Neumann Domains on Quantum Graphs

Abstract

The Neumann points of an eigenfunction f on a quantum (metric) graph are the interior zeros of $f'$. The Neumann domains of f are the sub-graphs bounded by the Neumann points. Neumann points and Neumann domains are the counterparts of the well-studied nodal points and nodal domains. We prove bounds on the number of Neumann points and properties of the probability distribution of this number. Two basic properties of Neumann domains are presented: the wavelength capacity and the spectral position. We state and prove bounds on those as well as key features of their probability distributions. To rigorously investigate those probabilities, we establish the notion of random variables for quantum graphs. In particular, we provide conditions for considering spectral functions of quantum graphs as random variables with respect to the natural density on ${\mathbb {N}}$.

Appendices

Appendix A. Nodal and Neumann Surpluses of Particular Examples

An intensive numerical investigation led us to believe that the Neumann surplus bounds in Theorem 2.2 are not optimal and to propose better bounds in Conjecture 2.4. In this appendix, we provide analytic evidence supporting the conjectured bounds, by calculating the support of the Neumann surplus for three families of graphs:

• (1) Stowers—Proposition A.2.

• (2) Mandarins—Proposition A.6.

• (3) Trees—Proposition A.12.

See Fig. 6 for examples of a stower and a mandarin. Using the support of the stowers we deduce that the bounds on $\sigma -\omega$, as presented in (3.1), are optimal in terms of $\beta$ and $\left| \partial \Gamma \right|$.

Fig. 6

On the left, a stower graph with $n=3$ loops and $m=4$ tails. On the right, a mandarin graph with $E=7$ edges

Remark A.1

Lemmas A.5 and A.9 are developed in collaboration with Gregory Berkolaiko during our joint work on [4].

1.1 A.1. Stowers

We say that $\Gamma$ is an $\left( n,m\right)$-stower graph with n loops (petals) and m tails (leaves), if it has only one interior vertex v (the central vertex), m boundary vertices, each connected to the central vertex, and n loops connecting the central vertex to itself. See Fig. 6 for example. In such case the first Betti number is $\beta =n$ and the boundary size is $\left| \partial \Gamma \right| =m$.

Proposition A.2

Let $\Gamma$ be a stower, then its Neumann surplus $\omega$ is bounded by

$$\begin{aligned} 1-\left| \partial \Gamma \right| \le \omega \le \beta -1. \end{aligned}$$

(A.1)

Moreover, if $\Gamma$ has rationally independent edge lengths, then its Neumann surplus and nodal surplus distributions satisfy

$$\begin{aligned} \forall j\in \left\{ 1-\left| \partial \Gamma \right| ,\ldots ,\beta -1\right\}&\qquad P\left( \omega =j\right)>0\\ \forall j\in \left\{ 1-\beta ,\ldots ,\beta +\left| \partial \Gamma \right| -1\right\}&\qquad P\left( \sigma -\omega =j\right) >0, \end{aligned}$$

and if $\left| \partial \Gamma \right| >0$, then

$$\begin{aligned} \forall j\in \left\{ 0,\ldots ,\beta \right\} \,\,\,\,\,P\left( \sigma =j\right) >0. \end{aligned}$$

(A.2)

In the case of $\left| \partial \Gamma \right| =0$, namely $\Gamma$ is a flower graph, the nodal surplus is bounded by

$$\begin{aligned} 1\le \sigma \le \beta -1, \end{aligned}$$

(A.3)

and

$$\begin{aligned} \forall j\in \left\{ 1,\ldots ,\beta -1\right\} \,\,\,\,\,P\left( \sigma =j\right) >0. \end{aligned}$$

(A.4)

This result shows that for any possible choice of $\beta$ and $\left| \partial \Gamma \right|$ there is a corresponding stower graph such that its $\sigma -\omega$ sequence achieves both upper and lower bounds in (3.1). This result supports Conjecture 2.4 and shows that the bounds in (3.1) are optimal.

To prove Proposition A.2, let us first state the following.

Definition A.3

 Let $\Gamma$ be a stower with m tails and n loops. For distinction, denote points in ${\mathbb {T}}^{E}$ by $\left( \vec {y},\vec {z}\right)$ such that $\vec {y}\in \mathbb {T}^{n}$ corresponds to loops and $\vec {z}\in \mathbb {T}^{m}$ corresponds to tails. Call a tail coordinate $z_{j}$ “bad” if either $\sin (z_{j})=0$ or $\cos (z_{j})=0$. Similarly, a loop coordinate $y_{i}$ is “bad” if either $\sin (\frac{y_{i}}{2})=0$ or $\cos (\frac{y_{i}}{2})=0$.

Denote the set of points in ${\mathbb {T}}^{E}$ which have at least one “bad” coordinate by ${\varvec{B}}^{\left( 1\right) }$, and the similarly denote the set of points with at least two “bad” coordinates by ${\varvec{B}}^{\left( 2\right) }$.

Remark A.4

Notice that $\dim \left( {\varvec{B}}^{\left( j\right) }\right) =E-j$.

Lemma A.5

Let $\Gamma$ be a stower with m tails and n loops and consider the notations of Definition (A.3). Then,

• (1) $\Sigma ^{\mathrm {gen}}$ is given by

  $$\begin{aligned} \Sigma ^{\mathrm {gen}}=\left\{ \left( \vec {y},\vec {z}\right) \in {\mathbb {T}}^{E}\setminus {\varvec{B}}^{\left( 1\right) }\,\,:\,\,\sum _{j=1}^{m}\tan (z_{j})+2\sum _{i=1}^{n}\tan \left( \frac{y_{i}}{2}\right) =0\right\} . \end{aligned}$$

  (A.5)

• (2) We denote for $\left( \vec {y},\vec {z}\right) \in \Sigma ^{\mathrm {gen}}$

  $$\begin{aligned} \varvec{i}_{\text {tails}}&(\vec {z}):=\left| \left\{ 1\le j\le m\,\,:\,\,\tan \left( z_{j}\right)<0\right\} \right| ,\\ \varvec{i}_{\text {loops}}&(\vec {y}):=\left| \left\{ 1\le i\le n\,\,:\,\,\tan \left( \frac{y_{i}}{2}\right) <0\right\} \right| . \end{aligned}$$

  Then, the Neumann surplus and nodal surplus functions (introduced in Lemma 5.2) satisfy,

  $$\begin{aligned} \varvec{\omega }\left( \vec {y},\vec {z}\right)&=n-(\varvec{i}_{\text {tails}}\left( \vec {z}\right) +\varvec{i}_{\text {loops}}\left( \vec {y}\right) ),\\ \varvec{\sigma }\left( \vec {y},\vec {z}\right)&=\varvec{i}_{\text {loops}}\left( \vec {y}\right) . \end{aligned}$$

The proof of Lemma A.5 appears after the proof of Proposition A.2:

Proof of Proposition A.2

According to Theorem 2.2 (2b), Theorem 4.8 and Lemma 5.2, it is enough to prove that the functions $\varvec{\sigma }$ and $\varvec{\omega }$, defined on $\Sigma ^{\mathrm {gen}}$, satisfy

$$\begin{aligned} \mathrm {Image}\left( \varvec{\omega }\right)&=\left\{ 1-\left| \partial \Gamma \right| ,\ldots ,\beta -1\right\} ,\end{aligned}$$

(A.6)

$$\begin{aligned} \mathrm {Image}\left( \varvec{\sigma }-\varvec{\omega }\right)&=\left\{ 1-\beta ,\ldots ,\beta +\left| \partial \Gamma \right| -1\right\} , \end{aligned}$$

(A.7)

and

$$\begin{aligned} \mathrm {Image}\left( \varvec{\sigma }\right) ={\left\{ \begin{array}{ll} \left\{ 0,\ldots \beta \right\} &{} \left| \partial \Gamma \right| >0\\ \left\{ 1,\ldots \beta -1\right\} &{} \left| \partial \Gamma \right| =0 \end{array}\right. }. \end{aligned}$$

(A.8)

Let $\Gamma$ be a stower with m tails and n loops and consider the notations of Definition (A.3). For every point $\left( \vec {y},\vec {z}\right) \in {\mathbb {T}}^{E}$, define $t_{j},s_{i}\in {\mathbb {R}}\cup \left\{ \infty \right\}$ by $t_{j}:=\tan \left( z_{j}\right)$ and $s_{i}:=\tan \left( \frac{y_{i}}{2}\right)$ for all coordinates $z_{j}$ and $y_{i}$. Notice that $\left( \vec {y},\vec {z}\right) \in {\mathbb {T}}^{E}\setminus {\varvec{B}}^{\left( 1\right) }$ if and only if all $t_{j}$’s and $s_{j}$’ lie in ${\mathbb {R}}\setminus \left\{ 0\right\}$. According to Lemma A.5 (1), $\left( \vec {y},\vec {z}\right) \in \Sigma ^{\mathrm {gen}}$ if and only if $\left( \vec {y},\vec {z}\right) \in {\mathbb {T}}^{E}\setminus {\varvec{B}}^{\left( 1\right) }$ and

$$\begin{aligned} \sum _{j=1}^{m}t_{j}+2\sum _{j=1+m}^{m+n}s_{j}=0. \end{aligned}$$

(A.9)

The $\varvec{i}_{\text {tails}}\left( \vec {z}\right)$ and $\varvec{i}_{\text {loops}}\left( \vec {y}\right)$ indices, defined in Lemma A.5 (2), are equal to the number of negative $t_{j}$’s and $s_{j}$’s correspondingly. Observe that

$$\begin{aligned} 0&\le \varvec{i}_{\text {tails}}\le m,\,\text {and} \end{aligned}$$

(A.10)

$$\begin{aligned} 0&\le \varvec{i}_{\text {loops}}\le n, \end{aligned}$$

(A.11)

however a solution to (A.9) with nonzero $t_{j}$’s and $s_{j}$’ must have at least one positive and one negative summands, and therefore

$$\begin{aligned} 1\le \varvec{i}_{\text {tails}}+\varvec{i}_{\text {loops}}\le m+n-1. \end{aligned}$$

(A.12)

Using Lemma A.5 (2) and the bounds in (A.10), (A.11) and (A.12), proves the corresponding bounds on $\varvec{\omega },\,\varvec{\sigma }$ and $\varvec{\sigma }-\varvec{\omega }$ and hence inclusion in (A.6), (A.7) and (A.8).

