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On Positive Braids, Monodromy Groups and Framings

We associate to every positive braid a braid monodromy group, generalizing the geometric monodromy group of an isolated plane curve singularity. If the closure of the braid is a knot, we identify the corresponding group with a framed mapping class group. In particular, this gives a well defined knot invariant. As an application, we obtain that the geometric monodromy group of an irreducible singularity is determined by the genus and the Arf invariant of the associated knot.


Introduction
Singularity theory is a genuine source of examples and inspiration for knot theory. Since the topological type of an isolated plane curve singularity is determined by an associated link, it is possible to understand properties of the singularity from a knot theoretical viewpoint, and knot theory has been successfully applied to solve algebraic questions. In another direction, links of singularities form an interesting class of links, with special properties and invariants that follow from the whole machinery of singularity theory. It is often unclear which of those properties are inherently algebraic and which ones could be generalized to wider classes of knots and links. Among other invariants, the fundamental group of the discriminant complement and the geometric monodromy group have drawn much attention but have proved to be hard to investigate.
In [5], Baader and Lönne associate to any positive braid an abstract group defined by generators and relations, which they call geometric braid group. Their motivation comes from the similarities between the combinatorial structure of positive braids and that of isolated plane curve singularities. In particular, they prove that for braids of type ADE and for braids whose closure is a torus link T p,q the geometric braid group is isomorphic to the fundamental group of the discriminant complement of the corresponding singularities (simple singularities in the former case, Brieskorn-Pham singularities f (x, y) = x p + y q in the latter). However, because of difficulties in dealing with conjugation in the positive braid monoid, they can prove that the geometric monodromy group is a well defined link invariant only for positive braids whose closure contains a positive half twist.
Inspired by their construction and in analogy with the definition of the geometric monodromy group of a singularity, in this article we associate to any positive braid β a group MG(β), which we call braid monodromy group, defined as a subgroup of the mapping class group of the unique genus minimizing Seifert surface of the closureβ, generated by the Dehn twists around some natural family of curves. This braid monodromy group is a quotient of Baader's and Lönne's geometric braid group which contains the monodromy diffeomorphism of the positive braid. Moreover, it is a generalization of the geometric monodromy group of an isolated plane curve singularity to the setting of positive braids.
Theorem 1. Let f : C 2 → C define an isolated plane curve singularity and L(f ) be the link of f . Then there exists a positive braid β representing L(f ) such that the geometric monodromy group of f is equal to MG(β).
In [20], Cuadrado and Salter proved that the geometric monodromy group of any singularity of genus at least 5 and not of type A n and D n is a framed mapping class group, i.e. the stabilizer of some canonical framing on the Milnor fibre associated to the singularity and, among other things, they use this result for deducing the non-injectivity of the geometric monodromy representation.
Following their approach, the main result of this paper is an identification of the monodromy group of a positive braid β whose closure is a knot with a framed mapping class group on the genus minimizing surface Σ β . In particular, it follows from the general theory of framed mapping class groups that, at least for knots of big enough genus, the braid monodromy group is a well defined knot invariant, which is actually determined by the genus and the Arf invariant of the knot, see Section 4.
Theorem 2. Let β be a prime positive braid not of type A n and whose closure is a knot. Then, for all but finitely many such braids, there exists a framing φ β on Σ β such that the monodromy group MG(β) is equal to the framed mapping class group Mod(Σ β , φ β ).
Of course, as a consequence of Theorem 1 and 2 we immediately obtain that the geometric monodromy group of an irreducible singularity is controlled by a framing, thus partially recovering the results of [20]. Working in the more general context of positive braids forces us to restrict to irreducible singularities, because of the increased difficulty of the underlying combinatorics. In particular, there are some infinite families of positive braid links for which our methods do not seem to work, see Remark 5.1. However this also has some advantages, as we will now explain.
Since the topological type of a singularity is completely determined by its link, a priori every topological invariant of a singularity could be expressed in terms of link invariants. For instance, the Milnor number corresponds to the minimal first Betti number, while the multiplicity corresponds to the braid index [24]. However, this translation is often far from straightforward. Notice that, although not explicitly stated therein, a consequence of the results of Cuadrado and Salter in [20] is that for irreducible singularities the geometric monodromy group is determined by the genus of the Milnor fibre and some Arf invariant associated to the framing. Working with positive braids allows us for explicit calculations that lead to the following corollary, expressing the geometric monodromy group entirely in terms of knot invariants. More precisely, we are able to identify the Arf invariant associated to the framing with the Arf invariant of the knot.
Corollary 3. Let f define an irreducible isolated plane curve singularity that is not of type A n and K(f ) be the knot of the singularity. For all but finitely many such singularities, the geometric monodromy group of f is determined by the genus and the Arf invariant of K(f ).
From a purely knot theoretical viewpoint, Theorem 2 might seem disappointing. It implies that the braid monodromy group is an invariant, but a rather useless one: it is hard to compute, but determined by two classical and much easier invariants, a natural number and a mod 2 class. Its interest lies in negative results such as Corollary 3. The geometric monodromy group, which was typically considered a rich yet hard to investigate invariant of a plane curve singularity, turns out (in the case of irreducible singularities) to be determined by two simple knot invariants, and the question whether two irreducible singularities have the same geometric monodromy group can be answered by a direct and easy computation, using existing formulas for the Arf invariant of a knot. Of course, for each fixed genus there are many different irreducible singularity, so there will be different singularities with the same geometric monodromy group. We believe that for big enough genus both values of the Arf invariant are realized, so that there would be exactly two geometric monodromy groups.
The study of the braid monodromy group has naturally its place in the context of finitely generated subgroups of the mapping class group, and in particular subgroups generated by Dehn twists around a family of curves with prescribed intersection pattern. Those subgroups are interesting by themselves from a mapping class group theoretical viewpoint, but also appear naturally in different contexts, such as singularity theory or in the study of Lefschetz fibrations. The question of what groups can arise in this way is completely solved in the case of two Dehn twists, see for example Chapter 3 of [9], but is in general widely open. In [19] Perron and Vannier, interested in the geometric monodromy group of singularities, proved that if the intersection pattern of the curves is a Dynkin diagram of type A n or D n , the group generated by the Dehn twists is isomorphic to the Artin group of corresponding type, and conjectured this to be true for general graphs. This was later disproved by Labruère [16] and Wajnryb [23], whose results show that the only Artin groups that geometrically embed in the mapping class group are precisely the ones of type A n and D n . The theory of framed mapping class groups seems to suggest that, at least if the intersection pattern is in some sense rich enough, those finitely generated subgroups are controlled by a framing on the surface. Theorem 2 is an example of such a result.
As a final remark, although in this paper we concentrate only on positive braids, they are not the only natural class of links generalizing links of singularities to which one could try to associate a monodromy group. A'Campo's divide links form another such interesting family, see Section 3. More generally, it is known that the Milnor fibre of an isolated plane curve singularity can be constructed by a sequence of positive Hopf plumbings such that the core curves of the Hopf bands coincide with the vanishing cycles of the singularity, the Dehn twist around which generate the geometric monodromy group, see [14]. For a general sequence of positive Hopf plumbings, one could define a monodromy group as the group generated by all the Dehn twists around the core curves of the Hopf bands. We expect that, at least for knots, results similar to Theorem 2 should hold in this more general setting. This is not difficult to see for Hopf plumbings with intersection pattern a tree and whose boundary is a knot of sufficiently big genus.
