Spectral diameter estimates for k-regular graphs

Transcription

1 Spectral diameter estimates for k-regular graphs Gregory Quenell May 1992 An undirected graph is called k-regular if exactly k edges meet at each vertex. The eigenvalues of the adjacency matrix of a finite, k-regular graph Γ assumed to be undirected and connected) satisfy λ i k, with k occurring as a simple eigenvalue. Let λγ) denote the maximum of { λ i : λ i = k}, and let denote the number of vertices in Γ. The diameter of Γ can be bounded in terms of and λγ). Chung [5]) and Sarnak [7]) have derived the estimate diamγ) arccosh 1) ) + 1, arccosh k λγ) which will be our Theorem 2. Following Brooks [2]), we define graph-theoretic spherical functions S λ on the universal cover of Γ, and apply the techniques of the Selberg pre-trace formula to write a spectral expression for the number of paths of length n joining two vertices x and y in Γ. Calling this number K n x, y), we have, as Lemma 9.4, K n x, y) = kk 1) n 1 i S λi n)ϕ i x)ϕ i y) where {ϕ i } is an orthonormal set of eigenfunctions for the adjacency operator, in correspondence with the eigenvalues {λ i }. 1

2 We then use the spectral expression for K r x, y), where r 1 is the injectivity radius of Γ, along with the ideas used to prove Theorem 2, to derive a new diameter estimate in Theorem 3: diamγ) ) arccosh 1 kk 1) r 1 ) + 2r + 1. arccosh k λγ) Finally, in Section 12, we argue that the bound given by Theorem 3 is usually stronger than the bound in Theorem 2. 1 The adjacency matrix Let Γ be a k-regular, connected, undirected graph with a finite number of vertices. We assume Γ has no loops or multiple edges, so that its adjacency matrix A is a symmetric, irreducible zero-one matrix with no ones on the diagonal and exactly k ones in each row and column. We will use the notation x y to mean the vertices x and y are joined by an edge. Also x Γ will be taken to mean x is a vertex in Γ. A walk in Γ is a sequence x 0, x 1,..., x n of vertices such that x i 1 x i for i = 1,..., n. The vertices x 0 and x n are the endpoints of the walk and the number n is its length. Let W n x, y) denote the number of walks of length n whose endpoints are x and y. Then it is easy to see that W n x, y) = [A n ] xy 1) where we have used the notation [M] xy for the entry of the matrix M in the row associated with the vertex x and the column associated with the vertex y. A path in Γ is a walk x 0, x 1,..., x n such that x i x i 2 for i = 2,..., n. That is to say, a path does not double back by traversing the same edge twice in succession. We will let K n x, y) denote the number of paths of length n whose endpoints are x and y. An expression for K n in terms of A is somewhat more complicated than 1), and will be the subject of a later section. The distance between two vertices x and y, denoted distx, y), is the length of the shortest walk whose endpoints are x and y. We are assuming 2

3 that Γ is connected, so all distances are finite. The diameter of Γ is the maximum of distx, y) over all pairs x, y). A walk or path x 0, x 1,..., x n is closed if x 0 = x n. A graph Γ is said to be bipartite if it has no closed walks of odd length. In terms of our present notation, Γ is bipartite if and only if W n x, x) = 0 for all odd n and all vertices x. Because Γ is connected, the matrix A is irreducible, and the condition just stated is equivalent to the period of the matrix A being even. Since we have for all vertices x, we can say even more: [A 2 ] xx = k 0 2) Lemma 1.1 The period of A is 2 if Γ is bipartite, and 1 if Γ is not bipartite. Because A is symmetric, all its eigenvalues are real. Furthermore, the k-regularity of Γ means that all the row sums of A are equal to k. Lemma 1.1 and the Perron-Frobenius theorem then imply Lemma 1.2 Let A be the adjacency matrix of a finite, connected, k-regular graph Γ. Then a) the number k is a simple eigenvalue of A; b) the number k is a simple eigenvalue of A if and only if Γ is bipartite; c) all other eigenvalues λ of A satisfy λ < k. Let λγ) denote the absolute value of the next-largest eigenvalue of A, that is λγ) = max{ λ : λ SpecA), λ = k}. 3) 3

4 2 The adjacency operator By a function on Γ, we will mean a real-valued function on the vertices of Γ. If Γ has vertices, then space of such functions is isomorphic, as a real inner-product space, with R. If ϕ and ψ are two functions on Γ, we will write their inner product with angle brackets: ϕ, ψ = x Γ ϕx)ψx). 4) If ϕ is a function on Γ, then we can write ϕ as a column vector ϕx) = ϕx 1 ) ϕx 2 ) ϕx )) t 5) where x i is the vertex associated with the i th row of the adjacency matrix A. We then obtain a new function Aϕ by matrix multiplication. In fact, the i th entry of the column vector Aϕ is given by [Aϕ] i = In more operator-like notation, we have x j x i ϕx j ). 6) Aϕ)x) = y x ϕy). 7) We will take expression 7) as our definition of the adjacency operator A on the space of functions on a graph. We have already remarked that the spectrum of A as a matrix, but also as an operator) is a subset of the real line. In the sequel, we will write the spectrum of A as {λ 0, λ 1,..., λ 1 } with k = λ 0 > λ 1 λ 2 λ 1. 8) Furthermore, because A is self-adjoint it is represented by a symmetric matrix), there exists a corresponding orthonormal basis of eigenfunctions of A, which we will write {ϕ 0, ϕ 1,..., ϕ 1 }. 9) 4