In order to prove the actual equalities in (A.6),(A.7) and (A.8), namely that every value is attained, we show that for any integers $i_{t}$ and $i_{s}$ satisfying

$$\begin{aligned} 0&\le i_{t}\le m,\\ 0&\le i_{s}\le n,\,\text {and}\\ 1&\le i_{t}+i_{s}\le m+n-1, \end{aligned}$$

there exist $\left( \vec {y},\vec {z}\right) \in \Sigma ^{\mathrm {gen}}$ for which $\varvec{i}_{\text {tails}}\left( \vec {z}\right) =i_{t}$ and $\varvec{i}_{\text {loops}}\left( \vec {y}\right) =i_{s}$. Clearly, one can construct a solution to (A.9) for which all $t_{j}$’s and $s_{j}$’ lie in ${\mathbb {R}}\setminus \left\{ 0\right\}$ and there are exactly $i_{t}$ negative $t_{j}$’s and $i_{s}$ negative $s_{j}$’s. Consider $\tan ^{-1}:{\mathbb {R}}\setminus \left\{ 0\right\} \rightarrow \left( 0,\frac{\pi }{2}\right) \cup \left( \frac{\pi }{2},\pi \right)$ and define $z_{j}=\tan ^{-1}\left( t_{j}\right)$ for all $j\le m$ and $y_{i}=2\tan ^{-1}\left( s_{i}\right)$ for $1\le i\le n$. By construction, $\left( \vec {y},\vec {z}\right) \in \Sigma ^{\mathrm {gen}}$ and satisfies $\varvec{i}_{\text {tails}}\left( \vec {z}\right) =i_{t}$ and $\varvec{i}_{\text {loops}}\left( \vec {y}\right) =i_{s}$. $\square$

It remains to prove Lemma A.5:

Proof of Lemma A.5

The first part of the lemma is an explicit characterization $\Sigma ^{\mathrm {gen}}$. To do so consider the standard graph $\Gamma _{\left( \vec {y},\vec {z}\right) }$, i.e., with edge lengths $\vec {l}=\left( \vec {y},\vec {z}\right)$. In the following, we construct an eigenfunction f of $\Gamma _{\vec {\kappa }}$ with eigenvalue $k=1$, providing the conditions on $\left( \vec {y},\vec {z}\right) \in {\mathbb {T}}^{E}$ for which such an eigenfunction exists.

Let v be the central vertex of $\Gamma$, and consider a parametrization of each tail $e_{j}$ by arc-length parametrization $x_{j}\in \left[ 0,z_{j}\right]$ with $x_{j}=0$ at v. Consider a similar parametrization for each loop $e_{i}$ but let $x_{i}\in \left[ -\frac{y_{i}}{2},\frac{y_{i}}{2}\right]$ such that $x_{i}=\pm \frac{y_{i}}{2}$ at v. Let f be an eigenfunction of eigenvalue $k=1$, then its restriction to every tail $e_{j}$ can be written as:

$$\begin{aligned} f|_{e_{j}}\left( x_{j}\right)&=A_{j}\cos \left( z_{j}-x_{j}\right) , \end{aligned}$$

(A.13)

and its restriction to every loop $e_{i}$ can be written as:

$$\begin{aligned} f|_{e_{i}}\left( x_{i}\right)&=C_{i}\cos \left( x_{i}\right) +B_{i}\sin \left( x_{i}\right) . \end{aligned}$$

(A.14)

For every loop $e_{i}$, consider the inversion defined by $x_{i}\mapsto -x_{i}$ on $e_{i}$ while fixing $\Gamma \setminus e_{i}$. It is an isometry of $\Gamma _{\left( \vec {y},\vec {z}\right) }$ and so we may choose all eigenfunctions of $\Gamma _{\left( \vec {y},\vec {z}\right) }$ to be either symmetric or anti-symmetric with respect to this inversion. Anti-symmetric eigenfunctions are loop-eigenfunctions (as defined in Sect. 1.3) and are not generic eigenfunctions (in the sense of Definition 1.3). A loop-eigenfunction which is supported on $e_{i}$ exists if and only if $y_{i}=2\pi$, in which case $\left( \vec {y},\vec {z}\right) \in {\varvec{B}}^{\left( 1\right) }$.

We call f symmetric if it is symmetric on every loop. If f is symmetric, then all $B_{i}$’s in (A.14) are zero. For symmetric eigenfunctions, the Neumann vertex condition at v can be written as:

$$\begin{aligned} \forall j\le m\qquad&A_{j}\cos \left( z_{j}\right) =f\left( v\right) , \end{aligned}$$

(A.15)

$$\begin{aligned} \forall i\le n\qquad&C_{i}\cos \left( \frac{y_{i}}{2}\right) =f\left( v\right) ,\,\,\text { and} \end{aligned}$$

(A.16)

$$\begin{aligned} -k\Bigg (\sum _{j=1}^{m}A_{j}\sin \left( z_{j}\right)&+2\sum _{i=1}^{m}C_{i} \sin \left( \frac{y_{i}}{2}\right) \Bigg )=0. \end{aligned}$$

(A.17)

In particular, for (A.17) to hold there must be at least two nonzero amplitudes among all $A_{j}$’s and $C_{i}$’s. We may deduce from (A.15) and (A.16) that if f is symmetric with $f\left( v\right) =0$, then $\left( \vec {y},\vec {z}\right) \in {\varvec{B}}^{\left( 2\right) }$.

A symmetric eigenfunction f, with $f\left( v\right) \ne 0$, implies that $\cos \left( z_{j}\right) \ne 0$ for every tail and $A_{j}=\frac{f\left( v\right) }{\cos \left( z_{j}\right) }$ by (A.15). Similarly, for every loop, $\cos \left( \frac{y_{i}}{2}\right) \ne 0$ and $C_{i}=\frac{f\left( v\right) }{\cos \left( \frac{y_{i}}{2}\right) }$ by (A.16). Dividing (A.17) by $-kf\left( v\right) \ne 0$, gives

$$\begin{aligned} \sum _{j=1}^{m}\tan \left( x_{j}\right) +2\sum _{i=1}^{n}\tan \left( \frac{y_{i}}{2}\right) =0. \end{aligned}$$

(A.18)

In such case, the outgoing derivatives of f at v do not vanish if and only if,

$$\begin{aligned} \forall j\le m\,\,\,\sin \left( z_{j}\right) \ne 0\,\,\text { and }\,\,\forall 1\le i\le n\,\,\sin \left( \frac{y_{i}}{2}\right) \ne 0. \end{aligned}$$

(A.19)

We may now deduce that f is not generic if $\left( \vec {y},\vec {z}\right) \in {\varvec{B}}^{\left( 1\right) }$. Moreover, if $\left( \vec {y},\vec {z}\right) \notin {\varvec{B}}^{\left( 1\right) }$ then $\Gamma _{\left( \vec {y},\vec {z}\right) }$ has an eigenfunction f of eigenvalue $k=1$ if and only if (A.18) holds, and in such case the construction implies that f is unique (up to a scalar multiplication), $f\left( v\right) \ne 0$, and none of the outgoing derivatives of f at v vanish. Namely, f is generic. This proves Lemma A.5 (1), that is:

$$\begin{aligned} \Sigma ^{\mathrm {gen}}=\left\{ \left( \vec {z},\vec {y}\right) \in {\mathbb {T}}^{E}\setminus {\varvec{B}}^{\left( 1\right) }\,\,:\,\,\sum _{j=1}^{m}\tan \left( x_{j}\right) +2\sum _{i=1}^{n}\tan \left( \frac{y_{i}}{2}\right) =0\right\} .\quad \end{aligned}$$

(A.20)

To prove Lemma A.5 (2), let $\left( \vec {y},\vec {z}\right) \in \Sigma ^{\mathrm {gen}}$ and let f be the corresponding generic eigenfunction of eigenvalue $k=1$. Since f must be symmetric, all $B_{i}$’s in are zero, and we may derive the nodal and Neumann counts from (A.13) and (A.14):

$$\begin{aligned} \phi \left( f\right)&=\left| \left\{ j\le m\,\,:\,\,z_{j}>\frac{\pi }{2}\right\} \right| +\left| \left\{ j\le m\,\,:\,\,z_{j}>\frac{3\pi }{2}\right\} \right| +\ldots \end{aligned}$$

(A.21)

$$\begin{aligned}&\quad \ldots +2\left| \left\{ 1\le i\le n\,\,:\,\,y_{i}>\pi \right\} \right| , \end{aligned}$$

(A.22)

$$\begin{aligned} \xi \left( f\right)&=\left| \left\{ j\le m\,\,:\,\,z_{j}>\pi \right\} \right| +n. \end{aligned}$$

(A.23)

The nodal and Neumann surpluses are given by subtracting the spectral position $N\left( \vec {z},\vec {y}\right)$ from the above. The spectral position is defined by

$$\begin{aligned} N\left( \vec {y},\vec {z}\right) :=&\left| \left\{ k\in \left( 0,1\right] \,\,:\,\,k\,\,\text {is an eigenvalue of }\Gamma _{\left( \vec {y},\vec {z}\right) }\right\} \right| ,\\ =&\left| \left\{ t\in \left( 0,1\right] \,\,:\,\,\left( t\vec {y},t\vec {z}\right) \in \Sigma \right\} \right| , \end{aligned}$$

both counts including multiplicity. In order to avoid multiplicities and to ease the computations define the “good” set:

$$\begin{aligned} M=\left\{ \left( \vec {y},\vec {z}\right) \in \Sigma ^{\mathrm {gen}}\,\,:\,\,\forall t\in \left( 0,1\right] \,\,\,\,\,\,\left( t\vec {y},t\vec {z}\right) \notin {\varvec{B}}^{\left( 2\right) }\cup \Sigma ^{\mathrm {sing}}\right\} . \end{aligned}$$

(A.24)

Let $\left( \vec {y},\vec {z}\right) \in M$ and let $t\in \left( 0,1\right)$ such that $\left( t\vec {y},t\vec {z}\right) \in \Sigma$, then $\left( t\vec {y},t\vec {z}\right) \in \Sigma ^{\mathrm {reg}}$ and corresponds to an eigenfunction that satisfies $f\left( v\right) \ne 0$ since $\left( t\vec {y},t\vec {z}\right) \notin {\varvec{B}}^{\left( 2\right) }$ and $ty_{i}<2\pi$ for all $i\le n$. Therefore, $N\left( \vec {y},\vec {z}\right)$ for $\left( \vec {y},\vec {z}\right) \in M$ is given by

$$\begin{aligned} N\left( \vec {y},\vec {z}\right) =\left| \left\{ t\in \left( 0,1\right] \,\,:\,\,\sum _{j=1}^{m}\tan \left( tz_{j}\right) +2\sum _{i=1}^{n}\tan \left( t\frac{y_{i}}{2}\right) =0\right\} \right| . \end{aligned}$$

(A.25)

