    On a fractional version of Haemers' bound

In this note, we present a fractional version of Haemers' bound on the Shannon capacity of a graph, which is originally due to Blasiak. This bound is a common strengthening of both Haemers' bound and the fractional chromatic number of a graph. We show that this fractional version outperforms any bound on the Shannon capacity that could be attained through Haemers' bound. We show also that this bound is multiplicative, unlike Haemers' bound.

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1. Introduction

For graphs , the strong product of , denoted , is the graph on vertex set where if and only if for every , either or . For brevity, we write .

The Shannon capacity of a graph , introduced by Shannon in , is

 Θ(G):=supnα(G⊠n)1/n=limn→∞α(G⊠n)1/n,

where denotes the independence number of . Despite the fact that Shannon defined this parameter in 1956, very little is known about it in general. For example, is still unknown.

There are two general upper bounds on . Firstly, the theta function, , is a bound on

which is the solution to a semi-definite program dealing with arrangements of vectors associated with

. Introduced by Lovász in , the theta function was used to verify that . Secondly, Haemers’ bound, , is a bound on which considers the rank of particular matrices over the field associated with the graph . Introduced by Haemers in [4, 3], was used to provide negative answers to three questions put forward by Lovász in .

In this paper, we present a strengthening of Haemers’ bound by defining a parameter , to which we refer to as the fractional Haemers bound. After we wrote the paper, we learned from Ron Holzman, that this parameter previously appeared in a thesis of Anna Blasiak [2, Section 2.2]. We defer the definition of this parameter to Section 2. We show the following results:

Theorem 1 (First proved by Blasiak ).

For any graph and field ,

 Θ(G)≤Hf(G;F)≤H(G;F).

The following results are new.

Theorem 2.

For any field of nonzero characteristic, there exists an explicit graph with

 Hf(G;F)

for every field .

Therefore, is a strict improvement over both and for some graphs.

Recall that Lovász showed for any graphs ; the fractional Haemers bound shares this property.

Theorem 3.

For graphs and a field ,

 Hf(G⊠H;F)=Hf(G;F)⋅Hf(H;F).

This is in contrast to . As we will show in Proposition 9, for any field , , yet .

does not improve upon the known bounds for the Shannon capacity of odd cycles.

Proposition 4.

For any positive integer and any field , .

The organization of this paper is as follows: in Section 2, we will define ; in fact, we will provide four equivalent definitions, each of which will be useful. In Section 3, we will prove Theorems 1 and 3 and Proposition 4. In Section 4, we will prove Theorem 2 and also show that and can differ when and are different fields. We will then briefly look at two attempts to “fractionalize” the Lovász theta function in Section 5. We conclude with a list of open problems in Section 6

2. The fractional Haemers bound

For a graph and a field , a matrix is said to fit if for all and whenever . Define to be the set of all matrices over that fit . The Haemers bound [4, 3] of is then defined as

 H(G;F):=min{rank(M):M∈MF(G)}.

Haemers introduced as an upper bound on in order to provide negative answers to three questions put forward by Lovász in . When the field is understood or arbitrary, we will condense the notation and write .

A drawback of Haemers’ bound is that it is always an integer. To combat this, we introduce a fractional version.

Let be a field, be a positive integer and be a graph, and consider matrices over whose rows and columns are indexed by . We can consider as a block matrix where for , the block, , consists of the entries with indices for all . We say that is a -representation of over if

1. for all , and

2. for all .

Define to be the set of all -representations of over . We then define the fractional Haemers bound to be

 Hf(G;F):=inf{rank(M)d:M∈MdF(G),d∈Z+}.

Notice that , so . Again, when the field is understood or arbitrary, we will condense the notation and simply write . More specifically, if we were to write, e.g., , it is assumed that the field is the same in both instances.

2.1. Alternative formulations

We now set out three equivalent ways to define , each of which will be useful going forward.

For positive integers , consider assigning to each a pair of matrices . We say that such an assignment is an -representation of over if

1. for every , and

2. whenever .

Proposition 5.

For a graph and field ,

 Hf(G;F)=infn,d{nd:G has an (n,d)-representation over F}.
Proof.

A matrix has if and only if it is of the form where . Let be the submatrix of consisting of all entries with indices for and , and let be defined similarly for . With this, for any , . Therefore, if and only if for every and whenever . ∎

A second way to understand is by considering the lexicographic product, , which is formed by “blowing up” each vertex of into a copy of . More formally, and in whenever either or and . In this context, it easy to verify that , so:

Proposition 6.

For a graph ,

 Hf(G)=infdH(G⋉¯¯¯¯¯¯¯Kd)d.

