On the Slice-Ribbon Conjecture for Montesinos knots

We establish the slice-ribbon conjecture for a large family of Montesinos' knots by means of Donaldson's theorem on the intersection forms of definite 4-manifolds.


Introduction
A celebrated open question in knot theory is the slice-ribbon conjecture due to Fox [Fo] in 1962. Recall that a knot K ⊆ S 3 is called (smoothly) slice if it bounds a smoothly and properly embedded disc D 2 ֒→ D 4 in the 4-ball D 4 , and ribbon if it bounds an immersed disc D 2 S 3 with only ribbon singularities (see [Ka,p. 70] for the definition). It is not difficult to prove that every ribbon knot is slice: simply push the ribbon singularities into the fourth dimension to obtain an embedded D 2 ֒→ D 4 . The converse, whether every slice knot is ribbon, is the well known Fox's slice-ribbon conjecture.
In recent works [Li, GJ] the conjecture has been proved for all 2-bridge knots and for infinitely many 3-stranded pretzel knots. Both families of knots are particular cases of the significantly broader family of Montesinos links (first constructed in [Mo2]). In [Wi] it is shown that no member of a five parameter family of Montesinos knots is slice. In the present work, in order to introduce Montesinos links we follow [Si,p. 18,Theorem (c)], where these links are defined as the boundary of 2-dimensional plumbings with star-shaped plumbing graphs. A star-shaped graph is a connected tree with a distinguished vertex v 0 (called the central vertex) such that the degree of any vertex other than the central one is ≤ 2. In a weighted star-shaped graph Γ each vertex represents a twisted band, that is a D 1 -bundle over S 1 , embedded in S 3 , with the number of half-twists given by the weight of the vertex. Bands are plumbed together precisely when the corresponding vertices are adjacent (see Figure 3 for an example). The result of this plumbing construction is a surface B Γ ⊂ S 3 whose boundary ML Γ is, by definition, a Montesinos link. Since S 3 = ∂D 4 , we can push the interior of B Γ into the interior of D 4 . It follows that the double covering of D 4 branched over B Γ is the 4-dimensional plumbing M Γ , obtained by plumbing D 2 -bundles over S 2 according to the graph Γ, which defined the Montesinos link. The boundary Y Γ := ∂M Γ is a Seifert space (see [Ra] for a proof) with as many singular fibers as legs of the graph Γ. A leg of a star-shaped graph is any connected component of the graph obtained by removing the central vertex. The involution u that defines the covering M Γ → D 4 ≃ M Γ /u, turns the Seifert space Y Γ into the double covering of S 3 branched along the Montesinos link ML Γ . Restricting our attention to three-legged star-shaped graphs Γ, it is well known [BZ,Theorem 12.29] that the Seifert space Y Γ is the double covering of S 3 branched along exactly one Montesinos link (up to link isotopy).
In the present work, following in part the approach of [Li], we study the family P of all three-legged connected plumbing graphs Γ such that: • I(Γ) := n i=0 (a i − 3) < −1, where by −a 0 , ..., −a n we denote the weights of the vertices of Γ; and • the central vertex has weight less or equal to −3 and every non central vertex has weight less or equal to −2.
Our choice of the family P is motivated by what follows. In [Li], the fact that a linear graph Γ has a dual graph Γ ′ such that Y Γ = −Y Γ ′ and M Γ , M Γ ′ are both negative definite, is strongly used. Indeed, this property allows one to assume, without loss of generality in the case of 2-bridge links, that I(·) < 0. In our case, a "dual" graph Γ ′ still exists, but it is always the case that one of M Γ , M Γ ′ is indefinite. This forces us to restrict to the case of plumbing graphs Γ such that I(Γ) < 0. In the present work, we deal with the case I(Γ) < −1. For the case I(Γ) = −1 so far we have obtained partial results [Le]. We hope to return to this case in a future paper. The second condition defining the family P is due to technical reasons. More specifically, we will show that, for every Γ ∈ P such that the Seifert space Y Γ bounds a rational homology ball, we have I(Γ) ∈ {−4, −3, −2} and we will study separately the three possible cases. It turns out that, allowing the central vertex of the graph Γ to have weight −2, the fact that Y Γ bounds a rational homology ball does not imply that I(Γ) is bounded from below. We note that, for every Montesinos knot ML P with P ∈ P we have that ML P is neither a 3-stranded pretzel knot nor a member of the family studied in [Wi], since these two families of knots have both associated negative graphs with central vertex of weight −2.
Our main result is the following.
Theorem 1.1. Consider Γ ∈ P. The Seifert space Y Γ is the boundary of a rational homology ball W if and only if there exist a surface Σ and a ribbon immersion Σ S 3 such that ∂Σ = ML Γ and χ(Σ) = 1.
Our analysis gives a complete list, Theorems 2.5 and 2.6, of the Seifert spaces Y Γ with Γ ∈ P which bound rational homology balls, providing a partial anser to a question of Andrew Casson [Ki,Problem 4.5].
Theorem 1.1 immediately implies, Corollary 1.2. The slice-ribbon conjecture holds true for all Montesinos knots ML Γ with Γ ∈ P.
Proof. Let Γ ∈ P be such that the knot ML Γ ⊂ S 3 is slice. Let D 2 ֒→ D 4 be a smooth slicing disc for ML Γ and W the 2-fold cover of D 4 branched along D 2 . It is well known [Ka,Lemma 17.2] that W is a rational homology ball and that ∂W = Y Γ . It follows immediately from Theorem 1.1 that the knot ML Γ is ribbon.
In [BS, CH, Fi, FS, St] other families of Seifert spaces bounding rational homology balls were constructed. A case by case comparison shows that the intersection between these families and the family studied in this work is essentially empty.
The strategy to approach Theorem 1.1 can be sketched as follows. All the plumbing graphs Γ ∈ P give rise to negative definite 4-manifolds M Γ with boundary ∂M Γ = Y Γ . Therefore, if we assume the existence of a rational homology ball W with ∂W = Y Γ , we can build a closed, oriented, negative definite, 4-manifold X Γ as M Γ ∪ YΓ (−W ). Donaldson's celebrated theorem [Do] implies that the intersection lattice (Z n , Q XΓ ) is isomorphic to the standard negative diagonal intersection lattice (Z n , −Id), where n := b 2 (X Γ ) = b 2 (M Γ ). Therefore, the intersection lattice (Z n , Q MΓ ) must embed in the standard negative definite intersection lattice of equal rank; that is, there must exist a monomorphism ι : Z n → Z n such that Q MΓ (α, β) = −Id(ι(α), ι(β)) for every α, β ∈ Z n ∼ = H 2 (M Γ ; Z)/Tors. The first step in the proof of Theorem 1.1 consists of a careful analysis of this obstruction, determining which among the intersection lattices (Z n , Q MΓ ), with Γ ∈ P, admit an embedding into the standard negative diagonal lattice. This analysis leads to a list of candidates P ∈ P (Theorems 2.5 and 2.6) such that the associated Seifert spaces Y P may bound rational homology balls. In the first part of this paper we use techniques and results from [Li].
In the second step of the proof, we find explicitly, for each ML P with P in the list of candidates, the ribbon surface claimed in Theorem 1.1. In [Li, GJ], once they arrive to their respective lists of candidates, the construction of the surface is not directly related to the analysis done in the first part of the proof. Our approach to the construction of the ribbon surfaces is different. In a first attempt we tried to find systematically the bands that describe the ribbon surfaces on the diagrams of our candidates, but we were soon discouraged after realizing how complicated and random they seemed to be in the standard projection of a Montesinos link (see Figure 1 for an example). To overcome this difficulty, instead of working with the link diagrams corresponding to the candidates P , we focus our attention on the corresponding 4-dimensional plumbings M P . The analysis done in the first step of the proof suggests how to modify M P with the addition of a 2-handle, yielding a 4-manifold M ′ whose boundary is the double cover of S 3 branched over a link. It turns out that this link bounds a surface with Euler characteristic equal to 2. A theorem due to Montesinos [Mo] implies that the added 2-handle corresponds to a ribbon move on the initial link ML P and this concludes the proof. (−3, 1)(−3, 1)(−3, 1)), with the notation inherited from the classical notation for Seifert spaces, is ribbon. In fact, performing a ribbon move along the gray band we obtain two unlinked unknots.
The paper is organized as follows: in Section 2 we give a quick overview of some basic facts on Seifert spaces and Montesinos links and we introduce the necessary definitions in order to state Theorems 2.5 and 2.6. These give a list of candidates P ⊆ P such that the knots ML P may be slice. We postpone the long and technical proof of Theorems 2.5 and 2.6 to Sections 4 to 7. A brief sketch of the proof can be found at the end of Section 2. In Section 3 we construct a ribbon surface with Euler characteristic 1 for every Montesinos' link stemming from the list of candidates and we prove Theorem 1.1.

The candidates
In this section we recall some terminology and well-known results concerning Seifert spaces and Montesinos links. Furthermore, we introduce the necessary concepts to state Theorems 2.5 and 2.6, which will be proved in Sections 4 to 7.
Seifert spaces and Montesinos links. Let Γ be a plumbing graph, that is, a graph in which every vertex v i carries an integer weight a i , i = 1, ..., n. Associated to each vertex v i is the 4-dimensional disc bundle X → S 2 with Euler number a i . If the vertex v i has d i edges connected to it in the graph Γ, we choose d i disjoint discs in the base of X → S 2 and call the disc bundle over the jth disc B ij = D 2 × D 2 . When two vertices are connected by an edge, we identify B ij with B kl by exchanging the base and fiber coordinates and smoothing the corners. This pasting operation is called plumbing (for a more general treatment we refer the reader to [GS]), and the resulting smooth 4-manifold M Γ is said to be obtained by plumbing according to Γ. The Kirby diagram of M Γ has an unknot for each vertex of the tree, and whenever two vertices are joined by an edge the corresponding unknots will be linked forming a Hopf link. Each framing will be the Euler number of the corresponding D 2 -bundle.
The group H 2 (M Γ ; Z) has a natural basis represented by the zero-sections of the plumbed bundles. We note that all these sections are embedded 2-spheres, and they can be oriented in such a way that the intersection form of M Γ will be given by the matrix Q Γ = (q ij ) i,j=1,...,n with the entries 1 if i is connected to j by an edge; 0 otherwise.
We will call (Z n , Q Γ ) the intersection lattice associated to Γ.