Define $F_{\left( \vec {y},\vec {z}\right) }:\left( 0,1\right] \rightarrow {\mathbb {R}}\cup \left\{ \infty \right\}$ by

$$\begin{aligned} F_{\left( \vec {y},\vec {z}\right) }\left( t\right) =\sum _{j=1}^{m}\tan \left( tz_{j}\right) +2\sum _{i=1}^{n}\tan \left( t\frac{y_{i}}{2}\right) . \end{aligned}$$

(A.26)

If $\left( \vec {y},\vec {z}\right) \in M$, then only one of the summands of $F_{\left( \vec {y},\vec {z}\right) }\left( t\right)$ can diverge at each time $t\in \left( 0,1\right]$ since $\left( t\vec {y},t\vec {z}\right) \notin {\varvec{B}}^{\left( 2\right) }$. Therefore, the poles of $F_{\left( \vec {y},\vec {z}\right) }\left( t\right)$ are simple. Since $\frac{d}{dt}F_{\left( \vec {y},\vec {z}\right) }\left( t\right) >0$ whenever t is not a pole, then the zeros of $F_{\left( \vec {y},\vec {z}\right) }$ are simple and they interlace with the poles. Since the interlacing sequence of zeros and poles ends with a zero at $F_{\left( \vec {y},\vec {z}\right) }\left( 1\right) =0$, we can find $N\left( \vec {y},\vec {z}\right)$ by counting poles:

$$\begin{aligned} N\left( \vec {y},\vec {z}\right)&=\left| \left\{ j\le m\,\,:\,\,z_{j}>\frac{\pi }{2}\right\} \right| +\left| \left\{ j\le m\,\,:\,\,z_{j}>\frac{3\pi }{2}\right\} \right| +\ldots \\&\quad \ldots +\left| \left\{ 1\le i\le n\,\,:\,\,y_{i}>\pi \right\} \right| . \end{aligned}$$

Subtracting $N\left( \vec {y},\vec {z}\right)$ from (A.21) and (A.23) gives the nodal and Neumann surplus for any $\left( \vec {y},\vec {z}\right) \in M$:

$$\begin{aligned} \varvec{\omega }\left( \vec {y},\vec {z}\right)&=n+\left| \left\{ j\le m\,\,:\,\,z_{j}>\pi \right\} \right| -N\left( \vec {z},\vec {y}\right) ,\\ \varvec{\sigma }\left( \vec {y},\vec {z}\right)&=\left| \left\{ 1\le i\le n\,\,:\,\,y_{i}>\pi \right\} \right| , \end{aligned}$$

which can be rewritten, using $\varvec{i}_{\text {tails}}$ and $\varvec{i}_{\text {loops}}$ (as defined in Lemma A.5 (2)), by

$$\begin{aligned} \varvec{\omega }\left( \vec {y},\vec {z}\right)&=n-\left( \varvec{i}_{\text {tails}}\left( \vec {z}\right) -\varvec{i}_{\text {loops}}\left( \vec {y}\right) \right) . \end{aligned}$$

(A.27)

$$\begin{aligned} \varvec{\sigma }\left( \vec {y},\vec {z}\right)&=\varvec{i}_{\text {loops}}\left( \vec {y}\right) . \end{aligned}$$

(A.28)

This proves the statement of Lemma A.5 (2) for points in M.

In order to extend this result from M to $\Sigma ^{\mathrm {gen}}$ we recall that $\varvec{\sigma }$ and $\varvec{\omega }$ are constant on each connected component of $\Sigma ^{\mathrm {gen}}$ (Lemma 5.2 and Theorem 5.1). Observe that $\varvec{i}_{\text {tails}}$ and $\varvec{i}_{\text {loops}}$ are constant on connected components of ${\mathbb {T}}^{E}\setminus {\varvec{B}}^{\left( 1\right) }$ and are therefore constant on connected components of $\Sigma ^{\mathrm {gen}}\subset {\mathbb {T}}^{E}\setminus {\varvec{B}}^{\left( 1\right) }$ as well. It is thus enough to show that M intersects each of the connected components of $\Sigma ^{\mathrm {gen}}$ to conclude that Lemma A.5 (2) holds. To do so, we prove that M is dense in $\Sigma ^{\mathrm {gen}}$.

Consider the open set $O=\Sigma ^{\mathrm {gen}}\setminus {\overline{M}}$ so that we need to prove that $O=\emptyset$. Consider the cone $PO:=\left\{ \left( t\vec {y},t\vec {z}\right) \,\,:\,\,t\in \left( 0,1\right] ,\,\left( \vec {y},\vec {z}\right) \in O\right\}$. By the definition of PO, it is a union of lines of the form $\left\{ \left( t\vec {y},t\vec {z}\right) \,\,:\,\,t\in \left( 0,1\right] \right\}$ for some $\left( \vec {y},\vec {z}\right) \in O$. Each such line intersect ${\varvec{B}}^{\left( 2\right) }\cup \Sigma ^{\mathrm {sing}}$, by the definition of O, and each two such lines are either disjoint or one is contained in the other. It follows that the dimension of PO is bounded by

$$\begin{aligned} \dim \left( PO\right) \le \dim \left( {\varvec{B}}^{\left( 2\right) }\cup \Sigma ^{\mathrm {sing}}\right) +1=E-1. \end{aligned}$$

(A.29)

In the last equality, we use Remark A.4 which states that $\dim \left( {\varvec{B}}^{\left( 2\right) }\right) =E-2$ and the fact that $\dim \left( \Sigma ^{\mathrm {sing}}\right) \le E-2$. On the other hand, by Lemma A.5 (1),

$$\begin{aligned} \Sigma ^{\mathrm {gen}}=\left\{ \left( \vec {y},\vec {z}\right) \in {\mathbb {T}}^{E}\setminus {\varvec{B}}^{\left( 1\right) }\,\,:\,\,F_{\left( \vec {y},\vec {z}\right) }\left( 1\right) =0\right\} , \end{aligned}$$

(A.30)

and since $\frac{d}{dt}F_{\left( \vec {y},\vec {z}\right) }\left( t\right) |_{t=1}>0$ on every $\left( \vec {y},\vec {z}\right) \in \Sigma ^{\mathrm {gen}}$, then each of the lines in PO intersect $\Sigma ^{\mathrm {gen}}$ transversely. It follows that

$$\begin{aligned} \dim \left( PO\right) =\dim \left( O\right) +1. \end{aligned}$$

(A.31)

It follows from (A.29) that $\dim \left( O\right) \le E-2$. However, O is open in $\Sigma ^{\mathrm {gen}}$ which is of $\dim \Sigma ^{\mathrm {gen}}=E-1$ so $O=\emptyset$. Therefore, M is dense in $\Sigma ^{\mathrm {gen}}$. $\square$

1.2 A.2. Mandarins

A mandarin graph $\Gamma$ is a graph with only two vertices and $E\ge 3$ edges such that every edge is connected to both vertices. See Fig. 6, for example. In such case $\beta =E-1$ and $\partial \Gamma =\emptyset$.

Proposition A.6

If $\Gamma$ is a mandarin, then its Neumann surplus is bounded by

$$\begin{aligned} 0\le \omega \le \beta , \end{aligned}$$

(A.32)

its nodal surplus is bounded by

$$\begin{aligned} 1\le \sigma \le \beta -1, \end{aligned}$$

(A.33)

and the difference $\sigma -\omega$ is supported inside the set $\left\{ 1{-}\beta ,3{-}\beta ,5{-}\beta ,\ldots ,{-}1{+}\beta \right\}$.

Moreover, if $\Gamma$ has rationally independent edge lengths, then its nodal surplus and Neumann surplus distributions satisfy

$$\begin{aligned} \forall j\in \left\{ 0,1,\ldots \beta -1\right\} \qquad&P\left( \omega =j\right) +P\left( \omega =j+1\right)>0,\\ \forall j\in \left\{ 1,\ldots \beta -2\right\} \qquad&P\left( \sigma =j\right) +P\left( \sigma =j+1\right)>0,\,\,\text {and}\\ \forall j\in \left\{ 1-\beta ,3-\beta ,5-\beta ,\ldots ,-1+\beta \right\} \qquad&P\left( \sigma -\omega =j\right) >0. \end{aligned}$$

This example supports Conjecture 2.4 and shows that the bounds in (3.1) can be achieved. To prove Proposition A.6, let us first state the following.

Remark A.7

The bound $1\le \sigma \le \beta -1$ for mandarin graphs was already shown in [10].

Definition A.8

 Let $\Gamma$ be a mandarin graph with E edges. Call a coordinate $\kappa _{j}$ “bad” if either $\sin (\frac{\kappa _{j}}{2})=0$ or $\cos (\frac{\kappa _{j}}{2})=0$. Denote the set of points in ${\mathbb {T}}^{E}$ which have at least one “bad” coordinate by ${\varvec{B}}^{\left( 1\right) }$, and the similarly denote the set of points with at least two “bad” coordinates by ${\varvec{B}}^{\left( 2\right) }$.

Lemma A.9

Let $\Gamma$ be a mandarin with E edges. Define two functions on ${\mathbb {T}}^{E}$ by,

$$\begin{aligned} F_{s}\left( \vec {\kappa }\right)&:=\sum _{j=1}^{E}\tan \left( \frac{\kappa _{j}}{2}\right) ,\,\text { and}\end{aligned}$$

(A.34)

$$\begin{aligned} F_{a}\left( \vec {\kappa }\right)&:=\sum _{j=1}^{E}\cot \left( \frac{\kappa _{j}}{2}\right) . \end{aligned}$$

(A.35)

Then, $\Sigma ^{\mathrm {gen}}$ has a disjoint decomposition $\Sigma ^{\mathrm {gen}}=\Sigma ^{\mathrm {gen}}_{s}\sqcup \Sigma ^{\mathrm {gen}}_{a}$ such that

• (1) $\Sigma ^{\mathrm {gen}}_{s}$ is given by

  $$\begin{aligned} \Sigma ^{\mathrm {gen}}_{s}=\left\{ \vec {\kappa }\in {\mathbb {T}}^{E}\setminus {\varvec{B}}^{\left( 1\right) }\,\,:\,\,F_{s}\left( \vec {\kappa }\right) =0,\,\text { and }\,F_{a}\left( \vec {\kappa }\right) \ne 0\right\} , \end{aligned}$$

  (A.36)

  and $\Sigma ^{\mathrm {gen}}_{a}$ is given by

  $$\begin{aligned} \Sigma ^{\mathrm {gen}}_{s}=\left\{ \vec {\kappa }\in {\mathbb {T}}^{E}\setminus {\varvec{B}}^{\left( 1\right) }\,\,:\,\,F_{s}\left( \vec {\kappa }\right) \ne 0,\,\text { and }\,F_{a}\left( \vec {\kappa }\right) =0\right\} . \end{aligned}$$

  (A.37)

  In particular, the inversion $T\left( \vec {\kappa }\right) := \left[ \vec {\kappa }+\left( \pi ,\pi ,\pi \ldots .\right) \right]$, is an isometry between the two components, $T\left( \Sigma ^{\mathrm {gen}}_{s}\right) =\Sigma ^{\mathrm {gen}}_{a}$.

