Drorbn - User contributions [en]

06-1350/Class Photo

2006-10-05T00:12:25Z

Oivrii: /* Who We Are */

{{06-1350/Navigation}}

Our class on September 28, 2006:

[[Image:06-1350-ClassPhoto.jpg|thumb|center|500px|Class Photo: click to enlarge]]

Please identify yourself in this photo! There are two ways to do that:

* [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.
* Send [[User:Drorbn|Dror]] an email message with this information.

The first option is more fun but less private.

===Who We Are===

{| align=center border=1
|-
!First name
!Last name
!UserID
!Email
!In the photo
!Comments
{{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn@ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank}}
{{Photo Entry|last=Dancso|first=Zsuzsi|userid=Zsuzsi|email=zsuzsi@ math.toronto.edu|location=first row, second from the right|comments=}}
{{Photo Entry|last=Ivrii|first=Oleg|userid=Oivrii|email=oleg@math.toronto.edu|location=last row, middle of the sitting |comments=}}
|}

06-1350/Class Photo

2006-10-04T23:51:54Z

Oivrii: /* Who We Are */

File:LYMPHOTU.png

2006-09-22T02:32:42Z

Oivrii:

This treasure -- evidently is unfair to two Polish mathematicians Przytycki and Traczik who made the discovery at the same time, but published later, not mentioning some Russian mathematicians who saw in this polynomial a variation of Jones polynomial and never thought of publication. Another acronym, LYMPHOTU was later offered by Israeli mathematician Dror Bar-Natan. It awarded fairness to the Polish and others (U = unknown discoverers), but did not get popular.

Above is shown two knots with same LYMPHOTU polynomials.

File:LYMPHOTU.png

2006-09-22T02:30:39Z

Oivrii:

This treasure -- evidently is unfair to two Polish mathematicians Przytycki and Traczik who made the discovery at the same time, but published later, not mentioning some Russian mathematicians who saw in this polynomial a variation of Jones polynomial and never thought of publication. Another acronym, LYMPHOTU was later offered by Israeli mathematician Dror Dar-Natan. It awarded fairness to the Polish and others (U = unknown discoverers), but did not get popular.

Above is shown two knots with same LYMPHOTU polynomials.

File:LYMPHOTU.png

2006-09-22T02:28:05Z

Oivrii: This treasure -- evidently is unfair to two Polish mathematicians Przytycki and Traczik who made the discovery at the same time, but published later, not mentioning some Russian mathematicians who saw in this polynomial a variation of Jones polynomial and

This treasure -- evidently is unfair to two Polish mathematicians Przytycki and Traczik who made the discovery at the same time, but published later, not mentioning some Russian mathematicians who saw in this polynomial a variation of Jones polynomial and never thought of publication. Another acronym, LYMPHOTU was later offered by Israeli mathematician Dror Dar-Natan. It awarded fairness to the Polish and others (U = unknown discoverers), but did not get popular.

User:Oivrii

2006-09-19T21:54:49Z

Oivrii:

I sit on the second row, second last spot to the left in BA 6183. Please don't steal my chair!

06-1350/Class Notes for Tuesday September 12

2006-09-19T21:49:55Z

Oivrii:

{{06-1350/Navigation}}

===Introduction===

We wish to define a knot as a continuous injective map from the circle to 3-dimensional Euclidean space
up to continuous homotopy.
 
Unfortunately this definition doesn’t quite work as most of the knots that we wish to
distinguish will actually be equivalent (we may shrink down the knotted part to a point). For this
reason, we replace continuous with smooth or piecewise linear (this time we may not shrink down
the knotted part to the point as we will get a singularity). We will not discuss this in detail, but
rather state the results:
 
Every knot can be represented by a finite diagram, e.g.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/e/e1/Trifoil.png
</center>
 

Two such diagrams represent the same knot if they differ by a sequence of
Reidemeister moves. When we draw diagrams, we often don’t draw the entire knot,
only the parts which are to be changed.
 
An invariant is a map from diagrams to say formal power series Z((q)) which respects R1 - R3. Thus, the invariant descends to equivalence classes of diagrams, i.e to knots.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/8a/Reid.png
 
R1: top left, R2: top right, R3: bottom middle
</center>
 

===3 Colouring===

We give an example of an invariant. Define I3 : {diagrams}→{true, false},
I3 is true
whenever its arcs may be coloured with red, green, blue (RGB), such that all colours appear and
at every crossing is mono or tri-chromatic.
 