Notice that, since M Γ is a 2-handlebody, any matrix representing the intersection form of M Γ is also a presentation matrix for H 1 (∂M Γ ; Z) (see e.g. Corollary 5.3.12 in [GS]). In particular, H 1 (∂M Γ ; Z) is finite if and only if det(Q Γ ) = 0, and in this case | det(Q Γ )| = |H 1 (∂M Γ ; Z)|. (2.1) The plumbing construction along a star-shaped plumbing graph Γ yields a 4manifold M Γ whose boundary Y Γ := ∂M Γ is a Seifert manifold (see [Ra] for a proof). Seifert manifolds are oriented, closed 3-manifolds admitting a fixed point free action of S 1 and they are classified by their "Seifert invariants" [OR, Se]. The unnormalized Seifert invariant of the manifold Y Γ is the collection of numbers, .., r}. This information can be read off from a star-shaped plumbing graph Γ as follows. First of all, the number r is precisely the number of legs of Γ. The number b is the weight of the central vertex. Finally, if the weights on the i-th leg are {−a 1,i , · · · , −a ki,i }, then the irreducible fraction αi βi is recovered from the continued fraction decomposition Notice that if a j,i ≥ 2 for every j, i, then we necessarily have 0 < β i < α i . Among the properties of continued fractions we will need Riemenschneider's point rule [Ri], which we now briefly recall. Let p > q > 0 be coprime integers, and suppose Then, the coefficients a 1 , ..., a ℓ and b 1 , ..., b k are related by a diagram of the form where the i-th row contains a i − 1 "points" for i = 1, ..., l, and the first point of each row is vertically aligned with the last point of the previous row. The point rule says that there are k columns, and the j-th column contains b j − 1 points for every j = 1, ..., k. For example for 17 7 = [3, 2, 4] and 17 10 = [2,4,2,2] the corresponding diagram is given by Remark 2.1. Given two strings of integers (a 1 , ..., a n ) and (b 1 , ..., b m ), we consider the following operations: (1) (a 1 , ..., a n ), (b 1 , ..., b m ) −→ (a 1 , ..., a n + 1), (b 1 , ..., b m , 2) (2) (a 1 , ..., a n ), (b 1 , ..., b m ) −→ (a 1 , ..., a n , 2), (b 1 , ..., b m + 1) It is straightforward to check that if we start with a 1 = (2), b 1 = (2) the strings obtained using the above described operations are related to one another by Riemenschneider's point rule.
The Kirby diagram of a star-shaped plumbing graph Γ consists of a link L in S 3 which is strongly invertible, i.e. there exists a π-rotation on S 3 that induces on each connected component of L an involution with two fixed points, see Figure 2. Particularizing the statement to 4-dimensional plumbings, we have the following result due to Montesinos [Mo,Theorem 3].

Theorem 2.2 (Montesinos). Consider the handle representation
where n is the number of vertices in Γ. If the n 2-handles are attached along a strongly invertible link in S 3 , then M Γ is a 2-fold cyclic covering space of D 4 branched over a 2-manifold.
The branching set in Theorem 2.2 is constructed as follows. Consider the strongly invertible link which represents the Kirby diagram of M Γ , take the half of the Kirby diagram under the symmetry axis, substitute the half-circles with bands with as many half-twists as indicated by the framing of the corresponding circle and glue all these bands to a rectangle as shown in Figure 3. This construction represents the Seifert space Y Γ = ∂M Γ as a double cover of S 3 branched over the boundary ML Γ = ∂B Γ . The link ML Γ is by definition a Montesinos link (first defined by Montesinos in [Mo2]). . Part (a) shows a star shaped plumbing graph and its associated Kirby diagram as a strongly invertible link. The bottom picture shows the branch surface in D 4 which is the image of Fix(u). This surface is homeomorphic to the result of plumbing bands according to the initial graph: the gray lines retract onto the sides of the rectangle. Remark 2.3. Note that, the surface B Γ is homeomorphic to the one obtained by plumbing twisted bands according to the plumbing graph Γ, which is how we introduced Montesinos links in Section 1.
The 2-bridge links are the Montesinos links arising when we consider connected linear plumbings, that is when the plumbing graph Γ has no distinguished central vertex because all vertices have valence ≤ 2. In this case the associated Seifert spaces Y Γ are lens spaces. We will use the classical notation L(p, q) for lens spaces and K(p, q) for the corresponding 2-bridge links. The numbers p > q > 0 are coprime integers and in this case the continued fraction expansion p q = [a 1 , ..., a ℓ ] gives the string of integers (a 1 , ..., a ℓ ) which are the weights of the vertices of the linear graph with opposite signs.
Definitions and notation concerning plumbing graphs. We consider Z n = Z ⊕ ... ⊕ Z as an intersection lattice with respect to the product · given in matrix form by where ·, · is the standard inner product of Z n . If we denote by e 1 , ..., e n the standard basis of Z n , we have e i · e j = −δ ij , ∀i, j = 1, ..., n.
We are interested in three-legged plumbing graphs P whose associated intersection lattice admits an embedding into Z n , where n = |P | is the number of vertices in the graph. The vertices of P , which from now on will be identified with their images in Z n and will also be called vectors, are indexed by elements of the set J := {(s, α)| s ∈ {0, 1, ..., n α }, α ∈ {1, 2, 3}}. Here, α labels the legs of the graph and L α := {v i,α ∈ P | i = 1, 2, ..., n α } is the set of vertices of the α-leg. We will write L α (P ) and n α (P ) when we want to point out the graph P to which these objects belong. The string associated to the leg L α is the n α -tuple of integers (a 1,α , ..., a nα,α ), where a i,α := −v i,α · v i,α ≥ 0. The three legs are connected to a common central vertex, which we denote indistinctly by v 0 = v 0,1 = v 0,2 = v 0,3 (notice that, with our notation, v 0 does not belong to any leg). We say that the legs L α , L β ⊆ P ⊆ Z n are complementary legs when their associated strings are related to one another by Riemenschneider's point rule.
The main assumption on P will be that the central vertex v 0 satisfies v 0 ·v 0 ≤ −3, while each other vertex v s,α satisfies v s,α · v s,α ≤ −2. Throughout the paper we will also deal with disconnected graphs with only one trivalent vertex. We will use the same notation for connected and disconnected graphs.
Given P ⊆ Z n we define the set of orthogonal matrices Υ P := {Ω ∈ O(n; Q)| Ωv s,α ∈ Z n for every v s,α ∈ P }.
This set has no group structure, nevertheless since its elements are orthogonal matrices, for every Ω ∈ Υ P the intersection graph ΩP := {Ωv s,α | v s,α ∈ P } is the same as the intersection graph of P and, by definition of Υ P , the set ΩP is again a subset of Z n . Notice that, for every P , the group O(n; Z) is a subset of Υ P and it contains the reflections across each hyperplane orthogonal to an e i , as well as all the transformations determined by the permutations of {e 1 , ..., e n }. The introduction of the set Υ P is due to the fact that we are interested in whether a plumbing graph admits an embedding, while the embedding itself is less relevant. Therefore, in the future we will usually identify P with ΩP where Ω ∈ Υ P . The number of connected components of the plumbing graph P will be denoted by c(P ) and we shall say that a vector v s,α ∈ P is isolated [resp. final] if it is an isolated vertex [resp. a leaf] of the plumbing graph. A vertex that is neither isolated nor final will be called internal. The only trivalent vertex in the graph is v 0 . We will assume it is internal and call it the central vertex.
Two vectors v, w ∈ Z n are linked if there exists e ∈ Z n with e · e = −1 such that v · e = 0 and w · e = 0. A set P ⊆ Z n is irreducible if, given two vectors v, w ∈ P , there exists a finite sequence v = v 1 , v 2 , ..., v k = w of vectors of P such that, for each i = 1, ..., k − 1, v i and v i+1 are linked. A set which is not irreducible is reducible.
Good and standard subsets of Z n . A possibly disconnected plumbing graph P ⊆ Z n is called good if it is irreducible and satisfies the following conditions.
• If it has one vertex of valence three, its incidence matrix has the form • if all its vertices have at most valence two (and in this case we call P a linear set), its incidence matrix has the form where again γ s,α ∈ {1, 0} and a s,α = −v s,α · v s,α ≥ 2. Furthermore, if P is a connected graph (i.e. all the γ s,α 's in its incidence matrix are equal to 1) we will say that P is standard. Standard and good linear sets were studied by Lisca in [Li].
The quantity I(·) and the family P. Given a subset P ⊆ Z n , the key quantity in our discussion will be the number Consider the family P containing all three-legged connected plumbing graphs such that: • The central vertex has weight less or equal to −3 and every non central vertex has weight less or equal to −2. • I(Γ) < −1 for every Γ ∈ P.
Remark 2.4. Notice that, for Γ ∈ P the matrix Q Γ is negative definite, since Γ is a canonical negative plumbing graph in the sense of [NR,Theorem 5.2].
As explained in Section 1 our first aim is to determine all the graphs P ∈ P whose associated intersection lattice admits an embedding in the standard negative diagonal intersection lattice. In the terminology that we have just introduced our aim is to determine all possible standard subsets P ⊆ Z n such that I(P ) < −1. For the linear case, the answer is known and can be found in Remark 3.2. For graphs with a trivalent vertex the complete list is given in Theorems 2.5 and 2.6 below. The proof of these theorems is developed in Sections 4 to 7.
The proof of Theorems 2.5 and 2.6, which is carried out in Sections 4 to 7, can be briefly sketched as follows. We start by defining an operation, the contraction, which given a good set P ⊆ Z n returns a subset of Z n−1 . Then we show that, under certain assumptions, the result of a contraction is again a good set P ′ and moreover, I(P ′ ) ≤ I(P ). Afterwards, we observe that complementary legs are preserved under contractions and that the contraction of a good set without complementary legs is again a set without complementary legs. Iterating contractions we prove that every good set with complementary legs can be contracted to one of the three good subsets of Z 5 which have I < −1. In turn, every good set without complementary legs can be contracted to one of the two good subsets of Z 3 with I < −1. Keeping track of the quantity I and of the number of connected components of the sets involved in the sequence of contractions leads to the complete classification of standard subsets.

Existence of Ribbon surfaces
Theorems 2.5 and 2.6 give the complete list of plumbing graphs P ∈ P whose associated intersection lattice admits an embedding in the standard negative diagonal intersection lattice. As explained in the introduction, this is a necessary condition for the corresponding Montesinos knots, ML P , to bound a slicing disc. In this section we shall find, for each ML P with P as in Theorem 2.5 or Theorem 2.6, a surface with boundary Σ such that χ(Σ) = 1 and a ribbon immersion i : Σ S 3 with i(∂Σ) = ML P . In this way we obtain that the slice-ribbon conjecture is true for all Montesinos knots ML P with P ∈ P. This section concludes with the proof of the main result of this work, Theorem 1.1.
3.1. Montesinos links with complementary legs. In this section we deal with the Montesinos links associated to the graphs P in Theorem 2.5. We start by showing that for each P there exists a cobordism between the Seifert space Y P and L(p, q)#(S 1 × S 2 ), where L(p, q) is a lens space bounding a rational homology ball. A theorem by Montesinos, Theorem 2.2, shows that this cobordism corresponds to a ribbon move (see [GS,p. 211] for the definition) on the link ML P .
Let Γ be a connected three-legged plumbing graph with two complementary legs L 2 and L 3 with associated strings (b 1 , ..., b k ) and (c 1 , ..., c l ) respectively. Let (a 1,1 , ..., a n1,1 ) be the string associated to the leg L 1 and −a 0 the weight of the central vertex. We recall that we write M Γ for the associated oriented 4-manifold and Y Γ := ∂M Γ for its oriented three dimensional boundary. For these graphs Γ we have the following result.
