Lecture 12: Prime Factorization of Knots
Notes by Zach Haney
February 25, 2016
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Introduction
In this lecture, we look at prime knots and “factoring" knots into a connected sum
of prime knots. Our goal for prime factorization is to see a connection to the integers:
unique factorization!
We recall the main definition here:
Definition 1.1. A nontrivial knot K is prime precisely when K = K1 #K2 implies
either K1 or K2 is trivial.
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Prime Factorization
Theorem 2.1. For any two knots K1 and K2 , g(K1 #K2 ) = g(K1 ) + g(K2 ).
Proof:
By the boundary-connected sum, we know that g(K1 #K2 ) ≤ g(K1 ) + g(K2 ). To
the equality, we must show the reverse inequality is true.
Let Σtot be a minimal genus Seifert surface for K1 #K2 . The strategy will be to
produce from Σtot another Seifert surface, Σ∗tot of minimal genus such that
Σ∗tot = Σ1 #Σ2
is a connected sum of Seifert surfaces for K1 and K2 . As seen in the figure:
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That is, we’ll see that g(K1 ) + g(K2 ) ≤ g(Σ1 ) + g(Σ2 ) = g(Σ∗tot ) = g(Σtot ), and
we’ll know that g(Σtot ) = g(K1 #K2 ).
So our main goal from above is to construct Σ∗tot .
To recognize the connected sum, we need to see the separating sphere, as shown
in the figure above, which intersects, without loss of generality, the knot transversally
in a single pair of points and the Seifert surface in an arc joining those points.
Our main issue to confront is that the minimal genus Seifert surface Σtot for
K1 #K2 may NOT represent a connected sum decomposition. We’ll see, of course,
that it in fact does using surgery.
So consider the PL sphere S 2 ⊂ S 3 which splits K1 #K2 into its summands. As
before, without loss, we know that Σtot intersects S 2 transversally, so that Σtot ∩ S 2 is
a compact 1-manifold with with exactly two boundary points where K1 #K2 pierces
the separating 2-sphere.
If Σtot ∩ S 2 is a single arc α as seen in the figure below, then this arc separates
Σtot into two components, namely Σ1 #Σ2 , bounding K1 and K2 individually. So we
are done.
Otherwise, Σtot ∩ S 2 consists of precisely one arc α and a finite number of disjoint,
simple closed curves. To remove the closed curves inductively, it suffices to construct
∗∗
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another Seifert surface Σ∗∗
tot for K1 #K2 of the same genus but with Σtot ∩S ⊂ Σtot ∩S .
Let C ⊂ Σtot ∩ S 2 be a closed curve which is “innermost” on S 2 \ α precisely when
the interior of the disk D2 ⊂ S 2 \ α, bounded by C, is disjoint from Σtot .
Perform surgery along C ⊂ Σtot , removing an annular neighborhood of C in Σtot
and capping the boundary circles by disks,
a shown in the figure
Claim: C separates Σtot (and must bound a disk D2 ⊂ Σtot ), so the new surface
F obtained by surgery is disconnected.
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Why is that true? Well, otherwise, F would b a new Seifert surface for K1 #K2 with
a strictly smaller genus as in the figure, and that’s not possible from our assumptions!
Let Σ∗∗
tot ⊂ F be the component bounding K1 #K2 . Surgery cannot increase the
∗∗
genus, so g(Σ∗∗
tot ) = g(Σtot ), and Σtot is a Seifert surface for K1 #K2 whose intersection
with S 2 has strictly fewer components than Σtot did.
Proposition 2.2. If K1 is nontrivial, so is the connected sum K1 #K2 for any other
knot K2 . If K1 ' K1 #K2 , then K2 must be trivial.
We’ve seen this before! This is the third proof, and in particular doesn’t involve
any wild knots such as those used by Mazur.
Proof:
If K1 #K2 is trivial, then g(K1 #K2 ) = 0 which means g(K1 ) = 0 = g(K2 ). Also,
g(K1 ) = g(K1 #K2 ) says that g(K2 ) = 0.
Corollary 2.3. A knot of genus 1 is prime. In fact, any nontrivial knot is a finite
sum of prime knots.
Proof:
If K is not prime, K = K1 #K2 for two nontrivial knots each of which have smaller
genus. Induct on the genus, and one works down to zero.
