Vertex and edge orbits in nut graphs

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Introduction
The main goal of the present paper is to find limitations on the numbers of orbits of vertices and edges of nut graphs under the action of the full automorphism group, and in particular to show that every nut graph has more than one orbit of edges.To substantiate this claim, we require some standard definitions.All graphs considered in this paper are simple and connected.By δ(G), d(G) and ∆(G) we denote the minimum, average and maximum degrees of a vertex in graph G (see [20,Section 1.2]).The adjacency matrix of graph G is A(G) and the dimension of the nullspace of A(G) is the nullity, η(G).Let Φ(M ; λ) denote the characteristic polynomial of square matrix M .The characteristic polynomial of graph G, denoted Φ(G; λ), is the characteristic polynomial of its adjacency matrix, i.e.Φ(G; λ) = Φ(A(G); λ) = det(A(G) − λI).The spectrum of graph G will be denoted σ(G).For a graph G of order n, we take V (G) = {1, 2, . . ., n}.The neighbourhood of a vertex v in graph G is denoted N G (v); where the graph G is clear from the context then we can simply write N (v).For other standard definitions we refer the reader to one of the many comprehensive treatments of graph spectra and related concepts (e.g.[14,15,8,13,5]).
Nut graphs [43] are graphs that have a one-dimensional nullspace (i.e., η(G) = 1), where the nontrivial kernel eigenvector x = [x 1 . . .x n ] ⊺ ∈ ker A(G) is full (i.e., |x i | > 0 for all i = 1, . . ., n).Nut graphs are connected, non-bipartite and have no leaves (i.e.δ(G) ≥ 2 for every nut graph G) [43].As the defining paper considered the isolated vertex to be a trivial case [43], the non-trivial nut graphs have seven or more vertices.Nut graphs of small order have been enumerated (see e.g.[4,11] and [12]).If G is a regular nut graph, then δ(G) = d(G) = ∆(G) ≥ 3. Note that there are no nut graphs with ∆(G) = 2, as it is known that cycles are not nut graphs.The case of ∆(G) = 3 is of interest in chemical applications of graph theory, as a chemical graph is a connected graph with maximum degree at most three.Chemical aspects of nut graphs are treated in [41].The nut graph is a special case of the core graph: a core graph is a graph with η(G) ≥ 1 for which it is possible to construct a kernel eigenvector in which all vertices of G carry a non-zero entry.Hence, a nut graph is a core graph of nullity one.Again, K 1 is presumably a trival core graph in the standard definition.Notice that a core graph may be bipartite or not, whereas a nut graph is not bipartite.
Let G and H be simple graphs.The cartesian product of G and H, denoted G □ H, is the graph with the vertex set {(u, v) | u ∈ V (G) and v ∈ V (H)} and the edge set {(u, v)(u ′ , v ′ ) | (uu ′ ∈ E(G) and v = v ′ ) or (u = u ′ and vv ′ ∈ E(H))}.For further details on graph products see [25].
An automorphism α of a graph G is a permutation α : V (G) → V (G) of the vertices of G that maps edges to edges and non-edges to non-edges.The set of all automorphisms of a graph G forms a group, the (full) automorphism group of G, denoted by Aut(G).The image of a vertex v ∈ V (G) under automorphism α will be denoted v α .Let u, v ∈ V (G).If there is an automorphism α ∈ Aut(G), such that u α = v, vertices u and v belong to the same vertex orbit.This relation partitions the vertex set V (G) into o v (G) vertex orbits.Let {u 1 , u 2 }, {v 1 , v 2 } ∈ E(G).If there is an automorphism α ∈ Aut(G), such that {u α 1 , u α 2 } = {v 1 , v 2 }, then edges u 1 u 2 and v 1 v 2 belong to the same edge orbit.This relation partitions the edge set E(G) into o e (G) edge orbits.See Figure 1 for examples.If o v (G) = 1 (i.e.all vertices belong to the same vertex orbit) then the graph G is said to be vertex transitive.Likewise, if o e (G) = 1, then the graph G is said to be edge transitive.A well known class of vertex-transitive graphs is the circulant graphs [24,Section 1.5].By Circ(Z n , S), where S ⊆ Z n , we denote the graph on the vertex set V (G) = Z n , where vertices u, v ∈ V (G) are adjacent if and only if |u − v| ∈ S. Let G be a subgroup of Aut(G).The stabiliser of a vertex v in G, denoted G v , is the subgroup of G that contains all elements α such that v α = v.For other standard definitions from algebraic graph theory, we refer the reader to textbooks, e.g.[24,21,3].
Example 1.There are three non-isomorphic nut graphs on 7 vertices.We denote them S 1 , S 2 and S 3 and call them the Sciriha graphs.For each Sciriha graph the numbers of vertex and edge orbits and the order of the (full) automorphism group are given; see Figure 1.⋄ (a) Ω(S 1 ) = (4, 5, 4) (b) Ω(S 2 ) = (4, 6, 4) (c) Ω(S 3 ) = (3,4,6) Figure 1: Vertex and edge orbits in the Sciriha graphs (i.e. the three nut graphs on 7 vertices).For brevity, we denote the triple (o v (G), o e (G), |Aut(G)|) by Ω(G).Vertices are coloured white or black to indicate equal and opposite entries in the kernel eigenvector.Note that for the Sciriha graphs, the kernel eigenvector is totally symmetric in each case (i.e. its trace is 1 for every automorphism).

