Trace Formula and Inverse Problems

Abstract

Our aim in this section is to establish an explicit formula allowing one to calculate the Euler characteristic of the metric graph directly from the spectrum of the standard Laplace operator.

9.1 Euler Characteristic for Standard Laplacians

Our aim in this section is to establish an explicit formula allowing one to calculate the Euler characteristic of the metric graph directly from the spectrum of the standard Laplace operator. Formula (8.21) shows that the Euler characteristic is determined by the spectrum alone: the left hand side is defined by $k_n$, whereas the right hand side is a sum of $\chi$, a delta function at the origin and a series containing delta functions with the supports at the lengths of periodic paths. It is clear that the lengths of periodic paths cannot be small—each path contains at least one edge, and therefore the constant term $\chi$ is uniquely determined.

The main idea how to get an explicit formula is to apply (8.21) to a test function with the support on the interval $[0, \mathrm {min}\; \{ \ell _n \}_{n=1}^N].$ Since we do not know the exact length of the shortest edge a priori, we are going to consider a sequence of test functions having smaller and smaller support.

Theorem 9.1

Let $\Gamma$ be a compact metric graph and $L^{\mathrm {st}} (\Gamma )$ be the standard Laplace operator. Then the Euler characteristic $\chi (\Gamma )$ is uniquely determined by the spectrum $\{ \lambda _n \}$ of the Laplacian $L^{\mathrm {st}} (\Gamma )$

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} \chi & = & \displaystyle 2 m_s(0) + 2 \lim_{t\rightarrow \infty} \sum_{k_n \neq 0 } \cos k_n/t \left( \frac{\sin k_n/2t}{k_n/2t} \right)^2\\ && \\ & = & \displaystyle 2 m_s(0) - 2 \lim_{t \rightarrow \infty} \sum_{k_n \neq 0} \frac{ 1- 2 \cos k_n/t + \cos 2 k_n/t}{(k_n/t)^2} , \end{array} \end{aligned} $$

(9.1)

where $k_n = \sqrt {\lambda _n} > 0.$

Proof

Our proof is based on the trace formula (8.20). The idea is very simple: find a test function $\varphi$ such that $\hat {\mu } [\varphi ] = \chi$. Then $\mu [\hat {\varphi }]$ is also equal to $\chi$ and provides the desired formula connecting $\chi$ and the spectrum.

Consider the function $\varphi$ defined by (Fig. 9.1)

$$\displaystyle \begin{aligned} {} \varphi ( \ell) = \left\{ \begin{array}{ll} \ell , & 0 \leq \ell \leq 1; \\ 2- \ell, & 1 \leq \ell \leq 2; \\ 0, & \mbox{otherwise}. \end{array} \right. \end{aligned} $$

(9.2)

Fig. 9.1

The test function $\varphi (\ell )$

This function and any scaled function $\varphi _t (x) = t \varphi (tx)$ are normalised as:

$$\displaystyle \begin{aligned} \int_{-\infty}^{+\infty} \varphi_t (x) dx = 1.\end{aligned}$$

To get the Euler characteristic we need to scale the test function so that

$$\displaystyle \begin{aligned} \min \{ \ell_j \} > 2/t\end{aligned}$$

holds. When applying $\hat {\mu }$ to the test function $\varphi _t(x)$ the contribution from all delta functions is zero and we have

$$\displaystyle \begin{aligned} \chi (\Gamma) = \lim_{t \rightarrow \infty} \hat{\mu}[\varphi_t], \end{aligned} $$

(9.3)

where we used the limit as the length of the shortest edge may be unknown.

To calculate $\mu [\hat {\varphi }]$ we need the Fourier transform of the test function

$$\displaystyle \begin{aligned} \hat{\varphi}_t(k) = e^{i k/t} \left( \frac{\sin k/2t}{k/2t} \right)^2 .\end{aligned}$$

Applying the spectral measure $\mu$ to the Schwartz test function $\hat {\varphi }_t$ we obtain

$$\displaystyle \begin{aligned} \begin{array}{ccl} \chi (\Gamma) & = & \displaystyle \lim_{t \rightarrow + \infty} \hat{\mu} [\varphi_t] \\ & = & \displaystyle \lim_{t \rightarrow + \infty} \mu [\hat{\varphi}_t] \\ & = & \displaystyle 2 m_s (0) + 2 \lim_{t \rightarrow + \infty} \sum_{k_n \neq 0} \cos k_n/t \left( \frac{\sin k_n/2t}{k_n/2t} \right)^2. \end{array} \end{aligned}$$

The second formula (9.1) follows from elementary trigonometry. □

Weyl asymptotic law (4.15) implies that $k_n$ grow linearly with $n$ and therefore the series in (9.1) is absolutely convergent. But the limit and summation signs cannot be exchanged. In fact there is no necessity to take the limit in formula (9.1): it is enough to consider sufficiently large values of $t.$ The series is equal to a constant for all $t > 2/\mathrm {min}\, \{ \ell _j \}.$ This is another interesting feature of the derived formula.

In Chap. 24 we are going to provide an alternative proof of the formula for Euler characteristic for equilateral graphs [333]. That proof does not use the trace formula and is based on the fact that the spectrum of the standard Laplacian on a metric graph with integer lengths is periodic in the k-scale.

It might be interesting to understand relations of the derived formula to indices of differential operators following [230].

We would like to present a few explicit examples illustrating formulas (9.1).

(1) Single Interval

Let the graph coincide with the interval $[0,\pi ]$ (with separated endpoints) (Fig. 9.2). The Euler characteristic is $\chi = 1.$ The spectrum of $L^{\mathrm {st}} (\Gamma _{(1.1)} )$ is $\Sigma (L) = \{ n^2, n=0,1,2,\ldots \}.$ Substituting $k_n = n, \, n=0,1,2,\ldots$ into formula (9.1) we get

$$\displaystyle \begin{aligned} {} \chi = 2- 2 \lim_{t \rightarrow \infty} \sum_{n=1}^\infty \frac{ 1- 2 \cos n/t + \cos 2n/t}{(n/t)^2} = 2- 1 = 1, \end{aligned} $$

(9.4)

where we used

$$\displaystyle \begin{aligned} {} \sum_{n=1}^{\infty} \frac{1-2 \cos n/t + \cos 2n/t}{(n/t)^2} = \frac{1}{2}. \end{aligned} $$

(9.5)

This formula can be proven using the sum ((1.443.3) from [245])

$$\displaystyle \begin{aligned} \sum_{m=1}^\infty \frac{\cos mx}{m^2} = \frac{\pi^2}{6} - \frac{\pi x}{2} + \frac{x^2}{4}, \; \; x \in [0, 2 \pi]\end{aligned}$$

leading to

$$\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{1-2 \cos n/t + \cos 2n/t}{(n/t)^2} = t^2 \left( \sum_{n=1}^\infty \frac{1}{n^2} - 2 \sum_{n=1}^\infty \frac{\cos n \frac{1}{t}}{n^2} + \sum_{n=1}^\infty \frac{\cos n \frac{2}{t}}{n^2}\right)\end{aligned}$$

$$\displaystyle \begin{aligned} = t^2 \left\{ \frac{\pi^2}{6} - 2 \left(\frac{\pi^2}{6} - \frac{\pi}{2} \frac{1}{t} + \frac{1}{4} \frac{1}{t^2} \right) + \frac{\pi^2}{6} - \frac{\pi}{2} \frac{2}{t} + \frac{1}{4} \frac{4}{t^2} \right\} = \frac{1}{2}. \end{aligned}$$

Fig. 9.2

Single interval graph $\Gamma _{(1.1)}$ of length $\pi$

(2) Simple Circle

Let the graph be the circle $\Gamma _{(1.2)}$ having length $\pi ,$ i.e. it can be treated as the interval $[0, \pi ]$ with the endpoints identified (Fig. 9.3). The Euler characteristic is $\chi = 0.$ The spectrum of $L^{\mathrm {st}} (\Gamma _{(1.2)} )$ is $\Sigma (L) = \{ (2n)^2, n=0,1,1,2,2,\ldots \}.$ Substitution $k_n$ into formula (9.1) gives

$$\displaystyle \begin{aligned} {} \chi = 2- 4 \lim_{t \rightarrow \infty} \sum_{n=1}^\infty \frac{ 1- 2 \cos 2 n/ t + \cos 4n/t}{(2n/t)^2} = 2- 2 = 0, \end{aligned} $$

(9.6)

where we again used formula (9.5).

