University of Nevada, Reno
Knot Groups and Their Homomorphisms into SU(2)
A thesis submitted in partial fulfillment of the requirements for the degree of Master of Science in Mathematics
By Rashelle DeBolt
Dr. Christopher Herald/Thesis Advisor
August, 2019 We recommend that the thesis prepared under our supervision by
RASHELLE DEBOLT
entitled
Knot Groups and their Homomorphisms into SU(2)
be accepted in partial fulfillment of the requirements for the degree of
MASTER OF SCIENCE
Christopher Herald, Ph.D., Advisor
Stanislav Jabuka, Ph.D., Committee Member
James Winn, Ph.D., Graduate School Representative
David W. Zeh, Ph.D., Dean, Graduate School
August, 2019 i
Abstract
We begin with an introduction to algebraic topology, knot theory, and SU(2) matrices as a subset of the quaternions, then proceed to introduce a technique of finding homomorphisms of knot complement fundamental groups into SU(2) and illustrate it by finding homomorphisms for the fundamental groups of the complement of the trefoil and the complement of the Whitehead link. Finally, the Seifert-van Kampen theorem allows us to find pairs of those homomorphisms, with nonabelian image, which give rise to homomorphisms from the knot group of the Whitehead double and therefore prove that the Whitehead double of the trefoil knot is not the trivial knot. ii
Acknowledgments
The author is grateful to Dr. Christopher Herald for his advice and encouragement and also to her thesis committee, professors, classmates, and family for all of their support. iii
Contents
1 An Introduction to Knots1
2 Free Groups and Finitely Presented Groups5
3 Fundamental Groups 11
4 The Seifert-van Kampen Theorem 20
5 The Wirtinger Presentation 27
6 Meridians and Longitudes of a Knot 33
7 The Whitehead Link 35
8 The Quaternions and SU(2) Matrices 37
9 Homomorphisms of Knot Groups Into SU(2) 45
10 The Whitehead Double of a Knot 51
Bibliography 63 1
1 An Introduction to Knots
A knot is an embedding K : S1 → S3. The complement of a point in S3 is
homeomorphic to R3; the homeomorphism S3 \{p} → R3 can be computed through stereographic projection. Since S3 \{p} is homeomorphic to R3, it is sufficient to picture knots in R3.
Definition 1.1 (Definition 1.1 of [2]). Two embeddings f0, f1 : X → Y are isotopic if there is an embedding F : X ×I → Y ×I such that F (x, t) = (f(x, t), t)
for x ∈ X, t ∈ I with f(x, 0) = f0(x) and f(x, 1) = f1(x). We call a level-
preserving map F an isotopy and frequently use the notation ft(x) for f(x, t) so the map f can be thought of as a path of maps from X to Y .
The general notion of Isotopy is not good enough, though, because any two embeddings can be isotopic even if they are different in knottedness. For example, the diagram of the trefoil in Figure1 can be contracted continuously so that the section with all of the crossings is contracted to a point, making it isotopic to the unknot.
Figure 1:
Definition 1.2 (Definition 1.2 of [2]). Two embeddings f0, f1 : X → Y are ambient isotopic if there is an isotopy H : Y × I → Y × I such that H(y, t) =
(ht(y), t) with f1 = h1 ◦ f0 and h0 = idY .
The difference between the two definitions is that an isotopy moves the set 2
of f0(X) continuously to f1(X) in Y ignoring the neighboring points of Y while
an ambient isotopy requires Y to move continuously along with ft(X).
In this thesis, we want to restrict our definition of knots to only include to tame knots.
Definition 1.3 (Definition 1.3 of [2]). A knot K is called tame if it is ambient isotopic to a simple closed polygon in S3. A knot is wild if it is not tame.
Definition 1.4. If two knots in S3 are ambient isotopic, we say they are equiv- alent.
The geometric description of a knot in S3 is complicated, but we can use
the orthogonal projection of a knot onto a plane in R3, which we will represent in R2 in the following way.
Definition 1.5. A knot projection is the image of K : S1 → S3 \{p} = R3, {p} sufficiently far from K(S1), projected onto a linear subspace E ⊂ R3. The notation K is used to denote the projection of K(S1).
Definition 1.6. A knot projection is regular if the map is not injective at only finitely many points and for each of those points, there are no more than two elements in the domain. We call the points at which the regular projection is not injective double points. We also want to insist that at each double point, the arcs actually cross; to do this, we won’t allow corners of any ambient isotopic polygonal knots to intersect in the projection.
An example of the image of a regular projection of the trefoil is presented in Figure2. 3
Figure 2:
In order to represent the image of a knot with a projection in a way that
helps us reconstruct the knot in R3 or S3, we need to keep track of the under- and over-crossings at each double point. To do this, we will use the knot diagram:
Definition 1.7. A knot diagram is an image of a regular projection which preserves over-crossings and under-crossings in the canonical way so that it is clear how to reconstruct the knot from the two dimensional image.
The knot diagrams of a few common knots are shown in Figure3. For
Figure 3:
more images of common knot diagrams, see appendix D in [2].
Proposition 1.8. For any knot in R3 = S3 \{p}, the set of vectors in S2 for which the orthogonal projection yields a regular projection is open and dense in S2. 4
Proof. The proof of this theorem is outside the scope of this thesis, so we will sketch it out. Sard’s theorem states that the set of critical values of a smooth function from one Euclidean space or manifold to another is a null set, which in the case of the set of regular projections, means the set of planes which do not yield regular projections have Lebesgue measure 0. Then, using differential topology, we can show that the set of regular projections is open and dense in S2.
Reidemeister was the first to find a finite collection of moves which generate the equivalence relation of ambient isotopy.
Theorem 1.9 (Reidemeister Moves). Two knots K0 and K1 are equivalent if and only if they possess diagrams which are related by finitely many Reidemeister
moves, Ωi, i = 1, 2, 3. These moves are seen in Figure4
Figure 4: Reidemeister Moves
Figure5 walks through an example that shows the figure-eight knot is ambient isotopic to its mirror image. (Note that this is not true in general.) 5
Figure 5:
2 Free Groups and Finitely Presented Groups
Like many questions in topology, determining whether two knots are equivalent is difficult without using other mathematical techniques. In particular, we need to introduce some algebraic tools. Free groups are an important ingredient in algebraic topology; however, before we define a free group, we need to talk about algebraic words.
Definition 2.1. Let G be a group and let {Gα}α∈J be a family of subgroups of G. We say these groups generate G if every element g ∈ G can be written as a finite product of elements of the groups Gα.A word, w, is a finite sequence of elements from the groups Gα. Furthermore, for any 1 ≤ i ≤ n and αi ∈ J, if
gi ∈ Gαi , the ordered list w = (g1, g2, ..., gn) is a word of length n. We include the empty list, w = ( ), which we call the empty word. 6
Definition 2.2. If {Gα}α∈J is a family of subgroups which generates G, there is
a finite word (g1, ..., gn) of elements of the groups Gα such that g = g1 ··· gn; we
say the word (g1, ..., gn) represents the element g ∈ G.
To combine two words we use the operation of concatenation. So, if
0 0 0 0 w = (g1, g2, ..., gn) and w = (g1, g2, ..., gm) are two words of lengths n and m
0 0 0 0 0 0 0 respectively, then ww = (g1, g2, ..., gn)(g1, g2, ..., gm) = (g1, g2, ..., gn, g1, g2, ..., gm) is also a word of length n + m.
We now introduce two operations which will be called reductions of words.
If gi and gi+1 are two consecutive entries of a word w and both belong to the same subgroup, then we can combine them to make a word of length n− 1. For example,
if w = (g1, g2, ..., gi, gi+1, ..., gn) and gi, gi+1 ∈ Gαi , we can define a shorter word w0 = (g , g , ..., g g , ..., g ). Moreover, if any g = e is an identity, we can 1 2 i i+1 n i Gαi delete it from the sequence to obtain an n − 1 length word. This can be seen if we define a word w = (g , g , ..., e , ..., g ). Then the identity in the i-th place can 1 2 Gαi n 0 be deleted to obtain the shorter word, w = (g1, g2, ...gi−1, gi+1, ..., gn).
Definition 2.3. A reduced word is a word in which no reductions can be made so it is of minimal length. Also, given any word, there is a unique reduced word it can be reduced to. We call two words equivalent if they can both be reduced to the same word.
Definition 2.4. Let G be a group and let {Gα}α∈J be a family of subgroups of G which generate G. Also suppose that Gα ∩ Gβ = 1 if α 6= β, then we say G is the
free product of the groups Gα if each element g ∈ G can be represented as a
unique reduced word, w = (g1, ..., gn), for all gi ∈ G.
This definition gives rise to the following lemma. 7
Lemma 2.5 (Lemma 68.1 of [6]). Let G be a group and let {Gα} be a family of
subgroups of G. If G is the free product of the groups Gα, then given any group H
and any family of homomorphisms hα : Gα → H, there is a unique homomorphism
h : G → H whose restriction to Gα equals hα, for each α.
The converse of the previous lemma also holds and is proved as a direct consequence of Lemma 68.5 in [6].
We also need to be able to talk about free products of groups which are not subgroups of a larger group G. In order to do this, we will need the following theorem.
Theorem 2.6 (Free Product, Theorem 68.2 of [6]). Given a family of groups
{Gα}α∈J , there is a group G and a family of injective homomorphisms iα : Gα → G such that G is the free product of the groups iα(Gα).
The elements of the group G from the above theorem can be taken to be
reduced words in which the entries are elements of the Gα. Thus the example below shows how we can reduce words in the free product of a family of groups.
r Example 2.7. Let G1 = Q8 = {±1, ±i, ±j, ±k} and G2 = 2Z = {a | r is even}
10 4 with G = G1 ∗ G2. If we have the word w = (g1, g2, g3, g4, g5) = (−i, k, a , a , 1),
14 we can reduce it to w = (j, a ) since g1 and g2 were elements of G1, g3 and g4
were elements of G2, and g5 = eG1 can be deleted.
As a direct consequence of Lemma 2.5, we have the following Lemma.
Lemma 2.8 (Lemmas 68.3 and 68.5 of [6]). Let {Gα}α∈J be a family of groups and
let G be a group. Let iα : Gα → G be a corresponding family of homomorphisms.
If each iα is an injective homomorphism and G is the free product of the groups 8
iα(Gα), then given any group H and any family of homomorphisms hα : Gα → H,
there is a unique homomorphism h : G → H such that h ◦ iα = hα, for each α. Furthermore, the converse also holds.