Structure of the paper: In Section 2 we define the braid monodromy group of a positive braid and prove some basic invariance properties. In Section 3 we recall some basics of singularity theory and, using A'Campo's theory of divides, we prove Theorem 1. In Section 4 we discuss the general theory of framed mapping class groups and construct the framing appearing in Theorem 2. Finally, Section 5 is the technical part of the paper, in which we prove Theorem 2. This basically consists of a lengthy case distinction that allows us to apply general results about framed mapping class groups.
Acknowledgements: I wish to thank Sebastian Baader for suggesting the topic and guiding me through this project. I am also very grateful to Livio Liechti for the several interesting discussions, and in particular for pointing out the connection to framed mapping class groups. Finally, thanks to Nick Salter and Pablo Portilla Cuadrado for their interesting comments.

Positive braids monodromy group
Let B + N be the monoid of positive braids on N strands and β ∈ B + N . We will usually represent such a braid with a brick diagram, a plane graph with N vertical lines connected by horizontal segments corresponding to the crossings. Since all the crossings are positive, one can reconstruct the braid from the brick diagram. It is well known that, if β is nonsplit, its closureβ is a fibred link, whose fibre surface can be constructed by taking a disk for each strand of β and, for each generator σ i in β, gluing a half twisted band between the i-th and (i + 1)-th disks. The brick diagram of β naturally embeds in this surface as a retract. Let us denote this fibre surface by Σ β , and let g be its genus and r the number of boundary components. On Σ β there is a standard family of 2g + r − 1 curves γ i , oriented counterclockwise, which are in one-to-one correspondence with the bricks of the brick diagram of β and form a basis of the first homology of Σ β . See Figure 1 for an example of Σ β with the corresponding curves for β = σ 3 σ 1 σ 2 σ 2 1 σ 3 σ 2 . The intersection pattern of those standard curves can be read off directly from the brick diagram, in the so called linking graph. One just needs to take a vertex for each brick and connect two vertices with and edge if the corresponding bricks are arranged as in the braids Figure 1: The fibre surface of σ 3 σ 1 σ 2 σ 2 1 σ 3 σ 2 or σ i+1 σ i σ i+1 σ i . That is, two vertices are connected with an edge if and only if the corresponding curves intersect each other. Here it is worth mentioning that since positive braid links are visually prime by [8], a positive braid link is prime if and only if the linking graph is connected. In what follows, we will say that a positive braid link is of type A n (resp. D n ) if it isotopic to the closure of the braid σ n+1 1 (resp. σ n−2 1 σ 2 σ 2 1 σ 2 ). Those braids have as linking graph the simply laced Dynkin diagram of type A n or D n . Definition 2.1. Let β be a positive braid. The monodromy group MG(β) is the subgroup of the mapping class group of Σ β generated by all the Dehn twists around the curves γ i , i = 1, · · · , 2g + r − 1, i.e.
Example 2.1. As already mentioned, it follows from [19] that if β = σ n+1 1 then MG(β) is isomorphic to the Artin group of type A n . Similarly, for β = σ n−2 1 σ 2 σ 2 1 σ 2 , MG(β) is isomorphic to the Artin group of type D n From the definition, it is clear that MG(β) is invariant under far-commutativity (i.e. σ i σ j = σ j σ i for |i − j| ≥ 2) and positive Markov move.  Figure 2: The isotopy between Σ βσ i and Σ σ i β Proof. Consider the fibre surfaces of βσ i and σ i β. Those surfaces are isotopic, by sliding the topmost band between the i-th and (i + 1)-th disks along the back of the disks and bringing it in the lowermost position. Note that this isotopy restricts to the identity outside of the i th -column. The surfaces Σ βσ i and Σ σ i β can hence be schematically represented as in Figure 2 , where we drew the i th −column and the light grey boxes on the two sides represent the remaining parts of the surface.
Let us number the standard curves of the i th column as in Figure 2. The isotopy will send each γ i , i = 1, · · · , n − 1 to the correspondingγ i , i = 1, · · · , n − 1 and transform γ n into the red curveγ n+1 . All what we have to prove is then that we can generate the Dehn twists around the curvesγ 1 , · · · ,γ n−1 ,γ n+1 usingγ 1 , · · · ,γ n−1 ,γ n , and vice-versa. But we note thatγ n+1 = Tγ n−1 · · · Tγ 2 Tγ 1 (γ n ), so that for h = Tγ n−1 · · · Tγ 2 Tγ 1 we have Tγ n+1 = hTγ n h −1 and the result is proved.   Proof. Up to elementary conjugation, we can suppose that α = ωσ i σ i+1 σ i and β = ωσ i+1 σ i σ i+1 , where ω is a positive braid on N strands and 1 ≤ i ≤ N − 2. At the level of surfaces Σ α and Σ β the braid relation can be realized by an isotopy as in Figure 3. It is clear that all the standard curves γ i are fixed by this isotopy but the ones (at most two) passing through the slidden band.
1. There is a generator σ i in ω: in this case, there are two curves on Σ α which are modified by the isotopy. Let us call them γ 1 and γ 2 , as in Figure 4. We see that, after the isotopy, γ 1 is transformed into the correspondingγ 1 , while γ 2 becomes T −1 γ 1 (γ 2 ). All the other standard curves are fixed. Therefore, we get that MG(α) MG(β).

2.
There is no σ i in ω: in this case, the only curve modified by the isotopy is γ 1 , which as before is transformed intoγ 1 . Again, we directly have that MG(α) MG(β).
The following corollary now follows directly by an observation of Orevkov about Garside's solution of the conjugacy problem in the braid group, saying that, in the presence of a positive half twist, two conjugate positive braids can be related by a sequence of braid relations and elementary conjugations, see [5], section 6.

Divides and monodromy of singularities
The monodromy group of a positive braid is a generalization of the geometric monodromy group of an isolated plane curve singularity. In this section, we will make this statement more precise.
Let f : C 2 → C define an isolated plane curve singularity. For a suitably small radius r > 0, the ball B 4 r ⊂ C 2 intersects the singular curve C = f −1 (0) transversally, so that the intersection , called the link of the singularity. It is well known that the isotopy type of L(f ) completely determines the topological type of the singularity. Moreover, in [18] Milnor proved that the map f |f | : S 3 \ L(f ) → S 1 is a fibration. Singularity links are therefore fibred links, with fibre a surface Σ(f ) called the Milnor fibre. It turns out that all the singularity links are iterated torus links, and in particular positive braid links. The fibration induces a monodromy diffeomorphism of the fibre, which is only defined up to isotopy and therefore defines a mapping class in Mod(Σ(f )), called the geometric monodromy of the singularity. The geometric monodromy is an important invariant, which determines the topology of the singularity and has been intensively studied in the context of singularity theory.
By the study of the deformations of the singularity, the geometric monodromy can be "promoted" to the so called geometric monodromy group of the singularity. It is a subgroup of Mod(Σ(f )) generated by the Dehn twists around some specific curves on the Milnor fibre, called vanishing cycles. The geometric monodromy can be expressed as a product of those generators and is therefore an element of the geometric monodromy group. We will not discuss the original definition of the geometric monodromy group of a singularity since, although classic, it would require quite some background knowledge in singularity theory and will not be useful for our purposes. However, there exists an easy combinatorial model for the Milnor fiber of a singularity which allows us to directly define the geometric monodromy group in terms of explicit generators. This was constructed by A'Campo using the theory of divides.
Definition 3.1. A divide D is a generic relative immersion of finitely many intervals in the unit disk (D 2 , ∂(D 2 )).
Here, generic means that the only singularities are double points and that the intervals meet the boundary ∂(D 2 ) transversally. Examples of divides can be seen in Figures 5  and 6.
