Nathan Ryder- Calculating Knot Polynomials

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Calculating Knot Polynomials

Nathan Ryder

November 28th 2006

Nathan Ryder Calculating Knot Polynomials

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Overview


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Knot Theory

Knot theory is about developing tools for distinguishing knottedcurves.

Fundamentally, we are looking for invariants of diagrams of knots.These take different forms, from simple observations of diagrams,to considering homologies or group structures generated by theknots, to polynomials.


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This Talk

We will consider two of these knot polynomials:

The Kauffman two-variable polynomial

The HOMFLY polynomialWe will look at how they can be calculated, some properties thatwe can deduce from the polynomials, and some extensions to themethod that I outline.


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Skein relations

For a diagram D the Kauffman two-variable polynomial, F (D ) canbe defined by the following skein relations:

zz

v v−1


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Plait Presentation


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Plait Presentation


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Plait Presentation

A k -plait has k capsat the top, and k

cups at the bottom.


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Plait Presentation

Braid


cups at the bottom.

Braid in between isfrom the braid groupB2k .


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Plait Presentation


cups at the bottom.



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Plait Presentation


cups at the bottom.


k is the bridgenumber of thediagram.


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k -Tangles

We have 2k endpoints along a line, and k arcs that join pairs of endpoints. The arcs lie without any restriction in the half plane.


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Motivation


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Motivation


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Motivation


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Motivation


M i i

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Motivation


S k d k T l

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Stacked k -Tangles

As before, 2k endpoints and k arcs, but with the restriction thatarcs cannot wind around each other, i.e., they don’t link.


S k d k T l

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Stacked k -Tangles



St k d k T l

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Stacked k -Tangles



St k d k T l

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Stacked k -Tangles


1


Stacked k Tangles

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Stacked k -Tangles


1


Stacked k Tangles

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Stacked k -Tangles


1 4


Stacked k Tangles

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Stacked k -Tangles


1 4


Stacked k Tangles

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Stacked k -Tangles


1 4

2

3


Stacked k-Tangles

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Stacked k -Tangles


1 42

3


Stacked k-Tangles

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Stacked k -Tangles


1 4

2

3


Stacked k-Tangles

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Stacked k Tangles


1 4

2

3

41321 3 2 4


Stacked k-Tangles

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Stacked k Tangles


1 4

2

3

41321 3 2 4

So can write a number sequence (12314324) for this diagram.


Multiplying Stacked Tangles

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We want to multiply stacked tangles by crossings from beneath,with the restriction that stacked tangles multiply to stackedtangles.



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1 4

2

3



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p y g g


1 4

2

3



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p y g g




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p y g g


1 4

2

3



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p y g g


1 4

2

3


Multiplying Stacked Tangles Redux

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Question:What are we to do if a stacked tangle does not multiply to anotherstacked tangle due to the action of a certain crossing?

Answer:Use the Kauffman skein relations to rewrite it as a linearcombination of stacked tangles that will all allow the multiplication.

We do this in the context of the number sequences that describe

the tangles.


Rearranging Number Sequences 1

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zz

We can see that arcs 1 and 2 in diagrams are related by

(2121) − (1212) = z (1122) − z (1221)

For adjacent arcs a and b (with b = a + 1) this is realised as

(baba) − (abab ) = z (aabb ) − z (abba)

Note that (aabb ) ∼= (bbaa) and (abba) ∼= (baab ).



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As an example, consider the following multiplication:



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We can denote this as (13214324)σ1.



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(13214324) = (1 14 4) + z (1 14 4) − z (1 14 4)



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(13214324) = (12314234) + z (12214334) − z (12314324)



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(13214324) = (12314234) + z (12214334) − z (12314324)



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(13214324) = (12314234) + z (12214334) − z (12314324)

(12214334) = ( 4334)



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(13214324) = (12314234) + z (12214334) − z (12314324)

(12214334) = (21124334)



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(13214324) = (12314234) + z (12214334) − z (12314324)

(12214334) = (21124334)

(12314234) = ( 3 4 34) − z ( 3 4 34) + z ( 3 4 34)



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(13214324) = (12314234) + z (12214334) − z (12314324)

(12214334) = (21124334)

(12314234) = (21324134) − z (11324234) + z (21314234)



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(13214324) = (12314234) + z (12214334) − z (12314324)

(12214334) = (21124334)

(12314234) = (21324134) − z (11324234) + z (21314234)

(12314324) =



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(13214324) = (12314234) + z (12214334) − z (12314324)

(12214334) = (21124334)

(12314234) = (21324134) − z (11324234) + z (21314234)

(12314324) = (21324314) − z (11324324) + z (21314324)



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So

(13214324) = (21324134)− z (11324234) + z (21314234)

+z (21124334) − z (21324314)+z 2(11324324) − z 2(21314324)

Now, multiplying by σ1:



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(13214324)σ1 = (21324134)σ1 − z (11324234)σ1 + z (21314234)σ1

+z (21124334)σ1 − z (21324314)σ1

+z 2(11324324)σ1− z 2(21314324)σ

1

= (12324134)− vz (11324234) + z (12314234)

+z (12124334) − z (12324314)

+vz

2

(11324324) − z

2

(12314324)


The Algorithm

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Put diagram into plait presentation

