Eigenvalues and edge-connectivity of regular graphs

Eigenvalues and edge-connectivity of regular graphs Sebastian M. Cioabă University of Delaware Department of Mathematical Sciences Newark DE 19716, USA cioaba@math.udel.edu August 3, 009 Abstract In this paper, we show that if the second largest eigenvalue of a d-regular graph is less than d (k 1) d+1, then the graph is k-edge-connected. When k is or 3, we prove stronger results. Let ρ(d) denote the largest root of x 3 (d 3)x (3d )x = 0. We show that if the second largest eigenvalue of a d-regular graph G is less than ρ(d), then G is -edge-connected and we prove that if the second largest eigenvalue of G is less than d 3+ (d+3) 16, then G is 3-edge-connected. 1 Introduction Let κ(g) and κ (G) denote the vertex- and edge-connectivity of a connected graph G. If δ is the minimum degree of G, then 1 κ(g) κ (G) δ. Let L = D A be the Laplacian matrix of G, where D is the diagonal degree matrix and A is the adjacency matrix of G. We denote by 0 = µ 1 < µ µ n the eigenvalues of the Laplacian of G. The complement of G is denoted by G. A graph is called disconnected if it is not connected. The join G 1 G of two vertex-disjoint graphs G 1 and G is the graph formed from the union of G 1 and G by joining each vertex of G 1 to each vertex of G. A classical result in spectral graph theory due to Fiedler [7] states that κ(g) µ (G) (1) for any non-complete graph G. Fiedler called µ (G) the algebraic connectivity of G and his work stimulated a large amount of research in spectral graph theory over the last forty years (see [1, 11, 13, 15]). In [1], Kirkland, Molitierno, Neumann and Shader characterize the equality case in Fiedler s inequality (1). Research supported by a startup grant from the Department of Mathematical Sciences of the University of Delaware. 1

Theorem 1.1 (Kirkland-Molitierno-Neumann-Shader [1]). Let G be a non-complete connected graph on n vertices. Then κ(g) = µ (G) if and only if G = G 1 G where G 1 is a disconnected graph on n κ(g) vertices and G is a graph on κ(g) vertices with µ (G ) κ(g) n. Eigenvalue techniques have been also used recently by Brouwer and Koolen [5] to show that the vertex-connectivity of a distance-regular graph equals its degree (see also [, 4] for related results). In this paper, we study the relations between the edge-connectivity and the second largest eigenvalue of a d-regular graph. Let λ 1 (G) λ (G) λ n (G) denote the eigenvalues of the adjacency matrix of G. If G is d-regular graph, then λ i (G) = d µ i (G) for any 1 i n. Thus, λ 1 (G) = d and G is connected if and only if λ (G) < d. Chandran [6] proved that if G is a d-regular graph of order n and λ (G) < d 1 d, n d then κ (G) = d and the only disconnecting edge-cuts are trivial, i.e., d edges adjacent to the same vertex. Krivelevich and Sudakov [13] showed that λ (G) d implies κ (G) = d. If G is a d-regular graph on n vertices and n d + 1, then κ (G) = d regardless of the eigenvalues of G (see also the proof of Lemma 1.3). Both results are based on the following well-known lemma (see [15] for a short proof). Lemma 1.. If G = (V,E) is a connected graph of order n and S is a subset of vertices of G, then e(s,v \ S) µ S (n S ) n where e(s,v \ S) denotes the number of edges between S and V \ S. We extend and improve the previous results as follows. We prove the following sufficient condition for the k-edge-connectivity of a d-regular graph for any k d. Theorem 1.3. If d k are two integers and G is a d-regular graph such that λ (G) d (k 1)n, then (d+1)(n d 1) κ (G) k. When k = d, this result states that if λ (G) d (d 1)n, then (d+1)(n d 1) κ (G) = d. We get a small improvement for d even because κ (G) must be even in this case (see Lemma 3.1). (d )n When d is even, Theorem 1.3 shows that λ (G) d implies (d+1)(n d 1) κ (G) = d. A simple calculation reveals that these bounds improve the previous result of Chandran. When n d +, Theorem 1.3 implies that if G is a d-regular graph with λ (G) d + 4, then d+1 κ (G) = d. When d is even, the right hand-side of the previous inequality can be replaced by d + 6. These results improve the previous bound of Krivelevich and d+1 Sudakov. Note that there are many d-regular graphs G with d < λ (G) d + 4. d+1 An example is presented in Figure 1. For k {, 3} we further improve these results as shown below. Note that the edgeconnectivity of a connected, regular graph of even degree is even (see Lemma 3.1). Theorem 1.4. Let d 3 be an odd integer and ρ(d) denote the largest root of x 3 (d 3)x (3d )x = 0. If G is a d-regular graph such that λ (G) < ρ(d), then κ (G).