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Theorem 2.4. Suppose a knot K can be decomposed as K = P #Q, with P prime,
and also as K = K1 #K2 . Then either
a) K1 = P #K1∗ and Q = K1∗ #K2 for some K1∗ or
b) K2 = P #K2∗ and Q = K1 #K2∗ for some K2∗ .
Proof:
Let Σ be a separating S 2 , intersecting K transversally at two points for the decomposition K = K1 #K2 , as seen for example in the figure.
Simillarly, the decomposition K = P #Q implies that there is some 3-ball B ⊂ S 3
such that B ∩ K is an arc α, intersecting ∂B transversally at two points, so that the
ball-arc pair (B, α) becomes, after gluing the trivial ball-arc pair to its boundary, the
pair (S 3 , P ).
We recall that a ball-arc pair is a 3-ball with a tangle inside, intersecting its
boundary at two points; moreover, the trivial ball-arc pair is the 3-ball plus the
unknotted diameter inside.
Without loss, Σ intersects ∂B transversally in a union of simple, closed curves
disjoint from K. If Σ ∩ ∂B = ∅, then B is contained in one of the components of
S 3 \ Σ, and the result follows immediately. The figure here gives these situations.
Our goal is to remove curves in Σ ∩ ∂B via surgery on B.
As Σ ∩ K is two points, any oriented simple, closed curve in Σ \ K has linking
number 0 or ±1 with K. (Why ±1? The curves would be meridians of K!).
(i) Eliminate components of linking number zero of Σ ∩ ∂B. Select innermost
curve in Σ ∩ ∂B which does not link K. This curve bounds a disk D ⊂ Σ, with
D ∩ ∂B = ∂D. The curve ∂D must also bound a disk D0 ⊂ ∂B which does NOT
intersect K. That is, D0 ∩ K = ∅, since ∂D andd K are unlinked. Look to this figure
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below to see visually what is happening in this case
By Schoenflies, the sphere D ∪ D0 bounds a 3-ball. We “move” D0 across ∂B to
the other side of D by adding or subtracting a 3-ball to/from B, disjoint from K, as
seen in the figure above.
The new pair (B, α) corresponds to (S 3 , P ) as before, but Σ ∩ ∂B has fewer
components. Repeating this, we may assume all curves in Σ∩∂B have linking number
±1 with K, as see in this figure
(ii) Eliminate components of linking number 1 of Σ ∩ ∂B.
a) If Σ ∩ B has a component which is a disk D00 , then D00 intersects K in one point.
Because P is prime, one side of D00 in B is a trivial ball-arc pair.
Subtract from B a regular neighborhood of this trivial ball-arc pair to produce a
new (B, α) with fewer components than Σ ∩ B. Repeat until all components of Σ ∩ B
are annuli.
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b) Let A be an annular component of Σ ∩ B. Then ∂A also bounds an annulus
A in ∂B. Choose A “outermost”, i..e farthest from α, so that A0 ∩ Σ = ∂A0 .
, Then A ∪ A0 ' T 2 is a torus which bounds a solid region M ⊂ B. Let δ be the
disk which is one of the components of ∂B \ A0 . Then ∂δ is a circle of ∂A0 , and δ
intersects K at one point.
We “thicken" δ to a regular neighborhood Nδ in B \ M . Then (Nδ , Nδ ∩ α) is a
trivial ball-arc pair. The union M ∪ Nδ must be a ball, because its boundary is a
sphere. (Why? Schoenflies!)
Hence, (M ∪ Nδ , Nδ ∩ α) is a ball-arc pair as well. As P is prime, either (M ∪
Nδ , Nδ ∩ α) is trivial (CASE 1) or (M ∪ Nδ , Nδ ∩ α) is a copy of (B, α) itself (CASE
2).
Looking at case 1, we see M is a solid torus. Do surgery in B as before to remove
the annulus A from Σ ∩ ∂B. Repeat on the rest of the annuli.
In case two, M is the exterior of α in B, with meridian ∂δ.
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Case 1
Case 2
Without loss, Σ bounds a ball U ⊂ S 3 containing both the summand K1 and M,
as well as the boundary torus ∂M .
This torus contains K1 and is peripheral to P. This means P must be a summand
of K1 = P #K1∗ for some K1∗ . The same torus in S 3 contains K1∗ #K2 which is then
Q = K1∗ #K2 , as seen in the following figure.
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