Linear Algebra and its Applications 529 (2017) 355–373
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Linear Algebra and its Applications www.elsevier.com/locate/laa
On Jacobian group and complexity of the generalized Petersen graph GP (n, k) through Chebyshev polynomials Y.S. Kwon a,∗ , A.D. Mednykh b , I.A. Mednykh b a
Department of Mathematics, Yeungnam University, Republic of Korea Sobolev Institute of Mathematics, Novosibirsk State University, Siberian Federal University, Russian Federation b
a r t i c l e
i n f o
Article history: Received 7 December 2016 Accepted 24 April 2017 Available online 29 April 2017 Submitted by R. Brualdi MSC: 05C30 39A10 Keywords: Spanning tree Jacobian group Petersen graph Laplacian matrix Chebyshev polynomial
a b s t r a c t In the present paper we give a new method for calculating Jacobian group Jac(GP (n, k)) of the generalized Petersen graph GP (n, k). We show that the minimum number of generators of Jac(GP (n, k)) is at least two and at most 2k + 1. Both estimates are sharp. Also, we obtain a closed formula for the number of spanning trees of GP (n, k) in terms of Chebyshev polynomials and investigate some arithmetical properties of this number. © 2017 Elsevier Inc. All rights reserved.
1. Introduction The notion of the Jacobian group of a graph, which is also known as the Picard group, the critical group, and the dollar or sandpile group, was independently introduced by * Corresponding author.
E-mail address:
[email protected] (Y.S. Kwon).
http://dx.doi.org/10.1016/j.laa.2017.04.032 0024-3795/© 2017 Elsevier Inc. All rights reserved.
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many authors ([7,3,4,2,11,14]). This notion arises as a discrete version of the Jacobian in the classical theory of Riemann surfaces. It also admits a natural interpretation in various areas of physics, coding theory, and financial mathematics. The Jacobian group is an important algebraic invariant of a finite graph. In particular, its order coincides with the number of spanning trees of the graph, which is known for some simple graphs such as the wheel, fan, prism, ladder, and Möebius ladder [5], grids [18], lattices [19], Sierpinski gaskets [6,8], 3-prism and 3-anti-prism [21]. At the same time, the structure of the Jacobian is known only in particular cases [7,4,15,24,25,16] and [17]. We mention that the number of spanning trees for circulant graphs is expressed in terms of the Chebyshev polynomials; it was found in [26,27] and [23]. We show that similar results are also true for the generalized Petersen graph GP (n, k). The generalized Petersen graph GP (n, k) has vertex set and edge set given by V (P (n, k)) = {ui , vi | i = 1, 2, . . . , n}
E(P (n, k)) = {ui ui+1 , ui vi , vi vi+k | i = 1, 2, . . . , n},
where the subscripts are expressed as integers modulo n. It is important to note that the graph GP (n, k) is isomorphic to GP (n, n − k), and if n and k are relative prime, then GP (n, k) is also isomorphic to GP (n, r) with k r ≡ 1 mod n [20]. In what follows we will deal with 3-valent graphs only. So, we assume n ≥ 3 and 1 ≤ k < n/2. The classical Petersen graph is GP (5, 2). The number of spanning trees of the Petersen graph is calculated in [12] and the spectrum of the generalized Petersen graphs is obtained in [13]. Even though the number of spanning trees of a regular graph can be computed by its eigenvalues, it is not easy to obtain a closed formula for the number of spanning trees for GP (n, k) by the result of [13]. In this paper, we find a closed formula for the number of spanning trees for GP (n, k) through Chebyshev polynomials. Also, we suggest a new method to calculate the Jacobian group of GP (n, k). 2. Basic definitions and preliminary facts Consider a connected finite graph G, allowed to have multiple edges but without loops. We denote the vertex and edge set of G by V (G) and E(G), respectively. Given u, v ∈ V (G), we set auv to be equal to the number of edges between vertices u and v. The matrix A = A(G) = {auv }u,v∈V (G) is called the adjacency matrix of the graph G. ! The degree d(v) of a vertex v ∈ V (G) is defined by d(v) = u∈V (G) auv . Let D = D(G) be the diagonal matrix indexed by the elements of V (G) with dvv = d(v). The matrix L = L(G) = D(G) − A(G) is called the Laplacian matrix, or simply Laplacian, of the graph G. Recall [15] the following useful relation between the structure of the Laplacian matrix and the Jacobian of a graph G. Consider the Laplacian L(G) as a homomorphism Z|V | → Z|V | , where |V | = |V (G)| is the number of vertices in G. The cokernel coker (L(G)) = Z|V | /im (L(G)) is an Abelian group. Let
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coker (L(G)) ∼ = Zd1 ⊕ Zd2 ⊕ · · · ⊕ Zd|V | " be its Smith normal form satisfying the conditions di "di+1 , (1 ≤ i ≤ |V |). If the graph is connected, then the groups Zd1 , Zd2 , . . . , Zd|V |−1 are finite, and Zd|V | = Z. In this paper, we define Jacobian group Jac(G) as the torsion subgroup of coker (L(G)). In other words, Jac(G) ∼ = Zd1 ⊕ Zd2 ⊕ · · · ⊕ Zd|V |−1 . Let M be an integer n × n matrix, then we can interpret M as a homomorphism from Zn to Zn . In this interpretation M has a kernel kerM , an image im M , and a cokernel cokerM = Zn /imM . We emphasize that cokerM of the matrix M is completely determined by its Smith normal form. In what follows, by In we denote the identity matrix of order n. We call an n × n matrix circulant, and denote it by circ(a0 , a1 , . . . , an−1 ) if it is of the form ⎛
a0
⎜ ⎜ an−1 circ(a0 , a1 , . . . , an−1 ) = ⎜ ⎜ ⎝ a1
a1 a0 .. . a2
⎞ . . . an−1 ⎟ . . . an−2 ⎟ .. ⎟ .. ⎟. . . ⎠ ... a0
a2 a1 a3
Recall [9] that the eigenvalues of matrix C = circ(a0 , a1 , . . . , an−1 ) are given by the following simple formulas λj = p(εjn ), j = 0, 1, . . . , n − 1 where p(x) = a0 + a1 x + . . . + an−1 xn−1 and εn is an order n primitive root of the unity. Moreover, the circulant matrix C = p(T ), where T = circ(0, 1, 0, . . . , 0) is the matrix representation of the shift operator T : (x0 , x1 , . . . , xn−2 , xn−1 ) → (x1 , x2 , . . . , xn−1 , x0 ). For any i = 0, . . . , n − 1, let vi = (n−1)i t ) be a column vector of length n. We note that all n × n circulant (1, εin , ε2i n , . . . , εn matrices share the same set of linearly independent eigenvectors v0 , v1 , . . . , vn−1 . By ([13], lemma 2.1), the 2n × 2n adjacency matrix of the generalized Petersen graph GP (n, k) has the following block structure A(GP (n, k)) =
)
Cnk In
In Cn1
*
,
where Cnk is the n × n circulant matrix of the form Cnk = circ(0, . . . , 0, 1, 0, . . . , 0, 1, 0, . . . , 0). + ,- . + ,- . + ,- . k
n−2k−1
k−1
Denote by L = L(GP (n, k)) the Laplacian of GP (n, k). Since the graph GP (n, k) is three-valent, we have
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L = 3I2n − A(GP (n, k)) =
)
3In − Cnk −In
−In 3In − Cn1
*
.
