Topology in Molecular Biology

5 From Tangle Fractions to DNA 85

knots and links, we can find astate summationformula for thebracket of the
tangle, denoted〈T〉,by summing over the states obtained by smoothing each
crossing in the tangle. For this we define theremainder of a state, denoted
RS, to be either the tangle [0] or the tangle [∞]. Then the evaluation of〈T〉
is given by
〈T〉=

∑

S

<T|S>δ||S||〈RS〉,

where〈T|S〉is the product of the vertex weights (AorA−^1 ) of the stateS
ofT. The above formula is consistent with the formula for knots obtained by
taking the closureN(T)orD(T). In fact, we have the following formula:

〈N(T)〉=

∑

S

〈T|S〉δ||S||〈N(RS)〉.

Note that〈N([0])〉=δand〈N([∞])〉=1.A similar formula holds for〈D(T)〉.
Thus,〈T〉appears as a linear combination with Laurent polynomial coeffi-
cients of〈[0]〉and〈[∞]〉,i.e.〈T〉takes values in the free module overZ[A, A−^1 ]
with basis{〈[0]〉,〈[∞]〉}.Notice that two elements in this module are equal
iff the corresponding coefficients of the basis elements coincide. Note also that
〈T〉is an invariant of regular isotopy with values in this module. We have just
proved the following:

Lemma 4.LetT be any two-tangle and let〈T〉be the formal expansion of
the bracket on this tangle. Then there exist elementsnT(A)anddT(A)in
Z[A, A−^1 ],such that

〈T〉=dT(A)〈[0]〉+nT(A)〈[∞]〉,

andnT(A)anddT(A)are regular isotopy invariants of the tangleT.

In order to evaluate〈N(T)〉in the formula above we need only apply the
closureNto [0] and [∞].More precisely, we have:

Lemma 5.〈N(T)〉=dTδ+nTand〈D(T)〉=dT+nTδ.

Proof.Since the smoothings of crossings do not interfere with the closure
(NorD), the closure will carry through linearly to the whole sum of〈T〉.
Thus,

〈N(T)〉=dT(A)〈N([0])〉+nT(A)〈N([∞])〉=dT(A)δ+nT(A),

〈D(T)〉=dT(A)〈D([0])〉+nT(A)〈D([∞])〉=dT(A)+nT(A)δ.

We define now thepolynomial fraction,fracT(A), of the two-tangleTto
be the ratio

fracT(A)=

nT(A)

dT(A)

in the ring of fractions ofZ[A, A−^1 ] with a formal symbol∞adjoined.