Stable Commutator Length and Knot Complements
Tim Susse
CUNY Graduate Center
AMS Eastern Sectionals
October 12, 2013
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
1 / 10
Definitions
Let G be a group, and g ∈ [G, G].
cl(g) is the minimal number of commutators whose product is g.
Let G = F2 = ha, bi, then
cl([a, b]) = 1, cl([a, b]2 ) = 2, but cl([a, b]3 ) = 2
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
2 / 10
Definitions
Let G be a group, and g ∈ [G, G].
cl(g) is the minimal number of commutators whose product is g.
Let G = F2 = ha, bi, then
cl([a, b]) = 1, cl([a, b]2 ) = 2, but cl([a, b]3 ) = 2
(Culler, 1981) [a, b]3 = [aba−1 , b−1 aba−2 ][b−1 ab, b2 ]. So, cl is not
homogeneous with respect to taking powers.
Definition.
cl(g n )
. Since n 7→ cl(g n ) is subadditive, this limit always
n→∞
n
scl(g) = lim
exists.
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
2 / 10
Definitions
Theorem.
Let X be a topological space with π1 (X ) = G. Then for any g ∈ G
−χ− (S)
n(S)
: |∂S| = 1, f : S → X , f∗ ([∂S]) = g
scl(g) = inf
,
2n(S)
where χ− (S) the sum of the Euler characteristics of all components of
S with non-positive Euler characteristic.
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
3 / 10
Definitions
Theorem.
Let X be a topological space with π1 (X ) = G. Then for any g ∈ G
−χ− (S)
n(S)
: |∂S| = 1, f : S → X , f∗ ([∂S]) = g
scl(g) = inf
,
2n(S)
where χ− (S) the sum of the Euler characteristics of all components of
S with non-positive Euler characteristic.
We use this reformulation of stable commutator length, allowing S to
have multiple boundary components, to extend the definition to
1-boundaries of G. For convenience, we consider B1 (G; R) the set of
all real one-boundaries of G.
Proposition.
.
scl is a pseudonorm on B1H (G; R) = B1 (G; R) g n − ng, hgh−1 − g .
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
3 / 10
PQL Groups
Theorem. (Calegari, 2009)
Stable commutator length is a piecewise rational linear norm on
B1H (Fr ; R) for r ≥ 2. Further, there is an efficient algorithm to compute
scl on rational chains ([Walker, 2013])
Such a group is called PQL.
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
4 / 10
PQL Groups
Theorem. (Calegari, 2009)
Stable commutator length is a piecewise rational linear norm on
B1H (Fr ; R) for r ≥ 2. Further, there is an efficient algorithm to compute
scl on rational chains ([Walker, 2013])
Such a group is called PQL.
Theorem. (Calegari, 2011)
Let {Ai } be a collection of Abelian groups. Then ∗Ai is PQL and there
exists an algorithm to compute scl on rational chains.
Conjecture. (Calegari)
Let M be a 3-manifold. Then π1 (M) is PQL.
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
4 / 10
Results
Theorem. (Susse, 2013)
Let A and B be two free Abelian groups of rank at least k , then A ∗Zk B
is PQL. Further, there is an algorithm to compute scl on rational chains.
Theorem. (Susse, 2013)
Let {Ai } be a collection for free Abelian groups of rank at least k . Then
∗Zk Ai , their free product amalgamated over a single shared subgroup
is PQL. Further, there is an algorithm to compute scl on rational chains.
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
5 / 10
Results
Theorem. (Susse, 2013)
Let A and B be two free Abelian groups of rank at least k , then A ∗Zk B
is PQL. Further, there is an algorithm to compute scl on rational chains.
Theorem. (Susse, 2013)
Let {Ai } be a collection for free Abelian groups of rank at least k . Then
∗Zk Ai , their free product amalgamated over a single shared subgroup
is PQL. Further, there is an algorithm to compute scl on rational chains.
