The classification of partially symmetric 3-braid links

Abstract We classify 3-braid links which are amphicheiral as unoriented links, including a new proof of Birman- Menasco’s result for the (orientedly) amphicheiral 3-braid links. Then we classify the partially invertible 3-braid links.


Introduction
Links given by closures of 3-braids have been studied for some time. Ultimately such links were classified by , in terms of a description of conjugacy classes of 3-braids with the same closure. Still many properties of links are not visible from braid representations, and the classification of 3-braids with such properties remains open. We have done some work in this regard in [27,28]. Two properties of the 3-braids which are directly visible in a braid representation are amphicheirality and invertibility. The description of 3-braid links with such symmetries thus follows as a consequence of the Birman-Menasco theorem. In their braid approach both types of symmetries are assumed to either preserve or reverse orientation of all components of the link. (Components can also be permuted.) In this paper we offer the two corresponding results for the case that orientation of some but not all components is reversed. Although our two results are similar in nature, they require somewhat different approaches (which are also rather different from the one of Birman-Menasco).
In the paper [29], we initiated some study of edge coefficients of the Jones polynomial of semiadequate links. This was applied, among others, to 3-braid links, which we proved to be semiadequate. Another recent study of the Jones polynomial of 3-braid links was performed by Futer, Kalfagianni and Purcell [11]. We will first use the information we obtained from both works to classify those 3-braid links amphicheiral as unoriented links. The notion of unoriented amphicheirality is weaker than Birman-Menasco's and is obtained by dropping restrictions on component orientation that an isotopy to the mirror image should incur (cf. Definition 2.2). Our proof of Theorem 1.1 will reproduce the oriented case in an alternative way (see Remarks 3.2

and 3.3).
The second main result deals with the partially invertible 3-braid links. A link L is partially invertible if L is isotopic to itself with the orientation of some but not all components reversed. (Again, we will allow components to be permuted by the isotopy.) Theorem 1.2. Let L be a 3-braid link which is partially invertible. Then L conforms to one of the two patterns shown on Figure 1. Note that several noteworthy links are included in these families: the split links, the pretzel links .n; 2; 2/ (among which is the Whitehead link), and also the Borromean rings.
The proof of Theorem 1.2 uses in a fundamental way the genus of the 3-braid links as described by Xu [32], and some delicate (and partly computer-aided) case analysis. In contrast to the amphicheiral case, the derivation of the list of (fully) invertible 3-braid links remains untouched (and probably untouchable) with our tools. We thus do not discuss it here, and refer to [4] for the treatment.
Our proofs will require the combination of a considerable deal of knowledge about invariants of 3-braids, which we will review first.

Semiadequacy and Kauffman bracket
It is useful to define here the Jones polynomial via Kauffman's state model [18]. Recall, that the Kauffman bracket hDi of a link diagram D is a Laurent polynomial in a variable A, obtained by summing over all states S the terms where a state is a choice of splicings (or splittings) of type A or B for any single crossing (see Figure 2), #A.S / and #B.S/ denote the number of type A (respectively, type B) splittings and jSj the number of (disjoint) circles obtained after all splittings in S . We call the A-state A.D/ the state in which all crossings are A-spliced, and the B-state B.D/ is defined analogously. The Jones polynomial of a link L can be specified from the Kauffman bracket of some diagram D of L by (2) Fig. 2. The A-and B-corners of a crossing, and its both splittings. The corner A (respectively 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.
with w.D/ being the writhe of D (see subsection 2.2 below). Let S be the A-state of a diagram D and S 0 a state of D with exactly one B-splicing. If jSj > jS 0 j for all such S 0 , we say that D is A-adequate. Similarly one defines a B-adequate diagram D. See [21,30]. Then we set a diagram to be adequate D A-adequate and B-adequate ; semiadequate D A-adequate or B-adequate ; inadequate D neither A-adequate nor B-adequate : (Note that inadequate is a stronger condition than not to be adequate.) It is adequate if it has an adequate diagram. This property is stronger than being both Aand B-adequate, since a link might have diagrams that enjoy either properties, but none that does so simultaneously. (The Perko knot 10 161 in [25, appendix] is an example; this feature of it transpires from results in [30], where the knot is discussed. Two basic observations are that a reduced alternating diagram (and hence an alternating link) is adequate, and that a positive diagram (and link) is A-adequate, and similarly negative ones are B-adequate.
We will use more commonly another notation for coefficients. Definition 2.1. Let V 2 ZOEt˙1 or V 2 t˙1 =2 ZOEt˙1, and n 0 an integer. Let m D min deg V and M D max deg V (then 2m 2 Z). We write a n .V / WD OEV mCn and N a n .V / WD OEV M n for the n C 1-st or n C 1-last coefficient of V .
A basic observation in [21] is that when L is Aresp. B-adequate then ja 0 .V .L//j D 1 resp. j N a 0 .V .L//j D 1. Thus if L is adequate, and in particular alternating, both properties hold. We will use this fact continuously below without explicit reference.
When L is an A-adequate link, then a n .V .L// for n Ä 2 were studied in [8,9,29]. We will need the formulas below, so let us recall them briefly. Let D be an A-adequate diagram of L. (We will assume that D is connected.) Then, in the notation of [29], and Here jA.D/j is the number of loops in the A-state A.D/, and the quantities e, e CC , ı, and 4 are the number of pairs or triples of loops in A.D/ for which there exist crossing traces (obtained as in Figure 2) making them look (up to moves in S 2 ) like in: (We do not require that these be the only traces connecting the loops, only that such traces should exist.) In our case, whenever we use (5), D will be alternating, and then ı D 0.