Definition 2.9. Let {Gα}α∈J be a family of infinite cyclic groups and let {xα}α∈J
be a set of generators, i.e. Gα = hxαi. Also, let G be a group and let iα : Gα → G be a collection of injective homomorphisms. Then if G is the free product of the
groups iα(Gα), then we say G is the free group generated by the {xα}.
The objects we are talking about are now free groups, so each Gα is
freely generated by some xα, so we can write a word in the free group G as
k1 kn k1 kn w = (g1, ..., gn) = (xα1 , ..., xαn ) which we will write more succinctly as xα1 ··· xαn . This notation implies that everything in the group G can be written as product where each factor is a power of a generator of the Gα.
Theorem 2.10 (Universal Property of the Free Group, Lemma 69.1 of [6]). Let
G be a group and let {aα}α∈J be a family of elements of G. Then G is a free group with generators {aα} if and only if given any group H and any family {yα}α∈J of
elements of H, there is a unique homomorphism h : G → H such that h(aα) = yα for each α.
Proof. (⇒) Assume G is free, then each Gα = haαi is infinite cyclic, so there
is a homomorphism hα : Gα → H with h(aα) = yα. Then, by Lemma 2.5, the
conclusion holds. (⇐) Now, let Gβ be a fixed group from {Gα}. By assumption,
there is a homomorphism h : G → Z such that h(aβ) = 1 and h(aα) = 0 for α 6= β.
Thus, the group Gβ is infinite cyclic and thus statement holds by the converse of Lemma 2.8. 9
While not all groups are free, we will now show that every group is a quotient of a free group.
Given a group G which is generated by the family {aα}α∈J , we can call
F the free group on these generators. Now the map h : {aα} → G such that
h(aα) = aα extends to a surjective homomorphism h : F → G, and if N = ker h, by the first isomorphism theorem, F/N ∼= G. Given that N is normal in F , we can describe N more precisely by characterizing the generators which normally generate N. Specifically, we can denote a family {Rβ} of words in F such that these elements and their conjugates generate N. In this definition, N is the least
normal subgroup of F which contains the elements Rβ. Thus we’ll be able to describe N more efficiently.
Definition 2.11. Given a group G which is generated by the family {aα}1≤α≤n and
the normal subgroup N = ker h as defined above with the generators {Rβ}1≤β≤m, we can write the finite presentation of G as
G = ha1, a2, ..., an | R1,R2, ..., Rmi.
The normal subgroup N generated by {Rβ} contains each element Rβ, all
powers of each Rβ, all conjugates of each Rβ, and all products of the Rβ. It is easy to see that this subgroup is not very small or realistically understandable, so it is great that we have this finite presentation for G. Thus, by the universal property of the free group, we can determine a homomorphism by finding a map from the generators of G into any group which satisfies the relations defined in N.
As an illustration, let us realize a free abelian group as a quotient of a free group. 10
Definition 2.12. Let G be a group. If x, y ∈ G, then the element [x, y] = xyx−1y−1 is called the commutator of x and y. Furthermore, the subgroup [G, G] generated by the set of all commutators in G is called the commutator subgroup of G.
Lemma 2.13 (Lemma 69.3 of [6]). Given G, the subgroup [G, G] is a normal subgroup of G and the quotient group G/[G, G] is abelian.
Proof. First, to show that [G, G] is normal, let [x, y] ∈ [G, G] and g ∈ G \ [G, G]. Then we have
g[x, y]g−1 = g(xyx−1y−1)g−1
= g(x)(y)(x−1)(y−1)g−1
= g(x)g−1g(y)g−1g(x−1)g−1g(y−1)g−1
= (gxg−1)(gyg−1)(gx−1g−1)(gy−1g−1)
= (gxg−1)(gyg−1)(gxg−1)−1(gyg−1)−1
= [gxg−1, gyg−1] which is in [G, G], so all conjugates of commutators are in [G, G]. Now, let z ∈ [G, G]. Then z is a product of commutators by definition, so we can write z =
−1 z1z2 ··· zn where each zi is a commutator. Then we can show that gzg ∈ [G, G] because
−1 −1 −1 −1 gzg = (gz1g )(gz2g ) ··· (gzng ) which is a product of elements in [G, G] by the first statement and hence in [G, G]. Now, let a, b ∈ G; we want to show that (a[G, G])(b[G, G]) = (b[G, G])(a[G, G]). In- deed, (a[G, G])(b[G, G]) = ab[G, G] = ab(b−1a−1ba) = ba[G, G] = (b[G, G])(a[G, G]) 11
by the definition of the quotient, so G/[G, G] must be abelian.
that we know [G, G] is a normal subgroup of G, we can use the fact that, for a group H, there is a quotient map q : G/[G, G] → H such that [G, G] = ker q. Then xyx−1y−1 = 1 and some rearranging yields xy = yx, hence the group H must be abelian and therefore G/[G, G] must also be abelian.
Using the construction above, we can see that it is also possible to abelianize a free product of groups. Specifically, let G be the free product of a family of abelian groups. Then the abelianization of the free product will be the direct
product of those groups. Also, if we have a free group G = hx1, x2, ..., xni, then
−1 −1 the abelianization G/[G, G] = hx1, x2, ..., xn | {xixjxi xj | i, j ∈ I, i < j}i is the ∼ n free abelian group hx1i ⊕ · · · ⊕ hxni = Z . Also, if H is a free abelian group, then there is a free group G such that φ : G/[G, G] → H is an isomorphism.
3 Fundamental Groups
A fundamental question in topology is whether or not two spaces are homeomorphic to each other. The fundamental group is a tool which allows us to show two spaces are not homeomorphic to one another. Before we can define the fundamental group, though, we need to introduce the idea of a path homotopy.
Definition 3.1. Let I = [0, 1]. Let X and Y be spaces. If f : X → Y and g : X → Y are continuous maps, then we say f is homotopic to g if there is a continuous map F : X × I → Y such that F (x, 0) = f(x) and F (x, 1) = g(x) for every x. If f is homotopic to g, we write f ' g. Also, if f ' 0, we call f null-homotopic. 12
The above definition allows us to think of homotopy as a ‘‘continuous deformation,’’ as Munkres states in [6], of the map f to g via the parameter t as t goes from 0 to 1.
Now, let the maps be paths in X, where a path α : I → X is a continuous map such that α(0) = x0 and α(1) = x1. Then we can define a stronger definition of homotopy of paths as follows.
Definition 3.2. Two paths α : I → X and β : I → X are path homotopic if they have the same initial point, x0, and final point, x1, and there is a continuous map F : I × I → X such that
F (t, 0) = α(t) F (t, 1) = β(t)
F (0, s) = x0 F (1, s) = x1 for each s ∈ I and t ∈ I. We call F a path homotopy from α to β and if α is path homotopic to β, we denote it α ' β as before.
Lemma 3.3. The relation ' is an equivalence relation.
Proof. Let X be a space and let α, β, and γ be paths in X. Then Reflexive: for any path α in X, α ' α via the homotopy F (t, s) = α(t). Symmetric: given paths α and β in X, if α ' β, then we have a path homotopy F (t, s) from α to β, so we can define a path homotopy G(t, s) = F (t, 1 − s) which traverses the paths in the opposite direction. Since, F is a path homotopy, so is G. Transitive: now, given α, β, andγ in X, if α ' β and β ' γ, then we have 13
0 path homotopies F (t1, s1) and F (t2, s2). So, let us define a map
1 F (t, 2s), t ∈ [0, 2 ] H(t, s) = 0 1 F (t, 2s − 1), t ∈ [ 2 , 1].
1 0 H is well defined because for t = 2 , F (t, 2s) = β(t) = F (t, 2s − 1) and H is continuous, thus α ' γ.
It is also important to note that multiplication of paths is done via the operation of concatenation where if α and β are paths in X such that α is a path from x0 to x1 and β is a path from x1 to x2. Then
1 α(2t), t ∈ [0, 2 ] α ∗ β = 1 β(2t − 1), t ∈ [ 2 , 1].
In our discussion of the fundamental group, we will primarily use loops rather than paths. Thus we have the following definition.
Definition 3.4. If X is a space and α is a path in X which begins and ends at x0, then we call α a loop based at x0.
Note that multiplication of loops is defined via the operation of concatena- tion, but the set of all loops is not a group, since it does not have a clear identity element which commutes everything in the group. Thus, we need to introduce equivalence classes of loops.
Using the above definition, we can talk about the set of all loops based at a point x0—we will denote the basepoint by p hereafter—in a space X, but since it 14
is not a group, we will need to construct a group by taking the equivalence class of all loops based at p. In order to do this, we will take the homotopy class of all loops based at p, denoted by [α].
Lemma 3.5. Let α, β and γ be loops based at p. Then [(α ∗ β) ∗ γ] = [α ∗ (β ∗ γ)].
Proof. We know that
1 α(2t), t ∈ [0, 2 ] α ∗ β = 1 β(2t − 1), t ∈ [ 2 , 1]
and therefore 1 α(4t), t ∈ [0, ] 4 (α ∗ β) ∗ γ(t) = 1 1 β(4t − 1), t ∈ [ 4 , 2 ] 1 γ(2t − 1), t ∈ [ 2 , 1]
is a loop based at p because ((α∗β)∗γ)(0) = α(0) = p and ((α∗β)∗γ)(0) = γ(1) = p. Similarly, we can find
1 α(2t), t ∈ [0, ] 2 α ∗ (β ∗ γ)(t) = 1 3 β(4t − 2), t ∈ [ 2 , 4 ] 3 γ(4t − 3), t ∈ [ 4 , 1] which is also a loop based at p because ((α ∗ β) ∗ γ)(0) = α(0) = p and ((α ∗ β) ∗ γ)(0) = γ(1) = p. Now we need to show that these two loops are homotopic. 15
Define a homotopy H : I × I → X such that
4t s+1 α( ), 0 ≤ t ≤ s+1 4 H(t, s) = s+1 s+2 β(4t − 1 − s), 4 < t ≤ 4 4t−2−s s+2 γ( 2−s ), 4 < t ≤ 1
then H(0, s) = α(0), H(1, s) = γ(1), H(t, 0) = (α ∗ β) ∗ γ(t), and H(t, 1) = α ∗ (β ∗ γ)(t). Hence [(α ∗ β) ∗ γ] = [α ∗ (β ∗ γ)].