5 3 Aside: the Laplacian A more familiar operator in the field of spectral geometry is the Laplacian operator,, which appears so prominently in the heat equation and the wave equation. A reasonable way to define a Laplacian operator on a graph is ϕ)x) = y x ϕx) ϕy)), 10) where ϕ is a function on the graph. If Γ is k-regular, the right-hand side of 10) is easily seen to be kϕx) y x ϕy) 11) so that = ki A, 12) where I represents the identity operator. It follows that the spectrum of the adjacency matrix A is the image of the spectrum of under an affine transformation. Thus the number λγ), loosely speaking, represents the fundamental frequency of the graph Γ. The analogy is not perfect, because λγ) may correspond to either the lowest or the highest frequency among the nontrivial steady-state solutions to the wave equation on Γ. onetheless, we will treat λγ) as an observable, and base our diameter estimates on this number. 4 The spectrum on a bipartite graph If Γ is bipartite then, as the term suggests, we can partition the vertices of Γ into two sets with the property that no two vertices of the same set are joined by an edge. Let us arbitrarily assign a parity to each of these two sets: sgnx) will be even if x belongs to the even set and odd if x belongs to the odd set. 5

6 Lemma 4.1 Suppose Γ is bipartite. Let ϕ be a function on Γ such that Aϕ = λϕ for some real number λ. Define a new function ψ by ψx) = 1) sgnx) ϕx). 13) Then Aψ = λψ. Proof. Aψ)x) = ψy) 14) y x = y x 1) sgny) ϕy) 15) = y x 1) sgnx) ϕy) 16) = 1) sgnx) Aϕ)x) 17) = λ 1) sgnx) ϕx) 18) = λψx) 19) Corollary 4.2 Let Γ be a k-regular bipartite graph with adjacency spectrum k = λ 0 > λ 1 λ 2 λ 2 > λ 1 = k. 20) Let {ϕ 0, ϕ 1,..., ϕ 1 } denote an orthonormal set of eigenfunctions corresponding to {λ 0, λ 1,..., λ 1 }. Then for each i {0, 1,..., 1}, a) λ 1 i = λ i ; b) we may take ϕ 1 i x) = 1) sgnx) ϕ i x); c) in this case, ϕ 1 i x)ϕ 1 i y) = 1) distx,y) ϕ i x)ϕ i y). 21) 6

7 Proof. Parts a) and b) follow from Lemma 4.1 and the existence of an orthonormal basis of eigenfunctions. From part b), we have ϕ 1 i x)ϕ 1 i y) = 1) sgnx)+sgny) ϕ i x)ϕ i y). 22) To prove c), we observe that sgnx) = sgny) if and only if distx, y) is even. 5 W n x, y) We write the walk-counting function W n x, y) in terms of the numbers λ i and the functions ϕ i. Lemma 5.1 W n x, y) = 1 i=0 λ n i ϕ i x)ϕ i y). 23) Proof. We have already observed that W n x, y) = [A n ] xy. 24) Let Φ be the matrix whose columns are the eigenfunctions ϕ i. The following are immediate a) Φ is unitary: Φ 1 = Φ t. b) Φ diagonalizes A: ΦAΦ t = D, where D is the diagonal matrix with diagonal entries λ 0, λ 1,..., λ 1. Then [A n ] xy = [Φ t D n Φ] xy 25) = 1 i=0 7 λ n i ϕ i x)ϕ i y). 26)

8 6 A spectral diameter estimate In addition to what we know about the spectrum of A on a k-regular graph, we will make use of two facts about the eigenfunctions ϕ i. We state them as Lemma 6.1 For each vertex x Γ, a) ϕ 0 x) = 1 ; 27) b) 1 i=0 ϕ 2 i x) = 1. 28) Proof. The constant function 1 satisfies A1 = λ 0 1, where λ 0 = k is a simple eigenvalue. Thus ϕ 0 is a multiple of the constant function, and since ϕ 0, ϕ 0 = 1, we must have ϕ 0 1, establishing part a). Part b) follows from the orthonormality of the ϕ i. If Φ is the matrix whose columns are the ϕ i, then Φ is unitary, so that its rows, as well as its columns, are orthonormal. A row of ϕ takes the form ϕ 0 x) ϕ 1 x) ϕ 1 x)) 29) and part b) is just the assertion that this row vector has norm 1. Our first spectral diameter estimate, which will set the pattern for those to follow, is found in [4]. Theorem 1 Let Γ be a k-regular graph with vertices. a) If Γ is not bipartite then diamγ) log 1) log k log λγ) ) 8

