Everywhere Equivalent 3-Braids

A knot (or link) diagram is said to be everywhere equivalent if all the diagrams obtained by switching one crossing represent the same knot (or link). We classify such diagrams of a closed 3-braid.


Introduction
How does a diagram D of a knot (or link) L look, which has the following property: all diagrams D obtained by changing exactly one crossing in D represent the same knot (or link) L (which we allow to be different from L)? This suggestive question was possibly first proposed in this form by K. Taniyama, who called such diagrams everywhere equivalent (see Definition 2.1 in Section 2.6). Through the connection between links and braid groups (see, e.g., [7]), this question turns out related to the following group-theoretic question: given a group in a set of conjugate generators, which words in these generators have the property that reversing each individual letter gives a set of conjugate elements?
When by the crossing-changed diagrams D display the unknot, Taniyama's question was studied previously in [16], where D was called everywhere trivial . There some efforts were made to identify such diagrams, mostly by computationally verifying a number of low-crossing cases. The upshot was that, while there is a (hard to be described) abundance of diagrams for D unknotted, only 6 simple diagrams seem to occur when D is not; see (2.9).
Motivated by Taniyama's more general concept, we made in [19] a further study of such phenomena. We conjectured, as an extension of the everywhere trivial case, a general description of everywhere equivalent diagrams for a knot, and proved some cases of low genus diagrams. We also proposed some graph-theoretic constructions of everywhere equivalent diagrams for links.
In this paper we give the answer for 3-braids. A 3-braid diagram corresponds to a particular braid word in Artin's generators. This word can be regarded up to inversion, cyclic permutations, the interchanges σ i ↔ σ −1 i (reflection) and σ 1 ↔ σ 2 (flip). However, beyond this it will be of importance to distinguish between braids and their words, i.e., how a given braid is written. 3) the words σ l 1 σ l 2 k for k, l ≥ 1 (symmetric case), and 4) the words σ k 1 for k > 0 (split case). Using the exponent sum (2.6), one easily sees that the answer to the related group-theoretic question for the 3-braid group consists of the last three families in the theorem. This outcome, and even more so its derivation, have turned out more complicated than expected.
The first family mainly comes, except for σ 1 σ 2 σ −1 1 σ −1 2 and the trivial cases, from the diagrams of (2.9) (under the exclusion of the 5 crossing one, which is not a 3-braid diagram). The last two families are also quite suggestive, and given (in more general form for arbitrary diagrams) in [19]. However, we initially entirely overlooked the second family of diagrams. We did not explicitly ask in [19] whether our link diagram constructions are exhaustive, but we certainly had them in mind when approaching Theorem 1.1. These previous examples come in some way from totally symmetric (edge transitive) planar graphs. However, there is little symmetry in general here. One can easily construct examples of central (element) words lacking any symmetry. (One can also see that every positive word in the 3-braid group can be realized as a subword of a positive word representing a central element.) The second family does not yield (and thus answer negatively our original question for) knots, and it does not lead out of the positive case (whose special role was well recognized in [19]). Still we take it as a caution that everywhere equivalence phenomena, although sporadic, may occur in far less predictable ways than believed.
Our proof is almost entirely algebraic, and will consist in using the Jones polynomial to distinguish the links of various D except in the desired cases. Mostly we will appeal to the description of the Jones polynomial in terms of the Burau representation, but at certain points it will be important to use information coming from the skein relation and the Kauffman bracket.
The proof occupies Sections 3 and 4, divided by whether the braid word is positive or not. Both parts require somewhat different treatment. We will see that for 3-braids the non-positive case quickly recurs to the everywhere trivial one.
A final observation (Proposition 4.8) addresses the lack of interest of the situation opposite to everywhere equivalence: when all crossing-switched versions of a diagram are to represent different links. (This property was called everywhere different, and some constructions for such knot diagrams were given in [19].) In a parallel paper [21] we observe how to solve the classification of (orientedly) everywhere equivalent diagrams in another case, this of two components.

Preliminaries
It seems useful to collect various preliminaries, which will be used at different places later in the paper.