Proof. The first link of Figure 5 is the Kirby diagram of the graph Γ with an added thick circle with framing −1. Blowing down this (−1)-circle we obtain a split link which consists of two linear chains. Since L 2 and L 3 are complementary legs, the strings (b 1 , ..., b k ) and (c 1 , ..., c l ) are related to one another by Riemenschneider's point rule. Notice that then, whenever l + k > 2, we necessarily have either b 1 = 2 and c 1 > 2 or b 1 > 2 and c 1 = 2. By symmetry, let us suppose that the first case occurs. It is immediate to check, using the Riemenschneider's point diagram, that (b 2 , ..., b k ) and (c 1 − 1, ..., c l ) are again two strings related to one another by Riemenschneider's point rule. Moreover, if l + k − 1 > 2 then either b 2 = 2 and c 1 − 1 > 2 or b 2 > 2 and c 1 − 1 = 2. Therefore, in the third diagram of Figure 5 we have −b 1 + 1 = −1 and blowing down this circle produces a new diagram with a new (−1)-circle linked to the first circles of two complementary chains. After k + l blow downs (starting with the first diagram), we arrive to a diagram with two components: an unknot with framing 0 and the leg L 1 linked to a circle with framing −a 0 +1. This diagram represents a 4-manifold whose boundary is the connected sum of S 2 × S 1 and the lens space L(p, q) where p q = [a n1,1 , a n1−1,1 , ..., a 1,1 , a 0 − 1]. Observe that the 2-handle h, represented by the thick circle in the top link in Figure 5, is added to the Kirby diagram of a three-legged star-shaped plumbing graph P , in such a way that we obtain a strongly invertible link in S 3 (with respect to u). Before adding h, the involution u turns M P into the double cover of D 4 which preserves the boundary of the 4-manifold. Notice that all these operations are done equivariantly with respect to the involution u.
−an 1 ,1 −a0 +1 Figure 6. Branch surface of the involution u on the 4-manifold defined by the last Kirby diagram in Figure 5. The discontinuous lines show where is performed a connected sum between the disjoint union of two unknots and the 2-bridge link K(p, q), where p q = [a n 1 ,1 , ..., a 1,1 , a 0 − 1], yielding U ∪ K(p, q).
branched over a surface B P , which is obtained by plumbing bands according to P and satisfies ∂B P = ML P . Let us call M ′ the 4-manifold in the first Kirby diagram of Figure 5, obtained after adding h to M P . By Montesinos' theorem, Theorem 2.2, the involution u turns M ′ into the double cover of D 4 branched over a surface B ′ , which is obtained by adding a band with a half-twist to the surface B P . The addition of this band is a ribbon move on ML P . In order to understand the link obtained after this ribbon move we argue as follows. Given a tubular neighborhood of an unknotted component C of a link in S 3 , its complement is a solid torus T . As explained in [Ro,Chap. 9 Sect. H], a blow down operation done along C consists of applying a meridinal twist to T and this alters the rest of the link as shown in Figure 5. The important point here is that this operation as well as the isotopies in Figure 5 are done equivariantly with respect to u. Therefore, the surface B ′ can be thought of as obtained by plumbing twisted bands according to the last diagram in Figure 5 (instead of according to the first one). The result is ilustrated in Figure 6. Since the ribbon move turns ML P into the boundary of B ′ , we conclude that after this move we obtain a split link of the form U ∪ K(p, q), where U is the unknot and K(p, q) is the 2-bridge link given by the fraction p q = [a n1,1 , ..., a 1,1 , a 0 − 1]. Let P be a plumbing graph as in Theorem 2.5. Then, a case-by-case check shows that the graph S ⊆ Z n1+1 in Lemma 3.1 is one of the linear graphs listed in Remark 3.2.
Summarizing the discussion of this section we have the following.
Lemma 3.3. For a plumbing graph P as in Theorem 2.5, depending on the number of connected components of the links ML P and K(p, q), we have the following possibilities, where the ribbon move is the attachment of the band corresponding to the 2-handle added to M P in Lemma 3.1.
• If ML P is a knot, there exists a ribbon move that reduces it to a 2-bridge ribbon knot and an unknot. It follows that ML P is ribbon. • If ML P is a 2-component link, there exists a ribbon move that reduces it either to a 2-bridge ribbon knot and an unknot or to a 2-bridge ribbon link and an unknot. In any case it follows that ML P = ∂Σ, where Σ is a ribbon surface in S 3 , which is the disjoint union of a disc and a Möbius band. • If ML P is a 3-component link, there exists a ribbon move that reduces it either to a 2-bridge ribbon knot and an unknot or to a 2-bridge ribbon link and an unknot. In both cases it follows that ML P = ∂Σ, where Σ is a ribbon surface in S 3 . In the first case Σ is the disjoint union of a disc and an annulus and in the second case Σ is the disjoint union of either two Möbius bands and a disc or two discs and K \ D 2 , where K stands for the Klein bottle.
Proof. We write the details for the second case, and we leave the remaining two as a straightforward exercise for the reader. The proof is also sketched in Figure 7. If ML P is a 2-component link, there are 3 possibilities for a ribbon move on it: (i) If the band joins the two components, it yields a knot; , where U is the unknot and K(p, q) is a 2-bridge link. It follows that the first possibility for the ribbon move is excluded. The second possibility describes a cobordism, which is a surface with Euler characteristic −1, from ML P to U ∪K(p, q), where K(p, q) is a ribbon knot. Since U ∪ K(p, q) is the boundary of two disjoint discs, it follows that ML P is the boundary of the disjoint union of a Möbius band and a disc. The third possibility describes a cobordism, which is again a surface with Euler characteristic −1, from ML P to U ∪ K(p, q), where this time K(p, q) is a link, boundary of a ribbon surface consisting of the disjoint union of a Möbius band and a disc. It follows that U ∪ K(p, q) is the boundary of two disjoint discs and a Möbius band. Therefore, ML P is the boundary of the disjoint union of a Möbius band and a disc.

Montesinos links without complementary legs.
In this section we follow the same strategy as in the preceding one. We start by showing that adding one handle to the Kirby diagrams representing the 4-manifolds M P with P as in Theorem 2.6, we obtain a 4-manifold with boundary S 1 × S 2 . We show that this gives a presentation of S 1 × S 2 via a strongly invertible framed link, so Montesinos' theorem, Theorem 2.2, guarantees that there exists a surface with boundary Σ, with χ(Σ) = 1 and a ribbon immersion i : Σ S 3 with i(∂Σ) = ML P .
Lemma 3.4. All the 4-manifolds represented by the plumbing graphs in Theorem 2.6 can be changed into a 4-manifold with boundary S 1 × S 2 by adding a 2-handle along a circle with framing −1.
Proof. The families (a) with t > 0, (b), (c) and (e) of Theorem 2.6, are represented schematically in Figure 8, where a black square on a circle represents a possible linear plumbing linked to it. The thicker circle represents the added 2-handle. In this way, we can perform Kirby calculus on this general figure and then specialize it to the different families, by substituting x, y, z and the black squares with the corresponding framings and linear plumbings respectively. This is done in Figure 9, where (i) [resp. (ii) and (iii)] represents the last diagram in Figure 8 with the data from family (a) with t > 0 or from family (b) [resp. (c) and (e)]. Since the last diagram in Figure 8 is star-shaped, in Figure 9 we have used the graph notation to improve clarity. In each family the graph has a vertex with weight −1. Blowing down this −1 we obtain a new graph with a new vertex with weight −1. For every family this blowing down operation can be iterated (the first case in Figure 9 is done in full detail; the other two are approached analogously) until we are left with a graph with only one vertex of weight 0, which represents the 4-manifold D 2 × S 2 with boundary S 1 × S 2 . Since the blow downs do not change the boundary, the statement is proved for these families.
We are left with the families (a) when t = 0 and (d), which are represented schematically in Figure 10. The thicker circle represents, as before, the added 2handle. Family (a) with t = 0 has x = −3 and = ∅, while family (d) has x = −t − 3 and represents a linear plumbing of t circles, each of them with framing −2. Figure 10 shows that the addition of the thick handle with framing −1 turns the original 4-manifold into another one with boundary S 1 × S 2 . We have done in detail the case (a) when t = 0, the study of family (d) being analogous.
Remark 3.5. The 2-handle additions used in Lemmas 3.1 and 3.4 were suggested by the analysis of standard subsets of Z n done in order to proof Theorems 2.5 and 2.6. In fact, consider for example the graph P belonging to the family (a) in Theorem 2.6 with t = 1, s = 2, k = 2, and b 1 = b 2 = 2. One can easily find the following embedding in Z 9 .
Notice that, if we erase the basis vector e 3 we are left with a graph having a vertex, the one on the extreme left, with weight −1 and with corresponding generator sent to −e 5 by the embedding. By erasing this time the e 5 vector we obtain again a vertex with weight −1 in the resulting graph. It is possible to continue in this way until we are left, in this example, with the graph This "erasing"operation without a formal framework is meaningless, however it suggests a chain of blow downs, which turn the graph into a 4-manifold with boundary S 1 × S 2 . In fact, the thick circle added to the first diagram in Figure 8 intersects the Kirby diagram of the family (a) in Theorem 2.6 as suggested by the basis vector e 3 in this example. The same happens with all the plumbing graphs that we have studied.
Let M P be the 4-manifold corresponding to a plumbing graph P as in Theorem 2.6. We have seen in Section 2 that M P admits a Kirby diagram consisting of a strongly invertible link in S 3 with respect to an involution u. Let us call M ′ the x + 1 x + 1 x + 2 y + 1 z + 1 Figure 8. This diagram shows how to add a 2-handle with framing −1 to the families (a) with t > 0, (b), (c) and (e) of Theorem 2.6, in order to obtain a 4-manifold with boundary S 1 × S 2 . The framings x, y and z, and the linear plumbings represented by black squares differ in the four families. The arrows in the diagram represent either blow downs or isotopies. Figure 9. Graph (i) is obtained from the last diagram of Figure 8 by substituting x, y, z and the black squares with the data from the graph in family (a) with t > 0 or in family (b). Graphs (ii) and (iii) correspond respectively to considering in the last diagram of Figure 8 the data from families (c) and (e) respectively.
series of blow downs and Figure 6 x Figure 10. This diagram shows how to add a 2-handle with framing −1 (the thicker one) to the families (a) with t = 0, and (d) of Theorem 2.6, in order to obtain a 4-manifold with boundary S 1 × S 2 . Family (a) with t = 0 satisfies x = −3 and = ∅, while family (d) has x = −t − 3 and represents a (−2)-chain of length t. The vertical arrows in the diagram represent either blow downs or isotopies. The last two diagrams have been specialized to family (a) with t = 0. u u u Figure 11. These three strongly invertible links can be isotoped to the first diagrams in Figures 8 and 10 as follows (for sake of clarity we have omitted the framings). The first invertible link can be isotoped to the first diagram in Figure 8 when we consider on it the data from family (a) with t > 0 of Theorem 2.6. The second invertible link is easily isotoped to the first diagram in Figure 8 when we consider on it the data either from family (b), (c) or (e) of Theorem 2.6. Finally, the last invertible link is easily isotoped to the first diagram in Figure 10, when we consider on it the data either from family (a) with t = 0 or (d) from Theorem 2.6.