Natural questions about nut graphs
Nut graphs are found within several well known graph classes, such as fullerenes, cubic polyhedra [42] and more general regular graphs [22,2].Nut graphs within these classes tend to have low symmetry, but attention has also been paid to finding nut graphs with high symmetry (in the sense of having a small number of vertex orbits).Recently, those pairs (n, d) for which a d-regular nut circulant of order n exists have been characterised in a series of papers [2,19,18,17,16].It is known that there are infinitely many vertex-transitive nut graphs [22].It seems natural, therefore, to consider the possibility of edge-transitive nut graphs.(Recall that if G is edge transitive, this does not imply that G is vertex transitive, nor does it imply that G is regular.However, an edge transitive connected graph has at most two vertex orbits.)To give some background for our question, a preliminary computer search based on the census of connected edge-transitive graphs on orders n ≤ 47 [10,9] was conducted.It found no examples of nut graphs.The census contains 1894 graphs in total.Of these, 335 graphs are non-singular and 2 graphs have nullity 1 (these graphs are K 1 and P 3 ).In the census, there is at least one graph for every admissible nullity (i.e. for each 0 ≤ k ≤ 45, there exists a graph G with η(G) = k).There are 1312 core graphs in the census (not counting K 1 ).Amongst these, there are 1098 bipartite graphs (945 non-regular, 25 regular non-vertex-transitive, and 128 vertex-transitive graphs).The remaining 214 non-bipartite edge-transitive core graphs are necessarily vertex-transitive, but none of these are nut graphs.On this basis, it seems plausible to question whether edge-transitive nut graphs exist.This prompted us to look for the general relationship between the numbers of vertex and edge orbits in nut graphs that is proved in the next section.In later sections, special attention is paid to nut graphs with one and two vertex orbits and the minimum number of edge orbits (respectively, two and three).Finally, the implications of constructions for the symmetry properties of nut graphs are investigated; a useful byproduct is a simple proof that there exist infinitely many nut graphs for each even number of vertex orbits and any number of edge orbits allowed by the main theorem.

A relation between numbers of vertex and edge orbits
The main result is embodied in the following theorem.
This theorem immediately implies the following corollary.

Corollary 2.
Let G be a nut graph.Then G is not edge transitive.
It is, however, relatively easy to find infinite families of vertex-transitive nut graphs with few edge orbits.For example, in Sections 3.1 and 4.1, we provide infinite families of nut graphs for (o v , o e ) = (1, 2) and (o v , o e ) = (2,3).Moreover, as we saw from our examination of the census, many core graphs are edge transitive.
To prepare for the proof of Theorem 2, we recall some established results.
Lemma 3 ([24, Lemma 3.2.1]).Let G be an edge-transitive graph with no isolated vertices.If G is not vertex transitive, then Aut(G) has exactly two orbits, and these two orbits are a bipartition of G.
A similar statement appears in [3] as Proposition 15.1.A theorem from a previous investigation specifies necessary conditions relating order and degree of a vertex-transitive nut graph: (c) we can take the entries to be x i ∈ {+1, −1}; (d) d(G) and n are both even.
Proof.As G is vertex-transitive and n ≥ 7, G has a non-trivial automorphism group Aut(G), i.e.
As G is a nut graph, the kernel eigenvector x belongs to a one-dimensional eigenspace, and hence spans a one-dimensional irreducible representation of Aut(G).As the graph is vertextransitive, each element α ∈ Aut(G) sends vertex v to an image vertex v α (v α may be v).The action of α on the vertices of G can be extended to x by defining x α = [x 1 α . . .x n α ].As a one-dimensional irreducible representation has trace either +1 or −1 under any particular automorphism α ∈ Aut(G), it follows that x = ±x α .This proves claim (a).In particular, |x 1 | = |x 1 α |.Since for every vertex v there exists an α ∈ Aut(G) such that v = 1 α , the claim (b) follows.To show that the entries in the kernel eigenvector can be drawn from the set {+1, −1}, it is enough to normalise the vector to x 1 = 1, verifying claim (c). to establish claim (d), note that the local condition for entries of the vector x ∈ ker A(H) is x u = 0 for v = 1, . . ., n. ( As all entries of x are in {+1, −1}, every vertex v must be of even degree.Since G is a regular graph, the entries of the Perron vector y of G (i.e. the eigenvector that corresponds to the largest eigenvalue λ 1 ) are all equal to +1.As x is orthogonal to y, i.e. n i=1 x i = 0, there are equal numbers of +1 and −1 entries in x, and G must have even order n, completing claim (d).
We note that the arguments used in claims (a) to (c) in the proof of Lemma 5 can be applied orbitwise for graphs that are not vertex transitive.Lemma 5 thus generalises naturally to the following: Lemma 6.Let G be a nut graph and let x = [x 1 . . .x n ] ⊺ ∈ ker A(G).Then the following statements hold:

and j belong to the same vertex orbit;
(c) we can take the entries to be x i ∈ {+a j , −a j } if i ∈ V j , where a j is a non-zero constant for orbit V j .
This lemma will be used in the proof of Theorem 1.Before the proof, we introduce some definitions.The first of these deal with edges.Let G be a nut graph with k vertex orbits There are several types of edge (as indicated schematically in Figure 2 for the case k = 3): (a) intra-orbit edge types, where both endvertices of an edge are in the same orbit V i , denoted e i ; (b) inter-orbit edge types, where endvertices of an edge are in two different orbits It is clear that edges of different types cannot belong to the same edge orbit.Moreover, a single edge type may comprise several edge orbits.
An additional definition will also be useful for the proof.Let the vertex-orbit graph of G, denoted G(G), be the graph whose vertices are vertex orbits of G and vertices V i and V j are adjacent in G(G) if there exists at least one edge of type e ij in G.Note that in our case G(G) contains k vertices.Moreover, the graph G(G) is a simple graph, not to be confused with the orbit graph as defined in [31], which is typically a pregraph.and (ii) vertices V ℓ form an independent set in G.
Then for every orbit V i of G it holds that Moreover, for each orbit V i it holds that |{j ∈ V i : x j > 0}| = |{j ∈ V i : x j < 0}| and the size of the orbit V i is even.
Proof.Let V ℓ ′ be the neighbour of V ℓ in G(G).Let d ij be the number of neighbours of a vertex v ∈ V i that reside in V j .The local condition says that u∈N (v) Therefore, This implies that Hence, the orbit V ′ ℓ must contain at least one vertex v with x v > 0 and at least one vertex w with x w < 0. This implies that there exists a sign-reversing α ∈ Aut(G).Within each orbit, α maps vertices with positive entries in the kernel eigenvector to vertices with negative entries, and vice-versa.Therefore, the cardinalities of these two sets of vertices are equal.Equation ( 5) follows by Lemma 6(b), and the claim about the parity is evident.
We can now proceed to the proof of the theorem.[6], since a connected graph on k vertices has at least k − 1 edges and each edge of G(G) gives rise to at least one edge orbit of G. Suppose there are no intra-orbit edges in G.In this case G is bipartite if and only if G(G) is bipartite.But a bipartite graph is not a nut graph.Hence we have o e (G) ≥ o v (G).To avoid bipartiteness we can do one of two things: (I) We may add intra-orbit edges to one or more vertex orbits.We need only consider addition of one such edge type, as addition of two or more would already imply o e (G) ≥ o v (G) + 1.
(II) We may add another type of inter-orbit edge to make an odd cycle in G(G).Note that G(G) becomes a unicyclic graph.Again, we do not need to consider addition of more than one edge type.
V ℓ  Suppose that G(G) contains a leaf V ℓ that is an independent set in G (in other words, there are no intra-orbit edges in V ℓ ).Let V ′ ℓ be the neighbour of V ℓ in G(G).By Lemma 7, the numbers of positive and negative entries in the kernel eigenvector are equal within any given orbit.This implies the existence of a sign-reversing automorphism α ∈ Aut(G).Each edge of type e ℓℓ ′ can be assigned one of four signatures according to signs of the kernel eigenvector entries for its endvertices (shown schematically in Figure 3).We now consider the action of the automorphism α on edges of each signature.Since α is sign-reversing, edge signatures are swapped: (+, +) ↔ (−, −) and (+, −) ↔ (−, +).Hence, edges of type e ℓℓ ′ fall into at least two orbits, determined by relative sign of endvertex entries: a like edge is of signature (+, +) or (−, −), while an unlike edge is of signature (+, −) or (−, +).By the local condition at a vertex of V ℓ , the presence of a (+, +) edge implies the presence of a (+, −) edge and vice-versa, hence the two corresponding edge orbits are both non-empty.As there is at least one edge orbit included within e ℓℓ ′ , the number of edge orbits in G is greater than the number of its edge types.
This proves case (I) and also case (II) where G(G) is a unicyclic graph but not a cycle.If G(G) is a cycle (necessarily odd) and there are no inter-orbit edges, a different argument is needed.Recall that no automorphism maps a like to an unlike edge (or vice versa), so they cannot be in the same edge orbit.If e i,i+1 contains both like and unlike edges, this immediately implies o e (G) ≥ o v (G) + 1.So, for every i, we can assume that e i,i+1 contains only like or only unlike edges.Take any vertex u ∈ V i .Since there are no intra-edges it has to be connected to neighbours in V i−1 via like and neighbours in V i+1 via unlike edges or vice versa.Therefore, the edges of G(G) can be properly coloured with colours 'like' and 'unlike'.But G(G) is an odd cycle, so no such edge colouring exists.Hence, at least one type e i,i+1 contains edges of both kinds.
Lemma 7 implies that a sign-reversing automorphism exists in a nut graph if at least one vertex orbit V ℓ is a leaf in G(G) and V ℓ has no intra-orbit edges.A similar structural result can also be obtained if G(G) is an odd cycle and G has no intra-orbit edges.
Suppose that every V ℓ forms an independent set in G and G(G) is an odd cycle.Then for every orbit V i of G it holds that Moreover, for each orbit V i it holds that |{j ∈ V i : x j > 0}| = |{j ∈ V i : x j < 0}| and the size of the orbit V i is even.

Lemma 9.
Let n be an odd integer and let A = [a i,j ] 1≤i,j≤n be a n × n matrix such that a i,j = 0 unless Proof.Recall that by definition where S n is the set of all permutations of length n.Note that the product n i=1 a i,σ(i) necessarily contains a zero factor, unless σ ∈ {(1 2 3 . . .n), (1 n n − 1 . . .2)}.
Proof of Proposition 8.If necessary, relabel the orbits V 1 , V 2 , . . ., V k , so that V i and V i+1 are neighbours in the cycle G(G).Let d ij be the number of neighbours of a vertex v ∈ V i that reside in V j .The local condition gives us one equation for each orbit, namely where we consider indices modulo k.Let us define s i = u∈V i x u for i = 1, . . ., k.We have the matrix equation By Lemma 9, the determinant of the square matrix in Equation ( 8) is This already implies the existence of a sign-reversing automorphism and the fact that |{j ∈ V i :

Vertex transitive nut graphs
We have seen that o e (G) = o v (G) = 1 is not possible for a nut graph G.However, many other possibilities for o e (G) may exist.First, we filtered out all nut graphs from databases of small vertextransitive graphs on up to n ≤ 46 vertices [38,26].The counts are shown in Table 1.Recall that a vertex-transitive graph G is a nut graph if and only if η(G) = 1, so this search requires only computation of the nullity and moreover, by Theorem 4, can be limited to graphs of even order and degree.As the table shows, most of these vertex transitive graphs are connected and a significant proportion of vertex transitive graphs of even order are nut graphs.As a preliminary survey of symmetry aspects, we calculated the number of edge orbits for all vertex-transitive nut graphs; see Table 2, which has a number of interesting features.It has only zero entries for o e (G) = 1, as demanded by Theorem 2, but there is no apparent restriction on the values of o e (G) that can occur for large enough order of G.Note that a vertex-transitive nut graph G with a large o e (G) must have large degree.To place these results in context, we also calculated the number of edge orbits of connected vertex-transitive graphs of even order.See Table 3.We see some intriguing gaps in Table 1 for particular pairs (n, o e ), e.g.(n, o e ) ∈ {(22, 3), (22,5), (22,7), (22,10), (22,11)}, even though the numbers of vertex-transitive graphs for these pairs of parameters are 37, 115, 138, 50 and 23, respectively.

