On the Boolean dimension of a graph and other related parameters

We present the Boolean dimension of a graph, we relate it with the notions of inner, geometric and symplectic dimensions, and with the rank and minrank of a graph. We obtain an exact formula for the Boolean dimension of a tree in terms of a certain star decomposition. We relate the Boolean dimension with the inversion index of a tournament.


Presentation and preliminaries
We define the notion of Boolean dimension of a graph, as it appears in Belkhechine et al. (2010) (see also (Belkhechine, 2009;Belkhechine et al., 2012)).We present the notions of geometric and symplectic dimensions, and the rank and minrank of a graph, which have been considered earlier.When finite, the Boolean dimension corresponds to the inner dimension; it plays an intermediate role between the geometric and symplectic dimensions, and does not seem to have been considered earlier.The notion of Boolean dimension was introduced in order to study tournaments and their reduction to acyclic tournaments by means of inversions.The key concept is the inversion index of a tournament (Belkhechine, 2009;Belkhechine et al., 2010Belkhechine et al., , 2012) ) presented in Section 3. Our main results are an exact formula for the Boolean dimension of a tree in terms of a certain star decomposition (Theorem 2.9) and the computation of the inversion index of an acyclic sum of 3-cycles (Theorem 3.7).
Notations in this paper are quite elementary.The diagonal of a set X is the set ∆ X ∶= {(x, x) ∶ x ∈ X}.We denote by (X) the collection of subsets of X, by X m the set of m-tuples (x 1 , . . ., x m ) of elements in X, by [X] m the m-element subsets of X, and by [X] <ω the collection of finite subsets of X.The cardinality of X is denoted by X .We denote by ℵ 0 the first infinite cardinal, by ℵ 1 the first uncountable cardinal, and by ω 1 the first uncountable ordinal.A cardinal κ is regular if no set X of cardinal κ can be divided in strictly less than κ subsets, all of cardinality strictly less than κ.If κ denotes a cardinal, 2 κ is the cardinality of the power set (X) of any set X of cardinality κ.If κ is an infinite cardinal, we set log 2 (κ) for the least cardinal µ such that κ ≤ 2 µ .We note that for an uncountable cardinal κ the equality log 2 (2 κ ) = κ may require some set theoretical axioms, such as the Generalized Continuum Hypothesis (GCH).If κ is an integer, we use log 2 (κ) in the ordinary sense, hence the least integer µ such that κ ≤ 2 µ is ⌈log 2 κ⌉.We refer the reader to Jech (2003) and Kunen (2011) for further background about axioms of set theory if needed.
The graphs we consider are undirected and have no loops.They do not need to be finite, but our main results are for finite graphs.A graph is a pair (V, E) where E is a subset of [V ] 2 , the set of 2-element subsets of V .Elements of V are the vertices and elements of E are the edges.Given a graph G, we denote by V (G) its vertex set and by E(G) its edge set.For u, v ∈ V (G), we write u ∼ v and say that u and v are adjacent if there is an edge joining u and v.The neighbourhood of a vertex u in G is the set N G (u) of vertices adjacent to u.The degree d G (u) of a vertex u is the cardinality of N G (u).If X is a subset of V (G), the subgraph of G induced by X is G ↾X ∶= (X, E ∩ [X] 2 ).A clique in a graph G is a set X of vertices such that any two distinct vertices in X are adjacent.If X is a subset of a set V , we set K V X ∶= (V, [X] 2 ); we say also that this graph is a clique.

The Boolean sum of graphs and the Boolean dimension of a graph
Let (G i ) i∈I be a family of graphs, all with the same vertex set V .The Boolean sum of this family is the graph, denoted by +(G i ) i∈I , with vertex set V such that an unordered pair e ∶= {x, y} of distinct elements of V is an edge if and only if it belongs to a finite and odd number of E(G i ).If the family consists of two elements, say (G i ) i∈{0,1} we denote this sum by G 0 +G 1 .This is an associative operation (but, beware, infinite sums are not associative).If each E(G i ) is the set of edges of some clique C i , we say (a bit improperly) that +(G i ) i∈I is a sum of cliques.We define the Boolean dimension of a graph G, which we denote by dim Bool (G), as the least cardinal κ such that G is a Boolean sum of κ cliques.In all, dim Bool (G) = κ if there is a family of κ subsets (C i ) i∈I of V (G), and not less, such that an unordered pair e ∶= {x, y} of distinct elements is an edge of G if and only if it is included in a finite and odd number of C i 's.
A Boolean representation of a graph G in a set E is a map f ∶ V (G) → (E) such that an unordered pair e ∶= {x, y} of distinct elements is an edge of G if and only if the intersection f (x) ∩ f (y) is finite and has an odd number of elements.
The following result is immediate, still it has some importance.
Proposition 1.2.A graph G is a Boolean sum of κ cliques if and only if G has a Boolean representation in a set of cardinality κ.
We note that the Boolean dimension of a graph and of the graph obtained by removing some isolated vertices are the same.Hence dim Bool (G) = 1 if and only if it is of the form G = K V X with X ≥ 2. Since every graph G ∶= (V, E) can be viewed as the Boolean sum of its edges, the Boolean dimension of G is always defined, and is at most the number of edges, that is, at most the cardinality Belkhechine et al. (2010).By induction on n: ).In fact, paths on n vertices are the only n-vertex graphs with Boolean dimension n − 1, see Theorem 2.4, a result that requires some ingredients developed below.
