Ernst, C.; Sumners, D. W. A calculus for rational tangles: applications to DNA recombination. Math. Proc. Cambridge Philos. Soc. 108 (1990), no. 3, 489–515.

Vazquez, Mariel; Sumners, De Witt; Tangle analysis of Gin site-specific recombination. Math. Proc. Cambridge Philos. Soc. 136 (2004), no. 3, 565–582.

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math.GT/0501173; Title: Classification of Tangle Solutions for Integrases, A Protein Family that Changes DNA Topology; Authors: Dorothy Buck, Cynthia Verjovsky Marcotte

math.GT/0412467; Title: Tangle Solutions for a Family of DNA-Rearranging Proteins; Authors: Dorothy Buck, Cynthia Verjovsky Marcotte

Useful facts about tangles:

p. 491:

The tangle corresponding to \frac{a_{1}}{b_{1}} is the same as the tangle corresponding to \frac{a_{2}}{b_{2}} if and only if \frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} Parity:

In order to determine the parity of \frac{a}{b}, the Euler bracket function will be used. Let E[x_{1}, ..., x_{n}] be the Euler bracket function which equals the sum of products of the x_{i}’s where zero or more disjoint pairs of consecutive x_{i}’s are omitted Roberts. If n = 0 then E[x_{1}, ..., x_{n}] = E[] = 1. If n < 0 define E[x_{1}, ..., x_{n}] = 0. Let [x_{n}, ..., x_{1}] denote the continued fraction x_{n}+_{\frac{1}{x_{n−1}+_{\frac{1}{...+ \frac{1}{x_{1}}}}}} , the fraction corresponding to the tangle

(x_{1}, ..., x_{n}). The following useful facts for n ≥ 1 can be found in

1.) E[x_{1}, ..., x_{n}] = x_{1}E[x_{2}, ..., x_{n}] + E[x_{3}, ..., x_{n}].

2.) [x_{n}, ..., x_{1}] = E[x_{1}, ..., x_{n}]/E[x_{1}, ..., x_{n−1}].

3.) Let a = E[x_{1}, ..., x_{n}], b = E[x_{1}, ..., x_{n−1}]. If y = (−1)^{n+1}E[x_{2}, ..., x_{n−1}]

and x = (−1)^{n+1}E[x_{2}, ..., x_{n}], then bx − ay = 1.

Lemma: If a is even, then the tangle \frac{a}{b} has parity 0. If a is odd, then if b is odd, the tangle \frac{a}{b} has parity 1 and if b is even, the tangle \frac{a}{b} has parity ∞. Proof. Induction on n noting that if \frac{a}{b} = [x_{n}, ..., x_{1}], then a = E[x_{1}, ..., x_{n}] and b = E[x_{1}, ..., x_{n−1}] . Both N(a/b) and D(a/b) are 4-plats. The knot/link N(a/b) is the 4-plat S(a, −b). The knot/link D(a/b) is the 4-plat S(b, a). p. 493: The 4-plats N(a_{1}/b_{1}) and N(a_{2}/b_{2}), a_{i} ≥ 0, are are the same if and only if a_{1} = a_{2} and b_{1}b_{2}^{±1} ≅ 1 (mod a_{1}). i.e. b_{1} = b_{2} (mod a_{1}) or b_{1}b_{2} = 1 (mod a_{1}) or equivalently, b_{1} = b_{2} (mod a_{1}) or b_{1} = b_{2}^{−1} (mod a_{1}) p. 494: Lemma 2.1: N(\frac{a_{1}}{b_{1}} + \frac{a_{2}}{b_{2}} ) = N(\frac{a_{1}b_{2}+a_{2}b_{1}}{a_{1}b′_{2}+a′_{2}b_{1}} ) where a′_{2}b_{2}−a_{2}b′_{2} = 1 [5]. p. 496: Thm 2.2: If N(\frac{j}{p} + \frac{f}{g}) = N(\frac{a}{b}), then \frac{f}{g} = \frac{da−jb′}{pb′−qa} for some integers d, q, and b′ such that pd − qj = 1, b′b^{±1} = 1 mod a.