Fig. 9.3

Simple circle graph $\Gamma _{(1.2)}$ of length $\pi$

(3) Equilateral Star Graph

Let $\Gamma$ be the star graph formed by $m$ equal edges of the length $\pi$ joined at one endpoint (Fig. 9.4). The Euler characteristic is $\chi = 1.$ The spectrum consists of simple eigenvalues $n^2, n= 0,1,2,\ldots ,$ and eigenvalues $(1/2 + n )^2 , n= 0,1,2,\ldots$ having multiplicity $m-1$. The formula (9.1) gives then

$$\displaystyle \begin{aligned} {} \begin{array}{ccl} \chi & = & \displaystyle 2 - 2 \lim_{t\rightarrow \infty} \sum_{n=1}^\infty \frac{1-2 \cos n/t + \cos 2 n/t}{(n/t)^2} \\ & & \displaystyle - 2 (m-1) \lim_{t \rightarrow \infty} \sum_{n=1}^\infty \frac{1-2 \cos (n+1/2)/t + \cos 2 (n+1/2)/t}{((n+1/2)/t)^2} \\ & = & 2 - 1 -0 = 1, \end{array} \end{aligned} $$

(9.7)

where we used formulas (9.5) and (24.41) (see Chap. 24).

Fig. 9.4

Equilateral star graph

Formula (9.1) requires knowledge of all eigenvalues of the standard Laplacian which is impossible in practice. In order to reconstruct the Euler characteristic from a finite number of eigenvalues one may use the following observations:

• The Euler characteristic $\chi$ is an integer, hence it is enough to calculate it with the accuracy less than $1/2$.

• There is no need to take the limit in formula (9.1), which holds exactly for any sufficiently large $t> t_0 = 2/\min \{ \ell _j\}.$

• For any fixed t the terms in the series satisfy a uniform estimate containing a certain negative power of n, hence it is easy to estimate the tail of the series.

Then for a sufficiently large t choose any K so that the reminder in the series is less than $1/2$. This idea has already been implemented to determine experimentally the Euler characteristic of microwave networks without inspecting the network visually [364, 365]. Our common work seems to be a good example of collaboration between mathematicians and applied scientists, since it appeared that the original formula (9.1) requires knowledge of too many eigenvalues, not detectable in practice. Formula (9.1) was modified by using a certain continuously differentiable test function instead of $\varphi$ given by (9.2). The corresponding formula has a better convergence and therefore requires a smaller number of eigenvalues. It is enough to know about 30 lowest eigenvalues to determine Euler characteristic of simple graphs. Mathematical description of the method can be found in [367]. It happens not so often that a mathematical formula can be checked through an experiment.

Problem 35

Check calculations leading to formulas (9.4,9.6,9.7). What is the smallest value of $t$ that can be taken to get precise value of $\chi$?

Problem 36

Let $\Gamma$ be a connected graph without loops. How can one determine the length of the shortest edge from the spectrum of the standard Laplacian?

9.2 Euler Characteristic for Graphs with Dirichlet Vertices

Assume that the Laplacian on a graph $\Gamma$ is determined by standard and Dirichlet conditions at the vertices. It is enough to assume that Dirichlet conditions are introduced at degree one vertices only, hence let us denote by $M_D$ the number of Dirichlet vertices. We are interested in recovering the Euler characteristic of the graph from its spectrum generalising formula (9.1). It is clear that the number of Dirichlet vertices $M_D$ should be involved since we have examples of isospectral graphs having different Euler characteristic (see Fig. 2.10), hence formula (9.1) has to be modified. The main reason is that formulas for the spectral and algebraic multiplicities need to be revised. Let us prove a counterpart of Theorem 8.2 assuming for simplicity that the graph is connected.

Theorem 9.2

Let $\Gamma$ be a finite compact connected metric graph with Euler characteristic $\chi ,$ and let $L^{\mathrm {st,D}} (\Gamma )$ be the Laplace operator defined by $M_D \geq 1$ Dirichlet conditions at some degree one vertices and standard conditions at all other vertices. Then the spectral and algebraic multiplicities of $\lambda = 0$ are

• $$\displaystyle \begin{aligned} m_s (0) = 0, \end{aligned} $$

  (9.8)

  i.e. $\lambda = 0$ is not an eigenvalue;

• $$\displaystyle \begin{aligned} m_a(0) = - \chi + M_D. \end{aligned} $$

  (9.9)

Proof

To prove that $\lambda = 0$ is not an eigenvalue for $M_D \geq 1$ assume that $\psi$ is the corresponding eigenfunction, then it holds

$$\displaystyle \begin{aligned} 0 = \langle \psi, L^{\mathrm{st,D}} \psi \rangle = \int_\Gamma \vert \psi' (x) \vert^2 dx\end{aligned}$$

implying that $\psi$ is a constant function on each edge. Taking into account standard conditions¹ and connectivity of the graph we conclude that $\psi$ is constant on the whole $\Gamma .$ If $M_D \geq 1$, then the function is identically zero.

Let us turn to the algebraic multiplicity. We are going to modify the proof of Theorem 8.2. We again introduce the vectors of amplitudes $\vec {A} , \vec {B}$. The relation given by $S_{\mathrm {e}}^n(0)$ is the same and formula (8.13) is preserved

$$\displaystyle \begin{aligned} a_{2n-1} = b_{2n} \; \; \mbox{and} \; \; a_{2n} = b_{2n-1}, \; \; n =1,2, \dots, N , \end{aligned}$$

while formula (8.14)

$$\displaystyle \begin{aligned} \left\{ \begin{array}{l} a_{i} + b_{i} = a_{j} + b_{j}, \; \; x_{i}, x_{j} \in V^m, \\[3mm] \displaystyle \sum_{x_j \in V^m} (a_j - b_j) = 0, \end{array} \right. \end{aligned}$$

holds for standard vertices only and has to be modified for Dirichlet vertices as

$$\displaystyle \begin{aligned} a_{i} + b_{i} = 0. \end{aligned}$$

Eliminating coefficients $b_i$ using the first relation we get the new system of linear equations

$$\displaystyle \begin{aligned} {} \left\{ \begin{array}{ll} a_{2i-1} + a_{2i} = 0, & i = 1,2, \dots, N \\[3mm] \displaystyle \sum_{i: x_i \in V^m} (a_i - a_{i-(-1)^i} ) = 0, & V^m\mbox{ is a standard vertex}; \end{array} \right. \end{aligned} $$

(9.10)

where we have taken into account that $M_D \geq 1$ implying that at least one, and hence all, of $a_{2i-1} + a_{2i}$ is zero.