• (2) Denote for $\vec {\kappa }\in \Sigma ^{\mathrm {gen}}$

  $$\begin{aligned} \varvec{i}(\vec {\kappa })&:=\left| \left\{ 1\le j\le m\,\,:\,\,\tan (\frac{\kappa _{j}}{2})<0\right\} \right| ,\\ C\left( \vec {\kappa }\right)&:={\left\{ \begin{array}{ll} 1 &{} F_{a}\left( \vec {\kappa }\right) \le 0\\ 0 &{} F_{a}\left( \vec {\kappa }\right) >0 \end{array}\right. }. \end{aligned}$$

  The Neumann surplus and nodal surplus functions (introduced in Lemma 5.2 and Theorem 5.1) satisfy, for $\vec {\kappa }\in \Sigma ^{\mathrm {gen}}_{s}$,

  $$\begin{aligned} \varvec{\sigma }\left( \vec {\kappa }\right)&={\varvec{i}}\left( \vec {\kappa }\right) -C\left( \vec {\kappa }\right) ,\,\,\,\text {and}\\ \varvec{\omega }\left( \vec {\kappa }\right)&=E-{\varvec{i}}\left( \vec {\kappa }\right) -C\left( \vec {\kappa }\right) . \end{aligned}$$

  For $\vec {\kappa }\in \Sigma ^{\mathrm {gen}}_{a}$, $T\left( \vec {\kappa }\right) \in \Sigma ^{\mathrm {gen}}_{s}$ and

  $$\begin{aligned} \varvec{\sigma }\left( \vec {\kappa }\right)&=\varvec{\sigma }\left( T\left( \vec {\kappa }\right) \right) ,\,\,\,\text {and}\\ \varvec{\omega }\left( \vec {\kappa }\right)&=\varvec{\omega }\left( T\left( \vec {\kappa }\right) \right) . \end{aligned}$$

Remark A.10

The s, a labeling of $\Sigma ^{\mathrm {gen}}_{s},\Sigma ^{\mathrm {gen}}_{a}$ stands for symmetric and anti-symmetric, as these parts are defined according to a certain symmetry of the eigenfunctions. It will be introduced in the proof of the Lemma.

The proof of Lemma A.9 appears after the proof of the Proposition A.6:

Proof of Proposition A.6

Define $t_{j}:=\tan \left( \frac{\kappa _{j}}{2}\right)$ so that $\vec {\kappa }\in \Sigma ^{\mathrm {gen}}_{s}$ if and only if $t_{j}\in {\mathbb {R}}\setminus \left\{ 0\right\}$ for every j and the following conditions hold:

$$\begin{aligned} \sum _{j=1}^{E}t_{j}&=0,\,\text {and} \end{aligned}$$

(A.38)

$$\begin{aligned} \sum _{j=1}^{E}\frac{1}{t_{j}}&\ne 0. \end{aligned}$$

(A.39)

In such case ${\varvec{i}}\left( \vec {\kappa }\right)$ is the number of negative $t_{j}$’s, and it is not hard to show that for any $1\le i\le E-1$, one can construct a tuple of $t_{j}$’s that satisfy the above conditions such that the number of negative $t_{j}$’s is equal to i. Namely, there exist $\vec {\kappa }\in \Sigma ^{\mathrm {gen}}_{s}$ with ${\varvec{i}}\left( \vec {\kappa }\right) =i$ for any $1\le i\le E-1=\beta$. In such case, by Lemma A.9 (2),

$$\begin{aligned} \varvec{\sigma }\left( \vec {\kappa }\right)&=i-C\left( \vec {\kappa }\right) \in \left\{ i,i-1\right\} ,\\ \varvec{\omega }\left( \vec {\kappa }\right)&=E-i-C\left( \vec {\kappa }\right) \in \left\{ \beta +1-i,\beta -i\right\} ,\,\,\,\text {and so}\\ \varvec{\sigma }\left( \vec {\kappa }\right) -\varvec{\omega }\left( \vec {\kappa }\right)&=2i-E. \end{aligned}$$

It follows that the image of $\varvec{\sigma }-\varvec{\omega }$ is $\left\{ 2-E,4-E\ldots ,E-2\right\}$, and that the Neumann surplus is bounded by

$$\begin{aligned} 0\le \varvec{\omega }\le \beta . \end{aligned}$$

(A.40)

Moreover, it follows that for any $j\in \left\{ 0,1,\ldots \beta -1\right\}$, $\mathrm {Image}\left( \varvec{\omega }\right)$ contains at least one of $j,j+1$. Same for $\mathrm {Image}\left( \varvec{\sigma }\right)$. Due to the T invariance the restriction of each of the above functions to $\Sigma ^{\mathrm {gen}}_{a}$ has the same image as the restriction to $\Sigma ^{\mathrm {gen}}_{s}$. This ends the proof of Proposition A.6 by the same arguments used in the proof of Proposition A.2.

As already stated, the bound $1\le \sigma \le \beta -1$ was shown in [10], but for completeness we will prove that $\varvec{\sigma }\left( \vec {\kappa }\right) \ne 0$ from which the above bounds follows from symmetry arguments. Assume by contradiction that there is a point $\vec {\kappa }\in \Sigma ^{\mathrm {gen}}_{s}$ for which $\varvec{\sigma }\left( \vec {\kappa }\right) =0$. It follows that ${\varvec{i}}\left( \vec {\kappa }\right) =1$ and $C\left( \vec {\kappa }\right) =1$. Without loss of generality assume that the negative $t_{j}$ is $t_{1}$ so that the $t_{j}$’s satisfy

$$\begin{aligned} \left| t_{1}\right| =\sum _{j=2}^{E}t_{j}, \end{aligned}$$

(A.41)

and $C\left( \vec {\kappa }\right) =1$ implies that

$$\begin{aligned} \sum _{j=2}^{E}\frac{1}{t_{j}}\le \frac{1}{\left| t_{1}\right| }=\frac{1}{\sum _{j=2}^{E}t_{j}}. \end{aligned}$$

(A.42)

As the $t_{j}$’s for $j\ge 2$ are strictly positive, we get

$$\begin{aligned} \frac{\sum _{j=2}^{E}t_{j}}{E-1}\le \frac{1}{E-1}\frac{1}{\sum _{j=2}^{E}\frac{1}{t_{j}}}<\frac{E-1}{\sum _{j=2}^{E}\frac{1}{t_{j}}}, \end{aligned}$$

(A.43)

which contradicts the means inequality. $\square$

The proof of Lemma A.9 is very similar to that of Lemma A.5, and we will therefore provide less details.

Proof of Lemma A.9

Let $\Gamma$ be a mandarin graph, and let $v_{-}$ and $v_{+}$ denote its two vertices. Consider $\Gamma _{\vec {\kappa }}$ for some $\vec {\kappa }\in {\mathbb {T}}^{E}$ with arc-length parametrization $x_{j}\in \left[ -\frac{\kappa _{j}}{2},\frac{\kappa _{j}}{2}\right]$ for every edge $e_{j}$, such that $x_{j}=\pm \frac{\kappa _{j}}{2}$ at $v_{\pm }$. As discussed in [10], the inversion $x_{j}\mapsto -x_{j}$ for all edges simultaneously, is an isometry of $\Gamma _{\vec {\kappa }}$ and we can choose all eigenfunctions to be symmetric$\backslash$ anti-symmetric. In particular, define

$$\begin{aligned} \Sigma ^{\mathrm {gen}}_{s}&:=\left\{ \vec {\kappa }\in \Sigma ^{\mathrm {gen}}\,\,:\,\,f_{\vec {\kappa }}\,\text { is symmetric}\right\} ,\,\text { and}\\ \Sigma ^{\mathrm {gen}}_{a}&:=\left\{ \vec {\kappa }\in \Sigma ^{\mathrm {gen}}\,\,:\,\,f_{\vec {\kappa }}\,\text { is anti-symmetric}\right\} . \end{aligned}$$

A symmetric eigenfunction $f^{\left( s\right) }$ of eigenvalue $k=1$ has the structure:

$$\begin{aligned} f^{\left( s\right) }|_{e_{j}}\left( x_{j}\right) =A_{j}\cos \left( x_{j}\right) . \end{aligned}$$

(A.44)

As in the proof of Lemma A.5 if $f^{\left( s\right) }\left( v_{\pm }\right) =0$ then $\vec {\kappa }\in {\varvec{B}}^{\left( 2\right) }$, and if $f^{\left( s\right) }\left( v_{\pm }\right) \ne 0$ then

$$\begin{aligned} \forall j\le E\,\qquad \cos \left( \kappa _{j}\right) \ne 0, \end{aligned}$$

(A.45)

and the Neumann vertex condition (on each of the vertices) gives

$$\begin{aligned} \sum _{j=1}^{E}\tan \left( \kappa _{j}\right) =F_{s}\left( \vec {\kappa }\right) =0. \end{aligned}$$

(A.46)

In such case, the outgoing derivatives of $f^{\left( s\right) }$ does not vanish if and only if

$$\begin{aligned} \forall j\le E\,\qquad \sin \left( \kappa _{j}\right) \ne 0. \end{aligned}$$

(A.47)

In particular, as in Lemma A.5, in the case of $\vec {\kappa }\notin {\varvec{B}}^{\left( 2\right) }$ there is a unique (up to scalar multiplication) construction of a $k=1$ symmetric eigenfunction with $f^{\left( s\right) }\left( v_{\pm }\right) \ne 0$ is and only if $F_{s}\left( \vec {\kappa }\right) =0$. Moreover, if the eigenfunction is generic then (A.45) and (A.47) implies that $\vec {\kappa }\notin {\varvec{B}}^{\left( 1\right) }$.

An anti-symmetric eigenfunction $f^{\left( a\right) }$ of eigenvalue $k=1$ has the structure:

$$\begin{aligned} f^{\left( a\right) }|_{e_{j}}\left( x_{j}\right) =B_{j}\sin \left( x_{j}\right) . \end{aligned}$$

(A.48)

A similar argument would show that in the case of $\vec {\kappa }\notin {\varvec{B}}^{\left( 2\right) }$ there is a unique (up to scalar multiplication) construction of a $k=1$ anti-symmetric eigenfunction with $f^{\left( a\right) }\left( v_{\pm }\right) \ne 0$ is and only if $F_{a}\left( \vec {\kappa }\right) =0$. Moreover, if the eigenfunction is generic, then (A.45) and (A.47) implies that $\vec {\kappa }\notin {\varvec{B}}^{\left( 1\right) }$.

