Factor Groups of Knots and LOTs - Boise State UniversityFactor Groups of Knots and LOTs Renata...

Factor Groups of Knots and LOTs

Renata Gerecke, Dr. Jens Harlander, Ryan Manheimer,Bryan Oakley, Sifat Rahman

Department of MathematicsBoise State University

July 31, 2013

Introduction Tiles The Theorem Idea of Proof

1 Introduction

2 Tiles

3 The Theorem

4 Idea of Proof


Definition

Group: (G, ·)where G is a set and · is a binary operation. The followingproperties hold:

Identity

Inverses

Associativity

Closure

Groups often arise when symmetry is present (algebraicsymmetry, geometric symmetry, etc)


Definition

Group: (G, ·)where G is a set and · is a binary operation. The followingproperties hold:

Identity

Inverses

Associativity

Closure

Groups often arise when symmetry is present (algebraicsymmetry, geometric symmetry, etc)


Definition

Group Presentation:

< x1, x2, x3, . . . |r1 = 1, r2 = 1, · · · >

Ex: < x1, x2|x1x2 = x2x1 >Elements are words written using the “alphabet” x1

±1 andx2

±1.Ex: x1x2x1

−1x2x2x1x2The operation is concatenation.Ex. x1x2x1x2

−1 · x2x1x2 = x1x2x1x2−1x2x1x2 = x1x2x1x1x2

= x13x2

2.

Question

How can we tell if the group generated from group presentationis finite or infinite?


Definition

Group Presentation:

< x1, x2, x3, . . . |r1 = 1, r2 = 1, · · · >


±1 andx2

±1.Ex: x1x2x1



= x13x2

2.

Question



Definition

Group Presentation:

< x1, x2, x3, . . . |r1 = 1, r2 = 1, · · · >


±1 andx2

±1.Ex: x1x2x1



= x13x2

2.

Question



Definition

Group Presentation:

< x1, x2, x3, . . . |r1 = 1, r2 = 1, · · · >


±1 andx2

±1.Ex: x1x2x1



= x13x2

2.

Question



Definition

Group Presentation:

< x1, x2, x3, . . . |r1 = 1, r2 = 1, · · · >


±1 andx2

±1.Ex: x1x2x1



= x13x2

2.

Question



Definition

Group Presentation:

< x1, x2, x3, . . . |r1 = 1, r2 = 1, · · · >


±1 andx2

±1.Ex: x1x2x1



= x13x2

2.

Question



Definition

Group Presentation:

< x1, x2, x3, . . . |r1 = 1, r2 = 1, · · · >


±1 andx2

±1.Ex: x1x2x1



= x13x2

2.

Question



Question


We use Geometry to figure out if a group is finite or infinite.

Definition

Tiles: < x1, x2|x12 = 1, x22 = 1, x1x2 = x2x1 >

x1 x2 x1x1 x2

x2

x2

x1


Question


We use Geometry to figure out if a group is finite or infinite.

Definition

Tiles: < x1, x2|x12 = 1, x22 = 1, x1x2 = x2x1 >

x1 x2 x1x1 x2

x2

x2

x1


Question

Can we tile a surface with these tiles in an “essential way?”

x1

x1x1

x2

x2

x2

x1

x2

x1

x1

Theorem (Huebschmann 1979)

If one cannot tile a 2-sphere in an “essential way” using tilesderived from a group presentation, then the group defined bythat presentation is either Zn or infinite.


Definition (Label Oriented Tree)

A Label Oriented Tree (LOT) is a directed, cycle-free graph withvertices labelled [x1, ..., xn] and with edges labelled by a vertexa.

aJ. Harlander and S. Rosebrock, 2013

LOTs are one method of writing down a group presentation inwhich all of the elements are conjugate.


Definition (Label Oriented Tree)

A Label Oriented Tree (LOT) is a directed, cycle-free graph withvertices labelled [x1, ..., xn] and with edges labelled by a vertexa.

aJ. Harlander and S. Rosebrock, 2013

LOTs are one method of writing down a group presentation inwhich all of the elements are conjugate.