With every edge $E_n$ we associate the flux $f_n = a_{2n-1} - a_{2n}$ as before. Then conditions at standard vertices can be interpreted as the sum of fluxes is zero there. Dirichlet vertices determine no conditions on the fluxes. Let us construct basic fluxes:

• each independent cycle in $\Gamma$ determine the flux $\mathcal F^n$ described in the proof of Theorem 8.2;

• each two Dirichlet vertices $V^i$ and $V^j$ determine the flux $\mathcal F_{i,j}$ supported by the shortest path connecting the vertices.

Let $\mathcal F$ be any flux supported by $\Gamma$ and satisfying conservation conditions (9.10) above. We denote by $E_n, n = 1,2, \dots , \beta _1$ the edges on the independent cycles whose deletion turns $\Gamma$ into a tree $\mathbf T$. For each Dirichlet vertex $V^m, m= 1,2, \dots , M_D$ let us denote by $E_{\beta _1 +m}$ the corresponding degree one edge. Then the flux

$$\displaystyle \begin{aligned} \mathcal F - \sum_{n=1}^{\beta_1} \mathcal F (E_n) \mathcal F^n - \sum_{n=\beta_1+1}^{\beta_1+M_D-1} \mathcal F (E_n) \mathcal F_{1,M_D}\end{aligned}$$

is supported by the tree $\mathbf T\setminus \{ E_n \}_{n=\beta _1+1}^{\beta _1+ M_D-1}$ with one Dirichlet vertex. Note that in the last sum we use fluxes between the Dirichlet vertices $V^1, \dots , V^{M_D-1}$ and the Diriclet vertex $V^{M_D}$. As before any such flux is identically zero. We have proven that the number of independent fluxes is $\beta _1 + M_D -1$, which accomplishes the proof. □

It is now straightforward to generalise Theorem 9.1 allowing Dirichlet vertices:

Theorem 9.3

Let $\Gamma$ be a compact connected metric graph and $L^{\mathrm {st,D}} (\Gamma )$ be the Laplace operator defined by $M_D \geq 1$ Dirichlet conditions at certain degree one vertices and standard conditions at all other vertices. Then the Euler characteristic $\chi (\Gamma )$ is uniquely determined by the spectrum $\{ \lambda _n \}$ of the Laplace operator $L^{\mathrm {st,D}} (\Gamma )$

$$\displaystyle \begin{aligned} {} \chi = \displaystyle M_D + 2 \lim_{t\rightarrow \infty} \sum_{k_n } \cos k_n/t \left( \frac{\sin k_n/2t}{k_n/2t} \right)^2, \end{aligned} $$

(9.11)

where $k_n = \sqrt {\lambda _n} > 0.$

Proof

Repeating the proof of Theorem 9.1 but using trace formula (8.45) for scaling-invariant vertex conditions we obtain

leading to (9.11). □

The above formula shows that two isospectral graphs with Dirichlet and standard vertices have a common value

$$\displaystyle \begin{aligned} \chi- M_D.\end{aligned}$$

We check this in the case of isospectral graphs presented in Fig. 2.10. Their Euler characteristic and number of Dirichlet vertices are $(1, 2)$ and $(0, 1)$ respectively:

$$\displaystyle \begin{aligned} 1- 2 = 0 -1 .\end{aligned}$$

Problem 37

Is it possible to find examples of isospectral graphs with $(\chi , M_D)$ equal to $(1,1)$ and $(0,0)$?

Problem 38

Formulate and prove analogues of Theorems 9.2 and 9.3 for not necessarily connected graphs $\Gamma$.

Problem 39

Consider the case of arbitrary scaling-invariant conditions at the vertices. Study possible values of the spectral and algebraic multiplicities. (Paper [347] might help.)

Formula (9.11) can be proven directly using symmetry arguments. Let us double the graph by adding to $\Gamma$ another copy of the same graph and gluing them by joining pairwise the former Dirichlet vertices $V^i, \; i = 1,2, \dots , M_D$ introducing there standard conditions. Let us denote the metric graph obtained in this way by $\Gamma _2.$ This graph is symmetric with respect to the exchange of the respective points on the two copies of $\Gamma$. Hence all eigenfunctions and the spectrum can be divided into two classes:

• symmetric eigenfunctions satisfying Neumann (i.e. standard) conditions at $V^i,$$i = 1, \dots , M_D$, giving the spectrum of the standard Laplacian $\Sigma (L^{\mathrm {st}} (\Gamma ))$;

• antisymmetric eigenfunctions satisfying Dirichlet conditions at $V^i, i=1,2,\dots , M_D$, giving the spectrum of $\Sigma (L^{\mathrm {st,D}} (\Gamma ))$.

Let $\chi$ be the Euler characteristic of $\Gamma$, then $\Gamma _2$ has $2 \beta _1 + M_D-1$ independent cycles and its Euler characteristic is

$$\displaystyle \begin{aligned} 2 \chi - M_D .\end{aligned}$$

Applying formula (9.1) to standard Laplacians on $\Gamma$ and $\Gamma _2$ we get

$$\displaystyle \begin{aligned} \begin{array}{rcl} 2 \chi - M_D & = & \displaystyle 2 \lim_{t\rightarrow \infty} \sum_{k_n^2 \in \Sigma (L^{\mathrm{st}} (\Gamma_2)) } \cos k_n/t \left( \frac{\sin k_n/2t}{k_n/2t} \right)^2 \\ & = & \displaystyle 2 \lim_{t\rightarrow \infty} \sum_{k_n^2 \in \Sigma (L^{\mathrm{st}} (\Gamma)) } \cos k_n/t \left( \frac{\sin k_n/2t}{k_n/2t} \right)^2 \\ && \displaystyle + 2 \lim_{t\rightarrow \infty} \sum_{k_n^2 \in \Sigma (L^{\mathrm{st,D}} (\Gamma))} \cos k_n/t \left( \frac{\sin k_n/2t}{k_n/2t} \right)^2, \\[8mm] \chi & = & \displaystyle 2 \lim_{t\rightarrow \infty} \sum_{k_n^2 \in \Sigma (L^{\mathrm{st}} (\Gamma))} \cos k_n/t \left( \frac{\sin k_n/2t}{k_n/2t} \right)^2, \end{array}\end{aligned}$$

where $\Sigma (L)$ denotes the spectrum of $L$. Elementary calculations imply (9.11).

9.3 Spectral Asymptotics and Schrödinger Operators

9.3.1 Euler Characteristic and Spectral Asymptotics

In this section we are going to show that the Euler characteristic is determined entirely by the asymptotics of the spectrum. The limit of each term in the series (9.1) does not depend on $k_n$

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

Taking this into account it is clear that changing any finite number of eigenvalues does not affect the limit (9.1). We are going to prove that the same is true even if the number of perturbed eigenvalues is infinite, but the perturbation is relatively small.

Let us denote by $(k_n^0)^2$ the spectrum of the Laplacian and by $k_n^2$ its perturbation. Then the following Lemma shows that if the perturbation is small in the sense that $k_n^0$ and $k_n$ possess the same asymptotics, then substituting the perturbed sequence into formula (9.1) one obtains the correct value of $\chi .$ This result can be used in numerical computations, but it has another important implication: as the perturbed sequence one may take the spectrum of the Schrödinger operator on the same metric graph provided the potential is sufficiently regular. This implication will be discussed in the following subsection.

