06-1350/Class Notes for Tuesday September 12

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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).