Theorem 3.6. The set of all homotopy classes of loops based at a point p in a space X forms a group under the operation ‘‘·’’ such that, if α and β are loops based at p, then [α] · [β] = [α ∗ β]. This group is called the fundamental group
of X relative to the base point p. We denote this group by π1(X, p).
Proof. Let X be a space. Then we can show the following for the group of all homotopy classes of loops based at p. Associativity: The group is associative since for [α], [β], [γ] ∈ X,
([α] · [β]) · [γ] = [α ∗ β] · [γ] = [(α ∗ β) ∗ γ]
and [α] · ([β] · [γ]) = [α] · [β ∗ γ] = [α ∗ (β ∗ γ)]
which are equivalent by Lemma 3.5. Identity: We can also see that the identity loop is just a constant map in X, so if we denote the identity loop as 1 : [0, 1] → {p}, then for any class of 16
loops [α] ∈ X based at p,
[α] · [1] = [α ∗ 1]
and [1] · [α] = [1 ∗ α]
which are both equivalent to [α] by a similar proof to Lemma 3.5. Inverse: Finally, if [α] ∈ X is a class of loops based at p, then [α]−1 = [α−1] is the loop traversed in the opposite direction of α and clearly
[α] · [α−1] = [α ∗ α−1]
and [α−1] · [α] = [α−1 ∗ α]
which are also both equivalent to the identity loop by a similar proof to Lemma 3.5. Thus, the group has an inverse. Hence, the set of all homotopy classes of loops based at a point p is a group.
Definition 3.7. A space X is path connected if for any two points in X, there is a path between them.
In this thesis, we will mostly be talking about path connected spaces. These spaces are nice because we can pick any basepoint in the space to calculate the fundamental group, up to isomorphism, as long as the base point is fixed throughout the calculation. This is shown in the following theorem and corollary.
Theorem 3.8 (Theorem 52.1 of [6]). If α : [0, 1] → X is a path from p1 to p2 in
−1 X, then the map αˆ : π1(X, p1) → π1(X, p2) given by αˆ([f]) = [(α ∗ f) ∗ α] is an 17
isomorphism.
Proof. αˆ is clearly well-defined group homomorphism, since concatenation of loops
is well-defined and for two loops, f and g based at p1 in X,
αˆ([f ∗ g]) = [(α−1 ∗ (f ∗ g)) ∗ α]
= [(α−1 ∗ f) ∗ (g ∗ α)]
= [(α−1 ∗ f) ∗ (α ∗ α−1) ∗ (g ∗ α)]
= [((α−1 ∗ f) ∗ α) ∗ ((α−1 ∗ g) ∗ α)]
=α ˆ([f]) · αˆ([g]).
We can also show that αˆ is an isomorphism since we can define a map βˆ which is the homomorphism derived from β = α−1 such that βˆ([h]) = [(β−1 ∗ h) ∗ β] =
−1 [(α ∗ h) ∗ α ] for [h] ∈ π1(X, p2), thus
(ˆα ◦ βˆ)([h]) =α ˆ([(α ∗ h) ∗ α−1])
= [(α−1 ∗ ((α ∗ h) ∗ α−1)) ∗ α]
= [((α−1 ∗ α) ∗ h) ∗ (α−1 ∗ α)]
= [h]
and similarly,
(βˆ ◦ αˆ)([f]) = [f].
Henceα ˆ is a group isomorphism.
Corollary 3.9 (Theorem 52.2 of [6]). If X is path connected and p0 and p1 are 18
any two points of X, then π1(X, p0) is isomorphic to π1(X, p1).
Now that we have defined the fundamental group, the next question is how does a continuous map between spaces affect their fundamental groups.
Suppose that h : X → Y is a continuous map that such that h(px) = py (we will call this a continuous basepoint preserving map). If f is a loop in X based
at px, then h ◦ f is a loop in Y based at py.
Proposition 3.10. Let X and Y be spaces and let f, g : X → Y be continuous basepoint preserving maps, such that f ' g and f(p) = g(p) = q. Then for any loop, γ ∈ X based at p, f ◦ γ and g ◦ γ are homotopic as loops based at q in Y .
Proof. We are given that f and g are homotopic, thus we can define a homotopy H : X × I → Y such that H(x, 0) = f(x), H(x, 1) = g(x), and H(p, t) = q for all t. We want to define a homotopy from f ◦ γ to g ◦ γ, so let K : I × I → Y be the homotopy such that K(s, t) = H(γ(s), t). This is indeed a homotopy from f ◦ γ to g ◦ γ because K(s, 0) = H(γ(s), 0) = f(γ(s)) and K(s, 1) = H(γ(s), 1) = g(γ(s)). Now we just need to make sure the map which sends s to K(s, t) is a loop. Indeed, K(0, t) = H(γ(0), t) = f(γ(0)) = f(p) = q and similarly, K(1, t) = q. Therefore f ◦ γ and g ◦ γ are homotopic as loops based at q in Y .
Definition 3.11. Let X,Y be spaces such that px ∈ X and py ∈ Y . Let h : X → Y
be a continuous basepoint preserving map. Then we can define h∗ : π1(X, px) →
π1(Y, py) such that h∗([f]) = [h ◦ f] using the notation above. We call the map h∗ the homomorphism induced by h.
Proposition 3.12. Let X and Y be spaces and let f, g : X → Y be continuous
basepoint preserving maps such that f ' g. Then f∗ : π1(X, p) → π1(Y, q) is equal
to g∗ : π1(X, p) → π1(Y, q). 19
Proof. We have that f is homotopic to g so, by Proposition 3.10, f ◦ γ ' g ◦ γ for
any loop γ based at p, thus f∗([γ]) = [f ◦ γ] = [g ◦ γ] = g∗([γ]). Therefore f∗ is
equal to g∗.
Theorem 3.13. If h : X → Y and k : Y → Z are continuous basepoint preserving
maps with px ∈ X, py ∈ Y , and pz ∈ Z, then (k ◦ h)∗ = k∗ ◦ h∗.
Proof. Let f be a loop in X. Then
(k ◦ h)∗([f]) = [(k ◦ h) ◦ f]
and similarly,
k∗ ◦ h∗([f]) = k∗([h ◦ f]) = [k ◦ (h ◦ f)].
Hence, (k ◦ h)∗ = k∗ ◦ h∗.
Corollary 3.14. If h : X → Y is a basepoint preserving homeomorphism, then
h∗ : π1(X, px) → π1(Y, py) is an isomorphism
Proof. Since h is a homeomorphism it has an inverse, h−1 : Y → X. Now, we can
−1 −1 −1 −1 show that h∗◦(h )∗ = (h◦h )∗ = (idY )∗ = idπ1(Y,py) and (h )∗◦h∗ = (h ◦h)∗ =
−1 −1 (idX )∗ = idπ1(X,px), so h∗ is an isomorphism and in particular, (h∗) = (h )∗ is
the inverse of h∗.
Theorem 3.15. The fundamental group of a circle is isomorphic to the free group
1 ∼ on one generator; π1(S , p) = Z.
To prove the above statement, we need to define covering spaces which are covered in [6] and [3]. For the sake of this thesis, though, we will just use the statement above to find the fundamental group of some more interesting spaces. To do so, we will use the following definitions. 20
Definition 3.16. If A ⊂ X and there is a map r : X → A such that r|A is the identity map of A, then we call it a retraction of X onto A.
We also have the stronger condition of a deformation retraction.
Definition 3.17. If i : A → X is an inclusion map and H : X × I → X is a continuous map such that H(x, 0) = x, H(x, 1) ∈ A, and H(a, t) = a for all x ∈ X and a ∈ A, then H is a deformation retraction of X onto A via the retraction
r : X → A such that r(x) = H(x, 1) and i ◦ r is homotopic to idX : X → X.
Theorem 3.18. If there is a deformation retraction of X onto A, then the inclusion
map i : A → X induces an isomorphism of fundamental groups i∗ : π1(A, p) →
π1(X, p) if the basepoint p is in A.
Proof. Since there is a deformation retraction of X onto A, there is a map
r : X → A such that r ◦ i = idA and i ◦ r is homotopic to idX by the above
definition. So, by Theorem 3.14, since r◦i = idA the induced map (r◦i)∗ = idπ1(A,p).
Also, since i ◦ r is homotopic to idX , we can define a homotopy H : X × I → X
such that H(x, 0) = i ◦ r(x), H(x, 1) = idX , and H(a, t) = a for any a ∈ A. Then,
by Proposition 3.10,(i ◦ r)∗ = idπ1(X,p). Hence i∗ : π1(A, p) → π1(X, p) is an isomorphism.
4 The Seifert-van Kampen Theorem
We are now able to compute the fundamental group of some very important spaces, such as the circle, wedge of circles, and contractible spaces, but we need to be able to compute it for much more complicated spaces. The following theorem, originally proved by Herbert Seifert and Egbert van Kampen, helps us with spaces that can 21
be expressed as a pair of open, path connected spaces which have a nontrivial intersection.
Theorem 4.1 (Seifert-van Kampen Theorem, Theorem 70.1 of [6]). Let X = U∪V , where U and V are open in X and assume U, V , and U ∩ V are path connected with the basepoint p ∈ U ∩ V . Also, let G be any group and let φ1, φ2,(i1)∗,(i2)∗,
(j1)∗,(j2)∗ be the homomorphisms indicated in the following commutative diagram, with (i1)∗,(i2)∗,(j1)∗,(j2)∗ induced by inclusions. Then, if φ1 ◦ (i1)∗ = φ2 ◦ (i2)∗
there is a unique homomorphism φ : π1(X, p) → G such that φ ◦ (j1)∗ = φ1 and
φ ◦ (j2)∗ = φ2 and thus, the following diagram commutes.
π1(U1, p) (i ) 1 ∗ φ1 (j1)∗ ∃!φ π1(U1 ∩ U2, p) π1(X, p) G
(j ) 2 ∗ φ (i2)∗ 2
π1(U2, p)
Proof. The proof will be sketched here; for the full proof, see [6]. It is easy
to show that φ is unique since it is completely determined by φ1 and φ2. To show that φ exists, we need to take the homotopy class of loops based at
p ∈ U1 ∩ U2, but those loops may not be loops in the disjoint spaces U1 and
U2. By Theorem 3.8, since U1 ∩ U2 is path connected, we can define base-
points p1 and p2 in U1 and U2 respectively, such that π1(U1, p1) ∗ π1(U2, p2)
is a normal subgroup of π1(X, p) such that the kernel of the map π1(U1, x1) ∗
π1(U2, x2) → π1(X, p) is normally generated by elements of the free product
−1 ((i1)∗(g))((i2)∗(g)) for any g ∈ π1(U1 ∩U2, p). As a result, if g1, ..., gn generate the
normal subgroup, then if w is any word in π1(U1, x1) ∗ π1(U2, x2) and g = wgi then
−1 −1 −1 ((i1)∗(wgi))((i2)∗(wgi)) = ((i1)∗(w))((i1)∗(gi))((i2)∗(gi)) ((i2)∗(w)) which is 22
−1 a conjugate of ((i1)∗(gi))((i2)∗(gi)) .