For example:

This LOT represents a group with five generators whoserelations include x1 · x4 = x4 · x2, x3 · x1 = x1 · x2,x4 · x2 = x2 · x3, x4 · x3 = x3 · x5.


For example:



For example:



For example:



For example:



Note:

A group with n generators will have n− 1 relations, as aninterval with n vertices will have n− 1 edges.

LOT presentations are generalizations of Wirtingerpresentations of fundamental knot groups.


Note:




Note:




For example:


For example:


For example:


For example:


For example:


Each intersection of the knot has an associated tile–andtherefore, relation–which contributes to our knowledge of thegroup. The trefoil knot, as drawn previously, is associated withthis LOT:


Definition (Q-Series)

Given a LOT T with presentation P (T ) we may find Qk(T ) byadding the relation xk1 = 1. The sequence of groupsQ2(T ), Q3(T ), ... is called the Q-Series of that presentation.


Theorem

If a presentation P (T ) comes from a knot, then the Q-Series isa knot invariant.

Proof.

If we look at each of the Reidemeister moves, we can see thatthey have no affect on the presentation of the knot, and thus donot affect the Q-Series.


Theorem

If a presentation P (T ) comes from a knot, then the Q-Series isa knot invariant.

Proof.

If we look at each of the Reidemeister moves, we can see thatthey have no affect on the presentation of the knot, and thus donot affect the Q-Series.


For the Trefoil Group, T =< x1, x2 |x1 · x2 · x1 = x2 · x1 · x2 >

Q2 = S3

Q3 = SL23

Q4 = SL23 n Z4

Q5 = SL25× Z5

|Q≥6| =∞

This result was proved by Coxeter. Our work is an attempt togeneralize his results.


Cardinality of the Q-Series

While in the trefoil case we see that Qk(T ) is finite for k ≤ 5 weexpect that generically Qk(T ) will be an infinite group. This isbecause Qk(T ) has as many relations as it does generators, andgenerally finite groups have far more relations than generators.


Cardinality of the Q-Series

While in the trefoil case we see that Qk(T ) is finite for k ≤ 5 weexpect that generically Qk(T ) will be an infinite group. This isbecause Qk(T ) has as many relations as it does generators, andgenerally finite groups have far more relations than generators.


So when is Qk(T ) Infinite?

Theorem

If T is a LOT such that P (T ) is a non-positively curved squarepresentation then Qk(T ) is infinite for appropriately chosen k.


So when is Qk(T ) Infinite?

Theorem



Non-Positive Square Complex

Definition

A non-positive square presentation is one in which:

1 All the tiles are squares.

2 All essential tilings require 4 or more squares around everyvertex



Definition






Definition






Definition





So none of this stuff!


Or this!


Okay...but what is “appropriately chosen” k supposedto mean?

Theorem


Bryan will sketch how can show that the theorem holds for allodd k ≥ 5.


Okay...but what is “appropriately chosen” k supposedto mean?

Theorem


Bryan will sketch how can show that the theorem holds for allodd k ≥ 5.


But in fact we can do better!

If the the theorem holds for odd k ≥ 5 it follows therefore thatthe result must hold for all

k = pe11 · · · penn

such that ∃pi ≥ 5. This is because all prime numbers other than2 are odd and therefore the theorem must hold for them. Wecan then consider the map

ϕ : Qk(T ) −→ Qpi(T )

xi 7→ xi




k = pe11 · · · penn


ϕ : Qk(T ) −→ Qpi(T )

xi 7→ xi




k = pe11 · · · penn


ϕ : Qk(T ) −→ Qpi(T )

xi 7→ xi




k = pe11 · · · penn


ϕ : Qk(T ) −→ Qpi(T )

xi 7→ xi


All the relations in Qk(T ) are the same as the relations inQpi(T ) except for one. In Qk(T ) we have xk1 = 1 while inQpi(T ) we have xpi1 = 1. But since pi|k it follows that xk1 = 1 inQpi(T ). Therefore all the relations holding in Qk(T ) also holdin Qpi(T ) making ϕ a surjective group homomorphism. SinceQpi(T ) is infinte so to is Qk(T ). Thus, this will hold true for allnumbers that are not 2i3j where i ≥ 0 and j = {0, 1}.