Figure 1: A 3-regular graph with 3 < λ = 1+ 5 < 3 + 4 3+1 The value of ρ(d) is about d d+5. Theorem 1.5. Let d 3 be any integer. If G is a d-regular graph such that then κ (G) 3. λ (G) < d 3 + (d + 3) 16 The value of d 3+ (d+3) 16 is about d 4. d+3 We show that Theorem 1.4 and Theorem 1.5 are best possible in the sense that for each odd d 3, there exists a d-regular graph X d such that λ (X d ) = ρ(d) and κ (X d ) = 1. Also, for each d 3, there exists a d-regular graph Y d such that λ (Y d ) = d 3+ (d+3) 16 and κ (Y d ) =. The following result is an immediate consequence of Theorem 1.4 and Theorem 1.5. Corollary 1.6. If G is a 3-regular graph and λ (G) < 5.3, then κ (G) = 3. If G is 4-regular graph and λ < 1+ 33 3.37, then κ (G) = 4. Our results imply that if G is a 3-regular graph with λ (G) < 5, then κ(g) = 3. This is because for 3-regular graphs the vertex- and the edge-connectivity are the same. It is very likely that the eigenvalues results for κ(g) and κ (G) can be quite different. We comment on this in the last section of the paper. Proof of Theorem 1.3 In this section, we give a short proof of Theorem 1.3. Recall that a partition V 1 V l = V (G) of the vertex set of a graph G is called equitable if for all 1 i,j l, the number of neighbours in V j of a vertex v in V i is a constant b ij independent of v (see Section 9.3 of [8] or [9] for more details on equitable partitions). Proof of Theorem 1.3. We prove the contrapositive, namely we show that if G is a d-regular graph such that κ (k 1)n (G) k 1, then λ (G) > d. Let V (d+1)(n d 1) 1 V = V (G) be 3

a partition of G such that r = e(v 1,V ) k 1 d 1. Note that if V 1 d, then d 1 e(v 1,V ) V 1 (d V 1 + 1) d which is a contradiction. Thus, n i := V i d + 1 for each i = 1,. Since n 1 +n = n, this implies that n 1 n (d+1)(n d 1). The quotient matrix of the partition V 1 V = V (G) (see [8, 9, 10]) is [ d r A = r n n 1 r n 1 d r n The eigenvalues of A are d and d r n 1 r n. Eigenvalue interlacing (see [8, 9, 10]), r k 1 and n 1 n (d + 1)(n d 1) imply that ] () λ (G) d r n 1 r n d (k 1)n n 1 n d (k 1)n (d + 1)(n d 1) (3) We actually have strict inequality here. Otherwise, the partition V 1 V would be equitable. This would mean that each vertex of V 1 has the same number of neighbours in V which is impossible since there are vertices in V 1 without a neighbour in V. The proof is finished. 3 Proof of Theorem 1.4 In this section, we present the proof of Theorem 1.4. First, we show that any connected, regular graph of even degree has edge-connectivity larger than two. This follows from the next lemma. Lemma 3.1. Let G be a connected d-regular graph. If d is even, then κ (G) is even. Proof. Let V = A B be a partition of the vertex set of G such that e(a,b) = κ (G), where e(a,b) denotes the number of edges with one endpoint in A and one endpoint in B. Summing up the degrees of the vertices in A, we obtain that d A = e(a) + e(a,b) = e(a) + κ (G), where e(a) denotes the number of edges with both endpoints in A. Since d is even, this implies κ (G) is even which finishes the proof. For the rest of this section, we assume that d is an odd integer with d 3. We describe first the d-regular graph X d having κ (X d ) = 1 and λ (X d ) = ρ(d). If H 1,...,H k are pairwise vertex-disjoint graphs, then we define the product H 1 H H k recursively as follows. If k = 1, the product is H 1. If k =, then H 1 H is the usual join of H 1 and H, i.e. the graph formed from the union of H 1 and H by joining each vertex of H 1 to each vertex of H. If k 3, then H 1 H H k is the graph obtained by taking the union of H 1 H H k 1 and H k and joining each vertex of H k 1 to each vertex of H k. The extremal graph X d is defined as: X d = K M d 1 K 1 K 1 M d 1 K (4) where M d 1 is the unique (d 3)-regular graph on d 1 vertices (the complement of a perfect matching M d 1 on d 1 vertices). Figure shows X 3 and Figure 3 describes X 5. 4