Lemma 9.4

Let $k_n$ and $k_n^0$ be two real sequences satisfying the following conditions

$$\displaystyle \begin{aligned} {} k_n - k_n^0 = \mathcal O \left( \frac{1}{n}\right) , \end{aligned} $$

(9.12)

$$\displaystyle \begin{aligned} {} k_n = \frac{\pi}{\mathcal L} n + \mathcal O(1), \end{aligned} $$

(9.13)

(Weyl’s asymptotics), then the following two limits coincide

$$\displaystyle \begin{aligned} {} \lim_{t\rightarrow \infty} \sum_{n = 1 }^\infty \cos k_n/t \left( \frac{\sin k_n/2t}{k_n/2t} \right)^2 = \lim_{t\rightarrow \infty} \sum_{n=1 }^\infty \cos k_n^0/t \left( \frac{\sin k_n^0/2t}{k_n^0/2t} \right)^2. \end{aligned} $$

(9.14)

Proof

Without loss of generality we assume that $\mathcal L = \pi$. It will be convenient to write estimates (9.12) and (9.13) in the form

$$\displaystyle \begin{aligned} {} \vert k_n - k_n^0 \vert \leq A \frac{1}{n}, \; \; \vert k_n - n \vert \leq B, \; \; \vert k_n^0 - n \vert \leq B, \; \; n=1,2,\ldots, \end{aligned} $$

(9.15)

with certain positive constants $A$ and $B .$ In addition we shall use the following notations

$$\displaystyle \begin{aligned} a_n (t) := \cos k_n/t \left( \frac{\sin k_n/2t}{k_n/2t} \right)^2, \; a_n^0(t) := \cos k_n^0/t \left( \frac{\sin k_n^0/2t}{k_n^0/2t} \right)^2 .\end{aligned}$$

To prove the Lemma we are going to establish two estimates which will be suitable for terms with small and large indices respectively:

Estimate 1 (Suitable for Small Values of $n$)

$$\displaystyle \begin{aligned} {} \vert a_n (t) - a_n^0(t) \vert \leq C \frac{(n+B)^2}{t^2}, \end{aligned} $$

(9.16)

where $C$ is a certain positive constant $C > 0.$

Consider the function

$$\displaystyle \begin{aligned} f(\alpha) = \left\{ \begin{array}{ll} \displaystyle \cos 2 \alpha \left( \frac{\sin \alpha}{\alpha} \right)^2, & \alpha \neq 0, \\ 1, & \alpha = 0 . \end{array} \right.\end{aligned}$$

The derivatives of $f$ are

$$\displaystyle \begin{aligned} \begin{array}{ccl} \displaystyle f'(\alpha) & = & \displaystyle - 2 \sin 2 \alpha \left( \frac{\sin \alpha}{\alpha} \right)^2 + 2 \cos 2 \alpha \frac{\sin \alpha}{\alpha} \frac{ \alpha \cos \alpha - \sin \alpha}{\alpha^2}, \\ \displaystyle f''(\alpha) & = & \displaystyle - 4 \cos 2 \alpha \left( \frac{\sin \alpha}{\alpha} \right)^2 - 8 \sin 2 \alpha \frac{\sin \alpha}{\alpha} \frac{ \alpha \cos \alpha - \sin \alpha}{\alpha^2} \\ && \displaystyle + 2 \cos 2 \alpha \left( \frac{\alpha \cos \alpha - \sin \alpha}{\alpha^2} \right)^2 \\ && \displaystyle + 2 \cos 2 \alpha \frac{\sin \alpha}{\alpha} \frac{- \alpha^2 \sin \alpha - 2 \alpha \cos \alpha +2 \sin \alpha}{\alpha^3} , \end{array}\end{aligned}$$

and we see that $f'(0) = 0$ and $f''(\alpha )$ is uniformly bounded. Hence Taylor’s formula gives

$$\displaystyle \begin{aligned} f(\alpha) - f(0) - f'(0) \alpha = f''(\xi) \frac{\alpha^2}{2} \end{aligned}$$

and therefore

$$\displaystyle \begin{aligned} \vert f(\alpha)-1 \vert \leq \frac{1}{2} \mathrm{max}\,\vert f''(\alpha)\vert \, \alpha^2 .\end{aligned}$$

This implies that

$$\displaystyle \begin{aligned} \vert a_n (t) - 1 \vert \leq \frac{1}{2} \mathrm{max}\, \vert f''(\alpha) \vert \frac{(n+B)^2}{4 t^2}\end{aligned}$$

and similar estimate (9.16) for the difference $\vert a_n (t) - a_n^0 (t) \vert$ with $C = \frac {1}{4} \mathrm {max}\, \vert f''(\alpha ) \vert .$

Estimate 2 (Suitable for Large Values of $n$)

$$\displaystyle \begin{aligned} {} \left\vert a_n(t) - a_n^0 (t) \right\vert \leq D \frac{t}{(n-B)^3}, \; n > B, \end{aligned} $$

(9.17)

where $D$ is a certain positive constant $D > 0 .$

To prove the estimate we use that the function $\alpha ^2 f'(\alpha )$ is uniformly bounded. Using the first mean value theorem we get

$$\displaystyle \begin{aligned} a_n (t) - a_n^0 (t) = f(k_n/2t) - f(k_n^0/2t) = f'(\xi_n) (k_n/2t-k_n^0/2t),\end{aligned}$$

where $\xi _n$ satisfies the same estimate as $k_n$ and $k_n^0$ (see the second and third estimates in (9.15))

$$\displaystyle \begin{aligned} \vert \xi_n - n \vert \leq B.\end{aligned}$$

For $n >B,$ it follows that

$$\displaystyle \begin{aligned} \vert a_n (t) - a_n^0(t) \vert \leq \mathrm{max}\, \vert \alpha^2 f'(\alpha) \vert \frac{1}{(\frac{n-B}{2t})^2} A \frac{1}{2nt} \leq D \frac{t}{(n-B)^3}\end{aligned}$$

with $D = 2 A \,\mathrm { max}\, \vert \alpha ^2 f'(\alpha ) \vert ,$ which is exactly estimate (9.17).

To prove (9.14) we need to show that the following limit equals zero

$$\displaystyle \begin{aligned} {} & \lim_{t \rightarrow \infty} \sum_{n=1}^\infty \left\vert \cos k_n/t \left( \frac{\sin k_n(2t}{k_n/2t} \right)^2 - \cos k_n^0/t \left( \frac{\sin k_n^0(2t}{k_n^0/2t} \right)^2 \right\vert \\ & \quad \equiv \lim_{t \rightarrow \infty} \sum_{n=1}^\infty \vert a_n(t) - a_n^0 (t) \vert. \end{aligned} $$

(9.18)

Let us split the infinite series into the finite sum of the first $K$ elements and the remaining infinite series as

$$\displaystyle \begin{aligned} \sum_{n=1}^\infty = \sum_{n=1}^K + \sum_{n=K+1}^\infty .\end{aligned}$$

To prove that the limit is zero it is enough to show that for any $\epsilon > 0$ there exists $t_0 = t_0 (\epsilon )$, such that for any $t > t_0 (\epsilon )$ the number $K = K(\epsilon , t)$ can be chosen in such a way that both the finite sum and the series are less than $\epsilon /2.$

