Discrete Graphs

Abstract

The spectra of metric equilateral metric graphs are essentially determined by the spectra of the normalised or averaging Laplacian matrices associated with the corresponding discrete graphs.

The spectra of metric equilateral metric graphs are essentially determined by the spectra of the normalised or averaging Laplacian matrices associated with the corresponding discrete graphs. We are going to prove an explicit formula connecting these spectra, despite that the metric graphs have infinitely many eigenvalues but the spectra of Laplacian matrices are finite.

The second original goal was to check how does the idea of topological perturbations (developed originally for metric graphs, see Sect. 12.5) work for discrete graphs—a well established area of discrete mathematics. After the chapter was accomplished, we learned that this question for the normalised Laplacian matrices was studied earlier by H. Urakawa and collaborators [285, 413].¹ We follow our original presentation in order to give a possibility to compare the two approaches.

24.1 Laplacian Matrices: Definitions and Elementary Properties

Let $G$ be a discrete graph with $M$ vertices and $N$ edges connecting some of the vertices. As before we are going to consider mostly finite graphs, i.e. graphs with a finite number of vertices and edges. Moreover, for simplicity we assume that no loops and parallel edges are present. With such a graph one naturally associates the following matrices:

• the connectivity, or adjacency matrix $C = \{ c_{nm} \}$

  $$\displaystyle \begin{aligned} c_{nm} = \left\{ \begin{array}{ll} 1, &\mbox{the vertices }n\mbox{ and }m\mbox{ are neighbours}, \\ & \mathit{i.e.}\mbox{ connected by an edge}; \\[2mm] 0, & \mbox{otherwise}, \end{array} \right.\end{aligned}$$

• the (diagonal) degree matrix $D = \mathrm {diag}\, \{ d_1, d_2, \dots , d_M \},$ where $d_m$ are the degrees (valencies) of the corresponding vertices.

We shall be interested in so-called Laplacian matrices and their spectral properties. Laplacian matrices are certain generalisations of the (differential) Laplace operator for the case where the set of points is discrete. All these matrices are defined on the finite dimensional Hilbert space $\ell _2 (G) = \mathbb C^M \ni \psi = (\psi (1), \psi (2), \dots , \psi (M))$, but using different formulas:

• Combinatorial Laplacian$L(G)$ [152, 153, 391]

  $$\displaystyle \begin{aligned} {} \left(L(G) \psi \right) (m) = \sum_{n \sim m} \left( \psi(m) - \psi (n) \right), \end{aligned} $$

  (24.1)

  where the sum is taken over all neighbouring vertices.² This matrix can also be defined using the connectivity matrix $C$ and the degree matrix $D$:

  $$\displaystyle \begin{aligned} L (G) = D - C.\end{aligned}$$

• normalised Laplacian$L_N (G)$ [132, 146]

  $$\displaystyle \begin{aligned} \left(L_N (G) \psi \right) (m) = \psi (m) - \frac{1}{\sqrt{d_m}} \sum_{n \sim m } \frac{1}{\sqrt{d_n}} \psi (n), \end{aligned} $$

  (24.2)

  also given by

  $$\displaystyle \begin{aligned} {} L_N (G) = D^{-1/2} L(G) D^{-1/2} = I - D^{-1/2} C D^{-1/2}. \end{aligned} $$

  (24.3)

The normalised Laplacian is similar to another Laplacian matrix to be called averaging Laplacian

$$\displaystyle \begin{aligned} \left( L_A (G) \psi \right) (m) = \psi (m) - \frac{1}{d_m} \sum_{n \sim m} \psi (n), \end{aligned} $$

(24.4)

or in matrix form

$$\displaystyle \begin{aligned} L_A = I - D^{-1} C. \end{aligned} $$

(24.5)

The second term here gives the average value of $\psi$ over all vertices neighbouring to $m$. It follows that any solution to the Laplace equation $L_A \psi = 0$ possesses the following property:

The value of$\psi (m)$is equal to the average value of$\psi$over all neighbouring vertices:

$$\displaystyle \begin{aligned} L_A \psi = 0 \Rightarrow \psi (m) = \frac{1}{d_m} \sum_{n \sim m} \psi (n). \end{aligned} $$

(24.6)

This property reminds of the Poisson formula for the differential Laplace operator. The averaging and normalised Laplacians are similar and therefore their spectra coincide:

$$\displaystyle \begin{aligned} {} L_A (G) = D^{-1/2} L_N (G) D^{1/2}. \end{aligned} $$

(24.7)

Both the standard and the normalised Laplacians are Hermitian, since they are given by (finite) real symmetric matrices. The averaging Laplacian has real spectrum (since $L_N$ is Hermitian), but the corresponding matrix in general is not Hermitian. The operator associated with the matrix $L_A$ is self-adjoint in the weighted Hilbert space $\ell _2^D (G) = \mathbb C^M$ with the scalar product given by

$$\displaystyle \begin{aligned} {} \langle \psi, \phi \rangle_{\ell_2^D(G)} = \langle D \psi, \phi \rangle_{\mathbb C^M} = \sum_{m=1}^M d_m \psi_m \overline{\phi_m}. \end{aligned} $$

(24.8)

Problem 99

For which graphs the averaging Laplacian is given by a Hermitian matrix in the original space $\ell _2 (G) = \mathbb C^M$?

All three Laplacian matrices are generalisations of the second difference matrix. For example, consider the one-dimensional chain with the subsequent vertices connected to each other. Then the degrees of all vertices are equal to $2$ and all three Laplacians remind of the discrete approximation of $- \frac {d^2}{dx^2}$

$$\displaystyle \begin{aligned} \begin{array}{cl} & \displaystyle \frac{1}{2} (L \psi ) (m) = (L_N \psi )(m) = (L_A \psi ) (m) \\[3mm] = & \displaystyle \frac{1}{2} \left( \psi(m) - \psi (m+1) + \psi (m) - \psi (m-1) \right) \\[5mm] = & \displaystyle - \frac{\displaystyle \psi(m+1) -2 \psi (m) + \psi (m-1)}{2}. \end{array} \end{aligned} $$

(24.9)

Different Laplacian matrices are widely used in applications. The study of their spectral properties is a well-developed branch of modern discrete mathematics. It is not our goal to give an overview of these results, we shall focus instead on the relation between spectral properties of discrete Laplacian matrices and standard Laplacians on metric graphs. We shall also show how our methods originally developed for metric graphs, work in the discrete case.

All three Laplacian matrices are nonnegative as operators acting in the respective spaces. This can be seen from their quadratic forms

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} \langle L(G) \psi, \psi \rangle_{\ell_2 (G)} & = & \displaystyle \sum_{m=1}^M \left( \sum_{n \sim m} \overline{(\psi(m)- \psi(n))} \psi(m) \right) \\ & = & \displaystyle \frac{1}{2} \sum_{n,m: n \sim m} \vert \psi (m) - \psi(n) \vert^2, \\[5mm] \langle L_N (G) \psi, \psi \rangle_{\ell_2 (G)} & = & \displaystyle \frac{1}{2} \sum_{n,m: n \sim m} \left\vert \frac{1}{\sqrt{d_m}} \psi (m) - \frac{1}{\sqrt{d_n}} \psi(n) \right\vert^2, \\[5mm] \langle L_A(G) \psi, \psi \rangle_{\ell_2^D(G)} & = & \langle D L_A (G) \psi, \psi \rangle_{\ell_2 (G)} \\[3mm] & = & \displaystyle \langle L_N (G) D^{1/2} \psi, D^{1/2} \psi \rangle_{\ell_2 (G)} \\[3mm] & = & \displaystyle \frac{1}{2} \sum_{n,m: n \sim m} \vert \psi (m) - \psi(n) \vert^2. \end{array} \end{aligned} $$

(24.10)

Problem 100

Prove that the quadratic forms for all three Laplacian matrices are given by formulas (24.10).

Problem 101

The quadratic forms of $L(G)$ and $L_A(G)$ are given by the same expressions. Is it possible to conclude that these operators are isospectral? Explain the reason and provide explicit examples to support your conclusion.

24.2 Topology and Spectra: Discrete Graphs

In this section we derive a few elementary properties of Laplacian matrices connected to topological characteristics of discrete graphs such as the number of connected components and Euler characteristic.