We can now distinguish the trifoil from the unknot as the trifoil is 3-colourable while the
unknot is not. However, we cannot distinguish (yet) the trifoil from its mirror image, so later we
will look for more powerful invariants.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/9/93/Trifoil2.png
 
</center>
 
We now prove that I3 is an invariant, that is we need to show that I3 is preserved after R1, R2 and R3. The test for R1 is straight forward:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/2/25/3col-a.png
 
</center>
 
For R2, we have two possibilities. The second case for R2 requires further consideration: where did the green go? Here, we
remember that we are only drawing part of the diagram. Since the red and blue branches must
meet, at such a crossing, green will appear.
Showing that invariance under R3 is more tedious (and is left as an exercise).

===Jones Polynomial===

The simplest way to define the Jones polynomial is via the Kauffman bracket. The idea is to
eliminate all crossings using the rule:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/6/64/Kauffman.png
 
</center>
 
In the right hand side, the first bracket is called the 0-smoothing and the second
is called the 1-smoothing. To calculate the Kauffman bracket we must sum over all
possible smoothings. For instance, for the trifoil, we 23 = 8 summands, one of which will
be:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/8b/Trifoil-smoothing.png
 
</center>
 
Each summand will have no crossing and thus will be a union of (possibly nested) unknots.
We define the bracket polynomial of k unknots to be dk-1 for some indeterminate d. Our hopes
that our polynomial in ℤ[d,A,B] will be an invariant under the Reidmeister moves. We first
verify R2.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/c/cd/Jones-r2.png
 
</center>
 

Collecting like terms and comparing we find AB = 1 and A2 + B2 + dAB = 0. Thus, we must
have B = A-1 and d = -(A2 + A2). Things are looking bad, we still have two moves to verify
and we already lost two of our variables.
 
We now verify R3. For this we remark that
 
<center>
http://katlas.math.toronto.edu/drorbn/images/f/fb/Jones-r3.png
 
</center>
 

The two diagrams with coefficients B coincide; and the diagrams with coefficients A differ by
two R2 moves. Now we verify R1:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/9/99/Jones-r1.png
 
</center>
 
The right hand side evaluates to A + A-1(-A2 -A-2) = -A3 of the desired. This is
unfortunate. One could salvage something by taking A to be one of the cube roots of -1;
however, the right way out of this is to define another invariant which fails in the exactly the
same way, and multiply it with the bracket polynomial.

===Writhe===

The invariant we are looking for is called the writhe. If D is a diagram of an oriented knot, we
define
 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/88/Mathtext1.png
</center>
 
We choose +1 if the crossing is positive (the overpass goes over the underpass from left to right) and -1 if the crossing is negative (otherwise):
 
<center>
http://katlas.math.toronto.edu/drorbn/images/3/38/Crossings.png
 
</center>
 

Lets have an example. Notice that the orientation of the knot is actually irrelevant.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/c/c5/Writhe-example.png
 
</center>
 
We now show that the writhe is also invariant under R2 and R3, under R1 we gain ±1
depending which way we apply it. In R2, the two crossings have opposite signs; and in R3, the
diagonal crossing doesn’t change sign and the other two are reversed.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/f/f9/Writhe-reid.png
 
</center>
 

It follows that 〈D〉⋅(-A-3)w(d) is a knot invariant. This is a polynomial in A, we now
substitute q
http://katlas.math.toronto.edu/drorbn/images/2/23/Mathtext2.png for A and call this the Jones polynomial (which strictly speaking, is not really a
polynomial).

06-1350/Class Notes for Tuesday September 12

2006-09-19T21:49:25Z

Oivrii:

===Introduction===

We wish to define a knot as a continuous injective map from the circle to 3-dimensional Euclidean space
up to continuous homotopy.
 
Unfortunately this definition doesn’t quite work as most of the knots that we wish to
distinguish will actually be equivalent (we may shrink down the knotted part to a point). For this
reason, we replace continuous with smooth or piecewise linear (this time we may not shrink down
the knotted part to the point as we will get a singularity). We will not discuss this in detail, but
rather state the results:
 
Every knot can be represented by a finite diagram, e.g.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/e/e1/Trifoil.png
</center>
 

Two such diagrams represent the same knot if they differ by a sequence of
Reidemeister moves. When we draw diagrams, we often don’t draw the entire knot,
only the parts which are to be changed.
 