The graph X d is d-regular, has d + 4 vertices and its edge-connectivity is 1. It has an equitable partition of its vertices into six parts which is described by the following quotient matrix X d. The sizes of the parts in this equitable partition are,d 1, 1, 1,d 1,. 1 d 1 0 0 0 0 d 3 1 0 0 0 X d = 0 d 1 0 1 0 0 0 0 1 0 d 1 0 (5) 0 0 0 1 d 3 0 0 0 0 d 1 1 Next, we show that the second largest eigenvalue of X d equals the second largest eigenvalue of Xd. This will enable us to get sharp estimates for λ (X d ) which will be used later in this section. Figure : X 3 is 3-regular with λ (X 3 ) = ρ(3).7784 Lemma 3.. For each odd integer d 3, we have that λ (X d ) = λ ( X d ) = ρ(d). Proof. Since the partition with quotient matrix X d is equitable, it follows that the spectrum of X d contains the spectrum of Xd. Using Maple, we determine the characteristic polynomial of Xd : P Xd (x) = (x d)(x 1)(x + )(x 3 (d 3)x (3d )x ) (6) The roots of P Xd are d, 1, and the three roots of Q(x) = x 3 (d 3)x (3d )x. Denote by λ λ 3 λ 4 the roots of Q(x). Since these roots sum up to d 3 0 and their product is, it follows that λ is positive and both λ 3 and λ 4 are negative. Because Q(1) = 4d + 4 < 0, we deduce that λ > 1. Thus, λ ( X d ) = λ = ρ(d). Let W R d+4 be the subspace of vectors which are constant on each part of the six part equitable partition. The lifted eigenvectors corresponding to the six roots of P Xd form a basis for W. The remaining eigenvectors in a basis of eigenvectors for X d can be chosen to be perpendicular to the vectors in W. Thus, they may be chosen to be perpendicular to the characteristic vectors of the parts in the six-part equitable partition since these characteristic vectors form a basis for W. This implies that these eigenvectors will correspond to the non-trivial eigenvalues of the graph obtained as a disjoint union of K and M d 1. The corresponding eigenvalues will be ( ) (d 3), ( 1) (), 0 (d 1), where the exponent denotes the multiplicity of each eigenvalue. These d eigenvalues together with the six roots of P Xd form the spectrum of the graph X d. Thus, λ (X d ) = λ ( X d ) = ρ(d). 5

Figure 3: X 5 is 5-regular with λ (X 5 ) = ρ(5) 4.7969 Using Maple, we find that ρ(3).7784 and ρ(5) 4.7969. A simple algebraic manipulation shows that ρ(d) satisfies the equation d ρ(d) = d ρ (d) + 3ρ(d) + > d d + 3d +, (7) where the last inequality follows since ρ(d) < d. Using (7), we deduce that d d + 4 < ρ(d) < d d + 5 for d 5. We are now ready to prove Theorem 1.4. Proof of Theorem 1.4. We will prove the contrapositive, namely we will show that if G is a d-regular graph with edge connectivity 1 (containing a bridge), then λ (G) λ (X d ) = ρ(d). We also show that equality happens if and only if G = X d. Consider a connected d-regular graph G that contains a bridge x 1 x. Deleting the edge x 1 x partitions V (G) into two connected components G 1 and G such that x i V (G i ) for i = 1,. Let n i = V (G i ) for i = 1,. Without loss of generality, we assume from now on that n 1 n. The graph G 1 contains n 1 1 vertices of degree d and one vertex of degree d 1. Because d is odd, this implies that n 1 is odd. By symmetry, n is also odd. Thus, n n 1 d +. Note that G = X d if and only if n 1 = n = d +. Let A be the quotient matrix of the partition of V (G) into V (G 1 ) and V (G ). Then [ d 1 ] A = n 1 1 n 1 1 n d 1 n The eigenvalues of A are: d and λ (A ) = d 1 n 1 1 n. Let A 3 be the quotient matrix of the partition of V (G) into V (G 1 ), {x },V (G ) \ {x }. Then d a a 0 A 3 = 1 0 d 1 (10) 0 b d b where a = 1 n 1 and b = d 1 n 1 3 are d,λ (A 3 ) = d a b+ (d a+b) +4(a b),λ 3 (A 3 ) = d a b (d a+b) +4(a b). Taking partial derivatives with respect to a and to b respectively, we 6 (8) (9)