4-manifold with boundary S 1 × S 2 obtained from M P by adding a 4-dimensional 2-handle as in Lemma 3.4. As shown in Figure 11, this 2-handle can be added in such a way that we obtain again a strongly invertible link with respect to the involution u. As in Section 3.1, it follows from Montesinos' theorem, Theorem 2.2, that M P is the double cover of D 4 branched along a surface B P , which consists of bands plumbed according to the graph P . In turn, M ′ is the double cover of D 4 branched over the surface B ′ , which is B P with an additional band. The addition of this band corresponds to a ribbon move on the Montesinos link ML P = ∂B P . This ribbon move necessarily leads to two unlinked unknots, since ∂M ′ = S 1 × S 2 and, by [KT], whenever S 1 × S 2 double branch covers S 3 , the branch set is the unlink of two unknotted components. The discussion of this section proves the following result.
Lemma 3.6. For a graph P as in Theorem 2.6, depending on the number of connected components of ML P , we have the following possibilities: • If ML P is a knot, then it bounds a ribbon disc in S 3 .
• If ML P is a 2-component link, then it bounds a ribbon surface in S 3 , which is the disjoint union of a disc and a Möbius band. • If ML P is a 3-component link, then it bounds a ribbon surface in S 3 , which is the disjoint union of a disc and an annulus.
Proof. The result follows easily from elementary facts on the classification of surfaces with boundary, as in the proof of Lemma 3.3.
Having established the existence of the desired ribbon surfaces, we are now ready to prove the main result of this paper, modulo the technical proofs of Theorems 2.5 and 2.6, which will be carried out in the remaining sections.
Proof of Theorem 1.1. We start assuming that Y Γ = ∂W , where W is a rational homology ball. Since Γ ∈ P, the intersection form of the 4-dimensional plumbing M Γ is negative definite (see Remark 2.4) and hence, (2.3). Therefore, via the embedding M Γ ⊂ X Γ we can view the above basis, and hence Γ, as a standard subset of Z n . If Γ has two complementary legs [resp. no complementary legs] then it belongs to the list in Theorem 2.5 [resp. Theorem 2.6] and the existence of the surface Σ and of the ribbon immersion follows from Lemma 3.3 [resp. Lemma 3.6].
The arguments we use to prove the "only if" part of the statement coincide with those in [Li,Proof of Theorem 1.2,(2) implies (1)]. For the reader's convenience we include them here. Assume that there exist a surface Σ and a ribbon immersion Σ S 3 such that ∂Σ = ML Γ and χ(Σ) = 1. Let Σ ′ ⊂ D 4 be a smoothly embedded surface obtained by pushing the interior of Σ inside the 4-ball. The 2-fold covering Y Γ → S 3 branched over ML Γ extends to a 2-fold covering W → D 4 branched over Σ ′ (see [Ka,). We conclude by showing that W is a rational homology ball, note that Y Γ = ∂W . By definition of Σ ′ , we may assume that the function distance from the origin D 4 → [0, 1] restricted to Σ ′ is a proper Morse function with only index-0 and index-1 critical points. This implies that W has a handlebody decomposition with only 0-, 1-and 2-handles (see [CH,. Therefore, from

Contractions of good sets
In the rest of the paper, Sections 4 to 7, we carry out the proof of Theorems 2.5 and 2.6. In the current section we introduce some notation, define contractions of good sets and give some preliminary results.
Let P ⊆ Z n be a good set and suppose that E h (P ) = {(s, α), (t, β)} for some h ∈ {1, ..., n} and (s, α), (t, β) ∈ J. Then, we say that the subset P ′ ⊆ Z n−1 = e 1 , ..., e h−1 , e h+1 , ..., e n defined by is obtained from P by a contraction, and we write P ց P ′ . Moreover, we say that P is obtained from P ′ by an expansion, and we write P ′ ր P . For good subsets P ⊆ Z n with two complementary legs, L 2 and L 3 (their associated strings are related to one another by Riemenschneider's point rule), we extend the definition of contraction to the following operation. Suppose that for some i ∈ {1, ..., n} and for some (s, 1) ∈ J, we have E i (P ) = {0, (s, 1)} and let v t,1 be any final vector in L 1 . Since the leg L 2 is connected to the central vertex, we have v 1,2 · v 0 = 1, and therefore there exists k ∈ {1, ..., n} such that k ∈ V v0 ∩ V v1,2 and v 0 =ṽ 0 ± e k . We say that the subset P ′ ⊆ Z n−1 = e 1 , ..., e i−1 , e i+1 , ..., e n defined by is obtained from P by a contraction, and we write P ց P ′ . Moreover, we say that P is obtained from P ′ by an expansion, and we write P ′ ր P .
Example 4.1. The two examples in Figure 12 are extended contractions of sets with complementary legs. Following the above notation P is the set on the left of the arrow and P ′ the one on the right. Notice that the first example shows that the extended contraction of a standard set is again a standard set.
Once that we have settled all the necessary definitions, we introduce here several preliminary results concerning good subsets.
Lemma 4.2. The elements of a good subset P ⊆ Z n are linearly independent over Z.
Proof. If P is a linear set, the claim follows from [Li,Remark 2.1]. If, on the contrary P has a trivalent vertex, let us further assume that all the γ s,α 's in Q P are equal to 1 (the general case is a straightforward extension of this one). Then, Q P turns out to be the intersection matrix associated to the Seifert space By definition of Seifert invariants we have α i > 1; by definition of good subset it holds a 0 ≥ 3 and a k,i ≥ 2 for every k ∈ {1, ..., n i }, which implies β i /α i < 1 (see Section 2). It follows that the determinant of Q P is non zero, since by (2.2) Now, consider the n × n matrix M having as rows the coordinates of the elements of P with respect to the standard basis of Z n . This matrix satisfies Q P = −M M t and therefore | det M | = | det Q P | 1/2 = 0, which readily gives the claim.
Remark 4.3. Notice that Lemma 4.2 remains valid for reducible sets P ⊆ Z n whose incidence matrix Q P has the form (2.3) or (2.4). In fact, in this case the matrix M (see the proof of Lemma 4.2) is a diagonal block matrix, i.e.
Let us denote by P i ⊆ P the irreducible subset having incidence matrix Q Pi = −M i M t i . Then, by Lemma 4.2, the matrices M i are non singular, hence M and Q P = −M M t are nonsingular as well.
For the reader's convenience we now include two Lemmas from [Li]. More precisely, Lemma 4.4 corresponds to Lemma 2.5 in [Li], which gives important information on p 1 (P ) and p 2 (P ) coming from the assumption I(P ) < 0. On the other hand, Lemma 4.5 points out the most relevant properties of good linear sets with p 1 (P ) > 0. Its proof follows from Section 3 in [Li], although in Lemma 4.5 we have dropped the assumption I(P ) < 0.
In the proof of [Li,Proposition 3.3(3)] the assumption I(P ) < 0 is not used. Hence, we obtain (3). The proof of [Li,Corollary 3.5] goes through taking into account that the base case in the induction (n = 3) depends on the number λ, and it is immediate to check that the possible cases are the sets considered in (4).

Bad Components
In the forthcoming sections, in order to determine all possible standard subsets P ⊆ Z n with I(P ) < −1, we shall study acutely contractions of good sets. The idea is to choose these contractions in a suitable way, to obtain again good sets and this will be possible as long as the good sets have no "bad components". In this section we introduce this concept and establish some properties of bad components under set contractions.
Let n ≥ 3 and We perform on C, an arbitrary number of times and in any order, the following two expansions: • Right expansion with final (−2)-vector. This expansion, sketched below, can be performed on any connected linear set C ⊆ Z n , whenever there exists i ∈ {1, ..., n} such that E i (C) consists of the two final vertices in C.
A connected component C ⊆ Z n obtained from C in this way will be called a linear bad component, and we will denote by v * the vector v s,α ∈ C. The number of linear bad components in a set P will be denoted by b(P ).
Example 5.1. A linear bad component after the sequence "right expansion, left expansion, right expansion" is the following: In a similar fashion, we define bad components with a trivalent vertex. We start with an arbitrary linear bad component C = {v 1,α , ..., v k,α } ⊆ Z n and we attach two (−2)-vectors to v 1,α (or analogously to v k,α ), as shown in the following graph: As before, we now perform an arbitrary number of final (−2)-vector expansions on the length-one legs of the graph in (5.1). We call any connected component D ⊆ Z n obtained in this way a three-legged bad component and, as before, we denote by v * the vector v s,α ∈ D. Notice that every three-legged bad component can be thought of as obtained from the following set D, through a combination of the above described expansions: we erase e k from the central vertex and expand the horizontal leg in the graph D as if it was a linear bad component; afterwards, we add +e k to one of the final vectors and we proceed with the expansions of the two length-one legs.
We focus now our attention on the study of bad components. More precisely, in Lemma 5.2 we prove that standard sets have no bad components of any type and in Lemmas 5.3 and 5.4 we establish some properties of bad components under set contractions.
Lemma 5.2. Every three-legged standard subset P ⊂ Z n with n ≥ 5 and I(P ) < −1, has no bad components of any type.
Proof. We will argue by contradiction. Suppose that P has a bad component, then, since the graph P is standard and it has a valence three vertex, its only connected component must be a three-legged bad component. By definition, P is obtained from the set D in (5.2) by (−2)-vector expansions and it is immediate to check that I(P ) = I( D). The set P , being standard, is good and then D ⊂ Z 5 is good too (it is irreducible since c(P ) = c( D) = 1 and, by construction, the incidence matrix Q e D is of the form (2.3)). Therefore, with the notation of (5.2), there must exist e ℓ ∈ Z 5 and λ ∈ Z such that v * = e j + λe ℓ . Since, by definition, a * > 2 we have λ > 1 and hence we obtain I( D) ≥ −1, a contradiction.
In the following result we study under which conditions a good set with no bad components can develop a bad component after a contraction.
Lemma 5.3. Let P ⊆ Z n be a good set with no bad components of any type and suppose that there is a contraction P ց P ′ , where P ′ is a good set with bad components given by for some h ∈ {1, ..., n} and (s, α), (t, β) ∈ J. Then, P ′ has either one linear bad component or one three-legged bad component, but not both.
(3) Consider the restricted projection π e h : P \ v s,α → P ′ , let D ′ ⊆ P ′ be the bad component in P ′ and let D := π −1 e h (D ′ ). Then, v t,β ∈ D and v s,α is not orthogonal to D.