Proof. N(\frac{j}{p} + \frac{f}{g}) = N(\frac{jg+pf}{dg+qf}) = N(\frac{a}{b}) where d and q are any integers such that pd − qj = 1. Since \frac{f}{g} = \frac{−f}{−g} and N(\frac{j}{p} + \frac{−f}{−g}) = N(\frac{−jg−pf}{−dg−qf}) = N(\frac{jg+pf}{dg+qf}), jg + pf will be taken to have the same sign as a. Thus jg + pf = a and dg + qf = b′ where b′b^{±1} = 1 mod a. Multiplying the first equation by d, the second equation by j, and solving for f results in f = (pd − qj)f = da − jb′. Multiplying the first equation by q, the second equation by p, and solving for g results in g = (pd − qj)g = pb′ − qa .

Alternate proof use:

N(A ∘ (c_{1}, ..., c_{n}) + B) = N(A + B ∘ (c_{n}, ..., c_{1})) if A or B is rational, n odd.

N(A + C) = N(C + A)

A + C ≠ C + A

p. 498:

Corollary 2.3: Let B, E be distinct rational tangles, Then there are at most two distinct rational solutions to the system of tangle equations N(X + B) = N(\frac{a}{b}) N(X + E) = N(\frac{z}{v})

A tangle C is locally unknotted if every 3-ball which intersects the arcs of C in exactly 2 points contains an unknotted spanning arc. A tangle is locally knotted if not locally unknotted. A tangle is prime if it is neither rational nor locally knotted.

Lemma 2.4: If R is rational, then R + R is locally unknotted. If R + R is rational, then R is integral. If R is not integral and does not have parity infinity, then R + R is prime

If A is a prime tangle and C is locally unknotted and not the ∞-tangle, then A + C is a prime tangle [9][10].

D(A + C) = D(A)#D(C).

p. 499:

Lemma 3.1: N(A + C) = 4-plat implies at least one of A or C is rational or locally knotted [8][9].

Cyclic Surgery Theorem [CGLS]: Suppose that M is not a Seifert fibered space. If π_{1}(M(r)) and π_{1}(M(s)) are cyclic, then |i(r, s)| ≤ 1. Hence there are at most three slopes r such that π_{1}(M(r)) is cyclic.

R is a rational tangle if and only if its double branch cover is a solid torus.

M a ambient isotopic to a sum of rational tangles if and only if its double branch cover is a Seifert fiber space with orbit surface a disk and n > 0 exceptional fibers [11].

Lemmas 3.1 - 3.7 are used to prove that the double branch cover of X is a torus and hence X is a rational tangle.

We will instead use the following to solve equations where X is a generalized M-tangle. This includes the case when X is rational.

If N(U + \frac{f_{1}}{g_{1}} ) = N(\frac{a}{b}) and N(U + \frac{f_{2}}{g_{2}} ) = N(\frac{z}{v}) and if |t| > 1, then U is a generalized M-tangle or equivalently, U = A ∘ C where A is a finite sum of rational tangles and C is a rational tangle.

N(U + \frac{0}{1}) = N(\frac{a}{b}) and N(U + \frac{t}{w}) = N(\frac{z}{v}) where N(\frac{a}{b}) and N(\frac{z}{v}) are unoriented 4-plats and U is a generalized M-tangle if and only if the following hold:

(a) If w ≅ ±̸1 mod t, then there exists an integer, b′ such that b′b^{±1} = 1 mod a, and for any integers x and y such that b′x − ay = 1,

N(\frac{z}{v}) = N(\frac{tb′+wa}{ty+wx})

In this case, U = \frac{a}{b′} for all b′ satisfying the above.

(b) If w ≅ ±1 mod t, then there exists relatively prime integers, p and q, where p may be chosen to be positive, such that

N(\frac{z}{v} ) = N(\frac{tp(pb−qa)±a}{tq(pb−qa)±b})

In this case, the solutions for U are (\frac{da−jb}{pb−qa} + \frac{j}{p}) ∘ (h, 0) and (\frac{j}{p} + \frac{da−jb}{pb−qa}) ∘ (h, 0), for all p, q satisfying the above, d and j are any integers such that pd − qj = 1, and h = \frac{−w±1}{t} where the ± sign agrees with that in (e-w1) (note, the choice of j and d such that pd − qj = 1 has no effect on U).