9 b) If Γ is bipartite then diamγ) log 2) log 2 log k log λγ) ) Proof. Choose x and y such that distx, y) = diamγ). If n is a non-negative integer less than distx, y), then clearly W n x, y) = 0. By Lemma 5.1, then, 0 = 1 i=0 λ n i ϕ i x)ϕ i y). 32) We know that λ 0 = k and ϕ 0 x) = ϕ 0 y) = 1/, so we can write which implies that 0 = kn 1 + λ n i ϕ i x)ϕ i y) 33) k n = 1 λ n i ϕ i x)ϕ i y). 34) Assume now that Γ is not bipartite. Then λγ)) n is the greatest element of the set { λ n i : i = 1,..., 1}. Applying the triangle inequality, the observation just made about λγ), and finally the Cauchy-Schwarz inequality, we obtain, from 34), k n 1 λ n i ϕ i x)ϕ i y) 35) 1 λγ)) n ϕ i x)ϕ i y) 36) 1 λγ)) n ϕ 2 i x) ) ϕ 2 i y) ) ) 9

10 Lemma 6.1 implies that 1 ϕ 2 i y) = Putting this into 37), we get 1 ϕ 2 i x) = ) k n λγ))n 1 1 ) 39) which yields the upper bound n log 1) log k log λγ). 40) This is true for each n less than distx, y) = diamγ), and we conclude that diamγ) log 1) ) log k log λγ) ow suppose Γ is bipartite. If the parity of n is different from the parity of distx, y), then parts a) and c) of Corollary 4.2 imply that λ n iϕ i x)ϕ i y) = λ n i ϕ i x)ϕ i y) 42) for each i. In this case the right-hand side of equation 32) is trivially zero, and the equation yields no information. If, on the other hand, the parity of n agrees with the parity of distx, y), then for each i, λ n 1 iϕ 1 i x)ϕ 1 i y) = λ n i ϕ i x)ϕ i y) 43) and equation 32) may be decomposed thus: 0 = 2 kn 2 + λ n i ϕ i x)ϕ i y). 44) 10

11 Imitating our previous computation 35) - 37)), we write 2 kn = 2 λ n i ϕ i x)ϕ i y) 45) 2 λγ)) n ϕ i x)ϕ i y) 46) 2 λγ)) n ϕ 2 i x) ) ϕ 2 i y) ) ) Using the fact that ϕ 1 x) = ±ϕ 0 x) and both parts of Lemma 6.1, we find that 2 ϕ 2 i y) = 2 Plugging this into 47), we continue: 2 kn λγ))n k n 2 λγ)) n 2 n ϕ 2 i x) = ) 1 2 ) 49) 50) log 2) log 2 log k log λγ). 51) We had to assume that n was less than diamγ) and of the same parity as diamγ), so we can conclude that diamγ) log 2) log 2 log k log λγ) ) 11

12 7 Polynomial diameter estimates We can improve on the estimate in Theorem 1 by making better use of our observation that 0 = 1 i=0 λ n i ϕ i x)ϕ i y) 53) for any non-negative integer n less than distx, y). Let x and y be such that distx, y) = diamγ), and let {p n : n = 0, 1, 2,...} be a family of polynomials such that the degree of p n is n. Then it follows from 53) that 0 = 1 i=0 p n λ i )ϕ i x)ϕ i y) 54) for each n less than diamγ). Proceeding as in the proof of Theorem 1, we write this as 0 = p n λ 0 )ϕ 0 x)ϕ 0 y) + 1 p n λ i )ϕ i x)ϕ i y). 55) We know that λ 0 = k and ϕ 0 1, and let us assume that p n k) will be positive for each n. Then 55) implies that p n k) = 1 p n λ i )ϕ i x)ϕ i y). 56) ext we apply the triangle inequality to get p n k) 1 p n λ i ) ϕ i x)ϕ i y). 57) As in the proof of Theorem 1, our diameter estimate will turn out the be the least n for which this inequality does not hold. Thus it is to our 12

13 advantage to make the left side of 57) as large as possible for each n, while keeping the right side small. If Γ is not bipartite and for now we will assume that this is the case), then λ i λγ) for each λ i appearing on the right side of 57). We wish to select a family {p n } of polynomials, then, such that a) p n k) is large for each n, and b) p n λ) is bounded for λ λγ). The Chebychev polynomials see [1]) satisfy T n x) = coshn arccoshx)) 58) T n x) 1 for x 1 59) and increase quite rapidly for x > 1. We do reasonably well by our criteria above by setting ) λ p n λ) = T n λγ) = cosh n arccosh 60) )) λ. 61) λγ) Then p n λ i ) 1 in each term on the right side of 57), and the inequality implies that p n k) 1 ϕ i x)ϕ i y). 62) As before, we use the Cauchy-Schwarz inequality and Lemma 6.1 to get p n k) ϕ 2 i x) ) ϕ 2 i y) ) ) ), 64) 13