Recall that a module in a graph G is any subset A of V (G) such that for every a, a ′ ∈ A and b ∈ V (G) ∖ A, we have a ∼ b if and only if a ′ ∼ b.A duo is any two-element module (e.g., see Courcelle and Delhommé (2008) for an account of the modular decomposition of graphs).
Lemma 1.3.If a graph G has no duo then every Boolean representation is one to one.In particular, dim Bool (G) ≥ log 2 ( V (G) ).
Proof: Observe that if f is a representation and v is in the range of f , then f −1 (v) is a module and this module is either a clique or an independent set.
The answer may depend on some set theoretical hypothesis (see Example 1.9).But we do not known if the Boolean dimension of every graph on at most a continuum of vertices is at most countable.Same question may be considered for trees.
Let E be a set; denote by O(E) ¬ the graph whose vertices are the subsets of E, two vertices X and Y being linked by an edge if they are distinct and their intersection is finite and odd.If κ is a cardinal, we set O(κ) ¬ for any graph isomorphic to O(E) ¬ , where E is a set of cardinality κ.
Theorem 1.4.A graph G with no duo has Boolean dimension at most κ if and only if it is embeddable in O(κ) ¬ .The Boolean dimension of O(κ) ¬ is at most κ.It is equal to κ if κ ≥ 2 and κ is at most countable, or if κ is uncountable and (GCH) holds.
has no duo.This relies on the following claim.
Claim 1.5.If A, B are two distinct subsets of E, then there is a subset C of E, distinct from A and B, with at most two elements such that the cardinalities of the sets A ∩ C and B ∩ C cannot have the same parity.

Indeed, we may suppose that
In this case, either B is empty and C ∶= {x, y}, with y = x will do, or B is nonempty, in which case, we may set C ∶= {y}, where y ∈ B if B > 1, or C ∶= {y, z}, where B = {y} and z ∈ E ∖ (A ∪ B).
If κ is at most countable, or κ is uncountable and (GCH) holds, then this last quantity is κ.This completes the proof of the theorem.
We can obtain the same conclusion with a weaker hypothesis than (GCH).
Proof: The proof relies on the following claim, which is of independent interest.Claim 1.7.Let µ ω be the cardinality of the set of countable subsets of an infinite cardinal µ.Then the cliques in O(µ) ¬ have cardinality at most µ ω .
The proof relies on a property of almost disjoint families.Let us recall that an almost disjoint family is a family A ∶= (A α ) α∈I of sets such that the intersection A α ∩ A β is finite for α = β.Note that if C is a clique in O(µ) ¬ , then for every pair of distinct sets X, Y in C, the intersection X ∩ Y is finite and its cardinality is odd.Hence, C is an almost disjoint family.
To prove our claim it suffices to prove the following claim, well known by set theorists.
Claim 1.8.There is no almost disjoint family of more that µ ω subsets of an infinite set of cardinality µ.
Proof of Claim 1.8.Suppose that such a family A ∶= (A α ) α∈I exists, with I > µ ω .Since µ <ω = µ, we may suppose that each A α is infinite and then select a countable subset B α of A α .The family B ∶= (B α ) α∈I is almost disjoint, but since I > µ ω , there are α = β such that B α = B β , hence B α ∩ B β is infinite, contradicting the fact that B is an almost disjoint family.◻ Now the proof of the lemma goes as follows.Suppose that dim Bool (O(κ) ¬ = µ < κ.Then there is an embedding from the graph O(κ) ¬ into the graph O(µ) ¬ .Trivially, O(κ) ¬ contains cliques of cardinality at least κ.Hence O(µ) ¬ too.But since µ ω < κ, Claim 1.7 says that this is impossible.Thus dim Bool (O(κ) ¬ ) = κ.
Examples 1.9.For a simple illustration of Lemma 1.6, take κ = (2 ℵ0 ) + the successor of 2 ℵ0 .For an example, negating (GCH), suppose Question 1.2.Does the equality dim Bool (O(κ) ¬ ) = κ hold without any set theoretical hypothesis?Remark 1.10.Theorem 1.4 asserts that O(κ) ¬ is universal among graphs with no duo of Boolean dimension at most κ (that is embeds all graphs with no duo of dimension at most κ), but we do not know which graphs on at most 2 κ vertices embed in O(κ) ¬ .
In contrast with Claim 1.7 we have: Lemma 1.11.For an infinite cardinal κ, the graph O(κ) ¬ embeds a graph made of 2 κ disjoint edges.It embeds also some trees made of 2 κ vertices.
Proof: Let G be the graph made of 2 κ disjoint edges {a α , b α } with α ∈ 2 κ .We show that G is isomorphic to an induced subgraph of O(E) ¬ , where E is the set [κ] <ω of finite subsets of κ, augmented of an extra element r.Since E = κ, this proves our first statement.For the purpose of the proof, select 2 κ subsets X α of κ which are pairwise incomparable with respect to inclusion and contain an infinite subset X.For each α ∈ 2 κ , let That A α and B α form an edge is obvious: their intersection is the one element set {r}.Now, let α = β.We claim that the three intersections A α ∩ A β , A α ∩ B β and B α ∩ B β are all infinite.For the first one, this is obvious (it contains [X α ∩ X β ] <ω ), for the next two, use the fact that the A α are up-directed with respect to inclusion, hence the difference A α ∖ A β is cofinal in A α , thus must be infinite, and the union A α ∪ A β cannot cover [κ] <ω , hence its complement is infinite.It follows that the graph H contains no other edges than the pairs {A α , B α }'s.This proves that H is isomorphic to G, and yields our first statement.For the second statement, add R ∶= [X] <ω ∪ {r} to the set of vertices of H.We get a tree.Indeed, for each α, the vertices R and <ω hence is infinite), while for each β, the vertices R and A β form an edge (since R ∩ A β = {r}).