The Number of Connected Components

We see that all three operators have $\mu _1 = 0$ as an eigenvalue. For the standard and averaging Laplacians the corresponding eigenfunction is equal to a constant on each connected component of $G$. For the normalised Laplacian the zero-energy eigenfunction should be modified as

$$\displaystyle \begin{aligned} {} \psi_1 (m) = \sqrt{d_m} c, \end{aligned} $$

(24.11)

where $c$ is a constant which can be chosen different for each connected component. In fact no other eigenfunctions corresponding to zero eigenvalue are present (see Lemma 24.2 below). Therefore we conclude that the number $\beta _0$ of connected components in $G$ is equal to the multiplicity of the ground state $\mu _1 = 0$

$$\displaystyle \begin{aligned} \beta_0 (G) = m (0), \end{aligned} $$

(24.12)

where $m(0)$ is the multiplicity of the eigenvalue $\mu _1 = 0.$ In particular, if the graph $G$ is connected, then the ground state has multiplicity $1.$

In what follows we are going to consider connected graphs only, since all results can be easily reformulated including not necessarily connected graphs. For connected graphs it is then natural to denote the eigenvalues $\mu _j (L(G))$ of the Laplacian matrices as follows

$$\displaystyle \begin{aligned} {} 0 = \mu_1 < \mu_2 \leq \mu_3 \leq \dots \leq \mu_{M}. \end{aligned} $$

(24.13)

The Volume

The volume for a discrete graph $G$ is just the number $M$ of vertices. If we know the spectrum of any of the Laplacian matrices, then the volume of $G$ is equal to the number of eigenvalues counted with multiplicities:

$$\displaystyle \begin{aligned} \# \{ \mu_j \} = M. \end{aligned} $$

(24.14)

Euler Characteristic

The trace of a Hermitian matrix is equal to the sum of its eigenvalues. Consider the combinatorial Laplacian matrix and calculate its trace in two different ways: using the eigenvalues and summing the diagonal elements:³

$$\displaystyle \begin{aligned} \sum_{j=1}^M \mu_j (L(G)) = \mathrm{Tr}\, L(G) = \mathrm{Tr}\, D = d_1 + d_2 + \dots + d_M = 2 N.\end{aligned}$$

This formula allows one to calculate the number of edges from the spectrum leading to the formula for the Euler characteristic

$$\displaystyle \begin{aligned} {} \chi (G) = \#\{ \mu_m (L(G)) \} - \frac{1}{2} \sum \mu_m (L(G)). \end{aligned} $$

(24.15)

This result cannot be generalised for normalised Laplacians, since it is not hard to provide examples of graphs, which are isospectral with respect to $L_N$ but have different Euler characteristic [114,115,116, 146]. Consider for example the following two graphs: three-star and four-cycle, shown in Fig. 24.1.

Fig. 24.1

The three-star $G_{(3.2)}$ and the four-cycle graphs

Both graphs have 4 vertices and the corresponding normalised Laplacians are given by

$$\displaystyle \begin{aligned} L_N^1 = \left( \begin{array}{cccc} 1 & - 1/\sqrt{3} & - 1/\sqrt{3} & - 1/\sqrt{3} \\ - 1/\sqrt{3} & 1 & 0 & 0 \\ - 1/\sqrt{3} & 0 & 1 & 0 \\ - 1/\sqrt{3} & 0 & 0 & 1 \end{array} \right),\end{aligned}$$

$$\displaystyle \begin{aligned} L_N^2 = \left( \begin{array}{cccc} 1 & - 1/2 & 0 & -1/2 \\ -1/2 & 1 & -1/2 & 0 \\ 0 & -1/2 & 1 & -1/2 \\ -1/2 & 0 & -1/2& 1 \end{array} \right). \end{aligned} $$

(24.16)

The characteristic polynomials coincide and are given by

$$\displaystyle \begin{aligned} p(\lambda) = \det (L_N^j - \lambda) = (1-\lambda)^4 - (1-\lambda)^2 \end{aligned} $$

(24.17)

showing that both normalised Laplacians have eigenvalues $0, 1, 1, 2.$ Obviously the two graphs have different Euler characteristic. Note that the corresponding eigenfunctions are different

$$\displaystyle \begin{aligned} \begin{array}{cccc} \psi_1^1 = \left( \begin{array}{c} 1 \\ 1/\sqrt{3} \\ 1/\sqrt{3} \\ 1/\sqrt{3} \end{array} \right), & \psi_2^1 = \left( \begin{array}{c} 1 \\ -1/\sqrt{3} \\ -1/\sqrt{3} \\ -1/\sqrt{3} \end{array} \right), & \psi_3^1 = \left( \begin{array}{c} 0 \\ 1 \\ -1 \\ 0 \end{array} \right), & \psi_4^1 = \left( \begin{array}{c} 0 \\ 1 \\ 0 \\ -1 \end{array} \right) ; \\[8mm] \psi_1^2 = \left( \begin{array}{c} 1 \\ 1 \\ 1 \\ 1 \end{array} \right), & \psi_2^2 = \left( \begin{array}{c} 1 \\ 1 \\ -1 \\ -1 \end{array} \right), & \psi_3^2 = \left( \begin{array}{c} 1 \\ -1 \\ -1 \\ 1 \end{array} \right), & \psi_4^2 = \left( \begin{array}{c} 1 \\ -1 \\ 1 \\ -1 \end{array} \right). \end{array} \end{aligned} $$

(24.18)

Problem 102

Construct your own pair of discrete graphs with isospectral combinatorial Laplacians.

24.3 Normalised Laplacians and Equilateral Metric Graphs

We are going to discuss the relations between the eigenvalues of normalised Laplacians $L_N (G)$ on discrete graphs G and the spectrum of the standard Laplacians $L^{\mathrm {st}} (\Gamma )$ on metric graphs $\Gamma$. Recall that the normalised and averaging Laplacian matrices are isospectral, hence all our results apply to averaging Laplacians. Moreover, both matrices will be used in the proofs.

In order to be able to compare these spectra, we need a one-to-one correspondence between discrete graphs $G$ and metric graphs $\Gamma$. With any finite discrete graph $G$ we may associate unique metric graph $\Gamma$ by assigning unit lengths to all edges in $G$. This rule establishes a one-to-one correspondence between the discrete and equilateral metric graphs and will be used throughout this chapter. If the metric graph does not have degree two vertices, then the corresponding discrete graph $G$ is the same as the discrete graph used in Chap. 6 to get secular polynomials.

Consider any discrete graph $G$ with $M$ vertices $1, 2, \dots , M$ and the corresponding normalised Laplacian $L_N (G)$. The quadratic form is nonnegative as can be seen from formula (24.10). The quadratic form of $L_N - 2$ is nonpositive

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} \displaystyle \langle ( L_N (G) - 2)\psi, \psi \rangle_{\ell_2(G)} & = & \displaystyle \frac{1}{2} \sum_{n \sim m} \left\vert \frac{1}{\sqrt{d_m}} \psi (m) - \frac{1}{\sqrt{d_n}} \psi(n) \right\vert^2 - 2 \sum_m \vert \psi(m) \vert^2 \\ & = & \displaystyle - \frac{1}{2} \sum_{n \sim m} \left\vert \frac{1}{\sqrt{d_m}} \psi (m) + \frac{1}{\sqrt{d_n}} \psi(n) \right\vert^2. \end{array} \end{aligned} $$

(24.19)

It follows that the eigenvalues of the normalised Laplacian are always lying between $0$ and 2. Let us denote the eigenvalues of $L_N$ by $\mu _j (L_N)$ ordering them following (24.13)

$$\displaystyle \begin{aligned} 0 = \mu_1 (L_N) \leq \mu_2 (L_N) \leq \dots \mu_M (L_N) \leq 2. \end{aligned}$$

The number $\mu = 0$ is always an eigenvalue, while $\mu _M$ can be less than $2.$ The eigenvalues $\mu = 0$ and $\mu =2$ will be called extremal.

The spectrum of the metric graph is discrete tending to $+\infty$. It is easy to see that the spectrum is $2 \pi$-periodic if one uses the variable $k, \; k^2 = \lambda$ instead of $\lambda$ and ignores that the multiplicities of $\lambda = 0$ and $\lambda = (2 \pi )^2$ could be different. Moreover, the spectrum in k-scale is symmetric with respect to the origin, hence it is enough to study the eigenvalues between $0$ and $\pi ^2$. We shall use our standard convention to denote the eigenvalues of the standard Laplacian

$$\displaystyle \begin{aligned} 0= \lambda_1 \leq \lambda_2 \leq \dots \leq \lambda_n \leq \dots . \end{aligned} $$

(24.20)

It turns out that while the correspondence between the eigenvalues $\mu \neq 0, 2$ and $\lambda \neq (m\pi )^2, \, m \in \mathbb Z$ can be described by an explicit formula, the correspondence between the extremal eigenvalues is slightly more involved. This is related to the fact that $(m \pi )^2$ are the eigenvalues of the Dirichlet Laplacian on the unit interval. Therefore extremal and all other (to be called generic) eigenvalues will be considered separately.

Generic Eigenvalues

Let us discuss the relation between the eigenvalues $\mu _n \neq 0, 2$ and $\lambda _j \neq \pi ^2 m^2, m \in \mathbb Z$ first.

Theorem 24.1

Assume that $\mu _n$ are the eigenvalues of the normalised Laplacian $L_N$ on a discrete graph $G$ and $\lambda _j$ are the eigenvalues of the standard Laplacian $L^{\mathrm {st}} (\Gamma )$ on the corresponding equilateral metric graph $\Gamma$ with the common edge length one. Then $\lambda _j : \lambda _j \neq \pi ^2 m^2, m \in \mathbb Z$ is an eigenvalue of $L^{\mathrm {st}}(\Gamma )$ if and only if

$$\displaystyle \begin{aligned} {} 1 - \cos \sqrt{\lambda_j}= \mu_n, \end{aligned} $$

(24.21)

for a certain $\mu _n \neq 0, 2$ from the spectrum of $L_N(G).$ Moreover the multiplicities of the eigenvalues coincide.

Proof

It is much easier to prove the theorem using the average Laplacian $L_A$ instead of the normalised Laplacian $L_N$. These two matrices are isospectral due to (24.7). Consider any eigenvalue $\mu$ and any corresponding eigenvector $\psi$

$$\displaystyle \begin{aligned} L_A \psi = \mu \psi, \end{aligned} $$

(24.22)

or in more details

$$\displaystyle \begin{aligned} \psi (m) - \frac{1}{d_m} \sum_{l \sim m} \psi_n (l) = \mu \psi (m), \ \ m= 1,2, \dots, M. \end{aligned} $$

(24.23)

Let us construct an eigenfunction $\tilde {\psi } (x)$ of $L^{\mathrm {st}} (\Gamma )$ corresponding to a certain positive eigenvalue $\lambda = k^2$ so that it attains the same values at the vertices as $\psi$

$$\displaystyle \begin{aligned} {} \tilde{\psi} (V^m) = \psi (m), \; \, m=1,2, \dots, M. \end{aligned} $$

(24.24)

The eigenvalue $\lambda$ is not determined yet, we shall see which values are possible in a few steps.

