The core of a complementary prism

The complementary prism ΓΓ¯\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Gamma {\bar{\Gamma }}$$\end{document} is constructed from the disjoint union of a graph Γ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Gamma $$\end{document} and its complement Γ¯\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\bar{\Gamma }}$$\end{document} if an edge is added between each pair of identical vertices in Γ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Gamma $$\end{document} and Γ¯\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\bar{\Gamma }}$$\end{document}. It generalizes the Petersen graph, which is obtained if Γ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Gamma $$\end{document} is the pentagon. The core of the complementary prism is investigated for arbitrary simple graph Γ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Gamma $$\end{document}. In particular, it is shown that if Γ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Gamma $$\end{document} is strongly regular and self-complementary, then ΓΓ¯\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\Gamma {\bar{\Gamma }}$$\end{document} is a core, i.e. all its endomorphisms are automorphisms.


Introduction
The existence of a homomorphism ϕ : 1 → 2 between two graphs provides information on the relation between the two values of various graph invariants for 1 and 2 .For example, the clique number and the chromatic number of 1 cannot exceed the corresponding value for 2 .Since many of these invariants are difficult to compute, the study of graph homomorphisms has gained in importance in the last couple of decades, and some systematic treatment appeared in the literature [20,23,26].On the other hand it is also not surprising that exploring graph homomorphisms between two given graphs is often a hard task.Related to this research is a notion of a core, which is a graph with the property that each its endomorphism is an automorphism.Each graph has a subgraph , which is a core and such that there exists some homomorphism ϕ : → .It is referred as the core of and it is unique up to isomorphism.The and u v otherwise.We use K n and P n to denote the complete graph and the path on n vertices, respectively.A subset C ⊆ V ( ) is a clique if all its elements are pairwise adjacent.A subset I ⊆ V ( ) is independent if it is a clique in the complement ¯ of graph .The clique number ω( ) and the independence number α( ) are the orders of the largest clique and the largest independent set in , respectively.Hence, ω( ¯ ) = α( ).The chromatic number of a graph is denoted by χ( ).Clearly, χ( ) ≥ ω( ).
Example 2. 1 Recall that the Kneser graph K (n, r ), for 2r < n, has the vertex set formed by all r -subsets in {1, 2, . . ., n}, where two r -subsets form and edge if and only if they are disjoint.It is obvious that ω(K (n, r )) = n r .On the other hand the Erdős-Ko-Rado theorem states that ω(K (n, r )) = α(K (n, r )) = n−1 r −1 (cf.[20,Theorem 7.8.1]).The famous Lovász proof of the Kneser conjecture [29] implies that χ(K (n, r )) = n − 2r + 2 (see also [20,Theorem 7.11.4]).The chromatic number of the complement is computed in [2, Corollary 4, p. 93] and equals for all u, v ∈ V ( 1 ).In particular, ϕ(u) = ϕ(v) for the two endpoints of each edge in 1 .If in addition ϕ is bijective, and the implication in (1) is replaced by the equivalence ⇐⇒, then ϕ is a graph isomorphism and graphs 1 , 2 are isomorphic, which is denoted by 1 ∼ = 2 .If 1 = 2 , then a graph homomorphism/graph isomorphism is a graph endomorphism/automorphism, respectively.A graph is self-complementary if there exists some antimorphism (a.k.a.complementing permutation), which is a graph isomorphism ϕ : → ¯ .A graph is k-regular if each vertex v ∈ V ( ) has k neighbours.A k-regular graph on n vertices is strongly regular with parameters (n, k, λ, μ) if each pair of adjacent vertices has λ common neighbours, while each pair of distinct nonadjacent vertices has μ common neighbours.
Given a graph with the vertex set {v 1 , . . ., v n } let A be its adjacency matrix, i.e. a n × n real matrix with (i, j)-th entry equal to 1 if v i ∼ v j and 0 otherwise.In this paper its eigenvalues are referred as the eigenvalues of , and we number them in the decreasing order [4]).Consequently, the largest eigenvalue of a regular self-complementary graph with n vertices equals n−1 2 , while other eigenvalues occur in pairs {λ, −1 − λ} that are symmetric with respect to the value − 1 2 .More can be said in the case of strong regularity.In fact, the parameters (n, k, λ, μ) of a strongly regular self-complementary graph are of the form [36,Lemma 5.1.2]or [33,Theorem 2]).The spectrum of the adjacency matrix of a strongly regular graph is determined by its parameters (cf.[4,Theorem 9.1.3]).Consequently, we have the following lemma.

