Positive braid knots of maximal topological 4-genus

# Positive braid knots of maximal topological 4-genus

Livio Liechti Mathematisches Institut, Universität Bern, Sidlerstrasse 5, 3012 Bern, Schweiz
###### Abstract.

We show that a positive braid knot has maximal topological 4-genus exactly if it has maximal signature invariant. As an application, we determine all positive braid knots with maximal topological 4-genus and compute the topological 4-genus for all positive braid knots with up to 12 crossings.

The author is supported by the Swiss National Science Foundation ()

## 1. Introduction

The slice genus of a torus knot equals the ordinary genus by a theorem of Kronheimer and Mrowka [9]. By work of Rudolph, this equality extends to the more general class of links bounding quasipositive surfaces, in particular to positive braid knots [11]. However, the story is very different for the topological 4-genus , i.e. the minimal genus among surfaces which are properly, locally flatly embedded in the 4-ball and have a given knot as boundary (in contrast to the slice genus, where the embedding is required to be smooth). A first example is due to Rudolph [10]: for the torus knot , the inequality holds. More recently, a large proportional difference with respect to was found for all torus knots with non-maximal signature in [3]. On the other hand, there exists a lower bound, due to Kauffman and Taylor, for the topological 4-genus of knots: holds for any knot  [8]. We show that for positive braid knots, this bound is in fact the only obstruction to non-maximal topological 4-genus, i.e. .

###### Theorem 1.

For a positive braid knot , the equality holds exactly if .

Combining this result with Baader’s classification of prime positive braid links of maximal signature [1], we immediately get a full description of all prime positive braid knots of maximal topological 4-genus: they are exactly the torus knots of maximal signature.

###### Corollary 2.

The torus knots , and are the only prime positive braid knots with .

Our proof of Theorem 1 uses two main ingredients. The first one is a homological criterion from [3] using Freedman’s disc theorem [7], allowing us to conclude for certain positive braids. The second one is that genus defect is inherited from surface minors, i.e. incompressible subsurfaces. Similar to Baader’s four surface minors , , and obstructing maximal signature for positive braid links, we use enriched versions , , and to obstruct maximal topological 4-genus for positive braid knots.

Theorem 1 also allows us to compute the topological 4-genus for positive braid knots with . Combining the lower bound of Kauffman and Taylor with yields the exact result . This suffices to compute the topological 4-genus for prime positive braids knots with up to 12 crossings. Table 1 lists all these knots, except for the torus knots , and , which have maximal topological 4-genus.

This list is created with the help of the software Knotinfo [5]. Previously, the values of the topological 4-genus for all these examples except were marked as unknown. However, these values could also be deduced from work of Borodzik and Friedl on the algebraic unknotting number [4].

Acknowledgements. I warmly thank Sebastian Baader, Peter Feller and Lukas Lewark for many inspiring discussions and ideas that found their way into this article. Furthermore, I thank the referee for corrections and suggestions.

## 2. Positive braids and trees

A positive braid knot is a knot that can be obtained from a positive braid via the closure operation, an important example being torus knots. A positive braid on strands is a finite word in positive powers of the braid generators . By a theorem of Stallings, positive braid knots are fibred with the standard Seifert surface as fibre [12]. As Baader did in [1], we use brick diagrams to visualise the fibre surface of positive braid knots: each horizontal bar corresponds to a braid generator and each brick, i.e. each rectangle, corresponds to a positive Hopf band in the plumbing construction of the fibre surface. If two bricks link, it means that the core curves of the corresponding positive Hopf bands intersect once, see Figure 1. Let the linking pattern be the the plane graph obtained by putting a vertex into every brick and an edge between two vertices exactly if the corresponding bricks link. It can be easily seen that if the intersection pattern of a positive braid is not connected, then the positive braid link is not prime. In fact, the converse is also true since positive braids are visually prime by a theorem of Cromwell [6].

### 2.1. Trees

Let us for a moment consider the case where the linking pattern is a tree. There are many brick diagrams that yield the same tree as linking pattern. Since closures of positive braids corresponding to different brick diagrams might be equivalent as links in , it is natural to ask whether the plane tree of the linking pattern uniquely determines the positive braid link up to ambient isotopy. As we will see in the following remark, this is indeed the case.

