Comptes Rendus Mathématique

. Intersection norms are integer norms on the ﬁrst homology group of a surface. In this article, we give examples of polytopes which are not the dual unit balls of intersection norms, answering a question asked in [2]. On the way, we investigate the set of collections of curves on Σ 2 whose complement is a disk.


Introduction
Intersection norms on surfaces were first quickly introduced by Turaev [8, p. 143], and studied by M. Cossarini and P. Dehornoy [2].They use intersection norms to classify, up to isotopy, all surfaces transverse to the geodesic flow on the complement of special links in the unit tangent bundle of a closed oriented surface.Their result makes explicit Thurston's fibered faced theory for Thurston norms on compact oriented 3-manifolds.It tells us that an intersection norm on a surface (respectively the Thurston norm on a 3-manifold) encodes the open book decompositions of the unit tangent bundle of that surface (respectively the topology of foliated 3manifold).
Our purpose in this article is to study intersection norms for their own sake.Let Σ g be a closed oriented surface of genus g ≥ 1, and Γ a collection of closed curves on Σ g .Throughout this article, we consider collections in generic position: collections with transverse double intersection points.Let α be a closed curve on Σ g ; we define the number i Γ (α) as follows: where the symbol ∼ (respectively ) denotes the free homotopy relation (respectively transversality).
The function defines a semi-norm on H 1 (Σ g , R) and it takes integer values on the lattice H 1 (Σ g , Z) (see [2]).Using a standard basis for the homology, we shall identify H 1 (Σ g , R) and H 1 (Σ g , R) with R 2g .By a theorem of Thurston [7], the dual unit ball of N Γ is a lattice polytope, ie, the convex hull of finitely many integer vectors (integer vector here means a vector in the lattice H 1 (Σ g , Z) ≈ Z 2g ).Moreover, if Γ fills Σ g , that is Σ g − Γ is a union of topological disks, then N Γ defines a norm and as a consequence, its dual unit ball has non-empty interior in H 1 (Σ g , R).
One constraint on the dual unit balls of intersection norms is that their vertices are congruent modulo 2. This comes from the fact that geometric and algebraic intersections have the same parity.In genus 1, this constraint happens to be the only one.So, every symmetric convex lattice polygon with mod 2 congruent vertices is the dual unit ball of an intersection norm on the torus.The proof of this fact follows from an implicit argument in Thurston's paper [7].We will explain it in Section 2 for completeness.Now we raise the following problem: Question 1 (Pierre Dehornoy).Fix g ≥ 2, and let P ⊂ H 1 (Σ g , R) be a symmetric lattice polytope all of whose vertices are congruent mod 2. Is P the dual unit ball of an intersection norm on Σ g ?
This question is natural when we deal with integer norms coming from topology (for instance, we have an analogue of this question for the Thurston norm).

Definition 2. Let P be a polytope obtained by taking the convex hull of finitely many vectors {v i }.
A sub-polytope P of P is a polytope obtained by taking the convex hull of a subset of {v i }.
In this article, we give examples of lattice polytopes on R 4 with mod 2 congruent vertices, which are not dual unit balls of intersection norms.More precisely, we show that sub-polytopes (with eight vertices and non-empty interior) of the cube [−1, 1] 4 are not the dual unit balls of intersection norms.This means that in higher dimension, dual unit balls of intersection norms come with other constraints.
Let P 8 be the set of all symmetric sub-polytopes of [−1, 1] 4 having eight vertices and nonempty interior.The set P 8 is not empty; it contains the polytope generated by the following vectors (and their opposites): . Now, we state the main result of this article: Theorem 3. Elements of P 8 are not dual unit balls of intersection norms.
If Γ is a filling collection of curves on a surface, whose complement is a disk, we say that Γ is a one-faced collection.The proof of Theorem 3 relies on: Theorem 4.There are four orbits (under the action of orientation-preserving homeomorphisms) of one-faced collections on Σ 2 , whose dual unit balls are in the cube [−1, 1] 4 .
Thurston showed that every symmetric polygon with mod 2 congruent vertices is the dual unit ball of a Thurston norm on a 3-manifold (see [7]).His construction is closely related to intersection norms on the torus.Roughly speaking, he showed that from a filling collection Γ on the torus, one can construct a 3-manifold M Γ such that the dual unit ball of the Thurston norm on M Γ and the dual unit ball of the intersection norm N Γ are the same.Thurston's construction extends to higher genus surfaces (see [6]), and Theorem 3 suggests that there are probably polytopes in dimension greater than two which are not the dual unit balls of Thurston norms on 3-manifolds.
For the proof of Theorem 1, we define a partial order on the set of filling collections of closed curves and we use it to show that if an element of P 8 is the dual unit ball of an intersection norm N Γ , then Γ is a one-faced collection.Finally, we check that none of the four collections of Theorem 4 realizes an element of P 8 .