We estimate the summands using (9.16) and (9.17) as

$$\displaystyle \begin{aligned} \begin{array}{ccccc} \displaystyle \sum_{n=1}^K \vert a_n (t) - a_n^0(t) \vert & \leq & \displaystyle \sum_{n=1}^K C \frac{(K+B)^2}{t^2} & \leq & \displaystyle C \frac{(K+B)^3}{t^2}, \\ \displaystyle \sum_{n=K+1}^\infty \vert a_n (t) - a_n^0(t) \vert & \leq & \displaystyle \sum_{n=K+1}^\infty D \frac{t}{(n-B)^3} & \leq & \displaystyle \frac{D}{2} \frac{t}{(K-B)^2}. \end{array} \end{aligned}$$

Each of sum is less than $\epsilon /2$ if the following two inequalities are satisfied

$$\displaystyle \begin{aligned} K(\epsilon,t) \leq \left(\frac{\epsilon t^2}{2C}\right)^{1/3} - B \; \; \; \; \mbox{and} \; \; \; \; K(\epsilon,t) \geq \sqrt{\frac{Dt}{\epsilon}} + B .\end{aligned}$$

Hence the series in (9.18) is less than $\epsilon$ if

$$\displaystyle \begin{aligned} \sqrt{\frac{Dt}{\epsilon}} + B \leq \left(\frac{\epsilon t^2}{2C}\right)^{1/3} - B .\end{aligned}$$

For any $\epsilon > 0$ there exists $t_0$, such that for any $t > t_0$ the last inequality is satisfied and it is possible to choose integer $K(\epsilon , t)$, such that both the finite and infinite sums are less than $\epsilon /2.$ For such $t$ we have that the infinite series in (9.18) is less than $\epsilon .$ It follows that the limit in (9.18) is zero. □

9.3.2 Schrödinger Operators and Euler Characteristic of Graphs

Let $q$ be any essentially bounded real potential and let $L^{\mathrm {st}}_q (\Gamma )$ be the corresponding standard Schrödinger operator, then the difference between the eigenvalues is uniformly bounded

$$\displaystyle \begin{aligned} {} k_n^2 - (k_n^0)^2 = \mathcal O(1), \end{aligned} $$

(9.19)

as the Schrödinger operator is a bounded perturbation of the Laplacian. The same estimate will be proven in Chap. 11 assuming that the potential is just absolutely integrable (see (11.32)). Also the estimate (9.19) for not essentially bounded potentials will be justified later. Therefore in the following theorem we are going to assume that the potential is from $L_1 (\Gamma )$.

We are able to prove now that the formula for Euler characteristic (9.1) gives the correct result, provided the spectrum of the Laplacian is substituted with the spectrum of the Schrödinger operator.

Theorem 9.5

Let $\Gamma$ be a finite compact metric graph and $q$ be a real valued absolutely integrable function on $\Gamma .$ Let $L^{\mathrm {st}} (\Gamma )$ and $L^{\mathrm {st}}_q (\Gamma )$ be the standard Laplace and Schrödinger operators. Then the Euler characteristic $\chi (\Gamma )$ of the graph $\Gamma$ is uniquely determined by the spectrum $\lambda _n (S)$ of the operator $L_q^{\mathrm {st}}$ and can be calculated using the limit

$$\displaystyle \begin{aligned} {} \chi (\Gamma) = 2 \lim_{t \rightarrow \infty} \sum_{n = 0}^\infty \cos \sqrt{\lambda_n (L^{\mathrm{st}}_q) }/t \left( \frac{\sin \sqrt{\lambda_n (L^{\mathrm{st}}_q)}/2t}{\sqrt{\lambda_n (L^{\mathrm{st}}_q)}/2t} \right)^2, \end{aligned} $$

(9.20)

where we use the following natural convention

$$\displaystyle \begin{aligned} \lambda_m = 0 \Rightarrow \frac{\sin \sqrt{\lambda_m(L^{\mathrm{st}}_q)}/2t}{\sqrt{\lambda_m(L^{\mathrm{st}}_q)}/2t} = 1. \end{aligned} $$

(9.21)

Proof

The estimate (9.19) together with the Weyl asymptotics (9.13) imply that

$$\displaystyle \begin{aligned} k_n - k_n^0 = \mathcal O \left(\frac{1}{n}\right) ,\end{aligned}$$

and Lemma 9.4 can be applied. It follows that

$$\displaystyle \begin{aligned} \lim_{t\rightarrow \infty} \sum_{n = 1 }^\infty \cos k_n/t \left( \frac{\sin k_n/2t}{k_n/2t} \right)^2 = \lim_{t\rightarrow \infty} \sum_{n=1 }^\infty \cos k_n^0/t \left( \frac{\sin k_n^0/2t}{k_n^0/2t} \right)^2 = \chi, \end{aligned}$$

where we used (9.1) on the last step. The introduced convention allowed us to remove $m_s (0)$ from the formula for Euler characteristic of the Laplacian. □

Note that the limit cannot be substituted with considering $t \geq \frac {2}{\min \{\ell _j \} }$ as can be done for Laplacians.

Theorem 9.5 together with Weyl’s asymptotics (4.25) imply that two Schrödinger operators on graphs may have the same spectrum only if the underlying graphs have the same total length and Euler characteristic, in other words, if the graphs have the same size and complexity.

Uniqueness Theorem 9.6

Let the metric graphs $\Gamma _1$ and $\Gamma _2$ be finite and compact and let the corresponding real potentials $q_1$ and $q_2$ be absolutely integrable. Then the corresponding standard Schrödinger operators $L_{q_j} (\Gamma _j) , \, j=1,2,$ have close spectra

$$\displaystyle \begin{aligned} {} \lambda_n \big(L_{q_1} (\Gamma_1) \big) - \lambda_n \big(L_{q_2} (\Gamma_2) \big) = \mathcal O(1) \end{aligned} $$

(9.22)

only if the metric graphs have the same

• total length;

• Euler characteristic.

Proof

Condition (9.22) together with the Weyl asymptotics (4.25) imply that the metric graphs $\Gamma _1$ and $\Gamma _2$ have the same total length.

To show that the graphs have the same Euler characteristic one repeats the arguments used in the proof of Theorem 9.20. □

9.3.3 General Vertex Conditions: A Counterexample

Obtained results can be extended to the case of most general vertex conditions. This problem appears to be more sophisticated than it may be expected. The main reason is that the vertex scattering matrix in general is not energy independent but tends to a certain limiting matrix $\mathbf S_{\mathbf {v}} (\infty ).$ The limiting matrix in its turn corresponds to certain symmetric vertex conditions, but these conditions may be incompatible with the connectivity of the original graph $\Gamma .$ In other words these new vertex conditions may not connect all edges joined at a vertex despite the fact that the original conditions (corresponding to the energy dependent scattering matrix) do connect all these edges together.

Let us study the following elementary example. Consider the interval $[-\pi ,\pi ]$ turned into circle by joining together the end points $-\pi$ and $\pi$ with the help of the following vertex conditions

$$\displaystyle \begin{aligned} {} \left\{ \begin{array}{ccc} \psi (-\pi) & = & - \partial_n \psi(+\pi), \\ \psi (\pi) & = & - \partial_n \psi (-\pi); \end{array} \right. \end{aligned} $$

(9.23)

which are obviously properly connecting, i.e. connect together the boundary values of the functions from both endpoints. The corresponding vertex scattering matrix

$$\displaystyle \begin{aligned} \mathbf S_{\mathbf{v}} (k) = - \frac{I- ik B}{I+ikB}\end{aligned}$$

with $B = \left ( \begin {array}{cc} 0 & -1 \\ -1 & 0 \end {array} \right )$ is irreducible, but it tends to the unit matrix $\mathbf S_{\mathbf {v}} (\infty ) = \mathbf I$ as $k \rightarrow \infty .$ The vertex conditions corresponding to $\mathbf S_{\mathbf {v}} (k) = \mathbf I$ are just Neumann boundary conditions $\partial _n \psi (+\pi ) = 0 = \partial _n \psi (-\pi ) ,$ which are obviously reducible: the two endpoints are not connected to each other. Therefore it is natural to call the vertex conditions (9.23) by not asymptotically properly connecting. If the vertex conditions are not asymptotically properly connecting, then the asymptotics of the spectrum is determined by the Laplacian not on the original graph $\Gamma$, but on a certain new graph $\Gamma ^\infty$ obtained from $\Gamma$ by chopping some of the vertices. In other words spectral asymptotics is determined by a different topology.