14 so that Since p n k) = coshn arccosh k than 1, it follows that p n k) 1. 65) λγ) )) and everything in sight is greater ) k n arccosh λγ) arccosh 1), 66) and finally that n arccosh 1) ). 67) arccosh k λγ) This is true, in particular, for n = diamγ) 1, and we have proved part a) of the following theorem. Part b) depends on a minor modification of the same argument. Theorem 2 Let Γ be a k-regular, connected graph with vertices. a) If Γ is not bipartite then diamγ) arccosh 1) ) ) arccosh k λγ) b) If Γ is bipartite then diamγ) arccosh 1) 2 ) ) arccosh k λγ) Sarnak uses the same technique, but expresses the result slightly differently in [7]. Chung presents a more general form of this theorem, not requiring k-regularity, in [5]. 14

15 8 Polynomial estimates are better ote that the approach we have used to prove Theorem 2 yields exactly the result in Theorem 1 if, instead of using the Chebychev polynomials, we define the family {p n } by p n λ) = In fact, since this family of polynomials also satisfies ) n λ. 70) λγ) p n λ) 1 for λ λγ), 71) we could write a proof of Theorem 1 identical to our proof of Theorem 2 except in the last step, where we would use logarithms, rather than the inverse hyperbolic cosine, to dig out the number n. In Theorem 1, the diameter bound is one more than the greatest n such that ) n k 1 72) λγ) and in Theorem 2 the bound is one more than the greatest n such that )) k cosh n arccosh λγ) 1. 73) To verify that the estimate in Theorem 2 is actually stronger than the estimate in Theorem 1, then, we need only verify that )) k cosh n arccosh λγ) ) n k 74) λγ) for each n. Since k > λγ), this requirement may be phrased more succinctly as coshn arccoshx)) x n for x > 1, n 0. 75) This is true, and may be verified as follows. 15

16 Lemma 8.1 Let T n x) = coshn arccoshx)). Let x > 1. Then a) T 0 x) = x 0 ; T 1 x) = x 1. b) If n > 1 then T n x) > x n. Proof. Both statements in part a) are immediate from the definitions. For part b), we wish to show that, under the stated conditions, coshn arccoshx)) > x n. 76) Since both sides of this inequality are greater than 1 by hypothesis), we can take the inverse hyperbolic cosine and find that the required inequality is equivalent to We will show that the expression n arccoshx) > arccoshx n ). 77) n arccoshx) arccoshx n ) 78) is positive for x > 1 and n > 1. Recalling or looking up) the derivative of arccosh, we write 78) as x n 1 We combine the two integrals to get 1 x t2 1 dt n 1 t n 1 dt. 79) t 2n 1 x t 2n 1 t n 1 t n 2 1 dt. 80) 1 t2 1)t 2n 1) The denominator in the integrand is positive throughout the interval 1, x), so we need only verify that the numerator is positive. Since we already know the antiderivative, we may safely assume that the integral converges.) The numerator is positive if t 2n 1 > t 2n t 2n 2. 81) Since n > 1 and t > 1, the subtrahend t 2n 2 is greater than 1, so the inequality holds. 16

17 9 K n x, y) ext we turn to the function K n x, y), which counts the number of paths of length n whose endpoints are x and y. Let Γ be a k-regular graph with vertices, and let Γ k be the k-tree, which is the universal cover of Γ. The adjacency operator on the space of functions on Γ k will be denoted by A, and is defined analogously to the adjacency operator on Γ, by expression 7). Lemma 9.1 Let x Γ k, λ R. Then there is a unique function S λ,x on Γ k such that a) S λ,x x) = 1 b) AS λ,x = λs λ,x c) S λ,x y) = S λ,x z) if distx, y) = distx, z). Proof. Let Cx, r) denote the set of vertices {y Γ k : distx, y) = r}. If r = 0, then the set Cx, r) contains just the vertex x. If r 1 then Cx, r) contains exactly kk 1) r 1 vertices. By condition c), the function S λ,x is constant on each set Cx, r). We will use the notation S λ r) to denote the common value of S λ,x on Cx, r). Condition a) fixes S λ 0). The value of S λ 1) is determined by implying that S λ 1) = λ/k. If r 1 and y Cx, r) then we have λ = λs λ 0) 82) = AS λ,x )x) 83) = y x S λ,x y) 84) = ks λ 1) 85) AS λ,x y) = S λ,x z 1 ) + k 1)S λ,x z 2 ) 86) where z 1 is some vertex in Cx, r 1) and z 2 is some vertex in Cx, r + 1). Thus 17