Organization of this article
In Section 2, we recall some facts on intersection norms and we show that for the question of realizability, we can restrict our attention to minimal collections.
In Section 3, we show that any intersection norm is bounded from below by a norm defined by a one-faced collection.
In Section 4, we count orbits (under the action of orientation-preserving homeomorphisms) of one-faced collections (whose dual unit balls are sub-polytopes of the cube [−1, 1] 4 ) on Σ 2 and we prove Theorem 3. We finish this section by establishing a link between the realization problem and the determination of one-faced collections by proving Proposition 5.

Preliminaries on intersection norms
In this section, we first recall some facts about integer (semi)-norms.Then, we define the intersection (semi)-norm associated to a collection of curves and we recall some basic notions about them (For more details on intersection norms, see [2]).We end this section by proving that, concerning the realizability of polytopes, we can restrict our attention to minimal collections.
Let ∼ denote the free homotopy relation between curves; let be the transversality relation and [ • ] the homology class.

Integer norms
Let E be a vector space of dimension n and the lattice generate by the vectors (u i ) i =1,...,n .

Definition 6 (Integer norm). A norm N : E −→ R + is an integer semi-norm relatively to the lattice L if the restriction of N to L takes positive integer values.
The dual unit ball of N is the unit ball of the dual norm N * : E * −→ R + ; where E * is the dual space of E .
The following theorem states that the dual unit ball of an integer semi-norm has a combinatorial description.
Theorem 7 (W.Thurston).If N is an integer semi-norm relatively to a lattice L, then its dual unit ball is a convex hull of finitely many vectors in the lattice; One can find a sketch of the proof of Theorem 7 in [7, p. 107-112].For a more complete proof, see [3] (Exposé Fourteen by David Fried).More recently, M. de la Salle gives a new proof of Theorem 7 (see [5]).

Definition of intersection norms
We consider a genus g closed oriented surface Σ g and a collection Γ = {γ 1 , . . ., γ n } of closed curves on Σ g .The collection Γ is allowed to change only in its isotopy class.Let a ∈ H 1 (Σ g , Z) be a homology class and α an oriented multi-curve representing a. Then we define: If a multi-curve α representing a homology class a is such that One important thing is that Γ-minimizing multi-curves can be chosen to be simple.In fact, if α is a non-simple Γ-minimizing multi-curve then by smoothing all the self-intersection points of α with respect to their orientation, we get a new oriented multi-curve α in the same homology class as α (its a common fact on homology that smoothing singularities preserves homology classes) and i Γ (α ) = i Γ (α).This implies that α is a simple Γ-minimizing multi-curve as we claim.

Proposition 8 (Cossarini-Dehornoy). The function N
The proof of Proposition 8 is not trivial and one can see [2] for the proof.Linearity on rays implies that N Γ can be extended to homology with rational coefficients since for all a ∈ H 1 (Σ g , Z) and q ∈ N, we have: It follows by convexity that N Γ extends uniquely to a positive function on H 1 (Σ g , R).Moreover, the extended function N Γ : H 1 (Σ g , R) −→ R + is still linear on rays and convex.Therefore, N Γ defines a semi-norm on H 1 (Σ g , R) and it takes integer values on the lattice H 1 (Σ g , Z).So, N Γ is an integer semi-norm.Theorem 7 implies that the dual unit ball B N * Γ is a convex hull of finitely many integer vectors.
If the collection is filling, then N Γ defines an integer norm.

