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Topology in Shallow-Water Waves: A Violation of Bulk-Edge Correspondence


We study the two-dimensional rotating shallow-water model describing Earth’s oceanic layers. It is formally analogue to a Schrödinger equation where the tools from topological insulators are relevant. Once regularized at small scale by an odd-viscous term, such a model has a well-defined bulk topological index. However, in presence of a sharp boundary, the number of edge modes depends on the boundary condition, showing an explicit violation of the bulk-edge correspondence. We study a continuous family of boundary conditions with a rich phase diagram, and explain the origin of this mismatch. Our approach relies on scattering theory and Levinson’s theorem. The latter does not apply at infinite momentum because of the analytic structure of the scattering amplitude there, ultimately responsible for the violation.

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We thank J. Kellendonk for careful reading of the manuscript and valuable remarks. Furthermore, we would like to thank the referees for valuable suggestions and relevant references. The research of H.J. was supported by the Swiss National Science Foundation through SwissMAP.

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Correspondence to Clément Tauber.

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Communicated by R. Seiringer


Chern Number for Spin s Representations

Proposition 1 appears here and there in the literature for various values of s, see e.g. [25, Ex 11.4]. Here we provide a self-contained and general proof. It suffices to prove the case where \(M=S^2\) and we do so by induction in s in steps of 1/2, starting with \(s=0\) and \(s=1/2\). In the first case the bundle is trivial, \(S^2\times \mathbb {C}\), whence \(C(P_{0,0})=0\), where the eigenprojection on the band with labels (sm) is denoted by \(P_{s,m}\). In the case \(s=1/2\), the integrand of (2.6) is

$$\begin{aligned} {{\,\mathrm{tr}\,}}(P_\pm [dP_\pm ,dP_\pm ]) = \pm \frac{{\mathrm i}}{2}\mathbf {e}\cdot (d\mathbf {e}\wedge d\mathbf {e}\,)=\pm \frac{{\mathrm i}}{2}w \end{aligned}$$

with \(P_\pm = P_{\frac{1}{2},\pm \frac{1}{2}}\). This result follows from

$$\begin{aligned} P_\pm =&\frac{1}{2}(1\pm \mathbf {e}\cdot \mathbf {\sigma })\,,\\ {{\,\mathrm{tr}\,}}\,\mathbf {a}\cdot \mathbf {\sigma }\,\big [\mathbf {b}\cdot \mathbf {\sigma },\mathbf {c}\cdot \mathbf {\sigma }\big ] =&4{\mathrm i}\mathbf {a}\cdot (\mathbf {b}\wedge \mathbf {c}\,)\,, \end{aligned}$$

where \(\mathbf {\sigma }=(\sigma _1,\,\sigma _2,\,\sigma _3)^T\) denotes the vector of Pauli matrices \(\sigma _i\). This leads in turn to \(C(P_\pm ) = \pm 1\) by \(\int _{S^2}w = 4\pi \).

We next assume the claim to be true up to s and prove it for \(s+1/2\). Let \(\mathcal {D}_s\) be the irreducible representation of SU(2) of spin s, equipped with the standard basis \({{|{s,m}\rangle }}_{m=-s}^s\) with respect to the quantization axis \(\mathbf {e}\), i.e., \(\mathbf {S}\cdot \mathbf {e}\,{|{s,m}\rangle }=m{|{s,m}\rangle }\). We then have by the Clebsch-Gordan series

$$\begin{aligned} \mathcal {D}_{s+\frac{1}{2}}\oplus \mathcal {D}_{s-\frac{1}{2}} = \mathcal {D}_s\otimes \mathcal {D}_{\frac{1}{2}}\,. \end{aligned}$$

We first treat the case \(m=s+1/2\), for which we find

$$\begin{aligned} {|{s+\frac{1}{2},s+\frac{1}{2}}\rangle }={|{s,s}\rangle }\otimes \,{|{\frac{1}{2},\frac{1}{2}}\rangle }, \end{aligned}$$


$$\begin{aligned} P_{s+\frac{1}{2},s+\frac{1}{2}}=P_{s,s}\otimes P_{\frac{1}{2},\frac{1}{2}}\,. \end{aligned}$$

Since the vector bundles are line bundles, the Chern number is additive, meaning

$$\begin{aligned} C(P_{s+\frac{1}{2},s+\frac{1}{2}})=C(P_{s,s})+C(P_{\frac{1}{2},\frac{1}{2}})=2s+1 \,, \end{aligned}$$

as claimed. The case \(m=-(s+1/2)\) is similar. Finally, we consider the intermediate cases \(m=-(s-1/2),...,s-1/2\). The eigenspace of (total) \(\mathbf {S}\cdot \mathbf {e}\) acting on (A.1) for eigenvalue m has dimension 2 and can be represented as a span in two ways:

$$\begin{aligned} \Bigg [{|{s,m-\frac{1}{2}}\rangle }\otimes \,{|{\frac{1}{2},\frac{1}{2}}\rangle },\, {|{s,m+\frac{1}{2}}\rangle }\otimes \,{|{\frac{1}{2},-\frac{1}{2}}\rangle }\Bigg ] = \Bigg [{|{s-\frac{1}{2},m}\rangle },\,{|{s+\frac{1}{2},m}\rangle }\Bigg ]\,. \end{aligned}$$