The last equivalent formulation of is, in some sense, the most general. Consider matrices over whose rows and columns are indexed by where is some positive integer assigned to . As with -representations, we consider as a block matrix where for , the block, , consists of those entries with indices for all . We say that is a rank--representation of over if

1. for all , and

2. whenever .

Proposition 7.

For a graph and field ,

 Hf(G;F)=infM,r{rank(M)r:M is a rank-r-representation of G over F}.
Proof.

The lower bound is immediate as an -representation of is also a rank--representation.

For the other direction, let be a rank--representation of over for some . As for all , we can find an an submatrix of of full rank, call this submatrix . Let be the submatrix of induced by the blocks ; we index the rows and columns of by . For any fixed , has full rank, so we may perform row operations on using only the rows indexed by to transform into . Let be the matrix formed by doing this for every . Therefore, for every . Further, as all row operations occurred only between rows corresponding to the same vertex, if , then we also have . Thus, as was a rank--representation of , is an -representation of . We conclude that

 Hf(G)≤rank(M′′)r=rank(M′)r≤rank(M)r,

so the same is true of the infimum over all and . ∎

3. Proofs of Theorems 1 and 3 and Proposition 4

We now set out to prove that and explore some basic properties.

Recalling the lexicographic product of graphs, the fractional chromatic number of a graph is defined as

 χf(G):=infdχ(G⋉Kd)d.

In his original paper, Shannon  established .

In the same spirit, Lovász showed that . Similarly, Haemers established ; note that, in general, , e.g.  whereas .

Theorem 8.

For any graph ,

 α(G)≤Hf(G)≤χf(¯¯¯¯G).
Proof.

Notice that for any graphs , and . Thus, as , we find

 α(G)=infdd⋅α(G)d=infdα(G⋉¯¯¯¯¯¯¯Kd)d≤infdH(G⋉¯¯¯¯¯¯¯Kd)d=Hf(G),

and

 Hf(G)=infdH(G⋉¯¯¯¯¯¯¯Kd)d≤infdχ(¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯G⋉¯¯¯¯¯¯¯Kd)d=infdχ(¯¯¯¯G⋉Kd)d=χf(¯¯¯¯G).\qed

We now provide a proof of Theorem 3. Before we do so, recall that ; however, the same is not true of .

Proposition 9.

For any field , , yet .

Proof.

As shown by Lovász, . Thus, as is always an integer, .

On the other hand, it is not difficult to verify that ; indeed, Figure 1 provides such a coloring. Therefore, for any field . ∎

Proof of Theorem 3.

Upper bound. Let and , and set where

is the tensor/Kronecker product. We claim that

. Indeed, the rows and columns of are indexed by , so we may in fact suppose they are indexed by . Further, for any , the block of satisfies . As such, . Further, if in , then either or , so either or . In either case, , so we have verified .

Finally, , so

 Hf(G⊠H)≤rank(M∗)d1d2=rank(M)d1⋅rank(N)d2.

Taking infimums establishes .

Lower bound. Let for some . For , let denote the submatrix of consisting of all blocks of the form for . Certainly we can consider as a matrix with blocks . Additionally, for every , the submatrix can be considered a -representation of the graph , where the block is . Set , so .

On the other hand, if , then and as in for every . Thus, as for all , is a rank--representation of , so .

Putting the bounds on and together, we have

 Hf(G)⋅Hf(H)≤rank(M)r⋅rd=rank(M)d,

so taking infimums yields . ∎

Putting together Theorems 3 and 8, we arrive at a proof of Theorem 1.

Proof of Theorem 1.

We have already noted that . On the other hand, by Theorems 3 and 8,

 Θ(G)=supnα(G⊠n)1/n≤supnHf(G⊠n)1/n=supnHf(G)=Hf(G).\qed

Theorem 3 has another nice corollary. Certainly , but one could additionally attain bounds on by using Haemers’ bound on large powers of , i.e. . This could lead to improved bounds as in general , e.g. . It turns out that outperforms any bound attained in this fashion.

Corollary 10.

For any positive integer and graph , .

Proof.

By Theorem 3, we calculate . ∎

To end this section, we show that the fractional Haemers bound cannot improve upon the known bounds for the Shannon capacity of odd cycles. We require the following observation about -representations of a graph.

Proposition 11.

Let be a graph and be an -representation of . For , let denote the column space of . If are disjoint sets of vertices where is an independent set and there are no edges between and , then the subspaces and are linearly independent.

Proof.

Let be the columns of and be the columns of . As is an -representation of , we know that

1. for every ,

2. for every and , and

3. for every whenever .