Link diagrams and Jones polynomial
All link diagrams are considered oriented, even if orientation is sometimes ignored. We also assume here that we actually regard the plane in which a link diagram lives as S 2 , that is, we consider as equivalent diagrams which differ by the choice of the point at infinity.
The Jones polynomial V can be defined as the polynomial taking the value 1 on the unknot, and satisfying the skein relation In each triple as in (2.1) the link diagrams are understood to be identical except at the designated spot. The fragments are said to depict a positive crossing, negative crossing, and smoothed out crossing. The skein smoothing is thus the replacement of a crossing by the third fragment. The writhe w(D) of a link diagram D is the sum of the signs of all crossings of D. If all crossings of D are positive, then D is called positive. It is useful to recall here the alternative description of V via Kauffman's state model [8]. A state is a choice of splicings (or splittings) of type A or B (see Fig. 1) for any single crossing of a link diagram D. We call the A-state the state in which all crossings are A-spliced, and the B-state B(D) is defined analogously.
When for a state S all splicings are performed, we obtain a splicing diagram, which consists of a collection of (disjoint) loops in the plane (solid lines) together with (crossing) traces (dashed lines). We call a loop separating if both its interior and exterior contain other loops (regardless of what traces). We will for convenience identify below a state S with its splicing diagram for fixed D. We will thus talk of the loops and traces of a state. Figure 1. The A-and B-corners of a crossing, and its both splittings. The corner A (resp. B) is the one passed by the overcrossing strand when rotated counterclockwise (respectively clockwise) towards the undercrossing strand. A type A (resp. B) splitting is obtained by connecting the A (resp. B) corners of the crossing. The dashed line indicates the trace of the crossing after the split.
Recall, that the Kauffman bracket D [8] of a link diagram D is a Laurent polynomial in a variable A, obtained by a sum over all states S of D: Here #A(S) and #B(S) denote the number of type A (respectively, type B) splittings and |S| the number of (solid line) loops in the splicing diagram of S. The formula (2.2) results from applying the first of the bracket relations to each crossing of D (here traces are ignored), and then deleting (except one) loops using the second relation, at the cost of a factor −A 2 − A −2 per deleted loop. (The normalization is thus here that the diagram of one circle with no crossings has unit bracket.) The Jones polynomial of a link L can be determined from the Kauffman bracket of some diagram D of L by with w(D) being the writhe of D. This is another way, different from (2.1), to specify the Jones polynomial.
It is well-known that V ∈ Z[t ±1 ] (i.e., only integral powers occur) for odd number of link components (in particular, for knots), while V ∈ t 1/2 · Z[t ±1 ] (i.e., only half-integral powers occur) for even number of components.
For V ∈ Z[t 1/2 , t −1/2 ], the minimal or maximal degree min deg V or max deg V is the minimal resp. maximal exponent of t with non-zero coefficient in V . Let span V = max deg V −min deg V .

Semiadequacy and adequacy
Let S be the A-state of a diagram D and S a state of D with exactly one B-splicing. If |S| > |S | for all such S , we say that D is A-adequate. Similarly one defines a B-adequate diagram D (see [11] [14,Appendix] is an example. This property of the Perko knot follows from work of Thistlethwaite [22], and is explained, e.g., in Cromwell's book [3, p. 234], or (along with further examples) in [20].
A link diagram D is said to be split, if its planar image (forgetting crossing information) is a disconnected set. A region of D is a connected component of the complement of this planar image. At every crossing of D four regions meet; if two coincide, we call the crossing nugatory. A diagram with no nugatory crossings is reduced .
It is easily observed (as in [11]) that a reduced alternating diagram (and hence an alternating link) is adequate, and that the A-and B-state of an alternating non-split diagram have no separating loops.
The maximal degree of A of the summands in (2.2) is realized by the A-state S. However, its contribution may be cancelled by this of some other state. One situation when this does not happen is when D is A-adequate. Then the A-state gives in (2.2) the unique contribution to the maximal degree in A, and, via (2.3), the minimal degree of V . We call this the extreme A-term. Thus, for A-adequate diagrams, min deg V can be read off from the A-state directly, and the coefficient is ±1. For not A-adequate diagrams, the situation is a bit more subtle and studied in [1].
We will use the following important special case of that study. When D is not A-adequate, the A-state has a trace connecting a loop to itself, which we call self-trace. For a given loop, we call a pair of self-traces ending on it intertwined if they have the mutual position (2.5) A self-trace is isolated if it is intertwined with no other self-trace. Bae and Morton show (among others) that if in the A-state a self-trace is isolated, the contribution in the extreme A-term of V is zero.
Similar remarks apply on B-adequate diagrams and max deg V , and then on adequate diagrams and span V .

Braid groups and words
The n-string braid , or shortly n-braid, group B n is considered generated by the Artin standard generators σ i for i = 1, . . . , n − 1. An Artin generator σ i , respectively, its inverse σ −1 i will be called a positive, respectively, negative letter, and i the index of this letter. We will almost entirely focus on n = 3.
The Artin generators are subject to commutativity relations (for n ≥ 4; the bracket denotes the commutator below) and braid relations, which give B n the presentation For the sake of legibility, we will commonly use a bracket notation for braid words. The meaning of this notation is the word obtained by replacing in the content of the brackets every integer ±i for i > 0, not occurring as an exponent, by σ ±1 i , and removing the enclosing brackets. Thus, e.g., Although negative exponents will not be used much here, let us fix for clarity that for a letter they are understood as the inverse letter, and for a longer subword as the inverse letters written in reverse order. Thus Occasionally we will insert into the bracket notation vertical bars '|'. They have no influence on the value of the expression, but we use them to highlight special subwords. A word which does not contain a letter followed or preceded by its inverse is called reduced . In a reduced braid word, a maximal subword σ ±k i for k > 0, i.e., one followed and preceded by letters of different index, is called a syllable. The number i is called index of the syllable, and the number ±k its exponent, which is composed of its sign '±' and its length k > 0. According to the sign, a syllable is positive or negative. A syllable of length 1 (of either sign) will be called trivial .
Obviously every reduced braid word decomposes into syllables in a unique way: The sequence (p 1 , . . . , p n ) will be called exponent vector . Thus an entry '±1' in the exponent vector corresponds to a trivial syllable. A word is positive, if all entries in its exponent vector are positive (i.e., it has no negative syllable).
Often braid words will be considered up to cyclic permutations. In this case so will be done with the exponent vector. The length of the exponent vector considered up to cyclic permutations will be called weight ω(β) of β. If β ∈ B 3 , then ω(β) is even, since indices 1 and 2 can only interchange, except when ω(β) = 1 (and β is a single syllable).
The quantity |p i | is the length of the word β (and is, of course, different from its weight, unless all syllables are trivial). The length-zero word will be called trivial word . The quantity is called exponent sum of β, and is invariant of the braid (i.e., equal for different words of the same braid), and in fact its conjugacy class. The half-twist element ∆ ∈ B n is given by and its square ∆ 2 = (σ 1 σ 2 · · · σ n−1 ) n is the generator of the center of B n . We will need mostly the group B 3 , where ∆ has the two positive word representations [121] and [212]. We will use the notation · for the involution σ 1 ↔ σ 2 of B 3 induced by conjugacy with ∆.

