Topology in Molecular Biology

5 From Tangle Fractions to DNA 95

R=[1]+ S=[-3]

[2]

_^1

~

Fig. 5.20.An example of the special cut

T T

open

to obtain

K = N([1] +T) = = [1] +T

Fig. 5.21.A standard cut

has fractionP/Q,for|P|>|Q|.ThusRcan be written in the formR= [1]+T
orR=[−1] +T.We consider the rational knot diagramK=N([1] +T),see
Fig. 5.21. (We analyseN([−1] +T) in the same way.) The tangle [1] +Tis
said to arise as astandard cutonK.
Notice that the indicated horizontal crossing ofN([1] +T) could also be
seen as a vertical one. So we could also cut the diagramKat the two other
marked points (see Fig. 5.22) and still obtain a rational tangle, sinceT is
rational. The tangle obtained by cuttingKin this second pair of points is said
to arise as aspecial cutonK.Figure 5.22 demonstrates that the tangle of the
special cut is the tangle [−1]− 1 /T. So we haveN([1] +T)∼N([−1]− 1 /T).
Suppose nowF(T)=p/q.ThenF([1] +T)=1+p/q =(p+q)/q,while
F([−1]− 1 /T)=− 1 −q/p=(p+q)/(−p),so the two rational tangles that
give rise to the same knotKare not isotopic. Since−p≡qmod(p+q),this
equivalence is another example for Theorem 2. In Fig. 5.22 if we tookT=1/[2]
then [−1]− 1 /T=[−3] and we would obtain the example of Fig. 5.20.
The proof of Theorem 2 can now proceed in two stages. First, given a
rational knot diagram we look for all possible places where we could cut and
open it to a rational tangle. The crux of our proof in [17] is the fact that
all possible “rational cuts” on a rational knot fall into one of the basic cases
that we have already discussed, i.e. we have the standard cuts, the palindrome
cuts and the special cuts. In Fig. 5.23 we illustrate on a representative rational
knot, all the cuts that exhibit that knot as a closure of a rational tangle. Each
pair of points is marked with the same number. The arithmetic is similar to