Lecture 10: Proof of the Loop Theorem II
Notes by Nate Bade
February 16, 2016
In this lecture, we will finish up the proof of the Loop Theorem from last time.
We will then prove a pair of result due to Papa: first that all higher homotopy groups
of the a knot complement XK = S 3 \K vanish, and second that π1 (XK ) is torsion
free.
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Proof of the Loop Theorem - Continued
. Let us briefly recall what we are trying to prove
Theorem 1 (Loop Theorem). Let M be a 3-manifold and let B ⊂ ∂M be a surface
contained in the boundary. Suppose a normal subgroup N E π1 (B) does not contain
the kernel ι∗ : π1 (B) → π1 (M ). Then there exists a piecewise linear embedding
g : (D2 , ∂D2 ) → (M, B)
such that
[g(∂D2 )] ∈ π1 (B)
is not contained in N
Proof. (Continued)
Last time, we showed that the kernel of ι∗ : π1 (B) → π1 (M ) provides a simplicial
map f0 : (D2 , ∂D2 ) → (M, B) such that [f (∂D2 )] is not in N E π1 (B). Let K0 =
f0 (D2 ) ⊂ M and let V0 be a neighborhood of K0 that deformation retracts onto K0 .
f
0
D2 −−→
K0 ,→ V0 ,→ M
(1.1)
Suppose V0 admits a nontrivial connected double cover p1 : M1 → V0 . Then f0 lifts to
a new map f1 : (D2 , ∂D2 ) → (M1 , B1 ). Set K1 = f1 (D2 ) and pick a neighborhood V1
that deformation retracts to K1 . As we showed last time, if we repeat this construction
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each time Vi has a double cover, the tower of we form is finite.
/ Vn 

KI n Mn
pn
.. z
.
..
.
..
.
p2
fn

K0 
z
/ V1 
= K1
/
f0
/
M1
p1
f1
D0
/
y
/ V0 
/
M
We will now show that ∂Vn = S 2 t . . . t S 2 is a disjoint union of spheres.
Lemma 1.1. Let M be a compact 3-manifold with boundary which does not admit a
nontrivial connected double cover. Then each component of ∂M is a two-sphere.
Proof. Each nontrivial double cover of M corresponds to a nontrivial homomorphism π1 (M ) → Z2 . Any such homomorphism must factor through the abelianization
H1 (M ; Z2 ). Double covers of M are then parametrized by
Hom(H1 (M ; Z2 ), Z2 ) ∼
= H 1 (M ; Z2 ) .
If M does not admit any double covers, H 1 (M ; Z2 ) = 0. But by Poincare duality
H 1 (M ; Z2 ) ∼
= H2 (M, ∂M ; Z2 ) ,
so the exact sequence in homology
...
/
/ H2 (M, ∂M ; Z2 )
/
H1 (∂M ; Z2 )
H1 (M ; Z2 )
/ ...
becomes
...
/
0
/
H1 (∂M ; Z2 )
/
0
/ ...
.
Since H1 (∂M ; Z2 ) = 0, ∂M has no higher genus components.
Recall that we have constructed the map fi : (D2 , ∂D2 ) → (Mi , Bi ) so that
[fi (∂D2 )] is not contained in Ni E π1 (Bi ) for any i. Since ∂Vn = tSj2 , the boundary
surface Bn must be planar, which is to say a disk with h holes. This is simply the
observation that any subsurface of S 2 must be planar.
Since Bn is planar, it’s fundamental group π1 (Bn ) = Fh is the free group on h
generators, and at least on generator must not be contained in Nn . We will now
2
Bn
gn (D2 )
Figure 1: The planar surface Bn with two generators of the fundamental group shown.
proceed to push this generator down the tower.
Let gn : (D2 , ∂D2 ) → (Vn , Bn ) be a smooth embedding of a disk, such that the
boundary gn (∂D2 ) is identified with the boundary of Bn not contained in N . We
make sure to push the interior of gn (D2 ) into Vn to ensure properness.
0
Consider the covering projection gn−1
= pn ◦ gn : (D2 , ∂D2 ) → (Mn−1 , Bn−1 ), and
0
0
perturb gn−1
to be in general position. Then gn−1
is an immersion, with the only
0
is a
singularities being arcs or closed loops of double points. In particular, since gn−1
2-1 covering map there are no triple branch points.
0
can be surgered to an embedding gn−1 : (D2 , ∂D2 ) →
Claim 1.2. The map gn−1
(Mn−1 , Bn−1 ) such that the image [gn−1 (∂D2 )] is not an element of Nn−1 . After
proving this claim we will simply induct downward to produce g ≡ g0 .
We will produce the embedding gn−1 by performing a series of local surgeries on
each of the double point singularities. There are two cases to consider, the loop of
double points and the and the arc of double points. We will consider these cases in
order.
Loop of double points Consider the loop of double points whose preimage is a
pair of disjoint, non-nested loops in a disk.
Note that the left hand side is a rotational cross section. Such a singularity can
0
be resolved by replacing gn−1
with the map that swaps the disks A and B, smoothing
as necessary. Note that this is only a local alteration, it does not change the map
away from the disks A and B.
Alternatively, the preimage of the singular set could be a nested set of loops in a
disk. We can resolve this singularity by cutting out an annulus of A and sewing back
in a disk B.
This exhausts the possibilities for singularities coming from a loop of double points.
3
D2
S2
S2
A
B
A
0
gn−1
B
Figure 2: Non-nested Loop Singularity
A
gn−1
D2
A
B
B
Figure 3: Resolution of Non-nested Loop Singularity
S2
A
A
gn−1
D2
A
B
B
S2
Figure 4: Nested Loop Singularity
An arc of double points Since there are no triple branch points arcs of double
points must end on the boundary ∂D2 . There are then only two possible arcs of
−1
double points under gn−1
:
Both of these cusp singularities can be resolved in one of two possible ways, either
as
(+)
(−)
Regardless, it will always be true that either [gn−1 (∂D2 )] ∈
/ Nn−1 or that [gn−1 (∂D2 )] ∈
/
2
Nn−1 . This follows by rewriting the boundary ∂D in one of two way. For case (a),
4
0
gn−1
D2
A
B
}
Omit
B
Figure 5: Resolution of Nested Loop Singularity
β
β
S2
γ
γ
α
S2
S2
α
S2
δ
δ
Case (a)
Case (b)
−→
or
0
gn−1
(+)
(−)
gn−1
gn−1
we write
)]
(αγ)
=
n−1 (∂D
2
δ −1 (αβ −1 γδ −1 )−1
=



