Star complements for ±2 in signed graphs

Theorem 3.1

The spectrum of a connected signed graphG˙lies in(−2,2)if and only ifG˙is switching equivalent to an induced subgraph of (at least) one of the signed graphs illustrated in Figure 1.

Proof

First, we confirm by inspection that the spectrum of each of the first ten signed graphs in Figure 1 lies in (−2,2), and the same holds for their induced subgraphs by the eigenvalue interlacing argument. For the remaining two (the negative quadrangle with two hanging paths and the negative even cycle), we observe that both are signed line graphs and in both cases the corresponding root is an unbalanced unicyclic simply signed graph. Therefore, their least eigenvalues are greater than −2 by Lemma 2.2(ii). Moreover, since both are bipartite, their spectra are symmetric about 0, and we are done.

Conversely, let G˙ be a connected signed graph whose spectrum lies in (−2,2). First, G˙ does not contain an odd cycle as an induced subgraph, since depending on the signature for such a cycle we have either λ=−2 or Λ=2 (where λ and Λ are defined at the beginning of Section 2). Next, G˙ cannot contain a positive even cycle as an induced subgraph, for otherwise both ∣λ(G˙)∣ and Λ(G˙) are at least 2. Hence, all induced cycles in G˙ are negative and of even length. We proceed with more restrictions:

1. G˙ cannot contain an induced negative even cycle with at least eight vertices and a hanging edge. (Otherwise, G˙ would contain the Smith graph obtained by attaching a hanging edge at the central vertex of a 7-vertex path, with Λ=−λ=2.)

2. G˙ cannot contain an induced signed graph in which two negative even cycles share more than one edge. (Observe that such an induced graph necessarily contains a negative odd cycle or a positive even cycle as an induced subgraph.)

By examining connected signed graphs with negative even cycles of length at most six and taking into account the restrictions of (i) and (ii), we arrive at the desired solutions.□

Proposition 3.2

Fort≥9, ℒ(C¨t)is the unique maximal extension of the signed graph (a) for±2.

Proof

Since t≥9, the maximal extension, say G˙, is a signed line graph. We observe next that the star complement is also a signed line graph of a (t−1)-vertex path with a negative digon; we denote this signed root graph by H˙. Therefore, G˙ is obtained by inserting the maximum number of edges in H˙ and then taking the signed line graph of the resulting root graph. Note that the edges between non-adjacent vertices of H˙ are disallowed since their existence produces positive triangles in G˙ which cannot exist because the largest eigenvalue of a positive triangle with an attached edge is greater than 2. The remaining edges are the two edges (one positive and the other negative) between the pendant vertices of H˙ and the edges that form negative digons. By inserting all of them we obtain the signed graph of Figure 2. This is ℒ(C¨t), and its spectrum is in [2t,(−2)t] (from the formula in Section 2). Therefore, ℒ(C¨t) is indeed the maximal extension for −2 and 2. This extension is unique since it includes all possible good vertices.□

Lemma 3.3

IfZt(U)is good, then∣U∣≤3.

Proof

If ∣U∣≥5, then (as odd triangles are disallowed) Zt(U) necessarily contains K1,5 as an induced subgraph, with λ<−2. If ∣U∣=4, then by simple reasoning we conclude that, unless t=4 or t=8 and U={1,4,6,8}, Zt(U) contains a forbidden induced subgraph that is a negative odd cycle with a hanging edge, or a positive even cycle with a hanging edge, or K1,4 with an edge attached at one of its pendant vertices. The critical cases are considered by inspection, and the lemma follows.□

Lemma 3.4

With the notation above, a vertexuand a corresponding setUare good forZtas a star complement for±2if and only if one of the following holds:

1. (a)U={1,2,t}withσ(u1)=−σ(u2), (b)U={1,2,4}withσ(u1)=σ(u2)=−σ(u4),

2. U={i,i+2}with3≤i≤t−2andσ(ui)=−σ(u(i+2)),

3. U={t−1},

4. t=8, and eitherUis one of{i,4,6}, {i,4,8}, {i,6,8}, {i,4}, {i,6}, {i,8}, wherei∈{1,2}and the edges fromualternate in sign, orU={i}(i∈{1,2}).

Proof

We set bu=b=(b1,b2,…,bt)⊺. Evidently Zt(U) is good for 2 if and only if ⟨b,b⟩=2, where the bilinear form is determined by the matrix (1).

Suppose first that ∣U∣=3, and let u be adjacent to the vertices 1, 2, and i. We have

For i≠4, we have b1=−b2 (otherwise, Zt(U) contains the positive quadrangle with a hanging edge), and so without loss of generality we may assume that b1=1,b2=−1. This leads to 4⟨b,b⟩=4(t−i+2), and so ⟨b,b⟩=2 if and only if i=t. For i=4, we obtain ⟨b,b⟩=2 with σ(u1)=σ(u2)=−σ(u4). Since in both situations the spectrum of Zt(U) is symmetric, we also have that U is good for −2, and (i) is completed.