For infinite graphs with finite Boolean dimension, a straightforward application of Tychonoff's theorem yields the following result.
Theorem 1.12.Let n ∈ N.For every graph G, dim Bool (G) ≤ n if and only if dim Bool (G ↾X ) ≤ n for every finite subset X of V (G).
Proof: Suppose that the second condition holds.For every finite subset X of V (G) let U X be the set of maps f from V (G) into the powerset K ∶= ({1, . . ., n}) such that the restriction f ↾X is a Boolean representation of G ↾X in {1, . . ., n}.Each such set U X is nonempty and closed in the set K V (G) equipped with the product topology, the set K being equipped with the discrete topology.Every finite intersection G) ensures that the intersection of all of those sets is nonempty.Any map in this intersection is a Boolean representation of G.
Examples of graphs with finite Boolean dimension are given at the end of the next subsection.

Geometric notions of dimensions of graphs
We introduce three notions of dimensions: geometric, inner, and symplectic, all based on bilinear forms.We prove that if the Boolean dimension of a graph is finite, then it coincides with the inner dimension, and either these dimensions minus 1 coincide with the geometric and the symplectic dimension, or they coincide with the geometric dimension, the symplectic being possibly larger (Theorem 1.18).We note before all that in general, the Boolean dimension is not based on a bilinear form.It uses the map ϕ ∶ (E) → 2 ∶= {0, 1} defined by setting ϕ(X, Y ) ∶= 1 if X ∩ Y is finite and odd and 0 otherwise.But except when E is finite, it is not bilinear on (E) equipped with the symmetric difference.
Let F be a field, and let U be a vector space over F, and let ϕ be a bilinear form over U .We recall that this form is We set x instead of {x} .We recall that ϕ is degenerate if there is some x ∈ U ∖ {0} such that ϕ(x, y) = 0 for all y ∈ U .The form ϕ is said to be alternating if each x ∈ U is isotropic, in which case (U, ϕ) is called a symplectic space.The form ϕ is an inner form or a scalar product if U has an orthonormal basis (made of non-isotropic and pairwise othogonal vectors).
Definition 1.13.Let U be a vector space equipped with a symmetric bilinear form ϕ. Let G be a graph.We say that a map f ∶ is the least cardinal κ for which there exists a geometric representation of G in a vector space U of dimension κ equipped with a symmetric bilinear form ϕ. The symplectic dimension of G, denoted by dim symp (G), is the least cardinal κ for which there exists a symplectic space (U, ϕ) in which G has a geometric representation.The inner dimension of G, denoted by dim inn (G), is the least cardinal κ for which G has a geometric representation in a vector space of dimension κ equipped with a scalar product.
The notions of geometric and symplectic dimension were considered by several authors, for example, Garzon (1987); Godsil and Royle (2001a).There is an extensive literature about this subject (e.g.Fallat and Hogben (2007); Grout ( 2010)), and notably the role of the field.But apparently, the Boolean dimension was not considered.
Except in subsection 1.4, we consider these notions only for the 2-element field F 2 , identified with the set {0, 1}.If U has finite dimension, say k, we identify it with F k 2 , the set of all k-tuples over {0, 1}; the basis (e i ) i∶=1,...,k , where e i is the 0-1-vector with a 1 in the i-th position and 0 elsewhere, is orthonormal; the scalar product of two vectors x ∶= (x 1 , . . ., x k ) and y ∶= (y 1 , . . ., y k ) of F k 2 is then ⟨x y⟩ ∶= x 1 y 1 + ⋅ ⋅ ⋅ + x k y k .We recall the following dichotomy result.
Theorem 1.14.A nondegenerate bilinear symmetric form ϕ on a finite k-dimensional space U over the two-element field F 2 falls into two types.Either ϕ is non-alternating and (U, ϕ) is isomorphic to (F k 2 , ) with the scalar product, or ϕ is alternating, k is even, and (U, ϕ) is isomorphic to the symplectic space H(k) ∶= (1 ¬ , ↾1 ¬ ), where 1 ¬ is the orthogonal of 1 ∶= (1, . . ., 1) with respect to the scalar product on F k+1 2 .For reader's convenience, we give a proof.The proof, suggested by Christian Delhommé, is based on two results exposed in Algebra, Vol. 3, of Cohn (1991).Let (U, ϕ) be as stated in the above theorem.Case 1: ϕ is not symplectic, that is ϕ(x, x) ≠ 0 for some vector x.We apply Proposition 7.1 page 344 of Cohn (1991), namely: If U is a vector space of characteristic 2 and ϕ is a symmetric bilinear form which is not alternating, then U has a orthogonal basis.Since ϕ is nondegenerate and the field if F 2 , any orthogonal basis is orthonormal, hence ϕ is a scalar product.Case 2: ϕ is symplectic.In this case, Lemma 5.1, p.331 of Cohn (1991) asserts in particular that: Every symplectic space, (that is a space equipped with a bilinear symmetric form which is nondegenerate and alternating) on an arbitrary field is a sum of hyperbolic planes.Thus k is even and in our case U is isomorphic to any symplectic space with the same dimension, in particular to H(k).
When dealing with these notions of dimension, we may always consider nondegenerate forms, hence in the case of finite dimensional representation, Theorem 1.14 applies.In fact Lemma 1.3 and Theorem 1.4 extend.