The above statement of the Seifert-van Kampen theorem is the more modern interpretation. When calculating the fundamental group of a space, though, it is more common to use the traditional statement in the next Theorem.
Theorem 4.2 (Traditional Seifert-van Kampen Theorem, Theorem 70.2 of [6]). With the hypothesis of the preceding theorem, let
j∗ : π1(U, p) ∗ π1(V, p) → π1(X, p)
be the homomorphism of the free product that extends the homomorphisms (j1)∗
and (j2)∗ induced by inclusion. Then j∗ is surjective, and its kernel is the least normal subgroup of N of the free product that contains all elements represented
−1 by words of the form ((i1)∗(g), (i2)∗(g) ), for g ∈ π1(U ∩ V, p).
Proof. The proof will be sketched here; for the full proof, see [6]. It is easy to
see that since the image of j∗ is generated by the images of (j1)∗ and (j2)∗, j∗ is
surjective. Also, N ⊂ ker j∗ since (j1)∗ ◦(i1)∗ = (j2)∗ ◦(i2)∗ are mapped to the same
−1 element in π1(X, p) and thus j∗ ◦ (i1)∗ = j∗ ◦ (i2)∗ implies that ((i1)∗(g), (i2)∗(g) )
belong to the kernel of j for any g ∈ π1(U ∩V, p). It is harder to see that N = ker j,
but to do so we need to show that the map ϕ :(π1(U, p) ∗ π1(V, p))/N → π1(X, p)
−1 −1 −1 has an inverse, ϕ , such that ϕ ◦ ϕ = idπ1(X,p) and ϕ ◦ ϕ = id(π1(U,p)∗π1(V,p))/N , then use the first isomorphism theorem to show that N is the kernel of j.
The above theorem is commonly written as
π1(X, p) = π1(U, p) ∗π1(U∩V,p) π1(V, p). 23
This notation is called the amalgamated free product which is defined below.
Definition 4.3. The amalgamated free product is the free product of two groups modulo the least normal subgroup of their intersection. In other words, if
G1,G2 are groups such that G1 ∩ G2 = H, then G1 ∗H G2 = (G1 ∗ G2)/N where N is the least normal subgroup such that H is contained in N.
With this theorem, we can now calculate the fundamental group of a torus, but first, we will calculate the fundamental group of the wedge of two circles, shown in Figure6.
Figure 6:
Example 4.4. Let X be the wedge of two circles, S1 ∨ S1 shown in Figure6. To use the Seifert-van Kampen theorem, we will divide X into the two spaces, A and B with the intersection A ∩ B pictured in Figure7. Note that the fundamental
Figure 7: 24
groups of A and B are both isomorphic to the fundamental group of a circle via a deformation retraction which continuously deforms the ‘‘whiskers’’ onto the circles. Also, the intersection is contractible, so its fundamental group is trivial. Now, we can construct the following commutative diagram of inclusions
A i1 j1
A ∩ B X
i2 j2 B which has the following induced diagram on the fundamental groups with basepoint p ∈ A ∩ B
π1(A, p)
(i1)∗ (j1)∗
π1(A ∩ B, p) π1(X, p)
(i2)∗ (j2)∗
π1(B, p)
∼ ∼ 1 ∼ ∼ We know π1(A, p) = π1(B, p) = π1(S , p) = Z and π1(A ∩ B, p) = 0 so by the ∼ ∼ Seifert-van Kampen theorem, π1(X, p) = π1(A, p)∗π1(A∩B,p) π1(B, p) = Z∗Z. Hence 1 1 ∼ π1(S ∨ S , p) = Z ∗ Z.
Now that we have calculated the fundamental group of a wedge of two circles, we can find the fundamental group of a torus.
Example 4.5. Let X = S1 ×S1 be the standard torus. Define U to be a closed disk
in the torus, X, and V to be the torus minus an open disk such that U ∩ V ∼= S1 as seen in Figure8. Now, since the intersection U ∩ V is a circle, which is path 25
Figure 8:
connected, we can construct the following commutative diagram of inclusions
U i1 j1
U ∩ V X
i2 j2 V which has the following induced diagram on the fundamental groups with basepoint p ∈ U ∩ V
π1(U, p)
(i1)∗ (j1)∗
π1(U ∩ V, p) π1(X, p)
(i2)∗ (j2)∗
π1(V, p)
∼ ∼ ∼ 1 We know π1(U, p) = 0 since U is contractible and π1(U ∩V, p) = Z since U ∩V = S , we will call the generator a so that π1(U ∩ V, p) = hai. The space V is a little more complicated, so let us define a quotient map q : I ×I → X which identifies the side (s, 0) to (s, 1) and the side (0, t) to (1, t) as seen in Figure9. Now, we can remove an open disk from the space I×I which will deformation retract radially outwards from the center of the disk to the edges of the rectangle as seen in Figure 10. Therefore the space V is the image q(V˜ ) and the fundamental group of V˜ is generated by ˜ ∼ the word (s, t) hence π1(q(V ), p) = π1(V, p) = h(s, t)i = Z ∗ Z. Now we can define 26
Figure 9:
Figure 10:
(i1)∗ : π1(U∩V, p) → π1(U, p) and (i2)∗ : π1(U∩V, p) → π1(V, p) to be the maps such
−1 −1 that (i1)∗(a) = 0 and (i2)∗((s, t)) = q(s, t) = sts t respectively. Thus, by the ∼ ∼ Seifert-van Kampen theorem, π1(X, p) = π1(U, p)∗π1(U∩V,p)π1(V, p) = (0∗(Z∗Z))/N ∼ −1 −1 where N is the normal subgroup generated by ((i1)∗(a), (i2)∗(a)) = sts t . Note
1 1 ∼ that this is the abelianization of π1(V, p), hence π1(S × S , p) = Z × Z.
Note, we could also have found the fundamental group of the torus by using the fact that the fundamental group of a product and the product of the
1 1 ∼ 1 1 ∼ fundamental groups. Thus π1(S × S , p) = π1(S , p) × π1(S , p) = Z × Z.
We can see that the fundamental group of the torus is free abelian of rank two, so we can write it as a finitely presented group with two generators, s and t, and the commutator relation sts−1t−1 = 1 as follows:
1 1 −1 −1 π1(S × S ) = hs, t | sts t i
The generators of the torus are commonly denoted using an inclusion map of the torus into the standard solid torus. 27
5 The Wirtinger Presentation
We now have a few ways to compute the fundamental group of a space, but in
1 ∼ this thesis we we need to be able to compute it for knots. Since π1(S , p) = Z, the fundamental group of the knot itself is not interesting, thus we will use the complement of the knot to determine the fundamental group because it is a sufficiently nice, path connected space.
Definition 5.1. If we have a tame knot, K, we can let N(K) be the open tubular neighborhood of K, defined by the embedding of an open normal disk cross a circle into S3, N(K): Int(D2) × S1 → S3. The we can also define N(K) to be the closure of N(K) such that N(K) : D2 × S1 → S3. In a tubular neighborhood of K, the radius of any circle cross section of the torus must be sufficiently small so that the edges do not touch. This construction gives us a ‘‘Thickened up’’ knot, or essentially a solid knotted torus.
Definition 5.2. The space S3\K is called the complement of the knot K.
3 ∼ 3 ∼ 3 Theorem 5.3. π1(S \N(K), p) = π1(S \K, p) = π1(S \N(K), p) where p is in the intersection, S3\N(K) ∩ S3\K ∩ S3\N(K).
Proof. Since S3\N(K) ⊂ S3\N(K) ⊂ S3\K, we can construct the following commutative diagram of inclusions:
j S3\N(K) i S3\N(K) S3\K.
We can also define a homeomorphism h : S3\N(K) → S3\K such that for any 28
point τeiφ in each normal disc cross section of S3\N(K),
iφ τe , τ ≥ 2 h(τeiφ) = iφ 2(2 − τ)e , 1 < τ < 2
Thus, we just need to show that two of the maps are isomorphisms in the induced diagram on the fundamental groups:
k∗
3 i∗ 3 j∗ 3 π1(S \N(K), p) π1(S \N(K), p) π1(S \K, p).
Since S3\K is homeomorphic to S3\N(K), their fundamental groups are trivially
2 1 isomorphic, hence h∗ is an isomorphism. Identifying N(K) with D × S , we can define a retraction r : N(K) \ K → ∂(N(K)) such that
r(ρeiθ, eiθ) = (eiθ, eiθ) which can be extended to r : S3 \ K → S3 \ N(K) by defining r(x) = x for x ∈ S3 \N(K). Observe this defines a deformation retraction, R : S3 \K ×I → S3 \K. Now, since there is a deformation retraction, the fundamental groups of the two
spaces must be isomorphic by Theorem 3.18, therefore r∗ is an isomorphism. Now, we know that h∗ and r∗ are isomorphisms, so (h ◦ r)∗ = i∗ is also an isomorphism.
3 ∼ 3 ∼ 3 Thus, π1(S \N(K), p) = π1(S \K, p) = π1(S \N(K), p).
Since it is not easy to say the fundamental group of the complement of the knot, we will call this group the ‘‘Knot Group.’’
In a knot diagram, where there is a double point, the arc of the under- 29
crossing is cut off to show that it is behind the other arc in the actual construction of the knot (see Definition 1.7). By the nature of this construction, there are distinct curves on the diagram which we will call connected components. In the Wirtinger presentation, each connected component has a meridianal loop which starts and ends at a basepoint p ∈ S3\K which is chosen to be sufficiently far
3 from the knot K. Figure 11 shows the generators of π1(S \K, p) where K is the right-handed trefoil.
Figure 11: Note that the knot has been stretched vertically for clarity.
Given a regular projection for a knot K, choose an orientation on the knot and then give each connected component a generator using a consistently right-handed or left-handed direction around the knot. This process is shown on the trefoil in Figure 12.