Consider any edge in $\Gamma$, say connecting the vertices $V^{m_1}$ and $V^{m_2}$. Then the unique function satisfying the differential equation

$$\displaystyle \begin{aligned} - \tilde{\psi}'' (x) = k^2 \tilde{\psi} (x)\end{aligned}$$

and having values at the endpoints prescribed by (24.24) is given by

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle \tilde{\psi} (x) & = & \displaystyle \frac{\psi (m_1) - \cos k \, \psi(m_2)}{\sin^2 k} \cos \left( k\, \mathrm{dist}\{x, V^{m_1} \} \right) \\[3mm] && \displaystyle + \frac{\psi (m_2) - \cos k\, \psi (m_1)}{\sin^2 k} \cos \left( k \, \mathrm{dist}\{x, V^{m_2} \} \right) . \end{array} \end{aligned} $$

(24.25)

The normal derivative of $\tilde {\psi }_j$ at the endpoints can be calculated as well. For example we have:

$$\displaystyle \begin{aligned} {} \partial_n \tilde{\psi} (V^{m_1}) = \frac{k}{\sin k} \left( - \cos k \; \psi (m_1) + \psi (m_2) \right) . \end{aligned} $$

(24.26)

Note that the last formula may also be obtained directly from (5.55).

We repeat this procedure for every edge in the metric graph $\Gamma .$ The function $\tilde {\psi }$ obtained in this way is continuous at the vertices by construction, it satisfies the same differential equation on each edge. Hence in order to check that it is really an eigenfunction of $L^{\mathrm {st}} (\Gamma )$ it remains to show that the sum of the normal derivatives at each vertex is zero

$$\displaystyle \begin{aligned} \sum_{x_i \in V^m} \partial_n \tilde{\psi} (x_i) = 0.\end{aligned}$$

Using (24.26) this equation can be written as

$$\displaystyle \begin{aligned} \sum_{l \sim m} \frac{k}{\sin k} \left( \cos k \; \psi (m) - \psi (l) \right) = 0 \end{aligned} $$

(24.27)

$$\displaystyle \begin{aligned} \Leftrightarrow d_m \cos k \; \psi (m) = \sum_{\ell \sim m} \psi (\ell).\end{aligned}$$

It is easy to transform the equation into the following form

$$\displaystyle \begin{aligned} \psi (m) - \frac{1}{d_m} \sum_{l \sim m} \psi (l) = (1- \cos k) \psi (m), \end{aligned} $$

(24.28)

which is precisely the eigenfunction equation for $\psi$, provided $\lambda$ and $\mu$ satisfy (24.21). Of course if $\mu$ is fixed, then Eq. (24.21) possesses infinitely many solutions $k.$

The constructed mapping

$$\displaystyle \begin{aligned} \psi \mapsto \tilde{\psi}\end{aligned}$$

is one-to-one, provided $\lambda$ is fixed: the functions $\psi$ and $\tilde {\psi }$ have the same values at the vertices and $\tilde {\psi }$ on each edge is uniquely determined by its values at the endpoints. This implies that the multiplicities of the two eigenvalues connected via (24.21) coincide. □

Formula (24.21) is often refereed to as von Below formula as it appeared for the first time in [488]. One may find different generalisations of this formula for equilateral metric graphs in [418, 422].

Formual (24.21) implies that every eigenvalue $\mu _n$ of $L_A (G)$ determines an infinite series of eigenvalues of $L^{\mathrm {st}} (\Gamma )$, since equation (24.21) has infinitely many solutions with respect to $\lambda :$ if $\lambda _j$ is a solution, then any $\big (\pm \sqrt {\lambda _j}+ 2 \pi m \big )^2 , \; \, m \in \mathbb Z$ is also a solution. One may use the variable $k$ instead of $\lambda = k^2$ to describe the spectrum. The observation above means that the set of $k \notin \mathbb Z$ in the spectrum of $L^{\mathrm {st}} (\Gamma )$ is periodic and symmetric with respect to the origin. In the interval $(0, 2 \pi )$ the eigenvalues are symmetric with respect to the middle point $k = \pi .$ The number of eigenvalues of $L^{\mathrm {st}} (\Gamma )$ in the interval $(0, \pi )$ is equal to the number of eigenvalues of $L_N (G)$ different from 0 and $2.$

Extremal Eigenvalues

It remains to study the case where $\mu _n = 0, 2$ and $\lambda _j = \pi ^2 n^2, \; n \in \mathbb Z.$

We are going to prove four Lemmas describing the cases

• $\mu = 0,$$\lambda = 0,$

• $\mu = 2,$

• $\lambda = \pi ^2(1 + 2 m)^2, \; m \in \mathbb Z$

• $\lambda = 4 \pi ^2 m^2, \; m \in \mathbb Z, m\neq 0 .$

The following Lemma determines the multiplicity of the zero eigenvalue for both normalised and standard Laplacians.

Lemma 24.2

The point zero is an eigenvalue for both the normalised Laplacian $L_N (G)$ and the standard Laplacian $L^{\mathrm {st}} (\Gamma )$ . The multiplicity of the eigenvalue is equal to the number of connected components in the graph.

Proof

It is clear from the construction that the number of connected components in the discrete graph $G$ and in the corresponding metric graph $\Gamma$ are equal.

Consider first the quadratic form $L_N (G)$ given by (24.10). If $G$ is connected then the vector $\psi _1 (m) = \sqrt {d_m} c$ is an eigenvector corresponding to $\mu _1 = 0.$ This vector is unique (up to multiplication of course). For non-connected graphs the multiplicity is equal to the number of connected components, since the constant $c$ can be chosen different on each connected component.

For the standard Laplacian we repat the proof of Lemma 4.10. The quadratic form is given by formula (3.55), which simplifies as follows

$$\displaystyle \begin{aligned} \langle L^{\mathrm{st}} (\Gamma) u, u \rangle_{L_2 (\Gamma)} = \sum_{n=1}^N \int_{E_n} \vert u'(x) \vert^2 dx . \end{aligned} $$

(24.29)

The constant function $\psi _1 (x) = c$ is an eigenfunction corresponding to $\lambda _1 = 0.$ Every function that minimises the quadratic form is a constant function on every edge. Standard matching conditions imply that the function is equal to a constant on every connected component of $\Gamma .$ It follows that the multiplicity of the zero eigenvalue is equal to the number of connected components. □

This Lemma implies in particular that the multiplicities of the zero eigenvalue for the normalised Laplacian and standard Laplacian coincide.

Lemma 24.3

Let $G$ be a connected discrete graph. Then the point $\mu = 2$ is an eigenvalue of the normalised Laplacian if and only if the graph $G$ is bipartite. ⁴

Problem 103

Prove Lemma 24.3

If the graph $G$ is not connected, then $\mu = 2$ is an eigenvalue of the averaging Laplace matrix with the multiplicity equal to the number of bipartite components. Here multiplicity zero means that $\mu = 2$ is not an eigenvalue.

Lemma 24.4

Let the equilateral metric graph $\Gamma$ with the common edge length one be connected. Then the points $\lambda = (1 + 2m)^2 \pi ^2, \; m \in \mathbb Z$ are eigenvalues of the standard Laplace operator on $\Gamma$ with the multiplicities equal to

• $\beta _1 + 1$ if the corresponding discrete graph $G$ is bipartite,

• $\beta _1 -1$ if the corresponding discrete graph $G$ is not bipartite. ⁵

Here $\beta _1$ is the number of independent cycles either on the metric graph $\Gamma$ , or on the discrete graph G.

Proof

Let us prove the lemma for $k = \pi$, then any soluton to the eigenfunction equation on each edge is given by

$$\displaystyle \begin{aligned} \tilde{\psi} (x) = \alpha \cos \pi x + \beta \sin \pi x.\end{aligned}$$

It follows that the values of the function at all endpoints have the same absolute value, but differ by sign:

$$\displaystyle \begin{aligned} \tilde{\psi} (x_{2j-1}) = - \tilde{\psi} (x_{2j}).\end{aligned}$$

Let us denote by $a$ the common absolute value of the function at all vertices

$$\displaystyle \begin{aligned} a := \vert \tilde{\psi} (x_j) \vert.\end{aligned}$$

An eigenfunction with $a \neq 0$ exists if and only if the graph $\Gamma$ is bipartite. Such eigenfunction resembles the averaged Laplacian eigenfunction for $\mu =2$ (see Lemma 24.3). Let us denote the corresponding (unique up to a multiplier) eigenfunction by $\tilde {\psi }^0.$

If $a = 0$ then the eigenfunctions are given by $\beta _j \sin \pi (x-x_{2j-1})$ on each interval. Such an eigenfunction exists if and only if one is able to combine the sine functions on the edges, so that the sum of derivatives at each vertex is equal to zero (certain balance condition is fulfilled).