3. Cokernels of some linear operators Let A = ⟨αj , j ∈ Z⟩ be a free Abelian group freely generated by elements αj , j ∈ Z. ! Each element of A is a finite linear combination cj αj with integer coefficients cj . !j ! Define the shift operator T : A → A by T ( j cj αj ) = T is enj cj αj+1 . Then !s t+k domorphism of A. Note that T (αj ) = αj+1 for any j ∈ Z. Let P (z) = k=0 ck z be a Laurent polynomial with integer coefficients c0 , c1 , . . . , cs , where t is some integer and s is a positive integer. Now the mapping A : A → A defined by A = P (T ) is also an endomorphism of A. Since A is a finite linear combination of powers of T , we have s s ! ! A(αj ) = ck αj+t+k for any j ∈ Z. For our convenience, let βj = ck αj+t+k . Now k=0
k=0
im A is a subgroup of A generated by βj , j ∈ Z. Hence, coker A = A/im A is an abstract s ! Abelian group ⟨xi , i ∈ Z | ck xj+t+k = 0, j ∈ Z⟩, namely, coker A is generated by k=0
xi , i ∈ Z with the set of defining relations
s !
k=0
ck xj+t+k = 0, j ∈ Z. Note that for any
j ∈ Z, xj can be considered as the image of αj under the canonical homomorphism A → A/im A. Since T and A = P (T ) commute, the subgroup im A is invariant under the action of T . Hence, the actions of T and A on the factor group A/im A induced by s ! T : xj → xj+1 and A : xj → ck xj+t+k are well defined, respectively. This allows to k=0
present the group A/im A as follows ⟨xi , i ∈ Z| P (T )xj = 0, j ∈ Z⟩. In a similar way, given a set P1 (z), P2 (z), . . . , Ps (z) of Laurent polynomials with integer coefficients, one can define the group ⟨xi , i ∈ Z| P1 (T )xj = 0, P2 (T )xj = 0, . . . , Ps (T )xj = 0, j ∈ Z⟩. We will use the following lemma.
Lemma 3.1. Let T : A → A be the shift operator. Consider endomorphisms A and B of the group A given by the formulas A = P (T ), B = Q(T ), where P (z) and Q(z) are Laurent polynomials with integer coefficients. Then B : A → A induces an endomorphism B|coker A of the group coker A = A/im A defined by B|coker A (α + ImA) = B(α) + ImA, α ∈ A. Furthermore ⟨xi , i ∈ Z| A(T )xj = 0, B(T )xj = 0, j ∈ Z⟩ ∼ = coker A/im(B|coker A ) ∼ = coker (B|coker A ). Proof. The images im A and im B are subgroups in A. Denote by ⟨im A, im B⟩ the subgroup generated by elements of im A and im B. Since P (z) and Q(z) are Laurent polynomials, the operators A = P (T ) and B = Q(T ) do commute. Hence, subgroup im A is invariant under endomorphism B. Indeed for any y = Ax ∈ im A, we have By = B(Ax) = A(Bx) ∈ im A. This means that B : A → A induces an endomorphism of
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the group coker A = A/im A. We denote this endomorphism by B|coker A . We note that the Abelian group ⟨xi , i ∈ Z| A(T )xj = 0, B(T )xj = 0, j ∈ Z⟩ is naturally isomorphic to A/⟨im A, im B⟩. So we have A/⟨im A, im B⟩ ∼ = (A/im A)/im (B|coker A ) ∼ = coker A/im(B|coker A ) ∼ = coker (B|coker A ). The lemma is proved. ✷ 4. Jacobian group for the generalized Petersen graph GP (n, k) Consider two integers k, l with l > 0. Let P (z) = a0 z k +a1 z k+1 +. . .+al−1 z k+l−1 +z k+l be a monic Laurent polynomial with integer coefficients a0 , a1 , . . . , al−1 . We introduce the companion matrix A for the polynomial P (z) as follows: A=
)
0
Il−1
−a0 , −a1 , . . . , −al−1
*
,
where Il−1 is the identity matrix of order l − 1. The main result of this section is the following theorem. Theorem 4.1. Let L = L(GP (n, k)) be the Laplacian of the generalized Petersen graph GP (n, k). Then coker L ∼ = coker(An − I), where A is 2(k + 1) × 2(k + 1) companion matrix for the Laurent polynomial (3 − z −1 − z)(3 − z −k − z k ) − 1. Proof. Let L be the Laplacian matrix of the graph GP (n, k). Then, as it was mentioned before, L is a 2n × 2n matrix of the form L=
)
3In − Cnk −In
−In 3In − Cn1
where Cnk = circ(0, . . . , 0, 1, 0, . . . , 0, 1, 0, . . . , 0 ). + ,- . + ,- . + ,- . k times
n−2k+1
k−1 times
*
,
Consider L as a Z-linear operator L : Z2n → Z2n . In this case, coker(L) = Z2n /im L is an abstract Abelian group generated by elements x1 , x2 , . . . , xn , y1 , y2 , . . . , yn satisfying the linear system of equations 3xj − xj−k − xj+k − yj = 0 and 3yj − yj−1 − yj+1 − xj = 0 for any j = 1, . . . , n, where subscripts are considered modulo n. Hence we have
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∼ ⟨xi , yi , i ∈ Z | 3xj − xj−k − xj+k − yj = 0, coker(L) =
3yj − yj−1 − yj+1 − xj = 0, xj+n = xj , yj+n = yj , j ∈ Z⟩
= ⟨xi , yi , i ∈ Z | (3 − T −k − T k )xj = yj , (3 − T −1 − T )yj = xj , T n xj = xj , T n yj = yj , j ∈ Z⟩
= ⟨xi , i ∈ Z | (3 − T −1 − T )(3 − T −k − T k )xj = xj , T n xj = xj , j ∈ Z ⟩
= ⟨xi , i ∈ Z | ((3 − T −1 − T )(3 − T −k − T k ) − 1)xj = 0, (T n − 1)xj = 0, j ∈ Z⟩.