Families with (o v , o e ) = (1, 2)
From the line for o e = 2 in Table 2 it appears likely that vertex transitive nut graphs with two edge orbits exist for all feasible orders.This is confirmed by the next theorem.The final column is the ratio between the number of VT nut graphs and the number of connected VT graphs on a given order, expressed as a percentage.
Theorem 10.For every even n ≥ 8, there exists a nut graph To prove this, we provide three families of quartic vertex-transitive graphs, which together cover all feasible orders and are described in Propositions 11 to 13.For the first family, let A ℓ , where ℓ ≥ 3, be the antiprism on 2ℓ vertices.Gauci et al. [23] proved the following proposition.
The next family is composed of cartesian products.Proposition 12.The graph C 3 □ C ℓ of order 3ℓ is a nut graph for even ℓ ≥ 4 such that ℓ ̸ ≡ 0 (mod 6).[5,Section 1.4.6]).Moreover, let x G be an eigenvector for an eigenvalue λ ∈ σ(G) and let x H be an eigenvector for and when ℓ is even and ℓ ̸ ≡ 0 (mod 6), it is clear that σ(C ℓ ) contains −2 with multiplicity 1, but not 1.Therefore, C 3 □ C ℓ contains a 0 eigenvalue with multiplicity 1.As the eigenvector of C ℓ for the eigenvalue −2 is full and so is the eigenvector of C 3 for the eigenvalue 2, it immediately follows that C 3 □ C ℓ is a nut graph.
For the third family, a variation on the cartesian product is used.Suppose that vertices of C ℓ are labeled 0, 1, . . .ℓ − 1 such that i and i + 1 are adjacent (indices modulo ℓ).Then the twisted product In other words, the construction C k τ C ℓ , is similar to the cartesian product of C k and C ℓ , but with a twist introduced between the first two C k layers.Proposition 13.The graph C 3 τ C ℓ of order 3ℓ is a nut graph for even ℓ ≥ 6 such that ℓ ≡ 0 (mod 6).
Proof.The twisted product C 3 τ C ℓ is an example of a graph bundle [33,32].Kwak et al. [27,7] studied characteristic polynomials of some specific graph bundles [27,7].Here, we apply their Theorem 8 from [27]; in the language of [27, Theorem 8], our where ϕ is an Aut(C n )-voltage assignment and n = 3. Aut(C 3 ) contains Z 3 as a subgroup.In our case, ϕ maps every directed edge of ⃗ C ℓ to 0 of Z 3 , except for the directed edges (0, 1) and (1, 0) which are mapped to 1 and its inverse 2, respectively.
From matrix C we can obtain an equivalent matrix D by permuting columns, namely Note that matrix D is composed of three blocks of size ℓ × (ℓ/3).Block i, 0 ≤ i ≤ 2, contains non-zero entries only in rows j, 0 ≤ j < ℓ, such that j ≡ i (mod 3).Now, we can define a matrix E that is equivalent to matrix D by defining, for 0 ≤ i < ℓ/3, Matrix E has a single non-zero entry in each row and each column; 2ℓ/3 − 1 of these entries are ω − 1 and ℓ/3 + 1 of these entries are ω + 2, and the determinant is (−1) ℓ/6 (ω − 1) 2ℓ/3−1 (ω + 2) ℓ/3+1 = 3 ℓ/2 ω ℓ/6+2 ̸ = 0.But matrix E is equivalent to B which is therefore of full rank.Hence the nullity of B is 1 and therefore C 3 τ C ℓ is a nut graph.Note that the proof does not require explicit construction of the kernel eigenvector.However, it is easily obtained.Define x :  (4,3,32), (6,4,32) and (10,5,32), respectively.The quartic example is shown in Figure 5(a); it is a tetracirculant with vertex The second smallest quartic example is shown in Figure 5(b) and is one of 14 non-Cayley nut graphs of order 30; it has 2 edge orbits and its automorphism group is of order 120.This is a generalisation of Rose Window graphs; its vertex set is

Nut graph with two vertex orbits
Data are available for graphs with two vertex orbits [39] and Table 4 shows our analysis for small graphs of this class.We observe that (o v (G), o e (G)) = (2, 1) and (o v (G), o e (G)) = (2, 2) do not occur in the table.This observation is, of course, consistent with Theorem 1 from Section 2. We also observe that the number of edge orbits can be large.Here, we provide infinite families of nut graphs with two vertex orbits and three edge orbits.

Families with (o v , o e ) = (2, 3)
From the line for o e = 3 in Table 4 it appears that nut graphs with two vertex orbits and three edge orbits exist for all orders n ≥ 9 such that n is not a prime; see Conjecture 23.Here, we provide two families of such nut graphs; one that covers orders that are multiples of three, and one that covers orders that are multiples of two but not multiples of three.Proposition 14.Let T n be a n-cycle with a triangle fused to every vertex (see Figure 6(a) for an example).The graph T n is a nut graph for every n ≥ 3.
Proof.Let the vertices of the n-cycle be labeled 0, 1, . . ., n − 1.Let a 0 , a 1 , . . ., a n−1 denote the entries on the vertices of the n-cycle in a kernel eigenvector of T n .It is easy to see that both vertices of the triangle fused to vertex i of the cycle must then carry entry −a i .The local condition at vertices of the cycle is where indices are modulo n.Equation ( 9) in matrix form is where A(C n ) is the adjacency matrix of the n-cycle and x = a 0 a 1 a 2 . . .a n−1 .The cycle C n is a 2-regular connected graph, and thus has a unique eigenvalue 2 in its spectrum, with x = 1 1 1 . . . 1 and the solution to Equation ( 9) is In 2008, the family of Rose Window graphs was introduced [44].A Rose Window graph, denoted R n (a, r), is defined by where all indices are modulo n.We will consider the subset with a = 1 and r = 2 (see Figures 6(b     where , can be assigned a row vector ξ(v) ∈ R 6 , acting as proxy for . Solving the linear recurrence relations (11) we obtain and By using the local condition (1) at vertices v 0 , v n−1 , u n−2 , u n−1 we obtain the four linear equations There are three cases to consider.
Case n ≡ 0 (mod 3): The equations ( 14) and ( 15) can be written in matrix form where µ = n 3 .It is easy to see that the matrix in ( 16) is of rank 3.This implies that R n (1, 2) has nullity 3.
Case n ≡ 1 (mod 3): The equations ( 14) and ( 15) can be written in matrix form where µ = n−1 3 .Note that µ ≥ 2 as n ≥ 5. Using tedious but elementary linear algebra, the matrix in ( 17) can be reduced to its echelon form, from which it can be seen that it is of rank 5.This implies that R n (1, 2) has nullity 1.
Case n ≡ 2 (mod 3): The equations ( 14) and ( 15) can be written in matrix form where µ = n−2 3 .Note that µ ≥ 1 as n ≥ 5.As before, the matrix in ( 18) can be reduced to its echelon form, from which it can be seen that it is of rank 5.This implies that R n (1, 2) has nullity 1 also in the present case.
It is easily seen that there exists a full vector in ker A(R n (1, 2)) in all three cases.Simply take a i = 1 and b i = −1 for all i, hence R n (1, 2) is a nut graph if n ̸ ≡ 0 (mod 3) and merely a core graph if n ≡ 0 (mod 3).
Figure 8: Kernel eigenvectors for the graph R 6 (1,2).Vectors (a) to (c) form an orthogonal basis that includes the rotationally symmetric vector that is present in the nullspace of R n (1, 2) for every n ≥ 5.
An alternative basis consists of vector (d) and its rotations by ±60 • .Signs of eigenvector entries are indicated by colour and relative magnitudes by area of the circles, where the possible magnitudes are 0, 1 and 2.