Assume now that the graph is bipartite, then all cycles in it have even lengths and it is easy to construct eigenfunctions supported on such cycles using multiples of $\sin \pi (x-x_{2j-1}).$ Every graph $G$ can be turned into a tree $\mathbf T$ by cutting away certain $\beta _1$ edges. Let us denote these edges by $\Delta _i, \; i =1,2, \dots , \beta _1,$ and by $\tilde {\psi }^i$ the eigenfunctions supported on the (shortest) cycles contained in $\mathbf T \cup \Delta _i.$ We have constructed $\beta _1 +1$ eigenfunctions corresponding to $k = \pi .$ It follows that the multiplicity of the corresponding eigenvalue is at least $\beta _1 +1 .$

Let $\tilde {\psi }$ be any eigenfunction corresponding to $\lambda = \pi ^2.$ Consider then the function

$$\displaystyle \begin{aligned} \hat{\psi} (x) = \tilde{\psi} (x) - f_0 \tilde{\psi}^0 (x) - \sum_{i=1}^{\beta_1} f_i \tilde{\psi}^i (x),\end{aligned}$$

where the constants $f_0, f_i$ are chosen so that the function $\hat {\psi }$ is equal to zero at all vertices and on all edges $\Delta _i, i = 1,2, \dots , \beta _1.$ Then the function $\hat {\psi }$ is supported on the tree $\mathbf T$ and therefore it is identically equal to zero. To show this, let us look at the pendant edges in $\mathbf T$. Obviously the function $\hat {\psi }$ is equal to zero on every such edge. One may take away this edge from the tree $\mathbf T$ and repeat the argument. Since the tree is finite we conclude that $\hat {\psi }$ is identically zero.

It remains to study the case where the graph $G$ is not bipartite and the eigenfunction is equal to zero at all vertices. Consider again the edges $\Delta _i$ and the (shortest) cycles on $\mathbf T \cup \Delta _i$ introduced above. Among such cycles there exists at least one cycle of odd length, since otherwise the graph is bipartite. Without loss of generality we assume that this cycle corresponds to $\Delta _{\beta _1}.$ For cycles of even length there exists an eigenfunction supported on it. For cycles $\Delta _i$ of odd length, there exists an eigenfunction supported on $\mathbf T \cup \Delta _i \cup \Delta _{\beta _!}.$ Hence we have constructed $\beta _1-1$ linearly independent eigenfunctions to be denoted by $\tilde {\psi }^i.$

To prove that the multiplicity of the eigenvalue is really equal to $\beta _1-1$ consider an arbitrary eigenfunction $\tilde {\psi }$ and the function $\hat {\psi }$ given by

$$\displaystyle \begin{aligned} \hat{\psi} (x) = \tilde{\psi} - \sum_{i=1}^{\beta_1-1} f_i \tilde{\psi}^{\beta_1} (x),\end{aligned}$$

where as before the parameters $f_i$ are adjusted so that the function $\hat {\psi }$ is not only equal to zero at all vertices but on all edges $\Delta _i, \, i = 1,2, \dots , \beta _1-1$ as well. Then the function $\hat {\psi }$ is an eigenfunction supported on $\mathbf T \cup \Delta _{\beta _1}.$ This graph contains only one cycle and this cycle has an odd length. As before the function $\hat {\psi }$ is equal to zero on all pendant edges. Repeating the argument we deduce that $\hat {\psi }$ is supported by the unique cycle in $\mathbf T \cup \Delta _g,$ but this cycle has odd length and therefore $\hat {\psi } \equiv 0.$ It follows that the multiplicity of the eigenvalue is $\beta _1-1$ in this case.

The proof for $k = (1+2m) \pi , \, m \neq 0$ is almost identical. □

The Lemma can easily be generalised for the case of not connected graphs by repeating the argument for every connected component.

Lemma 24.5

The points$\lambda = 4\pi ^2 m^2, \; m=1,2,\ldots$are eigenvalues of$L^{\mathrm {st}} (\Gamma )$with multiplicities$\beta _1+1$.

Proof

We assume first that $k = 2 \pi .$ Then it is easy to construct $\beta _1+1$ linearly independent eigenfunctions corresponding to this particular $k ,$ where $g$ is the genus of $\Gamma$. Let us denote by $\tilde {\psi }^0 (x,k)$ the eigenfunction given by

$$\displaystyle \begin{aligned} \tilde{\psi}^0 (x,k) = \cos k (x-x_{2j-1}), \; \; x\in [x_{2j-1}, x_{2j}].\end{aligned}$$

This function satisfies the differential equation on each edge, is continuous at all vertices (in fact it is equal to $1$ at all vertices) and all normal derivatives are equal to zero, which implies that their sums at each vertex are also zero and the standard vertex conditions are satisfied.

If $\Gamma$ is not a tree, then it can be transformed to a tree $\mathbf T$ by removing exactly $\beta _1$ edges denoted without loss of generality by $\Delta _1, \Delta _2, \ldots , \Delta _{\beta _1}$. Let $C_i$ be the (shortest) cycle on $\mathbf T \cup \Delta _i$ passing $\Delta _i$ in the positive direction. Note that every such cycle comes across exactly one removed edge. Consider the functions $\tilde {\psi }^i$ defined by

$$\displaystyle \begin{aligned} \tilde{\psi}^i(x,k) = \left\{ \begin{array}{ll} \pm \sin k (x-x_{2j-1}), & \mbox{provided }x \in \Delta_j \subset C_i; \\ 0, & \mbox{otherwise}; \end{array} \right.\end{aligned}$$

where the sign depends on whether the path $C_j$ runs along $\Delta _j$ in the positive $(+)$ or in the negative $(-)$ direction. The function $\psi _i$ is not only continuous along the path $C_i$ but its first derivative is continuous as well.

Each function $\psi _i$ satisfies the eigenfunction equation, is continuous at all vertices (in fact equal to zero there) and the sum of normal derivatives at each vertex is zero (if the vertex is on the path $C_i$ then only two normal derivatives are different from zero but cancel each other, if the vertex is not on the path, then all normal derivatives are zero).

It is clear that the functions $\tilde {\psi }^0, \tilde {\psi }^1, \ldots , \tilde {\psi }^{\beta _1}$ are linearly independent and this implies that the multiplicity of the eigenvalue $k$ is not less than $1+\beta _1.$

Repeating the arguments used in the proof of Lemma 24.4 one shows that the multiplicity does not exceed $1+ \beta _1$. The proof for $\lambda = \left ( 2 \pi m \right )^2, \; m=2,3,\ldots ,$ follows the same lines. □

Our studies allow us to characterise the spectrum of any equilateral metric graph.

Theorem 24.6

Let $L^{\mathrm {st}} (\Gamma )$ be the standard Laplace operator on a connected compact equilateral metric graph $\Gamma$ obtained from the discrete graph $G$ by assigning unit length to each edge. Then the spectrum $\big \{ \lambda _n = k_n^2 \big \}_{n=1}^\infty$ of the standard Laplacian $L^{\mathrm {st}} (\Gamma )$ has the following properties:

• (1) Apart from the point $k_1= 0 ,$ the set $\{ k_n \}_{n=1}^\infty$ is invariant under right shifts by $2\pi .$

• (2) The points $k_n$ inside the interval $(0, 2\pi )$ are situated symmetrically with respect to the centre of the interval, i.e. $\lambda _n = k_n^2$ is an eigenvalue if and only if $\left ( 2\pi -k_n \right )^2$ is also an eigenvalue.

• (3) The point$\lambda _1 = 0$has multiplicity$1$.

• (4) The points $\lambda = (2\pi m)^2, \; m=1,2,\ldots$ have the multiplicity

  $$\displaystyle \begin{aligned} \beta_1 +1 = 2 - \chi = N-M +2 . \end{aligned}$$

• (5) The number of eigenvalues inside each interval

  $$\displaystyle \begin{aligned} ((m \pi)^2, ((m+1) \pi)^2) , \, m = 0, 1, \dots\end{aligned}$$

  counted with multiplicities is equal to

  $$\displaystyle \begin{aligned} \begin{array}{ll} M-2, & \mathit{\mbox{if }} G\ \mathit{is}\ \mathit{bipartite}; \\[3mm] M-1 , & \mathit{\mbox{otherwise.}} \end{array} \end{aligned} $$

  (24.30)

• (6) The eigenvalues

  $$\displaystyle \begin{aligned} \lambda = (\pi (2m+1))^2, \; m= 0, 1, \ldots\end{aligned}$$

  have multiplicity

  $$\displaystyle \begin{aligned} \begin{array}{ll} \beta_1 +1 = - \chi+2 = N-M+2, & \mathit{\mbox{if }} G \ \mathit{is}\ \mathit{bipartite.} \\[3mm] \beta_1 -1 = - \chi = N-M , & \mathit{\mbox{otherwise.}} \end{array} \end{aligned} $$

  (24.31)

Proof

All statements of the Theorem are straightforward corollaries of Theorem 24.1 and Lemmas 24.2–24.5.

• (1) This statement follows directly from formula (24.21) for generic eigenvalues of $k$ and Lemmas 24.4 and 24.5 for $k$ being integer multiplies of $\pi .$

• (2) This is again a direct corollary of (24.21), since all eigenvalues in question here are generic, i.e. different from $\pi ^2 n^2.$

• (3), (4), and (6) are just reformulations of Lemmas 24.4, 24.2, and 24.5, respectively.

• (5) Lemma 24.3 implies that the averaging Laplacian has precisely $M-1$ or $M-2$ eigenvalues inside the interval $(0,2)$ depending on whether the graph $G$ is bipartite ($M-2$) or not ($M-1$). Then formula (24.21) implies that the standard Laplacian has the same number of eigenvalues inside each interval $((m \pi )^2, ((m+1) \pi ^2), \; m=0, 1, \dots$

• (6) Lemma 24.4 describes the multiplicities of the eigenvalues $(\pi (2m+1))^2.$

□

Note that the theorem can be proved without any use of formula (24.21), this approach was developed in [333].