###### Remark 3.

The fibre surface of a positive braid retracts to its brick diagram. Since for a successive plumbing of positive Hopf bands, the monodromy is conjugate to the product (in the succession of plumbing) of positive Dehn twists along the core curves of the Hopf bands [12], the conjugacy class of the monodromy is completely determined by the plane tree given by the linking pattern of the brick diagram. Therefore, also the corresponding fibred link is determined by the linking pattern of the brick diagram. Indeed, the monodromy determines the mapping torus (up to homeomorphism fixing the boundary pointwise) and the fibredness condition dictates how to glue solid tori along the boundary of the mapping torus to obtain containing a copy of the link .

Furthermore, if the linking pattern is a tree, a matrix for the Seifert form of the corresponding fibre surface is particularly easy to describe: as a basis of take the core curves of the positive Hopf bands corresponding to the bricks. A matrix for the Seifert form is then given by and if and the curves and intersect (i.e. if the corresponding vertices of the linking pattern are connected by an edge). All other entries are equal to zero.

###### Example 4.

Let , , and be the canonical fibre surfaces

 ˜T =Σ(σ51σ2σ41σ2), ˜E =Σ(σ71σ2σ31σ2), ˜X =Σ(σ21σ22σ1σ3σ22σ3), ˜Y =Σ(σ41σ22σ31σ2),

see Figure 2 for the corresponding brick diagrams and the linking patterns. By exhibiting a two-dimensional subspace of which is Alexander-trivial, i.e.  is a unit for some matrix of the Seifert form, it is shown in [3] that the three-component link does not have maximal topological 4-genus.

More precisely, it is shown that the topological 4-genus equals one while the ordinary genus equals two. In this example, we carry out the same computation for , and . For reasons of self-containedness, we also repeat the computation for . Number the vertices of the linking patterns in Figure 2 from top to bottom (and from left to right if several vertices are on the same level, as indicated for in Figure 2). As a basis for the first homology, take the core curves of the corresponding Hopf bands with the chosen numbering. In this basis, consider the subspaces

 B˜T = ⟨(−1,2,−3,4,−2,−3,2,−1,1)⊤, e8⟩, B˜E = ⟨(2,−4,6,−3,−5,4,−3,2,−1,1)⊤, e9⟩, B˜X = ⟨(−1,−1,2,−1,−1,0)⊤, e6⟩, B˜Y = ⟨(1,−2,3,−2,1,−2,1,−1)⊤, e7⟩

of , , and , respectively. Using the matrix of the Seifert form described above, it is a straightforward computation to see that in all four cases, the given subspaces are Alexander-trivial. Writing for the first basis vector of and for the Seifert form corresponding to , one obtains

 v⊤A˜Tv=0,v⊤A˜Te8=1, e⊤8A˜Tv=0,e⊤8A˜Te8=1,

or, equivalently,

 A˜T|B˜T×B˜T=(0101).

Consequently, , a unit in . The computation for the other cases works analogously. Proposition 3 in [3] now implies non-maximality of the topological 4-genus. Since the signature does not allow for a genus defect greater than one, we conclude

 g4(∂˜T) =g(∂˜T)−1=3, g4(∂˜E) =g(∂˜E)−1=4, g4(∂˜X) =g(∂˜X)−1=1, g4(∂˜Y) =g(∂˜Y)−1=3.

In order to detect genus defect for a positive braid knot , we search for minors or in the fibre surface . This is always based on the fact that the linking pattern of contains the tree corresponding to or via deleting vertices and contracting edges. One can then see that also contains or , respectively, as a surface minor, implying . For example, Figure 3 shows how the tree corresponding to is contained in the linking pattern of the positive braid .

Before we prove it for the case of positive braid knots, we show an analogue of Theorem 1 for knots obtained as plumbing of positive Hopf bands along a tree . This notion generalises knots corresponding to brick diagrams having some plane tree as linking pattern. Starting from any finite plane tree , we plumb positive Hopf bands (which are in one-to-one correspondence with the vertices of the tree) such that their core curves intersect once exactly if the corresponding vertices of are connected by an edge. Otherwise, they do not intersect. Furthermore, they respect the circular ordering of the vertices given by the plane tree structure of . By the argument given in Remark 3, there is, up to ambient isotopy, only one way to do this. This construction is strictly more general than positive braid knots with a plane tree as linking pattern: vertices of a tree can have arbitrary valency, while for linking patterns associated with positive braid knots, this valency is bounded from above by .