Multiplier constructions
The family T n can in fact be substantially generalised by defining the triangle-multiplier construction.Unlike some constructions in the literature [23], the triangle-multiplier applies to parent graphs that are not necessarily nut graphs.

Proposition 16.
Let G be a connected (2t)-regular graph, where t ≥ 1.Let M 3 (G) be the graph obtained from G by fusing a bouquet of t triangles to every vertex of G. Then M 3 (G) is a nut graph.
Proof.We follow the pattern of the proof of Proposition 14, where Equation ( 9) is replaced by which in matrix form is A(G)x = 2t x.So, kernel eigenvectors of M 3 (G) are precisely eigenvectors of G for the eigenvalue 2t.But the graph G is 2t-regular, so the solution of Equation 19 is unique and x, i.e. the Perron eigenvector, is full.
The choice of name for the construction is justified by the fact that |V (M 3 (G))| = (2t + 1)|V (G)|.As Proposition 29 in Section 5 will show, the triangle-multiplier construction adds one vertex orbit and two edge orbits to the graph G, irrespective of the value t.We can define a pentagon-multiplier construction as follows.As in the case of the triangle-multiplier, this construction applies to graphs that are not necessarily nut graphs.

Proposition 17.
Let G be a bipartite connected (2p)-regular graph, where p ≥ 1.Let M 5 (G) be the graph obtained from G by fusing a bouquet of p pentagons (i.e.5-cycles) to every vertex of G. Then M 5 (G) is a nut graph.
Proof.We follow the pattern of the proof of Proposition 16.Consider a pentagon fused at a vertex v ∈ V (G).The vertices of the pentagon that are adjacent to v both carry entry +x(v), while the remaining two vertices carry entry −x(v).
Equation ( 19) is replaced by which in matrix form is A(G)x = −2p x.Since G is connected, bipartite and (2p)-regular, it has a unique eigenvalue −2p in its spectrum, and the corresponding eigenvector is full.
We note that in Proposition 16, any fused triangle could be replaced by a (4q + 3)-cycle for any q ≥ 0. Likewise, in Proposition 17, any fused pentagon may be replaced by a (4q + 5)-cycle for any q ≥ 0. In fact, these changes are just repeated applications of the subdivision construction on the triangles (resp.pentagons) of the graph M 3 (G) (resp.M 5 (G)).
It seems natural to ask, what would happen if we fuse a mixture of triangles and pentagons to some vertices of a graph?Consider the case where we fuse a triangle and a pentagon to a vertex in a graph.

Proposition 18.
Let G be a nut graph and let v ∈ V (G) be a vertex.Let P(G, v) be the graph obtained from G by fusing a triangle and a pentagon to vertex v. Then P(G, v) is a nut graph.
Proof.Let x be a kernel eigenvector of P(G, v).Consider the two vertices on the fused triangle that are adjacent to v. Their entries in x are −x(v).Now consider the two vertices of the fused pentagon that are adjacent to v. Their entries in x are x(v); the entries of the remaining two vertices are −x(v).

The local condition at vertex
This is a special case of the coalescence construction devised by Sciriha [40].Corollary 21 in [40] is equivalent to the statement that coalescence of any two nut graphs G 1 and G 2 at any pair of vertices v 1 ∈ V (G 1 ) and v 2 ∈ V (G 2 ) produces a nut graph.The fusion of a triangle and a pentagon is one of the three Sciriha graphs; see Figure 1(a).Note that the above construction is used on a single vertex of a nut graph G. Had we used it iteratively on all vertices of G, that would give us yet another multiplier construction.In fact, various sorts of mixed objects can be envisaged.After the initial application of M 3 or M 5 on an appropriate parent graph, which gives rise to a nut graph, the way lies open to application of the coalescence construction, locally or globally.See Figure 9 for examples.
The triangle-multiplier and pentagon-multiplier constructions may be generalised to a k-multiplier construction: Let G be a (2r)-regular graph and let k ≥ 3. Let M k (G) be the graph obtained from G by fusing a bouquet of r k-cycles to every vertex of G.In fact, Propositions 16 and 17 have natural generalisations to every M k , where {k ≥ 3 and k ≡ 3 (mod 4)} and {k ≥ 5 and k ≡ 1 (mod 4)}, respectively.These generalisations follow immediately by the subdivision construction (see Section 5).

Characterisation of orders for nut graphs with 2 vertex orbits
Observe that columns for prime values of n are absent from Table 4.This is because the search did not reveal any examples in the range.As the next theorem shows, this is no coincidence.

Theorem 19.
Let G be a nut graph of order n with precisely two vertex orbits.Then n is not a prime number.
The next proposition will be useful in the proof of the above theorem.

Proposition 20. Let G be a nut graph and let
If there exists a signreversing automorphism α ∈ Aut(G) then all orbits are of even size.Moreover, for every j, half of the entries {x i | i ∈ V j } are positive, and the other half are negative.
Proof.Think of the automorphism α as a product of disjoint cycles.Note that elements of any given cycle of α are contained in the same vertex orbit.Since α is a sign-reversing automorphism, every vertex i is mapped to a vertex i α carrying the opposite sign (i.e.x i • x i α < 0).Therefore, every cycle of α is of even length and contains a perfect matching whose edges join vertices with entries of opposite sign.Hence, {x i | i ∈ V j } contains the same number of positive and negative elements.