Relation between the vectors of the dual unit ball
If α and β are two transverse oriented closed curves, then the algebraic intersection number between α and β is given by i where ε(p, α, β) is the algebraic sign of the intersection at p, relatively to the orientation of Σ g .We recall that i depends only on the homology classes of α and β, and defines a non degenerate skew-symmetric 2-form on H 1 (Σ g , R).
Then, if α and α are two homologous curves, by taking an orientation of Γ, we have Thus, i Γ (α) = i Γ (α ) mod 2 for every orientation of Γ.Therefore, if v 1 and v 2 are two integer vertices in the dual unit sphere of N Γ , The relation above is a necessary condition for a symmetric lattice polytope to be the dual unit ball of an intersection norm.The following statement shows that it is sufficient in the genus one case.The idea of the proof is from Thurston [7].
Proposition 9.If P is a symmetric lattice polygon in R 2 with congruent mod 2 vertices, then P is the dual unit ball of an intersection norm.
Proof.First, if P is a symmetric lattice segment in R 2 , then there is a matrix M ∈ SL(2, Z) such that P := M (P ) is a vertical segment with extremities in Z 2 .Moreover, M has a geometric realization since the group of isotopy classes of preserving orientation homeomorphisms of the torus is isomorphic to SL(2, Z).That is there is a homeomorphism φ of T 2 such that Secondly, if Γ := {γ 1 , . . ., γ n } is a collection of closed geodesics on T 2 (with the flat metric), and if a is a homology class represented by a collection α of oriented simple closed curves which are pairwise disjoint then It follows that the dual unit ball of N Γ is equal to the Minkowski sum of the dual unit balls of N γ j , which are symmetric lattice segments: Finally, every symmetric lattice polygon of R 2 is the Minkowski sum of finitely many symmetric lattice segments.
Combining the three arguments above we construct, for any symmetric lattice polygon P with mod 2 congruent vertices, a geodesic collection Γ such that

Minimality of collections
Now, we show that we can restrict to collections in minimal position.
Definition 10.Let γ 1 and γ 2 be two transverse closed curves on Σ g .They are in minimal position if they realize the geometric intersection in their free homotopy classes: If a collection Γ is not minimal, it admits a minimal representative Γ in its homotopy class.Moreover, the isotopy classes of Γ and Γ differ by Reidemester's moves (see Figure 1).Lemma 11.Let Γ be a collection of closed curves in Σ g , then there is a minimal collection Γ min such that N Γ = N Γ min .
Proof.One can apply a generic homotopy so that we get a collection in minimal position.Such a generic homotopy consists of doing a finite number of Reidemester moves (1, 2 and 3 as depicted in Figure 1).By Hass and Scott [4], one can choose a decreasing homotopy with respect to the intersection number of the collection.Moves 1 and 3 do not change the norm, while Move 2 (deleting a bigon) changes the norm.
Then, we replace Move 2 by a crossing move (see Figure 1).This new move changes the homotopy class of Γ but it does not change the norm.The intersection decreases by one with the crossing move (Move 2bis).As we can choose a decreasing homotopy, we get a collection Γ min in minimal position after applying finitely many Reidemester's moves 1 and 3 and crossing moves.Doing so, the norm does not change; hence we built a new collection Γ min in minimal position such that N Γ = N Γ min .