Figure 12:
Once the generators are named, each double point will represent a relation 30
in the Wirtinger Presentation. We will calculate the relation as shown on the trefoil in Figure 13.
Figure 13:
It is easy to see that each relation is equivalent to the identity because, as in Figure 14, the loops given by each relation retract to the point p trivially.
Figure 14: Note that the knot has been stretched vertically for clarity.
Once we have the generators {x1, ..., xn} (we tend to use {a, b, c, ...} for smaller sets, like the trefoil) and the relations {R1, ..., Rn}, we write the Wirtinger 31
Presentation as a finitely presented group:
3 π1(S \K, p) = hx1, x2, ..., xn|R1,R2, ..., Rni
= hx1, x2, ..., xni/N
where N is the normal subgroup generated by {R1,R2, ..., Rn}.
Note that we used the Trefoil for all of our calculations above, so we now know that the Knot Group of the Trefoil is:
3 −1 −1 −1 −1 −1 −1 π1(S \trefoil, p) = ha, b, c | aca b , cbc a , bab c i.
3 ∼ Theorem 5.4. If two knots K0 and K1 are ambient isotopic, then π1(S \K0) =
3 π1(S \K1).
1 3 1 3 Proof. Let two knots K0 : S → S and K1 : S → S be ambient isotopic. Then
3 3 there exists an ambient isotopy H : S × I → S × I such that H(y, t) = (ht(x), t)
3 3 where ht : S → S is a homeomorphism which sends the image of K0 to the image
of K1. Since ht is a homeomorphism, any restriction of ht is a homeomorphism
3 3 3 onto its image. Thus ht|S \K0 : S \ K0 → ht(S \ K0) is a homeomorphism which
3 3 means the induced map on the fundamental groups, (ht|S \K0 )∗ : π1(S \ K0) → 3 3 ∼ 3 π1(ht(S \ K0)) is an isomorphism. Hence π1(S \ K0) = π1(S \ K1).
The converse of Theorem 5.4 is not true in general. Here is a counterexample which shows where it may fail.
Counterexample 5.5. Let K1 be the granny knot shown in Figure 15 and K2 be the square knot shown in Figure 16. Then we can calculate the fundamental 32
Figure 15: Granny Knot
Figure 16: Square Knot
groups of the two knots to be
3 −1 −1 −1 −1 −1 −1 π1(S \ K1) = ha, b, c, d, e, f | cab a , abf b , bfa f ,
ecd−1c−1, dec−1e−1, fde−1d−1i
and
3 −1 −1 −1 −1 −1 −1 π1(S \ K2) = ha, b, c, d, e, f | cab a , abf b , bfa f ,
dec−1e−1, dfd−1e−1, fef −1d−1i.
Note that these two fundamental groups have three relations which are equivalent and three which are different; however, by using other tools in such as the knot signature, we can wee that the two groups are actually isomorphic which means we cannot use Theorem 5.4 because the converse is not necessarily true. Hence the fundamental group does not help us determine that the granny knot is not equivalent to the square knot. 33
6 Meridians and Longitudes of a Knot
A knot is an embedding of the circle into S3 and, using the definition we gave in Section5, the closed tubular neighborhood of a knot, N(K), is an embedding of a closed solid torus into S3. This distinction is important because we want to be able to represent the generators of S3 \ N(K) in terms of Wirtinger generators. Sine ∂(S3 \ N(K)) is homeomorphic to a torus, we can specify generators of
3 π1(∂(S \ N(K))) as the image of the meridian and longitude on the embedding of N(K) as µ0 and λ0 respectively. Note that µ0 and λ0 are elements of the fundamental group, so we can easily write them in terms of the generators of the Wirtinger presentation of the knot.
Traditionally, the meridian, µ0, of the embedding of the torus is the generator which bounds a normal disk of N(K). To get a well-defined second generator for the fundamental group of the boundary torus, independent of the knot diagram, it is helpful to use the first homology group. For the sake of this thesis, we will use the following definition of the first homology group.
Theorem 6.1. The first homology group, H1, is isomorphic to the abelianization
of the first fundamental group, π1. More specifically, if X is a space, then
∼ H1(X) = π1(X, p)/[π1(X, p), π1(X, p)].
Using this theorem, let us assume the Wirtinger presentation of our knot, K, is
3 π1(S \ K, p) = hx1, ..., xn | R1, ..., Rni, 34
then the first homology group of our knot is
3 −1 −1 −1 −1 H1(S \ K) = hx1, ..., xn | R1, ..., Rn, x1x2x1 x2 , ..., xn−1xnxn−1xn i ∼ = hx1i = ... = hxni = Z.
So, our meridians, x1, ..., xn, from the Wirtinger presentation are all equivalent when the group is abelianized which means the meridian of S3 \ N(K) is any single
0 generator from the set {x1, ..., xn}, but we will denote it as µ = x1 for simplicity.
3 3 The boundary of S \ N(K) is homeomorphic to a torus so π1(∂(S \
∼ 3 N(K))) = Z ⊕ Z, thus we need a second generator for π1(∂(S \ N(K))) which,
0 0 along with µ , forms a basis. As a result, we just need to determine i∗(λ ) since we
0 already determined that i∗(µ ) = x1.
0 0 We want to choose λ ∈ π1(∂(N(K)), p) such taht λ maps to the identity
3 3 under the composition of i∗ : π1(∂(N(K)), p) → π1(S \ K, p) and q∗ : π1(S \
3 K, p) → H1(S \ K).
Figure 17 pictures a trefoil with what we call the blackboard parallel
longitude, λbb. This generator is not trivial because in order to stay parallel to the
Figure 17:
trefoil, the generator had to travel underneath c, then a, and finally b so i∗ would
0 3 send λbb to cab. Under the map i∗, λbb would be (µ ) in the case of the trefoil or 35
(µ0)n where n is the number of under-crossings, counted with sign, on an arbitrary knot diagram. So, to obtain the generator which maps to the identity, we need to
0 −3 −3 0 −n −n compensate by concatenating (µ ) = a ((µ ) = x1 for an arbitrary knot)
0 −3 on the end of λbb. Thus, we call the longitude i∗(λ ) = caba on the trefoil. The same construction can be done on an arbitrary knot to obtain the null-homologous generator.
From the example above, we can see that the generators µ0 and λ0 on the trefoil will be as seen in Figure 18 where µ0 = a and λ0 = caba−3 in terms of the Wirtinger generators of the trefoil.
Figure 18:
7 The Whitehead Link
So far we have only talked about knots which are an embedding of one copy of S1, but there isn’t limit to how many disjoint embeddings of S1 a diagram can contain. A link is an embedding of a disjoint union of circles, L : S1 t ... t S1 → S3 while a two component link is an embedding of two disjoint circles L : S1 t S1 → S3. The link which we will be using in this thesis is the Whitehead link (WL) pictured in
Figure 19. The two components have been labeled L1 and L2 for clarity, so that
the untwisted circle is always L1.
Using the Wirtinger presentation from Section5 and the generators labeled 36
Figure 19:
3 as in Figure 20, we can find π1(S \ WL) to be
Figure 20:
3 −1 −1 −1 −1 −1 −1 π1(S \ WL) = hx1, x2, x3, y1, y2 | x3x1x3 x2 , y1x2y1 x3 , x3y2x1 y2 ,
−1 −1 −1 −1 x2y2x2 y1 , x1y2x1 y1 i.
By following the Reidemeister moves in the diagram in Figure 21, we can see another way to look at the link which is isotopic to the one presented in
Figure 19. Note that the L1 component is still unknotted with itself, but the L2 component has been transformed into what we call a clasp.
Once we have the final link diagram in Figure 21, it is easy to see how the complement of the Whitehead link would be a solid torus with a clasped knot removed. As seen in Figure 22, the complement of L1 gives us the solid torus from which we remove L2. It will be convenient to have the generators µ and λ
3 0 0 of π1(∂(S \ N(L1)), p), similarly to how we defined µ and λ in Section6. Then one can see, from Figures 21 and 22, that the generators µ and λ will correspond 37
Figure 21:
−1 to y2 and x2 x1 respectively.
Figure 22:
8 The Quaternions and SU(2) Matrices
The knot group may have a finite presentation, but that doesn’t necessarily let us know anything about the structure of the knot itself. This is because the knot group presentation uses the projection of the knot which depends on the plane
of projection and the point {p} which was removed to get S3 \{p} = R3. As a result, we cant tell whether two different presentations of groups are isomorphic because of the word problem in algebra which is undecidable. 38
Since it is not easy to tell if two arbitrary finitely presented groups are isomorphic to each other, we need a different approach. If we can show that the knot group of a knot has a homomorphism with a nonabelian image into some other group, then we at least know that the knot is not trivial since the knot group of the unknot is free abelian of rank one.
A simple example of a nonabelian multaplicative group is Q8 = {±1, ±i, ±j, ±k} with identity 1 and multiplication defined by i2 = j2 = k2 = ijk = −1, ij = −ji = k, jk = −kj = i, ki = −ik = j. This group works well because it can yield a non-commutative ring structure on R4 = span {1, i, j, k}.
The construction is similar to how we can algebraically complete R to obtain the set of complex numbers, C using the form z = a + bi, where (a, b) is the point in R2 = span {1, i} which represents the complex number z in the space C. Starting with the vector space span {1, i, j, k}, we can define a multiplicative operation to develop the quaternions, H, where the point (a, b, c, d) ∈ R4 is written in the form q = a + bi + cj + dk such that a is the real part of the quaternion q and bi + cj + dk is the imaginary part of q. Note that we want to extend the
multiplication of Q8 to formal linear combinations by extending linearly to form a four dimensional vector space. Thus the multiplicative ring structure on the space
H is exactly what you would expect:
1 · q = q, for all q ∈ H
i2 = j2 = k2 = ijk = −1
ij = −ji = k, jk = −kj = i, and ki = −ik = j.
Now, similar to C, the quaternions form a ring with addition as in R4 and 39
multiplication as defined above from the structure of Q8. However, they do not form a multiplicative group unless you remove the origin. It is easy to check
that H\{0} = H∗ is indeed a group which is associative but not commutative. Also, now that we have a nonabelian group structure on H∗, we can see that if q = a + bi + cj + dk is multiplied by its conjugate q¯ = a − bi − cj − dk, the result is a2 + b2 + c2 + d2 = ||q||2. Hence, every non-zero quaternion, q, has a multiplicative
−1 q¯ inverse, namely q = ||q||2 .