Properties of graph homomorphisms and cores
In the paper we rely on some graph parameters that are homomorphism monotone, which means that the existence of a homomorphism between two graphs implies that the graph parameter of the first graph is not larger than the corresponding value on the second graph.Lemma 2.3 says that the clique number and the chromatic number are both homomorphism monotone.The claim (i) is obvious.The claim (ii) is also well known and can be found in [23].
Lemma 2.3 Let 1 and 2 be graphs such that there exists a homomorphism from 1 to 2 .Then The Lovász theta function ϑ( ) of a graph with vertex set {v 1 , . . ., v n } was introduced in [30], and it can be defined as where the set M consists of all n × n real symmetric matrices M = [m i j ] such that m i j = 1 whenever i = j or {v i , v j } / ∈ E( ), and λ 1 (M) is the largest eigenvalue of M (see [4,30]).The value ϑ( ¯ ) can be defined also in a different way.If n = |V ( )| and ν < 0 is a real number, then consider the infinite graph S(n, ν) with the unit sphere S n+1 := {x ∈ R n : x, x n = 1} as the vertex set and the edge set given by {{x, y} : x, y ∈ S n+1 , x, y n = ν}.Here, x, y n = n i=1 x i y i is the dot product of vectors x = (x 1 , . . ., x n ) and y = (y 1 , . . ., y n ).In [28,Theorem 8.2] it was shown that the Lovász theta function of the complement ϑ( ¯ ) is the same as the infimum among all values 1 − 1 ν , where ν ranges over all negative values such that there exists a homomorphism ϕ n,ν : → S(n, ν).Consequently it was observed in [32] that ϑ( 1 ) ≤ ϑ( 2 ) whenever there exists a homomorphism ϕ : 1 → 2 (see Lemma 2.4).In fact, if n 1 := |V ( 1 )|, n 2 := |V ( 2 )|, and ϕ n 2 ,ν is a homomorphism between 2 and S(n 2 , ν), then the image of the map ϕ n 2 ,ν • ϕ spans a vector subspace U ⊆ R n 2 of dimension dim U ≤ n 1 .If we choose any (linear) map g : that preserves the dot product, and the map f : R dim U → R n 1 that extends vectors by n 1 − dim U zero entries, we deduce that the map Lemma 2.4 Let 1 and 2 be graphs such that there exists a homomorphism from 1 to 2 .Then ϑ( 1 ) ≤ ϑ( 2 ).
In the proof of Theorem 3.7 we rely also on the properties of ϑ that are given by Lemmas 2.5 and 2.6, and which were proved already by Lovász [ A graph is a core if all its endomorphisms are automorphisms.Simple examples include complete graphs and odd cycles.In fact, Lemma 2.3(ii) implies that all vertexcritical graphs are cores, i.e. graphs, where a removal of any vertex decreases the chromatic number.A subgraph in a graph is a core of , if is a core and there exists some graph homomorphism from to .A core of a graph is an induced subgraph and is unique up to isomorphism [20,Lemma 6.2.2].By core( ) we denote any of the cores of .There always exists a retraction ψ : → core( ), i.e. a graph homomorphism that fixes each vertex in core( ).Namely, if ϕ : → core( ) is any graph homomorphism, then its restriction to the vertices of core( ) is an automorphism of core( ) and therefore (ϕ| V (core( )) ) −1 • ϕ is a retraction.The core of a graph is a complete graph if and only if χ( ) = ω( ).
The following result is well known, and it can be found in [20,Lemma 6.2.3].
Lemma 2.8 Let 1 and 2 be graphs.Then there exist homomorphisms from 1 to 2 and from 2 to 1 if and only if core( 1 ) ∼ = core( 2 ).

Remark 2.9
In particular, if ψ is a retraction from onto its image, then and the graph induced by the image of ψ have isomorphic cores.
Lemma 2.10 is also well known, but we did not find it in the literature.

Lemma 2.10 If graph is connected, then core( ) is connected too.
Proof Let u and v be arbitrary vertices in core( ).Since core( ) is a subgraph in the connected graph , we can find a walk in that joins u and v.If ψ is any retraction of onto core( ), then is a walk in core( ) that joins u and v.