Let U be a vector space over F 2 and ϕ a symmetric bilinear form defined on U with values in F 2 .Let O ¬ ϕ be the graph of the non-orthogonality relation on U , that is, the graph whose edges are the pairs of distinct elements x and y such that ϕ(x, y) = 1.If k is an integer, then we denote by O ¬ F2 (k) the graph on F k 2 of the non-orthogonality relation associated with the inner product .Similarly, for k even, let O ¬ H (k) be the graph on H(k), the orthogonal of 1 ∶= (1, . . ., 1) with respect to the scalar product on (F 2 ) k+1 , equipped with the symplectic form induced by the scalar product.Lemma 1.15.If dim(U ), the dimension of the vector space U , is at least 3, then the graph O ¬ ϕ has no duo if and only if ϕ is nondegenerate.Hence, dim geom (O ¬ ϕ ) = dim(U ) when ϕ is nondegenerate.
Proof: Suppose that ϕ is degenerate.Pick a nonzero element a in the kernel of ϕ.Then, as it is easy to check, the 2-element set {0, a} is a module of O ¬ ϕ .Conversely, let {a, b} be a duo of O ¬ ϕ .We claim that c ∶= a+b belongs to the kernel of ϕ, that is ϕ(x, c) = 0 for every x ∈ U .Indeed, if x ∈ {a, b}, then ϕ(x, a) = ϕ(x, b), hence ϕ(x, c) = 0 since {a, b} is a module.If x ∈ {a, b} (e.g.x ∶= a), then since dim(U ) ≥ 3, we may pick some z ∈ span{a, b} ∶= {0, a, b, a + b}, hence ϕ(z, c) = 0. Since z + a ∈ {a, b}, ϕ(z + a, c) = 0.It follows that ϕ(a, c) = 0, proving our claim.According to Lemma 1.3, every representation of O ¬ ϕ is one to one; since the identity map is a representation, we have dim geom (O ¬ ϕ ) = dim(U ).We give below an existential result.The proof of the second item is based on the ∆-system lemma (see (Kunen, 2011;Rinot) for an elementary proof) that we recall now.
Lemma 1.16.Suppose that κ is a regular uncountable cardinal, and A ∶= (A α ) α∈κ is a family of finite sets.Then there exist a subfamily B ∶= (A α ) α∈K , where the cardinality of K is κ, and a finite set R such that A α ∩ A β = R for all distinct α, β ∈ K.
1. Every graph has a symplectic dimension, and hence, it has a geometric one.However: 2. not every graph has an inner dimension, e.g., a graph with κ vertices, with κ regular, and no clique and no independent set of κ vertices, does not have an inner representation; on an other hand: 3. every locally finite graph has an inner dimension. Proof: 1. Let G be a graph, and κ ∶= V (G) .Let U be a vector space over F 2 with dimension κ (e.g., , the set of maps f ∶ V (G) → F 2 which are 0 almost everywhere).Define a symplectic form ϕ on a basis Then extend ϕ on U by bilinearity.Since the vectors of the basis are isotropic and F 2 has characteristic two, ϕ is symplectic.By construction, the map v → b v is a representation of G in (U, ϕ).Hence G has a symplectic dimension.
2. An inner representation of a graph G reduces to a map f from V (G) into the vector space [E] <ω of finite subsets of a set E equipped with the symmetric difference such that for every two-element subset e ∶= {u, v} of V (G), we have e ∈ E(G) if and only if f (u) ∩ f (v) is odd.Suppose that V (G) = κ and no subset of V (G) of cardinality κ is a clique or an independent set.According to Ramsey's theorem, κ is uncountable.Apply Lemma 1.16 to A ∶= (f (u)) u∈V (G) .Let B ∶= (f (u)) u∈K be a subfamily of A, where K has cardinal κ, and let R be given by this lemma.Since f (u) ∩ f (v) = R for all every u, v ∈ K, the set K is a clique or an independent set depending on the fact that the cardinality of R is odd or even.Hence, if G has no clique and no independent set of κ vertices, it cannot have an inner representation.A basic example on cardinality ℵ 1 is provided by the comparability graph G of a Sierpinskization of a subchain A of the reals of cardinality ℵ 1 with an order of type ω 1 on A.

Let E ∶= E(G).
Let [E] <ω be the collection of finite subsets of E; equipped with the symmetric difference ∆, [E] <ω is a vector space over F 2 ; the one-element subsets of E form a basis; the map otherwise is a bilinear form for which the one-element subsets of E form an orthonormal basis.Hence ϕ is an inner product.Let f ∶ V (G) → (E) be defined by setting f As noted by Delhommé (2021), the Boolean dimension can be strictly smaller than the geometric dimension.For an example, if κ is an infinite cardinal, the geometric dimension of O(κ) ¬ is 2 κ while its Boolean dimension is at most κ.Indeed, from Theorem 1.17, O(κ) ¬ has a geometric representation in a vector space U .As for any representation, Lemma 1.3 is still valid; since O(κ) ¬ has no duo (for κ ≥ 3) the cardinality of U is at least 2 κ , thus the dimension of the vector space U is 2 κ , while O(κ) ¬ has a Boolean representation in a set of cardinality κ.Problem 1.3.Does every countable graph has an inner dimension? (i)   1.3 Graphs with finite geometric dimension We give some examples when the graphs are finite.Claim 1.21.If + 1 subsets A i , i < + 1, of an -element set A have odd size, then there are i, j < + 1, i = j such that A i ∩ A j has odd size.