Divides were first introduced by A'Campo ([4] [2]) and Gusein-Zade ( [12], [11]), who independently proved that they could be associated in a natural way to singularities and used them for studying properties of the monodromy. Later on, in [3], [1] A'Campo associated to any divide D a link L(D), constructed as follows. Consider the tangent bundle of the unit disk, The sphere S 3 can be seen as the unit sphere in T D 2 , Now let D ⊂ D 2 be a divide, the link of D is defined as This gives a link whose number of components is equal to the number of intervals in the divide. In the same papers, A'Campo proved that if the divide is connected the link is fibred and that if the divide was obtained from a singularity the associated link L(D) is ambient isotopic to the link of the singularity. In this latter case, he also provided an easy graphical algorithm to construct a model of the Milnor fibre on which a system of vanishing cycles is visible. We say that a face of a divide D is a connected component of D 2 \ D which does not intersect the boundary of D 2 . Let n be the number of intervals in D, δ be the number of crossings and r the number of faces. The Milnor fibre will be a surface with first Betti number µ = δ + r and n boundary components. The distinguished vanishing cycles will be given by one curve per crossing and one curve per face. The surface is constructed as follows: first, replace every crossing of D with a small circle, to get a trivalent graph. Now, realize every edge of this new graph by a half-twisted band. This will give a surface composed of twisted cylinders, corresponding to the crossings of D, connected by half-twisted bands corresponding to the edges of D. The vanishing cycle associated to a crossing will be given by the core curve of the corresponding cylinder, the vanishing cycle of a face will be given by the core curves of the bands bounding the face. An example of this construction is given in Figure 5.
Remark 3.1. A'Campo's construction only leads to a combinatorial model of the Milnor fibre which is not embedded. A graphical procedure to construct a diagram of the link of a divide and the associated embedded fibre surface has been given by Hirasawa in [13].
Definition 3.2. Let f be an isolated plane curve singularity, D a divide associated to f and Σ(f ) the surface constructed from D with the previous procedure. The geometric monodromy group of f is the subgroup of Mod(Σ(f )) generated by the Dehn twists around the vanishing cycles constructed on Σ(f ). This does not depend on the choice of the divide D.
As we have already mentioned, links of singularities are closures of positive braids. Since fibre surfaces of fibred links are unique, the Milnor fibre of a singularity f is ambient isotopic to the fibre surface Σ β of any positive braid β representing L(f ). We therefore now have two a priori distinct subgroups of Mod(Σ β ) = Mod(Σ(f )), the geometric monodromy group of f and the braid monodromy group of β. Theorem 1 says that those two groups coincide for at least one choice of β.
To prove Theorem 1, we will explicitly find an isotopy between the Milnor fibre constructed from a divide and the surface of an appropriate positive braid and identify the vanishing cycles on this braid surface. In order to do so, we need to use a divide from which the positive braid is somehow visible.
such that the orthogonal projection on this diameter is Morse when restricted to D, all the local maxima (resp. minima) have the same critical value b (resp. a) with b > a and all the crossings are mapped in the open interval (a, b).
Basically, a divide is ordered Morse (w.r.t. a given direction) if no local maxima or minima lie in an interior face of the divide. Examples of such divides are given in Figure 6. Ordered Morse divides were introduced by Couture and Perron [7], who used a generalization of those to construct a representative braid for any divide link. In particular, ordered Morse divides give positive braid links. Notice that every singularity has an associated divide which is ordered Morse (in fact, the divides originally constructed by A'Campo and Gusein-Zade are ordered Morse, see [7]). The result of Couture and Perron can be obtained geometrically: if we apply the algorithm of [13] to an ordered Morse divide, we get exactly the fibre surface of a positive braid. This was done in [10] for Lissajous divides and torus links, but the same procedure works for an arbitrary ordered Morse divide. The construction of the fibre surface is shown in Figure 7: one just has to replace the crossings and minima/maxima of the divide with the corresponding pieces of surface and glue them together following the pattern of the divide. Here we use that all the minima and maxima of the divide are in the exterior face: for general divides the fibre surface is more complicated.
Proof of Theorem 1. Let f be an isolated plane singularity and D an associated ordered Morse divide. Let Σ be the embedded surface constructed following [13], as explained   above. It is an embedded fibre surface whose boundary is the link L(D) = L(f ). To see that this is indeed the fibre surface of a positive braid, we just need to perform the isotopies shown in Figure 8 (2a), getting a collection of disks connected by half-twisted bands, and slide all the bands to the front. Let us remark that an ordered Morse divide is formed of N parallel lines (where N is the number of points in the preimage of a regular value of the Morse projection) connected by the crossings and the minima/maxima. The braid obtained will have N strands, a crossing of D gives a pair of (non-consecutive) generators while every maximum/minimum gives one generator. By further performing the isotopies of Figure 8, (2b) around all the crossings of D corresponding to generators σ i for even i, we can now directly identify Σ with an embedded version of A'Campo's model of the Milnor fibre. A system of vanishing cycles is therefore visible on the braid surface Σ. Those cycles are not exactly the same as the generators of the braid monodromy group, but the same arguments as in the proof of Proposition 2.1 show that the two groups are indeed the same.

Framings
We will now briefly recall the basics of the theory of framed surfaces, concentrating in particular on the action of the mapping class group on such structures, as investigated in [6] and [21]. In what follows, we will adhere to the notations and conventions of [6], but we will restrict only to the case of surfaces with connected boundary. Let Σ = Σ g,1 be a connected, compact, oriented surface of genus g with one boundary component. A framing φ on Σ is a trivialization of the tangent bundle T Σ. With the fixed orientation (and a choice of a Riemannian metric), a framing is simply given by a nowhere-vanishing vector field ξ φ on Σ. Two framings are isotopic if the associated vector fields are isotopic through nowhere-vanishing vector fields.
To a framing one can associate a winding number function, computing the holonomy of a simple closed curve. If c : S 1 → Σ is a C 1 embedding, one can define This defines a map from the set of simple closed curves on Σ to Z, which is clearly invariant under isotopy of φ and c. It is not hard to see that the converse also holds: the isotopy class of a framing on Σ is determined by its winding number function, and actually by the value on finitely many curves (see [6] Lemma 2.2 and [21] Prop.2.4). Thanks to this, we will use the term "framing" indifferently to refer to the isotopy class of the vector field ξ φ or to the associated winding number function φ.
The mapping class group of Σ acts on the set of isotopy classes of framings by pullback, and c a simple closed curve.
is the stabilizer of the isotopy class of φ.
Of particular interest is the action of Dehn twists.
where ·, · denotes the algebraic intersection number.
We say that a nonseparating simple closed curve a on (Σ, φ) is admissible if φ(a) = 0. As a consequence of Lemma 4.1 we have that a nonseparating simple closed curve a ⊂ Σ is admissible if and only if the corresponding Dehn twist preserves φ. Calderon and Salter proved that, for big enough genus, the framed mapping class group is generated by those admissible twists: But more is true. The framed mapping class group is generated by finitely many admissible twists around curves with prescribed intersection pattern. Again following [6]: c k } be a collection of curves on a surface Σ, pairwise in minimal position and intersecting at most once. We say that such a configuration: • spans the surface if Σ deformation retracts onto the union of curves in C; • is arboreal if its intersection graph is a tree, and E-arboreal if moreover it contains the Dynkin diagram E 6 as a subtree.
Definition 4.3. Let C = {c 1 , · · · , c k , c k+1 , · · · , c l } be a collection of curves on a surface Σ and denote by S j a regular neighbourhood of {c 1 , · · · , c j }. We say that C is an h-assemblage of type E if: • {c 1 , · · · , c k } is an E-arboreal spanning configuration on a subsurface S ⊂ Σ of genus h; • For j > k, c j ∩ S j−1 is a single arc; • S l = Σ.
