Algebra of tangles

In what follows -L is the tangle L reflected in the plane of the tangle (diagram). Lower case letters are integral tangles and L is an arbitrary tangle. Products are associated from the left and -mn...pq means (-m)(-n)...(-p)(-q) etc. L- is L(-1)0,    L+ is L10, and L+T is defined as indicated in the diagram following.

 

 

The following is easily proved using only the Reidemeister move R2.

 Lemma 1
                    i)   Lp+q = L(p+q) = Lp0q,
                   ii)  1q = q+1, (-1)q = q-1, 0p = , p = p.

The next lemma is more subtle. It relies on a geometrical fact. Consider two perpendicular planes in 3-space and their line of intersection. Reflection in the planes and rotation through 1800 about the line define a 3 element cyclic subgroup of the group of isometries of 3-space.

 Lemma 2
                  i)  L1- = (-L)+(-1),
                 ii)  L+ = ((-L)+(-1))1.

Proof     i)

The isotopy arises by rotating the  square about the NW, SE diagonal and observing that this rotation is the composition of reflection in the (plane through the) diagonal and reflection in the plane of the diagram.

     ii)

This follows from i) after rotating both sides about the NW, SE diagonal. 

Remark    The isotopies of lemma 2 can be seen as a kind of flyping - call it flipping:

 

Proposition 3
                          i)         Lq1- = -L(q+1),
                         ii)    L(q+1)- = -Lq1,
                        iii)         (-L)q = (L+(-1))1(q-1).

 Proof
           i)     Lq1- = (-L)(-q)+(-1)                by 2 i)
                           = (-L)(-q-1)                     by 1 ii)
                           = -L(q+1).

          ii)     L(q+1)- = Lq01-                       by 1 i)
                               = (-L)(-q)(-0)+(-1)       by 2 i)
                               = -Lq1                          by 1 ii)

        iii)         (-L)q = (-L)1 + (q-1)               by 1 ii)
                             = (L+(-1))10+(q-1)         by 2 ii)
                             = (L+(-1))1(q-1)              by 1 ii) 

 Remarks    Suppose given an arbitrary rational tangle  L = mn...rs which is not just an integer or . We can get L into standard form (m,n,...,r all positive and m>1) as follows.By lemma 1 we can assume none of m,n,...,r are 0. Now work from the right to left. Apply Proposition 3 iii) repeatedly to change signs and use lemma 1 to eliminate any 0's which may arise as we go. Finally apply lemma 1 to get m>1. eg.

Formula 3 iii) is just the Lagrange formula of 9.8. See the end of lecture 24 for another example. Thus R2 plus flyping (to convert west to east and north to south twists) and flipping can get us from one representation of a rational tangle to another. Could flipping have a wider role?
The formulae given by Conway: mn...pq1- = -mn...p(q+1) and mn...p(q+1)- = -mn...pq1 follow from Proposition 3 i) and ii).

Definition     An algebraic tangle is composed from integer tangles by using the sum and product operations.
 

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