We illustrate this idea by calculating the spectra of the operators appearing in the example under consideration. Let us denote by $\tilde {L}$ the second derivative operator $- \frac {d^2}{dx^2}$ defined on the functions from $W_2^2 [-\pi , \pi ]$ and satisfying vertex conditions (9.23). These vertex conditions can be written as follows using the derivatives with respect to the variable $x \in [-\pi , \pi ]$

$$\displaystyle \begin{aligned} {} \left\{ \begin{array}{ccc} \psi (-\pi) & = & \psi'(+\pi), \\ \psi'(-\pi) & = & - \psi (\pi). \end{array} \right. \end{aligned} $$

(9.24)

It is easy to see that the vertex conditions are invariant under the change of the coordinate $x \mapsto - x$ and hence the operator $\tilde {L}$ commutes with the symmetry operator $\mathcal P \psi (x) = \psi (-x) .$ Therefore all eigenfunctions of $\tilde {L}$ are either even or odd. The dispersion equations for even and odd functions can be obtained by substituting the Ansätze$\psi _s (x) = \cos kx$ and $\psi _a (x) = \sin kx$ into the vertex conditions

$$\displaystyle \begin{aligned} \begin{array}{rcl} \tan k^s \pi & = &\displaystyle - \frac{1}{k^s}; \end{array} \end{aligned} $$

(9.25)

$$\displaystyle \begin{aligned} \begin{array}{rcl} \cot k^a \pi & = &\displaystyle - \frac{1}{k^a}. \end{array} \end{aligned} $$

(9.26)

The eigenvalues satisfy the following asymptotic conditions

$$\displaystyle \begin{aligned} {} k^s_n = n + \mathcal O(\frac{1}{n}), \; \; \; k^a_n = \frac{2n+1}{2} + \mathcal O(\frac{1}{n}), \; \; n=0,1,2,\ldots, \end{aligned} $$

(9.27)

where $(k^s_n)^2$ and $(k^a_n)^2$ denote the eigenvalues for even and odd eigenfunctions respectively. These eigenvalues are asymptotically close to the eigenvalues $(k_n^{s0})^2 = ( n)^2$ and $(k_n^{a0})^2 = \left (\frac {2n+1}{2} \right )^2$ for the Laplace operator on the interval $[-\pi , \pi ]$ (with Neumann boundary conditions at the endpoints $\psi '(-\pi ) = 0 = \psi ' (\pi )$).

Substituting the spectrum of the operator $\tilde {L}$ into the formula (9.20) one obtains the Euler characteristic of the interval, not of the circle

$$\displaystyle \begin{aligned} \lim_{t \rightarrow \infty} \sum_{k_n = k_n^s, k_n^a} \cos k_n/t \left( \frac{\sin k_n/2t}{k_n/2t} \right) = \lim_{t \rightarrow \infty} \sum_{k^0_n = k_n^{s0}, k_n^{a0}} \cos k^0_n/t \left( \frac{\sin k^0_n/2t}{k^0_n/2t} \right) = 1, \end{aligned} $$

(9.28)

where we used Lemma 9.4. It follows that formula (9.1) in general is not valid for Schrödinger operators on graphs with general (Hermitian) vertex conditions.

9.4 Reconstruction of Graphs with Rationally Independent Lengths

Formula (8.21) can be applied to solve the inverse spectral problem in the very special case of graphs with edges having rationally independent lengths.² Our studies will again be restricted to the case of standard Laplacians. Such operators are uniquely determined by the underlying metric graphs and therefore the corresponding inverse problem is equivalent to the problem of recovering the metric graph $\Gamma$ from the spectrum of the Laplacian $L^{\mathrm {st}} (\Gamma ).$ In this section we follow our paper [346] inspired by Gutkin and Smilansky [252].

Note that this reconstruction is not possible for graphs having vertices of degree two. The two edges connected at such vertex can be substituted with the single edge of the length equal to the sum of the lengths. This is because standard conditions at a degree two vertex imply that the function and its first derivative are continuous at the vertex.

The set $\mathbb L$ of lengths of all periodic paths for a metric graph $\Gamma$ is usually called the length spectrum. This is a set of positive real numbers all being linear combinations of the lengths $\ell _n$ of the edges with coefficients being natural numbers. But not all such linear combinations are present in $\mathbb L$, since not all edges are connected to each other directly.

We are going to assume that the lengths of the edges are rationally independent, i.e. if the equality

$$\displaystyle \begin{aligned} \sum_{n=1}^N \alpha_n \ell_n = 0\end{aligned}$$

holds with certain rational $\alpha _n \in \mathbb Q$, then all $\alpha _n$ are necessarily equal to zero. This assumption is very important, since we already know that even trees cannot be reconstructed from the spectra of their Laplacians, unless extra restrictions are imposed (See Sect. 2.2 (Problem 6)). If the lengths are rationally independent then knowing the length $\ell (p)$ of a periodic path we know which edges this path comes across and how many times, of course provided we know $\ell _n.$ Hence our first task should be to recover the lengths of edges.

Looking at formula (8.21) one may get the impression that the lengths of all periodic paths can be recovered directly as the (positive) points supporting the delta functions $\delta _{\ell (p)}$. But one should pay attention to the fact that complicated graphs may have several periodic paths with precisely the same lengths. Then contributions from all such paths may cancel each other out.

Example 9.7 ([410])

Consider the graph presented in Fig. 9.5 with the lengths of edges indicated. There exist precisely six periodic paths with the length $2 \ell _1 + \ell _2 + \ell _3 + \ell _4 + \ell _5$. These paths are indicated on the lower part of the figure. For each closed curve there exists precisely two paths that run along it in opposite directions. We assume that the vertices $V^2$ and $V^4$ have arbitrary degrees $d_2$ and $d_4$ and the degrees of the vertices $V^1$ and $V^3$ are equal to $3.$ Then the product of scattering coefficients for the left path is

$$\displaystyle \begin{aligned} T(V^4) \cdot T(V^3) \cdot R(V^1) \cdot T(V^3) \cdot T(V^2) \cdot T(V^1) = \frac{2}{d_4} \cdot \frac{2}{3} \cdot (-\frac{1}{3}) \cdot \frac{2}{3} \cdot \frac{2}{d_2} \cdot \frac{2}{3} = - \frac{32}{81} \frac{1}{d_2 d_4}.\end{aligned}$$

Here the scattering coefficients are taken from formula (3.40) determining the vertex scattering matrix for standard vertex conditions. The central path gives the same contribution, whereas contribution from the right path is

$$\displaystyle \begin{aligned} T(V^4) \cdot T(V^3) \cdot T(V^1) \cdot T(V^2) \cdot T(V^3) \cdot T(V^1) = \frac{2}{d_4} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{d_2} \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{64}{81} \frac{1}{ d_2 d_4}.\end{aligned}$$

Then the total contribution from all six paths is given by

$$\displaystyle \begin{aligned} \begin{array}{l} \displaystyle \sum_{\stackrel{\gamma \in \mathcal P}{\ell(\gamma) = 2 \ell_1 + \ell_2 + \ell_3 + \ell_4 + \ell_5}} l(\mathrm{prim} (\gamma)) S_{\mathrm{v}} (\gamma) \\[3mm] \displaystyle = 2 (2 d_1 + d_2 + d_3 + d_4 + d_5) \left( - \frac{32}{27 d_2 d_4} - \frac{32}{27 d_2 d_4} + \frac{64}{27 d_2 d_4} \right) = 0. \end{array}\end{aligned}$$

It follows that formula (8.21) contains no delta function supported at $\ell =2 \ell _1 + \ell _2 + \ell _3 + \ell _4 + \ell _5$.