An invariant is a map from diagrams to say formal power series Z((q)) which respects R1 - R3. Thus, the invariant descends to equivalence classes of diagrams, i.e to knots.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/8a/Reid.png
 
R1: top left, R2: top right, R3: bottom middle
</center>
 

===3 Colouring===

We give an example of an invariant. Define I3 : {diagrams}→{true, false},
I3 is true
whenever its arcs may be coloured with red, green, blue (RGB), such that all colours appear and
at every crossing is mono or tri-chromatic.
 
We can now distinguish the trifoil from the unknot as the trifoil is 3-colourable while the
unknot is not. However, we cannot distinguish (yet) the trifoil from its mirror image, so later we
will look for more powerful invariants.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/9/93/Trifoil2.png
 
</center>
 
We now prove that I3 is an invariant, that is we need to show that I3 is preserved after R1, R2 and R3. The test for R1 is straight forward:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/2/25/3col-a.png
 
</center>
 
For R2, we have two possibilities. The second case for R2 requires further consideration: where did the green go? Here, we
remember that we are only drawing part of the diagram. Since the red and blue branches must
meet, at such a crossing, green will appear.
Showing that invariance under R3 is more tedious (and is left as an exercise).

===Jones Polynomial===

The simplest way to define the Jones polynomial is via the Kauffman bracket. The idea is to
eliminate all crossings using the rule:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/6/64/Kauffman.png
 
</center>
 
In the right hand side, the first bracket is called the 0-smoothing and the second
is called the 1-smoothing. To calculate the Kauffman bracket we must sum over all
possible smoothings. For instance, for the trifoil, we 23 = 8 summands, one of which will
be:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/8b/Trifoil-smoothing.png
 
</center>
 
Each summand will have no crossing and thus will be a union of (possibly nested) unknots.
We define the bracket polynomial of k unknots to be dk-1 for some indeterminate d. Our hopes
that our polynomial in ℤ[d,A,B] will be an invariant under the Reidmeister moves. We first
verify R2.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/c/cd/Jones-r2.png
 
</center>
 

Collecting like terms and comparing we find AB = 1 and A2 + B2 + dAB = 0. Thus, we must
have B = A-1 and d = -(A2 + A2). Things are looking bad, we still have two moves to verify
and we already lost two of our variables.
 
We now verify R3. For this we remark that
 
<center>
http://katlas.math.toronto.edu/drorbn/images/f/fb/Jones-r3.png
 
</center>
 

The two diagrams with coefficients B coincide; and the diagrams with coefficients A differ by
two R2 moves. Now we verify R1:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/9/99/Jones-r1.png
 
</center>
 
The right hand side evaluates to A + A-1(-A2 -A-2) = -A3 of the desired. This is
unfortunate. One could salvage something by taking A to be one of the cube roots of -1;
however, the right way out of this is to define another invariant which fails in the exactly the
same way, and multiply it with the bracket polynomial.

===Writhe===

The invariant we are looking for is called the writhe. If D is a diagram of an oriented knot, we
define
 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/88/Mathtext1.png
</center>
 
We choose +1 if the crossing is positive (the overpass goes over the underpass from left to right) and -1 if the crossing is negative (otherwise):
 
<center>
http://katlas.math.toronto.edu/drorbn/images/3/38/Crossings.png
 
</center>
 

Lets have an example. Notice that the orientation of the knot is actually irrelevant.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/c/c5/Writhe-example.png
 
</center>
 
We now show that the writhe is also invariant under R2 and R3, under R1 we gain ±1
depending which way we apply it. In R2, the two crossings have opposite signs; and in R3, the
diagonal crossing doesn’t change sign and the other two are reversed.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/f/f9/Writhe-reid.png
 
</center>
 

It follows that 〈D〉⋅(-A-3)w(d) is a knot invariant. This is a polynomial in A, we now
substitute q
http://katlas.math.toronto.edu/drorbn/images/2/23/Mathtext2.png for A and call this the Jones polynomial (which strictly speaking, is not really a
polynomial).