Remark 2.11
We end this subsection by emphasizing that the core of a regular graph is not necessarily regular.Consider for example the regular graph in Fig. 1.The map ψ defined on the vertex set {1, . . ., 9} by ψ(8) = 4, ψ(9) = 5, and ψ(i) = i for i ≤ 7, is a retraction onto the bold part of the graph.Since the bold part is a vertex-critical graph, it is a core of .Clearly, it is not regular.

Complementary prisms
Let be a graph with the vertex set V ( ) = {v 1 , . . ., v n }.The complementary prism of is the graph ¯ , which is obtained by the disjoint union of and its complement ¯ , if we connect each vertex in to its copy in ¯ .More precisely, where and the edge set E( ¯ ) is given by the union In particular, Clearly, the complementary prism ¯ is regular if and only if is n−1 2 -regular (see also [7,Theorem 3.6]).Recall that the diameter of a connected graph is the length of the shortest path between the most distanced vertices.Lemma 2.12 is proved in [24,Theorem 2].Lemma 2.12 Let be any graph.Then ¯ is a connected graph of diameter at most three.Moreover, Lemma 2.13 is proved in [7,Corollary 3.4].

the eigenvalues of the adjacency matrix of . Then the set of all eigenvalues of the adjacency matrix of
Corollary 2.14 is essential in the proof of Theorem 3.7.

Corollary 2.14
If is as in Lemma 2.13, then the largest and the smallest eigenvalues of respectively.
Proof Obviously, (5) If the right-hand side is negative, then the inequality is obviously true.Otherwise we can square the inequality and rearrange it to deduce n 2 (m − 4k 2 − 4k − 1) ≥ 0, which is true since m = 4k 2 + 4k + 5.