Link diagrams, skein relations, and genus
An alternative description of V is to be the polynomial taking the value 1 on the unknot, and satisfying the skein relation (6) This is another way, different from (2), to specify the Jones polynomial. We will denote in each triple as in (6) the link diagrams (from left to right) by D C , D and D 0 ; they are understood to be identical except at the designated spot.
The writhe is a number (˙1), assigned to any crossing in a link diagram. A crossing as on the left in (6) has writhe 1 and is called positive. A crossing as in the middle of (6) has writhe 1 and is called negative. The writhe w.D/ of a link diagram D is the sum of writhes of all its crossings. In the case of a crossing, we call the writhe also the (skein) sign.
Let c˙.D/ be the number of positive, respectively negative crossings of a diagram D, so that c.D/ D c C .D/ C c .D/ and w.D/ D c C .D/ c .D/.
By c.D/ we denote the number of crossings of a diagram D, n.D/ the number of components of D (or K, 1 if K is a knot), and s.D/ the number of Seifert circles of D. The crossing number c.K/ of a knot or link K is the minimal crossing number of all diagrams D of K.
A crossing between two different components K 1 and K 2 in a diagram D of a link L is called mixed. The linking number lk.K 1 ; K 2 / is half the sum of the signs of all (mixed) crossings between K 1 and K 2 in any diagram of L; this is a link invariant.
The (Seifert) genus g.K/ resp. Euler characteristic .K/ of a knot or link K is said to be the minimal genus resp. maximal Euler characteristic of Seifert surface of K. For a diagram D of K, g.D/ is defined to be the genus of the Seifert surface obtained by Seifert's algorithm on D, and .D/ its Euler characteristic. We have .D/ D s.D/ c.D/ and 2g.D/ D 2 n.D/ .D/. The skein polynomial P [10,24] is a generalization of V in two variables l; m and satisfies (in the convention we use) the skein relation (7) The replacement of a positive or negative crossing by the (non-crossing) fragment on the right will be called smoothing out.
We denote by ŠD the mirror image of a diagram D, and ŠK is the mirror image of K. Clearly g.ŠD/ D g.D/ and g.ŠK/ D g.K/. When one treats links, it should be emphasized that here mirroring is meant to preserve orientation of all components of the diagram. Definition 2.2. We call an oriented link positively (orientedly) amphicheiral, if it is isotopic to its mirror image with the orientation of all components preserved, and components (possibly and arbitrarily) permuted. A link is negatively (orientedly) amphicheiral, if there is an isotopy to its mirror image reversing orientation of all components (and possibly permuting them). We call a link (orientedly) amphicheiral if it is positively or negatively (orientedly) amphicheiral. The word achiral will be used as a synonym for 'amphicheiral'. We call a link unorientedly achiral (or amphicheiral), if it is isotopic to its mirror image with the orientation of some (possibly none, but also not necessarily all) components reversed (and components possibly permuted).
It is clear that oriented achirality (of some sign) implies unoriented one, and for knots both notions coincide. The Hopf link is an example of a link which is unorientedly achiral, but not orientedly so.
It can be seen from (2)  and we call such a polynomial conjugate. Further one can see from (2) (or from [20]) that changing orientation of individual components multiplies V by a power of t . This implies that the Jones polynomial of an unorientedly amphicheiral link L satisfies for some k 2 Z. We will call a polynomial with such property weakly conjugate.

Schreier normal form and Jones polynomial
The n-string braid group B n is considered generated by the Artin standard generators i for i D 1; : : : ; n 1. These are subject to relations of the type OE i ; j D 1 for ji j j > 1, which we call commutativity relations (the bracket denotes the commutator) and iC1 i i C1 D i iC1 i , which we call Yang-Baxter (or shortly YB) relations. For a braidˇ, let Ǒ denote the closure ofˇand let OEˇ denote the exponent sum ofˇ. For a braid wordˇ, we denote by OEˇ C and OEˇ the number of positive resp. negative letters inˇ, so that OEˇ D OEˇ C OEˇ . We callp ositive (resp. negative) if OEˇ D 0 (resp. OEˇ C D 0). There is a common graphical representation of braids. We will mostly use it implicitly, but to facilitate understanding, let us fix our convention. Strings will be assumed numbered from left to right, oriented upward, and braid words will be composed from bottom to top. To i should correspond a positive (right-hand) crossing, and to 1 i a negative one. (This sign convention is now opposite to [11], after a temporary mix-up of either signs in a preliminary version has been cleaned up. We have then to slightly change some formulas there when we quote them here.) The 3-braid group B 3 falls into a category of groups studied by Schreier in the 1920s [26], who developed a conjugacy normal form for this group.
Let C D . 1 2 / 3 be the center generator of B 3 .
To save space and enhance readability, we will often use the compact notation for braid words of [27]: we write the indices of i into brackets, writing i for i , i p for p i and i for i 1 . E.g.  (1) C k OE1 p 1 2 q 1 1 p s 2 q s , This form is unique up to cyclic permutation of the word following C k .
Braids in form 1 above will be called generic, and the others non-generic. The vector S D .p 1 ; q 1 ; ; p s ; q s / of the generic form is called Schreier vector.
There is an obvious action of the dihedral group on the entries of the Schreier vector by cyclic permutation and inversion. We say that a Schreier vector S admits a dihedral (anti)symmetry, if S can be turned into S by such an action.
Schreier's normal form was used quite extensively by . A more recent treatment, from the point of view of the Jones polynomial, is given in a paper by Futer, Kalfagianni and Purcell [11]. We will use some of the work there quite essentially.
The Burau representation W B 3 ! GL.2; ZOEt; t 1 /, into the ring of 2 2 matrices with coefficients being Laurent polynomials in t, is defined by The Jones polynomial of a 3-braid is given by where 'tr' is the trace. See [2,16]. Futer, Kalfagianni and Purcell [11] notice the following interesting consequence of formula (8) (stated there under a slight, but inessential for the proof, restriction).
Lemma 2.4. Let˛2 B 3 be a 3-braid, and letˇD C k˛. Then This is particularly interesting to apply to the generic form in Theorem 2.3, where˛is the alternating part ofˇ: with positive integers p i ; q i and s. We set We call s the length ofˇ. Futer, Kalfagianni and Purcell [11] obtain the following expression for the degrees of V . Ǫ/, using a study of the Kauffman bracket: max deg V . Ǫ/ D 3p q 2 and min deg V . Ǫ/ D p 3q 2 :

The genus of 3-braids
We will describe now how to determine the genus of 3-braids, which requires us to introduce a modified presentation of the 3-braid group.
In [32], Xu considered the new generator of B 3 (where 1;2 still denote Artin's generators), with which B 3 gains the presentation Then for any representation of a braidˇ2 B 3 as word in 1;2;3 one obtains a Seifert surface of the closure Ǒ ofb y inserting disks for each braid strand and connecting them by half-twisted bands along each ˙1 i . An important feature of 3-braid links is that there is a Seifert surface of this type, which has minimal genus (a fact first proved by Bennequin in [1]). Such a band representation can be obtained by an effective algorithm given by Xu. It allows one to write eachˇ2 B 3 in one of the two forms (A) . 2 1 / k R or L 1 . 2 1 / k .k 0/; or where L and R are positive words with (cyclically) non-decreasing indices (i.e. each i is followed by i or iC1 , and all exponents are positive). The form (A) (called "strongly quasipositive" in the terminology coined by Rudolph) will not be very relevant to us. We will mainly look at form (B). We can, and will, assume it to be cyclically reduced, i.e. that L and R do not start or end with the same letter.