We prove now that the examples satisfy the stated conditions.Item (a).The first part is obvious.For the second part, we use Claim 1.21 and Lemma 1.20.Indeed, let f ∶ V (G) → H(2k).Composing with the involution h of F 2k+1 2 we get a representation in 1 + H(2k), where the involution h is defined by h(x) = x + 1, where 1 ∶= (1, 1, . . ., 1) ∈ F 2k+1 2 .The image of a clique of G yields subsets of odd size such that the intersection of distinct subsets has even size.Thus from Claim 1.21 above there are no more than 2k + 1 such sets.
With that in hand, we prove the desired equality dim symp (K ◻ The paper by Godsil and Royle (2001a) contains many more results on the symplectic dimension over F 2 of finite graphs.

Dimension and rank
We compute the symplectic dimension and the geometric dimension of a graph G in terms of its adjacency matrix.
Let n ∈ N. Let A be an n × n symmetric matrix with coefficients in a field F. We denote by rank F (A) the rank of A computed over the field F. The minrank of A, denoted by minrank F (A), is the minimum of rank F (A + D), where D is any diagonal symmetric matrix with coefficients in F. If F = F 2 , we denote these quantities by rank 2 (A) and minrank 2 (A).Let G ∶= (V, E) be a graph on n vertices.Let v 1 , . . ., v n be an enumeration of V .The adjacency matrix of G is the n × n matrix A(G) ∶= (a i,j ) 1≤i,j≤n such that a i,j = 1 if v i ∼ v j and a i,j = 0 otherwise.Theorem 1.22.If G is a graph on n vertices, then the symplectic and the geometric dimensions of G over a field F are respectively equal to the rank and the minrank of A(G) over F.
An n×n symmetric matrix B ∶= (b i,j ) 1≤i,j≤n over a field F is representable as the matrix of a symmetric bilinear form ϕ on a vector space U over a field F if there exists n vectors u 1 , . . ., u n in U , not necessarily distinct, such that b i,j = ϕ(u i , u j ) for all 1 ≤ i, j ≤ n.
The matrix B can be represented in U ∶= F n , where (u i ) 1≤i≤n is the canonical basis and ϕ(u i , u j ) = b i,j .According to the following lemma (see Corollary 8.9.2 p. 179 of Godsil and Royle (2001b)), there is a representation in a vector space whose dimension is the rank of the matrix B.
Lemma 1.23.An n × n symmetric matrix B of rank r has a principal r × r submatrix of full rank.
The following result shows that this value is optimum.
Lemma 1.24.The smallest dimension of a vector space in which a symmetric matrix B is representable is the rank of B.
Proof: It is an immediate consequence of the following facts, whose proofs are a simple exercise in linear algebra.
1) Let r ∶= rank(B).Then r ≤ dim(U ) for any vector space U in which B is representable.Let ϕ be a bilinear form on U , and let u 1 , . . ., u n be n vectors of U such that ϕ(u i , u j ) = b ij for all 1 ≤ i, j ≤ n, where (b i,j ) 1≤i,j≤n = B. Let B(j 1 ), . . ., B(j r ) be r linearly independent column vectors of B with indices j 1 , . . ., j r .We claim that the corresponding vectors u j1 , . . ., u jr are linearly independent in U .
Suppose that a linear combination r k=1 λ j k u j k is zero.Then, for every vector u ∈ U , ϕ(

This rewrites as
Since these column vectors are linearly independent, the λ j k 's are zero.This proves our claim.
2) Suppose that ϕ is nondegenerate and U is spanned by the vectors u 1 , . . ., u n .Then r ≥ dim(U ).The proof follows the same lines as above.Let s ∶= dim(U ).Then, among the u j 's there are s linearly independent vectors, say u j1 , . . ., u js .We claim that the column vectors B(j 1 ), . . ., B(j s ) are linearly independent.Suppose that a linear combination Since the vectors u j1 , . . ., u js are linearly independent, the λ k 's are all zero.This proves our claim.
3) Suppose that B is representable in a vector space U equipped with a symmetric bilinear form ϕ. Then B is representable in a quotient of U equipped with a nondegenerate bilinear form.
Theorem 1.22 follows immediately from Lemma 1.24.Remark 1.25.Theorem 1.22 for the symplectic dimension of graphs over F 2 is due to Godsil and Royle (2001a).The minrank over several fields has been intensively studied, see Fallat and Hogben (2007) for a survey.These authors consider the problem of minrank of graphs, and obtain a combinatorial description for the minimum rank of trees.In the next section, we only state that in case of trees, the Boolean dimension, geometric dimension and the minimum rank coincide, thus the formula given in Theorem 2.9 below for the Boolean dimension gives yet another combinatorial description for the minimum rank of a tree.

Boolean dimension of trees
In this section, we show that there is a nice combinatorial interpretation for the Boolean dimension of trees.We mention first the following result of Houmem Belkhechine et al.Belkhechine et al. (2012).
2 be a representation of G in the vector space F m 2 equipped with a symmetric bilinear form ϕ.
Proof: Let X be a non empty finite subset of A. We claim that Since this holds for every finite subset X of A, the conclusion follows.
This suggests the following definition.
A is said to be dependent (mod 2).Let ind 2 (G) be the maximum size of an independent set (mod 2) in G. From now, we omit (mod 2) unless it is necessary to talk about independence in the graph theoretic sense.
