be
the linear transformation taking v (t) to (1; 0; 0); v (t) to (0; 1; 0); and v (t)
to (0; 0; 1): Left multiplication by sAt + (1 ? s)I describes a deformation of
R taking Ht into a loop of hexagons with v = (0; 0; 0); v = (0; 0; 1); v =
(1; 0; 0); and v = (0; 1; 0); as in Figure 35. Since this transformation is non3
1
1
3
1
1
3
3
2
5
3
1
2
2
3
5
degenerate, we can assume that
Ht = h(0; 0; 0); (0; 0; 1); (1; 0; 0); v (t); (0; 1; 0); v (t)i:
4
6
Let be the smallest angle which the xz-plane makes with the plane through
the x-axis determined by v (t); and set P to be the plane at this angle. Then
the edge v v intersects this plane for all t: Since only the edge v v pierces
6
6
6
5 6
2 3
37
Fig. 35. Hexagonal trefoil Ht = h(0; 0; 0); (0; 0; 1); (1; 0; 0); v4(t); (0; 1; 0); v6(t)i:
through the interior of the triangular disc 4v v v ; we can deform Ht by
pushing v along the ray v??!
v until it lies on P : Thus we can assume that Ht
has v (t) on the plane P for all t: Let be the largest value of kv (t)k in this
deformation. For each t; we can push v along the ray ?
v?!
v until kv (t)k = ;
so that we can assume that v (t) only moves along part of some circular arc
on P : Finally, let ! be the smallest measure of the angle \v v v : We can
move v (t) inside the plane P along the circle of radius about the origin
until ]v v v = ! : In particular, there is no danger that v v will intersect
v v since the former is contained in P and the latter crosses P only at v :
Therefore, we can assume that v (t) remains constant for all t; so that the
loop Ht consists of trefoils with ve xed vertices.
1 5 6
6
6 5
6
6
6
6
6
6
1 6
6
6
6
6
6
6
3 1 6
6
3 1 6
6
6
1 6
3 4
6
6
3
6
Now, let be the largest angle which the xz-plane makes with the plane
containing v (t) and the x-axis, and let P be the plane realizing this angle.
For each t; v (t) can be pushed along the ray ?
v?!
v until v (t) lies on P : Let
be the smallest value that the function kv (t)k takes on. Then v (t) can
be pushed in a straight-line path towards the origin until kv (t)k = for
all t: This deformation will create no self-intersections since the edges v v
and v v will move through the interior of the triangular discs 4v v v and
4v v v ; respectively, but Ht never pierces either of these discs. Hence, we
can assume that v (t) moves along a piece of some circular arc of radius on P : A contraction of this arc to a point will x v (t) for all values of t; so
that Ht is a null-homotopic loop in T : Therefore (T =R o SO(3)) = 1 and
(T ) = Z ; as claimed.
4
4
4
4
5 4
4
4
4
4
4
4
4
3 4
4 5
1 3 4
1 4 5
4
4
4
4
1
1
3
2
Arguing by symmetry among the four trefoil components of Equ and Geo
we obtain the following result.
(6)
(6)
Theorem 17. Each of the four components of trefoil knots in Equ has in(6)
nite fundamental group. On the other hand, the fundamental group of each
38
of the four components of trefoil knots in Geo(6) is isomorphic to Z2: Therefore the inclusion map i : Equ(6) ,! Geo(6) is not injective at the level of
fundamental group.
Theorem 2 shows that geometric knottedness is dierent than topological knottedness, and that there are distinct geometric knot types corresponding to the
same topological knot type. The irreversibility of hexagonal trefoils in Corollary 6 demonstrates this dierence. The lesson taught by Theorem 17 is that,
although Theorem 3 implies that geometric and equilateral knottedness coincide in the case when n 6; the two types of knottedness are of a quite
dierent nature.
Acknowledgments
I wish to thank my thesis committee at the University of California, Santa
Barbara, for their helpful advice on the early versions of this paper. In particular, I am indebted to my thesis advisor, Prof. Kenneth Millett, who oered
great suggestions and many interesting questions. I would also like to thank
the Department of Mathematics at Williams College for their hospitality during the year 1998{1999, and the referee for many suggestions which greatly
improved this paper. This research was funded in part by a National Science
Foundation Graduate Research Fellowship (1995{1998).
39
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40