Fig. 9.5

Periodic paths of length $2 d_1 + d_2 + d_3 + d_4 + d_5$ Ⓒ Marlena Nowaczyk. Reproduced with permission

We believe that the constructed example is the simplest one, since any such example should contain paths with reflections—paths without reflections always lead to positive scattering coefficients. But it might be interesting to find an even simpler example.

Problem 40

Construct your own example of a metric graph with periodic paths giving zero contribution to the right hand side of trace formula (8.21).

Having the presented example in mind let us introduce the notion of reduced length spectrum$\mathbb L' \subset \mathbb L$ defined as

$$\displaystyle \begin{aligned} {} \mathbb L' = \{ \ell : \Big (\sum_{ \begin{array}{c} {\gamma \in \mathcal P} \\{ \ell (\gamma) = \ell} \end{array}} \ell (\mathrm{prim}(\gamma)) S_{\mathrm{v}} (\gamma) \neq 0 \Big) \}. \end{aligned} $$

(9.29)

The following Lemma proves that the reduced length spectrum always contains the shortest periodic paths associated with an edge or a pair of neighbouring edges.

Lemma 9.8

Let $\Gamma$ be a connected finite metric graph without degree two vertices and with rationally independent lengths of edges. The reduced length spectrum $\mathbb L'$ contains at least the following lengths:

• the length of the shortest path formed only by a certain edge $E_j$ (i.e. $\ell _j$ or $2 \ell _j$ depending on whether $E_j$ forms a loop or not);

• the length of the shortest path formed only by a certain pair of neighbouring edges$E_j$and$E_k$(i.e.$2(\ell _j+ \ell _k)$, $\ell _j+ 2 \ell _k$, $2\ell _j+ \ell _k$, $\ell _j+ \ell _k$depending on how these edges are connected to each other).

Proof

The two assertions will be proven separately by considering all possible cases. Let us first note that if a periodic path of a length $\ell$ is unique, then the coefficient in front of the delta function $\delta _\ell$ is always different from zero, since the product of scattering coefficients is always different from zero (the graph is assumed to contain no degree two vertices). The same holds true if there are several paths of the same length, but the corresponding products of scattering coefficients are equal.

Consider first the case of a single edge $E_j$. Possible cases are:

• $E_j$forms a loop.

  There are two periodic paths of length $\ell _j$ running along the loop in opposite directions. The corresponding products of scattering coefficients are equal and therefore $\ell _j$ is in $\mathbb L'.$

• $E_j$connects two different vertices.

  There is a unique path³ of length $2 \ell _j$ and it is present in the reduced length spectrum as explained above.

Let $E_j$ and $E_k$ be two neighbouring edges, consider all possible ways they are connected to each other:

• The edges$E_j$ and $E_k$have one common endpoint.

  The shortest path has length $2 (\ell _j + \ell _k)$ and is unique and therefore its length is in the reduced length spectrum.

• The edge$E_j$forms a loop connected to one of the endpoints of$E_k.$

  There are two shortest paths of length $\ell _j + 2 \ell _k.$ The corresponding products of scattering coefficients are equal. Hence $\ell _j + 2 \ell _k$ belongs to $\mathbb L'.$

• Both edges$E_j$and$E_k$form loops connected at one vertex.

  There are four paths having length $\ell _ j + \ell _k$ but the coefficients are again equal and therefore $\ell _j + \ell _k \in \mathbb L'.$

• The edges$E_j$and$E_k$form a double edge.

  There are two paths of length $\ell _j + \ell _k$ with equal products of scattering coefficients. It follows that $\ell _j + \ell _k \in \mathbb L'.$

□

We are now going to show that the knowledge of the reduced length spectrum together with the total length of the graph is enough to reconstruct the graph. The first step in this direction is to recover the lengths of the edges from the total length of the graphs and the set $\mathbb L'.$ The following result can be proven by refining the method of Gutkin-Smilansky [252].

Lemma 9.9

Let the lengths of the edges of a finite connected metric graph$\Gamma$without degree two vertices be rationally independent. Then the total length$\mathcal L$of the graph and the reduced length spectrum$\mathbb L'$(defined by (9.29)) determine the lengths of all edges independently of whether these edges form loops or not.

Proof

The set $\mathbb L'$ is infinite, but we are interested in reconstructing $N$ rationally independent lengths $\ell _n$. Therefore it is wise to restrict our consideration to a smaller, even finite, set containing for sure the lengths of all shortest paths described in the previous lemma. For example if we take all periodic paths with the lengths less than double the total length $\mathcal L$, then all $\ell _n$ or $2 \ell _n$ for sure belong to the set.

Consider the finite subset $\mathbb L''$ of $\mathbb L' \subset \mathbb L$ consisting of all lengths less than or equal to $2 \mathcal L$

$$\displaystyle \begin{aligned} \mathbb L'' = \{ \ell \in \mathbb L': \ell \leq 2 \mathcal L \}.\end{aligned}$$

This finite set contains at least the numbers $2 \ell _1, 2 \ell _2, \ldots , 2 \ell _N.$ Therefore there exists a basis $s_1, s_2, \dots , s_N ,$ such that every length $\ell \in \mathbb L''$ (as well as from $\mathbb L$) can be written as a half-integer combination of $s_j$

$$\displaystyle \begin{aligned} \ell = \frac{1}{2} \sum_{j=1}^N n_j s_j, \; \; n_j \in \mathbb N. \end{aligned}$$

Such basis is not unique especially if the graph has loops. Any two bases $\{s_j \}$ and $\{s^{\prime }_j \}$ are related as follows $s_j = n_j s^{\prime }_{i_j}, \; n_j = \frac {1}{2}, 1, 2,$ where $i_1, i_2, \dots , i_N$ is a permutation of $1, 2, \dots , N.$ Then among all possible bases consider the basis with the shortest total length $\sum _{j=1}^N s_j.$ This basis is unique up to a permutation.