We may conclude that

$$\begin{aligned} \Sigma ^{\mathrm {gen}}_{s}&=\left\{ \vec {\kappa }\in {\mathbb {T}}^{E}\setminus {\varvec{B}}^{\left( 1\right) }\,\,:\,\,F_{s}\left( \vec {\kappa }\right) =0,\,\text { and }\,F_{a}\left( \vec {\kappa }\right) \ne 0\right\} ,\,\text { and}\\ \Sigma ^{\mathrm {gen}}_{a}&=\left\{ \vec {\kappa }\in {\mathbb {T}}^{E}\setminus {\varvec{B}}^{\left( 1\right) }\,\,:\,\,F_{s}\left( \vec {\kappa }\right) \ne 0,\,\text { and }\,F_{a}\left( \vec {\kappa }\right) =0\right\} . \end{aligned}$$

Observe that $\begin{pmatrix}\cos \left( \frac{\kappa +\pi }{2}\right) \\ \sin \left( \frac{\kappa +\pi }{2}\right) \end{pmatrix}=\begin{pmatrix}-\sin \left( \frac{\kappa }{2}\right) \\ \cos \left( \frac{\kappa }{2}\right) \end{pmatrix}$, and therefore ${\varvec{B}}^{\left( 1\right) }$ is invariant to T and $F_{s}\circ T=-F_{a}$. It follows that

$$\begin{aligned} T\left( \Sigma ^{\mathrm {gen}}_{s}\right) =\Sigma ^{\mathrm {gen}}_{a}. \end{aligned}$$

(A.49)

This proves Lemma A.9 (1). To prove Lemma A.9 (2), consider ${\varvec{i}}\left( \vec {\kappa }\right) :=\left| \left\{ j\le E\,\,:\,\,\tan \left( \frac{\kappa _{j}}{2}\right) <0\right\} \right|$ and let $\vec {\kappa }\in \Sigma ^{\mathrm {gen}}_{s}$ so that $f_{\vec {\kappa }}$ is symmetric and generic. Using (A.44), we derive the nodal and Neumann counts:

$$\begin{aligned} \phi \left( f_{\vec {\kappa }}\right)&=2\left| \left\{ j\le E\,\,:\,\,\frac{\kappa _{j}}{2}>\frac{\pi }{2}\right\} \right| =2{\varvec{i}}\left( \vec {\kappa }\right) ,\,\text {and} \end{aligned}$$

(A.50)

$$\begin{aligned} \xi \left( f_{\vec {\kappa }}\right)&=E. \end{aligned}$$

(A.51)

Similarly, for $\vec {\kappa }\in \Sigma ^{\mathrm {gen}}_{a}$, using (A.48),

$$\begin{aligned} \phi \left( f_{\vec {\kappa }}\right)&=E,\,\text {and} \end{aligned}$$

(A.52)

$$\begin{aligned} \xi \left( f_{\vec {\kappa }}\right)&=2\left| \left\{ j\le E\,\,:\,\,\frac{\kappa _{j}}{2}>\frac{\pi }{2}\right\} \right| =2{\varvec{i}}\left( \vec {\kappa }\right) . \end{aligned}$$

(A.53)

As in the proof of Lemma A.5 (2), we define a “good” set:

$$\begin{aligned} M:=\left\{ \vec {\kappa }\in \Sigma ^{\mathrm {gen}}\,\,:\,\,\forall t\in \left( 0,1\right] \,\,t\vec {\kappa }\notin {\varvec{B}}\cup \Sigma ^{\mathrm {sing}}\right\} , \end{aligned}$$

(A.54)

such that the spectral position for $\vec {\kappa }\in M$ is given by

$$\begin{aligned} N\left( \vec {\kappa }\right) =&\left| \left\{ t\in \left( 0,1\right] \,\,:\,\,F_{s}\left( t\vec {\kappa }\right) =0\right\} \right| +\left| \left\{ t\in \left( 0,1\right] \,\,:\,\,F_{a}\left( t\vec {\kappa }\right) =0\right\} \right| . \end{aligned}$$

For $\vec {\kappa }\in M$ and $t\in \left( 0,1\right)$ the function $t\mapsto F_{a}\left( t\vec {\kappa }\right)$ is continuous, and decreasing from infinity and so

$$\begin{aligned} \left| \left\{ t\in \left( 0,1\right] \,\,:\,\,F_{a}\left( t\vec {\kappa }\right) =0\right\} \right| ={\left\{ \begin{array}{ll} 1 &{} F_{a}\left( \vec {\kappa }\right) \le 0\\ 0 &{} F_{a}\left( \vec {\kappa }\right) >0 \end{array}\right. }=C\left( \vec {\kappa }\right) \end{aligned}$$

For $\vec {\kappa }\in M$, the function $t\mapsto F_{s}\left( t\vec {\kappa }\right)$ is increasing and has interlacing poles and zeros as in the proof of Lemma A.2 and using the same argument we get:

$$\begin{aligned} \left| \left\{ t{\in }\left( 0,1\right] \,\,:\,\,F_{s}\left( t\vec {\kappa }\right) {=}0\right\} \right|&{=}\left| \left\{ j\le E\,\,:\,\,\tan \left( \frac{\kappa _{j}}{2}\right)<0\right\} \right| -1+{\left\{ \begin{array}{ll} 1 &{} F_{s}\left( \vec {\kappa }\right) \ge 0\\ 0 &{} F_{s}\left( \vec {\kappa }\right) <0 \end{array}\right. }\nonumber \\ =&{\varvec{i}}\left( \vec {\kappa }\right) -1+C\left( T\left( \vec {\kappa }\right) \right) , \end{aligned}$$

(A.55)

so that for any $\vec {\kappa }\in M$,

$$\begin{aligned} N\left( \vec {\kappa }\right) ={\varvec{i}}\left( \vec {\kappa }\right) -1+C\left( \vec {\kappa }\right) +C\left( T\left( \vec {\kappa }\right) \right) \end{aligned}$$

(A.56)

Observe that $C\left( \vec {\kappa }\right) \equiv 1$ on $\Sigma ^{\mathrm {gen}}_{a}$ and so $C\left( T\left( \vec {\kappa }\right) \right) \equiv 1$ on $\Sigma ^{\mathrm {gen}}_{s}$. Subtracting $N\left( \vec {\kappa }\right)$ from (A.50) and (A.51), for any $\vec {\kappa }\in M\cap \Sigma ^{\mathrm {gen}}_{s}$, gives

$$\begin{aligned} \varvec{\sigma }\left( \vec {\kappa }\right)&={\varvec{i}}\left( \vec {\kappa }\right) -C\left( \vec {\kappa }\right) ,\,\,\,\text {and}\\ \varvec{\omega }\left( \vec {\kappa }\right)&=E-{\varvec{i}}\left( \vec {\kappa }\right) -C\left( \vec {\kappa }\right) . \end{aligned}$$

Similarly for $\vec {\kappa }\in M\cap \Sigma ^{\mathrm {gen}}_{a}$,

$$\begin{aligned} \varvec{\sigma }\left( \vec {\kappa }\right)&=E-{\varvec{i}}\left( \vec {\kappa }\right) -C\left( T\left( \vec {\kappa }\right) \right) ={\varvec{i}}\left( T\left( \vec {\kappa }\right) \right) -C\left( T\left( \vec {\kappa }\right) \right) =\varvec{\sigma }\left( T\left( \vec {\kappa }\right) \right) ,\\ \varvec{\omega }\left( \vec {\kappa }\right)&={\varvec{i}}\left( \vec {\kappa }\right) -C\left( T\left( \vec {\kappa }\right) \right) =E-{\varvec{i}}\left( T\left( \vec {\kappa }\right) \right) -C\left( T\left( \vec {\kappa }\right) \right) =\varvec{\omega }\left( T\left( \vec {\kappa }\right) \right) . \end{aligned}$$

The same argument as in the proof of Lemma A.2 shows that M is dense in $\Sigma ^{\mathrm {gen}}$ and so the above holds for all $\vec {\kappa }\in \Sigma ^{\mathrm {gen}}$. $\square$

1.3 A.3. Trees

As a particular case of Proposition A.2, the Neumann surplus of a star graph is bounded and gets all integer values between $-1$ to $1-\left| \partial \Gamma \right|$. This result can be generalized to tree graphs using the following lemma:

Lemma A.11

Let $\Gamma ^{\left( 1\right) }$ and $\Gamma ^{\left( 2\right) }$ be tree graphs with edge lengths $\vec {l}^{\left( 1\right) }$ and $\vec {l}^{\left( 2\right) }$, and let $\Gamma$ be the tree graph obtained by gluing the two graphs at two boundary vertices $v_{1}\in \partial \Gamma ^{\left( 1\right) }$ and $v_{2}\in \partial \Gamma ^{\left( 2\right) }$. Let $f^{\left( 1\right) }$ and $f^{\left( 2\right) }$ be generic eigenfunctions of $\Gamma ^{\left( 1\right) }$ and $\Gamma ^{\left( 2\right) }$ with the same eigenvalue k and define the function f on $\Gamma$ by:

$$\begin{aligned} f|_{\Gamma ^{\left( 1\right) }}&=\frac{1}{f^{\left( 1\right) }\left( v_{1}\right) }f^{\left( 1\right) },\\ f|_{\Gamma ^{\left( 2\right) }}&=\frac{1}{f^{\left( 2\right) }\left( v_{2}\right) }f^{\left( 2\right) }. \end{aligned}$$

Then, f is a generic eigenfunction of $\Gamma$ with Neumann surplus:

$$\begin{aligned} \omega \left( f\right) =\omega \left( f|_{\Gamma ^{\left( 1\right) }}\right) +\omega \left( f|_{\Gamma ^{\left( 2\right) }}\right) +1. \end{aligned}$$

(A.57)

