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An invariant of free links valued in free groups

V.O.Manturov and S.Kim*

V.O.Manturov S.Kim (BMSTU) An invariant of free links valued in free groups August 24. 2015 1 / 51

Basic definitions and notations

Definition 1.1

By a framed 4-graph with endpoints we mean a graph satisfying the followings:

1

every vertex is a 4-valent vertex except for 2n vertices for some n ∈ N ∪ {0} which are 1-valent vertices.

2

for each 4-valent vertex we fix a way of splitting of the four emanating half-edges into two pairs of edges called (formally) opposite.

1 2 3 4 Figure: {1, 3} and {2, 4} are pairs of ‘opposite’

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An intersection of different edges of the framed 4-graph with endpoints in interior points are called a virtual crossing and it is denoted by an intersection inside a circle.

R 1 R

Figure: A framed 4-graph with endpoints

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Definition 1.2

By a unicursal component of a framed 4-graph with endpoints we mean an equivalence class on the set of edges of the graph: Two edges e, e′ are equivalent if there exists a collection of edges e = e1, · · · , ek = e′ and a collection of 4-valent vertices v1, · · · , vk−1 (some of them may coincide) of the graph such that edges ei, ei+1 are opposite to each other at the vertex vi. By a unicursal circle of a framed 4-graph with endpoints we mean a unicursal component such that v1 = vk−1.

V.O.Manturov S.Kim (BMSTU) An invariant of free links valued in free groups August 24. 2015 4 / 51

R 1 R

Figure: A component of a framed 4-graph with endpoints

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R 1 R

Figure: A component of a framed 4-graph with endpoints

V.O.Manturov S.Kim (BMSTU) An invariant of free links valued in free groups August 24. 2015 6 / 51

R 1 R

Figure: A component of a framed 4-graph with endpoints

V.O.Manturov S.Kim (BMSTU) An invariant of free links valued in free groups August 24. 2015 7 / 51

R 1 R

Figure: A component of a framed 4-graph with endpoints

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R 1 R

Figure: Framed 4-graph with endpoints with 4 (unicursal) components

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Definition 1.3

A virtual tangle diagram is a generic immersion of a framed 4-graph with endpoints in R × I with each 4-valent vertex endowed with a classical crossing structure and every 1-valent vertex is in R × {0} or R × {1}. A virtual (link) diagram is a virtual tangle diagram without endpoints. A virtual tangle is an equivalence class of virtual tangle diagrams by usual Reidemeister moves and the detour move.

(1) (2) (3)

Figure: Reidemeister moves Figure: Detour move

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Definition 1.4

A free tangle is an equivalence class of framed 4-graphs with endpoints modulo Reidemeister moves for free diagrams. A free link is a free tangle without endpoints. By an n − n free tangle we mean a free tangle with points {p0

1, · · · , p0 n} in

R × {0} and {p1

1, · · · , p1 n} in R × {1} such that each component has end

points p0

i and p1 i .

RM1 RM2 RM3

Figure: Reidemeister moves for free diagrams

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Definition 1.5

If the components of a free tangle are numbered and the numbers of components preserve under Reidemeister moves for free diagrams, then the free tangle is enumerated.

T

1

T

2

T

3

T

4

T

1

T

2

T

3

Free tangle 3-3 free tangle Figure: Free tangle diagrams

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Remark 1.6

Every notion(n − n tangle, link, knot, braid, etc.) is just a partial case of

• tangles. On the other hand, free tangles(n − n tangles, links, knots, etc.) are

equivalence classes of virtual tangles modulo the following relations.

Figure: Crossing change and virtualization

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Motivation

In [3], by V.O.Manturov, the following principle was established:

Proposition 2.1

If a diagram K is complicated enough then it realizes itself. In other words, if K is complicated enough and K and K ′ are equivalent, then K ⊂ K ′. This proposition is proved by parity bracket which is valued in group generated by diagrams.

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Remark 2.2

This effects very similarly to the case of free groups: if the word abcba in Z ∗ Z ∗ Z is irreducible then it appears in any word equivalent to it (for example, abaa−1b−1bcbc3c−3a). In [4],[5], by V.O.Manturov and I.M.Nikonov, consider homomorphism from pure braid group PBn to a group presentation Gk

n.

consider homomorphism from Gk

n to Z ∗N 2 .

That is, there is a homomorphism from PBn to Z ∗N

2 .