The orbit space of this action was studied by Randal-Williams in [21]. It is classified by the Arf invariant. More precisely, it follows from work of Johnson [15] that the function (φ + 1) mod 2 is a quadratic refinement of the mod 2 intersection form. We can therefore define A(φ) to be the Arf invariant of this quadratic form. More concretely, let us denote by i (·, ·) the geometric intersection number and take a collection of oriented simple closed curves We then have This is of course independent of the choice of the curves {x 1 , y 1 , . . . , x g , y g }. The action of the mapping class group on the set of isotopy classes of framings on Σ = Σ g,1 has exactly two orbits, distinguished by the Arf invariant.
As a consequence, for a given Σ there are exactly two conjugacy classes of framed mapping class groups as subgroups of Mod(Σ).
Remark 4.2 (Caveat). In this section we only stated results for surfaces with connected boundary, in terms of absolute framings. For general surfaces, the whole theory is still valid, but needs to be formulated for relative framings, i.e. only allowing isotopies that are trivial on the boundary. In this more general context, the framed mapping class group is the stabilizer of the relative isotopy class of a framing, and one needs to also take into account the action on arcs, getting generalized winding number functions. The orbit space is now classified by a generalized Arf invariant together with the values of the framing on the different boundary components. However, if the boundary is connected the absolute and relative theories are equivalent and we can use this slightly simpler formulation.
On the vertical disks it is horizontal with alternating directions, on the twisted bands it is parallel to the core.

A framing for positive braids
Let β be a non-split positive braid and Σ β its fibre surface. We can construct a framing φ β on Σ β as in Figure 9. An explicit and straightforward computation now shows that every standard curve γ i on Σ β is admissible for φ β . Therefore, the monodromy group of β is contained in the framed mapping class group of φ β : We will prove that, at least for positive braids whose closure is a knot of big enough genus, the monodromy group is equal to this framed mapping class group. Therefore, in view of the previous discussion, we now want to compute the Arf invariant of φ β . Proposition 4.4. Let β be a positive braid whose closure is a knot K. Then To prove Proposition 4.4, we will need to discuss a bit more in detail the Arf invariant. Let V be a finite dimensional vector space over Z 2 equipped with a non-singular, symmetric bilinear pairing ·, · : V × V → Z 2 . Recall that a quadratic refinement of the bilinear pairing ·, · is a function q : V → Z 2 such that for all x, y ∈ V q(x + y) = q(x) + q(y) + x, y .
To such a mod 2 quadratic form it is classically associated the Arf invariant A(q) ∈ Z 2 .
In our context, we will take V = H 1 (Σ β , Z 2 ) and ·, · the mod 2 intersection form. As we have already mentioned, the framing φ β induces a quadratic refinement of the intersection form, whose Arf invariant is A(φ β ). On the other hand, if the closure of β is a knot K, it is known that the Seifert form also induces such a quadratic refinement. More precisely, if S : It is a classical result that the Arf invariant of this quadratic form is indeed an invariant of K, that we denote by A(K).
Proof of Proposition 4.4. Let β be a positive braid whose closure is a knot K and Σ β its fibre surface, equipped with the framing φ β . The family of curves γ i form a basis of V = H 1 (Σ β , Z 2 ). Since by construction all the γ i are admissible for φ β , for every i we have the equality φ β (γ i ) + 1 = 1 = S (γ i , γ i ) mod 2.
Since {γ i } is a basis, it now follows from the defining equation of a quadratic refinement that for every x ∈ V φ β (x) + 1 mod 2 = q(x).
Therefore the two quadratic forms (φ β + 1) mod 2 and q coincide, so their Arf invariants also do.

Proof of the main theorem
In this section we will give the proof of Theorem 2. In the previous section we have seen that MG(β) Mod(Σ β , φ β ), so we only need to deal with the opposite inclusion. This will be done by applying Proposition 4.2. As a first step, we have to find appropriate subsurfaces supporting an E-arboreal spanning configuration.
Proposition 5.1. Let β be a prime positive 3-braid of genus g ≥ 5 which is not of type A n or D n . Then, excepted finitely many braids, up to positive braid isotopy its linking graph contains an induced subtree which is an E-arboreal spanning configuration on a subsurface of genus g ≥ 5.
Proof. Let β be a positive 3-braid which is not of type A n . Up to elementary conjugation and braid relation we can assume that β = σ a 1 1 σ b 1 2 · · · σ am 1 σ bm 2 , with a i ≥ 2 and b i ≥ 1 for all i ∈ {1, · · · , m}. First of all, notice that if we can find a suitable subtree for a braid σ a 1 1 σ b 1 2 · · · σ am 1 σ bm 2 , the result will also hold for any braid σ We will now prove the result by case distinction over m.
• m = 4 : We will treat several cases. Let us first assume that there is an i such that b i ≥ 2. If there are i = j such that b i , b j ≥ 2, then up to cyclic ordering we only have to deal with the two cases depicted in the left of Figure 12, where we see the sought subtrees. Similarly, if there is only one b i greater than 2 but there is one a j bigger than 3 we will find one of the trees in the right of Figure 12. Finally, if all the a j are equal to 2 and there is only one b i greater than 2, it is enough to consider the braid σ 2 1 σ 2 σ 2 1 σ 2 σ 2 1 σ 2 σ 2 1 σ 2 2 , for which we can find the subtree after applying some braid relations as in Figure 13.
We are now left with b i = 1 for all i. Notice that in that case there need to be at least one a i ≥ 3, otherwise the braid has genus less than 5. If there are two non-consecutive a i and a j greater than 3, it is enough to consider the braid σ 3 1 σ 2 σ 2 1 σ 2 σ 3 1 σ 2 σ 2 1 σ 2 , for which we find an appropriate subtree in the left of Figure 14. If not, up to cyclic ordering there must be two consecutive a i = a i+1 = 2, in which case we can apply a sequence of braid relations as we did in the right of Figure 14 and find our subtree.
• m = 3 : This will be the lengthier case, since there are many low genus braids that require special treatment. Let β = σ a 1 1 σ b 1 2 σ a 2 1 σ b 2 2 σ a 3 1 σ b 3 2 be a braid of genus g ≥ 5, then a simple argument implies that it is enough to consider the braids in Figure 15. Similarly, when a i ≥ 11 it is enough to consider the case when all the b i are equal to 1, and up to elementary conjugation we can assume that a 3 ≥ 3. In this case, by taking all the vertices in the left column and only the topmost of the right column we will always end up finding a tripod tree T (1, k, 9 − k) for k ≥ 2, which all correspond to subsurfaces of genus 5, see Figure 16 for some examples.