The bundle over \(S^2\ni \mathbf {e}\) having the eigenspaces as fibers is thus

$$\begin{aligned} \Big (P_{s,m-\frac{1}{2}}\otimes P_{\frac{1}{2},\frac{1}{2}}\Big )\oplus \Big (P_{s,m+\frac{1}{2}}\otimes P_{\frac{1}{2},-\frac{1}{2}}\Big )=P_{s-\frac{1}{2},m}\oplus P_{s+\frac{1}{2},m}\,. \end{aligned}$$

Arguing as before we get for \(c_{s,m}:=C(P_{s,m})\)

$$\begin{aligned} (c_{s,m-\frac{1}{2}}+c_{\frac{1}{2},\frac{1}{2}})+(c_{s,m+\frac{1}{2}}+c_{\frac{1}{2},-\frac{1}{2}})=c_{s-\frac{1}{2},m}+c_{s+\frac{1}{2},m}\,, \end{aligned}$$


$$\begin{aligned} ((2m-1)+1)+(2m+1-1) = 2m + c_{s+\frac{1}{2},m} \end{aligned}$$

by induction assumption. Thus \(c_{s+\frac{1}{2},m}=2m\) which proves Proposition 1.

Self-adjoint Boundary Conditions

In this appendix we will characterize self-adjoint boundary conditions for our model with domain \(\left\{ (x,y)\,|\, y \ge 0\right\} \subset \mathbb {R}^2\) which preserve translation invariance. After Fourier transformation along x with conjugate variable \(k_x\) (translation invariance in x-direction) the Hamiltonian is given by (cf. (2.9))

$$\begin{aligned} \widetilde{H}(k_x)=\begin{pmatrix} 0 &{}\quad k_x &{}\quad -{\mathrm i}\partial _y\\ k_x &{}\quad 0 &{}\quad -{\mathrm i}(f-\nu (k_x^2-\partial _y^2))\\ -{\mathrm i}\partial _y &{}\quad {\mathrm i}(f-\nu (k_x^2-\partial _y^2)) &{}\quad 0 \end{pmatrix}\,, \end{aligned}$$

which is an operator on the half-line \(y>0\). We shall drop \(k_x\) from the notation till further notice, because it remains fixed. We also drop the \(\tilde{\cdot }\) and denote states by

$$\begin{aligned} \psi = \psi (y) =\begin{pmatrix} \eta \\ u\\ v \end{pmatrix}\,. \end{aligned}$$

Lemma B.1

Self-adjoint realizations of the Hamiltonian H correspond to subspaces \(M\subset \mathbb {C}^6\) with

$$\begin{aligned} \Omega M = M^\perp \,, \end{aligned}$$


$$\begin{aligned} \Omega =\begin{pmatrix} 0 &{}\quad 0 &{}\quad -1 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad -\nu \\ -1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \nu &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad \nu &{}\quad 0 &{} \quad 0 &{}\quad 0\\ 0 &{}\quad -\nu &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{pmatrix}\,. \end{aligned}$$

The domain \(\mathcal D(H)\) is a subspace of the Sobolev space \(H^1\oplus H^2\oplus H^2\) characterized by:

$$\begin{aligned} \psi \in \mathcal {D}(H)\Longleftrightarrow \Psi \in M\,, \end{aligned}$$

where the (stacked) column vector

$$\begin{aligned} \Psi =\begin{pmatrix} \psi \\ \psi ' \end{pmatrix} \end{aligned}$$

stands for the boundary values at \(y=0\).


Without yet imposing any boundary condition, the fact that \(\psi \in H^1 \oplus H^2\oplus H^2\) implies that \(\eta \), \(u'\), \(v'\) are continuous and vanish at infinity. Given two such states \(\psi , \, \tilde{\psi }\), a partial integration of \({\langle }{\tilde{\psi }}{\rangle }\) thus yields boundary terms at \(y=0\) only:

$$\begin{aligned} -{\mathrm i}\left( {\langle }{\tilde{\psi }}{\rangle }-{\langle }{H\tilde{\psi }}{\rangle }\right)&= -(\bar{\tilde{\eta }}v + \bar{\tilde{v}}\eta ) + \nu (\bar{\tilde{v}}u'-\bar{\tilde{v}}'u) -\nu (\bar{\tilde{u}}v'-\bar{\tilde{u}}'v)\nonumber \\&=\tilde{\Psi }^*\Omega \Psi \end{aligned}$$

with \(\Psi \) and \(\tilde{\Psi }\) as in (B.2), \(\Omega \) as in the Lemma to prove and \('=\partial _y\). Let the domain be \(\{ \psi \Vert \Psi \in M \}\), where \(M\subset \mathbb {C}^6\) is some subspace. Then M should have the properties

$$\begin{aligned} \tilde{\Psi }^*\Omega \Psi =0\,, \, (\Psi \in M)&\Rightarrow \tilde{\Psi }\in M\,,\\ \Psi \in M&\Rightarrow \tilde{\Psi }^*\Omega \Psi =0 \,(\forall \tilde{\Psi }\in M)\,. \end{aligned}$$