Corollary 21.
Let G be a nut graph of order n.If n is odd then Aut(G) does not contain any sign-reversing automorphism.Moreover, kernel eigenvector entries are constant within a given orbit.

Proof of Theorem 19.
Let The number of inter-orbit edges is The proof proceeds by contradiction.Suppose that n is a prime.The case n = 2 is trivial, since there are no nut graphs on 2 vertices.As n is odd, kernel eigenvector entries within each orbit are constant by Corollary 21.Let a i be the entry in V i .The local conditions are First, we note that Equation 21 has a unique solution.Equation 21 implies Since  22) can be expressed in matrix form as The determinant of the matrix is n 1 n 2 − d 11 d 22 > 0 and therefore a 1 = a 2 = 0.This contradicts the fact that G is a nut graph.Therefore, n cannot be prime.
We have justified the claim that nut graphs with two vertex orbits cannot be of prime order n.We add to the picture by showing that a nut graph with two vertex orbits exists for all composite orders n ≥ 9.
Theorem 22.Let n ≥ 9 such that n is not a prime.Then there exists a nut graph G of order n with Proof.Let n = p 1 p 2 • • • p k be the decomposition of n into prime factors.Without loss of generality, we may assume that Case 1: Suppose that p 1 = 2.If n ̸ ≡ 0 (mod 3) then Proposition 15 guarantees a solution.If n ≡ 0 (mod 3) then Proposition 14 guarantees a solution.
Case 2: Now suppose that p 1 > 2. Clearly, p 1 is an odd integer.The strategy is to find a vertex transitive graph H of order n = n/p 1 and degree d = p 1 − 1.Then use the triangle-multiplier construction to obtain M 3 (H).Since n is not a prime n/p 1 ≥ p 1 and so n > d.Note that d is an even number.The circulant graph H = Circ( n, {1, 2, . . ., d/2}) has the prescribed order and degree and it is of course vertex transitive as required.The graph M 3 (H) is of order n.By Proposition 29, this graph has two vertex orbits.
Theorem 22 shows that there is at least one non-zero entry in every column of Table 4.However, we believe that row o e = 3 by itself consists of non-zero entries.Conjecture 23 holds for all even numbers (covered by Case 1 of the proof), all multiples of 3 (covered by Proposition 14) and all perfect squares (in that case graph H in the proof is a complete graph).The conjecture can also be validated for those values of n, such that there exists an edge transitive graph H of order n and degree d in the proof of Theorem 22. Table 6 shows the orders up to 300 that are not resolved by anything mentioned thus far.For some of these orders we were able to provide graph H (see proof of Theorem 22).N/A in the table indicates that no such graph H exists (based on the census by Conder and Verret [10,9]).Order 35, for example, cannot be resolved in this way, because there is only one vertex transitive graph of order 7 and degree 4, namely Circ(7, {1, 2}), but it is not edge transitive, and so a completely different approach is required.For order 295, H would have to be a 4-regular edge transitive graph of order 59, but no such graph is known (see [37,36,35,34]).The classes of edge transitive circulants are provided in [30] could be used to resolve some orders beyond Table 6.All graphs H provided in Table 6 are circulants.However, for some orders there are non-circulant alternative possibilities, e.g. the graph C 5 □ C 5 could be used for n = 125 and Cay(Z 5 × Z 5 , {(0, 1), (1, 0), (1, 1)}) for order n = 175.
the three edges uw 1 , w 1 w 2 and w 2 v.If there exists an element φ ∈ Aut(G) such that u φ = v and v φ = u, then using its extension φ we get w φ 1 = w 2 .This means that vertices w 1 and w 2 are in the same vertex orbit.Similarly, edges uw 1 and w 2 v are in the same edge orbit.The claim follows.Note that we define the action of Aut(F (G, V)) v on pairs (w i , x j ) by taking (w i , x j ) α = (w α i , x α j ), where α ∈ Aut(F (G, V)) v .The proof of the above proposition uses the same approach as that of Proposition 25 and is skipped here.
We are interested in the growth of the number of edge orbits under the construction.Let us define Φ(G, v) = o e (F (G, V)) − o e (G), where V is the orbit of the vertex v. Proposition 27 has the following corollary.