Semiadequacy of 3-braids
Letˇbe a braid word. We writeˇD with i the Artin generators, p j ¤ p j C1 and q j ¤ 0. We will assume that the condition p j ¤ p j C1 includes p k ¤ p 1 when k > 1. This is not a restriction when we (as we will do below) consider braids up to conjugacy, which includes cyclic permutation of the letters. The term q j p j is called a p j -syllable, or more commonly, omitting the index, just a syllable. The number k is called syllable length ofˇand written by sl.ˇ/. We callˇalternating if for all j D 1; : : : ; k the numbers . 1/ p j q j are positive, or all are negative.
We will focus on 3-braids. For 3-braids the subscripts p j alternate between 1 and 2, and become (up to conjugacy) irrelevant. We call the vector .q 1 ; : : : ; q k /, regarded up to cyclic permutations, the exponent vector ofˇ. The number k is called its length; if k > 1, then k is even. Note that whenˇis alternating, then its exponent vector is the same as its Schreier vector.
Any braid wordˇgives a link diagram Ǒ under closure, as a braid gives a link. We call a braid word A-adequate if the link diagram Ǒ is A-adequate. We define similarly for B-adequate, and (semi)adequate. We call a braid to be A-adequate, or B-adequate, or (semi)adequate, if it has a word representation with the same property. Let be the square root of the center generator C of B 3 . The following is shown by a careful observation:

The V -Q formula
The Brandt-Lickorish-Millett-Ho polynomial [5] Q.z/ is given by the properties It is a polynomial invariant of unoriented links.
We will use below a formula, due to J. Murakami [23] (and also Kanenobu [17,Theorem 2]), which relates the Jones and Q polynomial of a 3-braid link.
Let i D p 1, u D p t and x D u C u 1 . Let further 1 for a braidˇof exponent sum OEˇ, Then Murakami's formula is: Proof. Let us for simplicity exclude the cases of links of braid index Ä 2, which are easy. We assumeˇis a minimal length word up to conjugacy. The cases thatˇhas an exponent vector of length two or less are easy to deal with, and lead to case (b). So we assume the exponent vector ofˇis of length at least 4.

Oriented achirality and 3-component cases
Nowˇis, say, A-adequate. If L is orientedly achiral, there is a braid representation of L as a B-adequate 3-braid wordˇ0 (namelyˇwith i changed to 1 i ). Nowˇ0 has the same length asˇ, and moreover the same writhe (exponent sum). The latter can be concluded either from [4], or from the proof of the Jones conjecture for 3-braid links in [28] (which uses essentially also work in [29]). Thus OEˇ D OEˇ0 D OEˇ, and OEˇ D 0. This in particular rules out the case that L has 2 components, since then OEˇ is odd. Now by the work of Thistlethwaite [30], for two diagrams of the same writhe and crossing number of the same link, one is B-adequate if and only if the other one is. Thusˇis also B-adequate, and hence adequate.
In the case that L is a 3-component unorientedly achiral link, each crossing of the diagram D D Ǒ is mixed, i.e. involves two distinct components. If we take the diagram Ǒ0 DŠD (the mirror image of D with all component orientations preserved) and change orientation of some components to obtain a diagram D 0 of L, then still D 0 is Badequate (since semiadequacy is independent on component orientation). Furthermore, D and D 0 both represent L, and all their crossings are mixed. This implies w.D 0 / D w.D/, because it is equal to twice the sum of linking numbers between all (pairs of) components of L. From this the conclusion that D (andˇ) is adequate follows as above. (This argument does not work for 2 components because then w.D 0 / D w.D/˙2, as can be seen from (17) below.) So from now on we assume thatˇis adequate.
In caseˇis alternating, we can apply [22]. An easy observation shows that the diagram is determined by the exponent (and in this case Schreier) vector up to dihedral moves (cyclic permutations and reversal of order), and the only possible flypes in a diagram of a closed 3-braid occur at exponent/Schreier vector of length 4, with an entry˙1 (see [4]). In that case, however, the flypes just reverse the orientation of (all components of) the link. So we obtain no new symmetries. We arrive at case (a).
It remains to exclude the case thatˇis not alternating. In that event, we know thatˇis positive or negative, and has no as subword. Positive (or negative) links are not orientedly achiral, so the possibility that L is a knot, or that the achirality is oriented is easily ruled out.
It is clear that all components of an unorientedly achiral link are achiral knots, or mirror images in pairs. The latter option does not lead to anything new for closed 3-braids, so we ignore it.
We argued that L D Ǒ cannot have 2 components, so assume that it is a 3-component link. Let unoriented achirality manifest itself so that L can be obtained from ŠL either by reversing component K 1 , or components K 2 and K 3 . This implies that the linking number lk.K 2 ; K 3 / D 0 (see subsection 2.2). Since the diagram D D Ǒ is positive or negative, K 2 and K 3 must have no common (mixed) crossing there. This means that the exponent vector ofˇis even, i.e. all entries are even, and (up to mirroring) positive. Then reversal of component orientation (of We know that in positive and negative diagrams the canonical Seifert surface has minimal genus (see e.g. [6]). That is, Remark 3.2. In Birman-Menasco [4] the oriented achirality result was obtained (which leads to case (a) above), so we already have a slight extension of their result. However, latter follows in [4] from the extremely involved proof of the classification of closed 3-braids. It is therefore very useful to obtain simpler proofs of at least consequences of Birman-Menasco's work. Thus more important than the partial drop of orientations here is the opportunity to bypass the method in [4]. Still we must admit that the insight in [4] motivated the present proof, and with [28], [30], and ultimately [22], we have to invoke another quite substantial body of results. Thus, whether our proof is in the end "simpler" is to some extent a matter of personal view. . We originally claimed in [29] also a result on the unorientedly amphicheiral links, but a remark of the referee helped discovering a mistake in the proof. An attempt to remedy it showed that we had in fact overlooked the two most interesting examples. The proof of the correct version, stated as Theorem 1.1 above, became too long and led us too far aside to discuss in [29]. We decided then to move out the entire treatment of the theorem (including the orientedly amphicheiral links) to this separate paper.

