06-1350/Class Notes for Tuesday September 12 - Revision history

69.197.142.199: /* 3 Colouring */

2006-10-22T02:31:35Z

3 Colouring

← Older revision		Revision as of 22:31, 21 October 2006
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	at every crossing is <i>mono</i> or <i>tri-chromatic</i>.		at every crossing is <i>mono</i> or <i>tri-chromatic</i>.
	<br><br>		<br><br>
	We can now distinguish the ~~trifoil~~ from the unknot as the ~~trifoil~~ is 3-colourable while the		We can now distinguish the trefoil from the unknot as the trefoil is 3-colourable while the
	unknot is not. However, we cannot distinguish (yet) the ~~trifoil~~ from its mirror image, so later we		unknot is not. However, we cannot distinguish (yet) the trefoil from its mirror image, so later we
	will look for more powerful invariants.		will look for more powerful invariants.
	<br><br>		<br><br>

Drorbn at 00:07, 21 September 2006

2006-09-21T00:07:16Z

← Older revision		Revision as of 20:07, 20 September 2006
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	meet, at such a crossing, green will appear.		meet, at such a crossing, green will appear.
	Showing that invariance under R3 is more tedious (and is left as an exercise).		Showing that invariance under R3 is more tedious (and is left as an exercise).

	===Jones Polynomial===

	The simplest way to define the <i>Jones polynomial</i> is via the <i>Kauffman bracket</i>. The idea is to
	eliminate all crossings using the rule:
	<br><br>
	<center>
	http://katlas.math.toronto.edu/drorbn/images/6/64/Kauffman.png
	<br>
	</center>
	<br>
	In the right hand side, the first bracket is called the <i>0-smoothing</i> and the second
	is called the <i>1-smoothing</i>. To calculate the Kauffman bracket we must sum over all
	possible smoothings. For instance, for the trifoil, we 2<sup>3</sup> = 8 summands, one of which will
	be:
	<br><br>
	<center>
	http://katlas.math.toronto.edu/drorbn/images/8/8b/Trifoil-smoothing.png
	<br>
	</center>
	<br>
	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 d<sup>k-1</sup> 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.
	<br><br>
	<center>
	http://katlas.math.toronto.edu/drorbn/images/c/cd/Jones-r2.png
	<br>
	</center>
	<br>

	Collecting like terms and comparing we find AB = 1 and A<sup>2</sup> + B<sup>2</sup> + dAB = 0. Thus, we must
	have B = A<sup>-1</sup> and d = -(A<sup>2</sup> + A<sup>2</sup>). Things are looking bad, we still have two moves to verify
	and we already lost two of our variables.
	<br><br>
	We now verify R3. For this we remark that
	<br><br>
	<center>
	http://katlas.math.toronto.edu/drorbn/images/f/fb/Jones-r3.png
	<br>
	</center>
	<br>

	The two diagrams with coefficients B coincide; and the diagrams with coefficients A differ by
	two R2 moves. Now we verify R1:
	<br><br>
	<center>
	http://katlas.math.toronto.edu/drorbn/images/9/99/Jones-r1.png
	<br>
	</center>
	<br>
	The right hand side evaluates to A + A<sup>-1</sup>(-A<sup>2</sup> -A<sup>-2</sup>) = -A<sup>3</sup> 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 <i>writhe</i>. If D is a diagram of an oriented knot, we
	define
	<br><br>
	<center>
	http://katlas.math.toronto.edu/drorbn/images/8/88/Mathtext1.png
	</center>
	<br>
	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):
	<br><br>
	<center>
	http://katlas.math.toronto.edu/drorbn/images/3/38/Crossings.png
	<br>
	</center>
	<br>

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

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

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

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

Drorbn: 06-1350/week1notes moved to 06-1350/Class Notes for Tuesday September 12

2006-09-21T00:06:46Z

06-1350/week1notes moved to 06-1350/Class Notes for Tuesday September 12

← Older revision

Revision as of 20:06, 20 September 2006

Oivrii at 21:49, 19 September 2006

2006-09-19T21:49:55Z

← Older revision		Revision as of 17:49, 19 September 2006
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			{{06-1350/Navigation}}

	===Introduction===		===Introduction===

Oivrii at 21:49, 19 September 2006

2006-09-19T21:49:25Z

← Older revision		Revision as of 17:49, 19 September 2006
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	===Introduction===		===Introduction===

	We wish to define a <i>knot</i> as a continuous injective map from the circle to 3-dimensional Euclidean space		We wish to define a <i>knot</i> as a continuous injective map from the circle to 3-dimensional Euclidean space
	up to continuous homotopy.		up to continuous homotopy.
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	R1: top left, R2: top right, R3: bottom middle		R1: top left, R2: top right, R3: bottom middle
	</center>		</center>
	<br>		<br>


	===3 Colouring===		===3 Colouring===

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	<br><br>		<br><br>
	<center>		<center>
	"http://katlas.math.toronto.edu/drorbn/images/3/38/Crossings.png		http://katlas.math.toronto.edu/drorbn/images/3/38/Crossings.png
	<br>		<br>
	</center>		</center>

Oivrii: /* Introduction */

2006-09-19T21:48:50Z

Introduction

← Older revision		Revision as of 17:48, 19 September 2006
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	===Introduction===		===Introduction===