06-1350/Class Notes for Tuesday September 12

2006-09-19T21:48:50Z

Oivrii: /* Introduction */

===Introduction===

We wish to define a knot as a continuous injective map from the circle to 3-dimensional Euclidean space
up to continuous homotopy.
 
Unfortunately this definition doesn’t quite work as most of the knots that we wish to
distinguish will actually be equivalent (we may shrink down the knotted part to a point). For this
reason, we replace continuous with smooth or piecewise linear (this time we may not shrink down
the knotted part to the point as we will get a singularity). We will not discuss this in detail, but
rather state the results:
 
Every knot can be represented by a finite diagram, e.g.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/e/e1/Trifoil.png
</center>
 

Two such diagrams represent the same knot if they differ by a sequence of
Reidemeister moves. When we draw diagrams, we often don’t draw the entire knot,
only the parts which are to be changed.
 
An invariant is a map from diagrams to say formal power series Z((q)) which respects R1 - R3. Thus, the invariant descends to equivalence classes of diagrams, i.e to knots.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/8a/Reid.png
 
R1: top left, R2: top right, R3: bottom middle
</center>
 

===3 Colouring===

We give an example of an invariant. Define I3 : {diagrams}→{true, false},
I3 is true
whenever its arcs may be coloured with red, green, blue (RGB), such that all colours appear and
at every crossing is mono or tri-chromatic.
 
We can now distinguish the trifoil from the unknot as the trifoil is 3-colourable while the
unknot is not. However, we cannot distinguish (yet) the trifoil from its mirror image, so later we
will look for more powerful invariants.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/9/93/Trifoil2.png
 
</center>
 
We now prove that I3 is an invariant, that is we need to show that I3 is preserved after R1, R2 and R3. The test for R1 is straight forward:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/2/25/3col-a.png
 
</center>
 
For R2, we have two possibilities. The second case for R2 requires further consideration: where did the green go? Here, we
remember that we are only drawing part of the diagram. Since the red and blue branches must
meet, at such a crossing, green will appear.
Showing that invariance under R3 is more tedious (and is left as an exercise).

===Jones Polynomial===

The simplest way to define the Jones polynomial is via the Kauffman bracket. The idea is to
eliminate all crossings using the rule:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/6/64/Kauffman.png
 
</center>
 
In the right hand side, the first bracket is called the 0-smoothing and the second
is called the 1-smoothing. To calculate the Kauffman bracket we must sum over all
possible smoothings. For instance, for the trifoil, we 23 = 8 summands, one of which will
be:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/8b/Trifoil-smoothing.png
 
</center>
 
Each summand will have no crossing and thus will be a union of (possibly nested) unknots.
We define the bracket polynomial of k unknots to be dk-1 for some indeterminate d. Our hopes
that our polynomial in ℤ[d,A,B] will be an invariant under the Reidmeister moves. We first
verify R2.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/c/cd/Jones-r2.png
 
</center>
 

Collecting like terms and comparing we find AB = 1 and A2 + B2 + dAB = 0. Thus, we must
have B = A-1 and d = -(A2 + A2). Things are looking bad, we still have two moves to verify
and we already lost two of our variables.
 
We now verify R3. For this we remark that
 
<center>
http://katlas.math.toronto.edu/drorbn/images/f/fb/Jones-r3.png
 
</center>
 

The two diagrams with coefficients B coincide; and the diagrams with coefficients A differ by
two R2 moves. Now we verify R1:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/9/99/Jones-r1.png
 
</center>
 
The right hand side evaluates to A + A-1(-A2 -A-2) = -A3 of the desired. This is
unfortunate. One could salvage something by taking A to be one of the cube roots of -1;
however, the right way out of this is to define another invariant which fails in the exactly the
same way, and multiply it with the bracket polynomial.