We are going to use these results to prove that an inequality between the standard and Dirichlet eigenvalues holds true for all equilateral bipartite graphs and for the majority of the eigenvalues if the equilateral graph is not bipartite.

Equilateral Graphs and Dirichlet Eigenvalues

Consider the Laplace operator defined on the functions satisfying Dirichlet conditions at all vertices. This operator is equal to the orthogonal sum of Dirichlet Laplacians on $N$ independent intervals. Its spectrum will be denoted by $\mu _m^D (\Gamma ).$

Let us discuss whether the inequality

$$\displaystyle \begin{aligned} {} \lambda^{\mathrm{st}}_{n+1} (\Gamma) \leq \lambda^{D}_{n} (\Gamma), \end{aligned} $$

(24.32)

holds true, where $\lambda _m^{\mathrm {st}} (\Gamma )$ are the eigenvalues of the standard Laplacians on $\Gamma .$

Lemma 24.7

Let $\Gamma$ be an equilateral graph with $N$ edges of length one, then the spectrum of $L^D (\Gamma )$ is given by the eigenvalues $(\pi m )^2 , \; m= 1,2, \dots$ with multiplicity $N.$

The proof is straightforward since $L^D (\Gamma )$ is given as the orthogonal sum of $N$ Dirichlet Laplace operators on the intervals $[0,1].$ Each such Laplacian has the spectrum $(\pi m )^2 , \; m= 1,2, \dots .$

So the first $2 N$ Dirichlet eigenvalues are as follows

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle \lambda_1^D = \lambda_2^D = \dots = \lambda_N^D = \pi; \\[2mm] \displaystyle \lambda_{N+1}^D = \lambda_{N+2}^D = \dots = \lambda_{2N}^D = 2 \pi. \end{array} \end{aligned} $$

(24.33)

If the graph is not bipartite, then the first $2 N+1$ eigenvalues of the standard Laplacian satisfy the following inequalities (Theorem 24.6):

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle 0 = \lambda_1^{\mathrm{st}} < \lambda_2^{\mathrm{st}} \leq \dots \leq \lambda_M^{\mathrm{st}} < \lambda_{M+1}^{\mathrm{st}} = \dots = \lambda_N^{\mathrm{st}} = \pi; \\[2mm] \displaystyle \pi < \lambda_{N+1}^{\mathrm{st}} \leq \dots \leq \lambda_{N+M-1}^{\mathrm{st}} < \lambda_{N+M}^{\mathrm{st}} = \dots = \lambda_{2N+1}^{\mathrm{st}} = 2 \pi. \end{array} \end{aligned} $$

(24.34)

It follows that inequality (24.32) is satisfied for all $n = 1,2, \dots , N-1, N+1, \dots , 2N.$ Moreover, if $n = N$, then the inequality is violated.

Considering higher eigenvalues, we see that the structure repeats and inequality is violated only for $n = (2m+1) N.$ For all other eigenvalues the inequality holds.

Assume now that the graph $G$ is bipartite, then the first $2 N+1$ eigenvalues of the standard Laplacian satisfy:

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle 0 = \lambda_1^{\mathrm{st}} < \lambda_2^{\mathrm{st}} \leq \dots \leq \lambda_{M-1}^{\mathrm{st}} < \lambda_{M}^{\mathrm{st}} = \dots = \lambda_{N+1}^{\mathrm{st}} = \pi; \\[2mm] \displaystyle \pi < \lambda_{N+2}^{\mathrm{st}} \leq \dots \leq \lambda_{N+M-1}^{\mathrm{st}} < \lambda_{N+M}^{\mathrm{st}} = \dots = \lambda_{2N+1}^{\mathrm{st}} = 2 \pi. \end{array} \end{aligned} $$

(24.35)

We see that (24.32) holds for any $n = 1,2, \dots , 2N$ and hence for any $n \in \mathbb N.$ We have just proven the following theorem, first appeared in [349].

Theorem 24.8

Let$\Gamma$be a connected equilateral metric graph with$N$edges. Then inequality (24.32) between the eigenvalues for Dirichlet and standard Laplacians holds for any$n$if and only if the corresponding discrete graph$G$is bipartite. If$G$is not bipartite, then (24.32) holds for any$n \neq (2m+1) N , \, m = 0,1, \dots ,$moreover (24.32) is violated for$n = (2m+1) N , \, m = 0,1, \dots .$

The theorem implies that for not bipartite $G$ the portion of eigenvalues for which inequality is violated is $1/(2N) .$ Moreover, it is easy to see that the inequality turns into equality for $\lambda = (m \pi )^2, \, m= 1,2, \dots$ only. Approximately $2 (N-M)$ eigenvalues out of $2 N$ lead to equality in (24.32) ($2 (N-M)+1$ if $G$ is not bipartite and $2(N-M)+4$ if $G$ is bipartite).

24.4 Isospectrality of Normalised and Standard Laplacians

Let us discuss the relation between the isospectrality of standard Laplacians on equilateral graphs and the isospectrality of normalised Lapalcians on corresponding discrete graphs. It is clear that the standard Lapalcians are isospectral only if the generic eigenvalues of the normalised Laplacians coincide. On the other hand, isospectrality of standard Laplacians requires that the graphs have the same Euler characteristics, but normalised Laplacians could be isospectral on graphs with different number of cycles.

Theorem 24.9

Let $\Gamma _j$ and $G_j$ be certain equilateral connected metric graphs and their discrete counterparts, respectively $j =1,2$ . The standard Laplacians on $\Gamma _1$ and $\Gamma _2$ are isospectral if and only if the normalised Laplacians on $G_1$ and $G_2$ are isospectral and the Euler characteristics of $G_1$ and $G_2$ (equal to those of $\Gamma _1$ and $\Gamma _2$ ) are equal.

Proof

Assume that the standard Laplacians on $\Gamma _1$ and $\Gamma _2$ are isospectral. Theorem 24.1 states that the generic eigenvalues of $L_N (G_1)$ and $L_N(G_2)$ coincide. It remains to check the extremal eigenvalues of the normalised Laplacian. Theorem 24.6 implies that the Euler characteristics are equal, hence looking at the multiplicity of the eigenvalue $\lambda = ((2m+1)\pi )^2$ it is possible to determine whether the discrete graph is bipartite or not: multiplicity $\beta _1+1 = 2-\chi$ implies that $G$ is bipartite, multiplicity $\beta _1 -1 = - \chi$—that $G$ is not bipartite. In the first case we have $\mu _M (L_N(G)) = 2$, in the second case $\mu _M (L_N(G)) < 2$ is generic. The multiplicities of $\lambda = 0$ are given by the number of connected components and are equal as well. We conclude that $L_N (G_1)$ and $L_N(G_2)$ are isospectral.

Assume now that the two normalised Laplacians are isospectral and the graphs have the same Euler characteristics. As before we conclude that all generic eigenvalues are equal and focus on the extremal eigenvalues. The Euler characteristic determines the multiplicity of $\lambda = (2m \pi )^2$. If $\mu =2$ is an eigenvalue of the normalised Laplacians, then the graphs are bipartite and the multiplicity of $\lambda = ((2m+1)\pi )^2$ is $\beta _1 +1 = - \chi +2$, otherwise the multiplicity is $\beta _1 -1 = - \chi$. The multiplicities of $\lambda = 0$ are determined by the number of connected components and are equal again. It follows that the standard Laplacians are isospectral. □

The theorem implies that to get equilateral isospectral metric graphs it is enough to check all families of discrete graphs with the same spectrum of the normalised Laplacian and then keep only graphs with the same Euler characteristic. The list of graphs leading to isospectral normalised Laplacians can be found e.g. in [116, 479] (see also [141, 342]).

Analysing the proof of the theorem one may notice that equality of Euler characteristics was used only to get equal multiplicities of the extremal eigenvalues. The reason for that is that for equilateral graphs formula (9.1) can be proven directly using the structure of the spectrum prescribed by Theorem 24.6. Particular values of the generic eigenvalues plays no role.

Proof of Formula (9.1) for Equilateral Graphs

Assume first that the graph $\Gamma$ is connected. Let us denote by $\omega _j^2 , j = 1,2,\ldots ,J$ the eigenvalues of $L^{\mathrm {st}} (\Gamma )$ inside the interval $(0, 2 \pi )$. Then the limit on the right hand side of (9.1) can be written as follows:

$$\displaystyle \begin{aligned} {} \begin{array}{cl} & \displaystyle 2 - 2 \lim_{t\rightarrow \infty} \sum_{k_n\neq 0} \frac{1- 2 \cos k_n/t + \cos 2 k_n/t}{(k_n/t)^2} \\ & \\ = & \displaystyle 2 - 2 (1+\beta_1) \lim_{t\rightarrow \infty} \sum_{m=1}^\infty \frac{1-2 \cos 2 \pi m/t + \cos 4 \pi m/t}{(2 \pi m/t)^2} \\ & \displaystyle - 2 \lim_{t\rightarrow \infty} \sum_{m=0}^\infty \sum_{j=1}^J \frac{1-2 \cos (\omega_j+ 2 \pi m)/t + \cos 2 (\omega_j+ 2 \pi m)/t }{((\omega_j+ 2 \pi m)/t)^2} , \end{array} \end{aligned} $$

(24.36)

where we used that all points $k= 2 \pi m, m= 1,2,\ldots$ have multiplicity $1+\beta _1$ and the point $k= 0$ has multiplicity $1$. The first limit can be calculated using formula (9.5).