In fact the first one implies \(H^*\subset H\) and the second \(H\subset H^*\), whence \(H=H^*\) as required. Because of \(\Omega ^*=\Omega \) the two properties are summarized by

$$\begin{aligned} \tilde{\Psi }^*\Omega \Psi =0\,,\,(\Psi \in M) \iff \tilde{\Psi }\in M\,, \end{aligned}$$

which is in turn equivalent to \((\Omega M)^\perp = M\), i.e. to \(\Omega M = M^\perp \). \(\quad \square \)

We note that \({{\,\mathrm{rk}\,}}\Omega = 4\) and

$$\begin{aligned} \ker \Omega \oplus {{\,\mathrm{im}\,}}\Omega = \mathbb {C}^6 \end{aligned}$$

(orthogonal direct sum) by \(\Omega =\Omega ^*\). Dimensions are 2 and 4. Let \(\hat{\Omega }\) be a partial left inverse of \(\Omega \):

$$\begin{aligned} \hat{\Omega }\Omega = P\,, \end{aligned}$$

where P is the orthogonal projection on \({{\,\mathrm{im}\,}}\Omega \) associated to (B.4). It follows

$$\begin{aligned} \Omega \hat{\Omega }v= v ,\quad (v\in {{\,\mathrm{im}\,}}\Omega )\,. \end{aligned}$$

Lemma B.2

Let \(M\subset \mathbb {C}^6\). The following are equivalent:

  1. (a)

    \(\Omega M = M^\perp \)

  2. (b)

    There is a subspace \(\tilde{M}\subset \mathbb {C}^6\) such that

    1. (1)

      \(M=\ker \Omega \oplus \tilde{M}\), orthogonal direct sum, (whence \(\tilde{M}\subset {{\,\mathrm{im}\,}}\Omega \)),

    2. (2)

      \(\hat{\Omega }\tilde{M}^\perp =\tilde{M}\), where \(\tilde{M}^\perp \) is the orthogonal complement of \(\tilde{M}\) within \({{\,\mathrm{im}\,}}\Omega \).

  3. (c)
    1. (1)

      \(\ker \Omega \subset M\),

    2. (2)

      \(\dim M =4\),

    3. (3)

      \(\hat{\Omega }M^\perp \subset M\).


(a)\(\Rightarrow \)(b): By (a), \(M^\perp \subset {{\,\mathrm{im}\,}}\Omega \), and thus by (B.4) \(M\supset \ker \Omega \), proving (b1). Next we have \(M^\perp =\tilde{M}^\perp \) as we find by (b1)

$$\begin{aligned} v\in M^\perp&\iff v\perp \ker \Omega ,\, v\perp \tilde{M}\\&\iff v\in \tilde{M}^\perp \end{aligned}$$

because \((\ker \Omega )^\perp ={{\,\mathrm{im}\,}}\Omega \). With (B.5) we get from (a) \(PM=\hat{\Omega }M^\perp \), i.e. \(\tilde{M}=\hat{\Omega }\tilde{M}^\perp \).

(b)\(\Rightarrow \)(c): First we see directly that (c1) follows from (b1). Furthermore, since \(\hat{\Omega }\) is regular as a map \({{\,\mathrm{im}\,}}\Omega \rightarrow {{\,\mathrm{im}\,}}\Omega \), we have by (b2): \(4-\dim \tilde{M}=\dim \tilde{M}\), i.e. \(\dim \tilde{M}=2\), proving (c2). Property (c3) follows from (b2) and \(M^\perp =\tilde{M}^\perp \).

(c)\(\Rightarrow \)(a): By (c1), i.e. \({{\,\mathrm{im}\,}}\Omega \supset M^\perp \), and (B.6) we get from (c3)

$$\begin{aligned} M^\perp \subset \Omega M\,. \end{aligned}$$

By the rank-nullity theorem applied to \(\Omega : M\rightarrow \mathbb {C}^6\), i.e.

$$\begin{aligned} \dim M = \dim \ker (\Omega \restriction M) + \dim \Omega M\,, \end{aligned}$$

we get \(4=2+\dim \Omega M\) by (c1,c2). Hence equality in (B.7). \(\quad \square \)

Boundary conditions in terms of equations

In the following the self-adjoint boundary conditions will be characterized more explicitly in terms of equations. For that we observe that \(\ker \Omega \) is spanned by the columns of the matrix

$$\begin{aligned} N=\begin{pmatrix} \nu &{}\quad 0\\ 0 &{}\quad 0\\ 0 &{}\quad 0\\ 0 &{}\quad 1\\ 1 &{}\quad 0\\ 0 &{}\quad 0 \end{pmatrix}\,. \end{aligned}$$

The partial left inverse \(\hat{\Omega }\) is uniquely determined on \({{\,\mathrm{im}\,}}\Omega \), but is arbitrary on \(\ker \Omega \). For definiteness, let us choose \(\hat{\Omega }v=0\), \((v\in \ker \Omega )\). By that we have explicitly

$$\begin{aligned} \hat{\Omega } = \begin{pmatrix} 0 &{}\quad 0 &{}\quad -\lambda &{}\quad 0 &{}\quad 0 &{} \quad 0\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad - \nu ^{-1}\\ -\lambda &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \lambda \nu &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad \lambda \nu &{}\quad 0 &{}\quad 0 &{}\quad 0\\ 0 &{} \quad - \nu ^{-1} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{pmatrix}\,,\quad \lambda = \frac{1}{1+\nu ^2}\,. \end{aligned}$$

In fact the so chosen partial left-inverse fulfills \(\Omega \hat{\Omega } = P\) and \(\hat{\Omega }N =0\) (cf. (B.5)).