find that when d > 1, the eigenvalue λ (A 3 ) is strictly monotone decreasing with respect to both a and b and so is strictly monotone increasing with respect to both n 1 and n. We consider now a partition of V (G) into the following six parts: V (G 1 )\(x 1 N(x 1 )),N(x 1 )\ {x }, {x 1 }, {x },N(x ) \ {x 1 } and V (G ) \ (x N(x )). Here N(u) denotes the neighbourhood of vertex u in G. Let e 1 denote the number of edges between N(x 1 ) \ {x } and V (G 1 ) \ (x 1 N(x 1 )). Let e denote the number of edges between N(x ) \ {x 1 } and V (G ) \ (x N(x )). Note that e i (d 1)(n i d) for i = 1,. Let A 6 be the quotient matrix of the previous partition of V (G) into six parts. Then d e 1 e 1 n 1 d n 1 0 0 0 0 d e 1 d 1 e 1 1 0 0 0 d 1 d 1 A 6 = 0 d 1 0 1 0 0 0 0 1 0 d 1 0 (11) 0 0 0 1 d 1 e 0 0 0 0 Eigenvalue interlacing (see [8, 9, 10]) implies that d 1 e d 1 e n d d e n d λ (G) max(λ (A ),λ (A 3 ),λ (A 6 )). (1) We will use this as well as the inequalities (7) and (8) to show that λ (G) λ (X d ) = ρ(d) and that equality happens if and only if G = X d. Recall that n n 1 d + and that d,n 1 and n are all odd. If n 1 d+6 and d 5, then we use the partition of V (G) into two parts whose quotient matrix A is given in (9). Using inequality (8), we have that λ (G) λ (A ) = d 1 n 1 1 n d d + 6 > ρ(d). If n 1 d+6 and d = 3, then we use the partition of V (G) into three parts whose quotient matrix A 3 is given in (10). We have that (3 λ (G) λ (A 3 ) 3 1 + 9 8 1 ( 9 8) + 4 1 ) 1 9 4 = 95 + 1049 >.84 > ρ(3). 7 If n n 1 = d + 4, then we use the partition of V (G) into three parts whose quotient matrix A 3 is shown in (10). We have that (d λ (G) λ (A 3 ) d 1 d 1 + d+4 d+3 1 + ) d 1 ( d+4 d+3 + 4 1 ) d 1 d+4 d+3 = d3 + 6d + 8d + 1 + d 6 + 16d 5 + 88d 4 + 174d 3 + 8d 96d + 385 (d + 7d + 1) = d3 + 6d + 8d + 1 + (d 3 + 8d + 1d 9) + 8d + 10d + 304 (d + 7d + 1) 7

If d = 3, the right hand side of the previous inequality equals 106+ 1661 84 >.796 > ρ(3). When d 5, from the previous inequality and (8) we obtain that λ (G) d3 + 6d + 8d + 1 + (d 3 + 8d + 1d 9) + 8d + 10d + 304 (d + 7d + 1) > d3 + 6d + 8d + 1 + (d 3 + 8d + 1d 9) = d3 + 7d + 10d 4 (d + 7d + 1) d + 7d + 1 d + 4 = d d + 7d + 1 > d d + 5 > ρ(d) If n 1 = d + and n d + 6, then we use the partition of V (G) into three parts whose quotient matrix is is given in (10). We obtain that (d λ (G) λ (A 3 ) d 1 d 1 + d+ d+5 1 + ) d 1 ( d+ d+5 + 4 1 ) d 1 d+ d+5 = d3 + 6d + 8d 3 + d 6 + 16d 5 + 80d 4 + 118d 3 4d + 56d + 39 (d + 7d + 10) When d = 3, the right hand side equals 10+ 14564 80 >.7835 > ρ(3). If 5 d 17, then from the previous identity, we deduce that λ (G) d3 + 6d + 8d 3 + (d 3 + 8d + 8d 5) 8d + 136d + 304 (d + 7d + 10) > d3 + 6d + 8d 3 + (d 3 + 8d + 8d 5) (d + 7d + 10) = d3 + 7d + 8d 4 d + 7d + 10 = d d + 5 > ρ(d). When d 19, from the previous inequality we obtain that λ (G) d3 + 6d + 8d 3 + (d 3 + 8d + 8d 6) + d 3 + 8d + 15d + 93 (d + 7d + 10) > d3 + 6d + 8d 3 + (d 3 + 8d + 8d 6) = d3 + 7d + 8d 4.5 (d + 7d + 10) d + 7d + 10 d + 4.5 = d d + 7d + 10 > d d d + 3d + > ρ(d). The only case which remains to consider is n 1 = d + and n = d + 4. In this case, e 1 = d 4,n d = 4 and e is an even integer with 4d 1 e 4d 4. Thus, 1 d 1 0 0 0 0 d 3 1 0 0 0 A 6 = 0 d 1 0 1 0 0 0 0 1 0 d 1 0 0 0 0 1 d 1 e e d 1 d 1 e 0 0 0 0 4 d e 4 8