Corollary 28.
Let G be a nut graph and let v ∈ V (G) be a vertex in G. Let V be the orbit of the vertex v under Aut(G).Let F (G, V) be as in Proposition 27 and also let Aut(G) Proof.To get the upper bound, assume that each vertex of N (v) is in its own orbit and therefore t = d.Similarly, each element of S is in its own orbit and therefore τ = d 2 − d.
For parameters t and τ from Proposition 27, it holds that t ≥ 1 and τ ≥ 1.Therefore, 2 ≤ o e (F (G, V)) − o e (G).By Lemma 6, α ∈ Aut(G) v cannot be sign-reversing.There must be at least one vertex in N (v) with a positive entry in the kernel eigenvector and at least one with a negative entry.Therefore, the group Aut(G) v partitions N (v) in at least 2 orbits, say U and U cannot be in the same orbit under Aut(G) v as any other vertex of F (G, v); see Figure 10.The same is true for w 1 , . . ., w d .Since Aut(G) ∼ = Aut(F (G, V)), {x 1 , . . ., x d } and {w 1 , . . ., w d } are partitioned into orbits under Aut(G) v in the same way as {u 1 , . . ., u d } (i.e., x i and x j belong to the same orbit if and only if u i and u j belong to the same orbit).Orbits U and U ′ induce orbits X and X ′ on {x 1 , . . ., x d } and orbits W and W ′ on {w 1 , . . ., w d }.There exists at least one edge connecting U to X , at least one edge connecting U ′ to X ′ , at least one edge connecting X to W ′ and at least one edge connecting X ′ to W. Each of these four edges must be in a distinct new edge orbit, hence 4 ≤ o e (F (G, V)) − o e (G).
If deg(v) ≥ 3 then either U or U ′ contains at least 2 vertices.Without loss of generality assume that |U| ≥ 2. Then there exists at least one edge connecting X to W. This edge cannot share the orbit with any of the above four edges, hence 5 ≤ o e (F (G, V)) − o e (G).The upper bound is best possible, because the equality in (24) can be attained if we take G to be any asymmetric nut graph (i.e.| Aut(G)| = 1).This bound is attained even within the class of vertex transitive graphs, when we take G to be a GRR [28,29] nut graph, such as the one in Figure 11.graphs of orders n ≤ 12 yields examples for d ∈ {5, 6, 7} for which Φ(G, v) = 6; see [1].Examples of graph with small Φ(G, v) can also be found in the class of regular graphs; see [1].Moreover, examples with small Φ(G, v) can be found in the class of vertex transitive graphs.For example, K 3,3 □ K 4 is a sextic vertex transitive nut graph with Φ(K 3,3 □ K 4 , v) = 6.
It is important to note that in Propositions 25, 26 and 27 the respective construction was applied to all edges/vertices within a given edge/vertex orbit.This ensures that the graph obtained by the construction inherits all the symmetries of the original graph G.The requirement that the order of the automorphism group of the graph does not increase upon applying the given construction, i.e.Aut(G) ∼ = Aut(F (G, V)), is also crucial to the propositions.Figure 13 illustrates some of the complications that can arise.
Proof.Every element α ∈ Aut(G) can be extended in a natural way to an element α ∈ M 3 (G); the element α moves the vertices of the original graph in the same way as α, and also moves the corresponding attached triangles.Therefore, | Aut(M 3 (G))| ≥ | Aut(G)|.Now, let us consider action of the stabiliser within Aut(M 3 (G)) that fixes the subgraph G. Consider the triangles attached to an arbitrary vertex v ∈ V (G).Clearly, the stabiliser permutes the t triangles; this contributes t! to the order of the stabiliser.In addition, there exist involutions that swap two degree-2 endvertices in any attached triangle; this contributes 2 t to the order of the stabiliser.Finally, all these operations can be done independently at every vertex v ∈ V (G).Therefore, the order of the stabiliser is (2 Now that the full automorphism group of M 3 (G)) is known, counting the vertex-and edge-orbits is straightforward.
Example 2. Consider graphs K 7 , Circ(12, {1, 5}) and the hypercube Q 6 .All these graphs are vertex and edge transitive.Let us determine the order of the automorphism group of M 3 (G) for G from the above list.It is easy to see Note that even though the automorphism group of M 3 (G) might be absurdly large, the numbers of vertex and edge orbits remain small, and in determining them we can ignore the extra symmetries.⋄ Proposition 29 has a natural generalisation.Proposition 30.Let k ≥ 3 be an odd integer and let G be a connected (2t)-regular graph, where t ≥ 1.If k ≡ 1 (mod 4) then the graph G is further required to be bipartite.Then shown in (e) is progressively subdivided; in (f) subdivision in a part orbit gives broken symmetry; in (g) subdivision in a part orbit of (f) preserves the order of the automorphism group; in (h) subdivision of the final edge causes merging of orbits and restoration of the original symmetry.Parts (i) to (k): the highly symmetric graph (i) has four edge orbits; in (j) the symmetry is considerably reduced by applying subdivision in a part of the green orbit of (i); the graph in (k) is obtained from (j) by applying subdivision on one of two equivalent edges but the symmetry increases; subdivision of the two remaining green edges would restore the full symmetry of (i).The signature Ω given for each graph G denotes the triple (o v (G), o e (G), |Aut(G)|) as in Figure 1.
Proof of Proposition 30 follows the same pattern as the proof of Proposition 29 and is left as an exercise to the reader.

Future work
The present paper gives a theorem for the relationship between vertex-orbit and edge-orbit counts for nut graphs.The result (Theorem 1) that o e ≥ o v + 1, compares to Buset's result o e ≥ o v − 1 for all connected graphs [6, Theorem 2].Edge-transitive nut graphs are therefore impossible objects.
We also provided a complete characterisation of the orders for which nut graphs with (o v , o e ) = (1, 2) exist.A partial answer was also found for the pair (o v , o e ) = (2, 3) (see Conjecture 23 and Question 24).It was possible to provide infinite families of nut graphs for the pairs (o v , o e ), where o v is an even number and o e > e v .The case where o v is an odd number remains to be completed.The ultimate goal is, of course, the complete characterisation of orders for all (o v , o e ) pairs.
During this work we encountered smallest examples of several interesting classes, including the non-Cayley nut graphs (see Figure 5), and GRR nut graphs (see Figure 11), which suggest directions for future explorations.The three infinite families used to prove Theorem 10, and the Rose Window family (see Proposition 15) are all quartic, but the problem of characterising orders for (o v , o e ) pairs is also a natural one for regular graphs, or graphs of prescribed degree.It is planned to investigate the cubic case first, because of its significance for chemical graph theory.
A substantial part of the paper was devoted to constructions of nut graphs and their effects on symmetry, which can be complicated.In some cases, the automorphism group of a constructed nut graph can be impressively large (see Example 2).The multiplier constructions (Subsection 4.2) give access to highly symmetric graphs with controlled number of vertex orbits.This prompts the question: For a given n, what is the most symmetric nut graph on that order, where by 'most symmetric' we mean in the sense of order of the automorphism group?From this perspective it is interesting that the graph with 288 automorphisms shown in Figure 13(i) is the nut graph with the largest full automorphism group amongst all nut graphs on 10 vertices and yet it is not vertex transitive.

1 Figure 2 :
Figure 2: Schematic representation of a graph with three vertex orbits (represented by the three bags of vertices).There are six edge types, denoted e 1 , e 2 , e 3 (intra-orbit), e 12 , e 13 and e 23 (inter-orbit).

Figure 3 :
Figure 3: Vertex orbits V ℓ and V ′ ℓ as defined in the proof of Theorem 1, showing the four possible signatures for edges of type e ℓℓ ′ .

(a) A 4 (b) C 3 □ C 4 (c) C 3 τ C 6 Figure 4 :
Figure 4: The smallest examples of each of the families described in Propositions 11 to 13.They are shown embedded on a circular strip; the end blue curves are to be identified.Entries in the kernel eigenvector in each graph are all of equal magnitude and represented by circles colour-coded for sign.