Note that the independent sets (mod 2) of a graph do not form a matroid in general.Indeed, let G be made of six vertices, three, say {a, b, c} forming a clique, the three others, say a ′ , b ′ , c ′ being respectively connected to a, b and c.Then {a ′ , a, b, c} is independent (mod 2), hence 4 ≤ ind 2 (G).Also, {a ′ , b ′ , c ′ } is independent (mod 2) but cannot be extended to a larger independent set (mod 2).Since G is the Boolean sum of a 3-vertex clique and three edges, From Corollary 2.3 above, we deduce the following result.
Theorem 2.4.The Boolean dimension of a path on n vertices (n ∈ N, n > 0) is n−1.Every other n-vertex graph, with n ≥ 2, has dimension at most n − 2.
Proof: Let P n be the path on {0, . . ., n − 1}, whose edges are pairs {i, i + 1}, with i < n − 1. Suppose n ≥ 2. Since P n is the Boolean sum of its edges, dim Bool (P n ) ≤ n − 1.Let A ∶= {0, . . ., n − 2}.Then A is independent (mod 2).Indeed, let X be a nonempty subset of A and x be its largest element, then the vertex of trivial stars in Σ (stars that are isomorphic to K 1,1 ), and let s(Σ) be the number of nontrivial stars in Σ (stars that are isomorphic to K 1,m for some m > 1).We define the parameter m(T ) ∶= min Σ {t(Σ) + 2s(Σ)} over all star decompositions Σ of T .A star decomposition Σ of T for which t(Σ)+2s(Σ) = m(T ) is called an optimal star decomposition of T .
The Boolean dimension of a graph counts the minimum number of cliques needed to obtain this graph as a Boolean sum.If Σ ∶= {S 1 , . . ., S k } is a star decomposition of a tree T , one has dim Bool (T ) ≤ , it is the Boolean sum of a clique on m + 1 vertices and a clique on a subset of m vertices), hence we have Here is our result.Theorem 2.9.For all trees T , we have ind 2 (T ) = dim Bool (T ) = m(T ).
We introduce the following definition.Definition 2.10.A cherry in a tree T is a maximal subtree S isomorphic to K 1,m for some m > 1 that contains m end vertices of T .We refer to a cherry with m edges as an m-cherry.Proposition 2.11.Let T ∶= (V, E) be a tree that contains a cherry.If all proper subtrees T ′ of T satisfy ind 2 (T ′ ) = m(T ′ ), then ind 2 (T ) = m(T ).
Proof: Let x ∈ V be the center of a k-cherry in T , with N T (x) = {u 1 , . . ., u k , w 1 , . . ., w }, where d T (u i ) = 1 for all i, and d T (w i ) > 1 for all i.For each i = 1 to , let T i be the maximal subtree that contains w i but does not contain x.
First, we show that any optimal star decomposition of T in which x is not the center of a nontrivial star can be transformed into an optimal star decomposition in which x is the center of a nontrivial star.Consider an optimal star decomposition Σ in which x is not the center of a nontrivial star.Therefore, edges xu i are trivial stars of Σ.Now if k > 2 or if there is a trivial star xw i in Σ, then we could have improved t(Σ) + 2s(Σ) by replacing all trivial stars containing x by their union, which is a star centered at x. Hence, assume that k = 2 and each w i is the center of a nontrivial star S i , which contains the edge xw i .Now replace each S i by S ′ i ∶= S i − xw i , and add a new star centered at x with edge set {xw 1 , . . ., xw , xu 1 , xu 2 }.The new decomposition is also optimal.Now consider an optimal star decomposition Σ in which x is the center of a nontrivial star.The induced decompositions on T i are all optimal since Σ is optimal.For each i ∈ {1, . . ., }, let A i be a maximum size independent set in T i .Hence Consider a non-empty set X ⊆ A. We show that there exists Proposition 2.12.Let T ∶= (V, E) be a tree that contains a vertex y of degree 2 adjacent to a vertex z of degree 1.If ind 2 (T − z) = m(T − z), then ind 2 (T ) = m(T ).
Proof: First, we show that m(T ) = m(T − z) + 1.If there is an optimal star decomposition of T − z − y in which some vertex x is the center of a star, then m(T Now we consider a maximum sized independent set A ′ in T −z.We have Since B ′ contains x and y, we conclude that u is not adjacent to any of y and z, hence N T (u) ∩ B is odd.
Thus we have shown that A is independent.We have ind 2 (T ) ≥ A = A ′ + 1 = m(T − z) + 1 = m(T ).Since ind 2 (T ) cannot be more than m(T ), we have ind 2 (T ) = m(T ).
Proof Proof of Theorem 2.9: If a tree T has two vertices, then ind 2 (T ) = m(T ) = 1.Each tree with at least 3 vertices contains a cherry or a vertex of degree 2 adjacent to a vertex of degree 1. (This is seen by considering the second-to-last vertex of a longest path in T .)Now, induction on the number of vertices, using Propositions 2.11 and 2.12, implies the result.
3 Inversion index of a tournament and Boolean dimension

Inversion index of a tournament
Let T be a tournament.Let V (T ) be its vertex set and A(T ) be its arc set.An inversion of an arc a ∶= (x, y) ∈ A(T ) consists to replace the arc a by a ⋆ ∶= (y, x) in A(T ).For a subset X ⊆ V (T ), let Inv(T, X) be the tournament obtained from T after reversing all arcs (x, y) ∈ A(T ) ∩ (X × X).For example, Inv(T, V ) is T * , the dual of T .For a finite sequence (X i ) i<m of subsets of V (T ), let Inv(T, (X i ) i<m ) be the tournament obtained from T by reversing successively all the arcs in each of the subsets X i , i < m, that is, the tournament equal to T if m = 0 and to Inv(Inv(T, (X i ) i<m−1 ), X m−1 ) if m ≥ 1. Said differently, an arc (x, y) ∈ A(T ) is reversed if and only if the number of indices i such that {x, y} ⊆ X i is odd.The inversion index of T , denoted by i(T ), is the least integer m such that there is a sequence (X i ) i<m of subsets of V (T ) for which Inv(T, (X i ) i<m ) is acyclic.