===Writhe===

The invariant we are looking for is called the writhe. If D is a diagram of an oriented knot, we
define
 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/88/Mathtext1.png
</center>
 
We choose +1 if the crossing is positive (the overpass goes over the underpass from left to right) and -1 if the crossing is negative (otherwise):
 
<center>
"http://katlas.math.toronto.edu/drorbn/images/3/38/Crossings.png
 
</center>
 

Lets have an example. Notice that the orientation of the knot is actually irrelevant.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/c/c5/Writhe-example.png
 
</center>
 
We now show that the writhe is also invariant under R2 and R3, under R1 we gain ±1
depending which way we apply it. In R2, the two crossings have opposite signs; and in R3, the
diagonal crossing doesn’t change sign and the other two are reversed.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/f/f9/Writhe-reid.png
 
</center>
 

It follows that 〈D〉⋅(-A-3)w(d) is a knot invariant. This is a polynomial in A, we now
substitute q
http://katlas.math.toronto.edu/drorbn/images/2/23/Mathtext2.png for A and call this the Jones polynomial (which strictly speaking, is not really a
polynomial).

06-1350/Class Notes for Tuesday September 12

2006-09-19T21:43:50Z

Oivrii:

===Introduction===

Unfortunately this definition doesn’t quite work as most of the knots that we wish to
distinguish will actually be equivalent (we may shrink down the knotted part to a point). For this
reason, we replace continuous with smooth or piecewise linear (this time we may not shrink down
the knotted part to the point as we will get a singularity). We will not discuss this in detail, but
rather state the results:
 
Every knot can be represented by a finite diagram, e.g.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/e/e1/Trifoil.png
</center>
 

Two such diagrams represent the same knot if they differ by a sequence of
Reidemeister moves. When we draw diagrams, we often don’t draw the entire knot,
only the parts which are to be changed.
 
An invariant is a map from diagrams to say formal power series Z((q)) which respects R1 - R3. Thus, the invariant descends to equivalence classes of diagrams, i.e to knots.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/8a/Reid.png
 
R1: top left, R2: top right, R3: bottom middle
</center>
 

===3 Colouring===

We give an example of an invariant. Define I3 : {diagrams}→{true, false},
I3 is true
whenever its arcs may be coloured with red, green, blue (RGB), such that all colours appear and
at every crossing is mono or tri-chromatic.
 
We can now distinguish the trifoil from the unknot as the trifoil is 3-colourable while the
unknot is not. However, we cannot distinguish (yet) the trifoil from its mirror image, so later we
will look for more powerful invariants.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/9/93/Trifoil2.png
 
</center>
 
We now prove that I3 is an invariant, that is we need to show that I3 is preserved after R1, R2 and R3. The test for R1 is straight forward:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/2/25/3col-a.png
 
</center>
 
For R2, we have two possibilities. The second case for R2 requires further consideration: where did the green go? Here, we
remember that we are only drawing part of the diagram. Since the red and blue branches must
meet, at such a crossing, green will appear.
Showing that invariance under R3 is more tedious (and is left as an exercise).

===Jones Polynomial===

The simplest way to define the Jones polynomial is via the Kauffman bracket. The idea is to
eliminate all crossings using the rule:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/6/64/Kauffman.png
 
</center>
 
In the right hand side, the first bracket is called the 0-smoothing and the second
is called the 1-smoothing. To calculate the Kauffman bracket we must sum over all
possible smoothings. For instance, for the trifoil, we 23 = 8 summands, one of which will
be:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/8b/Trifoil-smoothing.png
 
</center>
 
Each summand will have no crossing and thus will be a union of (possibly nested) unknots.
We define the bracket polynomial of k unknots to be dk-1 for some indeterminate d. Our hopes
that our polynomial in ℤ[d,A,B] will be an invariant under the Reidmeister moves. We first
verify R2.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/c/cd/Jones-r2.png
 
</center>
 

Collecting like terms and comparing we find AB = 1 and A2 + B2 + dAB = 0. Thus, we must
have B = A-1 and d = -(A2 + A2). Things are looking bad, we still have two moves to verify
and we already lost two of our variables.
 
We now verify R3. For this we remark that
 
<center>
http://katlas.math.toronto.edu/drorbn/images/f/fb/Jones-r3.png
 
</center>
 

The two diagrams with coefficients B coincide; and the diagrams with coefficients A differ by
two R2 moves. Now we verify R1:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/9/99/Jones-r1.png
 
</center>
 
The right hand side evaluates to A + A-1(-A2 -A-2) = -A3 of the desired. This is
unfortunate. One could salvage something by taking A to be one of the cube roots of -1;
however, the right way out of this is to define another invariant which fails in the exactly the
same way, and multiply it with the bracket polynomial.