The non-generic cases
The remaining part of the proof of Theorem 1.1 must be done by proving the following: Proposition 3.4. Let L be a 3-braid link of 2 components which is unorientedly achiral, but not orientedly so. Then L is the Hopf link, the (3,3)-rational link, or the link 9 2 61 .
Proof. If L D Ǒ is a 2-component link K 1 [ K 2 , two of the strands x; y ofˇform (under closure) an achiral knot of braid index at most 2. This knot must be trivial, and so x; y have linking number where linking number is meant to be now the sum of signs of their common crossings inˇ(this is not a linkcomponent-wise linking number, as defined in subsection 2.2). We fix the (oriented) mirroring ofˇfor the whole proof so that lk.x; y/ D 1 : Let z be the third strand ofˇ. We name the components of L so that K 2 is the closure of x [ y, i.e. the 2-string component ofˇ, while K 1 is the closure of z.
Let l WD lk.K 1 ; K 2 / be the linking number of the two components of L. Then with (18), We will assume that we obtain the oriented mirror image ŠL of L by reversing orientation of the component K 1 .
(The argument will apply the same way if K 2 had to be reversed, since we will use only polynomial invariants and the genus in the proof, which are preserved when both components are reversed.) We write by D D Ǒ the link diagram before component reversal, and by D 0 the one after it. Let us here record the following simple test for unoriented achirality: Proof. It is well-known (and can be seen from (2) for example, or see [20]) that when reversing a component K of a link L, then V shifts by t 3l , where l is the total linking number of K, i.e. the sum of the linking numbers with all other components of L. It is similarly well-known that taking the (oriented) mirror image changes t to t 1 .
Recall (see end of subsection 2.2) that a polynomial V 2 ZOEt˙1 =2 is weakly conjugate if V .t 1 / D t l V .t / for some l 2 Z. This property transcribes now to mean that a i D N a i for all i (see Definition 2.1). We use now the Schreier normal form, and settle a few simple cases first. The following is a simple observation.
Then when reversing K 1 in the diagram Ǒ , we have at most 3 C 2v Seifert circles.
Proof. Let D 0 be D D Ǒ with K 1 reversed. Then D 0 is supposed to be a diagram of ŠL. But In some situations we will need to estimate v from above. A well-known inequality proved by Bennequin [1,Theorem 3] can be paraphrased to state that v.ˇ/ Ä min .OEˇ C ; OEˇ / : Moreover, it is well-known that v.ˇ/ D 0 whenˇis alternating. We will combine below Lemma 3.6 with the following lower bound on s.D 0 /. : Proof. It is directly observed that whenˇD C k , then s.D 0 / D 2jkj C 1. If we look at the tangle T (with 3 in-and outputs) that corresponds to˛in D 0 forˇD C k˛, then we see also that s.D 0 / 2jkj C 1, and equality holds only if the Seifert circles in T form vertical lines. This in turn implies that the reversed strand z cannot have a common crossing in T with x or y, so that˛D p i . That p is odd finally follows for component number reasons. Lemma 3.8. Let L D Ǒ andˇbe non-generic. Then L is the Hopf link or the link 9 2 61 .
Proof. We look at the four non-generic forms in Theorem 2.3. Forms 3 and 5 give knots under closure, so they are excluded.
In form 4 we may assume using C D . 1 2 1 / 2 , that the word is positive (keeping in mind (18)). In this case v.ˇ/ D 0 in (20), which follows from (21). Then by Lemma 3.6, the diagram D 0 must have (at most) 3 Seifert circles. It is easy to check that thenˇD 1 2 1 , and L is the Hopf link.
In form 2 we use (18) to see thatˇD k 1 for some k 2 Z, where D 2 2 1 2 (note that C D ). The case k D 0 gives a trivial 2-component link, so assume k ¤ 0. Now we have from (21) that v.ˇ/ Ä 1 for k < 0 and v.ˇ/ D 0 for k > 0. By Lemmas 3.6 and 3.7, we see that we need to consider only k D 2; 1; 1.
For k D˙1 we have the .2; 4/-torus link (with parallel and reverse orientation) which is not unorientedly achiral. For k D 2 we have the link 9 2 61 , which is known to be unorientedly achiral (see p. 45 last paragraph of [19]).