Proposition B.3

  1. (i)

    Subspaces \(M\subset \mathbb {C}^6\) as in Lemma B.1 of dimension 4 are determined by \(2\times 6\) matrices A of maximal rank, i.e. \({{\,\mathrm{rk}\,}}A=2\), by means of

    $$\begin{aligned} M= \{\Psi \in \mathbb {C}^6 | A\Psi =0 \} = \ker A\,, \end{aligned}$$

    and conversely. Two such matrices A, \(\tilde{A}\) determine the same subspace if and only if \(A=B\tilde{A}\) with \(B\in \mathrm {GL}(2)\).

  2. (ii)

    Self-adjoint boundary conditions are determined precisely by matrices as in (i) with

    $$\begin{aligned} AN=0\,,\qquad A\hat{\Omega }A^*=0\,. \end{aligned}$$


  1. (i)

    Only the last sentence deserves proof, and in fact only the necessity of \(A=B\tilde{A}\). For that consider a map \(B:\,\mathbb {C}^2\rightarrow \mathbb {C}^2\) which is well-defined by \(Av\mapsto \tilde{A}v\), \(v\in \mathbb {C}^6\) because of \(\ker A=\ker \tilde{A}\).

  2. (ii)

    By the Lemma B.2 and in particular by the equivalence between (a) and (c), M is as in (B.9) by (c2). By (c1) and (B.8) the first equation (B.10) applies. Equation (B.9) states \(M^\perp ={{\,\mathrm{ran}\,}}A^* = \{A^*v|v\in \mathbb {C}^2\}\). Thus by (c3),

    $$\begin{aligned} {\langle }{A^*v_1}{\rangle }=0\,,\quad (v_1,\,v_2\in \mathbb {C}^2)\,, \end{aligned}$$

    i.e. the second equation (B.10).

\(\square \)

Example B.4

Dirichlet boundary conditions \(u=0\), \(v=0\) correspond to

$$\begin{aligned} A= \begin{pmatrix} 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{pmatrix} \end{aligned}$$

which is seen to satisfy (B.10).

Local boundary conditions

In this section we will study local boundary conditions which contain the ones studied in the main text of the paper.

For \(\psi (x,y\)), let us consider local boundary conditions at \(y=0\) of the form

$$\begin{aligned} B_0\psi +B_1\partial _x\psi +B_2\partial _y\psi = 0 \end{aligned}$$

with \(l\times 3\)-matrices \(B_i\) (\(l=2\) suffices). After using translation invariance

$$\begin{aligned} \psi (\varvec{x}) = \psi (y)e^{{\mathrm i}k_x x} \end{aligned}$$

they reduce to

$$\begin{aligned} (B_0+{\mathrm i}k_x B_1)\psi + B_2\psi ' =0\,, \end{aligned}$$

(\('=\partial /\partial y\)), i.e. to \(A\Psi =0\) as in Proposition B.3 with

$$\begin{aligned} \begin{aligned} A=(B_0+{\mathrm i}k_xB_1,\, B_2) \equiv A_{0}+{\mathrm i}k_xA_1 \equiv A(k_x)\\ A_{0}=(B_0,\, B_2)\,,\quad A_1=(B_1,\, 0)\,. \end{aligned} \end{aligned}$$

Remark B.5

Quite generally, the condition \({{\,\mathrm{rk}\,}}A \ge 2\) is equivalent to \(A\wedge A \ne 0\). So \({{\,\mathrm{rk}\,}}A(k_x) \ge 2\) (and hence \(=2\), cf. Proposition B.3) for a.e. \(k_x\) means that at least one among

$$\begin{aligned} A_{0}\wedge A_{0}\,,\quad A_{0}\wedge A_{1}+A_{1}\wedge A_{0}\,,\quad A_1\wedge A_1 \end{aligned}$$

does not vanish.

The conditions (B.10) now mean

$$\begin{aligned}&A_{0}N=0\,,\quad A_1N=0,\\&A_{0}\hat{\Omega }A_{0}^*=0\,,\quad A_{0}\hat{\Omega }A_{1}^*-A_{1}\hat{\Omega }A_{0}^*=0\,,\quad A_1\hat{\Omega }A_1^*=0\,. \end{aligned}$$

Example B.6

The boundary conditions \(v=0\), \(\partial _xu+a\partial _yv=0\) correspond to

$$\begin{aligned} A_{0}=\begin{pmatrix} 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad a \end{pmatrix}\,,\quad A_1=\begin{pmatrix} 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{pmatrix}\,, \end{aligned}$$

which are seen to fulfill the conditions stated above.