Eulerian Co-orientation
Let Γ be a collection of closed curves on Σ g such that Σ g − Γ is a union of topological disks.The collection Γ defines a filling graph on Σ g .We denote by V (Γ) the set of its vertices, defined as self-intersection points of Γ.Let E (Γ) be the set of edges and F (Γ) the set of faces.The Euler characteristic of Σ g is given by: Definition 13.A co-orientation ν of Γ is a choice of a positive direction to cross (transversally) every edge of Γ.We denote by Γ ν the collection Γ together with the co-orientation ν.
A co-orientation is Eulerian if a small oriented circle centered at a vertex crosses positively two edges and negatively the other two, relatively to the co-orientation.• Up to rotation, we distinguish two types of Eulerian co-orientations around a vertex (see Figure 2).A vertex is non-alternating if the arcs emanating from it, and belonging to the same curve are co-oriented in the same direction; otherwise it is alternating.• A co-orientation of an arc corresponds to an orientation of it.
• A collection Γ with c curves has at least 2 c Eulerian co-orientations given by all the different ways to co-orient Γ only by non-alternating vertices (this is equal to the number of possibilities to orient Γ).
Let α be an oriented closed curve on Σ g transverse to Γ, and let ν be a co-orientation of Γ.We define ν(α where ε(p, α, Γ ν ) = ±1 depending on whether α crosses Γ at p in the direction of the coorientation ν or not.Moreover, if ν is an Eulerian co-orientation, An Eulerian co-orientation ν defines a map Hence The proof of Theorem 15 is well explained in [2].

Partial order on the set of intersection norms of Σ g
In this section, we define a topological operation on filling collections of closed curves.That operation induces a partial order on the set of filling collections.
We finish by proving that every (non-even) intersection norm is bounded from below by an intersection norm induced by a one-faced collection.
Let p be a self-intersection point of a filling collection Γ in minimal position, such that two different and opposite faces F 1 and F 2 have p as a common point.
By smoothing the point p (see Figure 3) in such a way that F 1 and F 2 merge to a unique face, we obtain a new filling collection Γ .We call this operation merging of faces.Definition 16.Let Γ and Γ two filling collections.We say that Γ ≤ Γ if there is a sequence  If Γ is a bi-colorable collection, then N Γ restricted to the lattice H 1 (Σ g , Z) takes only even values.In fact, a closed curve α alternates between black and white faces.That is i Γ (α) is even; hence N Γ is even.In this case, the coordinates of the vectors defining the dual unit ball of N Γ are all even.If Γ is obtained by merging faces on a bi-colorable collection Γ, then Γ is also a bi-colorable.
For a graph defined by a collection of curves with only double points, we have |E | = 2|V |.Then, the Euler characteristic of the surface is given by: It follows that for a filling collection, we have |V | = |F | + 2g − 2 ≥ 2g − 1; the minimum is obtained for one-faced collections.In particular, in genus two, a one-faced collection has self-intersection number equal to 3.
The following lemma is one of the cornerstones of this article.

Lemma 19 (Lower bound for intersection norms).
If Γ is not bi-colorable, then there is a finite sequence Γ 0 = Γ −→ • • • −→ Γ n of merging of faces such that Γ n is one-faced.
Proof.Let Γ be a filling collection on Σ g and Γ 0 = Γ → • • • → Γ n be a sequence of merging of faces such that no merging is possible on Γ n .Let p be a double point of Γ n and e := (v 1 = p, v 2 , . . ., v n = p) an Eulerian cycle of Γ.Since all the vertices of Γ have degree 4, such cycles exist.
As Γ n is not reducible by a merging of faces, there are at most two different faces around p; let F a and F b be the faces (eventually equal) around p. When we turn around p, we read The faces F a and F b are also faces around v 2 since v 1 and v 2 share an edge.The configuration around v 2 is also F a − F b − F a − F b , since a merging of faces is not possible around v 2 .
Step by step, following the Eulerian cycle, we show that all the vertices of Γ n have the same configuration of faces: Therefore, Γ n has at most two faces in its complement depending on whether F a = F b or not.If Γ n has two faces in its complement, then two faces with a common edge have different color.So, Γ n is bi-colorable and so is Γ.
By taking the contrapositive, we obtain the proof.

Corollary 20. Every intersection norm with dual unit ball in the cube [−1, 1]
2g is bounded from below by a norm defined by a one-faced collection.

Proof. If B N *
Γ is a sub-polytope of [−1, 1] 2g , then N Γ is non-even and then not bi-colorable.Applying Lemma 19, we obtain the result.

One-faced collections with dual unit ball in the cube [−1, 1] 4
Now, we show that there are four one-faced collections (up to orientation-preserving homeomorphisms) with dual unit ball in the cube [−1, 1] 4 (Theorem 4).
The following lemma gives a canonical partial configuration for one-faced collections.

