Quandles

The Alexander Polynomial

Lable the knot with x1, (Subscript[x, 2]), ...satisfying crossing rules:

If knot has n crossings, there are n equations in n unknowns, giving an n×n matrix M.

Theorem. The matrix M is singular. However, its minors (determinates) are all equal up to a multiplicative factor of ±(t^k) for some k. 

Definition. If we normalize these determinates (by multiplying by an approprate ±(t^k) ) so that p(1)  = ±1 or 0 in the case of links, then the result is the Alexander polynomial for the knot.

Important Property. p(t) = (t^k)p(1/t). This means that reversing the order of the coefficients in p(t) has no effect.

Example

<<KnotTheory`

Show[Graphics[{AbsoluteThickness[3], KnotData[{3, 1}, "KnotDiagramData"]}]]

a = tb + (1-t)c
b = tc+ (1-t)a
c = ta + (1-t)b

The Matrix M

MatrixForm[{{1-t, -1, t}, {t, 1-t, -1}, {-1, t, 1-t}}]

Choose a minor.

p[t_]  = Det[{{1-t, -1}, {t, 1-t}}]

Alexander[Knot[3, 1]][t]

Expand[t Alexander[Knot[3, 1]][t]]

Examples

Some others.

(Subscript[4, 1])

Show[Graphics[{AbsoluteThickness[3], KnotData[{4, 1}, "KnotDiagramData"]}]]

Expand[ -t  Alexander[Knot[4, 1]][t]]

(Subscript[5, 1])

Show[Graphics[{AbsoluteThickness[3], KnotData[{5, 1}, "KnotDiagramData"]}]]

Expand[ t^2  Alexander[Knot[5, 1]][t]]

(Subscript[8, 10])

Show[Graphics[{AbsoluteThickness[3], KnotData[{8, 10}, "KnotDiagramData"]}]]

Expand[ t^3  Alexander[Knot[8, 10]][t]]

Knots from Braids

Torus Knot Example

br1 = BR[3, {-1, -2}];
Show[Graphics[{AbsoluteThickness[3], BraidPlot[br1]/.Graphics->List}]]

br3 = BR[3, {-1, -2, -1, -2, -1, -2}];
Show[Graphics[{AbsoluteThickness[3], BraidPlot[br3]/.Graphics->List}]]

Clear[m]

m[t_] = {{0, 1, 0}, {0, 0, 1}, {t^2, t(1-t), 1-t}}

MatrixForm[m[t]]

MatrixForm[m [t]- IdentityMatrix[3]]

Clear[alexander]

alexander[k_][t_] := Det[MatrixPower[m[t], k][[{1, 2}, {1, 2}]] - IdentityMatrix[2]]

alexander[2][t]

alexander[3][t]

Table[alexander[k][t], {k, 1, 8}]//TableForm

Torus Knot Fact. It is known that the Alexander polynomial of the torus knot T(r, k) is 
                                p(t) = ((((t^rk)-1)(t-1))/(((t^r) - 1)((t^k)-1))).

A Few Alexander Polynomial Facts

Theorem. Any polynomial p(t) that satisfies p(1) = ±1 and p(t) = (t^k)p(1/t)
is the Alexander polynomial of some knot.

Theorem. If X and Y are oriented knots with respective Alexander polnomials (Subscript[p, X])(t) and (Subscript[p, Y])(t) , then 
(Subscript[p, X#Y])(t) = (t^s)(Subscript[p, X])(t) (Subscript[p, Y])(t) , where X#Y is the connected sum X and Y, for some s.

Quandles

Definition. A quandle (K, ⊲, (⊲^(-1))) is a set K with two binary operations ⊲ and (⊲^(-1)) that satisfy (for all a, b, c in K)
1. a ⊲ a = a
2. (a ⊲ b)  (⊲^(-1))b  = a and (a  (⊲^(-1)) b)  ⊲ b = a
3. (a ⊲ b) ⊲ c = (a ⊲ c) ⊲ (b ⊲ c)

Note. If  (K, ⊲, (⊲^(-1))) is a quandle, its inverse (K,(⊲^(-1)),⊲) is also a quandle.

Examples

Linear Quandles from a Commutative Ring

If q is a unit in the commutative ring K, the linear quandle is (Subscript[K, q]):
a ⊲ b  = q a + (1-q) b.
a (⊲^(-1))b = (q^(-1)) a + (1-(q^(-1))) b.

For instance, K = (Subscript[Z, n]) with q a unit, gives the quandle (Subscript[Z, n, q]).

Another for example is the Alexander quandle, K[t, (t^(-1))]/(p(t)) for a polynomial p(t), where q is t.

Group Conjugation

If G is a group and q is an integer, 
a ⊲ b  = (b^q)a (b^(-q))
a (⊲^(-1)) b  = (b^(-q))a (b^q)

For instance, the set of  n-cycles in a group forms a conjugation quandle.

David Joyce's Quandle

If G is a group and σ is an automorphism of G, then
a ⊲ b  = σ(a(b^(-1))) b
a (⊲^(-1)) b  = (σ^(-1))(a(b^(-1))) b

Knot Quandles

Knot Labeling

Label the strands of the knot diagram with symbols (Subscript[x, 1]), (Subscript[x, 2]), ... from the quandle (K, ⊲, (⊲^(-1))) so that the following rules are satisfied.
1. At least 2 elements of K are used and
2. At every crossing, the above labeling rules must be satsified.

Linear Quandle Labeling

Theorem. A knot X can be labeled with a linear quandle (Subscript[K, q]) if, and only if, p(q) is not a unit in K, where p(t) is the Alexander polynomial of X.
Proof. Follows from the definition of the Alexander polynomial and accompanying discussion.