Proof

First we show that f is an eigenfunction of $\Gamma$. Let $x_{0}$ be the point in $\Gamma$ which is identified with both $v_{1}$ and $v_{2}$. As $\mathrm {deg}(x_{0})=2$ we will consider it as an interior point $x_{0}\in \Gamma \setminus \partial \Gamma$ and denote the edge containing $x_{0}$ by $e_{0}$. By construction, the Neumann vertex conditions are satisfied on all vertices of $\Gamma$, and the restrictions of f on every edge $e\ne e_{0}$ satisfies the ODE, $f|_{e}''=-k^{2}f|_{e}$. It is also clear that the latter ODE holds for f restricted to $e_{0}\setminus \left\{ x_{0}\right\}$, and we only need to show that both f and $f'$ are continuous at $x_{0}$ in order to conclude that $f|_{e_{0}}''=-k^{2}f|_{e_{0}}$. The normalization of the two parts of f was chosen such that, $f|_{\Gamma ^{\left( 1\right) }}\left( x_{0}\right) =\frac{1}{f^{\left( 1\right) }\left( v_{1}\right) }f^{\left( 1\right) }\left( v_{1}\right) =1$, and in the same way, $f|_{\Gamma ^{\left( 2\right) }}\left( x_{0}\right) =1$. The two directional derivatives at $x_{0}$ are

$$\begin{aligned} f|'_{\Gamma ^{\left( 1\right) }}\left( x_{0}\right)&\propto \partial _{e_{1}}f^{\left( 1\right) }\left( v_{1}\right) =0,\\ f|'_{\Gamma ^{\left( 2\right) }}\left( x_{0}\right)&\propto \partial _{e_{2}}f^{\left( 2\right) }\left( v_{2}\right) =0. \end{aligned}$$

This proves that f is an eigenfunction of $\Gamma$. By its construction, it satisfies both conditions (2) and (3) of the genericity in Definition 1.3, and is therefore generic (see Remark 1.4). Since f is generic, we may consider its nodal count $\phi \left( f\right)$ and Neumann count $\xi \left( f\right)$. By construction, $\phi \left( f\right) =\phi \left( f^{\left( 1\right) }\right) +\phi \left( f^{\left( 2\right) }\right)$ and since $x_{0}$ is an interior point on which $f'\left( x_{0}\right) =0$ then $\xi \left( f\right) =\xi \left( f^{\left( 1\right) }\right) +\xi \left( f^{\left( 2\right) }\right) +1$. Since $\Gamma ,\,\Gamma ^{\left( 1\right) }$ and $\Gamma ^{\left( 2\right) }$ are trees then $\sigma \left( f\right) ,\,\sigma \left( f^{\left( 1\right) }\right)$ and $\sigma \left( f^{\left( 2\right) }\right)$ are zero (see (3.2)). Therefore,

$$\begin{aligned} \omega \left( f^{\left( 1\right) }\right)&=\omega \left( f^{\left( 1\right) }\right) -\sigma \left( f^{\left( 1\right) }\right) =\xi \left( f^{\left( 1\right) }\right) -\phi \left( f^{\left( 1\right) }\right) , \end{aligned}$$

and in the same way

$$\begin{aligned} \omega \left( f^{\left( 2\right) }\right)&=\xi \left( f^{\left( 2\right) }\right) -\phi \left( f^{\left( 2\right) }\right) ,\,\text { and}\\ \omega \left( f\right)&=\xi \left( f\right) -\phi \left( f\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} \omega \left( f\right)&= \xi \left( f\right) -\phi \left( f\right) \\&= \left( \xi \left( f^{\left( 1\right) }\right) -\phi \left( f^{\left( 1\right) }\right) \right) +\left( \xi \left( f^{\left( 2\right) }\right) -\phi \left( f^{\left( 2\right) }\right) \right) +1\\&= \omega \left( f^{\left( 1\right) }\right) +\omega \left( f^{\left( 2\right) }\right) +1. \end{aligned}$$

$\square$

Proposition A.12

If $\Gamma$ is a tree graph, then its Neumann surplus is bounded by

$$\begin{aligned} -\left| \partial \Gamma \right| +1\le \omega \le -1. \end{aligned}$$

(A.58)

Moreover, if $\Gamma$ has rationally independent edge lengths, then its Neumann surplus distribution satisfies:

$$\begin{aligned} \forall j\in \left\{ -\left| \partial \Gamma \right| +1,-\left| \partial \Gamma \right| +2\ldots ,-1\right\} \qquad P\left( \omega =j\right) >0. \end{aligned}$$

(A.59)

Proof

Theorem 2.2 (1) ensures that the Neumann surplus of a tree is bounded between $-\left| \partial \Gamma \right| +1$ and $-1$. According to Theorem 2.2 (2b), Theorem 4.8 and Lemma 5.2, given a fixed value j, if there exist some $\vec {l}\in {\mathbb {R}}_{+}^{E}$ and some generic eigenfunction f of $\Gamma _{\vec {l}}$ such that $\omega \left( f\right) =j$, then for any rationally independent $\vec {l}$ the Neumann surplus distribution has $P\left( \omega =j\right) >0$.

We will prove that for any tree $\Gamma$ and any $j\in \left\{ -\left| \partial \Gamma \right| +1,\ldots -1\right\}$, there exist some $\vec {l}\in {\mathbb {R}}_{+}^{E}$ and some generic eigenfunction f of $\Gamma _{\vec {l}}$ for which $\omega \left( f\right) =j$. The proof is done by induction on the number of interior vertices $V_{in}:=\left| {\mathcal {V}}\setminus \partial \Gamma \right|$, where the case $V_{in}=1$ is a star graph, for which the statement is true by Lemma A.2.

Let $m>1$ and assume that the statement holds for any tree with $V_{in}<m$. Let $\Gamma$ be a tree with $V_{in}=m$, then there is an edge $e_{0}$ which is not a tail. For $x_{0}\in e_{0}$ which is an interior point, $\Gamma \setminus \left\{ x_{0}\right\}$ has two connected components, $\Gamma ^{\left( 1\right) }$ and $\Gamma ^{\left( 2\right) }$, which are both trees. Clearly, both $\Gamma ^{\left( 1\right) }$ and $\Gamma ^{\left( 2\right) }$ have less than m interior vertices. By the assumption, the statement holds for both $\Gamma ^{\left( 1\right) }$ and $\Gamma ^{\left( 2\right) }$. That is, for each $i\in \left\{ 1,2\right\}$, for any $j_{i}\in \left\{ -\left| \partial \Gamma _{i}\right| +1,\ldots -1\right\}$, there exist $\vec {l}^{\left( i\right) }$ and a generic eigenfunction $f^{\left( i\right) }$ of $\Gamma _{\vec {l}^{\left( i\right) }}^{\left( i\right) }$ with Neumann surplus $\omega \left( f^{\left( i\right) }\right) =j_{i}$. We may scale $\vec {l}^{\left( i\right) }$ such that $f^{\left( i\right) }$ has eigenvalue $k=1$. Consider $\Gamma$ with edge lengths according to the gluing of $\Gamma _{\vec {l}^{\left( 1\right) }}^{\left( 1\right) }$ and $\Gamma _{\vec {l}^{\left( 2\right) }}^{\left( 2\right) }$ at $x_{0}$. By Lemma A.11, $\Gamma$ has a generic eigenfunction f with

$$\begin{aligned} \omega \left( f\right) =j_{1}+j_{2}+1. \end{aligned}$$

(A.60)

By construction $\left| \partial \Gamma \right| =\left| \partial \Gamma _{1}\right| +\left| \partial \Gamma _{2}\right| -2$ and so $j:=j_{1}+j_{2}+1$ ranges over all values between

$$\begin{aligned} -\left| \partial \Gamma \right| +1=\left( -\left| \partial \Gamma _{1}\right| +1\right) +\left( -\left| \partial \Gamma _{2}\right| +1\right) +1\le j\le -1-1+1=-1. \end{aligned}$$

(A.61)

By induction, the statement is true for any choice of $V_{in}$. This proves the lemma. $\square$

Appendix B. Proofs of Lemma 4.2 and Lemma 4.12

Proof of Lemma 4.2

We start by providing a basic tool in the spectral analysis of metric graphs. Let $\Gamma$ be a standard graph with $E:={\left| {\mathcal {E}}\right| }$ edges. For every $\vec {\kappa }\in {\mathbb {T}}^{E}$ we abuse notation and denote by $\mathrm {e}^{\mathrm {i}\vec {\kappa }}\in U(2E)$ the following unitary diagonal matrix

$$\begin{aligned} \mathrm {e}^{\mathrm {i}\vec {\kappa }}:=\mathrm {diag}\left( \mathrm {e}^{\mathrm {i}\kappa _{1}},~\mathrm {e}^{\mathrm {i}\kappa _{1}},~\mathrm {e}^{\mathrm {i}\kappa _{2}},~\mathrm {e}^{\mathrm {i}\kappa _{2}},~\ldots ,e^{\mathrm {i}\kappa _{E}},~e^{\mathrm {i}\kappa _{E}}\right) , \end{aligned}$$

(B.1)

such that every exponent $\mathrm {e}^{\mathrm {i}\kappa _{j}}$ appears twice. Given a choice of edge lengths $\vec {l}$ and an eigenvalue k, it is convenient to denote the matrix $\mathrm {e}^{\mathrm {i}\vec {\kappa }}$ for $\vec {\kappa }= \left[ k\vec {l}\right]$ by $\mathrm {e}^{\mathrm {i}k\vec {l}}$. There exists an orthogonal matrix $S\in O(2E)$, uniquely determined by the connectivity of $\Gamma$, which is independent of the edge lengths and has the following property. A positive $k>0$ is an eigenvalue of $\Gamma$ if and only if $\ker ({\mathbb {I}}-\mathrm{e}^{\mathrm {i}k\vec {l}}S)$ is nontrivial [5, Section 3.2]. This relation is a consequence of an isomorphism between $\ker ({\mathbb {I}}-\mathrm{e}^{\mathrm {i}k\vec {l}}S)$ and the k-eigenspace of $\Gamma$. Explicitly, this isomorphism sends ${\varvec{a}}\in \ker ({\mathbb {I}}-\mathrm{e}^{\mathrm {i}k\vec {l}}S)$ to an eigenfunction f of the eigenvalue k such that the restriction to every edge $e\in {\mathcal {E}}$, is given by

$$\begin{aligned} \left. f\right| _{e}\left( x\right) =a_{e}\mathrm {e}^{\mathrm {i}kx}e^{-ikl_{e}}+a_{{\hat{e}}}\mathrm {e}^{-\mathrm {i}kx}, \end{aligned}$$

(B.2)

where $x\in [0,l_{e}]$ is an arc-length parametrization of the edge $e\in {\mathcal {E}}$, and $a_{e},a_{{\hat{e}}}$ are the two entries of the vector ${\varvec{a}}$ which correspond to edge e (according to the indexing of $\mathrm {e}^{\mathrm {i}\vec {\kappa }}$ which was introduced above). As an implication we get for each $\vec {l}$ an isomorphism between the k-eigenspace of $\Gamma _{\vec {l}}$ and the 1-eigenspace of $\Gamma _{\vec {\kappa }}$, where $\vec {\kappa }= \left[ k\vec {l}\right]$. This, together with (4.1) proves part (1) of the Lemma.