In general, at each stage rewrite so that we have a linearcombination of tangles that all allow multiplication

Perform multiplication, which is essentially just a case of moving coefficients

At the bottom we close off to (k − 1)-tangles, and thenrepeat until we have a 1-tangle whose coefficient is theKauffman polynomial of the knot


Complexity, Notes and Results

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The number of k -sequences = (2k )!2k

Unlike previous algorithms there is a strict bound on the

number of “diagrams” considered, so not exponential in thatrespect - but only for a fixed k

Talk more on complexity and other results later


Skein relations

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Unlike the Kauffman two-variable polynomial, HOMFLY is a

polynomial for an oriented knot diagram.

z

−1v v


Previous HOMFLY Algorithms

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There are other algorithms that calculate HOMFLY and whichare polynomial

However these are concerned with other families of knots

The intersection of these families with the plaits is small

The algorithm that I outline can calculate much that previousalgorithms could not


Oriented Stacked Tangles

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We now have to consider stacked tangles which carry orientationinformation. We provide an ordering for the arcs as before, butnumber the endpoints a and a. Hence

has a numbering (1 2 314324), and this is distinct from(12314324).

We now have (2k )! number sequences rather than (2k )!/2k .


All Is Not Lost!

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Consider:


All Is Not Lost!

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Consider:

z


All Is Not Lost!

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Consider:

z


All Is Not Lost!

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Consider:

z


All Is Not Lost!

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Consider:

z


All Is Not Lost!

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Consider:

z


All Is Not Lost!

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Consider:

z

Logically the only sequences that can be included in rewriting atany stage will belong to the subset of the number sequences thatshare the same signs of endpoints.

Hence we are back to considering (2k )!/2k

number sequences,with a separate sequence of length 2k which carries the signinformation.


Algorithm

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Put diagram into plait presentation, so that initial tangle can

be represented by the sequence (1 12 2 . . . k k ) At each stage rewrite so that we have a linear combination of

tangles that all allow multiplication

Decision for rearrangement depends on numbering in the first

instance, and then a check on the sign sequence to determinethe exact nature of rearrangement

Move coefficients around with multiplication, and then switchrelevant signs in the sign sequence

Close off to (k − 1)-tangles as previously, repeating until wehave the 1-tangle whose coefficient is the HOMFLYpolynomial of the knot


Complexity: Kauffman

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In the case of the Kauffman polynomial we previously had anexponential algorithm, which we certainly don’t have any more.

There are a finite number of stacked k -tangles, and so there is a

bound on the number of diagrams which we consider as numbersequences.

The growth of coefficients is not exponential - it is definitelyquadratic.


Complexity: Kauffman

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Believe that the complete algorithm is degree 4 with respect to c

for a fixed k . Some “worst case scenario” calculations support this.

There is a strict maximum number of rearrangements that can berequired for a fixed k . This suggests computation time wouldincrease cubically after a certain point.


Complexity: HOMFLY

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By similar arguments we can claim that the algorithm forcalculating HOMFLY is not exponential either. We expect that theorder of complexity is similar, but it is also obvious that less workis performed in the HOMFLY algorithm.

In practice an implementation of the HOMFLY algorithm willcalculate the invariant more quickly than an implementation of theKauffman algorithm.

In both cases, the complexity would be reduced drastically if wecalculated coefficients mod p for some prime p .


Calculations

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Kauffman program tested on 3-plaits with 90 crossings and4-plaits with 70 crossings.

HOMFLY program tested on 3-plaits with over 200 crossingsand 4-plaits with 160 crossings.

Previous HOMFLY programs could cope with significantly fewercrossings. These programs worked from the braid presentation of aknot, with a strict limit on the number of strings that could beused. New algorithm has calculated HOMFLY polynomial for knots

with significantly higher braid index than possible before.


Limitations

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In both cases have looked at calculations with k = 4 at most.

In calculations the Kauffman program takes significantly longer toperform calculations than the HOMFLY program. Expected, but

surprising just how much of a difference there is.

Limitations on both, with respect to c and k could be lifted byimplementing in a compiled language.


Further Calculations

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Looking for interesting examples, a family or class of knots either

Interesting to study

Previously out of reachWould be interesting to look at HOMFLY 3-parallels of 2-plaits,but these are beyond the reach of the Maple implementation.


Extensions

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Implementation in a compiled language for both algorithms wouldallow

Greater values of k : k = 6 should be feasible.

Higher values for c due to better control of coefficient storagein memory.


Extensions

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“Local” versus “Global” width is something that could be

interesting to look at.


Extensions

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Extensions

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Extensions

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If an implementation

could allow for thisinformation we couldhave significantreductions of work insome cases.


Extensions

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Could the basic idea of the algorithm be improved?

|{number sequences}| > |{stacked k -tangles}|> |{basis of stacked k -tangles}|

Using multiple bases?Not obvious that either approach would in principle do anythingmore than make the situation more complicated!


Summary

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Outlined methods for calculating the Kauffman and HOMFLYpolynomials of knots using stacked k -tangles

These methods are an improvement on previous algorithms, as

they work in polynomial time with respect to c for a fixed k Using implementations of these algorithms we can calculate

invariants for knots that were not previously possible


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Documents

Nathan Ryder- Calculating Knot Polynomials