Braids and links
There is a well-known graphical representation for braids: Thus, a (positive/negative) letter of a braid word gives a (positive/negative) crossing, and smoothing a crossing corresponds to deleting a letter. In this sense we will feel free to speak of (switching) crossings and smoothings of a braid (word).
For braid(word)s β there is a closure operationβ: In this way, a braid closes to a knot or link and a braid word (which can be also regarded here up to cyclic permutations) closes to a knot or link diagram. Here the separation between the two levels of correspondence must be kept in mind. Thus β is a positive word if and only ifβ is a positive link diagram. For a further analogy, we say that β is split ifβ is a split diagram, which means that β contains neither σ i nor σ −1 i for some i.
For every link L there is a braid β ∈ B n withβ = L. The minimal n for given L is called braid index of L. (See, e.g., [5,12].)

Burau representation
Then for k ∈ Z we have For a closed 3-braid, there is a relation between the Burau representation and the Jones polynomial, which is known from the related Hecke algebra theory explained in [7]: (2.8) We will more often than the formula itself use an important consequence of it: for a 3-braid, two of the Burau trace, exponent sum and Jones polynomial (of the closure) determine the third. Note that ψ is faithful on B 3 . One proof which uses directly this relation to the Jones polynomial is given in [18]. This property of ψ is not mandatory for our work, but we use it to save a bit of exposition overhead at some places.
More importantly, there is a way to identify for given matrix whether it is a Burau matrix, and if so, of which braid. The Burau matrix determines (for 3-braids) along with the Jones polynomial also the skein and Alexander polynomial. These in turn determine (the skein polynomial precisely [15], the Alexander polynomial up to a twofold ambiguity [17]) the minimal length of a band representation of the braid. Thus one has only a finite list of band representations to check for given Burau matrix. We used this method in fact also here to identify certain braids from their matrix. No tools even distantly comfortable are available (or likely even possible) for higher braid groups.

Some properties of everywhere equivalent diagrams
We stipulate that in general D will be used for a link diagram and D for a diagram obtained from D by exactly one crossing change. If we want to indicate that we switch a crossing numbered as i, we also write D i . Similarly β will stand for a braid, usually a particular word of it, and β for a word obtained by inverting exactly one letter in β.
The central attention of this study is the following type of diagrams. A, potentially complete, list of knotted everywhere trivial diagrams was determined in [16]. These are given below and consist of two trefoil and four figure-8-knot diagrams: We saw this list compatible with the 3-braid case in Theorem 1.1. The list draws evidence for its exhaustiveness from various sources. For minimal crossing diagrams (and in particular that only the trefoil and figure-8-knot occur), the problem seems to have been noticed previously, and is now believed to be some kind of "knot-theory folklore". Yet, despite its very simple formulation, it is extremely hard to resolve. Apart from our own (previously quoted) efforts, we are not aware of any recent progress on it.
Another situation where everywhere equivalence can be resolved is for 2-bridge (rational) and Montesinos link diagrams. The classification of their underlying links gives a rather straightforward, albeit somewhat tedious, method to list up such EE diagrams. In particular, these diagrams without trivial clasps are as follows, agreeing also with (2.9): (2.10) • the rational diagrams in (2.9) (all except the 8-crossing one), • those with Conway notation C(p) and C(p, −p) for p ≥ 1, and • the pretzel diagrams (p, . . . , p q times ) = P (q, p) for q ≥ 3 and p ≥ 2 (see (2.10) for an example).
Note that in both the Montesinos and 3-braid case, it is not necessary to exclude unknotted everywhere trivial diagrams to formulate a reasonable statement. However, we know that such diagrams occur in multitude outside these two classes, which is another hint to why the classes are rather special.
The next lemma proposes a new family of EE diagrams for links, suggested by the 3-braid case. It identifies the second family in Theorem 1.1 in more general form. Beyond this point, we will from now on focus on 3-braids. The proof of Theorem 1.1 relies on several of their very special properties. Despite the various input, the length of the argument shows that there was some effort in putting the pieces together. In that sense, our initial optimism in carrying out, for example, a similar investigation of 4-braids, seems little reasonable.
In relation to our method of proof, we conclude the preliminaries with the following remark. The use of some algebraic technology might be occasionally (but not always) obsolete, for Birman-Menasco's work [2] has reduced the isotopy problem of closed 3-braids (mainly) to conjugacy. This fact provided a guideline for our proof. One place where the analogy surfaces is Lemma 4.2, which in some vague sense imitates, on the level of the Jones polynomial, a partial case of the combination of Birman-Menasco with the summit power in Garside's conjugacy normal form [6]. We will invoke Garside's algorithm at one point quite explicitly, in Lemma 3.2. On the other hand, the use of [2] would not make the proof much simpler, yet would build a heavy framework around it, which we sought to avoid. We will see that we can manageably work with the Burau representation and Jones polynomial (and that they remain essential to our argument).