The generic cases with length greater than 2
Lemma 3.9. No L D Ǒ withˇgeneric of length s 3 occurs.
In order to prove this, we go first back to the work of Futer, Kalfagianni and Purcell [11] reviewed in subsection 2.3.
Since L is an A-adequate link, we can use formula (4), and we easily obtain (as basically observed also in [11]) the following. A similar formula holds for a 1 .V . Ǫ// when replacing p i by q i .
It is somewhat important here, as in [11], to understand when the four extra monomials on the right of (9) can cancel an edge term in the unit-shifted V . Ǫ/. In [11] examples were constructed where such a cancellation occurs. We show now that it cannot occur at either side.
Lemma 3.11. Let in lemma 2.4, Proof. We have from (11) that max deg  Proof. For k D 0 the claim is clear from (11). Up to taking the mirror image, we may assume now k > 0. Then by (24), P 1 determines the maximal degree of V . Ǒ /. Moreover, min deg V . Ǒ / Ä min deg P 1 C 1. Otherwise, we must have, besides min deg P 1 D min deg P 2 and a 0 .P 1 / D a 0 .P 2 /, also a 1 .P 1 / D a 1 .P 2 /. But a 1 .P 2 / D 0, while a 1 .P 1 / is non-zero by Lemma 3.10.
Proof of Lemma 3.9. We use first (9) in Lemma 3.5 and get with (19) and the condition We would like to conclude from (26) that the two polynomials on the left and P 1 .t/ D t 6k V . Ǫ/.t /, have equal extremal degrees. (The name P 3 was chosen in order not to collide with (23).) When k D 0, then the r.h.s. vanishes, and P 1 D P 3 , so assume k ¤ 0.
Then it is easy to see that the 8 monomials on the right cannot form a polynomial R with ja 0 .R/j D 1 and ja 1 .R/j > 1. Contrarily we know that ja 0 .P 1 /j D ja 0 .P 3 /j D 1. Now if min deg P 1 ¤ min deg P 3 , then ja 0 .P 1 P 3 /j D 1 and ja 1 .P 1 P 3 /j ja 1 .P i /j 1 for some i . But by Lemma 3.10 we have that ja 1 .P 1 /j D ja 1 .P 3 /j D s > 2, so that ja 1 .P 1 P 3 /j > 1 and (26) cannot hold. Thus min deg P 1 D min deg P 3 , and then also max deg and using (11) On the other hand, we have (19), and so 2OEˇ D 2.2l C 1/ D 3l : We obtain l D 2 : Then from (19) we have OEˇ D 3 : We know that L has a 3-braid representation of exponent sum 3, while ŠL, whose Jones polynomial is given from  Substituting and simplifying this yields, after some simple (though not very pleasant) calculation, To see what polynomial V can satisfy such an identity, look first at the extreme degrees of the three summands on the right. In order the terms to cancel out, at least two must have the same minimal degree, and two the same maximal degree. Using that 2 min deg V and 2 max deg V are odd integers for a 2-component link, we have only the options min deg V 2 9 2 ; 11 2 ; and max deg V 2 3 2 ; 1 2 : From here there are two ways to see that no relevant V can occur. The simpler one is to observe that then span V Ä 5. By corollary 3.12, this means that p C q Ä 6. But if s 3, then the only option is˛D . 1 1 2 / 3 , which gives a 3-component link.
The other check is to solve for V . To simplify the expressions a bit, let Q V .u/ D u 3=2 V .u/, and make the ansatz The right '=' is the definition of a compact notation for polynomials we use below, whereby we write 2 ) has up to cyclic permutation the parities .even, odd, odd, odd/ or .even, even, even, odd/ : Let us abbreviate these two options as 'eooo' and 'eeeo'. (Note that with this freedom, we have to take into account either signs for p 1 ; the signs of p 2 and q i are then determined by p i p j > 0 and p i q j < 0.) If k D 0, then v.ˇ/ D 0, becauseˇis then alternating. Thus by Lemma 3.6, we have s.D 0 / Ä 3. It is easy to see that in the case 'eeeo' s.D 0 / 5, while in the case 'eooo' we have s.D 0 / D 3 only if S D .2; 1; 1; 1/ (assuming (18)). ThenˇD OE1 2 21 2, giving the Whitehead link, which is not unorientedly achiral.
We assume now that k ¤ 0. We go back to (26). If we use (23) and (27), then (26) can be written as Since˛¤ . 1 1 2 / 2 (otherwise L is a knot), we have by Lemma 3.10 that at least one of ja 1 .V . Ǫ//j and j N a 1 .V . Ǫ//j is 2.
Case 1. The case max deg P 1 D max deg P 3 was already discussed in the proof of Lemma 3.9. The argument, which among others leads to (28), can be repeated, except that in (the shorter) one of the options concluding the proof span V . Ǒ / Ä 5 leaves us to consider S D˙.2; 1; 1; 1/. Then OEˇ D 6k˙1, which contradicts (29).
Case 2. Consider the case j max deg P 1 max deg P 3 j > 1 : Now the polynomial P 2 P 4 on the right of (35) cannot have ja 1 j 2 or j N a 1 j 2. This means that one of ja 1 .V . Ǫ//j and j N a 1 .V . Ǫ//j should be 0 or 1. By Lemma 3.10, this can occur only if (in 'eooo'), in which case ja 1 .V . Ǫ//j or j N a 1 .V . Ǫ//j is 1. (The former is 1 for q i D 1 and the latter for q i D 1.) Then, we need to have ja 1 .P 2 P 4 /j D j N a 1 .P 2 P 4 /j D 1 for the polynomial P 2 P 4 on the right of (35). This can only occur if (Note that max deg P 2 and max deg P 4 vary with˙3k when k changes sign, but this variation is canceled out in the difference.) So either Now from (37) we have ja 2 .P 2 P 4 /j D j N a 2 .P 2 P 4 /j D 1. Combining this with (39) shows that (35) can hold only if Then either l D 0, and from (38) then (since k is an integer) k D 0, in contradiction to assuming k ¤ 0, or l D 4 and k D 2. Thus nowˇD Now from (18), and keeping in mind that we have the case 'eooo', we have lk.x; y/ D 2k C p 2 D 4 C p 2 D 1, in contradiction to the first condition in (40).
Case 3. Thus we assume in the following that j max deg P 1 max deg P 3 j D 1 : Case 3.1. If one of a 1 .P 1 P 3 / or N a 1 .P 1 P 3 / vanishes, then one of ja 1 .V . Ǫ//j and j N a 1 .V . Ǫ//j should be 1. By Lemma 3.10, the other one must be 2, and (36) holds, so we have a Schreier vector of type 'eooo'. Then we can look at a 2 and N a 2 as in the previous case, and see that now (40) modifies to jp 2 j > 1 and j max deg P 1 max deg P 3 j D jl C 2j D 1 : Then either l D 1 or l D 3, and in both cases from (38) we have k D 1 (since k is an integer). Now, for type 'eooo', the mixed crossings corresponding to letters in the Schreier vector are only those counted by p 2 , and then (18) gives lk.x; y/ D 2k C p 2 D 2 C p 2 D 1 ; so p 2 D 3. Then p 1 > 0 and the sign on the r.h.s. of (36) is negative. Thuš Next, the values of l determine with (19) that OEˇ D 5 or OEˇ D 1. The former leads to p 1 D 0, which cannot occur, so we look at the latter case, wherě D C 1 4 This braid can be checked directly not to have a weakly conjugate V (see end of subsection 2.2). Case 3.2. Thus now we can assume none of a 1 .P 1 P 3 / or N a 1 .P 1 P 3 / vanishes. This implies again (37) and (38), and we still have the second condition in (42). This leads us again to consider only l D 1; 3 and k D 1 : (44) But we have now, so far, no conditions on p i and q i , and have to treat either cases 'eooo' and 'eeeo'. Case 3.2.1. 'eeeo'. By direct drawing of the braid, we see that Using (44) we find p 1 C p 2 D˙2, but since p i ¤ 0 are of the same sign and even, this cannot occur. This finishes the case 'eeeo'. Case 3.2.2. 'eooo'. In this case we find Moreover, lk.x; y/ D 2k C p 2 D 1, so that p 2 D 3 : This determines then the signs p 1 > 0, q i < 0. If q 1 D q 2 D 1, then from (45) we must have p 1 D 4 (and the positive sign on the r.h.s.), and from (44) and (46) we obtain the braid in (43), which we already checked.
If some q i < 1, then with (46) and Lemma 3.10 we have: From (37) we have ja 2 .P 2 P 4 /j D j N a 2 .P 2 P 4 /j D 1, which combined with (35), (41) and (47) gives: (Take into account that a i a i˙1 Ä 0 by [31].) To see that this cannot occur, we use again (5). We have e CC D 2 because both p i > 1, but not both q i D 1, then ı D 0 in alternating diagrams, and (using that q i are odd) 4 D 1 if some q i D 1, and 0 otherwise. With (47) we get from (5) then ja 2 .V . Ǫ//j 2 f4; 5g ; in contradiction to (48) . This finishes the case 'eooo'. Here is the end of case 3, and the proof of Lemma 3.13.