Figure 5 :
Figure 5: The two smallest 4-valent non-Cayley vertex transitive nut graphs.Entries in the kernel eigenvector in each graph are all of equal magnitude and represented by circles colour-coded for sign.

(a) T 5 (b) R 5 ( 1 Figure 6 :
Figure 6: Small examples of each of the families described in Propositions 14 and 15.Entries in the kernel eigenvector in each graph are all of equal magnitude and represented by circles colour-coded for sign.

Table 5 : 3 Figure 7 :
Figure 7: Labelling scheme for the Rose Window graph R n (1, 2).Entries in the candidate kernel eigenvector are a i on vertices v i and b i on vertices u i , all indices taken modulo n.

Figure 9 :
Figure 9: These two nut graphs were obtained from (a) C 3 and (b) C 4 by an application of the M 3 resp.M 5 , followed by repeated application of Proposition 18. Entries in the kernel eigenvector in each graph are all of equal magnitude and represented by circles colour-coded for sign.

Conjecture 23 .
Let n ≥ 9 such that n is not a prime.Then there exists a nut graph G of order n with o v (G) = 2 and o e (G) = 3.

Proposition 26 .Definition 1 .
Note that since Aut(G) ≤ Aut(B(G, E)), the condition |Aut(G)| = |Aut(B(G, E))| automatically implies Aut(G) ∼ = Aut(B(G, E)).The proof of the following proposition is analogous to that of Proposition 25 and is skipped here.Let G be a nut graph and let e = uv ∈ E(G) be an edge in G. Let E be the orbit of the edge e under Aut(G).The graph obtained from G by applying the subdivision construction on every edge from E, denoted S(G, E), is a nut graph and Aut(G) ≤ Aut(S(G, E)).If, in addition, Aut(G) ∼ = Aut(S(G, E)) then the following statements hold.(i) If there exists an element φ ∈ Aut(G) such that u φ = v and v φ = u, then o v (S(G, E)) = o v (G)+2 and o e (S(G, E)) = o e (G) + 2. (ii) If there is no element φ ∈ Aut(G) such that u φ = v and v φ = u, then o v (S(G, E)) = o v (G) + 4 and o e (S(G, E)) = o e (G) + 4. Let G be a nut graph and let v ∈ V (G) be a vertex of degree d in G. Let N (v) = {u 1 , . . ., u d }.The graph F (G, v), is obtained from G in the following way: (a) edges incident to v are deleted; (b) let w 1 , . . ., w d and x 1 , . . ., x d denote 2d newly added vertices; (c) new edges are added such that x i ∼ u i for i = 1, . . ., d; and x i ∼ w j for i ̸ = j, 1 ≤ i, j ≤ d; and w i ∼ v for all i = 1, . . ., d.The construction F (G, v) has been called 'the Fowler construction' in the nut-graph literature[23].

Figure 10 illustrates
Figure 10 illustrates the definition.Note that u 1 , . . ., u d are at distance 3 from v in F (G, v).Moreover, degrees of all newly added vertices are d.Proposition 27.Let G be a nut graph and let v ∈ V (G) be a vertex in G. Let V be the orbit of the vertex v under Aut(G).The graph obtained from G by applying the Fowler construction on every vertex from V, denoted F (G, V), is a nut graph and Aut(G) ≤ Aut(F (G, V)).Suppose that, in addition, Aut(G) ∼ = Aut(F (G, V)).Let the vertices in the neighbourhood of v in G and in the first, second and third neighbourhood of v in F (G, V) be labeled as in Definition 1.The stabiliser Aut(G) v fixes N (v) set-wise in the graph G and partitions N (v) into t orbits.Let S = {(w i , x j ) | i ̸ = j; 1 ≤ i, j ≤ d}.Aut(F (G, V)) v partitions S into τ orbits.Then o v (F (G, V)) = o v (G) + 2t and o e (F (G, V)) = o e (G) + t + τ .

Figure 10 :
Figure 10: A construction for expansion of a nut graph G about vertex v of degree d, to give F (G, v).Panel (a) shows the neighbourhood of vertex v in G. Panel (b) shows additional vertices and edges in F (G, v).

Figure 11 :Figure 12 :
Figure 11: The smallest GRR nut graph has order 12 and degree 6.The graph contains three cliques represented by shaded regions; edges within cliques are not drawn.Entries in the kernel eigenvector are all of equal magnitude and represented by circles colour-coded for sign.

Figure 13 :
Figure13: Interplay of constructions, orbits and automorphism group.Parts (a) to (d): in the starting graph (a) two orbits are marked in red and blue; in (b) subdivision on an entire orbit preserves the automorphism group; in (c) the bridge construction on an entire orbit leads to doubling in order of the automorphism group; in (d) subdivision in a part orbit leads to halving.Parts (e) to (h): the Sciriha graph S 3 shown in (e) is progressively subdivided; in (f) subdivision in a part orbit gives broken symmetry; in (g) subdivision in a part orbit of (f) preserves the order of the automorphism group; in (h) subdivision of the final edge causes merging of orbits and restoration of the original symmetry.Parts (i) to (k): the highly symmetric graph (i) has four edge orbits; in (j) the symmetry is considerably reduced by applying subdivision in a part of the green orbit of (i); the graph in (k) is obtained from (j) by applying subdivision on one of two equivalent edges but the symmetry increases; subdivision of the two remaining green edges would restore the full symmetry of (i).The signature Ω given for each graph G denotes the triple (o v (G), o e (G), |Aut(G)|) as in Figure1.

Table 1 :
The number of nut graphs among vertex transitive (VT) graphs on even orders 8 ≤ n ≤ 46.

Table 2 :
The number of vertex-transitive nut graphs of the given order n and number of edge orbits o e .

Table 3 :
The number of connected vertex-transitive graphs of the given order n, 8 ≤ n ≤ 46 even, and number of edge orbits o e .

Table 4 :
The number of nut graphs with precisely two vertex orbits of the given order n and number of edge orbits o e .
n 1 + n 2 is a prime factor, it divides either d 21 + d 12 or n 2 .As it clearly cannot divide n 2 , it divides d 21 + d 12 .But d 21 + d 12 ≤ n 1 + n 2 .Divisibility is possible only in the case where d 21 = n 1 and d 12 = n 2 .For this case, Equation (