18 λs λ r) = S λ r 1) + k 1)S λ r + 1). 87) This determines S λ r+1) in terms of S λ r) and S λ r 1), and we conclude that S λ is uniquely determined for all integers r 0, and thus that S λ,x is determined on all of Γ k. We extract from this proof the following corollary: Corollary 9.2 S λ r) is an r th -degree polynomial in λ. Moreover, for r 1, kk 1) r 1 S λ r) is a monic r th -degree polynomial in λ. If r is even, then S λ r) is even, and if r is odd, then S λ r) is odd. Proof. In the preceding proof, we remarked that S λ 0) = 1 88) S λ 1) = 1 k λ 89) establishing our present claim for r = 0 and r = 1. For r 2, we refer to 87), and adjust the index to rewrite the equation this way: k 1)S λ r) = λs λ r 1) S λ r 2). 90) Multiplying through by kk 1) r 2, we obtain kk 1) r 1 S λ r) = λkk 1) r 2 S λ r 1) kk 1) r 2 S λ r 2). 91) By induction, kk 1) r 2 S λ r 1) is a monic, r 1) th -degree polynomial in λ with parity equal to that of r 1. Therefore the first term on the right in 91) is a monic, r th -degree polynomial in λ with the same parity as r. The second term on the right is a polynomial of degree r 2. Its parity given by r 2) is also the same as the parity of r. The sum of the two terms is thus monic of degree r, with parity equal to the parity of r, as claimed. We note that Brooks, in [2], solves the difference equation 87), and the resulting closed-form expression yields much information about the behavior 18

19 of the function S λ. We will not need such detailed information here, however, and will make do with the basic properties of S λ given in the preceding lemma and corollary. Let ϕ be a function on Γ such that Aϕ = λϕ. Let ϕ denote the lift of ϕ to the universal cover, Γ k. Let x Γ and pick some lift x of x. Let ϕ denote the spherical average of ϕ about x, that is ϕ x) = ϕ x) 92) ϕ 1 ỹ) = ϕỹ) for dist x, ỹ) = r 1. 93) kk 1) r 1 Then we have Lemma 9.3 ỹ C x,r) ϕ = ϕx)s λ, x. 94) Proof. Clearly the value of ϕ ỹ) depends only on the distance from x to ỹ, that is, ϕ is constant on each set C x, r). We claim that A ϕ = λ ϕ 95) where A is the adjacency operator on Γ k, as given by 7). To see this, consider A ϕ )ỹ), where dist x, ỹ) = r 1. Because Γ k is simply connected, we have A ϕ )ỹ) = ϕ z 1 ) + k 1) ϕ z 2 ) 96) where z 1 is any element of C x, r 1) and z 2 is any element of C x, r+1). We sum this equation, letting ỹ run through C x, r), which contains kk 1) r 1 vertices. ỹ C x,r) A ϕ )ỹ) = 97) kk 1) r 1 A ϕ )ỹ) = kk 1) r 1 ϕ z 1 ) + kk 1) r ϕ z 2 ) 98) 19

20 = k 1) ϕ z) + ϕ z) 99) = ỹ C x,r) z C x,r 1) z C x,r+1) A ϕ)ỹ) 100) = λ ϕỹ) 101) ỹ C x,r) = λkk 1) r 1 ϕ ỹ). 102) From 102) and the left side of 98), we conclude that for dist x, ỹ) 1. If dist x, ỹ) = 0, then ỹ = x, and we get A ϕ )ỹ) = λ ϕ ỹ) 103) A ϕ ) x) = and we have established our claim 95). If ϕ x) is not zero, then the function = ỹ C x,1) ỹ C x,1) ϕ ỹ) 104) ϕỹ) 105) = A ϕ) x) 106) = λ ϕ x) 107) = λ ϕ x) 108) fỹ) = 1 ϕ x) ϕ ỹ) 109) depends only on dist x, ỹ), and satisfies Af = λf and f x) = 1. Therefore, by Lemma 9.1, we have so f = S λ, x 110) 20

21 ϕ = ϕ x)s λ, x 111) = ϕx)s λ, x. 112) If ϕ x) = 0, then ϕ x) = 0. Using this fact and an induction argument based on 103), it is easy to show that ϕ is identically zero, and so the lemma holds in this case, as well. Consider the function k n on Γ k Γ k given by k n x, ỹ) = { 1 if dist x, ỹ) = n 0 otherwise 113) Let K n be the sum of the covering-map images of k n, that is to say, let K n : Γ Γ R by K n x, y) = g G k n x, gỹ) 114) where x is some lift of x, ỹ is some lift of y, and G is the group of covering transformations of Γ k over Γ. The sum on the right in 114) contains a 1 for each lift ỹ of y in C x, n). The unique path in Γ k from x to each ỹ projects down to a path of length n from x to y in Γ. Conversely, the lift of any path of length n from x to y in Γ, when the lift is started at x, will terminate at some lift of y in C x, n). This establishes that the function K n x, y) is equal to the number of paths of length n from x to y. We now have ready the machinery which will allow us to write a spectral expression, in the spirit of the Selberg pre-trace formula, for K n. We want to express K n in this form: K n x, y) = i,j a ij ϕ i x)ϕ j y) 115) for some coefficients a ij. Integrating both sides of 115) against the function ϕ k y), we get 21