===Writhe===

The invariant we are looking for is called the writhe. If D is a diagram of an oriented knot, we
define
 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/88/Mathtext1.png
</center>
 
We choose +1 if the crossing is positive (the overpass goes over the underpass from left to right) and -1 if the crossing is negative (otherwise):
 
<center>
"http://katlas.math.toronto.edu/drorbn/images/3/38/Crossings.png
 
</center>
 

Lets have an example. Notice that the orientation of the knot is actually irrelevant.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/c/c5/Writhe-example.png
 
</center>
 
We now show that the writhe is also invariant under R2 and R3, under R1 we gain ±1
depending which way we apply it. In R2, the two crossings have opposite signs; and in R3, the
diagonal crossing doesn’t change sign and the other two are reversed.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/f/f9/Writhe-reid.png
 
</center>
 

It follows that 〈D〉⋅(-A-3)w(d) is a knot invariant. This is a polynomial in A, we now
substitute q
http://katlas.math.toronto.edu/drorbn/images/2/23/Mathtext2.png for A and call this the Jones polynomial (which strictly speaking, is not really a
polynomial).

06-1350/Class Notes for Tuesday September 12

2006-09-19T21:40:43Z

Oivrii:

Introduction

Unfortunately this definition doesn’t quite work as most of the knots that we wish to
distinguish will actually be equivalent (we may shrink down the knotted part to a point). For this
reason, we replace continuous with smooth or piecewise linear (this time we may not shrink down
the knotted part to the point as we will get a singularity). We will not discuss this in detail, but
rather state the results:
 
Every knot can be represented by a finite diagram, e.g.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/e/e1/Trifoil.png
</center>
 

Two such diagrams represent the same knot if they differ by a sequence of
Reidemeister moves. When we draw diagrams, we often don’t draw the entire knot,
only the parts which are to be changed.
 
An invariant is a map from diagrams to say formal power series Z((q)) which respects R1 - R3. Thus, the invariant descends to equivalence classes of diagrams, i.e to knots.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/8a/Reid.png
 
R1: top left, R2: top right, R3: bottom middle
</center>
 

3 Colouring
 

We give an example of an invariant. Define I3 : {diagrams}→{true, false},
I3 is true
whenever its arcs may be coloured with red, green, blue (RGB), such that all colours appear and
at every crossing is mono or tri-chromatic.
 
We can now distinguish the trifoil from the unknot as the trifoil is 3-colourable while the
unknot is not. However, we cannot distinguish (yet) the trifoil from its mirror image, so later we
will look for more powerful invariants.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/9/93/Trifoil2.png
 
</center>
 
We now prove that I3 is an invariant, that is we need to show that I3 is preserved after R1, R2 and R3. The test for R1 is straight forward:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/2/25/3col-a.png
 
</center>
 
For R2, we have two possibilities. The second case for R2 requires further consideration: where did the green go? Here, we
remember that we are only drawing part of the diagram. Since the red and blue branches must
meet, at such a crossing, green will appear.
Showing that invariance under R3 is more tedious (and is left as an exercise).

 

Jones Polynomial
 

The simplest way to define the Jones polynomial is via the Kauffman bracket. The idea is to
eliminate all crossings using the rule:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/6/64/Kauffman.png
 
</center>
 
In the right hand side, the first bracket is called the 0-smoothing and the second
is called the 1-smoothing. To calculate the Kauffman bracket we must sum over all
possible smoothings. For instance, for the trifoil, we 23 = 8 summands, one of which will
be:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/8b/Trifoil-smoothing.png
 
</center>
 
Each summand will have no crossing and thus will be a union of (possibly nested) unknots.
We define the bracket polynomial of k unknots to be dk-1 for some indeterminate d. Our hopes
that our polynomial in ℤ[d,A,B] will be an invariant under the Reidmeister moves. We first
verify R2.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/c/cd/Jones-r2.png
 
</center>
 

Collecting like terms and comparing we find AB = 1 and A2 + B2 + dAB = 0. Thus, we must
have B = A-1 and d = -(A2 + A2). Things are looking bad, we still have two moves to verify
and we already lost two of our variables.
 