To calculate the second limit let us use that the points $\omega _j$ are situated symmetrically with respect to the center of the interval $(0, 2 \pi )$

$$\displaystyle \begin{aligned} {} \begin{array}{cl} & \displaystyle \sum_{m=0}^\infty \sum_{j=1}^J \frac{1-2 \cos (\omega_j+ 2 \pi m)/t + \cos 2 (\omega_j+ 2 \pi m)/t }{((\omega_j+ 2 \pi m)/t)^2} \\ &= \displaystyle \frac{1}{2} \sum_{m=0}^\infty \sum_{j=1}^{J} \left\{ \frac{1-2 \cos (m+\omega_j/2 \pi)/(t/2 \pi) + \cos 2 (m+\omega_j/2 \pi)/(t/2 \pi) }{((m+ \omega_j/2 \pi)/(t/2 \pi))^2} \right. \\ & \displaystyle \left. + \frac{1-2 \cos (m+(2 \pi-\omega_j)/2 \pi)/(t/2 \pi) + \cos 2 (m+(2 \pi-\omega_j)/2 \pi)/(t/2 \pi) }{((m+ (2 \pi-\omega_j)/2 \pi)/(t/2 \pi))^2} \right\} \\ &= \displaystyle \frac{1}{2} \sum_{m \in \mathbb Z} \sum_{j=1}^J \frac{1-2 \cos (m+\omega_j/2 \pi)/(t/2 \pi) + \cos 2 (m+\omega_j/2 \pi)/(t/2 \pi) }{((m+ \omega_j/2 \pi)/(t/2 \pi))^2}. \end{array} \end{aligned} $$

(24.37)

We are going to prove that the last sum is equal to zero using the formula

$$\displaystyle \begin{aligned} {} \sum_{m \in \mathbb Z} \frac{e^{i(m+\alpha)x}}{(m+\alpha)^2} = \frac{2\pi e^{2\pi i \alpha}}{1-e^{2\pi i \alpha}} x - \frac{(2\pi)^2 e^{2\pi i \alpha}}{(1-e^{2\pi i \alpha})^2} , \; \; \alpha \notin \mathbb Z. \end{aligned} $$

(24.38)

To prove this formula one may exploit the following idea: find a linear function $f(x) = ax + b, 0 \leq x \leq 2\pi$, such that the series on the left hand side of the formula is exactly the Fourier series for $f$ in the orthogonal basis $e^{i(n+\alpha ) x} .$ The function $f$ is represented by the following almost everywhere converging Fourier series

$$\displaystyle \begin{aligned} {} f (x) = \frac{1}{2\pi} \sum_{m \in \mathbb Z} f_m e^{i(m+\alpha)x}, \end{aligned} $$

(24.39)

where

$$\displaystyle \begin{aligned} {} f_m = \int_0^{2\pi} (ax+b) e^{-i(m+\alpha)x} dx. \end{aligned} $$

(24.40)

The function $f$ may be chosen equal to

$$\displaystyle \begin{aligned} f(x) = \frac{e^{2\pi i \alpha}}{1- e^{2\pi i \alpha}} x - \frac{2\pi e^{2\pi i \alpha}}{(1-e^{2\pi i \alpha})^2} .\end{aligned}$$

Then the Fourier coefficients are given by

$$\displaystyle \begin{aligned} f_m = \int_0^{2\pi} \left( \frac{e^{2\pi i \alpha}}{1- e^{2\pi i \alpha}} x - \frac{2\pi e^{2\pi i \alpha}}{(1-e^{2\pi i \alpha})^2} \right) e^{-i(m+\alpha)x} dx = \frac{1}{(m+\alpha)^2} .\end{aligned}$$

Thus formula (24.38) is proven and it implies in particular that

$$\displaystyle \begin{aligned} {} \sum_{m\in \mathbb Z} \frac{1-2 e^{i(m+\alpha) x} + e^{2i (m+\alpha) x}}{(m+\alpha)^2} = 0, \; \; \mbox{provided } \alpha \notin \mathbb Z . \end{aligned} $$

(24.41)

Indeed, using (24.38) we have:

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle \sum_{m\in \mathbb Z} \frac{1-2 e^{i(m+\alpha) x} + e^{2i (m+\alpha) x}}{(m+\alpha)^2} \\ \displaystyle = \sum_{m\in \mathbb Z} \frac{1}{(m+\alpha)^2} - 2 \sum_{m \in \mathbb Z} \frac{e^{i(m+\alpha)x}}{(m+\alpha)^2} + \sum_{m\in \mathbb Z} \frac{e^{2i(m+\alpha)x}}{(m+\alpha)^2} \\ \displaystyle = - \frac{(2\pi)^2 e^{2\pi i \alpha}}{(1-e^{2\pi i \alpha})^2} -2 \left( - \frac{(2\pi)^2 e^{2\pi i \alpha}}{(1-e^{2\pi i \alpha})^2} + \frac{2\pi e^{2\pi i \alpha}}{1-e^{2\pi i \alpha}} x \right) \\ \displaystyle\quad - \frac{(2\pi)^2 e^{2\pi i \alpha}}{(1-e^{2\pi i \alpha})^2} + \frac{2\pi e^{2\pi i \alpha}}{1-e^{2\pi i \alpha}} 2 x \\ \displaystyle = 0 , \end{array}\end{aligned}$$

where to calculate the first sum we used that the series (24.39) at $x= 0$ converges to $\frac {1}{2}(f(+0) + e^{-2\pi 1 \alpha } f(2\pi -0)) .$ It turns out that

$$\displaystyle \begin{aligned} {} \sum_{m\in \mathbb Z} \frac{1-2 \cos (m+\alpha) x + \cos 2 (m+\alpha) x}{(m+\alpha)^2} = 0, \; \; \mbox{provided } \alpha \notin \mathbb Z , \end{aligned} $$

(24.42)

and therefore the second sum in (24.36) (the sum (24.37) ) is equal to zero. Finally we get

$$\displaystyle \begin{aligned} \displaystyle 2 - 2 \lim_{t\rightarrow \infty} \sum_{k_n\neq 0} \frac{1- 2 \cos k_n/t + \cos 2 k_n/t}{(k_n/t)^2} = 2 - 2 (1+ \beta_1) \frac{1}{2} + 0 = 1-\beta_1 = \chi.\end{aligned}$$

Now it is straightforward to generalise this result to include not connected graphs to get (9.1). □

Thus we have proven the formula for the Euler characteristic for equilateral graphs without any use of the trace formula. It might be important to find a similar proof for arbitrary graphs. Such an alternative to the trace formula approach may provide a new insight on the structure of the spectral asymptotics for $L^{\mathrm {st}} (\Gamma ).$

24.5 Spectral Gap for Discrete Laplacians

The goal of this section is to derive elementary estimates for the spectral gaps of combinatorial and normalised Laplacians based on the methods developed for standard Laplacians on metric graphs (see Chaps. 12 and 13). Since the spectral theory of discrete Laplacians has a long history, one may say that we return back and check how our ideas work in the discrete case. Our presentation here shall be limited since we are planning to discuss these questions in full detail in our forthcoming book with Delio Mugnolo and James Kennedy. We restrict our presentation to discussing what happens when edges are added and vertices are cut. It turns out that the answers have simplest form for combinatorial and normalised Laplacians, respectively.

Adding Edges: Combinatorial Laplacian

Our aim is to understand how the spectral gap—the difference between the lowest two eigenvalues $\mu _2 - \mu _1$—changes when the discrete graph is getting larger. We are interested in “small” perturbation of the graph like adding one edge between two existing vertices or adding one pendant edge. All results will be proved for the combinatorial Laplacian $L(G)$ given by (24.1), also their generalisation for the normalised Laplacian is often straightforward.

The following statement is a direct analog of Theorem 12.11 for combinatorial graphs:

Proposition 24.10

Let$G$be a connected discrete graph and let$G'$be the discrete graph obtained from G by adding one edge between the vertices$m_1$and$m_2$. Let L denote thecombinatorialLaplacian defined by (24.1). Then the following holds:

• (1) The first excited eigenvalues satisfy the inequality:

  $$\displaystyle \begin{aligned} \mu_2(L(G))\leq \mu_2(L(G')).\end{aligned}$$

• (2) The equality$\mu _2(L(G)) = \mu _2(L(G'))$holds if and only if the second eigenfunction$\psi _2^G$on the graph G may be chosen attaining equal values at the vertices$m_1$and$m_2$

  $$\displaystyle \begin{aligned} \psi_2^G(m_1) = \psi_2^G(m_2).\end{aligned}$$

Proof

The first statement follows from the fact that

$$\displaystyle \begin{aligned} L(G') - L(G) = \left( \begin{array}{ccccc} & \vdots & & \vdots & \\ \ldots & 1 & \ldots & -1 & \ldots\\ & \vdots & & \vdots & \\ \ldots & -1 & \ldots & 1 & \ldots\\ & \vdots & & \vdots & \end{array} \right) \end{aligned} $$

(24.43)

is a matrix with just four non-zero entries. It is easy to see that the matrix is positive semi-definite, since the eigenvalues are 0 (with the multiplicity $M-1$) and 2 (simple eigenvalue) and therefore $L(G')-L(G)\geq 0$ which implies the first statement.