Proof of the non-positive case
We start now with the proof of Theorem 1.1.

Initial restrictions
There is also here a dichotomy as to whether we are allowed to switch crossings of either sign, i.e., whether the diagram is (up to mirroring) positive or not. Let us throughout the following call a braid word everywhere equivalent (EE) if the diagramβ is such.
We start by dealing with non-positive braids. The goal is to obtain the first family in Theorem 1.1. Notice here that for any (non-trivial) such word β, the closureβ is either the unknot or the figure-8-knot.
For non-positive braids β, a strong restriction enters immediately, which will play a central role in the subsequent calculations. Since for a non-positive diagram D =β, one can get (3-string) braids of exponent sum differing by ±4 representing the same link, it follows from the Morton-Williams-Franks inequality [12,5] that the skein (HOMFLY-PT) polynomial P of such a link must have a single non-Alexander variable degree. Then a well-known identity [10,Proposition 21] implies that P = 1. By [15] we can conclude then thatβ = D is the unknot. In particular, the closure of β is a knot, and its exponent sum must be zero: We remark that if [β ] = ±2 (and the closure is unknotted), then by (2.8) its Burau trace is Note that not only does this trace determine a trivial Jones polynomial, but also that the Jones polynomial detects the unknot for 3-braids (see [18]). Thus this trace condition is in fact equivalent to the braid having unknotted closure.
Since we know a priori that we expect a finite answer, it turns out helpful first to rule out certain subwords of β.
Let us first exclude the cases when β contains up to cyclic permutations subwords of the form σ ±1 i σ ∓1 i , i.e., that it is not cyclically reduced. It is clear that if such β is everywhere equivalent, then so is the word obtained under the deletion of σ ±1 i σ ∓1 i . When we have proved the exhaustiveness of family 1 in Theorem 1.1, we see that it is enough to show that all words obtained by inserting σ ±1 i σ ∓1 i cyclically somewhere into any of these words β gives no everywhere equivalent word.
Most cases can be ruled out right away. Note that in such a situation for some word β both βσ 2 i and βσ −2 i must have unknotted closure. In particular, [β] = 0, and the positive words β need not be treated.
The other (non-positive) words β can be ruled out by noticing that when βσ 2 i gives (under closure) the unknot and β the unknot or figure-8-knot, then by the skein relation (2.1) of the Jones polynomial, βσ −2 i will have a closure with some non-trivial polynomial, and so it will not be unknotted.
Thus from now on we assume that β is a reduced word and has an exponent vector. The case of exponent vector of length (i.e., weight) 2 is rather easy, and leads only to σ 1 σ −1 2 , so let us exclude this in the following.