In the sequel, we consider tournaments for which this index is finite.In full generality, the inversion index of a tournament T can be defined as the least cardinal κ such the Boolean sum of T and a graph of Boolean dimension κ is acyclic.The case κ finite is stated in Lemma 3.8 below.We leave tournaments with infinite inversion index to further studies.
The motivation for the notion of inversion index originates in the study of critical tournaments.Indeed, the critical tournaments of Schmerl and Trotter (1993) can be easily defined from acyclic tournaments by means of one or two inversions whereas the (−1)-critical tournaments, characterized in Belkhechine et al. (2007), can be defined by means of two, three or four inversions Belkhechine (2009).Another interest comes from the point of view of logic.
Results about the inversion index originate in the thesis of Belkhechine (2009).Some results have been announced in Belkhechine et al. (2010); they have been presented at several conferences by the first author and included in a circulating manuscript Belkhechine et al. (2012).The lack of answer for some basic questions is responsible for the delay of publication.
The inversion index is a variant of the Slater index: the least number of arcs of a tournament which have to be reversed in order to get an acyclic tournament (Slater, 1961).The complexity of the computation of the Slater index was raised by Bang-Jensen and Thomassen (1992).Alon (2006) and independently Charbit, Thomassé, and Yeo (2007) showed that the problem is NP-hard.An extension of the inversion index to oriented graphs is studied in Bang-Jensen et al. (2020).
Problem 3.1.Is the computation of the inversion index NP-hard?Question 3.2.Are there tournaments of arbitrarily large inversion index?
This last question has a positive answer.There are two reasons, the first one is counting, the second one, easier, is based on the notion of well-quasi-ordering.
For n ∈ N, let i(n) be the maximum of the inversion index of tournaments on n vertices.We have i For larger n a counting argument Belkhechine (2009); Belkhechine et al. (2010Belkhechine et al. ( , 2012) ) yields the following result.
It is quite possible that i(n) ≥ ⌊ n−1 2 ⌋, due to the path of strong connectivity (it is not even known if reverse inequality holds).
The path of strong connectivity on n vertices is the tournament T n defined on N <n ∶= {0, . . ., n − 1} whose arcs are all pairs (i, i + 1) and (j, i) such that i + 1 < n and i < j < n.

Well-quasi-ordering
Basic notions of the theory of relations apply to the study of the inversion index.These notions include the quasi order of embeddability, the hereditary classes and their bounds, and the notion of well-quasi-order.
Let I <ω m be the class of finite tournaments T whose inversion index is at most m.This is a hereditary class in the sense that if T ∈ I <ω m and T ′ is embeddable into T then T ′ ∈ I <ω m .It can be characterized by obstructions or bounds.A bound is a tournament not in I <ω m such that all proper subtournaments are in I <ω m .We may note that the inversion index of every bound of I <ω m is at least m + 1.Hence, the fact that I <ω m is distinct of the class of all finite tournaments provides tournaments of inversion index larger than m.This fact relies on the notion of well-quasi-ordering.
A poset P is well-quasi-ordered if every sequence of elements of P contains an increasing subsequence.
Theorem 3.2.The class of all finite tournaments is not well-quasi-ordered by embeddability.
This is a well known fact.As indicated by a referee, it has been mentioned by several authors.See e.g., Latka (1994) for a much stronger version of Theorem 3.2 and also subsection 3.1 of Cherlin and Latka (2000).For the convenience of the reader we give a proof.
Proof: Let T n be the path of strong connectivity on {0, . . ., n − 1} as defined above.Let C n be the tournament obtained from T n by reversing the arc (n − 1, 0).We claim that for n ≥ 7, the C n 's form an antichain.Indeed, to C n we may associate the 3-uniform hypergraph H n on {0, . . ., n − 1} whose 3element hyperedges are the 3-element cycles of C n .An embedding from some C n to another C m , m > n, induces an embedding from H n to H m .To see that such an embedding cannot exist, observe first that the vertices 0 and n − 1 belong to exactly n − 3 hyperedges, and the vertices 1 and n − 2 belong to exactly two hyperedges, the other vertices to three hyperedges, hence an embedding h will send {0, n − 1} on {0, m − 1}.The preservation of the arc (0, n − 1) imposes h(0) = 0 and h(n − 1) = m − 1.Then, the preservation of the arcs (i, i + 1) yields a contradiction since n < m.
Proof: The class L <ω m made of a finite linear order L with m unary predicates U 1 , . . ., U m (alias m distinguished subsets) and ordered by embeddability is well-quasi-ordered.This is a straightforward consequence of Higman's theorem on words (see Higman, 1952), in fact, an equivalent statement.Higman's result asserts that the collection of words on a finite alphabet, ordered by the subword ordering, is wellquasi-ordered.Since members of L <ω m can be coded by words on an alphabet with 2 m elements, the class L <ω m is well-quasi-ordered.The map associating to each (L, U 1 , . . ., U m ) the Boolean sum L +U 1 . . .+U m preserves the embeddability relation, hence the range of that map is well-quasi-ordered.This range being equal to I <ω m , this later class is well-quasi-ordered.