Proof. In P there are only four possible configurations for D, v s,α and v t,β , which we analyze separately: The vector v s,α is orthogonal to D, i.e. for all v ∈ D, v s,α · v = 0, and v t,β ∈ D. In this case D is a connected component and moreover D = D ′ , which contradicts the assumption that P has no bad components of any type.
The vector v s,α is orthogonal to D and v t,β ∈ D.
If v * = π e h (v t,β ) (see the definition of linear and three-legged bad components above) then we would have that D ⊆ P is a bad component and this contradicts the assumption of the lemma. Observe that v s,α · v t,β = 0 implies that |V vs,α ∩ V D ′ | ≥ 2. In fact, since v s,α and v t,β are orthogonal and h ∈ V vs,α ∩ V v t,β there must be another which by the above discussion has square v ′ s,α · v ′ s,α ≤ −2. It follows that the set (D ′ \ {v * }) ∪ {v ′ s,α } ⊂ Z |D ′ |−1 is a good set and therefore, by Lemma 4.2, its |D ′ | vectors are linearly independent, but this is not possible since they belong to the span of |D ′ | − 1 vectors.
The vector v s,α is not orthogonal to D and v t,β ∈ D. We use the same argument of the preceding case. The vector In fact, since there exists v ∈ D such that v · v s,α = 1, there must exist j ∈ V vs,α ∩ V v thus, taking into account that v * is internal in D (so v = v * ), the claim follows from the definition of bad component (either linear or three-legged). Now, considering as in the preceding case the good set (D ′ \ {v * }) ∪ {v ′ s,α } ⊂ Z |D ′ |−1 , we obtain |D ′ | linearly independent vectors in the span of |D ′ | − 1 vectors.
The vector v s,α is not orthogonal to D and v t,β ∈ D. Notice that this case is the only possibility left and that the three other cases lead to contradiction, implying (3). As a consequence we obtain that a single contraction P ց P ′ cannot produce neither two linear bad components, nor together a three-legged bad component and a linear bad component. Therefore, (2) holds.
Since v t,β ∈ D, the projection π e h (v t,β ) belongs to the bad component D ′ . By definition of bad component (linear or three-legged) we know that |V πe h (v t,β ) | ≥ 2 and then |V v t,β | > 2. It only remains to prove that |V vs,α | > 2 and to this aim we further consider two subcases: (i) Suppose first that v t,β · v s,α = 1 and let us put t = s − 1 (the case t = s + 1, possible only if D ′ is a linear bad component, can be handled analogously). If |V vs,α | = 2 and a s,α > 2 then v s,α = ±e h + λe j where j ∈ {1, ..., n} and λ ∈ Z, |λ| > 1. Thus, replacing the vectors v s,α and v t,β respectively with π e h (v s,α ) and π e h (v t,β ), we obtain a good set of n vectors whose associated incidence matrix is of the form (2.3) or (2.4). By Remark 4.3, this n vectors are linearly independent, but at the same time they belong to the span of the n−1 vectors {e 1 , ...,ê h , ..., e n }. This contradiction implies that if |V vs,α | = 2, then necessarily a s,α = 2. Let us consider the biggest ℓ ≥ 0 such that the set S := {v s,α , v s+1,α , ..., v s+ℓ,α } is connected. If a s,α = a s+1,α = ... = a s+ℓ,α = 2 then the set (P \ S) ∪ {π e h (v t,β )} is a good set of n − (l + 1) linearly independent vectors (see Lemma 4.2) in the span of n − (l + 2) vectors (notice that a length d chain of (−2)-vectors, which is connected to some other vector of square smaller that −2, is contained in the span of d + 1 basis vectors). This contradiction shows that there exists a smallest r ∈ {1, ..., ℓ} such that a s+r > 2 and it is easy to check that for some k ∈ {1, ..., n} V vs+r−1,α ∩ V vs+r,α = {e k } and |v s+r,α · e k | = 1.
} is a good set of n − r linearly independent vectors in the span of n − (r + 1) basis vectors. This contradiction yields |V vs,α | > 2 and therefore, if v s,α · v t,β = 1, then (1) holds.
(ii) Suppose now that v s,α · v t,β = 0. If V vs,α = {h, j} then, since v s,α is not orthogonal to D, it holds j ∈ V D ′ and we necessarily have that v s,α = ±e j ± e h . Thus, a s,α = 2 and V vs,α ∩ V v t,β = {h, j}. Now we have to distinguish if D ′ is a three-legged or a linear bad component. Let us begin assuming that D ′ is a threelegged bad component. Then, by its definition and since a s,α = 2, there are only two possibilities: the vector v s,α is attached to the final vector of one of the legs of D ′ which grow by final (−2)-vector expansions and v t,β is the final vector in the other one, or the vector v s,α is attached to the leg containing π −1 e h (v * ) and v t,β is the central vertex. In both cases we have that (D ∪ {v s,α }) ⊆ P is a three-legged bad component, contradicting the assumption of the lemma. In case D ′ is a linear bad component, by its definition and since a s,α = 2, there is only one possibility: the vector v s,α is attached to a final vector of D ′ and v t,β is the other final vector. Then, the set (D ∪ {v s,α }) ⊆ P is a linear bad component, contradicting again the assumption of the lemma. Therefore, if v s,α · v t,β = 0, we have that |V vs,α | > 2 as claimed.
The last result in this section shows how to overcome the difficulties with the bad components: if after a contraction a good set with no bad components develops a bad component, there is always another possible contraction that yields a good set with no bad components. It is a key result to prove Lemma 6.10.
Lemma 5.4. Let P ⊆ Z n be a good subset with no bad components of any type and I(P ) < −1. Assume that, for some i ∈ {1, ..., n} and (s, α), (t, β) ∈ J, the set obtained by the contraction is good with a bad component D ′ ⊆ P ′ , linear or three-legged. Then, interchanging the roles of v s,α and v t,β , the set is good with no bad components of any type.
Next, since the contraction P ց P ′ produces a bad component D ′ , we know by Lemma 5.3 that π ei (v t,β ) ∈ D ′ . We claim that π ei (v t,β ) = v * ∈ D ′ (see definition of bad component above). In fact, by Lemma 5.3 (3), v s,α · v = 1, for some v ∈ D, where D := π −1 ei (D ′ ), and hence there exists j ∈ {1, ..., n} such that j ∈ V vs,α ∩ V v . If π ei (v t,β ) = v * , then v t,β is internal in D and we conclude j = i. By definition of bad component (linear or three-legged) we have that |E j (D ′ )| ≥ 2, and so there exists w ∈ D ′ , w = v, v * , such that j ∈ V w and w · v s,α = 0. Thus, it follows that the vector has an associated incidence matrix ot the form (2.3) or (2.4), and by Remark 4.3 its |D ′ | vectors are linearly independent, contradicting the fact that they belong to the span of |D ′ | − 1 vectors. Therefore π ei (v t,β ) = v * , and consequently, by definiton of bad component, every vector v ∈ P linked to v t,β is also linked to v t−1,β or to v t+1,β . Thus, P ′′ is irreducible.
Once we have shown that P ′′ ⊆ Z n−1 is good, it remains to prove that it has no bad components of any type. Suppose by contradiction that there exists a bad component D ′′ ⊆ P ′′ , and consider first the case v s,α · v t,β = 0. Applying Lemma 5.3 to the contraction P ց P ′′ we obtain that π ei (v s,α ) ∈ D ′′ and that v t,β is not orthogonal to π −1 ei (D ′′ ); hence, D ′ ∩D ′′ = ∅. Notice that since π ei (v s,α ) ∈ D ′′ and v s,α ∈ D ′ it holds that D ′′ \ D ′ = ∅. Since in P there is only one trivalent vertex, at least one between D ′ and D ′′ is a linear bad component, say it is D ′′ . In D ′′ there are two final vectors: we denote by v be the one belonging to D ′′ ∩ D ′ and by w be the one in D ′′ \ D ′ . By definition of linear bad component, we know that there are two elements j, h ∈ V v ∩ V w and so the vector w := − j∈V D ′ (w · e j )e j has squarew ·w ≤ −2. Consider the vector v * ∈ D ′ and observe that, by definition of bad component (either linear or three-legged), we have that the set (D ′ \ {v * }) ∪ {w} ⊆ Z |D ′ |−1 is good. Thus, by Lemma 4.2, its |D ′ | vectors are linearly independent, which contradicts the fact that they belong to the span of |D ′ | − 1 basis vectors.
We are left with the case v s,α · v t,β = 1 and let us suppose, without loss of generality as explained before, that D ′′ is a linear bad component. If {i} = V vs,α ∩ V v t,β , then replacing v s,α and v t,β respectively with π ei (v s,α ) and π ei (v t,β ), we would obtain a set of n linearly independent vectors in the span of n − 1 basis vectors (recall that by Lemma 5.3, we have |V vs,α | > 2 and |V v t,β | > 2). Therefore {i} V vs,α ∩ V v t,β . Since the contraction P ց P ′ produces a bad component D ′ , we have, by Lemma 5.3, π ei (v t,β ) ∈ D ′ . Analogously, applying Lemma 5.3 to the contraction P ց P ′′ we obtain π ei (v s,α ) ∈ D ′′ . Then, the definition of bad component guarantees that for every j ∈ {1, ..., n} we have |e j · v s,α |, |e j · v t,β | ≤ 1.
Since v s,α · v t,β = 1, it follows that |V vs,α ∩ V v t,β | ≥ 3 and then, since π ei (v t,β ) ∈ D ′ we have that |V D ′ ∩ V vs,α | ≥ 2. Applying Lemma 5.3 to the contraction P ց P ′′ we obtain that π ei (v s,α ) ∈ D ′′ is a final vector (since v s,α · v t,β = 1). Therefore, in this case the vector v s,α has the same properties as the vector v of the preceding case. Calling w the other final vector in D ′′ we can repeat, word for word, the argument of the preceding case arriving to a contradiction. Therefore, the good set P ′′ has no bad components of any type and the lemma is proved.
Remark 5.5. The author of the present paper discovered that, although the statement of [Li,Proposition 5.3] is correct, the arguments used to prove the claim "S ′ has no bad components" (p. 450, line 8) are incorrect. Lemma 5.4 can be used to prove such claim.

Determination of good sets
The aim of this section is to determine all good sets P with I(P ) < −1. We start by reformulating the definition of complementary legs and by showing how to construct from a linear good set a three-legged good set with two complementary legs. Note that, by (4.1), good sets P with I(P ) < −1 satisfy that p 1 (P ) or p 2 (P ) is greater than zero. We consider in the first subsection the easier case of good sets satisfying p 1 (P ) > 0, while in the second one we analyze good sets with p 1 (P ) = 0. Finally in the last part we consider the general case and conclude with the main result of this section, which is Proposition 6.16.