Syllable types
We regard thus now β as cyclically reduced, and start by examining what type of syllables can occur in the exponent vector. For a syllable, we are interested in the exponent (i.e., sign and length) of the syllable, and the sign of the preceding and following syllable.
Up to mirroring and σ 1 ↔ σ 2 we may restrict ourselves to positive syllables of σ 2 , and up to inversion we have three types of signs of neighboring syllables (of σ 1 ). Case 1. Two positive neighboring syllables. Up to cyclic permutations we have for n ≥ 1, Let us call the subword starting after α the visible subword of β, and its three syllables (at least two of which are trivial) the visible syllables. We cannot assume that α ends on σ ±1 2 , so that the first visible syllable of β may not be a genuine syllable.
We try to find out now how M = ψ(α) should look like. First note that because of (3.1), we must have (Here (−t) −1 has to occur everywhere on the right, since we always switch a positive crossing in β with (3.1), and thus [β ] = −2.) These three equalities give affine conditions on M , which restrict it onto a line in the space of 2 × 2 matrices, regarded over the fraction field F of Z[t, t −1 ]. Then the quadratic determinant condition (3.4) will give two solutions. These live a priori only in a quadratic extension of F. For the existence of α we need, (1) that the entries of M lie outside a quadratic field over F (i.e., the discriminant is a square), that then (2) they lie in Z[t, t −1 ] (i.e., that denominators disappear up to powers of t), and that in the end (3) the entries build up a valid Burau matrix of a braid. Something that startled us is that we encountered redundant cases which survived until any of these three intermediate stages.
Putting the equations into MATHEMATICA TM [23], using the formulas (2.7), gives the solutions for the lower right entry d of M . We will use from now on the extra variable u = −t to simplify the complicated expressions somewhat (even just a sign affects the presentation considerably!).
Our attention is dedicated first to the discriminant (occurring under the root). Removing obvious quadratic factors, we are left with the polynomial We need that this polynomial becomes the square of a polynomial in Z[t, t −1 ]. For n ≥ 4 the edge coefficients are −3, and thus the polynomial is not a square. Similarly for n = 2, where the minimal and maximal degrees are odd.
For n = 1 the visible subwords we consider in (3.3) are just the half-center element ∆ (and, under various symmetries, its inverse). Their exclusion as subwords of β will be most important for the rest of the argument (see Lemma 3.2), but occurs here in the most peculiar way.
For the choice of negative sign we get for d The evidence that this is not a Laurent polynomial in t can be sealed, for example, by setting t = 1 2 . The expression evaluates to 42 19 , whose denominator is not a power of 2. For the '+' we get d = −1/t. This gives indeed a matrix in Z[t, t −1 ]: But there is no braid α with such Burau matrix. This can be checked via the Alexander polynomial of the (prospective) closure, but there is a direct argument (which appeals, though, to the faithfulness of ψ). The matrix in question is M = t −2 · ψ(σ 2 ), but scalar Burau matrices are only in the image of the center of B 3 , and these are powers of t 3 .
For n = 3 the polynomial (3.6) is 1 − 4u + 8u 2 − 10u 3 + 8u 4 − 4u 5 + u 6 , which is a square, and the rather complicated expressions (3.5) become giving d = t −3 and d = −t −2 . These lead to the matrices We used the Alexander polynomial to check that these indeed occur as Burau matrices, and to see what their braids are. (We remind that ψ is faithful on B 3 .) The answer is These solutions were unexpected, but can be easily justified: putting α i for α in (3.3) for n = 3, one easily sees that switching any of the last 5 crossings gives a braid with unknotted closure. Case 2. One positive and one negative neighboring syllable. In this case we have β = ασ −1 1 σ n 2 σ 1 . (Now in the first case we switch a negative crossing, thus [β ] = 2, and we need the trace −t.) The solutions for the lower right entry d of M are now even less pleasant than (3.5), and thus we do not reproduce them here. It is again important mainly to look at the discriminant, which was identified by MATHEMATICA as and becomes a square times the following polynomial: This polynomial is not a square for n ≥ 3 by the leading coefficient argument. It is, though, a square for n = 1, 2, and gives solutions for d, M , and α. The braids α are more easily identified by direct observation. For n = 2 we have The solution α 1 can be seen from the word [−2 − 1 − 1221] in family 1. The other solution (was guessed but) is also easily confirmed. For n = 1 both solutions stem from words in family 1: Case 3. Two negative neighboring syllables. Here we have with The solutions for the lower right entry of M look thus: .
Here the edge coefficients become −3 for n ≥ 2. Thus n = 1. Now M again becomes a Burau matrix, and there are the solutions (Again the first comes from [1 − 21 − 2] in family 1.) With the discussion in the preceding three cases, we have thus now obtained restrictions on how syllables in β can look like. There are four 'local' syllable types (up to symmetries), which can be summarized thus, and we will call admissible.
• No syllable has length at least 4.
• A syllable of length 3 has both neighbored syllables of the same sign.
• A syllable of length 2 has exactly one of its two neighbored syllables having the same sign.
• A syllable of length 1 has at most one of its two neighbored syllables having the same sign.
For each admissible syllable, we have also identified the two possible braids outside the syllable (although not their word presentations).

Words of small length
The next stage of the work consists in verifying a number of words of small length. One can easily argue (see [21]) that for more than one component, there are no crossings in an everywhere equivalent diagram between a component and itself. We notice that this observation specializes here to saying that (for non-split braids) the exponent sum (or word length) is even. This will not be essential in the proof, but helpful to avoid checking certain low crossing cases.
The test of small length words is done by an algorithm which goes as follows.
We start building a word β of the form where γ is a word known to us, and we know the Burau matrix .) The understanding is that switching any of the crossings in γ gives a braid with unknotted closure. We call γ an extendable word , in the sense that it can potentially be extended to a solution β. Whenever M is the identity matrix, we can take β = γ and have an everywhere equivalent braid, which we output.
Next we try to extend γ by one letter τ = σ ±1 i , so that it is not the inverse of the preceding letter, and the admissible syllable shapes are not violated. LetM = ψ(τ −1 ) · M . We test whether tr M · ψ γ · τ −1 = (−t) ∓1 , which is equivalent to whether a crossing change at the new crossing (also) gives the unknot. If this happens, we can continue the algorithm with γ replaced by γ · τ and M replaced byM .
This procedure can yield the solutions β up to given number of crossings (word length), and also produce the list of extendable braids γ up to that crossing number.
Note that, since potential EE solutions can be directly checked, it is often enough to work with particular values of t. (These can be a priori complex numbers, but in practice most helpfully should be chosen to be rational.) We did not feel confident about this in the three initial cases, because of the presence of variable exponents. Alternatively, we could have used a t with |t| < 1 and some convergence (and error estimation) argument, but whether that would have made the proof nicer is doubtful.
For rational t, we implemented the above outlined procedure in C++, whose arithmetic is indefinitely faster than MATHEMATICA. It, however, has the disadvantage of thoughtlessly producing over-/underflows, and some effort was needed to take care of that and to work with rational numbers whose numerator and denominator are exceedingly large.
With this problem in mind, it is recommended to use simple (but non-trivial) values of t. We often chose t = 2, but also t = 3, and a few more exotic ones like t = 4/5 whenever feasible. We were able to perform the test up to 15 crossings for t = 2 and up to 12 crossings (still well enough for what we need, as we will see below) for the other t.
This yielded the desired family 1, but also still a long list of extendable words even for large crossing number. Such words were not entirely unexpected: one can see, for example, that when β is a solution, then any subword γ of a power β k of β is extendable (with α being, roughly, a subword of β 1−k ). There turned out to be, however, many more extendable words, which made extra treatment necessary.
The following argument gives a mild further restriction. Assume β has a subword [12 − 1 − 2]. Switching the '2' will give [−2 − 1], while switching the '−1' will give [21]. Now, these two subwords can be realized also by switching either crossings in [−21], which is a word of the same braid as [12 − 1 − 2]. This means that if β is EE, then also a word is EE in which [12 − 1 − 2] was replaced by [−21]. Thus verifying words up to 10 crossings would allow us to inductively discard words containing [12 − 1 − 2] and its various equivalents.