We are now left with the low genus cases.
a i = 6 : If b i = 6 we always get a link with 3 components and genus 4. If b i = 7 and there is at least one of the b i equal to one, up to elementary conjugation we can assume that β = σ 2 1 σ 2 σ 2 1 σ b 2 2 σ 2 1 σ b 3 2 with b 2 + b 3 = 6. Using that σ 2 1 σ 2 σ 2 1 commutes with σ 2 we get σ b 2 2 σ 2 1 σ 2 σ 4 1 σ b 3 2 , which is conjugate to σ 2 1 σ 2 σ 4 1 σ 6 2 , whose intersection graph is a tree with the required properties. We are now left with b i = 2 for all i. Up to elementary conjugation there is only one such braid, σ 2 1 σ 2 2 σ 2 1 σ 2 2 σ 2 1 σ 3 2 . Here there are no possible braid relations to apply and it is not possible to find a subtree of big enough genus.
a i = 7 : Let us first assume that b i = 6. If there is at least one b i equal to one, we can directly find our subtrees. In Figure 17 we see some of the cases. The omitted ones are symmetric and will give the same subtrees. Notice that this will also cover all the braids with a i ≥ 7 and b i ≥ 7. If b i = 2 for all i, up to elementary conjugation there is only the braid σ 3 1 σ 2 2 σ 2 1 σ 2 2 σ 2 1 σ 2 2 , for which again we cannot find any subtree of big enough genus.
a i = 10 : The only case left is when b i = 3. If one of the a i is odd we can suppose that a 3 is odd, in which case by taking all the bricks excepted the lowermost of the right column we will get a tripod tree T (1, 2, 6) or T (1, 4, 4),  which both correspond to subsurfaces of genus 5. If all the a i are even, up to elementary conjugation we only have the braids σ 4 1 σ 2 σ 4 1 σ 2 σ 2 1 σ 2 and σ 6 1 σ 2 σ 2 1 σ 2 σ 2 1 σ 2 . Those are actually related by braid relations and elementary conjugations, and the very same argument used for a i = 9 and b i = 3 will yield the required tree.
• m = 2 : For a braid β = σ a 1 1 σ b 1 2 σ a 2 1 σ b 2 2 of genus at least 5 the intersection graph is always a tree with at least 10 crossings. Furthermore, by direct inspection we see that those trees will always contain E 6 unless they are of type D n .
• m = 1 : In this case we only get non-prime braids.
We will now consider the braids with big positive braid index. First, we need another invariance property of the braid monodromy group.
Lemma 5.1. Let β ∈ B + N be a prime positive braid on N ≥ 3 strands. If for some i the linking graph of the subword induced by all the generators σ i and σ i+1 is a path, then there exists a positive braid β ∈ B + N −1 such thatβ =β and MG(β) = MG(β ). Proof. Up to elementary conjugation and symmetry, we can assume that the subword induced by σ i and σ i+1 is of the form σ a i σ i+1 σ i σ b i+1 . Moreover, we can suppose that all the generators σ j for j < i appear before the last occurrence of σ i and all the generators σ j for j > i + 1 appear after the first occurrence of σ i+1 . In Figure 20 we see an isotopy between the fibre surface Σ β and the fibre surface Σ β of a new braid β with one strand less. A direct computation now shows that MG(β) = MG(β ).
Proposition 5.2. Let β be a prime positive braid on N ≥ 11 strands and whose closure is a knot not of type A n . Then, up to positive braid isotopy and excepted finitely many braids, its linking graph contains an induced subtree which is an E-arboreal spanning configuration on a subsurface of genus g ≥ 5.
The finitely many exceptions in Proposition 5.2 come from Prop. 5.1 and Prop. 5.3, in case we can reduce our braid to one of the exceptions therein by applying Lemma 5.1. To prove this proposition, we will also need the following lemma from [17]. We will briefly recall the algorithm for constructing such a path, since this will be used in what follows. Let us say that we want to connect v to a column to its right. Start at v and move up or down its column until reaching the closest brick linked to the right (potentially, already v). Now, move to the right and repeat the procedure. If at the moment of moving to the right there are several possibilities, choose the brick which is the closest to a brick in the same column linked again to its right. It is easy to see that those choices prevent the creation of cycles, so that the result will be a path.
Proof of Proposition 5.2. Let β be a prime positive braid on N ≥ 11 strands. By Lemma 5.1 we can assume that, for every pair of adjacent columns in the brick diagram, the linking graph restricted to those columns is not a path. Let us furthermore assume that no braid relation of the form σ i σ i+1 σ i σ i+1 σ i σ i+1 can be performed. Our strategy goes as follows: we will start considering an induced path connecting the leftmost column to the rightmost, constructed with the previous algorithm, and try to add to it one single vertex, in order to get a tripod tree containing E 6 . Since b ≥ 11, the tripod tree will have at least 11 vertices and hence correspond to a subsurface of genus at least 5. So, let us fix one such path and look at the third column of the brick diagram. If we can add a brick of this column to the path and get an (induced) tripod tree we are done. There are two reasons why this might not be possible: either because there are no leftover bricks in the third column or because every available brick is linked to more than one brick of the path and adding it would generate a cycle. We will now analyse those cases in detail. By symmetry, we can assume that in the third column our path arrives from the left to a brick v 3 , potentially moves down to a brick w 3 and then continues to the right.
If there are no leftover bricks in the third column, then by the construction rule of our paths we know that w 3 is the only brick of column 3 linked to the right. We can now apply elementary conjugations on the right-hand side of the diagram in order to have all the generators σ i for i ≥ 4 appear before the last occurrence of σ 3 , and perform again all the possible braid moves σ i σ i+1 σ i σ i+1 σ i σ i+1 . Those transformations will not affect the first 3 columns and the part of the path therein. We now get that the sub-braid generated by σ 3 and σ 4 is σ a 4 σ 3 σ b 4 σ c 3 , with c ≥ 1 and a, b ≥ 2 by Lemma 5.1. Let us denote by v 4 the only brick of column 4 linked to w 3 , and let us attach a path connecting v 4 to the rightmost column.
If at least one of the bricks immediately above or below v 4 is not linked to the portion of the path in the fifth column (in particular, if v 4 is itself linked to the right), it can safely be added to get a tripod tree. We directly see that we are left with the case of Figure 21. Notice that, up to modifying the path in the fourth and fifth columns, we can always choose whether it passes by x or y. Now, if there is a brick x above x, either it is not linked to the path in the fifth column, in what case we can directly connect it to x, or it is, in what case we can change our path to w 3 → v 4 → x → x → {path in the fifth column} (thus avoiding v 5 ) and connect y to v 4 . Similarly, we can assume that there is no brick below y.
Let us now consider the fifth column. Notice that there must be at least one brick immediately above and one immediately below v 5 that are not linked to the fourth column, otherwise we could apply one of the forbidden braid relations. By applying the same reasoning as before, we conclude that we can always obtain a tripod tree, unless there are no other bricks in the column. In the latter case, however, the closure of the braid is not a knot.
We can now suppose that there are some leftover bricks in the third column.
1. If there are bricks above v 3 and below w 3 , we will be in one of the two cases of Figure 22.
(a) In the left-hand case of Figure 22, recalling that the path was constructed with the algorithm of Lemma 5.2, we know that either v 3 and w 3 are adjacent or they coincide. We will analyse those cases in great detail, since they serve as example of the kind of reasoning applied also to the rest of the proof.
If v 3 and w 3 are distinct and adjacent, again by the construction rule of our paths we know that w 3 is not linked to the right at all and v 3 is not linked to the path to the left. Now, if v 3 is linked to the path to the right above v 4 , we could change our path to w 2 → v 3 → v 3 → {path in the fourth column}, thus avoiding v 4 , and connect w 3 to v 3 to get a tripod. Otherwise, we can instead consider w 2 → w 3 → w 3 → v 4 → {path} and connect v 3 to v 4 .