Let P (z) = (3 −z −1 −z)(3 −z −k −z k ) −1. Now P (T ) = (3 −T −1 −T )(3 −T −k −T k ) −1 and we have coker(L) ∼ = ⟨xi , i ∈ Z | P (T )xj = 0, (T n − 1)xj = 0, j ∈ Z ⟩. To finish the proof, we apply Lemma 3.1 to the operators A = P (T ) and B = Q(T ) = T n − 1. First of all, we show that coker A ∼ = Zs , where Zs is the free Abelian group of rank s = 2k + 2. Then we describe the action of the endomorphism B|coker A on coker A. Since the Laurent polynomial P (z) is bimonic, it can be represented in the form P (z) = z −k−1 + a1 z −k + . . . + a2k+1 z k + z k+1 , where a1 , a2 , . . . , a2k+1 are integers. Hence, its companion matrix has the form A=
)
0
I2k+1
−1, −a1 , . . . , −a2k+1
*
,
where I2k+1 is the identity matrix of order 2k + 1. It is easy to see that det A = 1 and its inverse A−1 is also integer matrix. Note that for any j ∈ Z the relations P (T )xj = 0 can be rewritten as xj+s = −xj −a1 xj+1 −· · ·−as−1 xj+s−1 . Let xj = (xj+1 , xj+2 , . . . , xj+s )t be s-tuple of generators xj+1 , xj+2 , . . . , xj+s . Then the relation P (T )xj = 0 is equivalent to xj = A xj−1 . Hence, we have x1 = A x0 and x−1 = A−1 x0 , where x0 = (x1 , x2 , . . . , xs )t . So, xj = Aj x0 for any j ∈ Z. Conversely, the latter implies xj = A xj−1 and, as a consequence, P (T )xj = 0 for all j ∈ Z. Since coker A = A/im (A), it has the following presentation ⟨xi , i ∈ Z|P (T )xj = 0, j ∈ Z⟩. Now we have coker A = ⟨xi , i ∈ Z|P (T )xj = 0, j ∈ Z⟩
= ⟨xi , i ∈ Z| xj + a1 xj+1 + . . . + as−1 xj+s−1 + xj+s = 0, j ∈ Z⟩
= ⟨xi , i ∈ Z| (xj+1 , xj+2 , . . . , xj+s )t = A(xj , xj+1 , . . . , xj+s−1 )t , j ∈ Z⟩ = ⟨xi , i ∈ Z| (xj+1 , xj+2 , . . . , xj+s )t = Aj (x1 , x2 , . . . , xs )t , j ∈ Z⟩
= ⟨x1 , x2 , . . . , xs | ∅⟩ ∼ = Zs .
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Since the operators A = P (T ) and T commute, the action T |coker A on the coker A induced by the mapping xj → xj+1 , j ∈ Z is well defined. Now we describe the action of T |coker A on the set of generators x1 , x2 , . . . , xs . For any i = 1, . . . , s − 1, we have T |coker A (xi ) = xi+1 and T |coker A (xs ) = xs+1 = −x1 − a1 x2 − . . . − as−2 xs−1 − as−1 xs . Hence, the action of T |coker A on the coker A is given by the matrix A. Considering A as an endomorphism of the coker A, we can write T |coker A = A. Finally, B|coker A = Q(T |coker A ) = Q(A). Applying Lemma 3.1 we finish the proof of the theorem. ✷ Remark. It was noted by a referee that the) author’s method can be applied to any * f (C) −In 2n × 2n block matrix L of the form L = , where f (t) and g(t) are −In g(C) bimonic Laurent polynomials with integer coefficients and C = circ(0, 1, 0, . . . , 0) is the matrix representation of the shift operator. In this case, coker L ∼ = coker(An − I), where A is the companion matrix for the Laurent polynomial f (t)g(t) − 1. So the number of generators of the Smith Normal form of L is bounded by deg(f ) + deg(g). Below we have two immediate corollaries from Theorem 4.1. Corollary 4.2. The Jacobian group Jac(GP (n, k)) of the generalized Petersen graph GP (n, k) is isomorphic to the torsion subgroup of coker(An − I), where A is the companion matrix for the Laurent polynomial (3 − z −1 − z)(3 − z −k − z k ) − 1. The Corollary 4.2 gives a simple way to calculate Jacobian group Jac(GP (n, k)) for small values of k and sufficiently large numbers n. The numerical results are presented in Tables 1, 2 and 3. It is easy to see that the Jacobian group Jac(GP (n, k)) of the generalized Peterson graph on 2n vertices can be generated by n − 1 elements, since the Laplacian matrix contains an obvious n × n submatrix of determinant (−1)n . Theorem 4.1 implies that Jac(GP (n, k)) can be generated by smaller number of elements as the following corollary. Corollary 4.3. The Jacobian group Jac(GP (n, k)) of the generalized Peterson graph on 2n vertices can be generated by 2k + 1 elements. Proof. By Theorem 4.1, the Abelian group coker(L(G)) ∼ = Jac(G) ⊕ Z is generated by 2k + 2 elements. Since one of these elements generates the infinite cyclic group, Jac(G) can be generated by 2k + 1 elements. ✷ The upper bound in Corollary 4.3 is sharp. It is attained for graph GP (20, 3) with Jacobian Z2 ⊕ Z28 ⊕ Z16 ⊕ Z80 ⊕ Z11120 ⊕ Z33360 and for graph GP (30, 4) with Jacobian Z2 ⊕ Z34 ⊕ Z320 ⊕ Z16904300 ⊕ Z50712900 .