Now suppose that u is adjacent to 1, i, and j (3≤i<j≤t). In a very similar way, we find that 4⟨b,b⟩=t with σ(u1)=−σ(ui)=σ(uj). Hence t=8, with the solutions (iv) computed directly.

Finally, if u is adjacent to i,j,k (3≤i<j<k≤t), then we have ⟨b,b⟩=t−i+j−k+1 with σ(ui)=−σ(uj)=σ(uk). The equality t−i+j−k=1 holds only for k=t and i=j−1, but in this case Zt(U) contains a negative triangle with a hanging edge, and this implies λ(Zt(U))<−2.

Suppose now that ∣U∣=2. Then U≠{1,2} since in this case either Λ(Zt)>2 or Zt(U) is a signed line graph of an unbalanced unicyclic simply signed graph with λ(Zt(U))>−2. If u is adjacent to 1 and i (3≤1≤t), we find that 4⟨b,b⟩=t, and as before t=8 along with the given solutions. If u is adjacent to i and j (3≤i<j≤t), we have ⟨b,b⟩=j−i, and thus Zt(U) is good for 2 if and only if j−i=2. Simultaneously, Zt(U) is good for −2 since its spectrum is symmetric. This completes (ii).

For ∣U∣=1 we find easily that Zt(U) is good for ±2 if and only if U={1} or U={2}, in both cases along with t=8, and U={t−1} (for any t). In this way, we have completed (iv) and (iii).□

Lemma 3.5

A vertexuis good forνif and only ifu∗is good forν. Ifuis good, thenu,u∗are compatible if and only ifν∈{0,1,−1}.

Proof

With the notation of Theorem 2.1, the first part of this lemma follows from ⟨bu,bu⟩=⟨−bu,−bu⟩=⟨bu∗,bu∗⟩. For the second part, if u is good then we have ν=⟨bu,bu⟩=−⟨bu,bu∗⟩, which means that u,u∗ are compatible if and only if ν∈{0,1,−1}.□

Proposition 3.6

Fort≠8, ℒ(C¨t)is the unique maximal extension ofZt for ±2.

Proof

We first consider compatibility between the vertices of Lemma 3.4(i)–(iii). It is convenient to compute the vectors bu⊺M for M defined by (1). With the enumeration of this lemma we find that

1. 1. bu⊺M=4σ(ut)(1,0,1,1,…,1)forσ(u1)=1,4σ(ut)(0,1,1,1,…,1)forσ(u1)=−1;

   2. bu⊺M=4σ(u1)(1,1,1,0,0,…,0);

2. bu⊺M=4σ(ui)(1,1,2,2,…,2︸i,1,0,0,…,0);

3. bu⊺M=4σ(u(t−1))(1,1,2,2,…,2,1).

Observe now that a vertex u is never compatible with the vertex u∗ defined upon Lemma 3.5. Compatibility between the remaining vertices is given in the scheme below. For example, if u,v are of type (ii), then

which gives the assertion.

It follows that good vertices for Zt can be partitioned into two halves in such a way that for every good vertex u, its non-compatible pair u∗ belongs to different half. In addition, by the scheme above all vertices in one half are mutually compatible and comprise a maximal extension, while the vertices of the other half lead to a switching equivalent extension. In other words, there is a unique maximal extension of Zt (for t≠8). We construct it on the basis of the vertex compatibility, and obtain ℒ(C¨t).□

Theorem 4.1

A connected signed graph has spectrum[2t,(−2)t]if and only if it is one of the signed graphs illustrated inFigures 2and3.

Proof

With a slight modification in the proof of [8, Theorem 5.1.6] we get that every signed graph has a connected star complement for each of its eigenvalues. Therefore, the result follows unless there is a connected signed graph, say G˙, with given spectrum for which there is no common star complement for −2 and 2. In this case, G˙ has 2t vertices and contains a connected star complement, say H˙, for −2, but not for 2, which implies Λ(H˙)=2.

If G˙ is a signed line graph, then by Lemma 2.2H˙ is the signed line graph of either a tree or an unbalanced unicyclic simply signed graph.

In the former case, we find that H˙ contains a positive triangle with a hanging edge (along with a contradiction Λ(H˙)>2) unless H˙ is a complete unsigned graph, but in this case we have t≤3, which is easily resolved (a triangle in the role of H˙ leads to G˙≅ℒ(C¨3)).

In the latter case, we have the same situation (a positive triangle with a hanging edge) unless H˙ is the unbalanced quadrangle with at most two hanging paths (Figure 1) or H˙ is a positive cycle. In the first case, contrary to the assumption, we have Λ(H˙)<2. In the second case, by considering good vertices we find that H˙ can be transformed into a star complement common to both eigenvalues.

The situation in which G˙ is exceptional is resolved by inspecting potential star complements with six, seven, and eight vertices.□

Theorem 4.2

Every signed graph with spectrum[2t,(−2)t]is determined by one of the following star complements for±2: Zt, or the negative hexagon with a hanging edge, or the negative hexagon with two hanging edges attached at vertices at distance 3.

Proof

The result follows from Theorem 4.1 and the subsequent discussion.□