Corollary.  If there is an Alexander polynomial p(t) for which one, but not both, of p(q) and p(q'), is a unit in (Subscript[Z, n]), then the two linear quandles (Subscript[Z, n, q]) and (Subscript[Z, n, q']) are not isomorphic.

Example. Let p(t) = (t^2)- 3t + 1, the Alexander polynomial for the knot (Subscript[4, 1]). Then, p(2) = -1 is a unit in any (Subscript[Z, n]). However, p(4) = 5 is not a unit in any (Subscript[Z, n]) where n is a multiple of 5.Thus, if n is a multiple of 5, we know the quandles (Subscript[Z, n, 2]) and (Subscript[Z, n, 4]) cannot be isomorphic.

Note. If a quandle (K, ⊲, (⊲^(-1))) labels an oriented knot, then the inverse quandle (K,(⊲^(-1)),⊲) labels to same knot with opposite orientation.

Example

(Subscript[7, 6])

Show[Graphics[{AbsoluteThickness[3], KnotData[{7, 6}, "KnotDiagramData"]}]]

p[t_] =Expand[ -t^2  Alexander[Knot[7, 6]][t]]

Which quandles (Subscript[Z, 10, q]) can label this knot?

Select[Table[{q, GCD[p[q], 10]}, {q, 2, 9}], Not[#[[2]] ==1]&]

Answer: (Subscript[Z, 10, 2]), (Subscript[Z, 10, 3]), (Subscript[Z, 10, 7]), (Subscript[Z, 10, 8]).

Example

Theorem[?]. p(-1) ≡1 or 5 mod(8) for any knot. 

Corollary. (Subscript[Z, 8, -1])can never label a knot.

Connected Sum and Quandle Labeling

Theorem. The connected sum of two knots, X#Y, can be labeled by a linear quandle (Subscript[Z, n, q]) if, and only if, at least one of the components X or Y can be so labeled.
Proof.
(⟸) If, say, X can be labeled, then label X#Y by labeling all of Y one label.
(⟹) We prove the contrapositive. Assume neither X nor Y can be labeled.Then the respective Alexander polynomials evaluated at t = q, (Subscript[p, x])(q) and (Subscript[p, x])(q),  are units in (Subscript[Z, n]). So, the product (Subscript[p, X#Y])(q) = (Subscript[p, X])(q) (Subscript[p, Y])(q) is a unit in (Subscript[Z, n]). It follows that the connected sum X#Y cannot be labeled.

Torus Knot Revisited

(B^k) = T(3, k)

⋯

Clear[p,k, t]
p[k_, t_] := alexander[k][t]

Table[p[k, t], {k, 1, 5}]//TableForm

A (Subscript[Z, n])-labeling is a (Subscript[Z, n, -1])-labeling.

Table[p[k, -1], {k, 1, 20}]

Observation. T(3, k) can be (Subscript[Z, n, -1])-labeled if, and only if
1. k ≡ 0 mod 6, or
2. k ≡ 2, 4 mod 6 and gcd(3, n) > 1, or
3. k ≡ 3 mod 6 and n is odd.

What about (Subscript[Z, n, 2])-labeling? (Here, n must be odd.)

Table[p[k, 2], {k, 1, 10}]

Try factoring.

TableForm[Table[{k, FactorInteger[p[k, 2]]}, {k, 1, 21}]]

Observations. (n is odd.)
1. If k is even, T(3, k) can be (Subscript[Z, n, 2])-labeled if gcd(n, 3) > 1.
2. If k ≡ 0 mod 3 or k ≡ 0 mod 7, then T(3, k) can be (Subscript[Z, n, 2])-labeled if gcd(n, 7) > 1.
3. If k ≡ 0 mod 4,  then T(3, k) can be (Subscript[Z, n, 2])-labeled if gcd(n, 13) > 1.
4. etc.

Easy Theorm. If (B^k) can be (Subscript[Z, n, q])-labeled, then so can (B^km) for all m = 1, 2, 3, ....

A Non-Torus Braid Knot

br1 = BR[3, {-1, 2}];
Show[Graphics[{AbsoluteThickness[3], BraidPlot[br1]/.Graphics->List}]]

m[t_] = {{0, 1, 0}, {t-1, 2 - t - t^(-1), t^(-1)}, {t, 1-t, 0}}

MatrixForm[m[t]]

Clear[alexander]
alexander[k_][t_] := Expand[-(-t)^k Det[MatrixPower[m[t], k][[{1, 2}, {1, 2}]] - IdentityMatrix[2]]]

Table[alexander[k][t], {k, 1, 8}]//TableForm

Labeling by (Subscript[Z, n, -1]):

Table[alexander[k][-1], {k, 1, 8}]

This is the "alternate Lucas numbers-2". However...

Table[Sqrt[alexander[2k][-1]/5], {k, 0, 8}]

The alternate Fibonaccis.

Table[alexander[k][2], {k, 1, 20}]

Labeling by (Subscript[Z, n, 2]):

Table[alexander[k][2], {k,1, 20}]

Table[Sqrt[(-1)^(k+1)alexander[k][2]], {k, 1, 20}]

Recursion: (Subscript[s, n])=(Subscript[s, n-1])-2(Subscript[s, n-2]) then take absolute value of the resuting sequence.

Labeling by (Subscript[Z, n, 3]):

Table[alexander[k][3], {k,1, 20}]

Table[Sqrt[alexander[k][3]], {k,1, 20}]

Recursion: (Subscript[s, n])=(Subscript[s, n-1])+3(Subscript[s, n-2])

Labeling by (Subscript[Z, n, 4]):

Table[alexander[k][4], {k,1, 20}]

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