In Section 2, the definition of complementary legs is given in terms of their associated strings. For our purposes, it is now more convenient to bear in mind that L α , L β ⊆ P ⊆ Z n are complementary legs if they can be obtained as a sequence of final (−2)-vector expansions of the length-one legs L α := {v 1,α = −e k + e h } and L β := {v 1,β = −e k − e h }, where L α and L β are defined up to the action of an element of O(n; Z). Notice that L α and L β have associated strings (2) and (2) respectively. The final (−2)-vector expansions change the strings (2) and (2) as the operations described in Remark 2.1. Therefore, the strings associated to the legs L α and L β are related to one another by Riemenschneider's point rule. Thus, they are complementary legs according to the definition given in Section 2. Notice that, by definition, in every three-legged bad component there are two complementary legs. Moreover, the expansion of the set (5.2) along its horizontal leg corresponds to the expansion of a set with complementary legs defined in Section 2.
Given a good linear set P ′ , we can construct a good set P with a trivalent vertex by adding two complementary legs to P ′ . This construction produces a large family of good sets with a central vertex and, if we further require that P has no bad components of any type and that I(P ) < −1, we have a complete description of this family, as shown in the following lemma.
Lemma 6.1. Let n = n 1 + n 2 + n 3 + 1 and P n ⊆ Z n be a good set without bad components of any type, with two complementary legs, L 2 and L 3 , and such that I(P n ) < −1. Then, The set S n1+1 := L 1 ∪ {ṽ 0 } is a linear good subset of Z n1+1 , I(S n1+1 ) < 0, n 1 ≥ 2 and there exists a sequence of contractions S n1+1 ց S n1 ց · · · ց S 3 such that, for each k = 3, ..., n 1 , the set S k is good without bad components.
(5) If P n is a standard set, then every set in the sequence P n , P n−1 , ..., P 5 , built in (4) is standard too.
Proof. Since L 2 and L 3 are complementary legs, there exists a sequence of contractions to the legs L 2 = {v 1,2 = e j + e k } and L 3 = {v 1,3 = e j − e k } for some j, k ∈ {1, ..., n}. (1) holds. By definiton of complementary legs, a simple calculation yields |V L2∪L3 | = n 2 +n 3 and I(L 2 ∪ L 3 ) = −2. From the first equality it follows that S n1+1 = L 1 ∪ {ṽ 0 } ⊆ Z n1+1 is a good set. Since P n = L 1 ∪ {v 0 } ∪ L 2 ∪ L 3 and I(·) is additive under set union, we have I(P n ) = I(L 1 ∪ {v 0 }) − 2 < −1 and a straightforward computation gives then I(S n1+1 ) < 0. Since P n is a good set, we have n 1 ≥ 1. If n 1 = 1, then S 2 should be a standard set of Z 2 satisfying I(S 2 ) < 0, which is easily seen to be impossible. Since P n has no bad components of any type, the set S n1+1 has no bad components and hence it fulfills the hypothesis of [Li,Corollary 5.4] and (2) holds.
Assertions (4) and (5) follow from the above arguments. In fact, the sets P n , ..., P n1+3 are the contractions of the two complementary legs, which satisfy (I(P i+1 ), c(P i+1 )) = (I(P i ), c(P i )) for i ∈ {n 1 + 3, ..., n − 1}, and the set P n1+3 is L 1 ∪ {v 0 } ∪ L 2 ∪ L 3 . All these sets are good without bad components. The rest of the sets in the sequence, namely P n1−1+3 , ..., P 5 , are obtained using: • [Li, Theorem 6.4] applied to S n1+1 , if this set is standard, which implies that P n is standard and (5) follows; • [Li, Corollary 5.4] applied to S n1+1 , if this set is only good, and in this case (4) follows. In both cases we must take into account the following consideration: if in the contraction given by [Li,Theorem 6.4] or by [Li,Corollary 5.4] we discard the vectorṽ 0 , then we must add ±e j = v 0 −ṽ 0 to any final vector in the contracted set, that will then play the role of central vertex with two complementary legs attached (see Example 4.1). In this way we obtain the desired sequence of good sets, with no bad components of any type. 6.1. Good sets with p 1 > 0. Along this subsection we will say that a subset P ⊆ Z n satisfies the working assumptions when (w1) n = n 1 + n 2 + n 3 + 1 ≥ 5, (w2) P = {v 0 , v 1,1 , ..., v n3,3 } is a good set with a trivalent vertex and no bad components of any type, (w3) I(P ) < −1, (w4) p 1 (P ) > 0. The aim of this subsection is to prove Propositions 6.2 and 6.3, which imply that a set P fulfilling the working assumptions is necessarily standard, and can be obtained from the standard subset of Z 5 given in Proposition 6.2 by a finite sequence of expansions.
Proposition 6.2. Let n = 5 and P ⊆ Z n be a good subset with p 1 (P ) ≥ 1 and I(P ) < −1. Then, I(P ) = −4 and, up to replacing P with ΩP where Ω ∈ Υ P , the plumbing graph P with its embedding in the standard diagonal lattice is: Proposition 6.3. Let P n be a set satisfying the working assumptions. Then, P n is standard and there is a sequence of contractions P n ց P n−1 ց · · · ց P 6 ց P 5 such that P k is standard and I(P k ) = −4, for every k = 5, ..., n.
Proof. The set P , being good, is irreducible and therefore we have |V vs,α | ≥ 2. We claim that |V vs,α | = 2. In fact, assume by contradiction that |V vs,α | > 2, and consider the set P ′ obtained from P by replacing v s,α with π ei (v s,α ). Thus, P ′ is contained in the span of the n − 1 vectors e 1 , ..., e i−1 , e i+1 , ..., e n . However, since π ei (v s,α )·π ei (v s,α ) ≤ −2, the matrix Q P ′ still has the form (2.3) and, by Remark 4.3, P ′ consists of n linearly independent vectors, which gives a contradiction. Next, by assumption i ∈ V vs,α , and therefore there exists j = i such that V vs,α = {i, j}. Notice that |v s,α · e j | = 1. In fact, we have |v s,α · e j | ≥ 1, and if the last inequality is strict, replacing v s,α with π ei (v s,α ) in P we obtain the same contradiction as before.
On the other hand, since P is irreducible and E i (P ) = {(s, α)}, the vector v s,α is not isolated. Let us assume that v s,α is final, and in particular v s,α · v s+1,α = 0 (the case v s−1,α · v s,α = 0 can be handled analogously). Let ℓ ∈ {1, ..., s} be the largest number such that the set S := {v s−1,α , ..., v s−ℓ,α } has a connected intersection graph. Then, there exists some h ∈ {1, ..., l} such that a s−h,α > 2. In fact: • if ℓ = s, we can take h = s = ℓ since, by (2.3), we know a 0 = a s−h,α ≥ 3; • if ℓ < s and there is no h ∈ {1, ..., ℓ} such that a s−h,α > 2, we would have a s−1,α = ... = a s−ℓ,α = 2. Then, being P irreducible, there would exist an element of V S also belonging to V P \S . Since the graph associated to S is a chain of (−2)-vectors disconnected from the rest of P , we actually have V S ⊆ V P \S , and hence V vs,α ⊆ V P \S too, contradicting |E i (P )| = 1.
, by eliminating the vectors v s,α , v s−1,α , ..., v s−h+1,α and replacing v s−h,α with π e k (v s−h,α ), we obtain a set of n − h linearly independent vectors contained in the span of n − (h + 1) vectors. This contradiction shows that v s,α cannot be final, that is, it must be internal.
The fact that v s,α is not central is guaranteed by the inequality I(P ) < 0, as explained in the following. Indeed, if v s,α is the central vertex, then P := P \ {v s,α } is a set with at least three connected components and moreover, it is a good linear set with I( P ) < −1, (recall that, by (w3), I(P ) < −1 and that the central vertex has square at least −3). We claim that P has no linear bad components. In fact, if P has a linear bad component C = {v c1 , ..., v c k }, then, since, by (w2), P has no bad components, there must be an element of C connected to the central vertex v 0 ∈ P , let it be v c1 . Notice that j is the only possible element in the intersection V vc 1 ∩ V v0 . By the definition of C, the index j belongs to V vc k too, and we obtain the contradiction v 0 · v c k = 0. The fact that P has no linear bad components, in turn, contradicts the fact that c( P ) ≥ 3. In fact, the hypothesis of [Li2,Lemma 4.9] are fulfilled and we obtain that c( P ) ≤ 2. This proves that v s,α is not central and implies that j belongs to V vs−1,α ∩ V vs+1,α and, since |v s,α · e j | = 1, we conclude |v s−1,α · e j | = |v s+1,α · e j | = 1.
Given a set P n ⊆ Z n satisfying the working assumptions we show in Lemma 6.5 how to contract the leg L α , which contains the vector v s,α satisfying E i (P ) = {(s, α)}, in order to obtain a new good set, P n−1 ⊆ Z n−1 , that still satisfies the working assumptions. The notation used in Lemma 6.5 is the same as the one of Lemma 6.4. Lemma 6.5. Consider a subset P n ⊆ Z n satisfying the working assumptions and put λ := |e i · v s,α |. Then we have a s−1,α = 2, a s+1,α > 2, or a s−1,α > 2, a s+1,α = 2

Let us assign
Then, if n α > 2, the subset of Z n−1 = e 1 , ..., e i−1 , e i+1 , ..., e n P n−1 : where j is the index given by Lemma 6.4 (2), satisfies the working assumptions. Moreover, we have Proof. To begin with, notice that precisely one between a s−1,α and a s+1,α is equal to 2 (so that the other one is strictly greater than 2). In fact, if both a s−1,α and a s+1,α are equal to 2, then we have, since v s−1,α · v s+1,α = 0, that V vs−1,α = V vs+1,α and therefore that P n is reducible for n > 3, contradicting our working assumptions. On the other hand, if a s−1,α , a s+1,α > 2, erasing v s,α from P and replacing v s−1,α and v s+1,α respectively with π ej (v s−1,α ) and π ej (v s+1,α ), we get a set whose associated incidence matrix is of the form (2.3). Therefore, by Remark 4.3, its n − 1 elements are linearly independent, but at the same time they should belong to the span of n − 2 vectors, which gives a contradiction. Now, notice that the set has an associated intersection matrix Q Pn−1 of the form (2.3). Since E j (P n ) = {(p, α), (s, α), (q, α)}, we have that E j (P n−1 ) = {(p, α)} and, since π ej (v q,α ) and v p,α + (λ − v s,α · e j )e j are linked to each other, the set P n−1 is irreducible. The verification that I(P n ) = I(P n−1 ) is straightforward and therefore, P n−1 satisfies the working assumptions. We now show that V Lα ∩ V L β ∪Lγ = ∅. Since P n−1 still satisfies the working assumptions, by the above procedure we can construct P n−2 . We iterate this process until we obtain a set P whose leg L α ( P ) has length 2. At this point |V Lα( e P ) | = 3, and in fact, up to replacing P with Ω P where Ω ∈ Υ e P , we have L α do not belong to the other two legs. In order to conclude, we just need to remark that, along the reduction process, we have always discarded vectors that only appeared in the leg L α . The last assertion, namely |V Lα(Pn) | = n α (P n ) + 1 is also immediate from the reduction procedure: at each stage we have n α (P n ) = n α (P n−1 ) + 1 and |V Lα (P n )| = |V Lα (P n−1 )| + 1. Since |V Lα( e P ) | = 3 and n α ( P ) = 2 the claim follows.
