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06-1350/Class Notes for Tuesday October 3

2006-10-22T02:35:23Z

69.197.142.199:

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=== Attempted solution of the exercise ===

I'm talking about the one stating that

Z{Ribbon knots} = {<math>u\Psi: \Psi \in A(O-O-...-O), d\Psi = Z(O O...O)</math>}

''my dumbbel didn't come out perfectly, and there might be further ugliness due to my lack of latex ability- I apologize''

The inclusion of the left in the right is obvious. We're trying to do the reverse one.

Assume that we have a <math>\Psi \in A(O-...-O)</math> such that <math>d\Psi=Z(O...O)</math>.

What we need to show is that <math>\Psi=Z(\gamma)</math> for some <math>\gamma \in K(O-...-O),~ d\gamma=(O...O)</math>.

To do this, we first prove that <math>d:A(O-...-O) \to A(O...O)</math> is (almost) injective.

Let's just do this for a simple (two circles) dumbbell, it will generalize to longer dumbbells easily.

Injectivity means that for some <math>\Psi \in A(O-O)</math>, if <math>d\Psi \in A(O O)</math> is zero on all
invariants <math>\varphi \in Inv ~ K(O O)</math>, then <math>\Psi</math> has to be zero on all invariants in <math>Inv ~ K(O-O)</math>.

We use that <math>d \Psi ( \varphi)=\Psi ( \varphi \circ d)</math>

The key here is that the edge connecting the two circles carries no topological information, it's contractible. Therefore, an invariant of the knotted dumbbell cannot depend on how the edge wiggles- we can contract the edge to a point and get a vedge of two knotted circles, and there's only one way to vedge circles together. Therefore, all invariants of the knottings of the dumbbell are of the form <math>\varphi \circ d</math>, where <math>\varphi</math> is an invariant of the knottings of two circles. And hence we're done.

Of course there's a lie here: we are talking about framed graphs and so the edge does carry some information: which side it's attached to on each circle, and whether it's twisted or not. So there are <math>2^3=8</math> preimages of each value of d. (Still a lie: If the circles are linked, I think it's countably many, given by the sides and how many times the edge is twisted, but we only care about the unlinked case, and there two twists equal nothing.)

Now, still thinking about a simple dumbbell, we can show that <math>\Psi=Z\gamma</math> for some knotted dumbbell <math>\gamma</math>.

We have <math>d:A(O-O) \to A(O O), ~ d(\Psi)=Z(O O)</math>.
We know that <math>Z(O O)</math> has 8 preimages, and all of those are Z-images of the dumbbells we get from the 8 different ways of attaching d (think connected sum). Therefore, one of those has to be the <math>\gamma</math> we are searching for.

Clearly, this works just as well for the dumbbell with k circles, with <math>2^{3(k-1)}</math> in place of 8.

And obviously, for the <math>\gamma</math> we've found, <math>d\gamma=(O...O)</math>.

And we are done.

''Yesterday I posted something stupid about this same problem, then realized and took it off, so I'm not sure anymore that this really works... any comments welcome!
''

06-1350/Class Notes for Tuesday September 12

2006-10-22T02:31:35Z

69.197.142.199: /* 3 Colouring */

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===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 trefoil from the unknot as the trefoil is 3-colourable while the
unknot is not. However, we cannot distinguish (yet) the trefoil 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).

06-1350/Class Notes for Thursday September 14

2006-10-22T02:30:21Z

69.197.142.199: /* Jones Polynomial */

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The Mathematica notebook we wrote is [http://katlas.math.toronto.edu/svn/06-1350/ComputingTheJonesPolynomial.nb here].

===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 trefoil, 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).