From: lrudolph@panix.com (Lee Rudolph)
Newsgroups: sci.math
Subject: Re: What is a fundamental group?
Date: 12 Sep 1998 14:53:57 -0400

>Could anyone explain what a fundamental group is, in this context : 
>
>'A knot is an embedding K of the circle S1 into the 3-sphere S3.  An
>important invariant is the fundamental group of the complement S3-K, which
>we call the group of K'.

Depending on what you know already, various explanations will be
more suitable.  I will assume you know what a group (plain and
simple) is.

(1) If X is a sufficiently "nice" topological space (for instance,
an open subset of Euclidian 3-space E-3, like the complement S3-E
= E3-K of a knot K  which without loss of generality goes through
the "point at infinity" which you add to E3 to make its one-point
compactification S3), and * is a point of X, let L(X,*) be the
set of closed loops in X based at *: an element of L(X,*) is a
continuous map f: [0,1] -> X such that f(0)=f(1)=*.  Define an
equivalence relation ~ and a binary operation @ on L(X,*):
f ~ g if f is homotopic to g rel. * (that is, if there is 
a continuous map F from the square [0,1]x[0,1] into X
such that F(0,t)=f(t), F(1,t)=g(t), F(t,0)=F(t,1)=* 
for all t); f@g(t) is set equal to f(2t) for t in [0,1/2]
and to g(2t-1) for t in [1/2,1].  It turns out that the
~-equivalence classes in L(X,*), with the binary operation
that they (can be shown to) inherit from @, form a group.
This group is called the "fundamental group of X with basepoint *"
(or "the first homotopy group of X with basepoint *").
If X (always taken to be "nice") is connected, then
it also turns out that the fundamental group of X with 
basepoint * and the fundamental group of X with basepoint *'
are isomorphic for any basepoints *, *'.

(2) If K is "tame" (meaning that the "embedding" mentioned
somewhat sloppily in the definition you quote is a smooth,
or piecewise linear, embedding, up to homeomorphism of S3),
then there's a much more naive, combinatorial definition
of the "knot group" (and the group it defines can be proved
to be isomorphic to the fundamental group of the complement).
A tame knot can be described by a "knot diagram": assuming
for convenience that the knot is piecewise linear (i.e.,
it's a space polygon--with finitely many sides--in R^3),
then "almost every" orthogonal projection of K to a flat
2-plane is nice in the sense that (a) no point of the plane
is the image of more than two points of K, and there are
only finitely many "doublepoints" in the plane which do
come from two distinct points of K, (b) neither of the 
two points that project to a doublepoint is a vertex
of the polygon K.  Then the projection D(K) is a (non-simple)
closed polygon in the plane, and at each doublepoint you
can tell which of the two edges through the doublepoint comes
from  the "higher", and which from the "lower", of two edges
in K.  Conventionally you would draw D(K) not as an actual
closed polygon but rather as a broken polygon with "crossings"
indicated locally.  These crossings divide D(K) into a
set of pairwise disjoint "over-arcs", say x_1,...,x_n.
Choose one of the two orientations of K (it makes no 
difference which one).  At each crossing three over-arcs
come together, and by turning your head appropriately you're
in one of two cases: the over-arc that's actually *over*
crosses from left to right, while both it and the other arcs
go up; or that over-arc crosses from right to left, while
both it and the other arcs go up.

 x    y
  \  / 
    / 
   /
  /  \
 y    z

The picture illustrates case 1 (both arcs are going up!).
Now, corresponding to this crossing, write down the formal
equation yz = xy.  Do this for all the crossings.  The
group generated by x_1,...,x_n subject to the relations
just written down is the knot group.  (The only example
where this doesn't work quite as stated is the trivial
example K=O of an unknot which has a diagram with NO
crossings.  Then the group is infinite cyclic, with
one over*circle* instead of any arcs, therefore, one
generator, no relations.  A figure-8-like diagram of
the trivial knot has one crossing and one over-arc:
the arc appears at the crossing in all three roles,
so you get the relation xx=xx, again, infinite cyclic
group.  A trefoil diagram has three over-arcs, x, y, z,
and three relations xy=yz, yz=zx, zx=xy.  Finally we
get a group that isn't infinite cyclic: for if we send
x to the transposition (1 2), y to (2 3), and z to (1 3),
then the knot group I defined maps onto the 6-element
non-abelian group S_3.) Of course, showing that this
group is an *invariant* of the knot (and not just of
one particular diagram) takes work.

Lee Rudolph