Note that if A = B = Z, then
A ∗Z B = a, b : ap = bq .
If (p, q) = 1, then G = π1 (S 3 \ K ), where K is a torus knot.
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
5 / 10
The Set-up for Knot Complements
Let T be the boundary torus of the genus-1 splitting of S 3 , and let
K be embedded on T .
Let f : S → S 3 \ K be a map of a surface with (possible empty)
boundary.
Homotope f so that it is transverse to A = T \ K , which is an
annulus.
Consider f −1 (A), which is a properly embedded one-submanifold
of S (i.e., a disjoint collection of arc with end points on ∂S and
simple closed curves).
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
6 / 10
Parameterizing Surfaces
WLOG, components of S \ f −1 (A) are planar surfaces with
boundary, whose boundaries are made of arcs alternating
between:
Arcs from ∂S (τ -edges);
Arcs from f −1 (A) (σ-edges)
Some boundary components may be loops from f −1 (A) (σ-loops)
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
7 / 10
Parameterizing Surfaces
WLOG, components of S \ f −1 (A) are planar surfaces with
boundary, whose boundaries are made of arcs alternating
between:
Arcs from ∂S (τ -edges);
Arcs from f −1 (A) (σ-edges)
Some boundary components may be loops from f −1 (A) (σ-loops)
Using the σ- and τ -edges we can parameterize all of the surfaces
S → S3 \ K .
We do need to control contributions of σ-loops though.
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
7 / 10
Waldhausen’s Theorem
Theorem. (Waldhausen, 1967)
Let K be a torus knot, then the only embedded incompressible surface
in S 3 \ K is a boundary parallel torus
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
8 / 10
Waldhausen’s Theorem
Theorem. (Waldhausen, 1967)
Let K be a torus knot, then the only embedded incompressible surface
in S 3 \ K is a boundary parallel torus
New Proof.
Consider f : S → S 3 \ K , an incompressible surface map with S
closed. Isotope so that f is transverse to A and has the minimal
number of components of intersections with A. Since S is closed,
f −1 (A) consists only of σ-loops. Further, since S is incompressible, the
components of S \ f −1 (A) are planar surfaces and can only have one
or two boundary components – i.e., are discs or annuli.
By assumption, we can assume that none a discs. So S is formed as a
union of annuli, so χ(S) = 0 and S is a torus.
To see it is boundary parallel we look at the following diagram.
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
8 / 10
Waldhausen’s Theorem
Theorem. (Waldhausen, 1967)
Let K be a torus knot, then the only embedded incompressible surface
in S 3 \ K is a boundary parallel torus
New Proof.
Self-intersection
K
T
f (S) ∩ A
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
8 / 10
Outline of the proof
Let ν ∈ B1H (G; Z)
1
Control σ-loops by investigating gluing equations and winding
numbers.
2
Given a vector which can parameterize a surface (in one of the
two handle bodies), estimate the Euler characteristic in a
piecewise rational linear way.
3
Given vectors parameterizing surfaces which are compatible,
ensure that there is a way to glue them, after considering winding
numbers – do this by adding loops.
4
Show that the Euler characteristic estimate from the second step
is still good.
5
Maximize the Euler characteristic over a rational polyhedron
parameterizing all surfaces with boundary (a multiple of) ν using
linear programming.
Since we are looking at a piecewise rational linear function over a
rational polyhedron, the maximum is rational.
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
9 / 10
Step 1: Winding Numbers and Gluing Equations
Take a component of S \ f −1 (C). Attached to each σ-edge is an
integer, giving the class of the edge in H1 (A) ∼
= Z.
4
a
−1
−1
a4
a4
0
Figure : A disc in the (6, n) torus knot complement
For every component, the sum of the τ -edges and labels of the
σ-edges must be 0 in homology. Here, the τ -edges contribute 12a and
each −1 on the σ-edges contributed −6a.
Tim Susse CUNY Graduate Center
SCL and Knot Complements
October 12, 2013
10 / 10