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5. Counting the number of spanning trees for the generalized Petersen graph GP (n, k) Theorem 5.1. The number of spanning trees of the generalized Petersen graph GP (n, k) is given by the formula τ (GP (n, k)) = (−1)(n−1)(k−1) n
k / Tn (ws ) − 1 , ws − 1 s=1
where ws , s = 1, 2, . . . , k are roots of the order k algebraic equation 2Tk (w) − 3 = 0, and Tk (w) is the Chebyshev polynomial of the first kind.
Tk (w)−1 w−1
−
Proof. By the celebrated Kirchhoff theorem, the number of spanning trees τk (n) is equal to the product of nonzero eigenvalues of the Laplacian of a graph GP (n, k) divided by the number of its vertices 2n. To investigate the spectrum of Laplacian matrix, we note that matrix Cnk = T −k + T k , where T = circ(0, 1, . . . , 0) is the n × n shift operator. Hence, L=
)
3In − T −k − T k −In
−In 3In − T −1 − T
*
. 2πi
The eigenvalues of circulant matrix T are εjn , j = 0, 1, . . . , n − 1, where εn = e n . Since all of them are distinct, the matrix T is conjugate to the diagonal matrix T = diag(1, εn , . . . , εn−1 ) with diagonal entries 1, εn , . . . , εn−1 . To find spectrum of L, without n n loss of generality, one can assume that T = T. Then the n × n blocks of L are diagonal matrices. This essentially simplifies the problem of finding eigenvalues of L. Indeed, let λ be an eigenvalue of L and (x, y) = (x1 , . . . , xn , y1 , . . . , yn ) be the respective eigenvector. Then we have the following system of equations 0
(3In − T −k − T k )x − y −x + (3In − T −1 − T )y
= λx . = λy
From here we conclude that y = (3In − T −k − T k )x − λx = (3 − λ − T −k − T k )x. Substituting y into the second equation, we have ((3 − λ − T −1 − T )(3 − λ − T −k − T k ) − 1)x = 0. Recall the matrices under consideration are diagonal and the (j + 1, j + 1)-th entry of j −jk T is equal to εjn . Therefore, we have ((3 − λ − ε−j − εjk n − εn )(3 − λ − εn n ) − 1)xj+1 = 0 jk and yj+1 = (3 − λ − ε−jk − ε )x . j+1 n n So, for any j = 0, . . . , n − 1 the matrix L has two eigenvalues, say λ1,j and λ2,j , j −jk satisfying the quadratic equation (3 − λ − ε−j − εjk n − εn )(3 − λ − εn n ) − 1 = 0. The corresponding eigenvectors are (x, y), where x = ej+1 = (0, . . . , +,-. 1 , . . . , 0) and (j+1)−th
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y = (3 −λ −T −k −T k )ej+1 . In particular, if j = 0 for λ1,0 , λ2,0 we have (1 −λ)(1 −λ) −1 = λ(λ − 2) = 0. That is, λ1,0 = 0 and λ2,0 = 2. Since λ1,j and λ2,j are roots of the same quadratic equation, we obtain λ1,j λ2,j = P (εjn ), where P (z) = (3 − z −1 − z)(3 − z −k − z k ) − 1. Now we have n−1 n−1 n−1 / 1 1 / 1 / λ2,0 τk (n) = λ1,j λ2,j = λ1,j λ2,j = P (εjn ). 2n n n j=1 j=1 j=1
To continue the calculation of τk (n) we need the following two lemmas. Lemma 5.2. The following identity holds (3 − z −1 − z)(3 − z −k − z k ) − 1 = 2(w − 1)(2Tk (w) −
Tk (w) − 1 − 3), w−1
where Tk (w) is the Chebyshev polynomial of the first kind and w = 12 (z + z −1 ). Proof. Let us substitute z = eiϕ . It is easy to see that w = 12 (z + z −1 ) = cos ϕ, so we have Tk (w) = cos(k arccos w) = cos(kϕ). Then the statement of the lemma is equivalent to the following elementary identity (3 − 2 cos ϕ)(3 − 2 cos(kϕ)) − 1 = 2(cos ϕ − 1)(2 cos(kϕ) −
cos(kϕ) − 1 − 3) cos ϕ − 1
✷
By Lemma 5.2, P (z) = 2(w − 1)hk (w), where w = 12 (z + z −1 ) and hk (w) = 2Tk (w) − 2
(Tk (w) − 1)/(w − 1) − 3 is the polynomial of degree k. Note that 2(w − 1) = (z−1) . z Since hk (1) = 2Tk (1) − Tk′ (1) − 3 = −1 − k2 ̸= 0, the Laurent polynomial P (z) has the root z = 1 with multiplicity two. Hence, the roots of P (z) are 1, 1, z1 , 1/z1 , . . . , zk , 1/zk , where for all s = 1, . . . , k, zs ̸= 1 and ws = 12 (zs + zs−1 ) is a root of equation hk (w) = 0. k 1 1 We set H(z) = (z − zs )(z − zs−1 ). Then P (z) = zk+1 (z − 1)2 H(z). s=1
Lemma 5.3. Let H(z) =
k 1
s=1
(z − zs )(z − zs−1 ) and H(1) ̸= 0. Then n−1 / j=1
H(εjn ) =
k / Tn (ws ) − 1 , ws − 1 s=1
where ws = 12 (zs + zs−1 ), s = 1, . . . , k and Tn (x) is the Chebyshev polynomial of the first kind.