To prove the second assertion let us recall that $\mu _2(L(G'))$ can be calculated using the Rayleigh quotient

$$\displaystyle \begin{aligned} \mu_2 (L(G)) = \min_{\psi \bot {\mathbf 1}} \frac{\langle \psi, L(G) \psi \rangle}{\langle \psi, \psi \rangle} \leq \min_{\psi \bot {\mathbf 1}} \frac{\langle\psi, L(G')\psi\rangle}{\langle \psi,\psi\rangle}= \mu_2 (L(G')). \end{aligned}$$

Here the trial function $\psi$ should be chosen orthogonal to the ground state, i.e. having mean value zero. We have equality in the last formula if and only if $\psi$ minimizing the first and the second quotients can be chosen such that $(L(G')- L(G))\psi = 0,$ i. e. $\psi (m_1) = \psi (m_2)$. In other words, the vector $(\psi (m_1), \psi (m_2))$ can be chosen orthogonal to the eigenvector $(1,-1)$ corresponding to the nonzero eigenvalue of the matrix $\left ( \begin {array}{cc} 1 & -1 \\ -1 & 1 \end {array} \right ).$ □

Next we are interested in what happens if we add a pendant edge, that is an edge connected to the graph at one already existing node.

Proposition 24.11

Let G be a connected discrete graph and let $G'$ be another graph obtained from G by adding one vertex and one edge between the new vertex and the vertex $m_1$ . Then the following holds:

• (1) The first excited eigenvalues of the combinatorial Laplacian satisfy the following inequality:

  $$\displaystyle \begin{aligned} \mu_2(L(G))\geq \mu_2(L(G')).\end{aligned}$$

• (2) The equality $\mu _2(L(G)) = \mu _2(L(G'))$ holds if and only if every eigenfunction $\psi _2^G$ corresponding to $\mu _2 (L(G))$ on $G$ is equal to zero at $m_1$

  $$\displaystyle \begin{aligned} \psi_2^G (m_1) = 0.\end{aligned}$$

Proof

Let us define the following vector on $G'$:

$$\displaystyle \begin{aligned} \varphi(n):=\left\{ \begin{array}{ll} \psi_2^G(n), & \text{on } G,\\ \psi_2^G(m_1)& \text{on } G'\backslash G. \end{array}\right.\end{aligned}$$

This vector is not orthogonal to the zero energy eigenfunction ${\mathbf 1} \in \mathbb C^{M+1}$, where we keep the same notation ${\mathbf 1}$ for the vector build up of ones now on $G'$. Therefore consider the following nonzero vector $\gamma$, which is obtained from $\varphi$ by adding a certain constant c

$$\displaystyle \begin{aligned} \gamma(n):=\varphi(n)+c.\end{aligned}$$

Here c is chosen so that the orthogonality condition in $l_2(G') = \mathbb C^{M+1}$ holds

where $M' = M+1$ is the number of vertices in $G'$. This implies

$$\displaystyle \begin{aligned} {} c = -\frac{\psi_2^G(m_1)}{M'}.\end{aligned} $$

(24.44)

Using this vector the following estimate on the first exited eigenvalue holds

$$\displaystyle \begin{aligned} {} \mu_2(L(G'))&\leq \frac{\langle L (G') \gamma,\gamma\rangle_{l_2(G')}}{\|\gamma\|{}^2_{l_2(G')}} \\ &= \frac{\langle L (G) \psi_2^G, \psi_2^G\rangle_{l_2(G)}}{\|\psi_2^G\|{}_{l_2(G)}^2+c^2 M+|\psi_2^G(m_1)+c|{}^2}\leq \mu_2(L(G)), \end{aligned} $$

(24.45)

where we took into account (24.10). The last inequality follows from the fact that

$$\displaystyle \begin{aligned} \langle L (G) \psi_2^G, \psi_2^G\rangle_{l_2(G)} = \mu_2(L(G)) \|\psi_2^G\|{}^2_{l_2 (G)},\end{aligned}$$

and

$$\displaystyle \begin{aligned} \|\psi_2^G\|{}_{l_2(G)}^2+c^2 M+|\psi_2^G(m_1)+c|{}^2 \geq \|\psi_2^G\|{}_{l_2(G)}^2.\end{aligned}$$

Note that we have equality if and only if $c=0$ and $|\psi _2^G(m_1) + c|{ }^2=0$ which implies $\psi _2^G(m_1)=0$ (even without assuming (24.44)). If there exists an eigenfunction $\psi _2^G$, such that $\psi _2^G (m_1) \neq 0$, then the inequality in (24.45) is strict and we get

$$\displaystyle \begin{aligned} \mu_2 (L(G)) > \mu_2 (L(G')). \end{aligned}$$

□

We see that the first excited eigenvalue has a tendency to decrease if a pendant edge is attached to a graph. It is clear from the proof that gluing of any connected graph (instead of one edge) would lead to the same result, provided there is just one contact vertex. If the number of contact vertices is larger, then the spectral gap may increase as shown in Proposition 24.10.

Note that a different proof of the first part of Proposition 24.10 may be found in [221], Corollary 3.2. In the same paper, a bit weaker claim related to the first part of Proposition 24.11 is provided as Property 3.3.

Splitting Vertices: Normalised Laplacians

The first question we would like to answer is what happens to the spectral gap when a vertex in a graph $G$ is chopped into two vertices. The graph $\hat {G}$ obtained in this way has one vertex more than $G$. We shall answer this question for the normalised Laplacian $L_N$. Recall that this question has already been addressed for standard Laplacians on metric graphs in Chap. 12.

The same question for discrete graphs is slightly more sophisticated since chopping a vertex into two increases the number of vertices and therefore changes the Hilbert space where the discrete Laplacian is defined.

Proposition 24.12

Let G be a connected discrete graph and let $\hat {G}$ be another graph obtained from G by chopping one vertex into two, then the first excited eigenvalues of the normalised Laplacian satisfy the following inequality:

$$\displaystyle \begin{aligned} {} \mu_2(L_N(G))\geq \mu_2(L_N(\hat{G})).\end{aligned} $$

(24.46)

Proof

The theorem is an easy corollary of the following two theorems:

• Theorem 12.9 describing behaviour of the spectral gap when two vertices in a metric graph are joined together;

• Theorem 24.1 describing the relation between the generic eigenvalues of the standard Laplacian on an equilateral metric graph $\Gamma$ and of the normalised Laplacian on the corresponding discrete graph $G$.

Nevertheless, we present here an alternative direct proof.

Let us denote by $V^0$ the vertex that is chopped and by ${V^0}'$ and ${V^0}''$ the two new vertices in $\hat {G}$, so that the following relation for the corresponding degrees holds $d_0 = d_0^{\prime } + d_0^{\prime \prime }.$ Consider the eigenvector $\psi _2$ corresponding to $\mu _2(L_N(G)).$ Let us introduce the following vector on $\hat {G}$

$$\displaystyle \begin{aligned} \hat{u} (V^m) = \left\{ \begin{array}{ll} \psi_2 (V^m), & m \neq 0, \\ a', & V^m = {V^0}', \\ a'', & V^m = {V^0}''. \end{array} \right. \end{aligned} $$

(24.47)

The parameters $a'$ and $a''$ will be chosen so that the Rayleigh quotient does not change. It is natural to introduce the notation $\psi _2 (V^0) = a.$ Consider first the difference between the quadratic forms given by (24.10) on $G$ and $\hat {G}$ respectively, assuming without loss of generality that $\psi _2$ is real valued

$$\displaystyle \begin{aligned} \begin{array}{cl} & \displaystyle \langle \hat{L}_N \hat{u}, \hat{u} \rangle_{\ell_2(\hat{G})} - \langle L_N \psi_2, \psi_2 \rangle_{\ell_2 (G)} \\ &= \displaystyle \sum_{V^n \sim_{\hat{G}} {V^0}'} \left( \frac{1}{\sqrt{d_n}} \psi_2 (V^n) - \frac{1}{\sqrt{d_0^{\prime}}} a' \right)^2 + \sum_{V^n \sim_{\hat{G}} {V^0}''} \left( \frac{1}{\sqrt{d_n}} \psi_2 (V^n) - \frac{1}{\sqrt{d_0^{\prime\prime}}} a'' \right)^2 \\ & \displaystyle - \sum_{V^n \sim_{G} V^0} \left( \frac{1}{\sqrt{d_n}} \psi_2 (V^n) - \frac{1}{\sqrt{d_0}} a \right)^2 \\ & = \displaystyle (a')^2 + (a'')^2 - a^2 - 2 \frac{1}{\sqrt{d_0^{\prime}}} a' \sum_{V^n \sim_{\hat{G}} {V^0}'} \frac{1}{\sqrt{d_n}} \psi_2 (V^n) \\ &\quad - 2 \frac{1}{\sqrt{d_0^{\prime\prime}}} a'' \sum_{V^n \sim_{\hat{G}} {V^0}''} \frac{1}{\sqrt{d_n}} \psi_2 (V^n) \\ & \displaystyle + 2 \frac{1}{\sqrt{d_0}} a \sum_{V^n \sim_{G} V^0} \frac{1}{\sqrt{d_n}} \psi_2 (V^n) . \end{array} \end{aligned} $$

We use now that $\psi _2$ is an eigenvector and in particular the eigenvector equation holds for $m= 0$

$$\displaystyle \begin{aligned} a - \frac{1}{\sqrt{d_0}} \sum_{V^n \sim_{G} V^0} \frac{1}{\sqrt{d_n}} \psi_2 (V^n) = \mu_2(L_N(G)) a\end{aligned}$$

$$\displaystyle \begin{aligned} \Rightarrow \sum_{V^n \sim_{G} V^0} \frac{1}{\sqrt{d_n}} \psi_2 (V^n) = \sqrt{d_0} (1-\mu_2(L_N(G))) a .\end{aligned}$$