Definition 8.1. If q ∈ H is of the form q = ai + bj + ck and hence has no real part, then q is called a purely imaginary quaternion. Note that the set of
R3 = span {i, j, k}.
Definition 8.2. A quaternion is called a unit quaternion if it has distance from the origin equal to one. Any quaternion q can be normalized to have distance √ from the origin equal to one by dividing by the norm, qq, of q. Note that the set of all unit quaternions is homeomorphic to S3.
Definition 8.3. By the previous two definitions, a unit purely imaginary quaternion is a quaternion with no real part that is distance one from the origin. Note that the set of all unit purely imaginary quaternions is homeomorphic to S2.
Lemma 8.4. If p = ai + bj + ck and q = di + ej + fk in H are purely imaginary quaternions which can be represented as the three dimensional vectors ~u = (a, b, c) and ~v = (d, e, f), then Re(pq) = −~u · ~v and Im(pq) = ~u × ~v.
Proof. by computing pq using the multiplicative ring structure on H, we get
pq = (ai + bj + ck)(di + ej + fk)
= −(ad + be + cf) + (bf − ce)i + (af − cd)j + (ae − bd)k 40
so the real part of pq is precisely Re(pq) = −(ad+be+cf) = −(a, b, c)·(d, e, f) and the imaginary part is Im(pq) = (bf −ce, af −cd, ae−bd) = (bf −ce)i+(af −cd)j + (ae − bd)k which can be written as (bf − ce, af − cd, ae − bd) = (a, b, c) × (d, e, f). Therefore, Re(pq) = −~u · ~v and Im(pq) = ~u × ~v.
The Euler formula, reiθ = r(cos θ + i sin θ) where r > 0 and 0 ≤ θ < 2π, is commonly used to represent a non-zero complex number. For our purposes, though, we can replace i in this expression by any unit purely imaginary quaternion, uˆ = ai+
bj+ck. Then the Euler formula can represent a quaternion in the space H∗ as reuθˆ = r(cos θ+uˆ sin θ) = r(cos θ+(ai+bj+ck) sin θ) = r(cos θ+ai sin θ+bj sin θ+ck sin θ). Notice that since uˆ is of length one, the set {euθˆ | 0 ≤ θ < 2π and ||uˆ|| = 1} forms
the subgroup of unit quaternions in H∗; also, ||reuθˆ || = r since the length of euθˆ is one. So, for any arbitrary p, q ∈ H∗, we have ||p|| · ||q|| = ||p · q||. By applying this fact to the set of unit quaternions, we can see that any unit quaternion multiplied by another unit quaternion is, again, a unit quaternion. Hence the subgroup of unit quaternions is closed under multiplication. This construction is similar to
iθ 1 2 ∼ how {e | 0 ≤ θ < 2π} forms the unit circle, S ∈ R = C.
Lemma 8.5. The order of any unit purely imaginary quaternion in H∗ is four. Furthermore, the circle subgroup {euθˆ |0 ≤ θ < 2π} is commutative. (Note that i, j, and k are unit purely imaginary quaternions.)
Proof. Let uˆ = ai + bj + ck. Then uˆ2 = −(a2 + b2 + c2) + (bc − cb)i + (ac − ca)j + (ab − ba)k = −1, uˆ3 = −1 · uˆ = −uˆ, and uˆ4 = (−1) · (−1) = 1. Thus, the order of uˆ is four. Now, to show that the group {euθˆ |0 ≤ θ < 2π} is commutative, let 41
q = euθˆ and p = euφˆ . Then
qp = euθˆ euφˆ
= (cos θ +u ˆ sin θ)(cos φ +u ˆ sin φ)
= cos θ cos φ +u ˆ sin θ cos φ +u ˆ cos θ sin φ − sin θ sin φ
= cos(θ + φ) +u ˆ sin(θ + φ)
= cos(φ + θ) +u ˆ sin(φ + θ)
= euφˆ euθˆ
= pq.
So pq = qp and hence {euθˆ |0 ≤ θ < 2π} is commutative.
The set of unit quaternions as a subset of H∗ is isomorphic to the group SU(2) = {A ∈ GL(2, C) | det(A) = 1,AA¯T = I} of two dimensional special unitary matrices with coefficients in C. The term special means that for any A ∈ SU(2), det(A) = 1 and unitary means that AA¯T = I where I is the identity matrix.
Theorem 8.6. The isomorphism γ from the group of unit quaternions to the group SU(2) is given by the formula:
α β γ(α + βj) = −β¯ α¯ for α, β ∈ C.
Proof. Let q = a + bi + cj + dk be any unit quaternion. Then we can rewrite 42
q = a + bi + cj + dk = (a + bi) + (c + di)j, so
(a + bi)(c + di) γ((a + bi) + (c + di)j) = −(c + di) (a + bi) a + bi c + di = −c + di a − bi
a + bi c + di then, to see that A = in SU(2), we can do the following −c + di a − bi calculations.
a + bi c + di a − bi −c − di ¯T AA = −c + di a − bi c − di a + bi (a + bi)(a − bi) + (c + di)(c − di)(a + bi)(−c − di) + (c + di)(a + bi) = (−c + di)(a − bi) + (a − bi)(c − di)(−c + di)(−c − di) + (a − bi)(a + bi) a2 + b2 + c2 + d2 0 = 0 c2 + d2 + a2 + b2 1 0 = 0 1
= I 43
and
a + bi c + di det = (a + bi)(a − bi) − (c + di)(−c + di) −c + di a − bi
= a2 + b2 + c2 + d2
= 1
Therefore, AA¯T = I and det A = 1, Hence γ(q) is in SU(2). Now, to show that γ is a isomorphism, we need to show that it is injective and surjective. To show injectivity, let q = a + bi + cj + dk and p = e + fi + gj + hk be two distinct
elements in H∗. Then we have a + bi c + di γ((a + bi) + (c + di)j) = −c + di a − bi
and
e + fi g + hi γ((e + fi) + (g + hi)j) = −g + hi e − fi which are not equivalent since either the real or imaginary part of at least one component does not agree. Now, to show γ is surjective, we can see that if
α + βj ∈ H∗ such that αα¯ + ββ¯ = 1 has the property that
α β α + βj = , −β¯ α¯
so γ is surjective. Therefore, γ is an isomorphism. 44
This map yields the following representation of the unit quaternions 1, i, j, and k in SU(2):
1 0 i 0 γ(1) = γ(i) = 0 1 0 −i 0 1 0 i γ(j) = γ(k) = . −1 0 i 0
Proposition 8.7. The set of unit purely imaginary quaternions corresponds to the set of traceless SU(2) matrices.
Proof. All traceless matrices in SU(2) are of the form:
ia b α = , a ∈ R, b ∈ C. −¯b −ia
So, if you let b =u ˆ + ivˆ, then
ia uˆ + ivˆ α = −uˆ + ivˆ −ia i 0 0 1 0 i = a +u ˆ +v ˆ 0 −i −1 0 i 0 which corresponds to the unit purely imaginary quaternion ai +uj ˆ +vk ˆ .
Conjugations by SU(2) matrices are commonly used to represent rotations in three dimensions, so we may also use the unit quaternions to represent these rotations. For example, take the quaternions euαˆ and evβˆ where uˆ = i and 45
vˆ = ai + bj + ck are purely imaginary quaternions. Then if we conjugate evβˆ by euαˆ , we get the following calculation:
euαˆ evβˆ e−uαˆ = (cos α + i sin α)(cos β +v ˆ sin β)(cos α − i sin α)
= (cos α + i sin α)(cos β + ai sin β + bj sin β + ck sin β)(cos α − i sin α)
= cos β + ai sin β + sin β(bj cos 2α − cj sin 2α) + sin β(bk sin 2α + ck cos 2α).
Thus conjugation by euαˆ rotates evβˆ by 2α about the positive i-axis.
A more general calculation would be to conjugate evβˆ by euαˆ where uˆ is a unit purely imaginary quaternion and vˆ = auˆ + bwˆ + c(uˆ × wˆ) is also a unit purely imaginary quaternion for any unit purely imaginary quaternion wˆ which is perpendicular tou ˆ. Then we have:
euαˆ evβˆ e−uαˆ = (cos α +u ˆ sin α)(cos β +v ˆ sin β)(cos α − uˆ sin α)
= cos β + auˆ sin β + sin β(bwˆ cos 2α − cwˆ sin 2α)
+ sin β(b(ˆv × wˆ) sin 2α + c(ˆv × wˆ) cos 2α).
So conjugation of evβˆ by euαˆ rotated evβˆ by 2α aboutu ˆ.
9 Homomorphisms of Knot Groups Into SU(2)
The previous section shows us how there is a geometric relationship between conjugation of quaternions to rotations in the span of {i, j, k}. That relationship will allow us to compute homomorphisms of both the knot group of the trefoil and the knot group of the Whitehead link into SU(2). 46
3 1 1 First, since the knot group of a torus in S is abelian, namely π1(S ×S , p) =
−1 −1 1 1 hµ, λ | µλµ λ i, there is a clear homomorphism ρ : π1(S ×S , p) → SU(2) which
sends µ to euθˆ 1 and λ to euθˆ 2 in SU(2); up to conjugation, we can define ρ(µ) := eiθ1
and ρ(λ) := eiθ2 .
Note that if the knot group is not abelian, we do not want to choose a homomorphism which has an abelian image. The abelianization of a knot group
is Z for a knot and Z × Z for a link, thus an abelian homomorphism would not distinguish a knot or a link from the unknot or a torus respectively.
In Section5, we found the knot group of the trefoil, K, to be
3 −1 −1 −1 −1 −1 −1 π1(S \K, p) = ha, b, c | aca b , cbc a , bab c i.
This presentation is difficult to find a homomorphism for, though, so we will use
3 the fact that in π1(S \K, p), each relation is a consequence of the other two, so we can omit one of them to obtain the presentation
3 −1 −1 −1 −1 π1(S \K, p) = ha, b, c | aca b , bab c i.
Now, the generator c can be written in terms of the other two generators and then we can obtain a presentation with only two generators and one relation. To see this, let c = bab−1 and the insert it into the relation aca−1b−1 to get a(bab−1)a−1b−1. Now we have the following knot group of the trefoil
3 −1 −1 −1 π1(S \K, p) = ha, b | abab a b i.