Let be a basis for and consider a linear combination

 ∑v∈Sd∑i=1cv,iav,i+∑x∈Bdxx=0,

where for every . As is an independent set and there are no edges between and , we find that for any ,

 0=⟨∑v∈Sd∑i=1cv,iav,i+∑x∈Bdxx,bu,j⟩=cu,j.

Therefore for every , so we must have . As is a basis, this implies for every .

Thus, as and , we have shown that and are linearly independent subspaces. ∎

Proof of Proposition 4.

We will show that . It is well-known that , so we will focus only on the lower bound.

Identify the vertices of with in the natural way and let be an -representation of for any . Let denote the column space of . As , we observe that .

We observe that is an independent set in and further that the edge is not adjacent to any vertex in . By iterating Proposition 11, we find that

 dim((X0+X1)+∑i∈IXi)=dim(X0+X1)+dim(∑i∈IXi)=dim(X0+X1)+∑i∈Idim(Xi),

so

 n ≥dim((X0+X1)+∑i∈IXi) =dim(X0+X1)+∑i∈Idim(Xi) =dim(X0)+dim(X1)−dim(X0∩X1)+∑i∈Idim(Xi) =(k+1)d−dim(X0∩X1).

From this, we have , and so by symmetry, for any , .

Because in , by Proposition 11 it follows that . Since we also have , we conclude that

 d =dim(X0)≥dim((X0∩X1)+(X0∩X2k)) =dim(X0∩X1)+dim(X0∩X2k)≥2((k+1)d−n),

which implies that Taking the infimum yields . ∎

4. Proof of Theorem 2 and further separation

In this section, we first give a proof of Theorem 2; namely, for every field of nonzero characteristic, we need to find a graph for which for every field . After this, we provide further separation of over fields of different characteristics.

For the proof of Theorem 2, we need the following result. The first part of the following lemma was provided by Haemers in ; we provide a full proof for completeness.

Lemma 12.

For a prime and an integer , let be the graph with vertex set where in whenever .

1. If and is a field of characteristic , then

 H(Jpn;F)=α(Jpn)=n,
2. For fixed and all large ,

 ϑ(Jpn)=(p(p+1)2−o(1))n2.

Before continuing with the proof, it is important to point out a typo in  in which it is stated that . The correct formula is , though for brevity’s sake we prove only .

Proof.
1. [leftmargin=*]

2. Let be the incidence matrix of all -subsets of , i.e. the matrix with entries . Certainly the matrix fits over as has characteristic and . Thus, . On the other hand, as , partition into sets where for all . If are sets of size , then either or . Thus, the collection is an independent set in and has size .

3. We require the following fact which can be deduced quickly from  (see specifically items (12), (13) and (27)): provided ,

 ϑ(Jpn)=max1+a1+ap+1s.t.(p+1−u)(n−p−u−1)−u(p+1)(n−p−1)a1+(−1)u(n−p−u−1p+1−u)(n−p−1p+1)ap+1≥−1,for u∈{0,…,p+1}.

For large , the preceding inequality can be written as

 p+1−up+1a1+(−1)un−u(p+1)!(p+1−u)!ap+1 ≥−1+o(1)for u∈{0,…,p} (4.1) −1na1+O(n−p−1ap+1) ≥−1+o(1)for u=p+1. (4.2)

Set where the precise value of the term is chosen so that (4.2) is satisfied. Then set where the value of is chosen so the inequality (4.1) with is satisfied. The remaining inequalities are then satisfied as well. Indeed, for even , both terms on the left side of (4.1) are positive, whereas for odd the second term is for . Hence, .

On the other hand (4.2) implies that . Then the case of (4.1) implies that . Hence, .∎

Proof of Theorem 2.

Let be a field of characteristic and set where and . By Proposition 4 and Lemma 12, we calculate

 Hf(G;F)=Hf(Jpn;F)⋅Hf(C5;F)3=1258n.

For any field , as and , we have

 H(G;F′)≥Hf(G;F′)=Hf(Jpn;F′)⋅Hf(C5;F′)≥1258n.

However, as and is always an integer, we have .

Further, by Lemma 12 and the fact that , we have

 ϑ(G)=ϑ(Jpn)⋅ϑ(C5)3=53/2(p(p+1)2−o(1))n2.

Thus, for sufficiently large with and ,

 Hf(G;F)

for every field . ∎

We next show that the choice of field matters when evaluating . In particular, for any field of nonzero characteristic, we will show that there is an explicit graph for which for any field with .

First, we define a “universal graph” for . For a field and positive integers , define the graph as follows: where in if and only if . We require the following two facts.

Observation 13.

For graphs and , if there is a graph homomorphism from to , then .

Observation 14.

A graph has an -representation over if and only if there is a graph homomorphism from to