Topology in Molecular Biology

86 L.H. Kauffman and S. Lambropoulou

Lemma 6.fracT(A)is an invariant of ambient isotopy for two-tangles.

Proof.SincedT andnT are regular isotopy invariants ofT, it follows that
fracT(A) is also a regular isotopy invariant ofT. Suppose nowTγisTwith
a curl added. Then〈Tγ〉 =(−A^3 )〈T〉(same remark for ̄γ). So,nTγ(A)=
−A^3 nT(A)anddTγ(A)=−A^3 dT(A).Thus,nTγ/dTγ=nT/dT.This shows
thatfracT is also invariant under the Reidemeister move I, and hence an
ambient isotopy invariant.

Lemma 7.LetTandSbe two two-tangles. Then, we have the following for-
mula for the bracket of the sum of the tangles.

〈T+S〉=dTdS〈[0]〉+(dTnS+nTdS+nSδ)〈[∞]〉.

Thus

fracT+S=fracT+fracS+

nSδ

dTdS

.

Proof.We do first the smoothings inTleavingSintact, and then inS:

<T+S>=dT〈[0] +S〉+nT〈[∞]+S〉

=dT〈S〉+nT〈[∞]+S〉

=dT(dS〈[0]〉+nS〈[∞]〉)

+nT(dS〈[∞] + [0]〉+nS〈[∞]+[∞]〉)

=dT(dS〈[0]〉+nS〈[∞]〉)+nT(dS〈[∞]〉+nSδ〈[∞]〉)

=dTdS〈[0]〉+(dTnS+nTdS+nSδ)〈[∞]〉.

Thus,nT+S=(dTnS+nTdS+nSδ)anddT+S=dTdS.A straightforward
calculation gives nowfracT+S.

As we see from Lemma 4,fracT(A) will be additive on tangles if

δ=−A^2 −A−^2 =0.
Moreover,fromLemma2wehaveforδ=0,〈N(T)〉=nT,〈D(T)〉=dT.
This nice situation will be our main object of study in the rest of this section.
Now, if we setA=

√

iwherei^2 =− 1 ,then it is

δ=−A^2 −A−^2 =−i−i−^1 =−i+i=0.

For this reason, we shall henceforth assume thatAtakes the value

√

i.So
〈K〉will denote〈K〉(

√

i) for any knot or linkK.

We now define thetwo-tangle fractionF(T) by the following formula:

F(T)=i

nT(

√

i)

dT(

√

i)

.