Let P A6 (x) denote the characteristic polynomial of A 6. Since each row sum of A 6 is d, it follows that d is a root of P A6 (x). Thus, P A6 (x) = (x d)p 5 (x). The second largest root of P A6 (x) is the largest root of P 5 (x). Using the results of [3] page 130, we observe that P 5 (x) equals the characteristic polynomial of the following matrix: 1 1 0 0 0 0 1 0 0 0 d 1 d d 1 0 e 0 0 1 0 d 1 0 0 0 1 d e e d 1 n d Using Maple to divide P 5 (x) by the polynomial x 3 + (3 d)x + ( 3d)x, we obtain the quotient Q(x) = x 4d 4d de 3e x e and the remainder R(x) = (d e 4(d 1) d 1 )x + de +e +4d 4d. Thus, P 5 (x) = (x 3 + (3 d)x + ( 3d)x )Q(x) + (d e )x + de + 4d + e 4d. Because ρ(d) satisfies the equation x 3 + (3 d)x + ( 3d)x = 0, it follows that P 5 (ρ(d)) = (d e )ρ(d) + de + 4d + e 4d ( = e ρ(d) + d + 1 ) + (d )(ρ(d) d). Because d > ρ(d) > d 1 d+1, we deduce that P 5 (ρ(d)) < 0. (13) Assume that ρ(d) λ (A 6 ). Then ρ(d) is larger than any root of P 5 (x). This implies that P 5 (ρ(d)) = θ root of P 5 (ρ(d) θ) 0 which is a contradiction with (13). Thus, λ (G) λ (A 6 ) > ρ(d). Hence, λ (G) > ρ(d) whenever n d + 4. The last case to be considered is n 1 = n = d +. This means that G = X d and λ (G) = ρ(d) which finishes the proof. 4 Proof of Theorem 1.5 In this section we present the proof of Theorem 1.5. We describe first the d-regular graph Y d having κ (Y d ) = and λ (Y d ) = d 3+ (d+3) 16. Consider the graph H d = K d 1 K. It has d 1 vertices of degree d and two vertices of degree d 1. We construct Y d by taking two disjoint copies of K d 1 K and adding two disjoint edges between the vertices of degree d 1 in different copies of K d 1 K. Figure 4 describes Y 3 and Figure 5 shows Y 4. 9

Figure 4: Y 3 is 3-regular with λ (Y 3 ) = θ(3) = 5 The graph Y d is d-regular, has d + vertices and its edge-connectivity is. It has an equitable partition into four parts of sizes d 1,,,d 1 with the following quotient matrix: d 0 0 Ỹ d = d 1 0 1 0 0 1 0 d 1 (14) 0 0 d The characteristic polynomial of this matrix equals P Yd (x) = (x d)(x + 1)(x + (3 d)x + (4 3d)) Thus, the eigenvalues of Ỹd are d, 1 and d 3± (d+3) 16. To simplify our notation, let θ(d) = d 3+ (d+3) 16. Note that θ(d) is the largest root of and T(x) = x + (3 d)x + (4 3d) (15) d 4 d + < θ(d) < d 4 d + 3. (16) Lemma 4.1. The second largest eigenvalue of Y d equals θ(d). Proof. Since the previous partition of V (Y d ) into four parts is equitable, it follows that the four eigenvalues of Ỹ d are also eigenvalues of Y d. Obviously, the second largest of the eigenvalues of Ỹ d is θ(d). By an argument similar to the one of Lemma 3., one can show that the other eigenvalues of Y d are ( 1) (d ) which implies the desired result. We are ready now to prove Theorem 1.5. Proof of Theorem 1.5. We will prove the contrapositive, namely we will show that among all d-regular graphs with edge-connectivity less than or equal to, the graph Y d has the smallest λ. Let G be a d-regular graph with κ (G). We will prove that λ (G) θ(d) with equality if and only if G = Y d. If κ (G) = 1, then Theorem 1.4, (8) and (16) imply that λ (G) ρ(d) d d 4 d+3 > θ(d). 10 (d 1) (d+1)(d+)