The core of a complementary prism
We are now ready to state and prove the main results of this paper.We begin by studying the core of a complementary prism ¯ for general graph .Beside the three expected situations (i), (ii), (iii) in Theorem 3.1, two additional and very restricted possibilities for core( ¯ ) can occur.Recall that W 1 and W 2 denote the parts of the vertex set V ( ¯ ), which correspond to graphs and ¯ , respectively.
Theorem 3.1 Let be any graph that is not isomorphic to K 2 or K 2 .If core( ¯ ) is any core of ¯ , then one of the following five possibilities is true.
nonempty and the following statements are satisfied.
(a) The vertex set of core( ¯ ) equals (b) In , there is no edge with one endpoint in V 1 and the other endpoint in V 2 .
(c) In , each vertex in V 3 is adjacent to at most one vertex in V 2 .
(d) In , there exists a vertex in V 2 that has no neighbours in V 3 .
(e) If is any retraction from ¯ onto core( ¯ ), then nonempty and the following statements are satisfied.
(a) The vertex set of core( ¯ ) equals (b) In ¯ , there is no edge with one endpoint in V 1 and the other endpoint in V 2 .
(c) In ¯ , each vertex in V 3 is adjacent to at most one vertex in V 2 .
(d) In ¯ , there exists a vertex in V 2 that has no neighbours in V 3 .
(e) If is any retraction from ¯ onto core( ¯ ), then Proof The vertex set of core( ¯ ) can be written as where U 1 , U 2 are two subsets of V ( ) and at least one of them is nonempty.If U 2 = ∅, then the vertices of core( ¯ ) are contained in W 1 and any retraction of ¯ onto core( ¯ ) is a homomorphism from ¯ to .Since is isomorphic to an induced subgraph in ¯ we have also the obvious homomorphism in the other direction.From Lemma 2.8 we infer that core( ¯ ) ∼ = core( ), i.e. (ii) is true.If U 1 = ∅, then we deduce that (iii) is true in the same way.In what follows we assume that Moreover, Lemmas 2.12 and 2.10 imply that core( ¯ ) is connected.Hence, U 1 ∩ U 2 = ∅.We now separate three cases.
Next, observe that |U 1 | = 2 is not possible, since it would imply that core( ¯ ) is isomorphic to a path on four vertices, which is not a core.
Case 2 Let U 2 \U 1 = ∅.We prove that in this case (iv) is true.Denote Suppose that (b) is not true, i.e. there are vertices Next we prove (9).
i.e. (9).To prove the claim (a), assume the contrary, i.e.U 1 \U 2 = ∅.Then we show as in the previous two paragraphs that there are no edges in ¯ with one endpoint in V 1 := U 1 \U 2 and the other endpoint in V 2 , and for arbitrary v 1 ∈ V 1 we have we get in contradiction with (12) or (13), respectively.Hence, U 1 \U 2 = ∅ and the claim (a) is true.
To prove (c) assume the contrary, that is, there exist 1), which is not possible, as (v 2 , 1) and (v 2 , 1) do not have common neighbours.Hence, (c) is true.
Suppose that inclusion ( 7) is not true.Then by (a) there exist which is not possible.Assume now that i = 1, that is, It follows from (c) that (v 3 , 2) is adjacent to all vertices of the form (v, 2) where v ∈ V 2 , with one possible exception.Since (v, 2) = (v, 2) and Suppose that |V 2 | = 2, i.e.V 2 = {v 2 , v} for some v = v 2 .Then ( 14)-( 15) imply This is a contradiction, since (v 2 , 1) and (v, 1) do not have common neighbours in core( ¯ ) as they are its unique vertices that are contained in W 1 .On the other hand if |V 2 | = 1, i.e.V 2 = {v 2 }, then the graph with vertex set ( 6) is not a core.In fact, in such a case a map ˙ , which satisfies ˙ (v 2 , 1) = (v 1 , 2) for arbitrary v 1 ∈ V 1 and fixes all other vertices of core( ¯ ) is a nonbijective endomorphism of core( ¯ ).Hence inclusion ( 7) is correct.If ( 8) is not true, then there are ).Since (v 3 , 2) ∼ (v 3 , 1), we get in contradiction with (7), and ( 8) is correct.
It remains to prove the statement (d).Suppose that each vertex v 2 ∈ V 2 has at least one -neighbour in the set V 3 .Fix one of them and denote it by f (v 2 ).Choose an arbitrary retraction from ¯ onto core( ¯ ).Define a map ¨ on the vertex set of core( ¯ ), which is described in (6), by We claim that ¨ is a nonbijective endomorphism of core( ¯ ).This gives the desired contradiction that proves (d).Pick any v 2 ∈ V 2 .Then (7) which shows that the map ¨ is not bijective.Moreover, since To prove that ¨ is an endomorphism, suppose that (v, i) ∼ (u, j) are two adjacent vertices in core( ¯ ).
If i = 1 = j, then v, u ∈ V 2 , and (16) implies that which shows that ¨ is an endomorphism of core( ¯ ).Case 3 Let U 1 \U 2 = ∅.Then we proceed symmetrically as in Case 2 to deduce that (v) is true.

Remark 3.2
The set V 3 in Theorem 3.1 is allowed to be empty.In this case either or ¯ is disconnected.