We continue with the difference between the quadratic forms

$$\displaystyle \begin{aligned} \begin{array}{cl} & \displaystyle \langle \hat{L}_N \hat{u}, \hat{u} \rangle_{\ell_2(\hat{G})} - \langle L_N \psi_2, \psi_2 \rangle_{\ell_2 (G)} \\ = & \displaystyle (a')^2 + (a'')^2 - a^2 + 2 (1-\mu_2) a^2 - 2 \frac{\sqrt{d_0}}{\sqrt{d_0^{\prime\prime}}} (1-\mu_2) a a'' \\ & \displaystyle + 2 \left(- \frac{a'}{\sqrt{d_0^{\prime}}} + \frac{a''}{\sqrt{d_0^{\prime\prime}}} \right) \sum_{V^n \sim_{\hat{G}} {V^0}'} \frac{1}{\sqrt{d_n}} \psi_2 (V^n) . \end{array} \end{aligned} $$

(24.48)

Since it is hard to control the sum $\sum _{V^n \sim _{\hat {G}} {V^0}'} \frac {1}{\sqrt {d_n}} \psi _2 (V^n)$, let us assume that the coefficient in front of it vanishes:

$$\displaystyle \begin{aligned} {} - \frac{a'}{\sqrt{d_0^{\prime}}} + \frac{a''}{\sqrt{d_0^{\prime\prime}}} = 0 \end{aligned} $$

(24.49)

Under this condition the difference between the quadratic forms is given by

$$\displaystyle \begin{aligned} \displaystyle \langle \hat{L}_N \hat{u}, \hat{u} \rangle_{\ell_2(\hat{G})} - \langle L \psi_2, \psi_2 \rangle_{\ell_2 (G)} &= \displaystyle (a')^2 + (a'')^2 - a^2 + 2 (1-\mu_2) a^2 \\ &\quad - 2 \frac{\sqrt{d_0}}{\sqrt{d_0^{\prime\prime}}} (1-\mu_2) a a''. \end{aligned} $$

(24.50)

The second equation on $a', a''$ (in addition to (24.49))is obtained by requiring that $\hat {u}$ is orthogonal to the ground state eigenvector $\hat {\psi }_1 (V^n) = \sqrt {d_n}$

$$\displaystyle \begin{aligned} \langle \hat{u}, \hat{\psi}_1 \rangle_{\ell_2 (\hat{G})} = \langle \psi_2, \psi_1 \rangle_{\ell_2(G)} - \sqrt{d_0} a + \sqrt{d_0^{\prime}} a' + \sqrt{d_0^{\prime\prime}} a'', \end{aligned} $$

(24.51)

where $\psi _1$ is the ground state for $G.$ Taking into account that $\langle \psi _2, \psi _1 \rangle _{\ell _2(G)} = 0$ we shall require that

$$\displaystyle \begin{aligned} {} \sqrt{d_0^{\prime}} a' + \sqrt{d_0^{\prime\prime}} a'' = \sqrt{d_0} a. \end{aligned} $$

(24.52)

The system of linear equations (24.49) and (24.52) is easy to solve

$$\displaystyle \begin{aligned} \left\{ \begin{array}{ccl} a' & = & \displaystyle \frac{\sqrt{d_0^{\prime}}}{\sqrt{d_0}} a, \\ a'' & = & \displaystyle \frac{\sqrt{d_0^{\prime\prime}}}{\sqrt{d_0}} a. \end{array} \right. \end{aligned} $$

(24.53)

It follows in particular that

$$\displaystyle \begin{aligned} {} (a')^2 + (a'')^2 = a^2. \end{aligned} $$

(24.54)

With these values of $a'$ and $a''$ the vector $\hat {u}$ is admissible and its Rayleigh quotient gives an upper estimate for the $\mu _2(L_N(\hat {G}))$

$$\displaystyle \begin{aligned} \mu_2(L_N(\hat{G})) \leq \frac{ \langle \hat{L}_N \hat{u}, \hat{u} \rangle_{\ell_2 (\hat{G})}}{\parallel \hat{u} \parallel^2_{\ell_2 (\hat{G})}}. \end{aligned} $$

(24.55)

It turns out that with chosen $a'$ and $a''$ both the quadratic form and the norm remain the same (24.54):

$$\displaystyle \begin{aligned} \left\{ \begin{array}{lcl} \displaystyle \langle \hat{L}_N \hat{u}, \hat{u} \rangle_{\ell_2(\hat{G})} & = & \displaystyle \langle L_N \psi_2, \psi_2 \rangle_{\ell_2 (G)} ,\\ \displaystyle \parallel \hat{u} \parallel^2_{\ell_2 (\hat{G})} & = & \displaystyle \parallel \psi_2 \parallel^2_{\ell_2 (G)}. \end{array} \right. \end{aligned}$$

Since $\psi _2$ is an eigenfunction, it holds $\displaystyle \langle L _N\psi _2, \psi _2 \rangle _{\ell _2 (G)} = \mu _2(L_N(G)) \parallel \psi _2 \parallel ^2_{\ell _2 (G)}$ and (24.46) follows. □

Remark

if $\psi _2 (V^0) = 0$, then the constructed function $\hat {u}$ is an eigenvector for the Laplacian on $\hat {G}$ with the same eigenvalue. Most probably this statement can be generalised.

We are ready to prove the main result of this section

Theorem 24.13 (Fiedler)

Let $G$ be a connected discrete graph, then the spectral gap for the normalised Laplacian $L_N (G)$ satisfies the following lower estimate

$$\displaystyle \begin{aligned} {} \mu_2(L_N(G)) \geq 1 - \cos (\frac{\pi}{N}), \end{aligned} $$

(24.56)

where $N$ is the number of edges in $G.$

Proof

We start by doubling all edges in the original graph $G$. Let us denote the corresponding graph by $G^2$. The new normalised Laplacian $L_N (G^2)$ just coincides with $L_N(G)$ and therefore $\mu _2 (L_N(G^2 )) = \mu _2(L_N(G)).$

All vertices in $G^2$ have even degrees and therefore there exists an Eulerian path $P$ - any closed path going along each edge precisely once. This path can be seen as a loop obtained by chopping vertices in $G^2.$ As we have proven (Proposition 24.12) chopping the vertices does not increase the spectral gap. Thus we have

$$\displaystyle \begin{aligned} \mu_2 (L_N(P)) \leq \mu_2 (L_N(G^2)) = \mu_2(L_N(G)). \end{aligned} $$

(24.57)

To prove the theorem it remains to note that

$$\displaystyle \begin{aligned} {} \mu_2 (L_N(P)) = 1- \cos \frac{\pi}{N}. \end{aligned} $$

(24.58)

It is clear that every eigenfunction of $L_N(P)$ is quasi invariant:

$$\displaystyle \begin{aligned} \psi (V^n) = \psi (V^1) z^n,\end{aligned}$$

where $z$ is any $2N$-th root of $1:$$z_j = \exp \{i \frac {\pi }{N} j \}, j= 0,1, \dots , 2N-1.$ Substituting into the eigenfunction equation

$$\displaystyle \begin{aligned} (L_N \psi ) (V^n) = \psi (V^n) - \frac{1}{2} \left( z + 1/z \right) \psi (V^n) = \mu \psi (V^n)\end{aligned}$$

gives the following values of $\mu$

$$\displaystyle \begin{aligned} \mu (L_N(P)) = 1 - \cos \frac{\pi}{N} (j-1), \; \; j = 0, 1 , 2, \dots , N,\end{aligned}$$

where all eigenvalues except the lowest and the largest, that is $\mu = 0, 2$, have multiplicity $2.$ (All together there are $2N$ eigenvalues.) The two lowest eigenvalues are

$$\displaystyle \begin{aligned} \mu_1 (L_N(P)) = 0 \; \, \mbox{and} \; \, \mu_2 (L_N(P)) = 1- \cos \frac{\pi}{N}\end{aligned}$$

and the spectral gap is given by (24.56) □

Let us prove that the estimate is sharp. Consider the chain graph $G_N$ formed by $M+1$ vertices $V^0, V^1, \dots , V^N$ consequently connected by $N$ edges (like a chain). Then the first eigenfunction for the normalised Laplacian is given by

$$\displaystyle \begin{aligned} \psi_2 (V^m) = \left\{ \begin{array}{ll} 1/\sqrt{2}, & m = 0, \\ \displaystyle \cos \frac{\pi}{N} m, & m = 1,2, \dots, N-1, \\ - 1/\sqrt{2}, & m = N. \end{array} \right. \end{aligned} $$

(24.59)

The corresponding eigenvalue is $\mu _2 = 1 - \cos \pi /N$, where $N$ is the number of edges.

Problem 104

Calculate the spectrum of the complete graph $K_3$ formed by three vertices connected. Use both combinatorial and normalised Laplacians. What is the connection between their spectra?

Problem 105

Generalise the previous problem and calculate the spectrum of an arbitrary complete graph $K_M.$ What is the reason that the spectrum is highly degenerate? Both combinatorial and normalised Laplacians should be considered.

Problem 106

Prove counterparts of Propositions 24.10 and 24.11, now for the normalised Laplacian.

Problem 107

What is the analog of Theorem 24.13 for the combinatorial Laplacian?