06-1350/Class Notes for Tuesday September 12: Difference between revisions

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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>
Line 60: Line 60:
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 &#x2124;[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&#8217;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 &#x2329;D&#x232A;&#x22C5;(-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).

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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.

http://katlas.math.toronto.edu/drorbn/images/e/e1/Trifoil.png


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.



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


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.

http://katlas.math.toronto.edu/drorbn/images/9/93/Trifoil2.png


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:

http://katlas.math.toronto.edu/drorbn/images/2/25/3col-a.png


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).

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