Once we have the above presentation, we will denote x := ab and y := b−1a−1b−1. 47
Now, with the generators x and y, we can determine the relation abab−1a−1b−1 in terms of x and y to be x3y2. Thus we have the following presentation of the knot group of the trefoil
3 3 2 π1(S \K, p) = hx, y | x y i.
One can check that the words x = ab and y = b−1a−1b−1 generate because we can write a and b in terms of x and y, namely a = x2y and b = y−1x−1, which means any reduced word in ha, b | abab−1a−1b−1i can be written as a reduced word in hx, y | x3y2i.
Now that we have a simpler presentation for the knot group of the trefoil,
3 we can define a homomorphism σ : π1(S \K, p) → SU(2). Let us define xˆ and yˆ
to be the unit purely imaginary parts of the the Euler notations of σ(x) = exφˆ 1
and σ(y) = eyφˆ 2 . The relation x3 = y−2 implies that words x3 and y2 commute
3 3 xˆ(3φ1) with every element of the group π1(S \K, p), thus the elements σ(x ) = e
and σ(y2) = eyˆ(2φ2) must commute with every element in the image of σ.
The first option is to send both x and y to the same circle subgroup, namely xˆ =y ˆ = i up to conjugation, but this identification would make the entire image of σ an abelian group. Therefore, we want to send x and y to different circle subgroups of SU(2).
If x and y are sent to different circle subgroups in SU(2) under the map σ, σ(x3) and σ(y2) must be sent to ±1 since those are the only two elements of SU(2) which will commute with every element of the image of σ. If we choose to send y2 to 1, then either σ(y) = 1 or σ(y) = −1. In either case, though, σ(y) would commute with every element of the image of σ and we would again have an abelian image. Thus, in order for the homomorphism to have a nonabelian image, σ(y2) must be −1. Indeed, if σ(y2) = −1, then σ(y) could be any unit 48
yˆ π purely imaginary quaternion, or equivalently σ(y) = e 2 .
yˆ π 3 −2 Once we have defined σ(y) = e 2 , the relation x = y implies that
2 3 3 xˆ(3φ1) σ(y ) = σ(x ), thus σ(x ) = e = −1 which implies 3φ1 = π (mod 2π), thus
π 2π φ1 = 3 + k 3 . This means we have three possible solutions for φ1 (mod 2π). The 3π first solution would be 3 , but this would send σ(x) to −1 which is an abelian π 5π homomorphism. The other two solutions are 3 and 3 which would send σ(x) to xˆ π xˆ(− π ) e 3 and e 3 respectively. Since these solutions are negative of each other, we
xˆ π can use conjugation to make σ(x) = e 3 the nonabelian homomorphism. We may also assume, up to conjugation, that xˆ = i and yˆ = i cos ω + j sin ω for some angle
3 0 < ω < π. Hence, the homomorphism σ : π1(S \K, p) → SU(2) is given by
i π σ(x) = e 3
(i cos ω+j sin ω) π σ(y) = e 2 .
The following representation homomorphisms of the knot group of the Whitehead link, up to conjugation, are proved in greater detail by Klassen in [4].
3 We want to define a homomorphism τ : π1(S \ WL) → SU(2). In Section 7 we found the knot group of the Whitehead link to be
3 −1 −1 −1 −1 −1 −1 π1(S \ W L, p) = hx1, x2, x3, y1, y2 | x3x1x3 x2 , y1x2y1 x3 , x3y2x1 y2 ,
−1 −1 −1 −1 x2y2x2 y1 , x1y2x1 y1 i.
−1 The relation x1y2x1 = y1 implies that τ(y2) must be conjugate to τ(y1) as SU(2) elements, thus the real part of their images are equal. Therefore, we can let
yˆ1ψ1 yˆ2ψ1 τ(y1) = e and τ(y2) = e for some unit purely imaginary quaternions yˆ1 and 49
yˆ2 which correspond to the images of y1 and y2 under τ. Now, up to conjugation, we can let yˆ1 + yˆ2 be parallel to i and, up to additional conjugation, we can rotate
yˆ1 and yˆ2 about i so that they both lie in the ik-circle. Thus, if we let α be the
(i cos α−k sin α)ψ1 (i cos α+k sin α)ψ1 angle between i and yˆ1, then τ(y1) = e and τ(y2) = e .
Note that ψ1 is half the angle between the equator-like circle which intersects xˆ1
andy ˆ2 and the equator-like circle which intersectsx ˆ3 andy ˆ2.
−1 −1 Now, since we have the relations x1y2x1 = y1 and x2y2x2 = y1 which
imply both τ(x1) and τ(x2) must conjugate τ(y2) to τ(y1), xˆ1 and xˆ2 both must lie
on the ij-circle. These relations also imply that either xˆ1 = xˆ2 or they are reflected
−1 through the jk-plane; however, one can check that the relation x3x1x3 = x2
implies that if xˆ1 = xˆ2, then τ(x1), τ(x2), and τ(x3) all commute, so xˆ1 and xˆ2
−1 must be reflected through the jk-plane. The relation x3x1x3 = x2 also tells us
that τ(x3) must conjugate τ(x1) to τ(x2), so their real part must be equivalent.
Thus, if we denote the angle between the i-axis and each element xˆ1 and xˆ2 to be
(i cos β+j sin β)ψ2 (−i cos β+j sin β)ψ2 β, then τ(x1) = e and τ(x2) = e . Now, note that ψ2
is half the angle between the equator-like circle which intersects yˆ2 and xˆ1 and the
equator-like circle which intersectsy ˆ1 andx ˆ1.
−1 Finally, the relation x3x1x3 = x2 tells us that xˆ3 must be equidistant from xˆ1 and xˆ2 since τ(x3) conjugates τ(x1) to τ(x2), thus xˆ3 is on the jk-circle. We
−1 −1 −1 −1 can also see that the relations y1x2y1 x3 and x3y2x1 y2 tell us that the real part
of τ(x3) must be equivalent to that of τ(x1) and τ(x2). Therefore if we denote γ
(j cos γ+k sin γ)ψ2 to be the angle between the j-axis andx ˆ3, then τ(x3) = e .
We also need to determine the relationship between α, β, and γ. Let us take
−1 the relation y1x2y1 = x3; this relation can be rewritten to say that y1x2 = x3y1
and since rotating xˆ2 about yˆ1 can only yield xˆ3 if the angle between xˆ2 and yˆ1 50
equals the angle betweeny ˆ1 andx ˆ3, the following relationship must hold
yˆ1 · xˆ2 =x ˆ3 · yˆ1
(i cos α − k sin α)(−i cos β + j sin β) = (j cos γ + k sin γ)(i cos α − k sin α)
− cos α cos β = − sin γ sin α
cos α cos β = sin α sin γ.
3 Therefore the homomorphism τ : π1(S \ WL) → SU(2) is given by
(i cos β+j sin β)ψ2 (i cos α−k sin α)ψ1 τ(x1) = e τ(y1) = e
(−i cos β+j sin β)ψ2 (i cos α+k sin α)ψ1 τ(x2) = e τ(y2) = e
(j cos γ+k sin γ)ψ2 τ(x3) = e with the relation cos α cos β = sin α sin γ between the angles α, β, and γ. Figure 23 shows a picture of the relative positions of the projection of the images of all the generators in SU(2) onto S2 ⊂ span {i, j, k}.
Figure 23: 51
10 The Whitehead Double of a Knot
The Whitehead double of a knot, K, is a knot which follows the general path of the embedding of S1 into S3 but in place of the circle we send two strands through the embedding which clasp together, similar to the Whitehead link in the solid torus representation of Figure 22. Figure 24 roughly illustrates the Whitehead double of the trefoil, but there is some ambiguity about the twists in the omitted section.
Figure 24:
The reason a piece of the knot was omitted in Figure 24 is that we want define what we call the untwisted whitehead double which will, surprisingly, have some twists. The name untwisted Whitehead double seems to imply we want to use what we call the blackboard parallel Whitehead double pictured in Figure 25, but the blackboard parallel Whitehead double depends on the diagram.
All of the knots pictured in Figure 26 are Whitehead doubles, but we will determine the number of twists associated with the untwisted Whitehead double, Kˆ , of K by using some of the techniques which were introduced in Sections6 and 7.
First, we need to start with the complement of the knot and the complement 52
Figure 25:
Figure 26:
of the Whitehead link. In particular, let K be the knot which we want to find the Whitehead double of. Then we want to use U = S3 \ N(K) as the complement of K and V = S3 \ N(WL) as the complement of the Whitehead link. It is clear that the boundary of U is a torus and the ‘‘outer boundary’’ of V , namely the
3 boundary of V ∪ N(L2) = S \ N(L1), of the Whitehead link is also a torus. Now, we can identify these two tori together using any homomorphism of the meridians and longitudes, but we want to use the homomorphism which doesn’t depend on the knot diagram of K we are using. In order to do this, we will use the construction from Sections6 and7. Define the generators µ0 and λ0 on ∂U and µ
3 and λ on ∂(S \ N(L1)). Figure 27 shows the knot diagrams of K and WL with K = {trefoil}. Now that we have labeled the generators of the boundary tori, we are going to ‘‘glue’’ the complement of the Whitehead link to the complement of 53
Figure 27:
the knot K by identifying µ with λ0 and λ with µ0. Note that by identifying the
generators of the boundary tori in this way, the complement of L2 will appear twisted where the longitude of the complement of K winds around the knot. More
0 specifically, if λ runs around the knot, relative to λbb, n times, the untwisted Whitehead double will have n full twists (2n half twists).
This construction will give us the embedding of the untwisted Whitehead double into S3, but we also need to be able to draw the knot diagram of Kˆ . To do this, start by drawing the knot, K. Once that is done, draw the null-homologous longitude, λ0, as seen in Section6 and then draw another longitude which is parallel to λ0. Figure 28 shows the result on the trefoil. Once the diagram has two
Figure 28:
longitudes, the knot can be erased and the two longitudes can be connected via a clasp on the knot which is sufficiently far from the twisted section, as seen on the trefoil in Figure 29.
Figure 30 shows the result of the knot diagram of the Whitehead double of the trefoil using the construction above. 54
Figure 29:
Figure 30:
Since the untwisted Whitehead double has many crossings, applying the Wirtinger presentation method from Section5 will yield a very lengthy presentation for the knot group. Instead we’ll determine the knot group using the Wirtinger presentations of the knot and the Whitehead link, and the Seifert-van Kampen theorem. As an example, we will compute the knot group of the Whitehead double of the trefoil.