But this is NOT arranged for the case of general tangles.

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Remark 2.2

This effects very similarly to the case of free groups: if the word abcba in Z ∗ Z ∗ Z is irreducible then it appears in any word equivalent to it (for example, abaa−1b−1bcbc3c−3a). In [4],[5], by V.O.Manturov and I.M.Nikonov, consider homomorphism from pure braid group PBn to a group presentation Gk

n.

consider homomorphism from Gk

n to Z ∗N 2 .

That is, there is a homomorphism from PBn to Z ∗N

2 .

But this is NOT arranged for the case of general tangles.

V.O.Manturov S.Kim (BMSTU) An invariant of free links valued in free groups August 24. 2015 15 / 51

Remark 2.2

This effects very similarly to the case of free groups: if the word abcba in Z ∗ Z ∗ Z is irreducible then it appears in any word equivalent to it (for example, abaa−1b−1bcbc3c−3a). In [4],[5], by V.O.Manturov and I.M.Nikonov, consider homomorphism from pure braid group PBn to a group presentation Gk

n.

consider homomorphism from Gk

n to Z ∗N 2 .

That is, there is a homomorphism from PBn to Z ∗N

2 .

But this is NOT arranged for the case of general tangles.

V.O.Manturov S.Kim (BMSTU) An invariant of free links valued in free groups August 24. 2015 15 / 51

Goal : Construct an invariant of free n − n tangles and links

valued in free groups.

• Distinctions between general tangles and braids:

n − n tangles do not possess a group structure: there are no inverse elements. the existence of pure crossings (between a component and itself).

Figure: Free n − n tangle diagram with a reverse arc

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Adjusted goal : Construct an invariant of free n − n tangles and links without pure crossings valued in free groups.

i p

s i

p

e i

k k k k c 1 c 2 c 3 c 4

Figure: Crossings with respect to a component

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Statement If two diagrams with no pure crossings are equivalent then they

are equivalent by a sequence of moves where no intermediate diagram has a pure crossing.

Proposition 2.3 ([2], M.Goussarov, M.Polyak, O.Viro)

If two classical links are equivalent as virtual links, then they are equivalent as classical links.

Proposition 2.4 ([1], R.Fenn, R.Rimanyi, C.Rourke)

If two classical braids are equivalent as tangles, then they are equivalent as braids.

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Statement If two diagrams with no pure crossings are equivalent then they

are equivalent by a sequence of moves where no intermediate diagram has a pure crossing.

Proposition 2.3 ([2], M.Goussarov, M.Polyak, O.Viro)

If two classical links are equivalent as virtual links, then they are equivalent as classical links.

Proposition 2.4 ([1], R.Fenn, R.Rimanyi, C.Rourke)

If two classical braids are equivalent as tangles, then they are equivalent as braids.

V.O.Manturov S.Kim (BMSTU) An invariant of free links valued in free groups August 24. 2015 18 / 51

In section 3 we will prove the following statement. If two diagrams with no pure crossings are equivalent then they are equivalent by a sequence of moves where no intermediate diagram has a pure crossing. In section 4 we will construct invariant valued in Z ∗N

2

for free n − n tangles and free links without pure crossings.

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Tangles without pure crossings

Definition 3.1

Let L = L1 ∪ · · · ∪ Ln be an enumerated free tangle diagram. A crossing c is called a crossing of type (i, j) if two arcs of c are part of Li and Lj respectively. A classical crossing c of type (i, i) for some i ∈ {1, 2, · · · n} is called a pure crossing.

i j

L L

Figure: A crossing of type (i, j)

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Bracket for free tangles valued in Z2[T]

Define a g-equivalence relation ∼ = on enumerated free tangle diagrams without pure crossings as follows: T ∼ = T ′ - g-equivalent <=> ∃{Ti}n

i=0 such that

1

T0 = T, Tn = T ′,

2

Ti is obtained from Ti−1 by applying one of Reidemeister moves,

3

each Ti has no pure crossings. Let T be the set of equivalence classes of enumerated free tangle diagrams without pure crossings modulo ∼ =. Consider the linear space Z2[T].

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Let T be an enumerated free tangle diagram with n components. Let us consider pure crossings of T. For each pure crossing, there are two

• splicings. Herewith, the rest of the diagram remains unchanged. We may then

consider further splicings of T at several crossings. The result of splicings Ts would have at least n components.