Figure 5: Y 4 is 4-regular with λ (Y 4 ) = θ(4) = 1+ 33 If κ (G) =, then there exists a partition of V (G) into two parts V 1 and V such that e(v 1,V ) =. Let S 1 V 1 and S V denote the endpoints of the two edges between V 1 and V. We have that ( S 1, S ) {(1, ), (, ), (, 1)}. Let n i = V i for i {1, }. It is easy to see that n i d + 1 for each i {1, }. Note that n 1 = n = d + 1 is equivalent to G = Y d. Without loss of generality assume that n n 1 d + 1. If n 1 d + 3, consider the partition of V (G) into V 1 and V. The quotient matrix of this partition is [ d n 1 n 1 n d n and its eigenvalues are d and d n 1 n. Eigenvalue interlacing and n n 1 d+3 imply that λ (G) d n 1 n d 4. Using inequality (16), we obtain that λ d+3 (G) d 4 > θ(d) d+3 which finishes the proof of this case. If n 1 = d +, then we have a few cases to consider. If S 1 =, then e(s 1 ) = 0 or e(s 1 ) = 1. If e(s 1 ) = 0, consider the partition of V (G) into three parts: V 1 \ S 1,S 1,V. The quotient matrix of this partition is d d d 0 d d d 1 0 1 0 n d n Its characteristic polynomial equals ( ) P 3 (x) = (x d) x d n + n dn d x (d n + n + dn d) dn dn If P (x) = x d n +n dn d dn ( where R(x) = dn +n d d n +d n 4 dn dn +n x d x (d n +n + dn d) dn, then ] P (x) = T(x) + R(x) ). The expression d n +d n 4 dn +n d is decreasing with n and thus it attains its maximum when n = d+. This maximum equals d3 +d 8 d +d+4 = 11

d 4(d+) < d 4. Thus, P ( d +d+4 d+ (θ(d)) = R(θ(d)) < dn +n d dn d 4 θ(d)) < 0 where d+ the last inequality follows from (16). This fact and eigenvalue interlacing imply that λ (G) > θ(d). If e(s 1 ) = 1, then we consider the same partition into three parts: V 1 \ S 1,S 1,V. The quotient matrix is the following d d 4 d 4 0 d d d 1 1 0 n d n Its characteristic polynomial equals ( ) P 3 (x) = (x d) x d n + 4n dn d x d n + 4n + 8 6dn. dn dn If P (x) = x d n +4n dn d dn x d n +4n +8 6dn dn then P (x) = T(x) + R(x) ) ( where R(x) = (dn +n d) d n +dn n 4 dn dn +n x d. The expression d n +dn n 4 dn +n d decreases with n and thus, its maximum is attained at n = d +. This maximum equals d3 +3d 8 d +3d+4 = d 4(d+) < d 4. As before, the previous inequality and eigenvalue interlacing imply d +3d+4 d+ that λ (G) > θ(d). If S 1 = 1, then d must be even. Indeed, the subgraph induced by V 1 contains d + 1 vertices of degree d and one vertex of degree d. This cannot happen when d is odd. Thus, d is even and d 4. Consider the partition of G into three parts: V 1 \ S 1,S 1,V. The quotient matrix of this partition is d d d 0 d+1 d+1 d 0 0 n d n Its characteristic polynomial is ( ) P 3 (x) = (x d) x d n + n d x d n + d + 4n + 8 4dn (d + 1)n (d + 1)n If P (x) = x d n +n d (d+1)n x d n +d+4n +8 4dn, then where R(x) = dn +5n d d n +3dn 8n d 8 dn +5n d (d+1)n ( (d+1)n P (x) = T(x) + R(x) ) x. Because d 4, the expression d n +3dn 8n d 8 dn +5n d is decreasing with n and thus, its maximum is attained when n = d+. This maximum equals d3 +7d 4d 4 = d 1d+4 < d 4 d +7d+8 d +7d+8 interlacing imply λ (G) > θ(d). 1 d+. This fact and eigenvalue