Lemma 21. If Γ is a one-faced collection on Σ 2 with dual unit ball in the cube
Hence, up to diffeomorphism and outside A η , Γ looks like Figure 5. Proof.It is equivalent to show that, up to diffeomorphism, {α 1 , Up to diffeomorphism, we can take α = α 1 .
C. R. Mathématique, 2020, 358, n 8, 941-956 Now, let β be the Γ-minimizing simple curve in the homology class of b 1 .Then, Therefore, one can perform a surgery on β along α (See Figure 6) to get a new curve β such that β is a simple Γ-minimizing curve in the same homology class of β and such that Up to diffeomorphism, we can take β = β 1 .
If α and β are Γ-minimizing in the homology classes of α 2 and β 2 respectively, we have Again, by performing surgery on α and β, we get α and β such that Then, up to diffeomorphism α = α 2 and β = β 2 .This prove that up to diffeomorphism, (α 1 , β 1 , α 2 , β 2 ) are Γ-minimizing.Lemma 21 implies that, up to diffeomorphism, a one-faced collection with dual unit ball in the cube [−1, 1] 4 is obtained by connecting the extremities of the partial configuration by arcs in the annulus A η .Moreover, the self-intersection number of Γ is determined by the intersection between those arcs we used to complete the partial configuration.
Let a 1 , b 1 , a 2 and b 2 be the four oriented arcs in the partial configuration (see Figure 5).A closed curve from the partial configuration will be labelled by the arcs being used and the number of twists we make around η when we walk along that curve.For instance, a 1 η 2 b −1 1 b 2 is the closed curve depicted on Figure 7.As we are dealing with non oriented collections, the labeling of curves is defined up to cyclic permutation and reversing.For example, a 1 η 2 b −1 1 b 2 and a −1 1 b −1 2 b 1 η −2 are labels of the same curve.

Intersection of arcs in an annulus
As we said above, the geometric intersection of a one-faced collection in Σ 2 is completely determined by the intersection of arcs in an annulus.Here, the intersection number is computed over the homotopy class of arcs with fixed end points.Now, let λ be a simple oriented arc joining the two boundaries of A η .Cutting along λ, we obtain a rectangle with two opposite sides identified.Let X and Y be two points in the boundary components of A η .An oriented arc from X to Y will be denoted by X Y p where p ∈ Z is the algebraic intersection between X Y p and λ.Let A, B,C and D be four points in the boundaries of A η as in Figure 8.

Lemma 23. The following formulas give the intersection between two oriented arcs in A
Proof.Up to the Dehn twist τ −q η on the configuration of the arcs, one can assume that q is equal to 0 in all cases, meaning that one of the arc is untwisted.Therefore, we have: η (AB p ).Moreover, p = p − q.Hence, we obtain the result.
Again, for the second formula, we have: and p = p + q.The difference between the first two cases shows how crucial the orientation is for the computing of intersection.We next address the third case, and the remaining are handled in a similar way.We still have that i The appearance of −1 in this case comes from the cross configuration of the extremities.

List of one-faced collections with dual unit ball in the cube
Now, we are able to count all one-faced collections whose dual unit balls are sub-polytopes of the cube [−1, 1] 4 .Before that, we define some diffeomorphisms that will be useful for the counting.If γ is an oriented simple closed curve on Σ 2 , we recall that τ γ is the right-handed Dehn twist along γ.Let R 1 (respectively R 2 ) be the rotation of angle π along the axis D (respectively the horizontal axis) as depicted in Figure 9.The diffeomorphism R 1 (respectively R 2 ) is an involution and it maps α 1 to α 2 , β 1 to β 2 and η to η −1 (respectively α i to α −1 i , β i to β −1 i and η to η).We recall that α i and β i can be interchanged by a diffeomorphism.More precisely, there is a diffeomorphism sending α i to β i and β i to α −1 i .This fact implies that in the writing of the label of the curves, a i can be replaced by b i and b i by a −1 i ; we call this operation interchanging.Definition 24.Let Γ be a collection of closed curves on Σ 2 .A cycle γ in Γ ( Γ seen as a graph on Σ 2 ) is separating if Σ 2 − γ has more than one component.
The following lemma gives a necessary condition for a collection to be one-faced.
Lemma 25.If Γ is a one-faced collection, then Γ does not contain a separating cycle.
Proof.Assume that Γ contains a separating cycle γ, then Σ 2 − γ has at least two connected components.In this case, Γ has more than one disc in its complement.So if Γ is one-faced, it does not contain a separating cycle.Now, we can state the main result of this section which is an elaborate form of Theorem 4.