The above argument also implies that $\vec {\kappa }\in \Sigma ^{\mathrm {reg}}$ if and only if $\dim \ker (1-e^{i\vec {\kappa }}S)=1$. For any such $\vec {\kappa }$ value, the adjugate matrix $\mathrm {adj}({\mathbb {I}}-\mathrm{e}^{\mathrm {i}\vec {\kappa }}S)$ is a rank one matrix. Hence, there exists a non-trivial ${\varvec{a}}\in \ker ({\mathbb {I}}-\mathrm{e}^{\mathrm {i}\vec {\kappa }}S)$ such that $\mathrm {adj}({\mathbb {I}}-\mathrm{e}^{\mathrm {i}\vec {\kappa }}S)={\varvec{a}}{\varvec{a}}^{*}$. We use this $\varvec{a}$ to introduce a function $f_{\vec {\kappa }}$ by (B.2). This provides a family of canonical eigenfunctions $\{f_{\vec {\kappa }}\}_{\vec {\kappa }\in \Sigma ^{\mathrm {reg}}}$, as in the statement of the lemma. We should note that these functions are not uniquely determined. This is since the vectors $\varvec{a}$ above are determined only up to a multiplication by some unitary factor $c\in U(1)$; indeed $\mathrm {adj}({\mathbb {I}}-\mathrm{e}^{\mathrm {i}\vec {\kappa }}S)={\varvec{a}}{\varvec{a}}^{*}$ is invariant with respect to such a multiplication of $\varvec{a}$, and the factor might depend on $\vec {\kappa }$. Next, we show that this family of canonical eigenfunctions $\{f_{\vec {\kappa }}\}_{\vec {\kappa }\in \Sigma ^{\mathrm {reg}}}$ indeed satisfy the requirements in the lemma.

First, note that the arguments above already prove parts (2a) and (3) of the lemma (the $c\in {\mathbb {C}}$ factor appears in part (3) since the canonical eigenfunction, $f_{\vec {\kappa }}$, is not necessarily real). We proceed to prove part (2b) of the lemma.

Let $v\in {\mathcal {V}}$ and $e\in {\mathcal {E}}_{v}$ with arc-length parametrization $x\in [0,l_{e}]$ such that v is at $x=0$. By (B.2) we get that for each $\vec {\kappa }\in \Sigma ^{\mathrm {reg}}$,

$$\begin{aligned} f_{\vec {\kappa }}\left( v\right)&=f_{\vec {\kappa }}|_{e}\left( 0\right) =a_{e}\mathrm {e}^{-\mathrm {i}\kappa _{e}}+a_{{\hat{e}}}\end{aligned}$$

(B.3)

$$\begin{aligned} \partial _{e}f_{\vec {\kappa }}\left( v\right)&=f_{\vec {\kappa }}|_{e}'\left( 0\right) =\mathrm {i}\left( a_{e}\mathrm {e}^{-\mathrm {i}\kappa _{e}}-a_{{\hat{e}}}\right) . \end{aligned}$$

(B.4)

Observe that each term $f_{\vec {\kappa }}(v)$ or $\partial _{e}f_{\vec {\kappa }}(v)$ is linear in the ${\varvec{a}}$ entries with coefficients which are taken from $(\mathrm {e}^{-\mathrm {i}\kappa _{1}},\ldots ,\mathrm {e}^{-\mathrm {i}\kappa _{E}})$. Therefore, each product of the form $f_{\vec {\kappa }}\left( u\right) \overline{f_{\vec {\kappa }}\left( v\right) }$ or $f_{\vec {\kappa }}\left( u\right) \overline{\partial _{e}f_{\vec {\kappa }}\left( v\right) }$ is linear in the ${\varvec{a}}{\varvec{a}}^{*}$ entries with coefficients which are trigonometric polynomials (in $(\kappa _{1},\ldots ,\kappa _{E})$). But, the entries of ${\varvec{a}}{\varvec{a}}^{*}=\mathrm {adj}({\mathbb {I}}-\mathrm{e}^{\mathrm {i}\vec {\kappa }}S)$ are minors of ${\mathbb {I}}-\mathrm{e}^{\mathrm {i}\vec {\kappa }}S$ and so themselves trigonometric polynomials in $(\kappa _{1},\ldots ,\kappa _{E})$. Hence, $f_{\vec {\kappa }}\left( u\right) \overline{f_{\vec {\kappa }}\left( v\right) }$ is a trigonometric polynomial in $(\kappa _{1},\ldots ,\kappa _{E})$, which we denote by $p_{u,v}(\vec {\kappa })$; and $f_{\vec {\kappa }}\left( u\right) \overline{\partial _{e}f_{\vec {\kappa }}\left( v\right) }$ is a trigonometric polynomial in $(\kappa _{1},\ldots ,\kappa _{E})$, which we denote by $q_{u,v,e}(\vec {\kappa })$. This proves the relations (4.2) and (4.3) in the lemma, and it is left to show that $p_{u,v}$ and $q_{u,v,e}$ are actually real trigonometric polynomials. To do this, we redefine $p_{u,v}$ and $q_{u,v,e}$ to be equal to their real parts and argue that (4.2) and (4.3) are still valid after such a modification. Indeed, all products $f_{\vec {\kappa }}\left( u\right) \overline{f_{\vec {\kappa }}\left( v\right) }$ and $f_{\vec {\kappa }}\left( u\right) \overline{\partial _{e}f_{\vec {\kappa }}\left( v\right) }$ are real since $f_{\vec {\kappa }}$ is an eigenfunction of $\Gamma$ corresponding a simple eigenvalue and as such it is real up to a global multiplicative factor. $\square$

Proof of Lemma 4.12

We begin by referring to the proof of Lemma 4.2 above where it was argued that

$$\begin{aligned} \vec {\kappa }\in \Sigma \Leftrightarrow \dim \ker \left( {\mathbb {I}}-\mathrm{e}^{\mathrm {i}\vec {\kappa }}S\right) >0 \end{aligned}$$

(B.5)

and

$$\begin{aligned} \vec {\kappa }\in \Sigma ^{\mathrm {reg}}\Leftrightarrow \dim \ker \left( {\mathbb {I}}-\mathrm{e}^{\mathrm {i}\vec {\kappa }}S\right) =1. \end{aligned}$$

(B.6)

Combining this with the relation

$$\begin{aligned} \forall \vec {\kappa }\in {\mathbb {T}}^{E},\quad {\mathbb {I}}-\mathrm{e}^{\mathrm {i}{\mathcal {I}}\left( \vec {\kappa }\right) }S=\overline{\left( {\mathbb {I}}-\mathrm{e}^{\mathrm {i}\vec {\kappa }}S\right) }, \end{aligned}$$

(B.7)

gives

$$\begin{aligned} \vec {\kappa }\in \Sigma \Leftrightarrow {\mathcal {I}}(\vec {\kappa })\in \Sigma \quad \text {and }\quad \vec {\kappa }\in \Sigma ^{\mathrm {reg}}\Leftrightarrow {\mathcal {I}}(\vec {\kappa })\in \Sigma ^{\mathrm {reg}}, \end{aligned}$$

(B.8)

and proves that $\Sigma$ and $\Sigma ^{\mathrm {reg}}$ are each invariant under ${\mathcal {I}}$. We may further deduce that $\mathrm {adj}\left( {\mathbb {I}}-\mathrm{e}^{\mathrm {i}{\mathcal {I}}\left( \vec {\kappa }\right) }S\right) =\overline{\mathrm {adj}\left( {\mathbb {I}}-\mathrm{e}^{\mathrm {i}\vec {\kappa }}S\right) }$ and so if ${\varvec{a}}$ is a vector for which ${\varvec{a}}{\varvec{a}}^{*}=\mathrm {adj}\left( {\mathbb {I}}-\mathrm{e}^{\mathrm {i}\vec {\kappa }}S\right)$ then $\overline{{\varvec{a}}}$ is such that $\overline{{\varvec{a}}}\overline{{\varvec{a}}}^{*}=\mathrm {adj}\left( {\mathbb {I}}-\mathrm{e}^{\mathrm {i}{\mathcal {I}}\left( \vec {\kappa }\right) }S\right)$. These vectors define $f_{\vec {\kappa }}$ and $f_{{\mathcal {I}}\left( \vec {\kappa }\right) }$ by (B.3) and (B.4) and so we get

$$\begin{aligned} \forall v\in {\mathcal {V}},\quad \quad f_{{\mathcal {I}}\left( \vec {\kappa }\right) }\left( v\right)&=\overline{a_{e}}\mathrm {e}^{-\mathrm {i}\left[ {\mathcal {I}}\left( \vec {\kappa }\right) \right] _{e}}+\overline{a_{{\hat{e}}}}=\overline{a_{e}\mathrm {e}^{-\mathrm {i}\kappa _{e}}+a_{{\hat{e}}}}\quad \quad \text {and}\end{aligned}$$

(B.9)

$$\begin{aligned} \forall v\in {\mathcal {V}},~\forall e\in {\mathcal {E}}_{v}\quad \quad \partial _{e}f_{{\mathcal {I}}\left( \vec {\kappa }\right) }\left( v\right)&=\mathrm {i}\left( \overline{a_{e}}\mathrm {e}^{-\mathrm {i}\left[ {\mathcal {I}}\left( \vec {\kappa }\right) \right] _{e}}-\overline{a_{{\hat{e}}}}\right) =-\overline{\mathrm {i}\left( a_{e}\mathrm {e}^{-\mathrm {i}\kappa _{e}}-a_{{\hat{e}}}\right) }. \end{aligned}$$

(B.10)

Comparing the RHS above with (B.3) and (B.4) shows

$$\begin{aligned} \forall v\in {\mathcal {V}},~\forall e\in {\mathcal {E}}_{v}\quad \quad f_{{\mathcal {I}}\left( \vec {\kappa }\right) }\left( v\right) =c\overline{f_{\vec {\kappa }}\left( v\right) }\quad \text {and}\quad \partial _{e}f_{{\mathcal {I}}\left( \vec {\kappa }\right) }\left( v\right) =-{\overline{c}}\overline{\partial _{e}f_{\vec {\kappa }}\left( v\right) },\nonumber \\ \end{aligned}$$

(B.11)

where $c\in U(1)$ is a multiplicative factor which expresses a degree of freedom in determining $\varvec{a}$ (alternatively $f_{\vec {\kappa }}$) and c might depend on $\vec {\kappa }$; see also in proof of Lemma 4.2. From (B.11) together with the definition of $\Sigma ^{\mathrm {gen}}$, (4.6) we get that $\Sigma ^{\mathrm {gen}}$is invariant under ${\mathcal {I}}$, which finishes the proof of part (1) of the Lemma.