###### Proposition 5.

For a knot obtained by plumbing positive Hopf bands along a plane tree , the equality holds exactly if .

###### Proof.

If , then follows from the signature bound of Kauffman and Taylor [8]. If , we distinguish three different cases. If has at least three vertices of degree at least three, then the corresponding fibre surface contains as a minor and thus . If has two vertices of degree at least three, then at least one of the leaves has distance at least two from the closest vertex of degree at least three, since otherwise cannot be a knot. Again the corresponding fibre surface contains as a minor, since contains the tree corresponding to via deleting vertices and contracting edges. If has only one vertex of degree at least three, then holds if and only if contains the linking pattern of or as an induced subgraph. This can be calculated directly from the associated Seifert forms. Alternatively, it also follows from Baader’s classification of positive braid links of maximal signature [1]. Again, for to be a knot, cannot be equal to or . It follows that in fact contains the linking pattern of or as an induced subgraph. Hence, the corresponding fibre surface contains or as a surface minor. ∎

## 3. Proof of Theorem 1

The proof of Theorem 1 for positive braid knots is divided into two parts, depending on the positive braid index of , i.e. the minimal index of a positive braid representing the knot . For of positive index at most three, we can essentially reduce the problem to Proposition 5. For of positive index at least four, we show that the strict inequality always holds.

###### Proposition 6.

For a knot obtained as the closure of a positive -braid , the equality holds exactly if .

###### Proof.

We assume to have applied all possible braid relations to the braid , so, up to cyclic permutation, can be assumed to be of the form , where and . If , the linking pattern of the braid is a plane tree and we are done by Proposition 5. We now show that in the other cases we already have . For this, let and remark that at least one of the has to be odd and hence at least three, otherwise the permutation given by the braid leaves the third strand invariant and is not a knot.

Case 1: . Up to cyclic permutation, the braid contains the word as a subword, i.e. via reducing powers of occurrences of generators, and thus the fibre surface of contains as a minor, implying .

Case 2: , . If, up to cyclic permutation, equals , the Seifert form of is positive definite. If equals or , the second strand is left invariant by the permutation given by the braid, so we can assume that one of the is at least four. Furthermore, since one of the has to be odd, can be assumed to be at least or with respect to the product order. In both cases, contains the word as a subword and thus the fibre surface of contains as a minor, implying .

Case 3: , at least one . As before, one of the has to be at least three, say . If or is at least two, then contains, up to cyclic permutation, as a subword and thus the fibre surface of contains as a minor. Now assume and . We also assume , otherwise we are, up to cyclic permutation, in the case we already dealt with. Note that the permutation given by a braid of the form leaves the second strand invariant, so needs to be at least three in order for to be a knot. Now, up to cyclic permutation, must contain the word and the fibre surface of contains as a minor, implying . ∎

###### Lemma 7.

Let be a positive braid of index . If for some the linking pattern of the subword of induced by the generators and is a path, then is not of minimal positive index.

###### Proof.

We can assume the subword of induced by the generators and to be , for some positive numbers and . This can be achieved by cyclic permutation and possibly reversing the order of the word , operations that do not change the associated fibre surface. Similarly, we can assume that all occurrences of generators with index smaller than come before the last occurrence of and, likewise, all occurrences of generators with index greater than come after the first occurrence of . The situation is schematically depicted in Figure 4 on the left.

Now consider the brick diagram obtained by merging the two columns corresponding to the generators and as indicated in Figure 4 on the right. By definition, the corresponding positive braid has fewer strands than . To show that the closures of and are ambient isotopic in , we study the corresponding monodromy homeomorphism of their fibre surfaces and . Since the linking patterns of the two brick diagrams are equal, the corresponding monodromies are conjugate and the closures of and are ambient isotopic by the argument used in Remark 3. ∎

###### Proposition 8.

If is a prime knot obtained as the closure of a positive braid of minimal positive index , then .