Theorem 26 (Orbits of one-faced collections).
If Γ is a one-faced collection on Σ 2 with dual unit ball in the cube [−1, 1] 4 , then Γ has at most three closed curves.Moreover, up to diffeomorphism, Proof.If Γ is one-faced, then i (Γ, Γ) = 3 (this comes from an Euler characteristic argument; cf.Section 3).Now, if Γ has at least four closed curves, then the arcs a i , b i (i = 1, 2) belong to four different closed curves α i η p i , β i η q i ; otherwise Γ would contain a separating cycle.Therefore, i (Γ, η) = 0 which is absurd as Γ is filling.So, if Γ is one-faced |Γ| ≤ 3.

Case 1.
If |Γ| = 3, then two arcs of the partial configuration belong to the same closed curve and the other two belong to two different closed curves.Moreover, as Γ is filling, the two arcs contained in the same closed curve are in different handles.As one can interchange a i and b i , we can assume that the curve containing two arcs is γ := b 1 η p b −1 2 η q ; the other curves being α 1 η r and α 2 η s .Since Γ is one-faced, it does not contain a separating cycle meaning that r = s = 0. Up to a Dehn twist along η, one can take p=0; therefore, γ = b 1 b −1 2 η q .The fact that i (Γ, Γ) = 3 implies that i (γ, γ) = 1.
Figure 10.One-faced collection with three curves Case 2. If |Γ| = 2, then one of the curves of Γ is simple.Otherwise if the two curves are not simple, one can smooth intersection points of one of the curves in Γ (let g 1 be that curve) such that each smoothing separates g 1 in to two components.We obtain at least two simple curves λ j , j = 1, . . ., n.The curves λ j , as they are all parallels to g 1 , intersect g 2 .Then which is absurd since Γ is one-faced.Therefore, one of the two curves is simple, say g 1 .Moreover, Σ 2 does not admit a filling pair [1] (i.e a one faced-collection made of two simple closed curves).So, g 2 is non simple.
Up to diffeomorphism (interchanging and rotations), one can assume that a 1 is contained in g 1 .
Case 2.1.If g 1 does not contain another arc, then This implies that i (g 2 , g 2 ) = 2.The solution of this equation is p = 1 and ε = 1. So, which indeed is a one-faced collection (see Figure 11).
Case 2.2.If g 1 contains an arc other than a 1 , that arc cannot be in the same handle as a 1 (otherwise, Γ is not filling).Up to interchanging, we can assume that and again by applying a Dehn twist around η, one can take We have i (α 1 ∪α 2 , β 1 ∪β 2 ) ≡ i (g 1 , g 2 ) mod 2 since α 1 ∪α 2 (respectively β 1 ∪β 2 ) is homologous to g 1 (respectively g 2 ).It follows that 2 η r , we check that Γ is either not filling, or filling with more than one disk in its complement. For η r and we change a 1 to b 1 by a diffeomorphism (that diffeomorphism will map b 1 to a −1 1 ), g gets mapped to g = b 1 a −1 2 η p a 1 η q b 2 η r .Now, if we reverse the orientation of g starting at a 1 , we have The collections Therefore, up to diffeomorphism, we have one one-faced collection with one curve (see Figure 12), namely Γ = {a 1 a −1 2 b 1 b −1 2 η}.The collection Γ is one of the collections in Theorem 26.We check that dual unit balls of those collections are not in P 8 (see bellow for their dual unit balls); which finally proves that elements of P 8 are not realizable.
Computation of dual unit balls.We compute the dual unit ball of an intersection norm by evaluating all Eulerian co-orientations on the canonical homology basis [2].Doing so, we obtain the vertices (of the dual unit balls) below: The first collection has the whole unit cube as dual unit ball; the others three have dual unit balls with at least ten vectors.
We finish this article by giving a perspective in dimension greater than 4. Let P 4g be the set of symmetric non degenerate sub-polytopes of [−1, 1] 2g with 4g vertices.Since we are in dimension 2g , elements of P 4g are minimal (in term of number of vertices) among symmetric non degenerate sub-polytopes of [−1, 1] 2g .Lemma 27.If Γ is a one-faced collection obtained by merging faces of a filling collection Γ, then N Γ < N Γ .
Proof.For the proof, it is enough to show that if Γ is two-faced (F 1 and F 2 in Σ − Γ) and reduced to Γ , then N Γ < N Γ .
Since Γ reduced to a one-faced collection, there is a vertex v of Γ which is adjacent to F 1 and F 2 (see Figure 13).
Since Σ g is a (non singular) surface, there is an edge x in F 1 whose identified side x is in F 2 .Let α be the oriented closed curve on Σ g from x to x passing by v.The curve α is non trivial in homology since i (Γ, α) = 3.Moreover, α is Γ-minimizing in its homology class because α is not Proof of Proposition 5. Let P ∈ P 4g and let Γ be a filling collection on Σ g such that the dual unit ball of N Γ is equal to P .If Γ has more than two faces in its complement, then by Lemma 27 N Γ > N Γ ; where Γ is a one-faced collection.It follows that the dual unit ball of N Γ has less than 4g vertices, which is absurd.So, Γ is a one-faced collection.
Theorem 3 makes the general problem of realization for intersection norms a decision problem: given a symmetric non degenerate polytope P with congruent mod 2 vertices, state whether or not P is the dual unit ball of an intersection norm.On can restricted to minimal polytopes in dimension 2g : those non degenerate sub-polytopes of [−1, 1] 2g with 2g vertices, and compare their numbers (up to the action of the linear symplectomorphisms which fix [−1, 1] 2g ) with the number of one-faced collections in Σ g .

