Proof The idea is to turn c's into a's by flyping. A flype moves a crossing across a part of a diagram at the expense of turning the part over:
Notice that if we introduce two crossings using R2 then move one over with a flype we get:
Suppose c > 0 and consider the sequence of flypes:
Repeat c times to get:
Similarly if c < 0 . If b
is
odd K will have been rotated through pi about an horizontal
axis.
Use this procedure to eliminate cn , cn-1 , ... ,c2
in turn. Then finally:
We can be more efficient in unwrapping a daisy chain. For example:
After n - 2 double flypes we have:
We now use flips to eliminate the unshaded crossings. Here is
a flip:
Think of this as turning R over to change the sign of a crossing
between
middle and outer string at the expense of adding a crossing between
middle
and other outer string.
Combine a flip with R2 to get 3 new crossings all of the same sign
as
follows:
Apply R2 flips:
etc.
If n odd get:
which is isotopic to
and if n even get:
which is isotopic to:
Remark It follows that the classes of knots defined by
Daisy chains, 4-plats and 2-bridge knots are identical.
