Braid theory plays a fundamental role in knot theory since any oriented link can be represented by a closed braid (Alexander's Theorem [24]). The identification between links, braids, and the braid group allows us to pass back and forth between the geometric study of braids (and hence knots and links) and the algebraic study of the braid group. For some problems the original geometric study of braids is useful. For many other problems a purely algebraic approach provides the only intelligible solution.
Figure 5.14: (a) A braid on n-strands. (b) A trivial braid.
A geometric braid is constructed between two horizontal level lines with n base points chosen on the upper and lower level lines. From upper base points we draw n strings or strands to the n lower base points (Fig. 5.14(a)). Note that the strands have a natural orientation from top to bottom. The trivial braid is formed by taking the ith upper base point directly to the ith lower base point with no crossings between the strands (Fig. 5.14(b)). More typically, some of the strands will intersect. We say that the i+1st strand passes over the ith strand if there is a positive crossing between the two strands (see Crossing Convention, section 5.3.1).
Figure 5.15: Braid operators: (a) 
, overcross; (b) 
, undercross.
As illustrated in Figure 5.15(a), an overcrossing (or right-crossing)
between the i+1st and ith string is denoted by the symbol
. The inverse represents an undercross (or left-cross), i.e.,
the i+1st strand goes under the ith strand (Fig. 5.15(b)).
By connecting opposite ends of the strands we form a closed braid. Each closed braid is equivalent to a knot or link, and conversely Alexander's theorem states that any oriented link is represented by a closed braid.
Figure 5.16: (a) Braid of a trefoil knot. (b) Braid of a Hopf link.
Figure 5.16(a) shows a closed braid on two strands that is equivalent to the trefoil knot; Figure 5.16(b) shows a closed braid on three strands that is equivalent to the Hopf link. The representation of a link by a closed braid is not unique. However, only two operations on braids (the Markov moves) are needed to prove the identity between two topologically equivalent braids [24].
A general n-braid can be built up from
successive applications of the operators and .
This construction is illustrated for a braid on four strands in
Figure 5.17.
Figure 5.17: Braid on four strands whose braid word is 
.
The first crossing between the second and third strand is positive,
and is represented by the operator . The next crossing is negative,
, and is between
the third and fourth strands. The last positive crossing is represented
by the operator 
. Each geometric diagram for a braid is
equivalent to an algebraic braid word constructed from the
operators used to build the braid. The braid word for our example
on four strands is .
Two important conventions are followed in constructing a braid word.
First, at each level of operation ( , ) only
the ith and i+1st strands are involved.
There are no crossings between any other strands at a given
level of operation. Second,
it is not the string, but the base point which is numbered.
Each string involved in an operation increments or decrements
its base point by one. All other strings keep their base points fixed.
The braid group on n-strands, 
, is defined by the operators
. The identity element of is the
trivial n-braid. However, as previously mentioned, the expression
of a braid group element (that is, a braid word) is not unique.
The topological equivalence between seemingly different braid words is guaranteed by the braid relations (Fig. 5.18):
The braid relations are taken as the defining relations of the braid group. Each topologically equivalent class of braids represents a collection of words that are different representations for the same braid in the braid group. In principle, the braid relations can be used to show the equivalence of any two words in this collection. Finding a practical solution to word equivalence is called the word problem. The word problem is the algebraic analog of the geometric braid equivalence problem.