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Proof. It is easy to check that
n−1 1 j=1
(z − εjn ) =
z n −1 z−1
if z ̸= 1. Also we note that 12 (z n +
z −n ) = Tn ( 12 (z + z −1 )). By the substitution z = ei ϕ the latter follows from the evident identity cos(nϕ) = Tn (cos ϕ). Then we have n−1 /
H(εjn ) =
n−1 k //
(εjn − zs )(εjn − zs−1 )
k n−1 / /
(zs − εjn )(zs−1 − εjn )
j=1 s=1
j=1
=
s=1 j=1
=
Note that
3 / k 2 n k / Tn (ws ) − 1 zs − 1 zs−n − 1 = · −1 . zs − 1 zs − 1 ws − 1 s=1 s=1
n−1 1 j=1
(1 − εjn ) = lim
n−1 1
z→1 j=1
(z − εjn ) = lim
z n −1 z→1 z−1
= n and
j=1
a result, taking into account Lemma 5.3, we obtain τk (n) =
n−1 1
✷ εjn = (−1)n−1 . As
n−1 n−1 n−1 1 / 1 / (εjn − 1)2 (−1)(n−1)(k+1) n2 / j P (εjn ) = H(ε ) = H(εjn ) n n j=1 n j=1 (εjn )k+1 n j=1
= (−1)(n−1)(k−1) n
k / Tn (ws ) − 1 . ws − 1 s=1
✷
4 "1 "2 " k s " Corollary 5.4. τ (GP (n, k)) = n " s=1 Un−1 ( 1+w 2 )" , where ws , s = 1, 2, . . . , k are the same as in Theorem 5.1 and Un−1 (w) is the Chebyshev polynomial of the second kind. Proof. Follows from the identity
Tn (w)−1 w−1
4 2 = Un−1 ( 1+w 2 ).
✷
From now on, we aim to show that τk (n) = ck n ak (n)2 for some constant ck and integer sequence ak (n), n ∈ N. The proof of the following theorem is essentially based on the ideas by D. Lorenzini coming from his paper [15]. Theorem 5.5. Let τk (n) be the number of spanning trees for the generalized Petersen graph GP (n, k). Now there exists an integer sequence ak (n), n ∈ N such that τk (n) =
0
n ak (n)2 6n ak (n)2
if n is odd or k is even if n is even and k is odd.
Proof. Recall that for any j = 0, . . . , n − 1, the Laplacian matrix L of G(n, k) has two eigenvalues, say λ1,j and λ2,j , which are roots of the quadratic equation Qj (λ) = j −jk jk −j j −jk jk (3 −λ −ε−j n −εn )(3 −λ −εn −εn ) −1 = 0. So, λ1,j λ2,j = (3 −εn −εn )(3 −εn −εn ) −1.
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Note that λ1,0 and λ2,0 are roots of the equation (1 − λ)2 − 1 = −2λ + λ2 = 0, and hence one can take λ1,0 = 0 and λ2,0 = 2. Furthermore {λ1,j , λ2,j | j = 0, . . . , n − 1} is the set of all eigenvalues of L. The Kirchhoff theorem states the following 2n τk (n) = λ2,0
n−1 /
λ1,j λ2,j = 2
j=1
n−1 /
λ1,j λ2,j .
j=1
Since Qj (λ) = Qn−j (λ) for any j = 1, . . . , ⌊ n2 ⌋, we have {λ1,j , λ2,j } = {λ1,n−j , λ2,n−j } and so
n τk (n) =
n−1 /
λ1,j λ2,j
j=1
⎧ ⎛ n−1 ⎞2 ⎪ 2 / ⎪ ⎪ ⎪⎝ λ1,j λ2,j ⎠ ⎪ ⎪ ⎨ j=1 = ⎛n ⎞2 −1 ⎪ 2 ⎪ / ⎪ ⎪ ⎪ n n ⎝ λ1,j λ2,j ⎠ ⎪ ⎩ λ1, 2 λ2, 2
if n is odd if n is even.
j=1
The graph GP (n, k) admits a cyclic group of automorphisms isomorphic to Zn which acts freely on the set of spanning trees. Therefore, the value τk (n) is a multiple of n. So τk (n) is an integer and n ⎧⎛ ⎞2 n−1 ⎪ ⎪ 1 2 λ1,j λ2,j ⎪ ⎪ ⎝ j=1 ⎠ ⎪ ⎪ ⎨ n τk (n) = ) 1 n −1 *2 ⎪ n 2 ⎪ ⎪ λ1,j λ2,j j=1 ⎪ ⎪ ⎪ λ1, n2 λ2, n2 ⎩ n
if n is odd if n is even.
Since the algebraic number λi,j comes with all its Galois conjugate, the product 1⌊ n−1 2 ⌋ λ1,j λ2,j is an integer. j=1 1 n−1 2
Let n be an odd number and let ak (n) = j=1 n . Since τkn(n) is an integer, ak (n) is also an integer and τk (n) = n ak (n)2 . Now suppose that n = 2m is even. Since the eigenvalues λ1,m , λ2,m of L are roots of the equation (5 − λ)(3 − λ − 2(−1)k ) − 1 = 0, we have λ1,m λ2,m = 5(3 − 2(−1)k ) − 1. So λ1,m λ2,m is 4 if k is even; and 24 if k is odd. Let k be even. Now we get τk (n) = n 2
1m−1
λ1,j λ2,j
λ1,j λ2,j
) 1m−1 *2 2 j=1 λ1,j λ2,j
j=1 Let ak (n) = . Since n integer and τk (n) = nak (n)2 .
n
τ (n) n
.
is integer, so ak (n) =
2
1m−1 j=1
λ1,j λ2,j n
is also
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Finally, let k be an odd number. Now τk (n) =6 n 2
1m−1
) 1m−1 *2 2 j=1 λ1,j λ2,j n
.