Remarkable Edge Mode Branches

Coastal Kelvin wave.

In order to show the existence of a branch along (2.11), we note that such \((k_x,\omega )\) remain below the continuum, meaning that both solutions \(X_\pm \) of (3.1) correspond to evanescent waves of momenta \(\kappa _\pm \in {\mathrm i}\mathbb {R}\). As explained at the end of Sect. 3.3, they may superpose to a bound state iff \(g(k_x,\kappa _+)=0\). A condition that is in turn sufficient, but not necessary, for that one is

$$\begin{aligned} v_\infty (k_x,\kappa _\pm )=0\,, \end{aligned}$$

cf.  (3.9). We will confirm this in a few steps. The first one is that

$$\begin{aligned} f-\nu X_\pm >0\,. \end{aligned}$$

In fact \(2\nu (f-\nu X_\pm )=1\mp \sqrt{\Delta }\) by (3.2) with \(\Delta <1\), because of \(f>\omega ^2\nu \) by (2.11). In a second step we rewrite (3.1) as

$$\begin{aligned} \omega ^2=X_\pm +\omega ^2 q_\pm ^2\,,\quad X_\pm =k_x^2+\kappa _\pm ^2 \end{aligned}$$

with \(\omega q_\pm := f-\nu X_\pm \), so as to conclude that \(\kappa _\pm ^2+\omega ^2 q_\pm ^2 =0\). By \(\omega q_\pm >0\) the solution with positive imaginary parts is \(\kappa _\pm ={\mathrm i}\omega q_\pm = -{\mathrm i}k_x q_\pm \), which implies (C.1) by (3.6).

The case \(a=1\). There is a branch of edge states with flat dispersion

$$\begin{aligned} \omega =0\,,\quad (k_x\in \mathbb {R})\,, \end{aligned}$$

as seen in the middle image of Fig. 2: As a crosses 1, a line of edge modes changes the sign of its slope by going through a horizontal line along \(k_x\in \mathbb {R}\). Just as for the Kelvin waves, both solutions \(X_\pm \) of (3.1) correspond to evanescent modes; likewise we claim \(g(k_x,\kappa _+)=0\) on the basis of (3.9). Also in this case there is a sufficient condition, and it reads

$$\begin{aligned} f_\infty (k_x,\kappa _\pm )=0\,. \end{aligned}$$

Some caution is though to be taken first. Since the section \(\widehat{\psi }^\infty \), cf. (3.6), diverges at \(\omega =0\), we better multiply it by \(\omega \) first, so that it remains finite there. Then, a short computation yields

$$\begin{aligned} f_\infty (k_x,\kappa )=(k_x+{\mathrm i}\kappa )\omega +(a-1)\kappa v_\infty (k_x,\kappa )\,, \end{aligned}$$

where (3.1) is understood. Thus (C.2) holds for \(a=1\) at \(\omega =0\).

A further feature of Fig. 2 is confirmed by the following claim.

Proposition C.1

Given any \(\omega \) with \(\left|\omega \right|<f\) there is a branch of edge eigenvalues taking values in \([-\omega ,\omega ]\) at all \(k_x \in \mathbb R\), provided \(\left|a-1\right|\) is small enough.

We keep the normalization used for (C.3) and g as in (3.9) with \(\zeta =\infty \).

Lemma C.2

For \((k_x,\omega )\) in the gap, the function g is purely imaginary. We have

$$\begin{aligned} g(k_x,\omega ;a)=\omega g_1(k_x,\omega )+(a-1)g_2(k_x,\omega ) \end{aligned}$$

with asymptotic behavior at \(k_x\rightarrow \pm \infty \)

$$\begin{aligned} g_1(k_x,\omega )&= {\mathrm i}c(\omega )k_x^{-1}+O(\left|k_x\right|^{-3})\,, \end{aligned}$$
$$\begin{aligned} g_2(k_x,\omega )&= O(\left|k_x\right|^{-1})\,, \end{aligned}$$

where \(c(\omega )\ne 0\) for \(\omega +f>0\) and the estimates are locally uniform in \(\omega \).

The proposition is immediate from the lemma. The statement follows by continuity in a near \(a=1\) as long as \(k_x\) ranges over a compact interval. To be thus proven is existence of eigenvalues in the stated interval for large \(|k_x|\). Given \(\omega >0\), we have \(g(k_x,\pm \omega ;a)\ne 0\) for \(\left|a-1\right|\) small and \(\left|k_x\right|\) large enough; moreover the two values have opposite (imaginary) sign. There thus is \(\omega '\) in the interval with \(g(k_x,\omega ';a)=0\), i.e., an edge eigenvalue.