Corollary 3.4.There are finite tournaments with arbitrarily large inversion index.
We have the following result concerning the bounds.
Proof: From the proof of Theorem 3.3, the class I <ω m,1 made of tournaments of I <ω m , with one unary predicate added, is well-quasi-ordered.According to an adaptation of Proposition 2.2 of Pouzet (1972) translated in this case, I <ω m has finitely many bounds.
We thank the referee for observing that the well-quasi-ordering of I <ω m,1 suffices to yield the finiteness of the bounds of I <ω m .Question 3.4.What is the maximum of the cardinality of bounds of I <ω m ?Remark 3.6.It must be observed that the collection of graphs with geometric dimension at most m over a fixed finite field has finitely many bounds and an upper bound on their cardinality is given in Ding and Kotlov (2006).How the cardinality of these bounds relate to the cardinality of bounds of I <ω m is not known.
Theorem 3.7.The inversion index of the sum C 3 .n of 3-cycles over an n-element acyclic tournament is n.Proof of Theorem 3.7.Let T ∶= C 3 .n,V ∶= V (T ) and r ∶= i(T ).Clearly r ≤ n.Conversely, let H be a graph with vertex set V such that L ∶= T +H is an acyclic tournament and dim Bool (H) = r.Let U ∶= (F 2 ) r equipped with the ordinary scalar product and f ∶ V → U be a representation of H. Claim 3.9.For each i ∈ {0, . . ., n − 1}, we may enumerate the vertices of {0, 1, 2} × {i} into x i , y i , z i in such a way that (x i , y i ), (y i , z i ), (z i , x i ) are arcs of T , (f (x i ) f (z i )) = 1 and (f (x i ) f (y i )) = 0.
Claim 3.10.The set {f (x i ) ∶ i < n} is linearly independent in U .
Proof of Claim 3.10.This amounts to prove that ∑ i∈I f (x i ) = 0 for every non-empty subset I of {0, . . ., n − 1}.Let I be such a subset.Let m ∈ {0, . . ., n − 1} such that x m is the largest element of {x i ∶ i ∈ I} in the acyclic tournament L. Subclaim 3.11.(f (x i ) f (z m )) = (f (x i ) f (y m )) for each i ∈ I ∖ {m}.Proof of Subclaim 3.11.By construction, we have x m < L z m and x m < L y m , hence by transitivity x i < L z m and x i < L y m .If i < m in the natural order then, by definition of T , (x i , z m ) ∈ A(T ) and (x i , y m ) ∈ A(T ), thus (f (x i ) f (z m )) = 0 = (f (x i ) f (y m )), whereas if i > m in the natural order, then (z m , x i ) ∈ A(T ) and (y m , x i ) ∈ A(T ), thus (f (x i ) f (z m )) = 1 = (f (x i ) f (y m )), proving the subclaim.◻ Since (f (x m ) f (z m )) = 1 and (f (x m ) f (y m )) = 0, it follows that ∑ i∈I (f (x i ) f (z m )) = ∑ i∈I (f (x i ) f (y m )).That is ((∑ i∈I f (x i )) f (z m )) = ((∑ i∈I f (x i )) f (y m )).Thus the sum ∑ i∈I f (x i ) = 0 as claimed.
◻ We have n ≤ r.This proves the theorem.◻ Theorem 1.18.If the Boolean dimension of a graph G is finite, then it is equal to the inner dimension of G and either 1. the geometric dimension, the symplectic dimension and the Boolean dimension of G are equal, or 2. the geometric dimension and the symplectic dimension of G are equal to the Boolean dimension of G minus 1, or 3. the geometric dimension and the Boolean dimension of G are equal and are strictly less than the symplectic dimension of G, in which case the difference between these two numbers can be arbitrarily large.Proof: The first assertion is obvious.By definition, dim geom (G) ≤ min{dim Bool (G), dim symp (G)}.Apply Theorem 1.14.Let k ∶= dim geom (G).If k = dim Bool (G), then G is representable into H(k) and thus in F k+1 2 , hence (2) holds.If k = dim Bool (G), then dim symp (G) ≥ k.The examples given in (a) below show that the difference dim symp (G) − dim Bool (G) can be large.
(i) Norbert Sauer informed us on january 2022 that the answer is positive These examples are extracted fromBelkhechine et al. (2012).The paper being unpublished, we give a hint below.We use the following lemma.Lemma 1.20.If G ∶= (V, E) is a graph for which dim symp (G) = 2k ∈ N, then every clique of G has at most 2k + 1 elements.This fact is a straightforward consequence of the following claim which appears equivalently formulated invan Lint and Wilson (2001) as Problem 19O.(i), page 238.
The reverse inequality follows from Lemma 1.20.Item (b).If k = 1, the graph O ¬ F2 (k) is made of two isolated vertices, and if k = 2 the graph is a path on three vertices plus an isolated vertex, their respective Boolean dimensions are 1 and 2, as claimed.If k ≥ 3 the result follows from the conclusion of Lemma 1.15.Item (c) If k = 2, the graph O ¬ H (k) is made of a clique on three vertices plus an isolated vertex, hence its Boolean dimension is 1.If k ≥ 4, the equality dim geom (O ¬ H (k)) = dim symp (O ¬ H (k)) follows from the conclusion of Lemma 1.15.The number of edges of O ¬ H (k) and O ¬ F2 (k) are different, hence O ¬ H (k) cannot have a Boolean representation in (F k 2 , ).Since it has a representation in (F k+1 2 , ), the result follows.