Assume now that n 1 = d + 1. This implies that V 1 induces a subgraph isomorphic to K d 1 K and consequently, S 1 = and e(s 1 ) = 0. If n d + 3, then consider the partition of G into three parts: V 1 \ S 1,S 1,V. The quotient matrix of this partition is d 0 d 1 0 1 0 n d n Its characteristic polynomial is If P (x) = x which implies that P 3 (x) = (x d) (d n ) x + + n d, then ( ) x (d n x + + ) d. n P (x) = T(x) + dn + n (n )x n P (θ(d)) = dn + n (n )θ(d) n The expression dn + n (n )θ(d) n is decreasing with n and therefore, its maximum is attained when n = d + 3. Thus, P (θ(d)) d + d 4 (d + 1)θ(d) d + 3 = (d + 1)( d 4 θ(d)) d+1 < 0. d + 3 This inequality and eigenvalue interlacing imply that λ (G) > θ(d). Thus, the remaining case is n 1 = d + 1 and n d +. If d is odd, then n d +. Otherwise, the sum of the degrees of the graph induced by V would equal d V = d(d + ) which is an odd number. This is impossible and thus, n = d + 1. This implies G = Y d. If n = d + and d is even, then we have a few cases to consider. If S = 1, then both vertices in S 1 are adjacent to the vertex a of S. Thus, a has exactly d neighbours in V which means there are 3 vertices of V which are not adjacent to a. Each of these three vertices has degree d and the only way this can happen is they form a clique and each of them is adjacent to the d neighbours of a in V. Using a degree argument, it also follows that the d neighbours of a in V induce a subgraph isomorphic to the complement of a perfect matching on d vertices M d. Using the notation from the previous section, it follows that G = K d 1 K K 1 M d K 3. The graph G has an obvious equitable partition into five parts whose quotient matrix is d 0 0 0 d 1 0 1 0 0 0 0 d 0 0 0 1 d 4 3 0 0 0 d 13

Its characteristic polynomial equals P 5 (x) = (x d)p 4 (x) where P 4 (x) = x 4 + ( d + 4)x 3 + ( 4d+4)x 8x+6d 1. Dividing P 4 (x) by T(x) we get that P 4 (x) = T(x)Q 4 (x)+r 4 (x) where Q 4 (x) = x + x 3 and R 4 (x) = 3x 3d. It follows that P 4 (θ(d)) = R 4 (θ(d)) = 3θ(d) 3d < 0. This fact and eigenvalue interlacing imply that λ (G) > θ(d). The case remaining is S =. If e(s ) = 0, then each vertex of S has exactly d 1 neighbours in V. Let S = {x,y }. If x and y are adjacent to the same d 1 vertices of V, then because V = n = d +, there is exactly one vertex of V outside the vertices of S and their d 1 common neighbours. This vertex cannot have degree d. Thus, this case cannot happen and therefore, both x and y have exactly d common neighbours in V. We call this set U. Let {u,w } = V \ ({x,y } U). Because x and y have d 1 neighbours in V, we may assume that x is adjacent to u and y is adjacent to w. A degree argument implies that u and w must be adjacent and both of them are adjacent to each vertex of U. Finally, the subgraph induced by U must be (d 4)-regular. The following is a five-part equitable partition of G: V 1 \ S 1,S 1,S,U, {u,w }. The quotient matrix of this partition is the following: d 0 0 0 d 1 0 1 0 0 0 1 0 d 1 0 0 d 4 0 0 1 d 1 Its characteristic polynomial equals P 5 (x) = (x d)p 4 (x) where P 4 (x) = x 4 +(5 d)x 3 +(10 5d)x +( 6d+7)x d. Dividing P 4 (x) by T(x) we get that P 4 (x) = T(x)Q 4 (x)+r 4 (x) where Q 4 (x) = x + x and R 4 (x) = x d. It follows that P 4 (θ(d)) = R 4 (θ(d)) = θ(d) d < 0. This and eigenvalue interlacing show that λ (G) > θ(d). The final case of our proof is e(s ) = 1. Because V \ S = d and both x and y have d neighbours in V \ S, it follows that x and y have at least d 4 common neighbours in V \ S. If x and y have at least d 3 common neighbours in V \ S, we deduce that there exists at least one vertex z of V \ S that is not adjacent to x nor y. The vertex z cannot have degree d since its only possible neighbours are in V \ (S {z}) which has size d 1. We conclude that x and y must have precisely d 4 common neighbours in V \ S. There are 4 remaining vertices a,b,c,d in V \ S and without loss of generality, assume that x is adjacent to both a and b and y is adjacent to both c and d. Each of the vertices a,b,c,d will be adjacent to every other vertex of V \ S. The degree constraint implies that the remaining d 4 vertices of V \ S induce a (d 6)-regular subgraph of G. Obviously, this argument shows this case is only possible when d 6. The following is a five-part equitable partition of G: V 1 \ S 1,S 1,S, {a,b,c,d },V \ (S {a,b,c,d } ). Its quotient matrix is the following: d 0 0 0 d 1 0 1 0 0 0 1 1 d 4 0 0 1 3 d 4 0 0 4 d 6 14