Global conditions
All these 'local' conditions were still not enough to rule out all possibilities, and finally we had to invent a 'global' argument.
For this we use the following fact: if β ∈ B 3 has unknotted closure, then β is conju- The first proof appears to be due to Murasugi [13]. It was recovered by Birman-Menasco [2]. A different proof, based on the Jones polynomial, follows from (though not explicitly stated in) [18]. Namely, one can use relations of the sort which do not augment word length (together with their versions under the various involutions), and cyclic permutations to reduce β to a length-2 word. The non-conjugacy to σ ±1 1 σ ±1 2 of a crossing-switched version β of our braids β is detected by Garside's algorithm [6]. We adapt it in our situation as follows.
Lemma 3.1. Assume a braid β ∈ B 3 is written as ∆ k α with α a positive word with cyclically no trivial syllables and k ≤ −2. Then β is not conjugate to σ ±1 1 σ ±1 2 . Proof . This is a consequence of Garside's summit power in the conjugacy normal form. There is again an alternative (but a bit longer) way using span t tr ψ(β ). We refer for comparison to the proof of Lemma 4.2, but for space reasons only briefly sketch the argument here. For α one uses the relation (2.8) and that the span of V (α) is determined by the adequacy of the diagramα. The center multiplies tr ψ only by powers of t 3 . • k is at least half of the number of negative letters in β .
• Only the first and last syllable of α can be trivial. If β starts and ends with positive letters τ (not necessarily the same for start and end), which are not followed resp. preceded byτ , then α has no trivial syllable.
(For the bar notation recall the end of Section 2.3.) Proof . This is the result of the application of Garside's procedure on β . We manage, starting with trivial α and k = 0, a word presentation of β with the following property: α is a positive word with only the first and last syllable possibly trivial, and γ is a terminal subword of β . We also demand that if γ starts with a positive letter, this is the same as the final letter of α (unless α is trivial). We call this the edge condition. We apply the following iteration.
1. Move as many positive initial letters from γ as possible into the end of α , so as γ to start with a negative letter τ . This will not produce internal trivial syllables in α because of the edge condition and because β has no ∆ subword.
2. If γ has no negative letters, then we move it out into α entirely, and are done with k = k .
Then go back to step 1.
In the end we obtain the desired form. Since there is no ∆ −1 in β , each copy of ∆ −1 added compensates for at most two negative letters of β . Now we consider an EE word β of, say, more than 10 crossings. We switch, in a way we specify below, a crossing properly and apply the above procedure starting at a cyclically well-chosen point of the resulting word β . We obtain the shape of Lemma 3.2. This gives the contradiction that the closureβ is not unknotted by Lemma 3.1.
If β contains a syllable of length 2 or 3, then we have (up to symmetries) [12221] or [−1221]. Switch the final '1'. This does not create a ∆ −1 because there is no subword [21 − 2 − 1] in β. Then apply the procedure starting from the second last letter: With this we can exclude non-trivial syllables in β. Next, if there is a trivial syllable between such of opposite sign, up to symmetries [−2−12], we have with its two further neighbored letters (3.10) The right neighbor cannot be '1', because we excluded [−2 − 121] as subword. The left neighbor cannot be '−1', because we excluded ∆ −1 . Now we switch the middle '−1' in the portion (3.10), and start the procedure with the following (second last in that presentation) letter '2'. The only words that remain now are the alternating words β = [(1 − 2) k ]. There are various ways to see that they no longer create problems. In our context, one can switch a positive letter, group out a ∆ −1 built together with the neighbored letters, and then start the procedure right after that ∆ −1 .
This completes the proof of the non-positive braids.
4 Proof of the positive case