Let K be the trefoil and with the construction above, let U = S3 \ N(K)
3 3 and V = S \ N(WL). Let us also define Ve = S \ N(L1) to be the solid torus without N(L2) removed (note that V = Ve \ N(L2)). The desired homomorphism φ : ∂U → ∂Ve, which we’ll use to identify the boundary tori together so the resulting quotient is homeomorphic to the complement of Kˆ , is specified by the
0 condition that the induced homomorphism on π1 is determined by φ∗(µ ) = λ and
0 φ∗(λ ) = µ. Now, we can construct a space homeomorphic to the complement of the untwisted Whitehead double of the trefoil, S3 \ Kˆ , by defining a quotient 55
U t V/ ∼ with T 0 = ∂U, T = ∂Ve such that if u ∈ T 0 and v ∈ T , then u ∼ v if and only if v = φ(u).
To be explicit, the quotient map for the above quotient space is q : U tV → U t V/ ∼ such that
0 {u}, u∈ / T {v}, v∈ / T q(u) = and q(v) = . 0 −1 {u, φ(u)}, u ∈ T {v, φ (v)}, v ∈ T
We will use the notation U = q(U) and V = q(V ). As a result, we can also use the notation T = U ∩ V such that q(T 0) = q(T ) = T . As a result, we can
determine the induced map on the fundamental groups, q∗, to be the map such
0 0 that q∗(µ ) = q∗(φ∗(λ)) = µ and q∗(λ ) = q∗(φ∗(µ)) = λ.
Now, we have the following commutative diagram of inclusions
U j1 i1
T S3 \ Kˆ (10.1)
i 2 j2 V which has an induced commutative diagram on the fundamental groups with the basepoint p in the intersection T shown below.
π1(U, p)
(i1)∗ (j1)∗
3 ˆ (10.2) π1(T , p) π1(S \ K, p)
(i2)∗ (j2)∗
π1(V , p) 56
Then by the Seifert-van Kampen theorem,
3 ˆ π1(S \ K, p) = π1(U, p) ∗φ∗ π1(V , p).
Now, we know that π1(T , p), is the free abelian group generated by µ and λ. Also, by our work in previous sections, we have the following two calculations:
3 −1 −1 −1 −1 −1 −1 π1(S \trefoil, p) = ha, b, c | aca b , cbc a , bab c i
= hx, y | x3y2i. where x = ab and y = b−1a−1b−1 and
3 −1 −1 −1 −1 π1(S \ W L, p) = hx1, x2, x3, y1, y2 | x3x1x3 x2 , y1x2y1 x3 ,
−1 −1 −1 −1 −1 −1 x3y2x1 y2 , x2y2x2 y1 , x1y2x1 y1 i.
So, the free product π1(U, p) ∗ π1(V , p) is given by the following finite presentation:
3 2 −1 −1 −1 −1 π1(U, p) ∗ π1(V , p) = hx, y, x1, x2, x3, y1, y2 | x y , x3x1x3 x2 , y1x2y1 x3 ,
−1 −1 −1 −1 −1 −1 x3y2x1 y2 , x2y2x2 y1 , x1y2x1 y1 i.
Now we need to determine the normal subgroup in the kernel of the
amalgamated free product, so we need to calculate (i1)∗ and (i2)∗. Using our 57
results from Section6, we have the following calculation
(i1)∗ : π1(T , p) → π1(U, p)
(i1)∗(µ) = a
−3 (i1)∗(λ) = caba
so we need to find the meridian and longitude in terms of the generators x and y. We made the identifications x = ab, y = b−1a−1b−1, and c = bab−1 in Section9, thus one can check that a = x2y, b = y−1x−1, and c = y−1xyxy. Hence we have the following homomorphism
(i1)∗ : π1(T , p) → π1(U, p)
2 (i1)∗(µ) = x y
−2 2 −6 (i1)∗(λ) = y (x y) .
Now, using our results from section 10, we have the following homomorphism
(i2)∗ : π1(T , p) → π1(V , p)
−1 (i2)∗(µ) = x2 x1
(i2)∗(λ) = y2.
Therefore the normal subgroup is normally generated by the two words
2 −1 x yx1 x2
and
−2 2 −6 −1 y (x y) y2 . 58
Thus by the construction of the amalgamated free product, we can calculate the knot group of the untwisted Whitehead double to be
3 ˆ 3 2 −1 −1 −1 −1 −1 −1 π1(S \ K) = hx, y, x1, x2, x3, y1, y2 | x y , x3x1x3 x2 , y1x2y1 x3 , x3y2x1 y2 ,
−1 −1 −1 −1 2 −1 x2y2x2 y1 , x1y2x1 y1 , x yx1 x2,
−2 2 −6 −1 y (x y) y2 i.
Now that we have the knot group of the untwisted Whitehead double, Kˆ , of the trefoil, K, we’ll show how the techniques in this thesis can be used to show Kˆ is not the trivial knot. To do this, we will use the homomorphisms we found in Section9, namely
σ : π1(U, p) → SU(2)
τ : π1(V , p) → SU(2)
which determine the homomorphism η : π1(U, p) ∗ π1(V , p) → SU(2) given by
π i (i cos β+j sin β)ψ2 η(x) = e 3 η(x1) = e
π (i cos ω+j sin ω) (−i cos β+j sin β)ψ2 η(y) = e 2 η(x2) = e
(i cos α−k sin α)ψ1 (j cos γ+k sin γ)ψ2 η(y1) = e η(x3) = e
(i cos α+k sin α)ψ1 η(y2) = e
π −π π with the angles ω ∈ (0, π), α ∈ (0, 2 ], β ∈ [ 2 , 2 ], and γ ∈ [0, π] and the relation cos α cos β = sin α sin γ.
We want to show that η descends to the homomorphism η˜ :(π1(U, p) ∗
π1(V , p))/N → SU(2) where N is the normal subgroup generated by the words 59
−1 −1 (σ ◦ (i1)∗(µ), (τ ◦ (i2)∗(µ)) ) and (σ ◦ (i1)∗(λ), (τ ◦ (i2)∗(λ)) ).
Using our previous results, we can calculate
σ ◦ (i1)∗ : π1(T , p) → SU(2)
2 i 2π (i cos ω+j sin ω) π σ ◦ (i1)∗(µ) = σ(x y) = e 3 e 2
−2 2 −6 i 2π (i cos ω+j sin ω) π −6 σ ◦ (i1)∗(λ) = σ(y (x y) ) = −(e 3 e 2 ) .
Let us denote the real part of σ(x2y) to be cos θ and the imaginary part to be an arbitrary unit purely imaginary quaternion uˆ. This means that cos θ =
2π π 5π − sin 3 cos ω and since 0 < ω < π, we can determine that 6 < θ < 6 . Thus the
map σ ◦ (i1)∗ will be
σ ◦ (i1)∗ : π1(T , p) → SU(2)
2 uθˆ σ ◦ (i1)∗(µ) = σ(x y) = e
−2 2 −6 uˆ(π−6θ) σ ◦ (i1)∗(λ) = σ(y (x y) ) = e
π 5π where 6 < θ < 6 . We can also calculate
τ ◦ (i2)∗ : π1(T , p) → SU(2)
−1 (−i cos β+j sin β)(−ψ2) (i cos β+j sin β)ψ2 τ ◦ (i2)∗(µ) = τ(x2 x1) = e e
(i cos α+k sin α)ψ1 τ ◦ (i2)∗(λ) = τ(y2) = e .
Now, we know that µ and λ commute in π1(T , p), τ ◦ (i2)∗(µ) and τ ◦ (i2)∗(λ) must lie in the same circle subgroup of SU(2). Thus we can label their (normalized)
−1 imaginary part Im(τ(x2 x1)) = Im(τ(y2)) = vˆ. Once we have determined that, we 60
have the following homomorphism
τ ◦ (i2)∗ : π1(T , p) → SU(2)
−1 vρˆ τ ◦ (i2)∗(µ) = τ(x2 x1) = e
vψˆ 1 τ ◦ (i2)∗(λ) = τ(y2) = e
−1 −1 for some angle ρ = cos (Re(τ(x2 x1))).
Note that the unit purely imaginary quaternions uˆ and vˆ do not agree, but we can conjugate the homomorphism σ by a quaternion which rotates σ ◦ (i1)∗(µ)
and σ ◦ (i1)∗(λ) into the same circle subgroup as τ ◦ (i2)∗(µ) and τ ◦ (i2)∗(λ). After this change, we can assume that uˆ = vˆ and therefore we can compute the images of the normal generators for N are the quaternions
evθˆ evˆ(−ρ)
and
evˆ(π−6θ)evˆ(−ψ1).
The normal generators map to the identity if and only if the pair of angles
(θ, π − 6θ) agree, mod 2π with the pair (−ρ, ψ1). To compare these pairs, we’ll plot
−1 −1 the angle cos (Re(σ ◦ (i1)∗(µ))) vertically and cos (Re(σ ◦ (i1)∗(λ))) horizontally.
1 the pair (θ, π − 6θ) describes a line with a slope of − 6 as illustrated in Figure 31.
In [5], Klassen shows how to construct the angle pairs of the restriction
η˜(τ ◦ (i2)∗) as seen in Figure 32 where the homomorphisms are everything in the shaded region.
Thus, when we intersect the lines in Figure 31 with the shaded region in 61
Figure 31:
Figure 32:
Figure 32, we have a picture of the points which are hit by the images of both restrictions Thus, part of the image of η˜ :(π1(U, p) ∗ π1(V , p))/N → SU(2) can be represented by the torus in Figure 33. Therefore, there are at least two lines of
3 ˆ nonabelian homomorphisms of π1(S \ K) into SU(2) under the map η˜. Hence the 62
Figure 33:
knot group of the Whitehead double of the trefoil is not equivalent to the knot group of the unknot, so by Theorem 5.4, the Whitehead double of the trefoil is not ambient isotopic to the unknot. 63
References
[1] S. Akbulut, J. McCarthy, Casson’s invariant for oriented homology 3-spheres: an exposition, Princeton University Press, Princeton 1990.
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[3] A. Hatcher, Algebraic topology, Cambridge University Press, 2002.
[4] E. Klassen, Representations of knot groups in SU(2), Trans. Amer. Math. Soc., 326 (1991), 795-828.
[5] E. Klassen, Representations in SU(2) of the fundamental groups of the White- head link and of doubled knots, Forum Math, 5 (1993), 93-109.
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[8] D. Rolfsen, Knots and links, Publish or Perish, Berkeley, CA, 1976.
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