We next prove Lem. C.2. The function g is imaginary because in the regime considered \(u_\pm =u_\infty (k_x,\kappa _\pm )\) and \(v_\pm =v_\infty (k_x,\kappa _\pm )\) are real and imaginary respectively, cf. (3.63.9). From the latter equation and  (C.3) we get (C.4) with

$$\begin{aligned} g_1=\begin{vmatrix} k_x +{\mathrm i}\kappa _+&k_x-{\mathrm i}\kappa _-\\ v_+&v_- \end{vmatrix}\,,\quad g_2=\begin{vmatrix} \kappa _+ v_+&\kappa _- v_-\\ v_+&v_- \end{vmatrix}\,. \end{aligned}$$

We have the asymptotics for \(k_x\rightarrow \pm \infty \)

$$\begin{aligned} \kappa _\pm ={\mathrm i}\sqrt{k_x^2-X_\pm }={\mathrm i}k_x -{\mathrm i}\frac{X_\pm }{2k_x}+O(\left|k_x\right|^{-3}) \end{aligned}$$

since \(X_\pm \) are bounded here, just as \(\omega \) is. Thus

$$\begin{aligned} k_x+{\mathrm i}\kappa _\pm =\frac{X_\pm }{2k_x}+O(\left|k_x\right|^{-3}) \end{aligned}$$


$$\begin{aligned} v_\pm =\frac{1}{k_x-{\mathrm i}\kappa _\pm }\left( \kappa _\pm \omega +{\mathrm i}k_x(f-vX_\pm )\right) =v_{\pm ,\infty }(\omega )+O(k_x^{-2})\,,\\ v_{\pm ,\infty }(\omega )=\frac{{\mathrm i}}{2}(\omega +f-\nu X_\pm )\,. \end{aligned}$$

We then compute

$$\begin{aligned} g_1 =\frac{1}{2k_x}(X_+v_--X_-v_+)+O(\left|k_x\right|^{-3}) \end{aligned}$$

which fits (C.5) with

$$\begin{aligned} {\mathrm i}c(\omega )&=\frac{{\mathrm i}}{4}\left( X_+(\omega +f-\nu X_-)-X_-(\omega +f -\nu X_+)\right) =\frac{{\mathrm i}}{4}(\omega +f)(X_+-X_-)\,, \end{aligned}$$

which is \(\ne 0\) by \(\Delta \ne 0\), cf. (3.2). Likewise

$$\begin{aligned} g_2=(\kappa _+-\kappa _-)v_+v_-=O(\left|k_x\right|^{-1})\,, \end{aligned}$$

in line with (C.6).

The case \(a=0\). General results imply that \(\mathcal {H}\) is a self-adjoint operator on the half-plane iff \(H(k_x)\) is one on the half-line for a.e. \(k_x\), yet not necessarily for all \(k_x\in \mathbb {R}\). For \(a=0\) such an exception occurs, in that \(A(k_x=0)=A_0\), cf. (B.11), and \({{\,\mathrm{rk}\,}}A_0=1\), which misses the required rank of 2. This lack of self-adjointness at \(k_x=0\) goes along with the transition at \(a=0\) seen in Proposition 5: As a crosses 0, a line of edge modes changes the sign of its slope by going through a vertical line at \(k_x=0\).

Scattering Theory for General Boundary Conditions

The scattering states and the scattering amplitude can be defined for any self-adjoint boundary condition. In the main text we focused on the variables \((k_x,\kappa )\), so that \(\omega =\omega _+(k_x,\kappa )\). Here, we work instead with \((k_x,\omega )\). Asymptotic states are given in terms of plane wave solutions

$$\begin{aligned} \psi (x,t) = \hat{\psi } e^{{\mathrm i}(k_xx+k_yy-\omega t)}\,. \end{aligned}$$

Let \(k_x\) and \(\omega >\sqrt{k_x^2+(f-\nu k_x^2)^2}\) be given. There are two solutions \(X_\pm \equiv \varvec{k}^2\) of

$$\begin{aligned} \omega ^2=X+(f-\nu X)^2\,; \end{aligned}$$

they have \(\pm X_\pm >0\) because of \(\nu ^2 X_+X_-=f^2-\omega ^2<0\); and hence 4 solutions \(k_y\) of \(X=k_x^2+k_y^2\), namely two real ones, incoming (\(\kappa _\mathrm {in}<0)\), outgoing (\(\kappa _\mathrm {out}=-\kappa _\mathrm {in}>0\)); and two imaginary ones, decaying (\(\kappa _\mathrm {ev},\, {\mathrm i}\kappa _\mathrm {ev} <0\)) and diverging (\(\kappa _\mathrm {div}=-\kappa _\mathrm {ev},\, {\mathrm i}\kappa _\mathrm {div}>0\)). The first three are, up to multiples, cf. (3.6),

$$\begin{aligned} \widehat{\psi }_\mathrm {in} = \begin{pmatrix} (k_x^2+\kappa _\mathrm {in}^2)/\omega \\ k_x-{\mathrm i}\kappa _\mathrm {in} q_+\\ \kappa _\mathrm {in}+{\mathrm i}k_x q_+ \end{pmatrix}\,,\quad \widehat{\psi }_\mathrm {out} = \begin{pmatrix} (k_x^2+\kappa _\mathrm {out}^2)/\omega \\ k_x-{\mathrm i}\kappa _\mathrm {out} q_+\\ \kappa _\mathrm {out}+{\mathrm i}k_x q_+ \end{pmatrix}\,,\quad \widehat{\psi }_\mathrm {ev} = \begin{pmatrix} (k_x^2+\kappa _\mathrm {ev}^2)/\omega \\ k_x- {\mathrm i}\kappa _\mathrm {ev} q_-\\ \kappa _\mathrm {ev}+{\mathrm i}k_x q_- \end{pmatrix}\,, \end{aligned}$$