Adequate words
From now we start examining, and gradually excluding the undesired, positive braids in B 3 . The nature of this part is somewhat different. Here no electronic computations are necessary, but instead a delicate induction argument. The presence of the central braids and their realization of every positive word as subword explain that no 'local' argument can work as in the non-positive case. Thus from the beginning we must use the 'global' features of the braid words.
Our attitude will be that except for the stated words β, we find two diagrams D =β we can distinguish by the Jones polynomial. Because of the skein relation (2.1) of V , one can either distinguish the Jones polynomial of (the closure of) two properly chosen crossing-switched versions β , or of two smoothings of β. (In the crossing-switched versions, a letter of β is turned into its inverse, and in the smoothings it is deleted.) Moreover, one can switch back and forth between the Jones polynomial and the Burau trace, because of the consequence of (2.8) stated below it. Notice that for a positive word, length and exponent sum are the same.
Accordingly we call a word ψ-everywhere equivalent, if all crossing-switched versions (or equivalently, all smoothed versions) have the same Burau trace.
Trivial syllables will require a lot of attention in the following, and thus, to simplify language, we set up the following terminology.
Definition 4.1. We call a positive word adequate, resp. cyclically adequate, if it has no trivial syllable (the exponent vector has no '1'), resp. has no such syllable after cyclic permutations. Otherwise, the word is called (cyclically) non-adequate.
This choice of language is suggested by observing that cyclically adequate words β give adequate diagramsβ. Contrarily, for cyclically non-adequate words β the diagramsβ are not adequate: a trivial syllable of a positive word β always gives a self-trace in the B-state ofβ, i.e.,β is not B-adequate. (However, being positive,β is always A-adequate, and thus it is not inadequate in the sense of (2.4).) A useful application of adequacy is the following key lemma, which will help us carry out the induction without unmanageable calculations. We call below a trivial syllable isolated if it is not cyclically followed or preceded by another trivial syllable. Recall also the weight ω(β) from Section 2.3.
Proof . It is enough to argue with the Jones polynomial. The closure diagramγ is adequate, and by counting loops in the A-and B-states, we see span V (γ) = [γ] − ω(γ) + 1. If ω(γ) = ω(β), the right hand-sides of (4.1) and (4.2) agree, so we argue that the inequality (4.2) is strict. When a trivial syllable in β is isolated, so is its (self-)trace in the B-state ofβ, as defined below (2.5). In follows then from the explained work of [1] that the extreme B-degree term is zero, making (4.2) strict.
We use the following lemma to first get disposed of cyclically adequate braids. Let us from now on use the symbol ' . =' for equality of braid words up to cyclic permutations.
independent of α and i. We apply this argument for k = m − 1 once on the syllable σ m i and once on σ m j . Then we see again two positive words of equal length and weight that must have the same Jones polynomial, one of which has a (single, and thus isolated) trivial syllable, and the other has none. As before, Lemma 4.2 gives a contradiction.
For the rest of the proof, we consider a cyclically non-adequate word β, and use induction over the word length. We assume that ψ-everywhere equivalent braids of smaller length are in families 2, 3 and 4. It will be helpful to make the families disjoint by excluding the (central) cases of l = 1 and 3 | k in family 2.
Lemma 4.4. When β is positive and ψ-everywhere equivalent and β has a 6-letter subword representing ∆ 2 , then deleting that subword gives a ψ-everywhere equivalent braid word.
Proof . All the crossing switched versions of β outside the copy of ∆ 2 have the same Burau trace, and deleting that copy of ∆ 2 , the Burau trace multiplies by t −3 .
Proof . Note that the crossing changes outside the two copies of ∆ will commute with putting the two ∆ close using to form a ∆ 2 , and then apply Lemma 4.4.
The move (4.4) will be used extensively below, and will be called sliding. Obviously, one can slide any copy of ∆ through any subword α i .