Figure: Splicings of a crossing

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Example 3.2

T

1

T

2

T

3

T

4

Ts1 Ts2 Ts3 Ts4 T Ts5 Ts6 Ts7 Ts8

Figure: Splicings on pure crossings

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Remark 3.3

If Ts has exactly n components, then it is possible to enumerate components which agrees with enumeration of T because we splice only pure crossings and any two different components cannot be connected by the operation.

T

1

T

2

T

3

T

4

T

1

T

2

T

3

T

4

Ts1 T

Figure: Ts with n components

V.O.Manturov S.Kim (BMSTU) An invariant of free links valued in free groups August 24. 2015 24 / 51

Definition 3.4

Let us consider the following sum [T] = Σ{s−pure,n comp.}Ts ∈ Z2[T] (1) which is taken over all splicings in all pure crossings, and only those summands are taken into account where Ts has exactly n components.

Remark 3.5

If T has m pure crossings, then [T] will contain at most 2m summands, and if T has no pure crossings, then we shall have exactly one summand.

Remark 3.6

This bracket is similar to the parity bracket which is introduced in [3]. (We splice crossings of ‘specific’ type.)

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Example

T

1

T

2

T

3

T

4

Ts3 T s4 T

T

1

T

2

T

3

T

4

T

1

T

2

T

3

T

4

T

1

T

2

T

3

T

4

Ts1

Figure: Calculation of bracket for free links with 4 components

V.O.Manturov S.Kim (BMSTU) An invariant of free links valued in free groups August 24. 2015 26 / 51

Lemma 3.7

The bracket [ · ] for enumerated free tangles is an invariant under Reidemeister moves for free diagrams.

• Proof. Let T and T ′ be two enumerated free tangle diagrams such that T ′ is
• btained from T by applying one of Reidemeister moves for free diagrams.

We will show that [T] = [T ′].

Figure: Splicing for RM1

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j i j i i i Figure: Splicing for RM2 between a component and itself.

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i i i i i i

Figure: Splicing for the third Reidemeister move, i = j = k

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j i j j i j j i j i Figure: Splicing for the third Reidemeister move, i = j = k

V.O.Manturov S.Kim (BMSTU) An invariant of free links valued in free groups August 24. 2015 30 / 51

Theorem 3.8

If two enumerated free tangle diagrams with no pure crossings are equivalent then they are g-equivalent.

Proof.

For two enumerated free tangle diagrams T and T ′ without pure crossings if T ′ can be obtained from T by Reidemeister moves for free diagrams, then [T] = [T ′]. Since T and T ′ have no pure crossings, T = [T] = [T ′] = T ′, that is, T and T ′ are g-equivalent.

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An invariant valued in free group

An invariant for free n − n tangles valued in free group Definition 4.1

Let L = L1 ∪ L2 ∪ · · · ∪ Ln be an enumerated free tangle diagram. If the number of crossings of type (i, j) is even for all i, j such that i = j, then we call L a diagram in a good condition. This condition is preserved by Reidemeister moves.

Definition 4.2

For each classical crossing c of T of type (i, j) and for k ∈ {1, 2, · · · , n}\{i, j}, define lki

c(k) by the number of all crossings of type (i, k) on the Ti from the

points ps

i to the crossing c. Define lkc(k) = lki c(k) + lkj c(k) modulo Z2. Note

that lkc can be considered as a map from {1, 2, · · · , n}\{i, j} to Z2.

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EXAMPLE Let (i, j) = (1, 2). Calculate lkc and lkc′.

c 1 2 3 4 c’

Figure:

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EXAMPLE Let (i, j) = (1, 2). Calculate lkc and lkc′.

c 1 2 3 4 c’

Figure:

lk1

c (3) = 1, lk2 c (3) = 0, lkc(3) = lk1 c (3) + lk2 c (3) = 1 + 0 = 1

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EXAMPLE Let (i, j) = (1, 2). Calculate lkc and lkc′.

c 1 2 3 4 c’

Figure:

lk1

c′(3) = 1, lk2 c′(3) = 1, lkc′(3) = lk1 c′(3) + lk2 c′(3) = 1 + 1 = 0 (mod 2)

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Definition 4.3

Define a group presentation G(i,j)

n

generated by {σ | σ : {1, 2, · · · n}\{i, j} → Z2} with relations {σ2 = 1}. Let us define an action of aut(G(i,j)

n

) on G(i,j)

n

by f · g = f(g) for f ∈ aut(G(i,j)

n

) and for g ∈ G(i,j)

n

. For l ∈ {1, · · · , n}\{i, j}, let fl be an automorphism on G(i,j)

n

such that fl(σ)(x) =

• σ(x)

if x = l σ(x) + 1 if x = l. Define two word w and w′ in G(i,j)

n

are slide-equivalent in G(i,j)

n

if w′ = fl · w for some fl in the subgroup < {fl}l∈{1,2,···m} > of aut(G(i,j)

n

).