22 K n x, y)ϕ k y) = y Γ i,j = i,j = i a ij ϕ i x) y Γ ϕ j y)ϕ k y) 116) a ij ϕ i x)δ j,k 117) a ik ϕ i x). 118) Turning our attention to the expression on the left side of 116), we let ϕ k be the lift of ϕ k to Γ k and use the definition of K n, 114), to write the expression as a summation over vertices in Γ k : y Γ K n x, y)ϕ k y) = y Γ k n x, gỹ) ϕ k ỹ) 119) where x is some lift of x and ỹ denotes some lift of each y Γ. Because ϕ k is G-invariant, we can express the right side of 119) as g G k n x, gỹ) ϕ k gỹ). 120) y Γ g G As g runs through G and y through Γ, the image gỹ hits every vertex in Γ k exactly once, so we get K n x, y)ϕ k y) = k n x, ỹ) ϕ k ỹ). 121) ỹ Γ k y Γ By the definition of k n, 113), this is equal to ỹ C x,n) ϕ k ỹ). 122) to We now assume n 1 and apply definition 93) to see that this is equal kk 1) n 1 ϕ k ỹ) 123) 22

23 where ỹ is any vertex in C x, n). By Lemma 9.3, so that 122) is equal to ϕ k = ϕ kx)s λk, x, 124) Substituting this back into 116), we get kk 1) n 1 ϕ k x)s λk n). 125) kk 1) n 1 ϕ k x)s λk n) = i a ik ϕ i x) 126) which implies that a ik = 0 if i k and We have shown: Lemma 9.4 For n 1, a ii = kk 1) n 1 S λi n). 127) K n x, y) = kk 1) n 1 i S λi n)ϕ i x)ϕ i y). 128) 10 Broken paths Let x and y be two vertices in Γ, and consider the expression K n x, z)k m z, y). 129) z Γ The term corresponding to a single z in the sum is equal to the number of paths of length n from x to z times the number of paths of length m from z to y. In other words, it is the number of walks from x to y which are made up of a path of length n ending at z followed by a path of length m from z to y. The entire expression, then, is equal to the number of walks from x to y 23

24 of length m + n which do not double back, except possibly at the n th vertex, counting from x. We can use this idea to derive a spectral diameter estimate for k-regular graphs which depends explicitly on the injectivity radius. We define the girth of a graph to be the length of its shortest closed path excluding closed paths of length zero). Let g be the girth of Γ and let r = g 1. Then the ball of radius r about any vertex in Γ is simply 2 connected, and therefore isomorphic to a ball of radius r in the k-tree. We will call r the injectivity radius of Γ. Lemma 10.1 Let Γ be a k-regular graph with vertices, injectivity radius r, and adjacency spectrum {λ 0, λ 1,..., λ 1 }. Let {ϕ 0, ϕ 1,..., ϕ 1 } be corresponding orthonormal eigenfunctions. Then for each vertex x Γ, 1 i=0 Proof. Consider the expression S 2 λ i r)ϕ 2 i x) = ) kk 1) r 1 K r x, y)k r y, x) 131) y Γ This is equal to the number of walks, beginning and ending at x, comprising a path of length r followed by a second path of length r. Because the ball of radius r about x is simply connected, the second path must coincide except for orientation) with the first. The vertex where the two paths join together is at a distance r from x, and each vertex at this distance from x determines exactly one walk of length 2r, beginning and ending at x. Thus expression 131) is equal to the number of vertices at distance r from x. The ball of radius r about x is isometric with a ball of radius r in the k-tree, so the number of such vertices is kk 1) r 1. We apply Lemma 9.4: kk 1) r 1 = y Γ K r x, y)k r y, x) = k 2 k 1) 2r 1) y Γ 1 i=0 1 S λi r)ϕ i x)ϕ i y) j=0 S λj r)ϕ j y)ϕ j x) 132) 24

25 so that kk 1) r 1 1 = k 2 k 1) 2r 1) S λi r)s λj r)ϕ i x)ϕ j x) ϕ i y)ϕ j y). i,j=0 y Γ The functions ϕ i are orthonormal, so 133) ϕ i y)ϕ j y) = δ i,j, 134) y Γ and the right side of 133) becomes a summation over a single index: 1 kk 1) r 1 = k 2 k 1) 2r 1) i=0 S 2 λ i r)ϕ 2 i x). 135) We divide each side by k 2 k 1) 2r 1) to complete the proof. 11 Injectivity radius and diameter We are now ready to state and prove our third diameter estimate, which depends explicitly on the injectivity radius of the graph. Theorem 3 Let Γ be a k-regular, connected graph with vertices and injectivity radius r 1. a) If Γ is not bipartite, then ) arccosh 1 kk 1) diamγ) r 1 ) + 2r ) arccosh k λγ) b) If Γ is bipartite, then diamγ) ) arccosh 1 2kk 1) r 1 ) + 2r ) arccosh k λγ) 25

