061350/Class Notes for Tuesday September 12

Introduction
We wish to define a knot as a continuous injective map from the circle to 3dimensional 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 I_{3} : {diagrams}→{true, false},
I_{3} 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 trichromatic.
We can now distinguish the trefoil from the unknot as the trefoil is 3colourable 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 I_{3} is an invariant, that is we need to show that I_{3} is preserved after R1, R2 and R3. The test for R1 is straight forward:
http://katlas.math.toronto.edu/drorbn/images/2/25/3cola.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).