λ1,j λ2,j
j=1 Set ak (n) = . This is a rational number, so ak (n) = pq , where p and q are n coprime integers. Since 6p2 /q 2 is an integer, 6p2 is a multiple of q 2 . As p is coprime with q, 6 is multiple of q 2 . But 6 is square free number. Hence q is 1. So ak (n) is an integer and τk (n) = 6nak (n)2 . ✷
Remark. For all examples in the Tables 1, 2 and 3, if n is odd or k is even, then Jac(GP (n, k)) ∼ = K ⊕H ⊕H, where the orders of K and H are n and ak (n), respectively. This implies that Jac(GP (n, k)) is killed by nak (n). Furthermore when n is even and k is odd, Jac(GP (n, k)) is killed by 6nak (n) for all examples in the Tables 1, 2 and 3. We suggest the following question: Are these true in general? Note that the product of distinct non-zero eigenvalues of Laplacian matrix L is less 1m 1m−1 than or equal to 2 j=1 λ1,j λ2,j if n = 2m + 1 is odd; 8 j=1 λ1,j λ2,j if n = 2m is 1m−1 even and k is even; and 48 j=1 λ1,j λ2,j if n = 2m is even and k is odd. So for any cases, the product of distinct non-zero eigenvalues of Laplacian matrix L is at most 9 1n−1 96nτk (n) by the fact that 2n τk (n) = 2 j=1 λ1,j λ2,j . From now on, we aim to show that Jac(GP (n, k)) is a non-cyclic group. Proposition 5.6 ([15], Prop. 2.6). For a connected graph G, let λ1 , λ2 , . . . , λt denote the distinct non-zero eigenvalues of the Laplacian matrix of G. Then Jac(G) is killed by 1t i=1 λi . By Proposition 5.6, we have the following corollary.
Corollary 5.7. For a connected graph G, let λ1 , λ2 , . . . , λt denote the distinct non-zero 1t eigenvalues of the Laplacian matrix of G. If the order of Jac(G) is greater than i=1 λi , then Jac(G) is non-cyclic. Using Corollary 5.7, we have the following result. Theorem 5.8. For a generalized Petersen graph GP (n, k), the group Jac(GP (n, k)) is non-cyclic. Proof. For our convenience, let τk (n) and θ(n) be the number of spanning trees of GP (n, k) and the product of distinct non-zero eigenvalues of Laplacian matrix L of GP (n, k), respectively. For n ≤ 9, we checked that Jac(GP (n, k)) is non-cyclic and so we assume that n ≥ 10 in this proof. For our convenience, let d = (n, k) be the greatest common divisor of n and k.
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A generalized Petersen graph GP (n, k) is composed of outer rim, inner rim and n spokes. Outer rim is a cycle of length n and inner rim is composed of d cycles of length nd . : ;d So the number of maximal forests of outer rim and inner rim are n and nd , respectively. For a given maximal forests F1 of outer rim and F2 of inner rim, to make spanning tree of GP (n, k) containing both F1 and F2 , it suffices to choose d spokes such that any connected component of F2 is incident to exactly one spoke among d chosen spokes. : ;d Hence the number of spanning tree of GP (n, k) containing both F1 and F2 is nd . Therefore the number of spanning trees of GP (n, k) whose outer rim part and inner : ;2d rim part are maximal forests of outer rim and inner rim, respectively, is n nd . Since for any positive integers n1 , n2 with n1 , n2 ≥ 2, n1 n2 ≥ n1 + n2 , by induction we have : n ;d : ;2d ≥ n. Hence n nd ≥ n3 . This implies that the number of spanning d 9 trees of 3 GP (n, k), namely order of Jac(GP (n, k)), is at least n . Since θ(n) ≤ 96nτ (n) √ the9 √ and 96n < n3 ≤ τ (n) for any n ≥ 10, we have θ(n) < τ (n). By Corollary 5.7, Jac(GP (n, k)) is non-cyclic. ✷ Combining the obtained result with Corollary 4.3 we get the following important observation. Theorem 5.9. The minimum number of generators of the group Jac(GP (n, k)) is at least two and at most 2k + 1. Both estimates are sharp. 6. Applications and Examples 6.1. Prism graph GP (n, 1) As the first consequence of Theorem 5.1 we have the following formula for number of spanning trees of the n-prism graph GP (n, 1): τ1 (n) = n(Tn (2) − 1). This formula is well known and was independently obtained by many authors. For example, by J. Sedlácěk, J.W. Moon, N. Biggs and others ([5]). The structure of Jac(GP (n, 1)) was independently described in the papers [1,22] and [10]. 6.2. Graph GP (n, 2) Theorem 6.1. The number τ2 (n) of the spanning trees for the generalized Petersen graph 1 GP (n, 2) is equal to (−1)n 20 n(α2 − 29β 2 ), where the integers α and β are given by the √
√
equality: Tn ( 1+4 29 ) − 1 = α+β4 29 . Moreover τ2 (n) = na(n)2 , where the integer sequence a(n) satisfies the following recursive relation
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a(n + 4) = a(n + 3) + 3a(n + 2) − a(n + 1) − a(n), a(0) = 0, a(1) = 1, a(2) = 1, a(3) = 5.
Note a(n) is A192422 sequence in the On-Line Encyclopedia of Integer Sequences. Proof. Note that 2T2 (w) − equation 2T2 (w) − we obtain
T2 (w)−1 w−1
T2 (w)−1 w−1
− 3 = 4w2 − 2w − 7, and hence two roots of the
− 3 = 0 are w1 =
√ 1+ 29 4
and w2 =
√ 1− 29 . 4
By Theorem 5.1,
2 / Tn (ws ) − 1 τ2 (n) = (−1) n ws − 1 s=1 √ √ (−1)n−1 n(α + β 29)(α − β 29) (−1)n n(α2 − 29β 2 ) = = . 16(w1 − 1)(w2 − 1) 20 n−1
To prove the second statement, by Corollary 4 4 5.4, we have a(n) = |p(n)|, where p(n) =
1 2 Un−1 (θ1 )Un−1 (θ2 ), θ1 = 1+w and θ2 = 1+w 2 2 . Recall that Chebyshev polynomial Un (θ) satisfies the recursive relation Un+1 (θ) − 2θUn (θ) + Un−1 (θ) = 0 with initial data U0 (θ) = 1 and U1 (θ) = 2θ. To find the recursive relation for the sequence p(n), we will use the following lemma.