26 Proof. Let x 0 and y 0 be vertices in Γ such that distx 0, y 0 ) = diamγ). Let n be a positive integer such that n + 2r < distx 0, y 0 ). Consider the expression x,y Γ K r x 0, x)w n x, y)k r y, y 0 ). 138) This is equal to the number of walks from x 0 to y 0 comprising a path of length r followed by a walk of length n and then a second path of length r. The total length of such a walk is n + 2r. But the distance from x 0 to y 0 is greater than n + 2r, so there are no such walks, and the expression must be equal to zero. Using the spectral expansions of K r and W n in Lemmas 9.4 and 5.1 respectively, we conclude that 0 = 1 x,y Γ i=0 1 S λi r)ϕ i x 0 )ϕ i x) j=0 1 λ n j ϕ j x)ϕ j y) l=0 S λl r)ϕ l y)ϕ l y 0 ) 139) where we have divided out the constant factors kk 1) r 1. Sorting out this multiple summation, we get 0 = 1 S λi r)ϕ i x 0 )λ n j S λl r)ϕ l y 0 ) ϕ i x)ϕ j x) ϕ j y)ϕ l y). i,j,l=0 x Γ y Γ 140) The functions ϕ i are orthonormal, so ϕ i x)ϕ j x) = δ i,j and x Γ ϕ j y)ϕ l y) = δ j,l y Γ 141) and the entire summation collapses to a single index: 0 = 1 i=0 S 2 λ i r)λ n i ϕ i x 0 )ϕ i y 0 ). 142) 26

27 This holds for each exponent n less than distx 0, y 0 ) 2r, so, summing over n, we obtain 0 = 1 i=0 S 2 λ i r)p n λ i )ϕ i x 0 )ϕ i y 0 ) 143) where p n is any polynomial of degree n < distx 0, y 0 ) 2r. We set )) λ p n λ) = cosh n arccosh. 144) λγ) Assume that Γ is not bipartite. We write 143) as 0 = p n λ 0 )Sλ 2 0 r)ϕ 0 x 0 )ϕ 0 y 0 ) 1 + p n λ i )Sλ 2 i r)ϕ i x 0 )ϕ i y 0 ). 145) All the factors in the first term on the right are known: λ 0 = k, S λ0 is constantly equal to 1, and ϕ 0 1. Plugging these values in, we determine that p n k) = 1 p n λ i )Sλ 2 i r)ϕ i x 0 )ϕ i y 0 ). 146) Because Γ is not bipartite, we have λ i λγ) for i = 1, 2,..., 1, so that p n λ i ) 1 for i = 1, 2,..., ) By the triangle inequality, then, we get p n k) 1 S λi r)ϕ i x 0 )S λi r)ϕ i y 0 ). 148) 27

28 We apply the Cauchy-Schwarz inequality: p n k) 1 S 2 λ i r)ϕ 2 i x 0 ) ) S 2 λ i r)ϕ 2 i y 0 ) ) ) Using Lemma 10.1 and the known constant) values for S λ0 find that each factor on the right is equal to and ϕ 0, we thus 1 kk 1) r 1 1 ) 1 2, 150) cosh n arccosh k λγ) p n k) )) n 1 151) kk 1) r ) kk 1) r 1 ) arccosh 1 kk 1) r 1 ). 153) arccosh k λγ) This holds for n = distx 0, y 0 ) 2r 1, and result a) follows. Assume now that Γ is bipartite. Then for each i, λ 1 i = λ i. The polynomial p n λ) is even if n is even and odd if n is odd, so p n λ 1 i ) = 1) n p n λ i ) 154) for each i. The functions S λ r), considered as r th -degree polynomials in λ, have the same parity property Corollary 9.2), so that S 2 λ 1 i r) = S 2 λ i r) 155) for each i. Using these observations and part c) of Lemma 6.1, we conclude that, if the parity of n is different from the parity of distx 0, y 0 ), 28