The total length of the graph $\mathcal L$ can then be written as a sum of $s_j$ with the coefficients equal to $1$ or $1/2$

$$\displaystyle \begin{aligned} {} \mathcal L = \sum_{j=1}^N \alpha_j s_j, \; \; \alpha_j = 1, 1/2. \end{aligned} $$

(9.30)

The coefficients in this sum are equal to $1$ if $s_j$ is equal to the length of a certain edge $E_j$, i.e. when the edge forms a loop. The coefficient $1/2$ appears if $s_j$ is equal to double the length of an edge. In this case the edge does not form a loop. Therefore the lengths of the edges up to a permutation can be recovered from (9.30) using the formula $\ell _j = \alpha _j s_j, \; j=1,2, \dots , N.$ To check whether an edge $E_j$ forms a loop or not it is enough to check whether $\ell _j$ belongs to $\mathbb L'$ or not. □

Once the lengths of all edges are known the graph can be reconstructed from the reduced length spectrum. Lemma 9.8 implies that looking at the reduced length spectrum $\mathbb L'$ one can determine whether any two edges $E_j$ and $E_k$ are neighbours or not (have at least one common endpoint): the edges $E_j$ and $E_k$ are neighbours if and only if $\mathbb L'$ contains at least one of the lengths $\ell _j + \ell _k, 2 \ell _j + \ell _k, \ell _j +2 \ell _k,$ or $2(\ell _j+\ell _k).$

Lemma 9.10

Every finite connected metric graph$\Gamma$without degree two vertices can be reconstructed from the set$\{ \ell _n \}_{n=1}^N$of the lengths of all edges and the reduced length spectrum$\mathbb L'$defined by (9.29), provided that$\ell _n$are rationally independent.

Proof

Let us introduce the set of edges $\mathbb E=\{ E_n \}_{n=1}^{N}$ uniquely determined by the lengths $\ell _j$. We shall prove the lemma for simple graphs first. A graph is called simple if it contains no loops and no multiple edges. From an arbitrary graph one can obtain a simple graph by cancelling all loops and choosing only one edge from every multiple one:

• (1) If $\ell _k \in \mathbb L'$ then the corresponding edge is a loop. Then remove $E_k$ from $\mathbb E$ and all lengths containing $\ell _k$ from $\mathbb L'$.

• (2) If $\ell _k+ \ell _j \in \mathbb L'$ then there exists a double edge composed of $E_j$ and $E_k$ (since the loops have already been removed). Then remove either $E_j$ or $E_k$ from $\mathbb E$ and also all lengths containing the chosen length from $\mathbb L'$.

The new subsets $\mathbb E^* \subset \mathbb E$ containing $N^* \leq N$ elements and $\mathbb L^* \subset \mathbb L'$ obtained in this way correspond to a simple subgraph $\Gamma ^* \subset \Gamma$ which can be obtained from $\Gamma$ by removing all loops and reducing all multiple edges (Fig. 9.6). One obtains different $\Gamma ^*$ by choosing different edges to be left during the reduction, but all $\Gamma ^*$ have the same topology.

Fig. 9.6

A metric graph $\Gamma$ and its simple subgraph $\Gamma ^*$

The graph $\Gamma ^*$ has the same vertex set as $\Gamma$. Note that the reduced graph may have degree two vertices, but such vertices are not dangerous since the edges connected at such vertex are present in the reduced length spectrum $\mathbb L'$.

The reconstruction will be done iteratively and we will construct an increasing finite sequence of subgraphs such that $\Gamma _1 \subset \Gamma _2 \subset \ldots \Gamma _{N^*}=\Gamma ^*$. The corresponding subsets of edges will be denoted by $\mathbb E_k.$

For $k=1$ take the graph $\Gamma _1$ consisting of one edge, say $E_1.$ The endpoints are not connected as we have reduced all loops.

Suppose that connected subgraph $\Gamma _k$ consisting of k edges is reconstructed. Pick up any edge $E_{k+1}$ which is a neighbour of at least one of the edges in $\Gamma _k.$ Let us denote by $\mathbb E_k^{\mathrm {nbh}}$ the subset of $\mathbb E_k$ of all edges which are neighbours of $E_{k+1}.$ We have to identify one or two vertices in $\Gamma _k$ to which the new $E_{k+1}$ is attached. Every such vertex is uniquely determined by listing the edges joined at this vertex, since the subgraph $\Gamma _k$ is simple. Therefore we have to separate $\mathbb E_k^{\mathrm {nbh}}$ into two classes of edges attached to each of the endpoints of $E_{k+1}.$ (One of the two sets can be empty, which corresponds to the case when the edge $E_{k+1}$ is attached to $\Gamma _k$ at one vertex only.)

Take any two edges from $\mathbb E_k^{\mathrm {nbh}}$, say $E'$ and $E''.$ The edges $E'$ and $E''$ belong to the same class if and only if:

• $E'$ and $E''$ are neighbours themselves and

• $\ell '+\ell ''+ \ell _{k+1} \notin \mathbb L'$ i.e. the edges $E'$, $E''$ and $E _{k+1}$ do not build a cycle. Note that if $E', E''$ and $E_{k+1}$ form a cycle, then there are two periodic paths with the length $\ell ' + \ell '' + \ell _{k+1}$ and the corresponding $S_{\mathrm {v}} (p)$-coefficients are equal, which implies that $\ell ' + \ell '' + \ell _{k+1} \in \mathbb L'.$

In this way we either separate $\mathbb E_k^{\mathrm {nbh}}$ into two classes of edges or $\mathbb E_k^{\mathrm {nbh}}$ consists of edges joined at one vertex. In the first case the new edge $E_{k+1}$ connects the two unique vertices determined by the subclasses. In the second case $E_{k+1}$ is attached by one endpoint to $\Gamma _k$ at the vertex uniquely determined by $\mathbb E_k^{\mathrm {nbh}}.$ It does not play any role which of the two endpoints of $E_{k+1}$ is attached to the chosen vertex of $\Gamma _k$, since the two possible graphs are equivalent.

Denote the graph obtained in this way by $\Gamma _{k+1}.$

Since the graph $\Gamma ^*$ is connected and finite after $N^*$ steps one arrives at $\Gamma _{N^*}=\Gamma ^*.$

It remains to add all loops and multiple edges to reconstruct the initial graph $\Gamma$. Suppose that the reconstructed subgraph $\Gamma ^*$ is not trivial, i.e. consists of more than one edge. Then every vertex is uniquely determined by listing all edges joined at it. Check first to which vertex the loop $E_n$ is connected by checking if periodic paths of the length $\ell _n+2 \ell _j$ belongs to $\mathbb L'$ or not. All such edges $E_j$ determine the unique vertex to which $E_n$ should be attached. To reconstruct multiple edges check whether $\ell _m+\ell _j$ is from $\mathbb L'$, where $E_j \in \mathbb E^*.$ Substitute all such edges $E_j$ with corresponding multiple edges.

In the case $\Gamma ^*$ is trivial, the proof is an easy exercise. □

Our main result can be obtained as a straightforward implication of Lemmas 9.9 and 9.10.

Theorem 9.11

The spectrum of a Laplace operator on a metric graph determines the graph uniquely, provided that:

• the graph is finite and connected,

• the graph has no degree two vertices,

• the edge lengths are rationally independent.

Proof

The spectrum of the operator determines the left-hand side of the trace formula (8.20). Formula (8.21) shows that the spectrum of the graph determines the total length of the graph and the reduced length spectrum. The total length can also be determined using Weyl’s asymptotics (4.25) (or (9.13))

$$\displaystyle \begin{aligned} \mathcal L = \pi \lim_{n \rightarrow \infty} \frac{n}{k_n}. \end{aligned} $$

(9.31)

Having reconstructed the total length one may use Lemma 9.9 to conclude that the lengths of all edges can be extracted from the reduced length spectrum. Lemma 9.10 then implies that the whole graph can be reconstructed provided that edge lengths are rationally independent. □

One can easily remove the condition that the graph is connected. The result can be generalised to include more general differential operators on the edges and vertex conditions. Moreover, it is enough to require that only edges situated close to each other have rationally independent lengths [408, 409].