If v 3 = w 3 , then we know that w 2 has to be linked to the first column, otherwise we could perform one of the forbidden braid relations.   Let us suppose first that w 2 is linked to a brick v 1 below it, as in the left-hand side of Figure 23 . Notice that the brick denoted by w 2 needs to exist because of the condition on the possible braid relations. Hence, we can assume that in the first column there are at most two bricks and no bricks below v 1 , and that the brick immediately below w 2 (if any) is linked with v 1 , otherwise we could immediately find an appropriate tripod. If there is a linking between the second and third columns above v 3 , we could modify our path by starting from v 1 and w 2 , then moving upwards in the second column until we reach the first connection with the third column above v 3 and finally going down on the third column until the first connection to the original path in the fourth column (which occurs at the latest at v 3 ). This will give us a path avoiding v 3 . We can now safely connect w 3 to w 2 and get a tripod. If not, up to elementary conjugations on the first two columns, we can suppose that there are no bricks in the second column above v 3 , as in the central picture of Figure 23. In this case, we can assume that above w 2 there is at most one brick. Now, if in the first column there are two bricks, again by elementary conjugation we are back to the case where there is a brick below v 1 and we are done. We are hence left with just one brick in the first column, as in the right-hand side of Figure 23. Notice that in this case the brick w 3 is forced to be linked to w 2 , otherwise the closure of the braid is not a knot. This in turn forces the existence of the brick denoted by b below w 3 , otherwise we could apply a forbidden braid relation. If there is a brick a below w 2 , we can consider v 1 → a → w 2 → w 3 → v 3 → {path} and connect b to w 3 . On the other hand, if there are no bricks below w 2 we see that either the closure of the braid is not a knot, if there is a brick above w 2 , or we can reduce the braid with Lemma 5.1.
We can now suppose that w 2 is linked to a brick v 1 above it, but is not linked with any brick of the first column below it, as in the left image of Figure 24. If there are at least two bricks below w 2 we immediately find a tripod. If there is exactly one brick below w 2 , we can furthermore assume that v 1 is the only brick in the first column. Let us now consider how v 1 is connected with the second column. If it is only linked to w 2 , by applying an elementary conjugation we are back to the previous case where v 1 was below w 2 . Notice that the existence of a brick below w 2 ensures that the condition about the possible braid relations is still satisfied after the conjugation. If v 1 is linked to another brick of the second a? Figure 23: The dashed lines show where the following brick (if existing) would be. In the third column, there is still a brick w 3 as in Figure 22, which is linked to w 2 and may or may not be linked to w 2 .
∅ Figure 24: Below v 1 there can be bricks only until the dashed line. In the second column, there is at least one brick above w 2 but below v 1 .
column above w 2 , called a, and a is in the region denoted by X as in the central image of Figure 24, we immediately see that either we find a suitable tripod or the closure is not a knot, depending on how many bricks there are in the second column between w 2 and a (there is at least one by braid relation). Finally, if a is not in the region denoted by X , as in the right image of Figure 24, we know that there is a brick between a and w 2 linked to v 3 . We can now consider v 1 → a → {second column} → v 3 → {path} and connect w 3 to v 3 . The only case left now is when there are no bricks below w 2 . Again, if v 1 is linked to another brick of the second column above w 2 the exact same argument applies. If v 1 is linked only to w 2 , this time we cannot simply apply an elementary conjugation to reduce it to a previously treated case. However, if there are no bricks above v 1 (resp. below v 1 ) we could apply Lemma 5.1, whilst if there are bricks in the first column both above and below v 1 it is immediate to conclude that either we find a tripod or the closure is not a knot.
(b) In the right-hand case of Figure 22, we know that w 2 needs to be linked to a brick in the first column.
Suppose first that w 2 is linked to a brick v 1 above it. By excluding all the cases where one can immediately find a tripod, we are left with at most two bricks in the first column v 1 and we can reduce the study to one of the cases in the left-hand side of Figure 25. If there are two bricks in the first column, we either have a brick above v 2 , in which case we can find a tripod by simply starting our path from v 1 and adding two bricks above w 2 , or we can apply an elementary conjugation to the first column to get a brick below v 1 , which again immediately gives a tripod. If in the first column there is just one brick, we know that v 2 needs to be linked to the third column, otherwise the closure is not a knot. After removing all the trivial cases, we are left with the diagram on the right-hand side of Figure 25. Notice that now by Lemma 5.1 there needs to be at least one brick below w 2 . If none of the bricks below w 2 are linked to v 3 , we see that according to the number of those bricks we either get a tripod or the closure is not a knot. Hence we can suppose that there is a brick w 2 below w 2 linked to v 3 . If w 2 is connected to the original path in the third column below v 3 , we can instead consider v 1 → w 2 → · · · → w 2 → {path} and get a tripod by connecting to w 2 the bricks v 3 and v 3 . If not, we can simply take our original path starting from v 3 and connect to it w 2 , v 3 and v 3 .
Suppose now that w 2 is only linked to a brick v 1 below it. We immediately see that there can be at most one brick above w 2 , and if this brick exists then v 1 is the only brick of the first column. After excluding the trivial cases we are left with the diagrams of Figure 26. First, if w 2 is linked to the path in the third column below v 3 , we can take v 1 → w 2 → · · · → w 2 → path and add to it a brick in the third column (which will be at most v 3 ). Otherwise, if there is a brick w 2 below w 2 , we can simply take our original path from v 3 and add to it v 3 , w 2 and w 2 . Finally, let's assume that there are no bricks below w 2 . If there is a brick above w 2 we can apply an elementary conjugation to the first column and get back to the previous case. If not, Lemma 5.1 forces the existence of bricks above and below v 1 , in which case either we get a tripod or the closure is not a knot.
2. Let us now consider the case where there is at least one free brick v 3 above v 3 , but none below w 3 . First of all, if after w 3 our path moves to a brick v 4 of the fourth column which is below it, we are basically in the same situation as Case 1b, and the precise same arguments apply. We can hence suppose that the path moves upwards in the fourth column. We will now treat different cases according to how v 3 is linked to the neighbouring columns. Figure 26: Between w 2 and w 2 there can be at most one brick. In the leftmost diagram, the brick above w 2 is not necessarily linked with v 3 . Figure 27: Diagrams for Cases 2a and 2b.
(a) If v 3 is not linked to the right, we know that it needs to be linked to a brick w 2 in the second column, which in turns needs to be linked to a brick v 1 in the first column. Let us suppose first that v 1 is above w 2 , then by the first part of Case 1b, we are left with only one brick in the first column, as in the central image of Figure 25. Furthermore, if the brick immediately above v 3 is not linked to its right, all the arguments from Case 1b still apply. We are hence left with the leftmost diagram of Figure 27. Now if v 3 is not linked to the path above v 4 it can directly be added as additional vertex, otherwise we can instead consider the path v 1 → w 2 → v 3 → v 3 → {path} and add a brick to this new path in the fourth column. Finally, if v 1 is below w 2 , we conclude directly as in the second part of Case 1b.
(b) If v 3 is linked to the right (to v 4 ) and to the left (to a brick w 2 ), by the construction rules of the path we know that either v 3 and and w 3 are adjacent or they coincide, and by the assumption on the braid relations w 2 is linked to a brick v 1 in the first column.