To continue proving the second part of the lemma, we note that (B.11) implies in particular that all products $f_{\vec {\kappa }}\left( u\right) \overline{f_{\vec {\kappa }}\left( v\right) }$ are ${\mathcal {I}}$ invariant and all products $f_{\vec {\kappa }}\left( u\right) \overline{\partial _{e}f_{\vec {\kappa }}\left( v\right) }$ are ${\mathcal {I}}$ anti-symmetric. As a conclusion, we get that the real trigonometric polynomials $p_{u,v}$ and $q_{u,v,e}$ that are defined in the proof of Lemma 4.2 satisfy

$$\begin{aligned}&\forall \vec {\kappa }\in \Sigma ^{\mathrm {reg}}\,\,\,\,p_{u,v}\left( {\mathcal {I}}\left( \vec {\kappa }\right) \right) =p_{u,v}\left( \vec {\kappa }\right) ,\,\text { and}\nonumber \\&\qquad \forall \vec {\kappa }\in \Sigma ^{\mathrm {reg}}\,\,\,\,q_{u,v,e}\left( {\mathcal {I}}\left( \vec {\kappa }\right) \right) =-q_{u,v,e}\left( \vec {\kappa }\right) . \end{aligned}$$

(B.12)

Note that these are almost the required relations (1),(3). But (B.12) holds for $\vec {\kappa }\in \Sigma ^{\mathrm {reg}}$, whereas (1),(3) are stated for all $\vec {\kappa }\in {\mathbb {T}}^{E}$. Let us abuse notation once again and redefine $p_{u,v}$ as $\frac{p_{u,v}+p_{u,v}\circ {\mathcal {I}}}{2}$ and $q_{u,v,e}$ as $\frac{q_{u,v,e}-q_{u,v,e}\circ {\mathcal {I}}}{2}$. This ensures that $p_{u,v}$ and $q_{u,v,e}$ are ${\mathcal {I}}$ symmetric and anti-symmetric, respectively, while being real trigonometric polynomials whose restrictions are equal to the products $f_{\vec {\kappa }}\left( u\right) \overline{f_{\vec {\kappa }}\left( v\right) }$ and $f_{\vec {\kappa }}\left( u\right) \overline{\partial _{e}f_{\vec {\kappa }}\left( u\right) }$ for any $\vec {\kappa }\in \Sigma ^{reg}$. To show that $\Sigma ^{\mathrm {gen}}$ is ${\mathcal {I}}$ invariant, recall that $\Sigma ^{\mathrm {gen}}$ is the set of $\vec {\kappa }\in \Sigma ^{\mathrm {reg}}$ for which $f_{\vec {\kappa }}$ is generic. Definition 1.3 for genericity has three conditions, where condition (1) holds if and only if $\vec {\kappa }\in \Sigma ^{\mathrm {reg}}$. The other two conditions, (2) and (3), hold if and only if $p_{u,u}\left( \vec {\kappa }\right) \ne 0$ and $q_{v,v,e}\left( \vec {\kappa }\right) \ne 0$ for all $u\in {\mathcal {V}},\,v\in {\mathcal {V}}\setminus \partial \Gamma$ and $e\in {\mathcal {E}}_{v}$. As $\Sigma ^{\mathrm {reg}}$ is ${\mathcal {I}}$ invariant and $p_{u,u}$ and $q_{v,v,e}$ are ${\mathcal {I}}$ symmetric$\backslash$ anti-symmetric, then $f_{{\mathcal {I}}\left( \vec {\kappa }\right) }$ is generic if and only if $f_{\vec {\kappa }}$ is and therefore $\Sigma ^{\mathrm {gen}}$ is ${\mathcal {I}}$ invariant.

The third part of our lemma is stated and proved in [5, lem. 3.13]. $\square$

Appendix C. An Analogue of Proposition 2.7 for Nodal Domains

This appendix provides bounds on the wavelength capacity and the Neumann count of nodal domains.

Proposition C.1

Let $\Gamma$ be a standard graph with minimal edge length $L_{\mathrm {min}}$. Let $k^{2}$ be an eigenvalue of $\Gamma$ which satisfies $k>\frac{\pi }{L_{\mathrm {min}}}$. Let $\Omega _{D}$ be a nodal domain of a generic eigenfunction f which corresponds to k. Denote the total length of $\Omega _{D}$ by $\left| \Omega _{D}\right|$, its wavelength capacity by $\rho (\Omega _{D}):=\frac{\left| \Omega _{D}\right| k}{\pi }$, and its Neumann count by $\xi (\Omega _{D}):=\xi (f|_{\Omega _{D}})$. The following bounds hold

$$\begin{aligned} 1\le&\xi (\Omega _{D})\le \left| \partial \Omega _{D}\right| -1 \end{aligned}$$

(C.1)

$$\begin{aligned} 1\le \frac{1}{2}\left( \xi (\Omega _{D})+1\right) \le&\rho (\Omega _{D})\le \frac{1}{2}\left( \xi (\Omega _{D})+\left| \partial \Omega _{D}\right| -1\right) \le \left| \partial \Omega _{D}\right| -1 \end{aligned}$$

(C.2)

Remark C.2

The spectral position of a Neumann domain is equal to its nodal count. Hence, the bounds (2.7) in Proposition 2.7 may be perceived as the bounds on the nodal count of a Neumann domain. Therefore, we consider the bounds in (C.1) of the proposition above as the analogous bounds. Furthermore, there is no interest in bounds for the spectral position of a nodal domain, as it trivially equals to 1 (see Sect. 1.5).

Proof

We start by noting that the condition $k>\frac{\pi }{L_{\mathrm {min}}}$ guarantees that $\Omega _{D}$ is either a star graph or an interval (see also Lemma 3.1). If $\Omega _{D}$ is an interval, the bounds follow trivially as $\xi (\Omega _{D})=1$ and $\rho (\Omega _{D})=1$. We proceed by assuming that $\Omega _{D}$ is a star graph. From here, the main argument in the proof is a map from star graphs which are nodal domains (such as $\Omega _{D}$) to star graphs which are Neumann domains, as described next. Denote the edge lengths of $\Omega _{D}$ by $\{l_{j}\}_{j=1}^{\left| \partial \Omega _{D}\right| }$. We may write $f|_{\Omega _{D}}$ on every edge $e_{j}$ of $\Omega _{D}$ as

$$\begin{aligned} f|_{e_{j}}\left( x\right) =A_{j}\sin \left( k(x-l_{j})\right) \,\,\,x\in \left[ 0,l_{j}\right] , \end{aligned}$$

(C.3)

where $x=0$ at the central vertex and $x=l_{j}$ at the boundary vertex. The absence of nodal points in the interior of $\Omega _{D}$ together with the genericity of f imply that $kl_{j}\in \left( 0,\frac{\pi }{2}\right) \cup \left( \frac{\pi }{2},\pi \right)$ for each j. We use this to construct an auxiliary star graph, $\Omega _{N}$, which has the same number of edges, $\left| \partial \Omega _{N}\right| =\left| \partial \Omega _{D}\right|$ and its edge lengths are given by

$$\begin{aligned} \forall 1\le j\le \left| \partial \Omega _{N}\right| ,\quad {\tilde{l}}_{j}={\left\{ \begin{array}{ll} l_{j}-\frac{\pi }{2k} &{} kl_{j}\in \left( \frac{\pi }{2},\pi \right) \\ l_{j}+\frac{\pi }{2k} &{} kl_{j}\in \left( 0,\frac{\pi }{2}\right) . \end{array}\right. } \end{aligned}$$

(C.4)

Define a function ${\tilde{f}}$ on $\Omega _{N}$ by

$$\begin{aligned} {\tilde{f}}|_{{\tilde{e}}_{j}}\left( x\right) =A_{j}\sin \left( k(x-l_{j})\right) \,\,\,x\in \left[ 0,{\tilde{l}}_{j}\right] , \end{aligned}$$

(C.5)

for any edge ${\tilde{e}}_{j}$ of $\Omega _{N}$, where $A_{j}$ and $l_{j}$ are the same as in (C.3). This construction guarantees ${\tilde{f}}'|_{e_{j}}\left( {\tilde{l}}_{j}\right) =A_{j}\sin \left( \pm \frac{\pi }{2}\right) =0$ and so ${\tilde{f}}$ satisfies Neumann conditions at the boundary vertices of $\Omega _{N}$. In addition, ${\tilde{f}}$ and f share the same value and derivatives at the corresponding central vertex (of $\Omega _{N}$ and of $\Omega _{D}$). Hence, ${\tilde{f}}$ satisfies Neumann vertex conditions at the central vertex of $\Omega _{N}$. We conclude that ${\tilde{f}}$ is an eigenfunction of $\Omega _{N}$ with eigenvalue k.

Since f and ${\tilde{f}}$ share the same value and derivatives at the central vertex, and f is generic, then ${\tilde{f}}$ satisfies both conditions (2) and (3) of the genericity in Definition 1.3. In addition, k is a simple eigenvalue of $\Omega _{N}$ by an argument similar to the one which was given in the proof of Lemma 3.1 (this argument is based on [20, Corollary 3.1.9] and on $\Omega _{N}$ being a tree). Therefore, ${\tilde{f}}$ is generic. By the construction, ${\tilde{f}}$ has no interior Neumann points, so that $\Omega _{N}$ is a single Neumann domain of the generic eigenfunction ${\tilde{f}}$. In particular, the bounds on $N(\Omega _{N})$ and $\rho (\Omega _{N})$ in Proposition 2.7 apply to it. To finish the proof we just need to relate $N(\Omega _{N})$ and $\rho (\Omega _{N})$ to $\xi (\Omega _{D})$ and $\rho (\Omega _{D})$ and apply those bounds. The needed relations follow from a simple calculations based on (C.3) and (C.5):

$$\begin{aligned} \phi \left( \left. {\tilde{f}}\right| _{\Omega _{N}}\right) +\xi (\Omega _{D})&=\left| \partial \Omega _{D}\right| , \end{aligned}$$

and

$$\begin{aligned}&\rho \left( \Omega _{N}\right) =\frac{1}{\pi }\sum _{j=1}^{\left| \partial \Omega _{N}\right| }{\tilde{l}}_{j}=\frac{1}{\pi }\left( \sum _{j=1}^{\left| \partial \Omega _{D}\right| }\left( l_{j}+\frac{\pi }{2}\right) -\pi \xi \left( f|_{\Omega _{D}}\right) \right) \\&\qquad =\rho \left( \Omega _{D}\right) +\frac{\left| \partial \Omega _{D}\right| }{2}-\xi (\Omega _{D}). \end{aligned}$$

Using the relations above with $\phi ({\tilde{f}}|_{\Omega _{N}})=N(\Omega _{N})$ [see (3.11)] and applying the bounds on $N(\Omega _{N})$ and $\rho (\Omega _{N})$ in Proposition 2.7 yield all the desired bounds in (C.1) and (C.2). $\square$