Remark 4.4

For a crossing c of type (i, j), each lkc is one of the generators of G(i,j)

n

. And G(i,j)

n

is isomorphic to Z2n−2

2

.

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Fix i = j, i, j ∈ {1, . . . , n}. Let {c1, · · · , cm} be the set of classical crossings of type (i, j) such that for each k, l ∈ {1, 2, · · · , m}, k < l if and only if we meet ck earlier than cl when we follow i-th component from the point ps

i .

c1 c2 c3 1 2 3 4 d1 d2 d3 c4 d4

Figure: Enumerated crossings with respect to 1 and 2 components

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Fix i = j, i, j ∈ {1, . . . , n}. Let {c1, · · · , cm} be the set of classical crossings of type (i, j) such that for each k, l ∈ {1, 2, · · · , m}, k < l if and only if we meet ck earlier than cl when we follow i-th component from the point ps

i .

c1 c2 c3 1 2 3 4 d1 d2 d3 c4 d4

Figure: Enumerated crossings with respect to 1 and 2 components

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Definition 4.5

Define a word wi

(i,j)(T) in G(i,j) n

for T by wi

(i,j)(T) = lkc1lkc2 · · · lkcm.

Example 4.6

c1 c2 c3 1 2 3 4 c4

lkc1 =(0,0) lkc2 =(1,0) lkc3 =(1,0) lkc4 =(0,1)

Figure:

If lkck = (lkck(3), lkck (4)), then w1

(1,2) = (0, 0)(1, 0)(1, 0)(0, 1) = (0, 0)(0, 1).

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Theorem 4.7

For a positive integer n and for i, j ∈ {1, · · · , n} such that i = j, wi

(i,j) is

invariant for oriented enumerated n − n free tangles in a good condition without pure crossings.

• Proof. Let T and T ′ be oriented enumerated n − n free tangle diagrams in a

good condition without pure crossings. Fix a pair (i, j) such that i = j ∈ {1, · · · , n}. Suppose that T and T ′ are equivalent as free tangles. By Theorem 3.8 and by definition of tangles in a good condition, we may assume that T ′ is obtained from T by applying one of the second and the third Reidemeister moves. The proof consists of three parts:

1

If two crossings c and c′ are contained in the second Reidemeister move, then lkc = lkc′.

2

If a crossing c in T is not contained in the second and the third Reidemeister moves, then lkc = lkc′ where c′ is a crossing in T ′ associated to c with respect to the move.

3

wi

(i,j) is an invariant under the second and the third Reidemeister moves.

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1

If two crossings c and c′ are contained in the second Reidemeister move, then lkc = lkc′.

c c’

i j i j

RM2

Figure: First part of proof.

For two crossings c and c′ contained in the second Reidemeister move, since there are no other crossings between c and c′, by definition of lkc, (1) holds.

V.O.Manturov S.Kim (BMSTU) An invariant of free links valued in free groups August 24. 2015 41 / 51

then lkc = lkc′.

2

If a crossing c in T is not contained in the second and the third Reidemeister moves, then lkc = lkc′ where c′ is a crossing in T ′ associated to c with respect to the move.

c c i j k i j k c c i j k l i j k l

Figure: Second part of proof.

Consider a crossing c in T which is not contained in the second and the third Reidemeister moves. If we apply the second Reidemeister move, then the number of crossings of type (i, k) and (j, k) from ps

i to c and from ps j to c is

changed by 0 or +2 or −2. If we apply the third Reidemeister move, then the number of crossings of type (i, k) and (j, k) from ps

i to c from ps j to c is not

• changed. Hence lkc = lkc′ and it is proved that (2) holds.

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associated to c with respect to the move.