If v 1 is above w 2 , we are left with the two diagrams at the right-hand side of Figure 27. Let us first consider the case where v 3 and w 3 are distinct and adjacent. If v 3 is not linked to the path above v 4 , we can simply consider v 1 → w 2 → v 3 → v 4 → {path} and add v 3 (notice that this would also work if v 3 and w 3 did coincide). If v 3 is linked to the path in the fourth column above v 4 , take instead v 2 → v 3 → {path} and add v 3 and w 3 . Suppose now that v 3 and w 3 coincide. In this case, notice that no brick below w 2 can be linked to v 3 (otherwise we could perform a forbidden braid relation), and that therefore if there are Figure 28: Diagrams for Case 2c; v 3 is linked to the second column, but v 3 is not. On the right, we know that w 2 needs to be linked to some brick v 1 in the first column.
at least two bricks below w 2 we immediately get a tripod. It follows that there need to be a brick v 2 above v 2 , otherwise either we can apply Lemma 5.1 or the closure is not a knot. Now, if v 3 is not linked to the path in the fourth column above v 4 , we can find the same tripod as in the previous case. If v 3 is linked to the path above v 4 , we can instead consider we are left with one of the diagrams of Figure 26 (where the brick v 3 is now w 3 ). The case where v 3 and w 3 coincide is excluded by the condition on the braid relations. Furthermore, by what was done in the second part of Case 1b, we know that we can assume the existence of a brick w 2 below w 2 . Hence, if v 3 is not connected to the path above v 4 we can take v 1 → w 2 → v 3 → v 4 → {path} and add w 3 , if v 3 is connected to the path above v 4 we can instead take w 2 → w 2 → v 3 → v 3 → {path} and add w 3 .
(c) If v 3 is not linked to the left, either v 3 and w 3 are adjacent or they coincide, as in the left hand side of Figure 28. In both cases, if v 3 is connected to the path above v 4 , we can simply let our path pass by v 3 instead of w 3 (thus skipping v 4 ) and add a brick in the fourth column (which will be at most v 4 ).
Suppose now that v 3 is not connected to the path above v 4 and v 3 ,w 3 are distinct. If w 3 is linked to the left we are in the situation at the right-hand side of Figure 28 and we directly find a tripod by considering v 1 → w 2 → w 3 → v 4 → {path} and adding v 3 . If not, then we are in a situation analogous to Figure 21 and the same arguments apply.
We are left with the case where v 3 and w 3 coincide and v 3 is not connected to the path above v 4 .
Let us suppose first that there is a brick v 2 in the second column below v 3 . We know that v 2 needs to be linked to a brick in the first column. If there is a brick v 1 in the first column above v 2 , we are in one of the situations in the left of Figure 29. In the first one, we directly see that either we find a tripod (if there is at least another brick in the first column) or the closure is not a knot. In second one, using Lemma 5.1 we furthermore know that there must be a brick in the second column above v 1 . By excluding the direct cases, we end up with the diagram on the right-hand side of Figure 29. Notice that we can assume that there is no brick below v 2 , because otherwise by elementary conjugations we would get two bricks Figure 29: More diagrams for Case 2c, when v 3 and w 3 coincide and there is a brick v 2 below v 3 .
above v 2 and would find a tripod by taking {second column} → · · · → v 2 → v 3 → {path} and adding v 1 . With similar arguments we can conclude there are no bricks in the second column above v 2 and v 1 is the only brick of the first column. We now see that there need to be a brick in the third column above v 2 , otherwise the closure is not a knot. If there are at least two bricks of the third column above v 2 , we get a tripod by taking {third column} → · · · → v 3 → v 4 → {path} and adding v 2 and v 2 . Otherwise, we can consider v 1 → v 2 → · · · → v 2 → v 3 → {path} and add the other brick in the third column linked to v 2 (which now we know will not link to any other brick of the second column.) If there are no bricks in the first column above v 2 , but v 2 is linked to a brick v 1 below it, we can directly conclude that either the closure is not a knot or we find an appropriate tripod. Suppose now that there are no bricks in the second column below v 3 , which is therefore only linked to a brick v 2 above it. If in the second column there are bricks both above and below v 2 , noticing that if there are at least four bricks in the second column we are done, we are only left with the cases of Figure 30. For the leftmost diagram, if there is only one brick in the first column the result is not a knot, otherwise up to elementary conjugation we get a tripod. In the two central diagrams we directly find a tripod. In the rightmost diagram, if v 2 is not linked to the third column the closure is not a knot, otherwise we directly get a tripod. If in the second column there are only bricks under v 2 , by minimality of the number of strands there is a brick v 3 in the third column above v 2 (and similarly if there are only bricks above v 2 ). If v 2 is not linked to the first column or if there are at least two bricks in the first column we can simply take our original path and add to it v 3 . If there is only one brick in the first column and this brick is linked to v 2 , according to the number of bricks of the second column below v 2 we will either get a tripod or a link with at least two components. Finally, if in the second column there are only bricks above v 2 , let us consider v 2 the first brick of the second column linked to a brick v 1 of the first column (starting from v 2 upwards, potentially v 2 = v 2 ). If there is still a brick v 2 above it, up to elementary conjugation on the first column we can assume that v 1 is above v 2 . We now directly see that we are left with only one brick in the first column and that according to whether v 2 is linked to its right, we either get a tripod or a link with more than two components, as in the left of Figure 31. If there are no more bricks above v 2 , we are left  Figure 31: In the left, the case when there is a brick above v 2 which is linked to the right. In the right, when there are no bricks above v 2 .
with the diagram at the right-hand side of Figure 31. By Lemma 5.1, we know that there must be bricks above and below v 1 and we conclude with an usual argument.
3. We finally have to treat the case where there is a free brick w 3 below w 3 , but no brick above v 3 .
(a) Let us first suppose that w 3 is linked to a brick w 2 of the original path in the second column (which, by construction, will also be linked to v 3 ). Then either v 3 and w 3 are adjacent or they coincide, as in the left of Figure 32.
If v 3 and w 3 are distinct, we furthermore know that they are not linked to any brick of the fourth column. If v 3 is linked to a brick v 2 of the second column above w 2 , we know that v 2 need to be linked to the first column. In this case, we could simply connect the first column to v 3 via v 2 (thus skipping w 2 ), continue with our original path and add to it w 3 to get a tripod. Similarly, suppose that w 3 is linked to some brick in the second column below w 2 . If there is a connection between the first and second columns below w 2 , the previous argument still apply. Otherwise, all the bricks in the second column below w 2 are "free" and can be added to our path. If there are at least two bricks below Figure 32: In the left, the two main possibilities for case 3a. In the centre, the diagram left after treating all the direct cases, for v 3 and w 3 distinct. In the right, the last diagram for v 3 = w 3 . are done; if there is only one, according to the number of bricks in the first column we see that we either find a tripod or the closure is not a knot. We are therefore only left with the diagram in the centre of Figure 32. If in the second column there are bricks both above and below w 2 , we are basically in the situation of Figure 30 (with the appropriate changes in the third column) and the same arguments apply. If there are only bricks above w 2 , we end up with diagrams as the ones in Figure 31 and we conclude similarly. The case where there are only bricks below w 2 is symmetric.
If v 3 and w 3 coincide, we know that w 2 needs to be linked to a brick v 1 in the first column. If v 1 is above w 2 , after removing all the cases where one can directly find a tripod, we are left with diagrams as in the left-hand side of Figure 25, and the same arguments apply (remembering that now there are no bricks in the third column above v 3 ). If w 2 is not linked to any brick in the first column from above, we can reduce to a diagram as in the right-hand side of Figure 23. Now, if b is not linked to the path in the fourth column, the same argument as in the corresponding part of Case 1a works. If b is linked to the path in the fourth column from below, one can consider v 1 → a → w 2 → w 3 → b → {path} and add v 3 to get a tripod. Similarly if w 3 is linked to the path in the fourth column from below. Finally, if b is linked to the path in the fourth column but w 3 is not, we are in the case drawn in the right-hand side of Figure 32. If in the third column there are no bricks below a, one can simply perform an elementary conjugation on the second column to get a brick v 2 above w 2 , take the original path starting from v 3 and add to it w 3 → w 2 and v 2 to obtain a tripod. Finally, if in the third column there is a brick below a, in particular w 2 is linked to a brick b of the third column below b. One can hence take v 1 → a → w 2 → b → · · · → b → v 4 → {path} (or potentially skipping w 2 if b is also linked to a ) and connect v 3 to v 4 .