Figure 1 .
Figure 1.From left to right, we have the three Reidemeister's moves and the last one is the crossing move (2bis) on which a bigon is replaced by a transverse self-intersection.The crossing move changes the homotopy class of the collection.The curves in red color represent the local configuration of Γ, and curves in black are sub-arcs of curves in Σ g .One can see that Reidemeister's move 1 and 3 and crossing move do not change the intersection.Otherwise, the bigon deleting does change the intersection.
, an Eulerian co-orientation defines an integer cohomology class.We denote by Eulco(Γ) the set of all Eulerian co-orientations of Γ and by [Eulco(Γ)] the set of their cohomology classes (different co-orientations can give the same cohomology class).Theorem 15 (M.Cossarini & P. Dehornoy).The set [Eulco(Γ)] is a subset of the unit dual ball B N * Γ .Moreover, every integer vector in B N * Γ , mod 2 congruent to the vertices of B N * Γ belongs to [Eulco(Γ)].
where each step of the sequence is a Move 1, Move 2bis, Move 3 or a merging of faces.Lemma 17.If Γ ≤ Γ, then N Γ ≤ N Γ .Proof.Let Γ be a filling collection in minimal position and Γ a collection obtained by merging two faces at a point p.The collections Γ and Γ are different only in a small neighborhood of p.Then, for any closed curves α onΣ g i Γ (α) ≤ i Γ (α);and it follows that N Γ ≤ N Γ .Since Move 1, Move 2bis and Move 3 do not change the intersection norm, the inequality holds for Γ in minimal position.

F 1 F 2 Figure 3 .
Figure 3. Smoothing a self-intersection point and the merging of two faces.

Figure 6 .
Figure 6.Surgery along the vertical curve.

Figure 12 .
Figure 12.One-faced collection made of one curve
Let P ∈ P 4g .If P is the dual unit ball of an intersection norm N Γ , then Γ is a one-faced collection.