$$\begin{aligned} q_\pm = \frac{f-\nu X_\pm }{\omega }\,. \end{aligned}$$

The solutions (D.2) are to be seen in relation with (D.1). They contribute boundary values

$$\begin{aligned} \Psi _\mathrm {in}=\begin{pmatrix} \widehat{\psi }_\mathrm {in}\\ {\mathrm i}\kappa _\mathrm {in}\widehat{\psi }_\mathrm {in} \end{pmatrix}\,,\quad \Psi _\mathrm {out}=\begin{pmatrix} \widehat{\psi }_\mathrm {out}\\ {\mathrm i}\kappa _\mathrm {out}\widehat{\psi }_\mathrm {out} \end{pmatrix}\,,\quad \Psi _\mathrm {ev}=\begin{pmatrix} \widehat{\psi }_\mathrm {ev}\\ {\mathrm i}\kappa _\mathrm {ev} \widehat{\psi }_\mathrm {ev} \end{pmatrix}\,. \end{aligned}$$

We shall assume that there are no embedded eigenvalues. We conjecture this to be true for any self-adjoint boundary condition, and we show it for (2.8), which is of relevance for the rest of this paper. We do so at the end of this section.

Lemma D.1

For \(k_y\ne 0\) (cf. \(\omega >\sqrt{k_x^2+(f-\nu k_x^2)^2}\) above) the three vectors in (D.4) are linearly independent.


Inspection of \(H\psi =\omega \psi \) as a differential equation in y, cf. (2.9), shows that any initial values

$$\begin{aligned} \Psi = \begin{pmatrix} \psi \\ \psi ' \end{pmatrix}=\begin{pmatrix} \eta \\ \vdots \\ v' \end{pmatrix} \end{aligned}$$

determine an existing and unique solution provided

$$\begin{aligned} k_x u -{\mathrm i}v'=\omega \eta \,, \end{aligned}$$

which (D.4) do. A linear combination

$$\begin{aligned} f_\mathrm {in}\Psi _\mathrm {in} + f_\mathrm {out}\Psi _\mathrm {out}+f_\mathrm {ev}\Psi _\mathrm {ev}=0 \end{aligned}$$

then implies

$$\begin{aligned} f_\mathrm {in}\hat{\psi }_\mathrm {in}e^{{\mathrm i}\kappa _\mathrm {in} y}+f_\mathrm {out}\hat{\psi }_\mathrm {out}e^{{\mathrm i}\kappa _\mathrm {out} y}+f_\mathrm {ev}\hat{\psi }_\mathrm {ev}e^{{\mathrm i}\kappa _\mathrm {ev} y}=0\,. \end{aligned}$$

Linear independence of the functions \(e^{{\mathrm i}\kappa _\mathrm {in} y}\), \(e^{{\mathrm i}\kappa _\mathrm {out} y}\), \(e^{{\mathrm i}\kappa _\mathrm {ev} y}\) implies \(f_\mathrm {in}=f_\mathrm {out}=f_\mathrm {ev}=0\), as was to be shown. \(\quad \square \)

Lemma D.2

Let \(\psi \), \(\tilde{\psi }\) be two solutions of \(H\psi =\omega \psi \) that are bounded in \(y\ge 0\), but regardless of boundary conditions. If one of them vanishes at \(y\rightarrow +\infty \), then

$$\begin{aligned} \tilde{\Psi }^*\Omega \Psi =0\,, \end{aligned}$$

where \(\Psi \) is given by (D.3).


The terms on the r.h.s of (B.3) are supposed to be evaluated at \(y=0\). The equation itself was obtained because the same terms would vanish for \(y\rightarrow \infty \), and that is what they still do here, because in each of them one factor does while the other stays bounded. In fact, if a solution \(\psi \) is bounded (or even vanishes at infinity), i.e. \({{\,\mathrm{Im}\,}}{k_y}\ge 0\) in (D.1), then so does \(\psi '\). Moreover, the l.h.s. of (B.3) vanishes by \(H\psi =\omega \psi \). \(\square \)

Let a boundary condition M be determined by a matrix A as in (B.9B.10). It reads

$$\begin{aligned} A\Psi = 0\,,\quad \text {for}\quad \Psi =f_\mathrm {in}\Psi _\mathrm {in}+f_\mathrm {out}\Psi _\mathrm {out} +f_\mathrm {ev}\Psi _\mathrm {ev}\,. \end{aligned}$$

Lemma D.3

There is a unique solution \((f_\mathrm {in},\,f_\mathrm {out},\,f_\mathrm {ev})\) up to multiples.