The generic cases with length 1
Lemma 3.14. Let L D Ǒ andˇbe generic with s D 1. Then L is the (3,3)-rational link or the Hopf link.
Proof. We haveˇD C k p 1 q 2 with p q < 0. The case k D 0 easily leads to the Hopf link, so assume k ¤ 0. By connectivity we may further assume that p is even and q is odd, and then from (18), we have Note first that then k and q have opposite sign, and so k and p have the same sign. From (49) we also have OE˛ D p C 1 2k ; and so in (23) we have Now we look at the degrees on the right of (9). We have with (49), and similarly The degrees of P 2 depend on the sign of k. Let first k > 0. Then q < 0 and p > 0, and From (50) and (52) with k > 0 we see that min deg P 2 < min deg P 1 1 ; and so a 1 .V . Ǒ // D 0. Weak symmetry means then that N a 1 .V . Ǒ // D 0 : On the other hand from (51) and (52) we have max deg P 2 < max deg P 1 (because p C2k > 2). Combined with (53) this means that either max deg P 2 D max deg P 1 1 ; The option (54) says that 2k C p D 3, which for k; p > 0 means k D p D 1, but this contradicts our assumption that p is even. Thus consider the option (55). Using either the well-known polynomials of the .2; : /-torus knots, or formula (4), we find that the condition on N a 1 means that p Ä 2. Since p was assumed even, p D 2, and thusˇis of the formˇD (where : D means equality up to conjugacy). Next assume k < 0 (and p < 0). Then we have instead of (52) that Up to mirroring, we can combine (56) and (58) intǒ We want to use again Lemma 3.6, and need to estimate v.ˇ/. From (21) we get v.ˇ/ D 0 for 'C' and v.ˇ/ Ä 1 for ' ', where 'C' and ' ' refer to the ˙1 2 in (59). A direct look at D 0 shows that s.D 0 / D 3 C 2k. By Lemma 3.6, we see that only the ' ' is possible, for k D 1. This corresponds to the word OE12 2 1 3 2, which gives the (3,3)-rational link. Proposition 3.4 follows now from Lemmas 3.8, 3.13, 3.14, and 3.9.
With this the proof of Theorem 1.1 is finished.

Proof of the second main result
This section contains the proof of Theorem 1.2.

Genus estimate
We will assume nowˇto be a 3-braid. We want to find out when the link M D Ǒ is isotopic to one Q M obtained by reversing some component(s). Of course, for most links there will not be such an isotopy. Thus our goal is to distinguish M from Q M whenever possible. The only invariant I see that is generally effective yet controllable is the genus, which is equivalent to the Euler characteristic we will use below. The proof will thus center around establishing that for mostˇ. A few further cases are then subjected to tests using the skein polynomial P . What remains turns out to conform to one of the two diagrams of Figure 1.
Since both the genus and skein polynomial are invariant when all components are reversed, it is admissible throughout the proof to make the assumption that the component of M we reverse is an unknot O given by a 1-string subbraid S ofˇ.
Before we turn to the genus, we make one important unrelated observation. Let lk.O/ be the total linking number of O. It can be determined by one half of the sum of the writhes of all crossings ofˇinto which S enters, and this is how we will keep track of it below. Then if M and Q M are isotopic, we have There are several ways to see this, e.g. using the degree shift of V in (2). (It also follows from a formula of Hoste-Hosokawa [14,15] for the minimal coefficient of the Conway polynomial.) Apart from this (important) insight, the Jones polynomial becomes useless for the rest of the proof. In order to work with the genus, it is imperative that we use the form (13). Let us at this point exclude the split links M . They are partially invertible, and conform to the first pattern of Figure 1, where some of the boxes contains two trivial (uncrossing) strands. With this exclusion and condition (61), we see that we ruled out the forms (A) in (13), and so we will focus throughout on form (B).
We assume from now on thatˇD RL 1 : For convenience we will introduce i D 1 i in order to have only positive powers in the syllables, and we continue using the designation where we denote the i by their indices in a bracketed list, e.g.
following (12), and '˙3-syllables' are those of ˙3 . Let us fix that throughout this proof, except a few places where indicated otherwise, we always use this first (rather than the second) way in (12) to expand k 3 . When D is the closed braid diagram ofˇso expanded (into a word in Artin's generators), then the crossings inˇare counted by c.D/.
Now we compare (63) to 1 . Q M /, which we estimate by the canonical surface coming from the diagram Q D obtained from D after the strand S is reversed. For this it is necessary to count the Seifert circles in Q D. This is essentially done by looking at the case that the middle strand is reversed in a braid of the form The result is illustrated in Figure 3: each syllable contributes 2a i 1 resp. 2b i 1 small Seifert circles, and there is one 'global' Seifert circle running along the whole braid. Thus the number of Seifert circles is

Fig. 3. Counting the Seifert circles after reversing a braid strand
Let us below take care of the quantity l in the following way. We call a flip a subword ofˇ(regarded as a cyclic word in 1 , 2 ) of the form ˙1 We observe now that this formula applies also for generalˇ. Every diagram Q D can be turned into one of the type in Figure 3 by smoothing out (see below (7)) crossings between the two strands different from S (unless S is an isolated left or right strand). This smoothing procedure does not alter any of the quantities in (66) (and S will be a split component in Q D after this transformation if and only if it is so before it).
In certain cases below we will be able to see that Q D is not a minimal genus diagram of Q M , and then With c.D/ D c. Q D/, we obtain from (63) and (65):