Lemma 6.2. Let P (z) and Q(z) be polynomials without multiple roots and let T u(n) = u(n+1) be the shift operator. Suppose the sequences u(n) and v(n) satisfy the recursive relations P (T )u(n) = 0 and Q(T )v(n) = 0 respectively. Then the sequence p(n) = u(n)v(n) satisfies the recursive relations R(T )p(n) = 0, where R(z) is the resultant of polynomials P (ξ) and ξ deg Q Q( zξ ) with respect to ξ and deg Q is the degree of Q(z). Proof. Let λ1 , λ2 , . . . , λs and µ1 , µ2 , . . . , µt be distinct roots of polynomials P (z) and Q(z) respectively. Then each solution u(n) of the recursive equation P (T )u(n) = 0 is a linear combination of the functions λn1 , λn2 , . . . , λns , while solution v(n) of the recursive equation Q(T )v(n) = 0 is a linear combination of µn1 , µn2 , . . . , µnt . Hence, their product p(n) is a linear combination of the functions λnj µnk , j = 1, 2, . . . , s, k = 1, 2, . . . , t. By definition of resultant, we have R(λj µk ) = 0 for all j, k and the proof of the lemma follows. ✷
Now we apply Lemma 6.2 to the polynomials P (z) = z 2 − 2θ1 z + 1 and Q(z) = z − 2θ2 z + 1. The resultant R(z) of polynomials P (ξ) and ξ 2 Q( zξ ) with respect to ξ is 2
2
θ1 +
2
θ1 −
4 4
θ12
32 3 4 4 4 2 2 2 − 1 − (θ2 + θ2 − 1)z θ1 + θ1 − 1 − (θ2 − θ2 − 1)z ×
θ12 − 1 − (θ2 +
32 3 4 4 4 θ22 − 1)z θ1 − θ12 − 1 − (θ2 − θ22 − 1)z .
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If we expand the equation, we get the equality R(z) = z 4 − iz 3 + 3z 2 − iz + 1. Hence, we have the following recursive relation for p(n) = Un−1 (θ1 )Un−1 (θ2 ): p(n) − i p(n + 1) + 3p(n + 2) − i p(n + 3) + p(n + 4) = 0. Observing that a(n) = |p(n)| = (−i)n−1 p(n) and the initial values a(0) = 0, a(1) = 1, a(2) = 1, a(3) = 5, we get the result. ✷
6.3. Graph GP (n, 3)
Theorem 6.3. The number τ3 (n) of the spanning trees for the generalized Petersen graph GP (n, 3) is given by the formula 3 / Tn (ws ) − 1 , τ3 (n) = n ws − 1 s=1
and w1 , w2 , w3 are roots of the equation 4w3 − 2w2 − 5w − 2 = 0. Moreover, τ3 (2n) = 12n a(n)2 and τ3 (2n + 1) = (2n + 1)b(n)2 , where the integer sequences a(n) and b(n) satisfy the recursive relation u(n) − 4u(n + 1) − u(n + 2) − 24u(n + 3) + 65u(n + 4)
− 24u(n + 5) − u(n + 6) − 4u(n + 7) + u(n + 8) = 0
with the following initial data a(0) = 0, a(1) = 1, a(2) = 4, a(3) = 9, a(4) = 72, a(5) = 320, u(6) = 1332, a(7) = 6889 and b(0) = 1, b(1) = 1, b(2) = 20, b(3) = 83, b(4) = 289, b(5) = 1693, b(6) = 7775, b(7) = 34820. Proof. The first statement of the theorem directly follows from Theorem 5.1. To prove 13 2 the second, 4 by Corollary 5.4, we have τ3 (n) = n|c(n)| , where c(n) = s=1 Un−1 (θs ) and θs =
1+ws 2 ,s
= 1, 2, 3. Then the sequence u(n) = Un−1 (θs ) satisfies the following
recursive relation Ps (T )u(n) = 0, where Ps (z) = z 2 − 2θs z + 1. Suppose that λ, µ, ν are roots of the equations P1 (λ) = 0, P2 (µ) = 0, P3 (ν) = 0. By applying Lemma 6.2 twice, we obtain that η = λ µ ν is a root of the equation 1 + η 2 + 11η 4 + η 6 + η 8 =
√
6η(1 + 2η 2 + 2η 4 + η 6 ).
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Denote by η1 , η2 , . . . , η8 distinct roots of this equation. Then the sequence c(n) = 8 ! Un−1 (θ1 )Un−1 (θ2 )× Un−1 (θ3 ) can be written in the form c(n) = cj ηjn , where c1 , c2 , . . . , c8 are suitable constants. Let ζj = ηj2 . Now c(2n) = 8 !
j=1
(cj ηj )ζjn . Since
8 !
j=1
j=1
cj ζjn and c(2n + 1) =
(1 + η 2 + 11η 4 + η 6 + η 8 )2 = 6η 2 (1 + 2η 2 + 2η 4 + η 6 )2 ,
all ζj , j = 1, 2, . . . , 8 are the roots of the equation (1 + ζ + 11ζ 2 + ζ 3 + ζ 4 )2 − 6ζ(1 + 2ζ + 2ζ 2 + ζ 3 )2 = 0. Hence, 1 − 4ζ − ζ 2 − 24ζ 3 + 65ζ 4 − 24ζ 5 − ζ 6 − 4ζ 7 + ζ 8 = 0 and both sequences c(2n) and c(2n + 1) are solution of the difference equation u(n) − 4u(n + 1) − u(n + 2) − 24u(n + 3) + 65u(n + 4)
− 24u(n + 5) − u(n + 6) − 4u(n + 7) + u(n + 8) = 0.