3

wi

(i,j) is an invariant under the second and the third Reidemeister moves.

Clearly, wi

(i,j) does not change under the second Reidemeister move because

• f the relations {σ2 = 1}.

RM3

c c i j k i j k

Figure: Third part of proof.

Let us consider the third Reidemeister move. The number of crossings of type (i, z) and (j, z) from ps

i to c and from ps j to c is changed by +2 or −2 and it is

not changed modulo Z2.

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An invariant valued in free group

An invariant for free links valued in free group

Now, let us consider an oriented enumerated free link diagram L = L1 ∪ · · · ∪ Ln in a good condition without pure crossings. For each i ∈ {1, · · · , n} let us fix a point pi on an edge of Li and ps

i and pe i in R × {0}

and R × {1}, respectively. We may assume that ps

i < ps i+1 and pe i < pe i+1 for all

i ∈ {1, · · · , n − 1} as elements in R. We cut each component Li at pi. Then we get two points p0

i and p1 i such that the orientaion of Li goes from p0 i to p1 i .

Let us connect two points p0

i and p1 i with ps i and pe i , respectively. Herewith, we

change new intersections to virtual crossings. Then we can get an oriented enumerated n − n free tangle diagram TL in a good condition without pure crossings from L.

Remark 4.8

Apparently, the word wi

(i,j) does depend on the cut points {pi}n

• 1. But it does

not depend on the points {ps

i , pe i } in R × {0} and R × {1} since every new

intersection is considered as a virtual crossing.

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Example 4.9

1 2 3 1 2 3

L T L R 1 R

p e

1

p e

2

p e

3

p s

1

p s

2

p s

3

p

1

p

2

p

3

Figure: An n − n free tangle from a free link

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Theorem 4.10

Let L = L1 ∪ · · · ∪ Ln be an oriented enumerated free link diagram in a good condition without pure crossings. For each pair (i, j), wi

(i,j) is an invariant for

• riented enumerated free links in a good condition without pure crossings with

respect to slide-equivalence relation in G(i,j)

n

and conjugation.

• Proof. Let L = L1 ∪ · · · ∪ Ln be an oriented enumerated free link diagram in a

good condition without pure crossings. If cut points {pk}n

k=1 are fixed, then for

a pair (i, j), by Theorem 3.8 and Lemma 4.7, wi

(i,j)(L) = wi (i,j)(L′). If one of the

fixed points pk is moved to another point p′

k on Lk, we may assume that

between pk and p′

k there is the only one crossing c of type (k, l).

pk p’

k

c k l

Figure: Move of point pk

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Say {c1, · · · , cm} and {c′

1, · · · , c′ m} are sets of crossings of type (i, j) on Li

such that crossings in {c1, · · · , cm} (in {c′

1, · · · , c′ m}) are ordered with respect

to the orientation of Li and to pk (to p′

k).

Case0 : k = i, l = j The move of pk does not effect to wi

(i,j).

Case1 : k = i, l = j

Then ca = c′

a for all a ∈ {1, · · · , m} and lkca(i) is changed from 0 to 1(or 1 to

0), that is, lkc′

a = fi(lkca). Therefore wi

(i,j)and w′i (i,j) are slide-equivalent in G(i,j) n

.

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Case2 : k = i, l = j

pk p’

k

c k l l

1

c2 c’

m

c’

1

Figure: Move of point pk

Since lkc′

a = lkca+1 for a ∈ {1, · · · m − 1} and lkc′ m = lkc1, the word w′i

(i,j) is

lk−1

c

wi

(i,j)lkc.

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Example 4.11

Let us calculate the word w1

(1,2). We enumerate crossings with respect to first

• component. If lkck = (lkck (3), lkck (4)), then we get lkc1 = (0, 0), lkc2 = (0, 1),

lkc3 = (1, 1), lkc4 = (0, 0). Therefore w1

(1,2) = (0, 0)(0, 1)(1, 1)(0, 0) = (0, 1)(1, 1). Since the word cannot

be reduced to the trivial, this link is not trivial.

1 2 3 4

c1 c2 c3 c4

p 1 p 2 p 3 p 4

Figure:

V.O.Manturov S.Kim (BMSTU) An invariant of free links valued in free groups August 24. 2015 49 / 51

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Thank you

V.O.Manturov S.Kim (BMSTU) An invariant of free links valued in free groups August 24. 2015 51 / 51