29 p n λ 1 i )S 2 λ 1 i r)ϕ 1 i x 0 )ϕ 1 i y 0 ) = p n λ i )S 2 λ i r)ϕ i x 0 )ϕ i y 0 ) 156) so that the right side of 143) is trivially zero, and the equation yields no information. If, however, the parity of n agrees with the parity of distx 0, y 0 ), then we have p n λ 1 )S 2 λ 1 r)ϕ 1 x 0 )ϕ 1 y 0 ) = p n λ 0 )S 2 λ 0 r)ϕ 0 x 0 )ϕ 0 y 0 ) = p nk) and equation 143) implies that 157) 2 p nk) = 2 p n λ i )Sλ 2 i r)ϕ i x 0 )ϕ i y 0 ). 158) Becuase λ i λγ) for i = 1, 2,..., 2, we have p n λ i ) 1 for i = 1, 2,..., ) Applying this and the triangle inequality to 158), we get 2 p nk) 2 S λi r)ϕ i x 0 )S λi r)ϕ i y 0 ). 160) As before, we apply the Cauchy-Schwarz inequality: 2 p nk) 2 S 2 λ i r)ϕ 2 i x 0 ) ) S 2 λ i r)ϕ 2 i y 0 ) ) ) 29

30 We use Lemma 10.1 to evaluate each summation on the right. The i = 0 and i = 1 terms are missing, so each factor is equal to and we have 1 kk 1) r 1 2 ) 1 2, 162) cosh n arccosh 2p n k) )) k λγ) n 2 163) kk 1) r ) 2kk 1) r 1 ) arccosh 1 2kk 1) r 1 ). 165) arccosh k λγ) This is true for each n less than distx 0, y 0 ) 2r, provided the parity of n is the same as that of distx 0, y 0 ). We set n = distx 0, y 0 ) 2r 2, and the proof is complete. 12 Using the injectivity radius is worth it We verify that Theorem 3 actually improves upon Theorem 2, at least in a very large number of cases. Specifically, we will show, in these cases, that ) arccosh 1 kk 1) r 1 ) + 2r < arccosh k λγ) arccosh 1) ). 166) arccosh k λγ) Clearly, this inequality will be more unequal, and therefore easier to establish, when λγ) is close to k, so that the denominators of the fractions are small. We will show that 166) holds provided λγ) 2 k 1. To show that this situation is quite general, we appeal to a result in the spectral geometry of manifolds. Cheng s theorem [3]) gives an upper bound on the first eigenvalue of the Laplacian on a compact, hyperbolic surface S. It says that if S is large, then the first eigenvalue of the Laplacian can t be 30

31 much greater than 1/4. There is an analogous theorem for k-regular graphs see [7], Proposition and [2] Theorem 4.4) giving a lower bound on λγ). This theorem states that λγ) 2 k 1 ɛγ) 167) where ɛ is some positive function which goes to zero as Γ gets large. Our proof of inequality 166) will begin with the hypothesis λγ) 2 k 1, but there will be a little slack in the computation that is, the result will really be proved for λγ) slightly less than 2 k 1 as well as λγ) 2 k 1. For simplicity, we will not try to determine exactly how much less, but, because of Cheng s theorem, any amount of slack allows us to infer the result for all sufficiently large graphs, in addition to all smaller graphs with λγ) 2 k 1. We will make use of the following lemma. Lemma 12.1 Let 1 r < s. Then arccoshs) arccoshr) > logs) logr). 168) Proof. arccoshs) arccoshr) = We now state 166) as s r dt s t2 1 > r dt t = logs) logr). 169) Lemma 12.2 For r 1, k 3, > kk 1) r 1, and λ 2 k 1, ) arccosh 1 kk 1) r 1 arccosh ) + 2r < k λ arccosh 1) arccosh ). 170) k λ Proof. From λ 2 k 1, we get k λ k 2 k ) 31

32 We observe that coshlog k 1) = both sides of 171): k 2 k 1 and take the arccosh of arccosh ) k λ log k ) It follows that 2r arccosh ) k λ r logk 1). 173) k From the hypotheses, k 1 and 1 are both greater than kk 1) r 1 1, so we can add their logarithms to the right side this is where the slack first comes in): ) k 2r arccosh λ ) k < r logk 1) + log k 1 = log 1) log + log kk 1) r 1 1 ) 1 ) kk 1) r 1. ow we apply Lemma 12.1 another source of slack), to get ) k λ 2r arccosh < arccosh 1) arccosh kk 1) r 1 1 ). 174) 175) Dividing through by arccosh k ) and rearranging the terms in the obvious λ way gives the result. References [1] Milton Abramowitz and Irene A. Stegun, editors. Handbook of Mathematical Functions. Dover, ew York,

33 [2] Robert Brooks. The spectral geometry of k-regular graphs. Journal d Analyse Mathématique, to appear). [3] S. Y. Cheng. Eigenvalue comparison theorems and its geometric applications. Mathematische Zeitschrift, 143: , [4] F. R. K. Chung. Diameters and eigenvalues. Journal of the American Mathematical Society, 22): , [5] F. R. K. Chung, V. Faber, and Thomas A. Manteuffel. An upper bound on the diameter of a graph from eigenvalues associated with its laplacian. SIAM Journal on Discrete Mathematics, to appear). [6] A. Lubotzky, R. Phillips, and P. Sarnak. Ramanujan graphs. Combinatorica, 83): , [7] Peter Sarnak. Some Applications of Modular Forms. Cambridge University Press,