√ We use the formulas a(n) = c(2n)/ 6 and b(n) = c(2n + 1) to calculate the initial elements of sequences a(n) and b(n) directly. ✷ 6.4. Graph GP (n, 4)
Theorem 6.4. The number τ4 (n) of the spanning trees for the generalized Petersen graph GP (n, 4) is given by the formula τ4 (n) = (−1)n−1 n
4 / Tn (ws ) − 1 , ws − 1 s=1
and w1 , w2 , w3 , w4 are roots of the equation 16w4 − 8w3 − 24w2 − 1 = 0. Moreover, τ4 (n) = n a(n)2 , where the integer sequences a(n) satisfies the recursive relation P (T )a(n) = 0, where P (T ) = T 16 − T 15 − 2T 13 − 16T 12 + 10T 11 − 2T 10 + 16T 9
+ 50T 8 − 16T 7 − 2T 6 − 10T 5 − 16T 4 + 2T 3 + T + 1
and T a(n) = a(n + 1) is the shift operator.
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Table 1 Graph GP (n, 2). n
Jac(GP (n, 2))
3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Z5 ⊕ Z15 Z7 ⊕ Z28 Z2 ⊕ Z10 ⊕ Z10 ⊕ Z10 Z35 ⊕ Z210 Z83 ⊕ Z581 Z161 ⊕ Z1288 Z355 ⊕ Z3195 Z2 ⊕ Z12 ⊕ Z60 ⊕ Z60 ⊕ Z60 Z1541 ⊕ Z16951 Z7 ⊕ Z7 ⊕ Z1365 ⊕ Z1820 Z6733 ⊕ Z87529 Z14027 ⊕ Z196378 Z5 ⊕ Z10 ⊕ Z10 ⊕ Z2950 ⊕ Z8850 Z61663 ⊕ Z986608 Z129403 ⊕ Z2199851 Z270865 ⊕ Z4875570 Z567911 ⊕ Z10790309 Z4 ⊕ Z24 ⊕ Z120 ⊕ Z49560 ⊕ Z49560
τ2 (n) = |Jac(GP (n, 2))| 75 196 2000 7350 48223 207368 1134225 5184000 26121491 121730700 583332757 2754594206 13053750000 60837209104 284667318953 1320621268050 6127935174499 28295350272000
The initial data of a(n) is given by the table below. n a(n)
−7
−83
−6
35
−5
−19
−4
1
−3 −5
−2
1
−1 −1
0
1
2
3
4
5
6
7
8
0
1
1
5
1
19
35
83
73
Proof. The first statement of the theorem directly follows from Theorem 5.1. Applying Lemma 6.2 several times and using the same arguments as in the proof of Theorem 6.1 we conclude that the sequence a(n) = |p(n)| = (−i)n−1 p(n), where 4 14 s p(n) = s=1 Un−1 ( 1+x 4 ) satisfies the recursive relation P (T )a(n) = 0. ✷ 7. Final Remarks and Tables
Theorem 4.1 is the first step to understand the structure of the Jacobian for GP (n, k). Also, it gives a simple way for numerical calculations of Jac(GP (n, k)) for small values of n and k. Theorems 6.1, 6.3 and 6.4 contain very convenient formulas for counting the number of spanning trees. The corresponding numerical results are given in Tables 1, 2 and 3. Acknowledgements The first author was supported by the 2015 Yeungnam University Research Grant. The second and the third authors were supported by the Grant of the Russian Federation Government at Siberian Federal University (grant no. 14.Y26.31.0006), by the Presidium of the Russian Academy of Sciences (project no. 0314-2015-0011), and by the Russian Foundation for Basic Research (projects no. 15-01-07906 and 16-31-00138).
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Table 2 Graph GP (n, 3). n
Jac(GP (n, 3))
4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Z2 ⊕ Z8 ⊕ Z24 Z2 ⊕ Z10 ⊕ Z10 ⊕ Z10 Z3 ⊕ Z9 ⊕ Z108 Z83 ⊕ Z581 Z3 ⊕ Z3 ⊕ Z12 ⊕ Z48 ⊕ Z48 Z289 ⊕ Z2601 Z4 ⊕ Z8 ⊕ Z40 ⊕ Z40 ⊕ Z120 Z1693 ⊕ Z18623 Z6 ⊕ Z2664 ⊕ Z7992 Z5 ⊕ Z5 ⊕ Z1555 ⊕ Z20215 Z83 ⊕ Z83 ⊕ Z83 ⊕ Z6972 Z2 ⊕ Z10 ⊕ Z52230 ⊕ Z17410 Z3 ⊕ Z3 ⊕ Z24 ⊕ Z21408 ⊕ Z21408 Z170917 ⊕ Z2905589 Z9 ⊕ Z148257 ⊕ Z1779084 Z802141 ⊕ Z15240679 Z2 ⊕ Z8 ⊕ Z8 ⊕ Z16 ⊕ Z80 ⊕ Z11120 ⊕ Z33360
τ3 (n) = |Jac(GP (n, 3))| 384 2000 2916 48223 248832 751689 6144000 31528739 127744128 785858125 3986498964 18186486000 98993332224 496614555113 2373854909292 12225173493739 60778610688000
Table 3 Graph GP (n, 4). n
Jac(GP (n, 4))
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Z19 ⊕ Z95 Z35 ⊕ Z210 Z83 ⊕ Z581 Z73 ⊕ Z584 Z355 ⊕ Z3195 Z779 ⊕ Z7790 Z1693 ⊕ Z18623 Z2555 ⊕ Z30660 Z5 ⊕ Z5 ⊕ Z1555 ⊕ Z20215 Z17513 ⊕ Z245182 Z2 ⊕ Z2 ⊕ Z2 ⊕ Z10 ⊕ Z10 ⊕ Z10 ⊕ Z950 ⊕ Z2850 Z71321 ⊕ Z1141136 Z103 ⊕ Z1751 ⊕ Z1751 ⊕ Z1751 Z405055 ⊕ Z7290990 Z37 ⊕ Z37 ⊕ Z23939 ⊕ Z454841 Z1823639 ⊕ Z36472780
τ4 (n) = |Jac(GP (n, 4))| 1805 7350 48223 42632 1134225 6068410 31528739 78336300 785858125 4293872366 21660000000 81386960656 552962478353 2953251954450 14906272578931 66513184046420
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