The defect
We are thus led to consider (68). The idea we follow now is that for sufficiently long wordsˇ, the l.h.s. of (68) will become large, and thus (68) cannot hold. Let us here fix some more terminology on extensions. We call a syllable kCn i an n-extension of k i (where k; n > 0). We use the same name also for an iterated n-extension. A 1-extension is simply an extension. A word w 0 is an extension of another word w, if w 0 is obtained by (possibly repeatedly) extending syllables in w. A syllable k 1 or k 2 is inner if it involves the to-reverse strand S , otherwise it is outer. A syllable k 3 is inner if S enters and exits as the first or third strand, otherwise (second strand) it is outer. An extension or n-extension is inner/outer if it is performed at an inner/outer syllable. Now we look at the form (62). We can writě where R 0 D Q k iD1 R 0 i and R 0 i is an extension of OE1; 2; 3, and L 0 D Q k 0 iD1 L 0 i with L 0 i extensions of OE 3; 2; 1. Moreover,˛, Q and are either empty, or extensions of OE1.; 2/, OE. 2/; 1 and OE 3.; 2/ resp. (with the parenthesized syllable present or not). We have a freedom to cyclically permute indices in L and R, which we use here to fix that R starts with 1 . This means that ı is empty. (We will nevertheless see later why we need to care about a ı standing at that place in (69).) We return attention to (68). Let us call the l.h.s. of (68) the defect d.ˇ/ of a braid wordˇ. We assume this definition made for every word w written in 1;2;3 , and not necessarily of the form (69): d.w/ WD #fcrossings of S in wg #fflips in wg 2#f˙3-syllables in wg : We will now analyze the defect ofˇusing the structure (69). In doing so, we observe that it is often enough to look at words reduced under inner 2-extension and outer extension. (This means, all exponents of outer syllables should be 1, and for inner syllables 1 or 2.) Outer extension does not change the defect, and so with every word satisfying (68) all its outer extensions do as well. (That the minimal genus property is not changed can be argued by results of Gabai [12,13] we will invoke more substantially later.) Inner 2-extension augments the defect by two. It may change the r.h.s. of (68) if the genus non-minimizing property is used, but this change occurs only once (if inner 2-extension is applied iteratedly).
When we expand 3 as in (64), then the defect ofˇcan be calculated by those of the parts on the right of (69), with R 0 resp. L 0 further splitted into R 0 i resp. L 0 i . This is because no flips can occur where the words are composed within R and L. Flips can, though, occur at the two places R and L 1 are (cyclically) composed, and we call these at most two flips extra flips.
With this explanation, (68) becomes now We next look at the individual defects that occur in (71). Proof. Since the defect increases by two under an inner 2-extension, it is enough to look first at words reduced under inner 2-extension and outer extension. We try out the 8 words in which the exponents of OE1; 2; 3 are 1 or 2.
We separate the R 0 i and L 0 i into three types. type 0: all the words of positive defect, type 1: a word of zero defect in which the strand S enters/exits as second or third strand (and has 3 crossings), type 2: a word of zero defect in which the strand S enters and exits as first strand (and has 4 crossings).
The R 0 i of non-zero type look, without outer extension, like this: type 1 type 1 type 2 (The corresponding L 0 i are exactly the mirror images thereof.) Note that the two type 1 words have (one or two) syllables that admit outer extension: in the first type 1 word, the 1and the 3-syllable is extendable, in the second one the 2-syllable. However, a type 2 word admits no extension (since S is involved in all syllables).
This means now that the only way in which one can make the number of syllables ofˇindefinitely large is by introducing R 0 i and L 0 i of type 1 and 2. We will now describe ways to control such words. Type 1 words. We first consider type 1 words. Let us note that if R 0 i 1 and R 0 iC1 are type 1, then R 0 i is not type 2. That is, a sequence of consecutive R 0 i of type 1 is terminated on either side by a type 0 word (or the start/end of R 0 ). A similar remark applies for R 0 i of type 2, and for L 0 i . In order to limit the consecutive R 0 i (and L 0 i ) of type 1, we use the observation shown on the left. If we have two consecutive R 0 i of type 1, such that the second is entered by S as the second strand, we can modify S within these two words (as shown by the dashed lines) so as to avoid a pair of flips. That is, two such R 0 i of type 1 can be counted to have a total defect of 2. We call this procedure type 1 cancellation. In particular, this means that a defect will always occur when at least 3 consecutive R 0 i are of type 1. This gives a way to bound the number of type 1 words. Type 1 words allow us also to detect the non-minimality of the genus of Q D. We call this argument below shortly type 1 reduction. It is easily observed that when S is reverted in a type 1 word, then Q D has a pair of crossings of opposite writhe which connect the same two Seifert circles (the 'global' Seifert circle in Figure 3 and one of the small ones). Then by the work of Gabai [12,13], Q D's canonical surface is not of minimal genus. This is because the two opposite crossings enter into a (special) Murasugi summand of Q D, and the canonical surface of this summand in compressible. However, this argument does not work when R 0 i is involved in a type 1 cancellation. Moreover, the argument may fail if an inner 2-extension is applied.
Type 2 words. We next consider type 2 words. These are equal to (and not extensions of) OE1; 2; 3 and OE 3; 2; 1, entered (and exited) by S being the first strand. As explained above, they are not intermixed by type 1 words. Now we observe that OE1; 2; 3 D OE1; 2 3 OE 2 3 , with the first factor being central in B 3 . Similarly OE 3; 2; 1 D OE1; 2 3 OE2 3 . Consequently, OE1; 2; 3 and OE 3; 2; 1 of type 2 can be cancelled against each other, at the cost of replacing them by powers of ˙3 2 . If we allow this, the only way in whichˇcan have arbitrary many syllables remains when R has many more type 2 words than L (or vice versa). This, however, is controlled from (61), since the number of type 0 and type 1 words (and their contributions to lk.O/) is already bounded.
The type 2 cancellation we just described, of course, requires us to modify the presentation (69). We will discuss this in more detail below. Let us notice, however, that it preserved the word length (in 1;2;3 ), and so we still have a minimal genus surface spanned by the new word. Moreover, (71) still holds: we replaced a subword of zero defect with another one, and there are still no flips created within R or L 1 . (Note that where ˙3 2 is inserted, S is always the first strand.) After type 2 cancellation, we have now gained control over the syllable length ofˇ. We will derive an explicit (but crude) bound below, on the way of obtaining the list of relevantˇ.