			We wish to define a <i>knot</i> as a continuous injective map from the circle to 3-dimensional Euclidean space
			up to continuous homotopy.
			<br><br>
	Unfortunately this definition doesn’t quite work as most of the knots that we wish to		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		distinguish will actually be equivalent (we may shrink down the knotted part to a point). For this
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	R1: top left, R2: top right, R3: bottom middle		R1: top left, R2: top right, R3: bottom middle
	</center>		</center>
	<br>		<br>


	===3 Colouring===		===3 Colouring===

Oivrii at 21:43, 19 September 2006

2006-09-19T21:43:50Z

← Older revision		Revision as of 17:43, 19 September 2006
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	~~<b><font size~~=~~"+1">~~Introduction~~</font></b>~~		===Introduction===

	Unfortunately this definition doesn’t quite work as most of the knots that we wish to		Unfortunately this definition doesn’t quite work as most of the knots that we wish to
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	<br>		<br>

	~~<b><font size~~=~~"+1">~~3 Colouring~~</font></b>~~		===3 Colouring===
	<br><br>

	We give an example of an invariant. Define I<sub>3</sub> : {diagrams}→{true, false},		We give an example of an invariant. Define I<sub>3</sub> : {diagrams}→{true, false},
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	Showing that invariance under R3 is more tedious (and is left as an exercise).		Showing that invariance under R3 is more tedious (and is left as an exercise).

			===Jones Polynomial===
	<br><br>

	<b><font size="+1">Jones Polynomial</font></b>
	<br><br>

	The simplest way to define the <i>Jones polynomial</i> is via the <i>Kauffman bracket</i>. The idea is to		The simplest way to define the <i>Jones polynomial</i> is via the <i>Kauffman bracket</i>. The idea is to
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	however, the right way out of this is to define another invariant which fails in the exactly the		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.		same way, and multiply it with the bracket polynomial.

	<br><br>
			===Writhe===
	<b><font size="+1">Writhe</font></b>

	<br><br>
	The invariant we are looking for is called the <i>writhe</i>. If D is a diagram of an oriented knot, we		The invariant we are looking for is called the <i>writhe</i>. If D is a diagram of an oriented knot, we
	define		define
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	</center>		</center>
	<br>		<br>
	We now show that the ~~write~~ is also invariant under R2 and R3, under R1 we gain ±1		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		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.		diagonal crossing doesn’t change sign and the other two are reversed.

Oivrii at 21:40, 19 September 2006

2006-09-19T21:40:43Z

← Older revision		Revision as of 17:40, 19 September 2006
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	</center>		</center>
	<br>		<br>
		⚫	Each summand will have no crossing and thus will be a union of (possibly nested) unknots.

⚫	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 d<sup>k-1</sup> for some indeterminate d. Our hopes		We define the bracket polynomial of k unknots to be d<sup>k-1</sup> for some indeterminate d. Our hopes
	that our polynomial in ℤ[d,A,B] will be an invariant under the Reidmeister moves. We first		that our polynomial in ℤ[d,A,B] will be an invariant under the Reidmeister moves. We first
	verify R2.		verify R2.

	<br><br>		<br><br>
	<center>		<center>
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	define		define
	<br><br>		<br><br>
	<center>		<center>
	http://katlas.math.toronto.edu/drorbn/images/8/88/Mathtext1.png		http://katlas.math.toronto.edu/drorbn/images/8/88/Mathtext1.png
	</center>		</center>
Line 130:		Line 128:
	<br><br>		<br><br>
	<center>		<center>
	~~src=~~"http://katlas.math.toronto.edu/drorbn/images/3/38/Crossings.png		"http://katlas.math.toronto.edu/drorbn/images/3/38/Crossings.png
	<br>		<br>
	</center>		</center>
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	</center>		</center>
	<br>		<br>
		⚫	We now show that the write is also invariant under R2 and R3, under R1 we gain ±1

⚫	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		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.		diagonal crossing doesn’t change sign and the other two are reversed.
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	<br><br>		<br><br>
	<center>		<center>
	~~src="~~http://katlas.math.toronto.edu/drorbn/images/f/f9/Writhe-reid.png		http://katlas.math.toronto.edu/drorbn/images/f/f9/Writhe-reid.png
	<br>		<br>
	</center>		</center>
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	It follows that 〈D〉⋅(-A<sup>-3</sup>)<sup>w(d)</sup> is a knot invariant. This is a polynomial in A, we now		It follows that 〈D〉⋅(-A<sup>-3</sup>)<sup>w(d)</sup> is a knot invariant. This is a polynomial in A, we now
	substitute q<sup>~~<img~~		substitute q<sup>
	~~src="~~http://katlas.math.toronto.edu/drorbn/images/2/23/Mathtext2.png~~" alt="14" class="frac" align="middle">~~</sup> for A and call this the <i>Jones polynomial</i> (which strictly speaking, is not really a		http://katlas.math.toronto.edu/drorbn/images/2/23/Mathtext2.png</sup> for A and call this the <i>Jones polynomial</i> (which strictly speaking, is not really a
	polynomial).		polynomial).

Oivrii at 21:37, 19 September 2006

2006-09-19T21:37:32Z

New page

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).