We now verify R3. For this we remark that
 
<center>
http://katlas.math.toronto.edu/drorbn/images/f/fb/Jones-r3.png
 
</center>
 

The two diagrams with coefficients B coincide; and the diagrams with coefficients A differ by
two R2 moves. Now we verify R1:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/9/99/Jones-r1.png
 
</center>
 
The right hand side evaluates to A + A-1(-A2 -A-2) = -A3 of the desired. This is
unfortunate. One could salvage something by taking A to be one of the cube roots of -1;
however, the right way out of this is to define another invariant which fails in the exactly the
same way, and multiply it with the bracket polynomial.
 
Writhe
 
The invariant we are looking for is called the writhe. If D is a diagram of an oriented knot, we
define
 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/88/Mathtext1.png
</center>
 
We choose +1 if the crossing is positive (the overpass goes over the underpass from left to right) and -1 if the crossing is negative (otherwise):
 
<center>
"http://katlas.math.toronto.edu/drorbn/images/3/38/Crossings.png
 
</center>
 

Lets have an example. Notice that the orientation of the knot is actually irrelevant.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/c/c5/Writhe-example.png
 
</center>
 
We now show that the write is also invariant under R2 and R3, under R1 we gain ±1
depending which way we apply it. In R2, the two crossings have opposite signs; and in R3, the
diagonal crossing doesn’t change sign and the other two are reversed.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/f/f9/Writhe-reid.png
 
</center>
 

It follows that 〈D〉⋅(-A-3)w(d) is a knot invariant. This is a polynomial in A, we now
substitute q
http://katlas.math.toronto.edu/drorbn/images/2/23/Mathtext2.png for A and call this the Jones polynomial (which strictly speaking, is not really a
polynomial).

06-1350/Class Notes for Tuesday September 12

2006-09-19T21:37:32Z

Oivrii:

Introduction

Unfortunately this definition doesn’t quite work as most of the knots that we wish to
distinguish will actually be equivalent (we may shrink down the knotted part to a point). For this
reason, we replace continuous with smooth or piecewise linear (this time we may not shrink down
the knotted part to the point as we will get a singularity). We will not discuss this in detail, but
rather state the results:
 
Every knot can be represented by a finite diagram, e.g.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/e/e1/Trifoil.png
</center>
 

Two such diagrams represent the same knot if they differ by a sequence of
Reidemeister moves. When we draw diagrams, we often don’t draw the entire knot,
only the parts which are to be changed.
 
An invariant is a map from diagrams to say formal power series Z((q)) which respects R1 - R3. Thus, the invariant descends to equivalence classes of diagrams, i.e to knots.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/8a/Reid.png
 
R1: top left, R2: top right, R3: bottom middle
</center>
 

3 Colouring
 

We give an example of an invariant. Define I3 : {diagrams}→{true, false},
I3 is true
whenever its arcs may be coloured with red, green, blue (RGB), such that all colours appear and
at every crossing is mono or tri-chromatic.
 
We can now distinguish the trifoil from the unknot as the trifoil is 3-colourable while the
unknot is not. However, we cannot distinguish (yet) the trifoil from its mirror image, so later we
will look for more powerful invariants.
 
<center>
http://katlas.math.toronto.edu/drorbn/images/9/93/Trifoil2.png
 
</center>
 
We now prove that I3 is an invariant, that is we need to show that I3 is preserved after R1, R2 and R3. The test for R1 is straight forward:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/2/25/3col-a.png
 
</center>
 
For R2, we have two possibilities. The second case for R2 requires further consideration: where did the green go? Here, we
remember that we are only drawing part of the diagram. Since the red and blue branches must
meet, at such a crossing, green will appear.
Showing that invariance under R3 is more tedious (and is left as an exercise).

 

Jones Polynomial
 

The simplest way to define the Jones polynomial is via the Kauffman bracket. The idea is to
eliminate all crossings using the rule:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/6/64/Kauffman.png
 
</center>
 
In the right hand side, the first bracket is called the 0-smoothing and the second
is called the 1-smoothing. To calculate the Kauffman bracket we must sum over all
possible smoothings. For instance, for the trifoil, we 23 = 8 summands, one of which will
be:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/8b/Trifoil-smoothing.png
 
</center>
 

Each summand will have no crossing and thus will be a union of (possibly nested) unknots.
We define the bracket polynomial of k unknots to be dk-1 for some indeterminate d. Our hopes
that our polynomial in ℤ[d,A,B] will be an invariant under the Reidmeister moves. We first
verify R2.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/c/cd/Jones-r2.png
 
</center>
 

Collecting like terms and comparing we find AB = 1 and A2 + B2 + dAB = 0. Thus, we must
have B = A-1 and d = -(A2 + A2). Things are looking bad, we still have two moves to verify
and we already lost two of our variables.
 