Its characteristic polynomial equals P 5 (x) = (x d)p 4 (x) where P 4 (x) = x 4 + ( d + 4)x 3 + ( 4d+6)x +( d 1)x+3d 6. Dividing P 4 (x) by T(x) we get that P 4 (x) = T(x)Q 4 (x)+ R 4 (x) where Q 4 (x) = x +x 1 and R 4 (x) = x. It follows that P 4 (θ(d)) = R 4 (θ(d)) = θ(d) < 0. This and eigenvalue interlacing imply that λ (G) > θ(d) which finishes our proof. 5 Some Remarks Any strongly regular graph of degree d 3 satisfies the condition λ d and thus, is d-edge-connected. The fact that the edge-connectivity of a strongly regular graph equals its degree, was observed by Plesńik in 1975 (cf. []). As mentioned in the introduction, much more is true, namely the vertex-connectivity of any distance-regular graph equals its degree (see [5]). It is known that any vertex transitive d-regular graph whose second largest eigenvalue is simple has λ (G) d and consequently, is d-edge-connected. In fact, any vertex transitive d-regular graph is d-edge-connected as shown by Mader in 1971 (see [14] or Chapter 3 of [8]). I expect that Theorem 1.4 and Theorem 1.5 can be extended to other values of edgeconnectivity and vertex-connectivity. For example, it seems that λ (G) d 1 implies κ(g). Note however that in many cases, Fiedler s bound κ(g) d λ cannot be improved. When k d and dk is even, consider the graph K d k+1 H where H is a (k d)-regular graph on k vertices. This graph is d-regular, has vertex connectivity k and its second largest eigenvalue equals d k. 6 Acknowledgments I am grateful to David Gregory, Edwin van Dam, Willem Haemers and an anonymous referee for many suggestions that have greatly improved the original manuscript. I thank Professor Nair Abreu, the organizers and the participants in the Spectral Graph Theory in Rio Workshop for their support and enthusiasm. References [1] N.M.M. de Abreu, Old and new results on algebraic connectivity of graphs, Linear Algebra and its Applications 43 (007), 53-73. [] A.E. Brouwer, Spectrum and connectivity of a graph, CWI Quarterly 9 (1996), 37-40. [3] A.E. Brouwer, A.M. Cohen and A. Neumaier, Distance Regular Graphs, Springer (1989). [4] A.E. Brouwer and W.H. Haemers, Eigenvalues and perfect matchings, Linear Algebra and its Applications, 395 (005), 155-16. [5] A.E. Brouwer and J. Koolen, The vertex-connectivity of a distance-regular graph, European Journal of Combinatorics 30 (009), 668-673. 15

[6] S.L. Chandran, Minimum cuts, girth and spectral threshold, Inform. Process. Lett. 89 (004), no. 3, 105 110. [7] M. Fiedler, Algebraic connectivity of graphs, Czech. Math. J. 3 (1973), 98-305. [8] C. Godsil and G. Royle, Algebraic Graph Theory, Springer 001. [9] W.H. Haemers, Eigenvalue techniques in design and graph theory, Ph.D. Thesis (1979), Eindhoven University of Technology. [10] W. Haemers, Interlacing eigenvalues and graphs, Linear Algebra and its Applications (1995) 6-8, 593-616. [11] S. Hoory, N. Linial and A. Wigderson, Expander graphs and their applications, Bull. Amer. Math. Soc. (006) 46, 439-561. [1] S. Kirkland, J.J. Molitierno, M. Neumann and B.L. Shader, On graphs with equal algebraic and vertex connectivity, Linear Algebra and its Applications 341 (00), 45-56. [13] M. Krivelevich and B. Sudakov, Pseudo-random graphs, in More Sets, Graphs and Numbers, Bolyai Society Mathematical Studies 15 (006), 199-6. [14] W. Mader, Minimale n-fach kantenzusammenhängende Graphen, Math. Ann. 191 (1971), 1-8. [15] B. Mohar, Some applications of Laplace eigenvalues of graphs, Graph Symmetry: Algebraic Methods and Applications, Eds. G. Hahn and G. Sabidussi, NATO ASI Ser. C 487, Kluwer, (1997), 5-75. 16