In the previous result we have shown how to contract the leg L α with the vector e i . The following one, in turn, explains how to contract the other two legs of a set satisfying the working assumptions.

Proof. Consider the vectorŝ
We claim that |Vv 0 | = 1. In fact, since the legs L 2 and L 3 are connected to the center of the graph we have v 0 · v 1,2 = v 0 · v 1,3 = 1 and therefore |Vv 0 | ≥ 1. If the last inequality were strict, the linear set S := L 2 ∪ L 3 ∪ {v 0 } would be good and it would consist of n 2 + n 3 + 1 (linearly independent) vectors lying in the span of n 2 + n 3 vectors 2 , which gives a contradiction. Denoting by j the only element in Vv 0 , sincev 0 · v 1,2 = 1 we have |v 0 · e j | = 1.
In order to conclude we just have to notice that the contraction performed on the set S ′ can be done on P n , obtaining the set P n−1 in the statement that fulfills the working assumptions: we have not changed the leg L α (P n ) and therefore p 1 (P n−1 ) ≥ 1 and I(P n−1 ) = I(P n ) < −1.
We now have all the elements that we need to prove Propositions 6.2 and 6.3.
Notice that, if we impose that the vector v s,α is not the central vertex, then we can relax assumption (w3) and the non existence of bad components in (w2). Indeed, the proofs of Lemmas 6.5 and 6.6 do not use these hypotheses, while in Lemma 6.4 they are used only to guarantee that v s,α = v 0 . Therefore, we readily obtain the following statement.
Lemma 6.8. Let n ≥ 5 and P n ⊆ Z n be a good set with a trivalent vertex such that there exist i ∈ {1, ..., n} and (s, α) ∈ J that satisfy E i (P n ) = {(s, α)}. If v s,α is not the central vertex, then • P n has two complementary legs, L β ,L γ = L α , and V L β ∪Lγ ∩ V Lα = ∅.
• There exists j ∈ {1, ..., n}, j = i, such that the central vertex v 0 can be written v 0 =ṽ 0 + e j , where e j ∈ V L β ∪Lγ and Vṽ 0 ⊆ V Lα . • The set L α ∪ṽ 0 is a good linear set that satisfies the assumptions of Lemma 4.5.
6.2. Good sets with p 1 = 0. It follows from Lemma 4.4 that if a subset P ⊆ Z n of cardinality n satisfies I(P ) < 0 and p 1 (P ) = 0, then necessarily p 2 (P ) > 0.
Having already dealt with the case p 1 (P ) > 0, I(P ) < −1, now we approach the more difficult case of a good subset with p 1 (P ) = 0, p 2 (P ) > 0 and I(P ) < −1.
(3) v s,α · v t,β = 0, v s,α is not internal, |V v t,β | > 2, the set is a good set with no bad components of any type and I(P ′ ) ≤ I(P ).
If v s,α is internal in P , then (2) holds. Suppose now that v s,α is isolated and notice that in this case we have v t,β = v 0 . In fact, suppose by contradiction that v t,β = v 0 . Then, since E i (P ) = E j (P ) = {(s, α), 0}, the set P 2 := P \ {v s,α , v 0 } ⊆ Z n−2 is a good linear set: its incidence matrix obviously satisfies (2.4) and it is irreducible because |V v0 | = 3. Hence, every v ∈ P different from v s,α and linked to v 0 = v t,β is also linked to v 1,δ for each δ ∈ {1, 2, 3}. Moreover, the set P 2 has no linear bad components, since b(P ) = 0 and P has no three-legged bad components. Moreover, it satisfies I(P 2 ) < 0, but at the same time, it must have at least three connected components, which contradicts [Li2,Lemma 4.9]. Now, since v t,β = v 0 and |V v t,β | = 3 we can define P ′ := (P \ {v s,α , v t,β }) ∪ {π ei (v t,β )} whose incidence matrix Q P ′ is of the form (2.3) or (2.4). Since every v ∈ P linked to v s,α must satisfy v · e j = 0, P ′ is irreducible. A straightforward computation gives I(P ′ ) ≤ I(P ) and, since a s,α = 2, by Lemma 5.3, P ′ has no bad components of any type and therefore (3) holds. Now, we analyze the case in which v s,α is a final vector and let us suppose v s−1,α · v s,α = 1 and v s,α · v s+1,α = 0 (the other case is analogous). To begin with, let us further assume that v t,β = v 0 . In this case we can define the set P ′ := (P \ {v s,α , v t,β }) ∪ {π ei (v t,β )} that has all the desired properties, just like when we considered v s,α isolated, and so (3) holds. We are now left with the study under the assumption v t,β = v 0 . This time, since v 0 · v 1,β = 1 for some β = α and V v0 = {i, j, h}, we have |v 0 · e h | = 1. We analyze separately the following two possibilities: (i) s = 2 and V v2,α ∩ V v1,α ∩ V v0 = j. If furthermore a 0 > 3, we necessarily have that |v 0 · e i | = |v 0 · e j | ≥ 2, and then the set P ′ := (P \ {v 2,α , v 0 }) ∪ {π ei (v 0 )} has, just like before, all the desired properties and so (3) holds. Suppose now that a 0 = 3, then P has a three-legged bad component, as explained in what follows. If V v1,α = {j, k} then E k (P ) = {(1, α)} and hence p 1 (P ) > 0, which contradicts our assumptions. Therefore, |V v1,α | > 2 and so a 1,α > 2. Consider the set P 3 := (P \ {v 2,α , v 1,α , v 0 }) ∪ {π ei (v 0 )} ⊆ Z n−1 . The associated incidence matrix Q P3 is of the form (2.4) (P 3 is a linear set). Thus, p 1 (P 3 ) = 1 (since E j (P 3 ) = {(π ei (v 0 ))}) and a direct calculation gives I(P 3 ) < −1. On the one hand, if P 3 is an irreducible set it is good, and therefore Lemma 4.5 applies and (4) gives that P 3 is obtained by final (−2)-vector expansions and therefore P has a three-legged bad component. On the other hand, if P 3 is reducible, there exist P 1 3 and P 2 3 such that P 3 = P 1 3 ∪ P 2 3 with V P 1 3 ∩ V P 2 3 = ∅. Let us suppose, without loss of generality, that π ei (v 0 ) ∈ P 1 3 . Since P is irreducible we have that P 1 3 is a linear irreducible set, and therefore it is good. As before, we apply Lemma 4.5 to the set P 1 3 and by (4) we get that it is obtained by final (−2)-vector expansions. Hence, we conclude again that P has a three-legged bad component, contradicting the assumption of the lemma.
(ii) E j (P ) = {(s − 1, α), (s, α), 0} and v s−1,α · v 0 = 0. In this case we have that |V vs−1,α | > 2. In fact, since v s−1,α ·v 0 = 0 and j ∈ V vs−1,α ∩V v0 , we need h ∈ V vs−1,α . Moreover, since for some β = α we have h ∈ V v 1,β and v s−1,α · v 1,β = 0, there must be some k ∈ {1, ..., n}, k = j, h such that k ∈ V vs−1,α . Therefore, |V vs−1,α | ≥ 3 and the set P 4 := (P \ {v s,α , v s−1,α , v 0 }) ∪ {π ej (v s−1,α )} is a linear good set with I(P 4 ) < −1 and at least three connected components. Let us check that the set P 4 has no linear bad components. Indeed, arguing as in the proof of Lemma 6.4, if P 4 has a linear bad component C = {v c1 , ..., v c k }, then, since P has no bad components, there must be an element of C connected to the central vertex v 0 ∈ P , let it be v c1 . Notice that h is the only possible element in the intersection V vc 1 ∩ V v0 . By the definition of C, h belongs to V vc k and we obtain the contradiction v 0 · v c k = 0. Thus, the assumptions of [Li2,Lemma 4.9] are fulfilled and and it follows c(P 4 ) ≤ 2, a contradiction.
Fourth case: v s,α · v t,β = 1 and a t,β > 2. Again in this case we have β = α and by symmetry we may assume t = s − 1. Since p 1 (P ) = 0, the vector v s,α is not final and this implies j ∈ V vs−1,α . In fact, if j ∈ V vs−1,α we get a contradiction as in the previous case by considering the biggest l ≥ 0 such that {v s,α , ..., v s+l,α } has connected plumbing graph.
Notice that in this case we do not need to worry about v s−1,α being the central vertex. In fact, if this was the case, then v s,α would be v 1,α and hence P ′ would be a linear good set, for which π ei (v s−1,α ) · π ei (v s−1,α ) ≤ −2 is guaranteed from the above discussion.
The vector v s,α is the central vector, i.e. v s,α = v 0 , and in P there are two complementary legs.
Remark 6.11. The argument used in (ii) in the proof of Lemma 6.10 can be used to fix a wrong claim in the proof of [Li,Lemma 4.3]. More precisely, the vector v s+1 , considered in [Li,last line,p. 445,Lemma 4.3] is internal in S ′′ l , contrary to what is claimed in [Li], and hence the claimed contradiction is not achieved. The argument used in (ii) provides the desired contradiction, since in the case "v s is not final" [Li,line 20,p. 445,Lemma 4.3], it implies that the set S of the statement of [Li,Lemma 4.3] has a bad component, contrary to the assumptions.
6.3. The general case. In this part we use all the work done in Sections 6.1 and 6.2 in order to prove that any good set P with no bad components of any type and I(P ) < −1 has I(P ) ∈ {−2, −3, −4} and is obtained by a sequence of expansions from a subset of Z k , where k ∈ {3, 5}. The main result is Proposition 6.16, however, a considerable part of its proof is developed before in Proposition 6.14. The proof of Proposition 6.16 works by induction and the initial case is studied in the following lemma.
Lemma 6.13. Let P ⊆ Z 5 = e 1 , e 2 , e 3 , e 4 , e 5 be a good set with a trivalent vertex and I(P ) < −1. Then, P is, up to replacing P with ΩP where Ω ∈ Υ P , one of the following graphs: In [Li] the statement of the Lemma refers to linear good sets, nevertheless the proof does not use the linearity of the set and it holds word for word in our case. (3) Moreover, I(P ) ∈ {−4, −3, −2}.
Third, a 0 ≥ 5. Since I(P ) < −1 and n = 5, we have a 0 ≤ 5 and therefore we need only to consider the case where a 0 = 5 and all the other vectors have square equal to −2. These conditions are incompatible with the fact that P ⊆ Z 5 is a good set. In fact, in this case for every v ∈ P it holds |v · v 0 | = 1.