Idea of Proof of Theorem

To any LOT T, we associate a presentationP (T ) =< x1, . . . , xn|{re = 1} >, as mentioned before by Renata.Let this presentation be a non-positively curved squarepresentation.Now, to examine the Q Series, we need to considerPk(T ) =< x1, . . . , xn|{re = 1}, xk1 = 1 >However, we want to work withP̄k(T ) =< x1, . . . , xn|{re = 1}, xk1 = 1, . . . , xkn = 1 >The only difference between Pk(T ) and P̄k(T ) is that in thelatter we are allowed to use k-gons associated to xk2 = 1 all theway up to xkn = 1.From Pk(T ) to P̄k(T ), we have changed the presentation butnot the group. Thus, any result we get for the latter must holdfor the former.

















Now, in P̄k(T ) =< x1, . . . , xn|{re = 1}, xk1 = 1, . . . , xkn = 1 >,consider each of the relations re = 1.Each of these relations corresponds to a non-trivial tiling of the2-sphere in the following way:

Note that for each edge, we call the corresponding 2-sphere ame-sphere.


Now, in P̄k(T ) =< x1, . . . , xn|{re = 1}, xk1 = 1, . . . , xkn = 1 >,consider each of the relations re = 1.Each of these relations corresponds to a non-trivial tiling of the2-sphere in the following way:

Note that for each edge, we call the corresponding 2-sphere ame-sphere.


Given any non-empty spherical diagram over P̄k(T ), it sufficesto show that we can reduce the number of tiles in said diagram.I.e. we will show that at least one of the following must occur inany non-empty spherical diagram over P̄k(T ):

1 There exists a mirror image cancellation.

2 There exists an allowable me-sphere rewriting.














Recall that a mirror image cancellation occurs when we have atile and its mirror image sharing an edge.Example:


Question

What is an allowable me-sphere rewriting?

Best explained by picture example, here is an allowable sphererewriting:


Question




Question




Recall that inP̄k(T ) =< x1, . . . , xn|{re = 1}, xk1 = 1, . . . , xkn = 1 >, we will beworking with the following two types of tiles:

Now, to prove the claim that there must be a mirror imagecancellation or an allowable me-sphere reduction, we will beworking with combinatorial curvature.


Recall that inP̄k(T ) =< x1, . . . , xn|{re = 1}, xk1 = 1, . . . , xkn = 1 >, we will beworking with the following two types of tiles:

Now, to prove the claim that there must be a mirror imagecancellation or an allowable me-sphere reduction, we will beworking with combinatorial curvature.


Given any non-empty spherical diagram over P̄k(T ), it followthat:Combinatorial Gauss-Bonnet∑κ(v) +

∑κ(t) = 4π.

Where κ(v) is the curvature at a vertex v and κ(t) is thecurvature of a tile t.We will begin by assuming that no such mirror imagecancellation or allowable me-sphere reduction occur in our givenspherical diagram.We will use these facts, indeed, to show that∑κ(v) +

∑κ(t) ≤ 0, which gives us an immediate

contradiction to our assumption.



∑κ(t) = 4π.






∑κ(t) = 4π.






∑κ(t) = 4π.





How do we begin? We want∑κ(v) +

∑κ(t) ≤ 0, so let us start

with∑κ(t).

To make things more simple, we will force each tile to have zeroface curvature. This immediately gives that

∑κ(t) = 0.

Easily enough, we accomplish this by letting each tile be aregular, convex n-gon.Example: Let β be the interior angle.

For our square tiles, β =π

2

And for our k-gons, then β =(k − 2)π

k

In particular, if k = 5, then β =3π

5




with∑κ(t).


∑κ(t) = 0.



2


k


5




with∑κ(t).


∑κ(t) = 0.



2


k


5




with∑κ(t).


∑κ(t) = 0.



2


k


5




with∑κ(t).


∑κ(t) = 0.



2


k


5




with∑κ(t).


∑κ(t) = 0.



2


k


5


We want∑κ(v) +

∑κ(t) ≤ 0. We have just seen that∑

κ(t) = 0, so it suffices to show∑κ(v) ≤ 0.