###### Proof.

Let be a positive braid of minimal braid index whose closure is a prime knot. We assume to have applied all possible braid relations to . This process terminates: it increases the sum of all indices of generators (counted with multiplicity) while not changing the number of generators. In other words, the crossings of are as far to the right as possible. We can furthermore assume that still contains, up to cyclic permutation, the subword , since otherwise would not be of minimal index.

We first delete, without disconnecting the linking pattern, a minimal amount of occurrences of so that the induced subword of in the first two generators is, after a possible cyclic permutation, of the form , where and are greater or equal to two. For example, if the induced subword of in the first two generators is , we delete one occurrence of (to the power two), yielding, after a possible cyclic permutation, . Note that in case , no generators have to be deleted to achieve the desired form.

Case 1: . In this case, we did not have to delete any occurrence of and the induced subword of in the first two generators is exactly . Both occurrences of have to be split by an occurrence of , since otherwise the permutation given by would leave the first or second strand invariant, see Figure 5.

Furthermore, these occurrences have to be to the power at least two, since we ruled out the possibility of a braid relation . If one of the occurrences of is to some power at least three, contains, up to cyclic permutation, the subword and thus the fibre surface of contains the minor , implying . If the power of both occurrences of is equal to two, we repeat the same argument: both occurrences of have to be split by an occurrence of , otherwise the permutation given by would leave the first or second strand invariant. As before, we distinguish cases depending on the powers of the occurrences of . We repeat this argument and case distinction with increasing index as long as necessary. Eventually, some splitting occurrence has to be of power at least three and contains, up to cyclic permutation, the subword .

Case 2: . In this case, we did not have to delete any occurrence of and the induced subword of in the first two generators is exactly . As in Case 1, the second occurrence of has to be split by an occurrence of (otherwise the permutation given by would leave the second strand invariant), so must contain a subword of the form . Note that must be greater or equal to two, since we applied all possible braid relations .

Figure 6 depicts the brick diagram and intersecting pattern of this subword for and . Since the intersection pattern is not connected, there has to be another occurrence of in , otherwise the closure would not be prime. What are the possibilities for the other occurrences of ? If the first occurrence of is split by an occurrence of , again the occurrence of has to be to the power at least two. Hence, contains, up to reversing order and cyclic permutation, the subword and the fibre surface of contains the minor , implying . Similarly, if contains, up to reversing order and cyclic permutation, the subword , again the fibre surface of contains the minor , implying . If we exclude these cases, the only two possibilities for the induced subword of in the first three generators are and , which are, up to cyclic permutation, reverse to each other. If is greater or equal to two, the fibre surface of again contains the minor , implying , so we assume the induced subword of in the first three generators to be, up to reversing order and cyclic permutation, . But in this case, restricted to the second and third generator has a path as linking pattern and is not minimal by Lemma 7.

Case 3: . The only possibility not considered in Case 2 is the following: contains, up to cyclic permutation, the subword , thus also and the fibre surface of contains as a minor, implying . However, when reconsidering our discussion of Case 2, the powers of appearing could be greater, so we get as possibilities for the induced subword of in the first three generators, where . Again, note that if or is greater or equal to two, the fibre surface of contains the minor , implying , so we assume the induced subword of in the first three generators to be, up to reversing order and cyclic permutation, . Again, restricted to the second and third generator has a path as linking pattern and is not minimal.

Case 4: . We can apply the same arguments as in the cases above. From this it follows that if the fibre surface of contains no minor , then the induced subword in the first three generators is, after the described process of deleting some generators , either or . As before, these two words are, up to cyclic permutation, reverse to each other. But since we might have deleted some generators to obtain them, we should consider them separately. Again as before, if or is greater or equal to two, the fibre surface of contains the minor , implying . If we restrict to the second and third generator, the linking pattern is a path. Note that reinserting the deleted generators would split or . In any case, the linking pattern of restricted to the second and third generator is still a path and is not minimal. This does not necessarily hold for the other possibility . However, note that if is greater or equal to two, then contains the subword and the fibre surface of contains the minor , implying . So we are left with the possibility . If all the deleted occurrences of appeared before the first occurrence of in , after a cyclic permutation the linking pattern of restricted to the second and third generator again is a path and is not minimal. If some deleted occurrence of appeared after the first occurrence of in , then contains the word as a subword and, as before, .