Example 3.3
The possibilities (iv),(v) in Theorem 3.1 can occur even if the graph is regular.For example, if is a (disjoint) union of a triangle and a pentagon, then the core of ¯ is the graph, which is induced by all vertices in W 2 together with the triangle part of W 1 .The verification of this fact is left to the reader.On the other hand it seems mysteriously complicated to provide an example of a regular graph , where both and ¯ are connected and the possibility (iv) or (v) occurs for core( ¯ ).Below Now, consider the complement of the complete graph on 99 vertices, K 99 , the graph K (10, 4) − e, and the Cayley graph Cay(F 49 × F 4 , S), where F 49 and F 4 = {0, 1, ı, 1 + ı} are finite fields with 49 and 4 elements, respectively, and S = {(x, y) : 0 = x ∈ F 49 , y ∈ {0, 1}}.Recall that Cay(F 49 × F 4 , S) has the additive group F 49 × F 4 as the vertex set and two vertices (x, y), (x , y ) form an edge if and only if (x, y) − (x , y ) ∈ S.Moreover, the graph Cay(F 49 × F 4 , S) has two connected components, namely Fix any (x 1 , y 1 ) ∈ C 1 and (x 2 , y 2 ) ∈ C 2 and number the vertices {w 1 , . . ., w 196 } of Cay(F 49 × F 4 , S) in such way that w 195 = (x 1 , y 1 ) and w 196 = (x 2 , y 2 ).Further, number the vertices {v 1 , . . ., v 99 } of K 99 in some order and let u 1 , u 2 be the two endpoints of the deleted edge e in K (10, 4) − e.We form the graph from the disjoint union of graphs K 99 , K (10, 4) − e, Cay(F 49 × F 4 , S) by adding the following edges: • {w j , v 98 }, {w j , v 99 } where j ∈ {1, . . ., 194}, • {v i , w s } where i ∈ {1, . . ., 97} and s ∈ {1, . . ., 194}\{i, i + 97}.
Observe that the graph is 194-regular.The connectedness of K (10, 4) − e implies the connectedness of .The connectedness of ¯ is obvious.We next prove that the vertices of a core of ¯ are contained in the set X 1 ∪ X 2 , where (10,4) − e ∪ {(v 1 , 2), . . ., (v 99 , 2)}, and none of the possibilities (i), (ii), (iii) of Theorem 3.1 can occur.Observe that the sets C = {(x, 0) : x ∈ F 49 } and I = {(0, y) : y ∈ F 4 } are a clique and an independent set in Cay(F 49 × F 4 , S), respectively.Since each Cayley graph is vertex-transitive it follows from [17, Corollaries 2.1.2,2.1.3]that the product of its clique and independence numbers does not exceed the number of its vertices.Consequently, the orders of the clique C and the independent set I are the largest possible and the maps p 1 : (x, y) → (x, 0) and p 2 : (x, y) → (0, y) are endomorphisms of Cay(F 49 × F 4 , S) and Cay(F 49 × F 4 , S) onto their maximum cliques, respectively (compare with [41,Proposition 1]).Let f 1 : C → {v 1 , . . ., v 49 } and f 2 : I → {v 50 , v 51 , v 52 , v 53 } be any bijections and consider the map on ¯ , which fixes the vertices in X 1 ∪ X 2 and satisfies It is straightforward to check that is a retraction.Since it is not bijective, the claim (i) in Theorem 3.1 is not true.Moreover, Remark 2.9 implies that core( where X 1 ∪ X 2 is the subgraph, which is induced by the set and consequently Lemmas 2.8, 2.3(i), and ( 17)- (18) imply that 101 ≤ ω(K 99 ) + ω(K (10,4) and consequently Lemmas 2.8, 2.3(ii), ( 17)- (18), and (19) imply that Next we consider the case of a regular complementary prism ¯ .
Corollary 3.4 Let be any graph on n vertices that is n−1 2 -regular.Then only the statements (i), (ii), (iii) in Theorem 3.1 are possible.
Proof Suppose the claim (iv) from Theorem 3.1 is true.Then V core( ¯ ) equals (6).
Consequently, each vertex v ∈ V 2 has some neighbour f (v) ∈ V 3 , viewed in graph .In fact, the opposite would imply that v has at least n 2 neighbours in ¯ , which is not possible.By statement (c), the map f : V 2 → V 3 is injective.Hence, Moreover, each vertex v ∈ V 2 has zero and at most n Further observe that since the equality n 2 = 1 would imply that the graph with vertex set ( 6) is not a core.We claim also that If n 1 = 1, then we infer from inclusion (7) that in graph ¯ there are no edges inside the set V 3 and there are no edges connecting the vertex in V 1 with some vertex in V 3 .Consequently, we deduce by the statement (b) in Theorem 3.1 (iv) that a vertex f (v) ∈ V 3 as above, has in graph (at least) neighbours.This is a contradiction that proves bound (22).Consequently, the equality n 2 = 2 is not possible by ( 21)-( 22), while the inequality n 2 > 2 transforms (21) into which contradicts (20).
In the same way we see that the claim (v) in Theorem 3.1 is not true.
In Corollary 3.6 we obtain the same conclusion as in Corollary 3.4 if we move the regularity assumption from ¯ to core( ¯ ) and if we exclude some small examples.
Proof of Corollary 3.6 Suppose the claim (iv) from Theorem 3.1 is true.Let 1 , 2 , 2 be the subgraphs in ¯ , which are induced by the vertex sets respectively.Since core( ¯ ) is regular, it follows from (6) and the statement (b) in Theorem 3.1 (iv) that all three graphs 1 , 2 , 2 are regular.If k1 , n 1 and k2 , n 2 are the degree and the order of 1 and 2 , respectively, then it follows from (6) that the degree of a vertex in ¯ equals

Fig. 2 Corollary 3 . 6
Fig. 2 The adjacency matrix of the graph in Example 3.5 and the graph ¯