Note that κ(v) is the curvature around a vertex v and is definedas κ(v) = 2π −

∑αi where αi are angles assigned to corners

around the vertex v.


We want∑κ(v) +







We want∑κ(v) +







Example of finding κ(v): Let one 5-gon and 2 squares meet at v:

Then κ(v) = 2π −(π

2+π

2+

3π

5

)=

2π

5.


The key idea is to classify two types of vertices: either v istouching a k-gon or it is not.Simply, ∑

κ(v) =∑

k-gon at v

κ(v) +∑

no k-gon at v

κ(v)

Recall that, because P (T ) was a non-positively curved squarepresentation, ∑

no k-gon at v

κ(v) ≤ 0

(since the only tiles involved are, indeed, square tiles).Thus, it suffices to consider, now, the vertices v at k-gons.



κ(v) =∑

k-gon at v

κ(v) +∑

no k-gon at v

κ(v)


no k-gon at v

κ(v) ≤ 0




κ(v) =∑

k-gon at v

κ(v) +∑

no k-gon at v

κ(v)


no k-gon at v

κ(v) ≤ 0




κ(v) =∑

k-gon at v

κ(v) +∑

no k-gon at v

κ(v)


no k-gon at v

κ(v) ≤ 0



From here, for simplicity of the argument, let k=5.Looking at the below configurations, notice that the vertex v1on the left has positive curvature and the vertex v2 on the righthas negative curvature.

Numbers:

The valency 3 vertex v1 is s.t. κ(v1) =2π

5, as seen before.

And the valency 5 vertex v2 is s.t.

κ(v2) = 2π −(π

2+π

2+π

2+π

2+

3π

5

)= −3π

5.

Indeed, it is true that the only configuration where we havepositive curvature is the valency 3 configuration above.



Numbers:


5, as seen before.


κ(v2) = 2π −(π

2+π

2+π

2+π

2+

3π

5

)= −3π

5.




Numbers:


5, as seen before.


κ(v2) = 2π −(π

2+π

2+π

2+π

2+

3π

5

)= −3π

5.




Numbers:


5, as seen before.


κ(v2) = 2π −(π

2+π

2+π

2+π

2+

3π

5

)= −3π

5.



Consider the below tiling:

The three red vertices are valency 3 and the two blue verticesare valency 5.Thus, we sum the curvature around our 5-gon:∑κ(v) = 3

(2π

5

)+ 2

(−3π

5

)= 0

It takes a significant number of these valency 3 configurationsto force the curvature around a k-gon to be positive.In fact, even having 3 of 5 vertices on our 5-gon having positivecurvature was not enough to force the sum of curvature aroundthe 5-gon to be positive.




(2π

5

)+ 2

(−3π

5

)= 0





(2π

5

)+ 2

(−3π

5

)= 0





(2π

5

)+ 2

(−3π

5

)= 0





(2π

5

)+ 2

(−3π

5

)= 0



Even better, if we happen to find a long enough gallery of thesevalency 3 configurations, then we are able to apply an allowableme-sphere rewriting, as we have seen.


Thus, using these and various other tools, we are able to showthat, given any spherical diagram over P̄k(T ), we can alwaysremove tiles.I.e. Given any spherical diagram, we will be able to arrive at anempty diagram.Recall the original theorem from Heubschmann: if you cannottile in an essential way, then the group defined by thatpresentation is either Zn or infinite.






In fact, our result is close enough to this theorem that we canapply similar topological techniques to find a similar conclusion.For those topologists in the crowd, we do this by looking atgenerators of π2, the set of functions mapping the 2-sphere tothe 2-complex defined by our presentation P̄k(T ).


In fact, our result is close enough to this theorem that we canapply similar topological techniques to find a similar conclusion.For those topologists in the crowd, we do this by looking atgenerators of π2, the set of functions mapping the 2-sphere tothe 2-complex defined by our presentation P̄k(T ).


Acknowledgements

Thanks to our faculty advisor Jens Harlander, funding from theNational Science Foundation, the Boise State UniversityMathematics Department, and the work done by the 2012 REUteam.


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