[g 0
(αγ)
=
=
αβγδ
=
Case (a)





(+)
[gn−1 (∂D2 )]
(−)
[gn−1 (∂D2 )]
(+)
[gn−1 (∂D2 )]
5
δ
.
For case (b) we can similarly write
)]
(αγ −1 )−1
(αδγβ)
=
n−1 (∂D
2
(γδ)−1
=



[g 0
(αγ −1 )
=
=
αβγδ
=
Case (b)





(+)
[gn−1 (∂D2 )]
(+)
[gn−1 (∂D2 )]
(−)
[gn−1 (∂D2 )]
(γδ)
.
(±)
0
Therefore, if both [gn−1 (∂D2 )] were members of Nn−1 , the class [gn−1
(∂D2 )] would also
0
be contained in Nn−1 . But this contradicts our assumptions on gn−1
. Therefore the
0
map gn−1
can be resolved to a map gn−1 , in such a way as preserves the requirement
that the image of the boundary [gn−1 (∂D2 )] does not lie in the normal subgroup Nn−1
downstairs.
This completes the proof of the Loop Theorem.
We will now discuss some result about the homology groups of the knot
complements.
2
Homology groups of the Knot Complement
Recall the Sphere Theorem:
Theorem 2 (Sphere Theorem). Let M be a three-manifold and let N ( π2 (M ) be a
proper, π1 (M )-invariant subgroup. Then there is a piece-wise linear embedding
g : S 2 → M such that the homotopy class [g] is not a member of N .
For N = {1}, the Sphere Theorem implies that if π2 (M ) 6= 0, then M contains an
embedded, noncontractible S 2 . The Sphere Theorem immediately implies the
asphericity of the knot complement.
Theorem 3 (Papa, 1957). Let XK = S 3 \VK be a knot exterior. Then (a)
πn (XK ) = 0 for all n ≥ 2 and (b) Γk := π1 (XK ) is torsion free.
Proof. (a) Assume that π2 (XK ) 6= 0. Then there exists a piecewise linear sphere
S 2 ⊂ XK which is not contractible. By the Schoenflies Theorem, this sphere divides
S 3 into a pair of three-balls B1 and B2 . One of them, say B1 , contains the knot K,
and so B2 ∈ XK . But then S 2 = ∂B2 would be contractible in XK . Therefore
π2 (XK ) = 0.
f be the universal cover of X and recall that
For higher πn (XK ), let X
K
K
f
f
πn (XK ) = πn (XK ). Since ΓK has infinite order, XK is a non-compact
three-manifold. However, if M is any non-compact n-manifold, then Hi (M ) = 0 for
f ) = 0 for i ≥ 3.
i ≥ n, so in particular Hi (X
K
6
f (X) = 0 for all
The Hurewicz Theorem says that if πi (X) = 0 for 1 ≤ i ≤ n, then H
i
f
i < n, and that πn (X) = Hn (X). Applying the Hurewicz Theorem to the universal
f ) = 0 for all n ≥ 2. Note that
cover of the knot complement gives us πn (Xk ) = πn (X
k
this implies that Xk = K(ΓK , 1) is the Eilenberg-Maclaine space for the knot group.
(b) Assume that g ∈ ΓK has finite order, it that g m = 1 for some m > 1. Then g
generates a cyclic subgroup Zm ∈ ΓK , with an associated cover YK → XK , such that
π1 (YK ) = Zm .
By part (a), the higher homotopy groups of YK vanish πn (YK ) = 0 for n > 1. Hence
YK = K(Zm , 1) is an Eilenberg-Maclaine space for the finite group Zm . We recall
here a fact about the Eilenberg-Maclaine spaces for cyclic groups:
f (K(Z , 1)) =
H
i
m


Zm
for i odd

0
for i even
But since YK has homotopy groups in degree i > 3, it cannot be a three-manifold.
Therefore ΓK contains no elements of finite order.
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