Induction for words with trivial syllables
Case 1. β has a ∆ 2 subword. We can apply Lemma 4.4, and use induction.
We have a central ∆ 2 word inserted somewhere in a remainder , which is either (a) a central word, (b) a split word σ k 1 or (c) a symmetric word [1 l 2 l ] k . In (a) we have a central word β, and this case is clear. So consider (b) and (c).
Case 1.1. The split remainders. Then β = ∆ 2 σ k 1 for k > 0. So we have the words 1 (up to reversal, cyclic permutations and σ 1 ↔ σ 2 ) We distinguish them directly by two smoothings: for the letters indicated in bold, we obtain a (2, n)-torus link, and smoothing a letter in 1 k gives a positive braid with ∆ 2 , and thus under closure a link of braid index 3. (For ' . =' recall above Lemma 4.3. Of course, δ represents ∆ 2 , but we separate both symbols as subwords of β.) If the whole word β is ψ-everywhere equivalent, then by Lemma 4.4 (since δ represents ∆ 2 ) so is ∆ 2 α, obtained after deleting δ in the remainder. Note for later, when we insert δ back, that this insertion must be done so that the last letter of δ and the first letter of α are not the same.
Iterating this argument, we can start by testing α = [12] and α = [1212]. Thus it is enough to see that when one inserts a ∆ 2 word into (or before or after) [12] or [1212], the result β is not ψ-everywhere equivalent, unless it is β . = [12] k for k = 4 or 5. These are all knot diagrams of ≤ 10 crossings, and they can be checked directly. Then (iteratedly) inserting back δ must be done so as to yield only β . = [12] k for higher k. Case 1.2.2. Consider next l ≥ 2. Then β is up to cyclic permutations of the form ∆ 2 α with the exponent vector of α having a '1' possibly only at the start and/or end.
We want to apply Lemma 4.2 to exclude these cases. Case 1.2.2.1. Let α be adequate (i.e., have no trivial syllable even at its start and end). We compare two crossing-switched versions of β = ∆ 2 α. First we switch a crossing in ∆ 2 , turning β into σ 2 1 σ 2 2 α or σ 2 2 σ 2 1 α, which is a cyclically adequate word. Another time we switch a crossing in (any non-trivial syllable of) α, yielding ∆ 2 α for a positive α , whereby the weight decreases by 2.
Let us make this argument more precise. We look at the three words for ∆ 2 in ( In a similar way one checks for the other seven cases in (4.5) that one can apply Lemma 4.2. It is important to notice for later that in the first four cases, we do not need in fact that any of the neighboring syllables of ∆ 2 is trivial.
Case 2. There is no ∆ 2 subword in β, but there are two ∆ subwords, i.e., we can apply Lemma 4.5. Thus withᾱ 1 α 2 a ψ-everywhere equivalent word. Ifᾱ 1 α 2 is central, the situation is clear. Case 2.1.ᾱ 1 α 2 is split. These are the words β (up to symmetries) They are distinguished by smoothings of the indicated boldfaced letter (giving as in Case 1.1 a (2, n)-torus link) and some letter outside the copies of ∆ (giving a link of braid index 3). Case 2.2. We have We can obviously assume, by excluding ∆ 2 subwords, that none of α i is the trivial (empty) word.
Recall the sliding (4.4) we used to bring two subwords ∆ together to form a ∆ 2 : Case 2.2.1. If one of the ∆ in (4.6) has neighbored letters of different index, after the sliding the other ∆ close, will have neighbored letters to ∆ 2 of the same index.
Look at the words in the first row of (4.5). These are built around the the four words for ∆ 2 factoring into two subwords of ∆. In all four cases, the indicated to-change crossings lie entirely in one of the copies of ∆ in ∆ 2 . By symmetry, it can be chosen in either of them, in particular in the one we did not slide.
Thus one can undo in the same way bringing together the two copies of ∆ after either crossing changes (on the right of (4.8)), and realize the two crossing changes in the original braid word β (on the left of (4.8)). The crossing changes in (4.5) thus apply also in β to give positive words satisfying the assumptions of Lemma 4.2, and we are again done.
Case 2.2.2. Now either of ∆ has neighboring syllables in α i of the same index. We assumed (by excluding ∆ 2 subwords in the present case) that none of α i is a trivial word. Case 2.2.2.1. Let us first exclude that an α i has length 1. We will return to this situation later.
In either the first and third case, none of the neighboring syllables to ∆ can be trivial, because otherwise we have a ∆ 2 subword up to cyclic permutations, and we dealt with this case. Thus permuting back all of α's letters to the right, we have the shape we wanted.
In the second case, if the neighboring syllable '2' is trivial, then we have [1|121|21]. Now choosing a better ∆: [11|212|1], one arrives (up to σ 1 ↔ σ 2 ) in the third case above. Thus if we cannot obtain the desired shape of α, the syllable after ∆ must be non-trivial: [1|121|22]. Then compare the smoothings of the first and last '2': [1|11|22] and [1|121|2]. The first subword gives a cyclically adequate word of smaller weight, the second one a word of equal weight.
This finishes the proof of Theorem 1.1.
Remark 4.6. The use of the Jones polynomial means a priori that we distinguish the links of D i as links with orientation, up to simultaneously reversing orientation of all components. However, this restriction is not necessary, and in fact, we may see the links of D i non-isotopic as unoriented links. Namely, in the non-positive case (family 1 in Theorem 1.1), we have only knot diagrams D, where the issue is irrelevant. In the positive case (families 2 and 3), one can easily see that all β can be simplified to a positive braid word of two fewer crossings. It is a consequence of the minimal degree of the Jones polynomial (see, e.g., [4]) and its reversing property (see, e.g., [9]), that if the closures of two positive braids of the same exponent sum (and same number of strings) are isotopic as unoriented links, then they are isotopic (up to reversing all components simultaneously) with their positive orientations.
Remark 4.7. Note also that, for links, our method does not restrict us to (excluding) isotopies between the links of D i mapping components in prescribed ways. For example, the diagram D gives a natural bijection between components of D i and D j , but this correspondence never played any role.

On everywhere dif ferent diagrams
As an epilogue, we make a useful remark that by modifying Lemmas 4.4 and 4.5, one can easily see that the construction of everywhere different diagrams (where all crossing-switched versions represent different links, and here even in the strict, oriented, sense) is meaningless for 3-braids. Proof . We show that for each 3-braid word β, there are two crossing-switched versions giving conjugate braids.
We show this by induction over the word length. Assume β is an everywhere different 3-braid word.
Obviously, by induction, we can restrict ourselves to braid words β with no σ ±1 i σ ∓1 i (whose deletion preserves the everywhere different property).
Thus β has an exponent vector, and for evident reasons, all syllables must be trivial. (In particular the word length is even, and the closure is not a 2-component link.) If β were an alternating word, then β = (σ 1 σ −1 2 ) k , which is obviously not everywhere different (at most two different links occur after a crossing change).
Since β is thus not alternating, it must contain cyclically a word for ∆ = [121] or [212], or ∆ −1 . Moreover, one easily sees that in a subword [1212] the edge crossing changes give the same braid, so that both syllables around a ∆ (resp. ∆ −1 ) must be negative (resp. positive). In particular, different subwords of β representing ∆ (or ∆ −1 ) are disjoint.
Again by symmetry reasons, there must be more than one ∆ ±1 word. (The braids [121(−21) k −2] are not a problem to exclude: switch the two crossings cyclically neighbored to [121]; if k = 0 there are two unknot diagrams.) Thus The wordᾱ 1 α 2 has now by induction two crossing changes (of the same sign) giving conjugate braids. Since ∆ ±1 ∆ ±1 is central (if not trivial), these crossing changes will remain valid in ∆ ±1 ∆ ±1ᾱ 1 α 2 , and then also, by the sliding argument, in β.