We now verify R3. For this we remark that
 
<center>
http://katlas.math.toronto.edu/drorbn/images/f/fb/Jones-r3.png
 
</center>
 

The two diagrams with coefficients B coincide; and the diagrams with coefficients A differ by
two R2 moves. Now we verify R1:
 
<center>
http://katlas.math.toronto.edu/drorbn/images/9/99/Jones-r1.png
 
</center>
 
The right hand side evaluates to A + A-1(-A2 -A-2) = -A3 of the desired. This is
unfortunate. One could salvage something by taking A to be one of the cube roots of -1;
however, the right way out of this is to define another invariant which fails in the exactly the
same way, and multiply it with the bracket polynomial.
 
Writhe
 
The invariant we are looking for is called the writhe. If D is a diagram of an oriented knot, we
define
 
<center>
http://katlas.math.toronto.edu/drorbn/images/8/88/Mathtext1.png
</center>
 
We choose +1 if the crossing is positive (the overpass goes over the underpass from left to right) and -1 if the crossing is negative (otherwise):
 
<center>
src="http://katlas.math.toronto.edu/drorbn/images/3/38/Crossings.png
 
</center>
 

Lets have an example. Notice that the orientation of the knot is actually irrelevant.

 
<center>
http://katlas.math.toronto.edu/drorbn/images/c/c5/Writhe-example.png
 
</center>
 

We now show that the write is also invariant under R2 and R3, under R1 we gain ±1
depending which way we apply it. In R2, the two crossings have opposite signs; and in R3, the
diagonal crossing doesn’t change sign and the other two are reversed.

 
<center>
src="http://katlas.math.toronto.edu/drorbn/images/f/f9/Writhe-reid.png
 
</center>
 

It follows that 〈D〉⋅(-A-3)w(d) is a knot invariant. This is a polynomial in A, we now
substitute q<img
src="http://katlas.math.toronto.edu/drorbn/images/2/23/Mathtext2.png" alt="14" class="frac" align="middle"> for A and call this the Jones polynomial (which strictly speaking, is not really a
polynomial).

User:Oivrii

2006-09-19T21:24:46Z

Oivrii:

Weeeee

File:Mathtext2.png

2006-09-19T21:14:26Z

Oivrii: needed formula

needed formula

File:Mathtext1.png

2006-09-19T21:14:07Z

Oivrii: needed formula

needed formula

File:Writhe-reid.png

2006-09-19T21:09:28Z

Oivrii: Writhe reid image

Writhe reid image

File:Writhe-example.png

2006-09-19T21:09:10Z

Oivrii: Writhe example mage

Writhe example mage

File:Trifoil2.png

2006-09-19T21:08:43Z

Oivrii: Trifoil 2 image

Trifoil 2 image

File:Trifoil.png

2006-09-19T21:08:23Z

Oivrii: Trifoil image

Trifoil image

File:Trifoil-smoothing.png

2006-09-19T21:07:43Z

Oivrii: Trifoil smoothing image

Trifoil smoothing image

File:Reid.png

2006-09-19T21:07:11Z

Oivrii: Reid image

Reid image

File:Kauffman.png

2006-09-19T21:06:51Z

Oivrii: Kauffman image

Kauffman image

File:Jones-r3.png

2006-09-19T21:06:34Z

Oivrii: Jones R3 image

Jones R3 image

File:Jones-r2.png

2006-09-19T21:06:17Z

Oivrii: Jones R2 image

Jones R2 image

File:Jones-r1.png

2006-09-19T21:06:04Z

Oivrii: Jones R1 image

Jones R1 image

File:Crossings.png

2006-09-19T21:05:30Z

Oivrii: Crossings image

Crossings image

File:3col-a.png

2006-09-19T21:02:56Z

Oivrii: 3 Colouring image

3 Colouring image