A syllable length bound
We now derive a bound on the syllable length ofˇafter type 2 cancellation.
With (71), and at most 2 extra flips, we have With (72) we have and 4 d.
We also get this way Now let Q R resp. Q L be R 0 resp. L 0 with all type 2 words R 0 i resp. L 0 i removed. With what we said about type 1 words R 0 i and L 0 i , we have that at least every third of R 0 i and L 0 i in Q R resp. Q L contributes to a defect. Thus with (76), Q R resp. Q L have at most 17 R 0 i resp. L 0 i , and together at most 19 . : We notice that e lk is non-negative for every letter in R and non-positive for every letter in L 1 . Then an easy estimate shows that 2 e Note that this quantity is not affected by the syllables k 2 introduced by type 2 cancellation. Next, we know that after type 2 cancellation only one of R 0 and L 0 contains type 2 words R 0 i or L 0 i . Since for such words we have e lk D˙2, this means that after type 2 cancellation, there are at most 15 type 2 words R 0 i or L 0 i . Combining with (77) we find: R 0 and L 0 have at most 34 R 0 i and L 0 i altogether, i.e. k C k 0 Ä 34.
We still have to take care of the new syllables, extensions of ˙3 2 , obtained by type 2 cancellation. Such syllables can occur only between R 0 i , or at the start or end of R 0 , and similarly for L 0 i and L 0 . Thus there are at most kCk 0 C2 Ä 36 such syllables. Then with (80) we obtain that the syllable lengths satisfy and with sl.˛/; sl. Q /; sl. / Ä 2, we finally get This bound is of course rather rough. It is clear that only a fraction of the words with at most so many syllables are relevant, but it is also evident that still these are too many to be obtainable by a manual enumeration. We thus felt compelled to write a computer program in that realm, and the above bound became central in allocating data resources for the underlying implementation.

The computer enumeration
We will now construct the list of patterns. A pattern is a family of 3-braid words (which potentially give under closure a partially invertible link), obtained from one fixed word w by outer extensions. We often call for simplicity the minimal word w itself the pattern.
In our algorithm for building the patterns we fixed first S by its number at the start of R. All three choices must be tested, and the procedure below depends essentially on each choice.
We have now to modify the meaning of the presentation of (69) to account for the syllables ˙3 2 introduced after type 2 cancellation.
To this vein, we introduce a new generator OE4 D 4 D 3 2 ; OE 4 D 4 D 1 4 D 3 2 ; and regard R and L as words with indices now cyclically non-decreasing in 1; 2; 3; 4. The new˙4-syllables (S1) should occur only if S is the first strand, and (S2) their total exponents must add up to 0 .
Note that cannot be empty or a single 4-syllable, because in this case L 1 would end on 1 prior to type 2 cancellation. This contradicts the cyclical reducedness property under our assumption that R starts with 1.
To facilitate work, we will start building the patterns reduced under inner 2-extensions. We fix also that we always cancel the innermost type 2 words in R with the outermost type 2 words in L 1 . Thus we can assume that (U1) no 4-syllable occurs after a type 2 word OE1; 2; 3 in R, (U2) no 4-syllable occurs before a type 2 word OE 3; 2; 1 in L 1 , (U3) a 4-syllable occurs in R if and only if a 4-syllable occurs in L 1 , (U4) type 2 words do not occur in both R and L 1 .
Property (U3) is what condition (S2) above blows down to, when we regard˙4-syllables as extendable. Property (U4) just means that we cancel the maximum of type 2 words. Properties (U1), (U2) and (U4) must in fact hold after inner 2-extension, and inner 2-extension of a type 2 word is no longer type 2 (a fact we had first overlooked). This can still be accounted for dynamically, while the words R and L are generated, by counting how many type 2 words must forcibly be subjected to inner 2-extension in order not to violate properties (U1), (U2) and (U4).
With this we have the following options for exponents: -exponent of˙1,˙2 and˙3 should be 1 if syllable is outer, and 1 and 2 if syllable is inner -exponent of˙4 should be zero if syllable is inner, or preceded by type 2 word OE1; 2; 3 in R resp. followed by a type 2 word OE 3; 2; 1 in L 1 . The exponent should not be zero for the 4-syllable in ı. (Thus ı can be non-empty only if S D 1.) With this choice of exponent, we build the patterns by choosing first˛, Q , , ı, and then recursively expanding R and L until the defect becomes too large (note that each new R 0 i and L 0 i can only increase the defect). We take care of type 1 reduction and type 1 cancellation (and that we apply them on disjoint words), and that not both R 0 and L 0 contain type 2 words.
In order to obtain an inner 2-extension with (61) from some pattern w, we have to make sure that we can create at least ( 2j e lk.w/j if type 1 cancellation was not used max.j2 e lk.w/j 2; 0/ if type 1 cancellation was used ) more defect. This is the minimal additional defect produced by the 2-extensions needed to establish (61). The reason for treating extra the type 1 cancellation was already mentioned: type 1 cancellation may become invalid after one inner 2-extension. Then the defect this 2-extension creates is compensated by the vanishing reduction on the bound on the right of (71). (As we report in subsection 4.4.2, we later removed, for verification purposes, part of the type 1 cancellation test. This still led to the same result, but after considerably longer calculation.) Once we determined how much more defect can be created by inner 2-extensions (in fact, at most one single extension turned out possible), we created the words under this 2-extension which satisfy (61). Leaving all outer syllables extendable, we obtain then a list of patterns, which potentially satisfy (71). (We say 'potentially', because type 1 cancellation is not a thorough test for the genus non-minimality of Q D. Moreover, in case of genus nonminimality we used in (67) and (68) only . Q M / . Q D/ 2, while the difference could be larger.) The further details of our algorithm are rather technical and are skipped here. Let us summarize the outcome: after 3 days of programming work and about 600 lines of C++ code, the program needed about 5 minutes to render a list of 398 patterns which require further attention.
In this way the extendable Artin generator i behaves very similarly to the element i of the singular 3-braid monoid (see [3] or [7, Proposition 2.1]), but satisfies the additional gobbling/yielding relations # 54: This completes the proof of Theorem 1.2.
We have displayed Table 1 also to illustrate how many different representations (13) the braids in Figure 1 can have. Still the simplicity of the figure does not suggest that such a lengthy (and partly computerized) case treatment is a natural approach. For better or for worse, we have yet nothing else to offer.