Case 5: . In this case, there is one last new possibility: as in Case 1, the word could be a subword of (without directly yielding as a subword). Again, since we applied all possible braid relations , and are greater or equal to two. If should, up to cyclic permutation, neither contain nor as subword, then and are both equal to two and the induced subword of in the first two generators is exactly . In particular, we again did not have to delete any occurrence of in the deletion process described above. If the induced subword of in the first three generators was , the permutation given by would leave the third strand invariant and would not be a knot. Thus, there has to be at least one more occurrence of a generator . This gives the last two possibilities of induced subwords of in the first three generators: and , which are, up to cyclic permutation, reverse to each other. If is of index four, then actually would have to equal . But the closure can never be a knot, since the last two strands get permuted among themselves independently of and . Now let be of index at least five. If is the induced subword of in the first three generators, one of the occurrences of has to be separated by an occurrence of to the power at least two (recall that we ruled out the possibilty of a braid relation ), since otherwise the first two strands would get permuted among themselves by and would not be a knot. One can then see that contains, up to reversing order and cyclic permutation, one of the subwords or , each of which guarantees the existence of a minor in the fibre surface of , implying . ∎

## 4. Characterisation by forbidden minors

In the previous section, we established a characterisation of positive braid knots with maximal topological 4-genus by the forbidden surface minors and . More precisely, we used all minors in the case of braid index 3, while we only needed the minor in the case of minimal positive braid index . Another way to think of this is that for positive braid knots, genus defect is characterised by these four forbidden minors. A natural question to ask is whether a similar result holds for larger genus defect.

###### Question 9.

For positive braid knots, can genus defect be characterised by finitely many forbidden surface minors for any ?

As noted by Baader and Dehornoy [2], this is implied by Higman’s Lemma if we restrict ourselves to positive braids of index bounded by some natural number . A possible way of dealing with Question 9 in the unbounded case could be to give a positive answer to the following question, which can be thought of as a strengthening of Proposition 8.

###### Question 10.

For positive braid knots, is bounded from below by an increasing affine function of the positive braid index?

## References

• [1] S. Baader: Positive braids of maximal signature, Enseign. Math. 59 (2013), no. 3–4, 351–358.
• [2] S. Baader, P. Dehornoy: Minor theory for surfaces and divides of maximal signature, http://arxiv.org/abs/1211.7348.
• [3] S. Baader, P. Feller, L. Lewark, L. Liechti: On the topological 4–genus of torus knots, http://arxiv.org/abs/1509.07634.
• [4] M. Borodzik, S. Friedl: The unknotting number and classical invariants, I, Algebr. Geom. Topol. 15 (2015), no. 1, 85–135.
• [5] J. C. Cha, C. Livingston: KnotInfo: Table of Knot Invariants, http://www.indiana.edu/~knotinfo, 12.11.2015.
• [6] P. R. Cromwell: Positive braids are visually prime, Proc. London Math. Soc. (3) 67 (1993), no. 2, 384–424.
• [7] M. H. Freedman: The topology of four-dimensional manifolds, J. Differential Geom. 17 (1982), no. 3, 357–453.
• [8] L. H. Kauffman, L. R. Taylor: Signature of links, Trans. Amer. Math. Soc. 216 (1976), 351–365.
• [9] P. B. Kronheimer, T. S. Mrowka: The genus of embedded surfaces in the projective plane, Math. Res. Lett. 1 (1994), no. 6, 797–808.
• [10] L. Rudolph: Some topologically locally–flat surfaces in the complex projective plane, Comment. Math. Helv. 59 (1984), no. 4, 592–599.
• [11] L. Rudolph: Quasipositivity as an obstruction to sliceness, Bull. Amer. Math. Soc. (N.S.) 29 (1993), no. 1, 51–59.
• [12] J. Stallings: Constructions of fibred knots and links, Algebraic and Geometric Topology, 55–60, Proc. Sympos. Pure Math. 32 (1978) Amer. Math. Soc., Providence, R.I.
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