Proposition 6.14. Let n ≥ 5 and P ⊆ Z n be a good set with a trivalent vertex, no bad components of any type and such that I(P ) < −1. Then, the following hold: (1) There exists a sequence of contractions P n := P ց P n−1 ց · · · ց P k , where k ∈ {3, 5}, such that for all i ∈ {k, k + 1, ..., n} the set P i ⊆ Z i is good with no bad components of any type and, moreover, (3) For every j ∈ {1, ..., n} and every (s, α) ∈ J, we have |v s,α · e j | ≤ 1.
Proof. We argue by induction on n ≥ 5. If n = 5 the whole statement is a straightforward consequence of Lemma 6.13. Therefore, from now on we assume n > 5. Since I(P ) < −1, by Lemma 4.4 inequality (4.1) holds and therefore, either p 1 (P ) > 0 or p 2 (P ) > 0. If p 1 (P ) > 0 then P satisfies the working assumptions of Section 6.1 and then, (1) and (2) follow from Proposition 6.3 (recall that by Lemma 5.2 a standard set has no bad components), while (3) holds by Remark 6.7 (1). Thus, we can assume that p 1 (P ) = 0 and p 2 (P ) > 0 and so Lemma 6.9 or Lemma 6.10 apply. If either Lemma 6.9 (1) or Lemma 6.10 (3) holds, then in P there are two complementary legs, the assumptions of Lemma 6.1 are fulfilled and (1) and (3) follow. Suppose now that in P there are no complementary legs. If such is the case, by Lemma 6.12 there is a contraction of P that gives a good set P ′ ⊂ Z n−1 with no bad components of any type and such that −1 > I(P ) ≥ I(P ′ ).
Next, starting with P n := P we shall define a decreasing sequence of contractions of good sets without bad components of any type P n ց P n−1 ց · · · ց P k where P i ⊆ Z i for every i ∈ {k, k + 1, ..., n}, k ∈ {3, 5} and I(P i+1 ) ≥ I(P i ) for every i = k, k + 1, ..., n − 1. In this way, we shall obtain (1). We define P n−1 := P ′ and continue the sequence as follows: If P n−1 is a linear set and n − 1 ≥ 3, then the assumptions of [Li,Corollary 5.4] are satisfied and we get the contractions P n−1 ց P n−2 ց · · · ց P 3 . All the sets involved are good with no bad components and moreover, it holds I(P i+1 ) ≥ I(P i ) for i = 3, 4, ..., n − 1.
If P n−1 ⊆ Z n−1 has a trivalent vertex then, in order to define P n−2 and the rest of the sequence we must take into account: (i) If n − 1 = 5 we stop and the sequence finishes with P n−1 , which by Lemma 6.13 satisfies I(P n−1 ) ∈ {−4, −3, −2}.
(ii) If n − 1 ≥ 6 and p 1 (P n−1 ) > 0 then, by Proposition 6.3, P n−1 is a standard set, I(P n−1 ) = −4 and we can finish the sequence with the contractions of standard sets P n−1 ց P n−2 ց · · · ց P 5 . Thus, we have a sequence of good sets with no bad components of any type (recall that by Lemma 5.2 a standard set has no bad components) and such that I(P i+1 ) ≥ I(P i ) for i = 5, 6, ..., n − 1.
• If P n−1 has two complementary legs then, since it has no bad components of any type, by Lemma 6.1 (4) we can finish the sequence with the contractions P n−1 ց P n−2 ց · · · ց P 5 . In this way we have built the desired sequence of contractions of good sets without bad components of any type and such that I(P i+1 ) ≥ I(P i ) for i = n − 1, n − 2, ..., 5. • If P n−1 has no complementary legs then, since it has no bad components of any type, by Lemma 6.12 there is a good subset with no bad components P n−2 ⊆ Z n−2 such that I(P n−1 ) ≥ I(P n−2 ). In order to define the set P n−3 ⊆ Z n−3 and the rest of the sequence we make P n−2 play the role of P n−1 in the above argument.
Observe that, in the sequence of good sets without bad components P n ց P n−1 ց · · · ց P k there are two possibilities for k. If all the sets P i have a trivalent vertex, then k = 5 and thus, by Lemma 6.13, I(P 5 ) ≥ −4. If, on the contrary, there is a contraction P i+1 ց P i such that P i is a linear set, then k = 3 and thus, by [Li,Corollary 5.4], I(P 3 ) ≥ −3. In either case, the following inequality holds, n−1 i=k (I(P i+1 ) − I(P i )) = −I(P k ) + I(P n ) ≤ 4 − 2 = 2. (6.1) Notice that, by construction, the sets in the sequence satisfy I(P i+1 ) ≥ I(P i ) for each i = k, k + 1, ..., n − 1. Therefore, on the one hand, since by assumption I(P = P n ) < −1 and we have shown that I(P k ) ≥ −4, we have (2), i.e. I(P ) ∈ {−4, −3, −2}. On the other hand, I(P i+1 ) − I(P i ) ≥ 0 for every i, and inequality (6.1) implies 0 ≤ I(P i+1 )−I(P i ) ≤ 2 for every i, and in particular I(P n )−I(P n−1 ) ≤ 2.
The following Lemma 6.15 is an important consequence of Proposition 6.14 (2). It allows us to divide three-legged good subsets P ⊆ Z n with I(P ) < −1 into two subclasses closed under contraction, namely those with complementary legs and those without them. In fact, the following lemma guarantees that complementary legs are"invariant"under contractions: we already know that they can be contracted to length-one complementary legs and we are about to prove that if in a set P there are no complementary legs they will not appear after a contraction. Lemma 6.15. Let P ⊆ Z n be a good subset with no bad components of any type and I(P ) < −1. Furthermore, suppose that P has no complementary legs. Then, for every contraction P ց P ′ with P ′ good with no bad components of any type, we have that P ′ has no complementary legs.
Proof. First of all, note that, in any contraction P ց P ′ , the vertices of P ′ inherit naturally the indexes from P . Now, suppose by contradiction that the good set obtained by a contraction has two complementary legs L ′ 2 and L ′ 3 . Let us call L 2 and L 3 the corresponding legs in P and analogously, L 1 and L ′ 1 the third leg respectively in P and in P ′ . Throughout the proof the reader must keep in mind that, by Proposition 6.14 (3), for every k ∈ {1, ..., n} and every (r, γ) ∈ J(P ), we have |v r,γ · e k | ≤ 1 and that, by Lemma 6.1 (1), there exists j ∈ {1, ..., n} such that the central vertex If v s,α = v 0 , then P ′ is a linear set, and the statement follows trivially. In case the vector v s,α is isolated, we will consider, without loss of generality, that it belongs to L 1 . By definition of complementary legs, L ′ 2 and L ′ 3 are connected and therefore, if v s,α ∈ L 1 , it must be final in L 2 or L 3 . By symmetry we may suppose that the first case occurs.
We start dealing with the possibility v s,α final in L 2 and v t,β ∈ L 2 ∪ L 3 . It follows that there exists a vector v ∈ L 2 such that v · v s,α = 1 and therefore |V vs,α ∩ V L ′ 2 | ≥ 1. By definition of complementary legs and since v t,β ∈ L 2 ∪ L 3 , it holds that |V vs,α ∩ V L ′ 2 ∪L ′ 3 | ≥ 2. Consider now an auxiliary vector e aux with e aux · e aux = −1 and definē v s,α := − k∈V L ′ 2 ∪L ′ 3 (v s,α · e k )e k andv 0 := ±e j + e aux .
We are left with the last possibility, namely v s,α ∈ L 1 . This implies, since in P there are no complementary legs while in P ′ there are, that v t,β ∈ L 2 ∪ L 3 . From the equality v s,α · v t,β = 0, we deduce that |V vs,α ∩ V L2∪L3 | ≥ 2. Let us consider the vectorṽ s,α := − k∈VL 2 ∪L 3 (v s,α · e k )e k and the setS := {v 0 }∪L 2 ∪L 3 ∪{ṽ s,α }, wherev 0 is defined above. Note that inS the vertexṽ s,α is isolated. The setS is a good subset of Z n2+n3+2 . In fact, its incidence matrix has the form (2.4) and it is irreducible, since {v 0 } ∪ L 2 ∪ L 3 is connected andṽ s,α is linked to v t,β ∈ L 2 . By construction, it holds p 1 (S) > 0 and therefore by Lemma 4.5, the setS is standard and thus connected. This contradiction finishes the proof.
The following Proposition 6.16 shows that good subsets with no bad components of any type, possibly disconnected intersection graphs and sufficiently negative quantity I(P ) can be contracted to subsets having the same properties. This is the main result of the section and will be used in the proof of Theorem 7.2.
Remark 6.17. In the last paragraph of the proof of Proposition 6.16 we have proved that we cannot have a contraction of standard sets P i+1 ց P i such that P i+1 has a trivalent vertex, I(P i+1 ) = −4 and P i is a linear set with I(P i ) = −3.

Proof of Theorems 2.5 and 2.6
In this section we specialize the analysis done in Sections 4 to 6 to the case of standard subsets with I < −1 and we finally prove Theorems 2.5 and 2.6.
The graph of a standard set with a trivalent vertex has at least 4 vertices. The following lemma justifies the fact that in most of the statements of Section 6 we have assumed n ≥ 5.
Lemma 7.1. In Z 4 = e 1 , e 2 , e 3 , e 4 there are no standard subsets P with a trivalent vertex and I(P ) < −1.
In order to prove Theorems 2.5 and 2.6 we will apply Theorem 7.2 to identify the numbers {a 0 , ..., a n3,3 } corresponding to standard subsets P ⊆ Z n with a trivalent vertex and I(P ) < −1.
for cases (1) and (2) and k = 1, b 1 = 2 in case (3) it yields the same graphP ⊆ Z 4 , which, up to replacingP with ΩP where Ω ∈ ΥP , is the following. e 1 − e 2 e 2 − e 3 −e 2 − e 1 + e 4 e 2 + e 3 + e 4 If we expandP by final −2 vectors we obtain (3) in Remark 3.2 (II). It is immediate to check that no graph in (3) can play the role of P j−1 described above. This is so because there is no pair of vectorsṽ 0 ,ṽ t,β , such thatṽ 0 is an internal vector with aṽ 0 ≥ 3 and such that there exists k with E k (P j−1 ) = {0, (t, β)}.
If we expandP as described in (1) in Remark 3.2 (II), there are two possible pairs of vectorsṽ 0 ,ṽ t,β such thatṽ 0 is an internal vector with aṽ 0 ≥ 3 and such that there exists k with E k (P j−1 ) = {0, (t, β)}. In both pairsṽ t,β is final. By symmetry, these two pairs yield graph (a) in the statement. The condition s > 0 in the statement comes from the requirement aṽ 0 ≥ 3 withṽ 0 ∈ P j−1 .
Finally, if we expandP as described in (2) in Remark 3.2 (II), there are different pairs of vectorsṽ 0 ,ṽ t,β such thatṽ 0 is an internal vector with aṽ 0 ≥ 3 and such that there exists k with E k (P j−1 ) = {0, (t, β)}. Now, it is not difficult to check that these pairs yield to the graphs (b), (c), (d) and (e) in the statement.