(b) We now suppose that w 3 is not linked to the original path in the second column (and therefore has to be linked to the path in the fourth column). By construction, we know that v 3 is linked to some brick in the second column.
Assume first that v 3 is linked to a brick w 2 above it. We know that w 2 will be linked to a brick of the first column. If it is linked to a brick above it, we conclude as in the first part of Case 1b, so we can assume that w 2 is only linked to a brick v 1 below it. By the second part of Case 1b, we are only left with the diagrams of Figure 26 (in this case, the brick denoted by v 3 does not exist). We furthermore know that the brick w 2 is not linked to the original path in the third column below v 3 and that there is at least one brick in the second column below v 1 . After removing all the cases where one can directly find a tripod, we are left with the leftmost diagram of Figure 33. Furthermore, if there are at least two bricks in the first column, up to elementary conjugation we can assume the existence of a brick below w 2 and can now obtain a tripod by taking w 2 → · · · → w 2 → v 3 → {path} and connecting to w 2 a brick below it and v 1 together with another brick of the first column. We can hence reduce to the central diagram of Figure 33. But now we observe that there needs to be a brick in the third column below w 2 , otherwise the closure is not a knot. In particular, w 2 is linked to a brick w 3 of the third column below v 3 . If v 3 is not linked to the fourth column we can simply take v 1 → w 2 → · · · → w 2 → w 3 → · · · → {path in the fourth column} and connect v 3 to v 2 . If v 3 is linked to the fourth column (so in particular v 3 = w 3 ), we have the rightmost diagram of Figure 33. If w 3 is linked to the path in the fourth column under v 4 , we can simply consider v 1 → w 2 → v 3 → w 3 → {path} and add v 4 . Finally, if w 3 is not linked to the path in the fourth column below v 4 , then all the bricks in the third column under w 3 also are not, and can hence be freely used. If there is still at least one brick in the third column under w 3 , we can take w 2 → w 2 → w 3 → · · · → w 3 → v 4 → {path} and add a brick below w 3 to get a tripod. If w 3 is the last brick of the third column, in particular it is not linked to any of the bricks below w 2 , so we can take v 1 → w 2 → · · · → w 2 → v 3 → {path} and connect w 3 to w 2 .
We can now suppose that v 3 is only linked to a brick v 2 of the second column below it. In particular, our original path was passing by v 2 , which is therefore not linked to w 3 . If v 3 and w 3 are distinct, we end up with a diagram similar to Figure 21 and the exact same arguments apply. If v 3 and w 3 coincide, we are in a situation perfectly symmetric to the final part of Case 2c, in particular as in Figures 30 and 31, and again the same arguments apply.
We still have to consider the braids of intermediate positive braid index. One could probably study those by hands, in a similar way to Prop. 5.1 and Prop. 5.2, but the computations would quickly get too complicated. Instead, we will treat them by directly applying Proposition 5.1, at the cost of loosing some low genus cases. Proposition 5.3. Let β be a prime positive braid on 4 ≤ N ≤ 10 strands whose closure is a knot not of type A n . Suppose that β has genus g(β) > 4(N − 1). Then there exists a family of curves on Σ β that is an E-arboreal spanning configuration on a subsurface of genus at least 5.
The curves appearing in Proposition 5.3 will not necessarily be vertices of the intersection graph, but we might need to do some "change of basis", i.e. modify some of the curves by applying appropriate Dehn twists. This will change the intersection pattern of the curves in question, but not the subsurface they span nor the subgroup that the corresponding Dehn twists generate in Mod(Σ β ).
Proof. Let β be such a positive braid. Since g(β) > 4(N − 1), there exists 1 ≤ i ≤ N − 2 such that the subword induced by all the generators σ i and σ i+1 has first Betti number at least 12, when seen as a 3-braid. Let us denote this subword by β i,i+1 . By Proposition 5.1, either β i,i+1 is positively isotopic to a 3-braid β i,i+1 containing the required spanning configuration, or it is of type A n or D n (the other finitely many exceptions have first Betti number 11).
In the first case, the required positive braid isotopy might not be realizable when β i,i+1 is seen as a subword of β. However, since at the level of curves the effect of braid relations and elementary conjugations is obtained by Dehn twists, we can still find a family of curves in Σ β i,i+1 ⊂ Σ β whose intersection pattern is equal to the linking graph of β i,i+1 , and the result follows.
If β i,i+1 is of type A n , since there are only three strands one can directly verify that up to elementary conjugation its linking graph is a path. We can therefore apply Lemma 5.1 to β and reduce it to a braid with less strands.
If β i,i+1 is of type D n , up to elementary conjugation and symmetry it is of one of three forms: σ n−3 i σ 2 i+1 σ i σ 2 i+1 , σ n−2 i σ i+1 σ 2 i σ i+1 or σ a i σ i+1 σ i σ b i+1 σ i σ i+1 with, a + b = n − 2. This follows from a direct computation, or can be seen by applying the classification of checkerboard graphs of type D n contained in Lucas Vilanova's PhD thesis [22]. In all the cases one can see that we can always add a brick in a neighbouring column and find the required subtree. We will do it for β i,i+1 = σ n−3 i σ 2 i+1 σ i σ 2 i+1 , the others are analogous. In this case, we know that i < N − 2, otherwise the closure is not a knot. Since β is prime, its intersection graph is connected, so at least one of the three bricks in the i + 1-th column needs to be linked to its right. After removing the cases where one directly finds an appropriate subtree, we are left with one of the three cases of Figure 34. The first one is excluded since the closure is not a knot; in the second one we can find a subtree after braid relation, as shown in the Figure; for the third one, up to elementary conjugation we can suppose that there are no generators σ i+2 above the last occurrence of σ i+1 . Now we see that if there are at least two bricks in the i + 2-th column we are done, otherwise either the closure is not a knot (if i + 2 = N − 1) or we can still add one brick further to the right and again find the required subtree.  Everything is now ready to prove our main Theorem.
Proof of Theorem 2. Let β be a prime positive braid not of type A n and whose closure is a knot. We want to prove that MG(β) = Mod(Σ β , φ β ) by using Proposition 4.2. Let V = {γ 1 , · · · , γ 2g } be the family of standard curves on Σ β corresponding to the vertices of the linking graph of β. In Propositions 5.1, 5.2 and 5.3 we have constructed the starting E-arboreal spanning configuration of genus h ≥ 5 for all but finitely many such prime positive braids. In general, this is obtained by taking a subfamily of curves V 0 ⊂ V and potentially modifying some of them by applying Dehn twists around other curves of V 0 , obtaining a family V 0 of curves in Σ β . In particular, the subsurface spanned by V 0 is the same as the subsurface spanned by V 0 . It is now clear that the remaining curves of V \ V 0 can be attached in an order that respects the definition of h-assemblage, so that Remark 5.1. The requirement of the closure of β being a knot is a consequence of our method of proof. Even excluding the two exceptional families A n and D n , that do not contain an E 6 subgraph but whose braid monodromy group is well understood, there are other infinite families, both with bounded and unbounded braid index, that most likely do not contain an E 6 . For example, we could not find such subtrees for the braids β n = σ 1 σ 2 2 σ 1 σ n−4 2 σ 3 σ 2 2 σ 3 ∈ B + 4 , whose linking graph is the extended Dynkin diagramD n , nor for β N = (σ 1 · · · σ N σ N · · · σ 1 ) 2 ∈ B + N . We do not know whether the corresponding braid monodromy groups are equal to the whole framed mapping class group.