There is at least one solution, since the span of (D.4), which has dimension 3 by Lemma D.1, must intersect non-trivially \(\ker A\) (of dimension 4) in view of \(3+4>6\). On the other hand, there are no two linearly independent solutions. In fact, if so, there would be a solution of (D.6) with \(f_\mathrm {in}=0\), i.e.

$$\begin{aligned} \Psi =f_\mathrm {out}\Psi _\mathrm {out}+f_\mathrm {ev}\Psi _\mathrm {ev}\,, \end{aligned}$$

which we shall rule out, unless trivial. Since \(\Psi \in M\) we have \(\Psi ^*\Omega \Psi =0\) by (B.1). Even though (D.5) does not allow to reach the same conclusion for \(\Psi _{\mathrm {out}}\), because \(\psi _{\mathrm {out}}\) is not vanishing at \(y\rightarrow \infty \), it does for \(f_{\mathrm {out}}\Psi _{\mathrm {out}}=\Psi -f_{\mathrm {ev}}\Psi _{\mathrm {ev}}\), because \(\psi _{\mathrm {ev}}\) does and so any contribution to (D.5) involving it:

$$\begin{aligned} \left|f_{\mathrm {out}}\right|^2\cdot \Psi _{\mathrm {out}}^*\Omega \Psi _{\mathrm {out}}=0\,. \end{aligned}$$

We claim that the second factor does not vanish. In fact, a straightforward computation based on (B.3) gives

$$\begin{aligned} \Psi _{\mathrm {out}}^*\Omega \Psi _{\mathrm {out}}&= -\frac{2X_+}{\omega }\kappa _{\mathrm {out}}+4\nu X_+\kappa _{\mathrm {out}}q_+\nonumber \\&= -\frac{2X_+\kappa _{\mathrm {out}}}{\omega }\left( 1-2\nu (f-\nu X_+)\right) <0 \end{aligned}$$

because of \(1-2\nu f>0\), \(2\nu ^2X_+>0\). \(\quad \square \)

Proposition D.4

For any self-adjoint boundary condition, \(k_x\in {\mathbb {R}}\) and \(\omega > \sqrt{k_x^2+(f-\nu k_x^2)^2}\) the scattering amplitude \(S(k_x,\omega )\equiv S = f_\mathrm {out}/f_\mathrm {in}\) is well-defined and satisfies \(|S|=1\).


A straightforward computation yields

$$\begin{aligned} \Psi _{\mathrm {out}}^*\Omega \Psi _{\mathrm {out}} =&-\Psi _{\mathrm {in}}^*\Omega \Psi _{\mathrm {in}}\ne 0\,,\\ \Psi _{\mathrm {out}}^*\Omega \Psi _{\mathrm {in}} =&0 = \Psi _{\mathrm {in}}^*\Omega \Psi _{\mathrm {out}}\,, \end{aligned}$$

where the last equality follows by \(\Omega ^*=\Omega \). Using \(\Psi ^*\Omega \Psi =0\) for \(\Psi \) as in (D.6) yields \(\left|f_{\mathrm {out}}\right|^2-\left|f_{\mathrm {in}}\right|^2=0\) by (D.5). \(\quad \square \)

Remark D.5

The boundary condition (2.8), and in particular \(v=0\), does not allow for embedded eigenvalues. In fact, in view of (D.2), that would amount to

$$\begin{aligned} \kappa _{\mathrm {ev}}+{\mathrm i}k_x q_-=0\,, \end{aligned}$$

and thus to

$$\begin{aligned} X_-&= k_x^2+\kappa _{\mathrm {ev}}^2=k_x^2(1-q_-^2)\\ {}&= \frac{k_x^2}{\omega ^2}( \omega ^2-(f-\nu X_-)^2) =\frac{k_x^2}{\omega ^2}X_- \end{aligned}$$

or equivalently \((\omega ^2-k_x^2)X_-=0\). Since both factors are known to be non-zero this is impossible.

Dirichlet Boundary Condition

The computation of the scattering amplitude is similar to the one in Sect. 3.3, but with (2.8) replaced by (2.19) we end up with the simpler form

$$\begin{aligned} S_\zeta (k_x,\kappa ) = - \dfrac{g_\zeta (k_x,-\kappa )}{g_\zeta (k_x,\kappa )},\qquad g_\zeta (k_x,\kappa ) = \begin{vmatrix} u_\zeta (k_x,\kappa )&u_\infty (k_x,\kappa _\mathrm {ev})\\ v_\zeta (k_x,\kappa )&v_\infty (k_x,\kappa _\mathrm {ev}) \end{vmatrix}. \end{aligned}$$

In particular, for \(\zeta =0\) one infers \(g_0(k_x,\kappa ) \rightarrow 2{\mathrm i}\) as \((k_x,\kappa ) \rightarrow \infty \), so that \(S_0 \rightarrow -1\). Hence the scattering amplitude has no zero or pole in the neighborhood of \(\infty \) and does not wind either. This is the regular situation where Levinson’s theorem applies. Apart from (2.18), the rest of Theorem 9 applies indeed similarly for Dirichlet boundary condition. This implies \(C_+=n_\mathrm {b}\) so that the bulk-edge correspondence is satisfied.

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Graf, G.M., Jud, H. & Tauber, C. Topology in Shallow-Water Waves: A Violation of Bulk-Edge Correspondence. Commun. Math. Phys. 383, 731–761 (2021).

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