Suppose bx−ay = 1, N(U + \frac{0}{1}) = N(\frac{a}{b}) and N(U + \frac{t}{w}) = N(\frac{z}{v}) where N(\frac{a}{b}) and N(\frac{z}{v}) are unoriented 4-plats. If w ≅ ±̸1 or if U is rational, then \frac{t}{w} = \frac{xz−av′}{bv′−yz−kt} and U = \frac{a}{b+ka} OR \frac{t}{w} = \frac{bz−av′}{xv′−yz−kt} and U = \frac{a}{x+ka} where v′ is any integer such that v′v^{±1} = 1 mod z. If w ≅ ±1 mod t, then t divides z ∓ a.

bottom of p. 502:

Theorem 7 [19] If N(U + \frac{f_{1}}{g_{1}} ) = unknot and N(U + \frac{f_{2}}{g_{2}} ) = N(2z/1) where f_{1}g_{2} − f_{2}g_{1} = ±1, then U is rational.

Theorems 8 and 10 can be used when not all the knots involved are 4-plats and are a generalization of [20] and [21], respectively.

Theorem 8 [16] If d_{t/w}(L, N(a/b)) = 1, then for some integer n, and s ≅ w mod t, |H_{1}(L̃)| ≅< m, L_{1} | tnL_{1} + sm, aL_{1} − bnm >.

Corollary 9 If d_{t/w}(N(a/b), K) = 1, then there exists an integer n such that |H_{1}(M_{K})| ≡ |aw + tbn^{2}| (mod ta).

Theorem 10 [22] If d_{t}(K, K′) = 1, then there exist a ∈ H_{1}(M_{K}) and a′ ∈ H_{1}(M_{K′}) such that λ(a, a) ≡ n/|H_{1}(M_{K})| and λ(a′, a′) ≡ m/|H_{1}(M′_{K})| (mod 1), where

m = n = \frac{±1}{t}(|H_{1}(M_{K})| − |H_{1}(M′_{K})|)

or

−m = n = \frac{±1}{t}(|H_{1}(M_{K})| + |H_{1}(M′_{K})|)

Note that theorem 10 also give information regarding \frac{t}{kt±1} moves since these moves are equivalent to ±t moves.

Theorem [5] Let U and R be tangles such that N(U + iR) =4-plat for some i ≥ 2, and N(U + jR) ≠ N(U + iR) for some j. Then R is a rational tangle. If i ≥ 3, then R is an integral tangle.

Proof [10].

If R were locally knotted, then N(U+iR), i ≥ 2 would be composite. Since 4-plats are prime, R cannot be locally knotted. Suppose R is a prime tangle. By tangle fact (9) U + (i − 1)R is rational or locally knotted. By tangle fact (10), R prime implies (i − 1)R prime and U must be the ∞-tangle or locally knotted.

Note that U cannot be the ∞-tangle. If U were the infinity tangle, then N(U +iR) = D(iR) = D(R)#...#D(R). Since 4plats are prime, D(R) = unknot. But N(U + iR) = D(iR) = unknot = D(jR) = N(U + jR), a contradiction. Thus if U is locally unknotted, R must be rational.

If i ≥ 3, then R does not have parity ∞ since 4-plats have at most two components. If i ≥ 3, U is locally unknotted, and R is not integral, then if U is not integral, U +R and (i−1)R are prime by tangle facts (5) and (10). But N(U + iR) =4-plat would then contradict tangle fact (9). If U is integral and if R is rational, then N(U + iR) = 4-plat, i ≥ 3, if and only if R is integral by tangle fact (8). Thus if U is locally unknotted and i ≥ 3, R must be integral.

Suppose U is locally knotted. Then if U′ is the tangle formed from U by removing the local knot, then N(U′+iR) =unknot, since 4-plats are prime. N(U +jR) ≠ N(U +iR) implies that N(U′ + jR) ≠ N(U′ + iR). Since the unknot is a 4-plat and U′ is locally unknotted, R is rational if i ≥ 2 and integral if i ≥ 3

Cor 6.2: Solve N(U + P) = N(1/0), N(U + R) = N(−2/1), N(U + R + R) = N((−2, −1, −1)) = N(−5/2), N(U + R + R + R) = N((−1, −1, −1, −1, −1)) = N(−8/5)