# Classifying bulk-edge anomalies in the Dirac Hamiltonian

Hansueli Jud  Lab42, Obere Strasse 22B, 7270 Davos, Switzerland  Clément
Tauber<EMAIL_ADDRESS>CEREMADE, CNRS, Université Paris-Dauphine,
Université PSL, 75016 PARIS, FRANCE

###### Abstract

We study the Dirac Hamiltonian in dimension two with a mass term and a large
momentum regularization, and show that bulk-edge correspondence fails. Despite
a well defined bulk topological index –the Chern number–, the number of edge
modes depends on the boundary condition. The origin of this anomaly is rooted
in the unbounded nature of the spectrum. It is detected with Levinson’s
theorem from scattering theory and quantified via an anomalous winding number
at infinite energy, dubbed ghost charge. First we classify, up to equivalence,
all self-adjoint boundary conditions, using Schubert cell decomposition of a
Grassmanian. Then, we investigate which ones are anomalous. We expand the
scattering amplitude near infinite energy, for which a dominant scale captures
the asymptotic winding number. Remarkably, this can be achieved for every
self-adjoint boundary condition, leading to an exhaustive anomaly
classification. It shows that anomalies are ubiquitous and stable. Boundary
condition with a ghost charge of $2$ is also revealed within the process.

##  1 Introduction

Bulk-edge correspondence is a profound result of topological insulators. It
has been used as a paradigm for more than 40 years to understand the exotic
properties of topological materials. The bulk index, defined for an infinite
sample without boundary, predicts the number of states appearing at the edge
of a half-infinite sample with a sharp boundary. Such edge states are confined
near the boundary and have remarkable properties, like robust unidirectional
propagation in two dimensions. Their number is counted by the edge index,
which equals the bulk index via the bulk-edge correspondence.

Such a result was established with various techniques from spectral analysis
to K-theory and for a very broad class of models, originally in the context of
the quantum Hall effect [16, 24] and then extended to various condensed matter
systems [2, 5, 33, 19, 8, 21, 1, 15, 29, 26, 6, 14, 17, 10, 9, 4] and actually
to almost any energy-preserving classical wave phenomenon [31, 11, 28, 13].
Yet, it has its limitation: Bulk-edge correspondence has been shown to fail in
the context of shallow-water waves [35, 20]. There, the system has a well-
defined bulk index and yet the number of edge modes depends on the choice of
boundary condition. The origin of the anomaly was identified in the unbounded
nature of the operator spectrum. Using scattering theory and a relative
version of Levinson’s theorem, a winding number captures the anomalous
asymptotic properties of the spectrum and compensates for the lack or excess
of edge modes in the finite spectrum. Such a winding number was dubbed “ghost”
charge in [35], with possible physical consequences.

The purpose of this paper is to show that this anomalous situation is not
isolated. First, it is not restricted to water waves. As noticed in [35], we
show that it also appears in the celebrated Dirac Hamiltonian, a canonical
model to describe graphene near Fermi energy [7]. Its massive version is used
as an effective theory for various topological materials, like the Haldane or
Kane-Mele model [23, 25]. However, because its bulk index is ill-defined, we
consider instead its regularized version which compactifies the problem at
large momenta. The regularized and massive Dirac Hamiltonian appears for
example in [38] in the context of superfluid Helium. Second, and more
importantly, we do not study a specific anomalous boundary condition as in
[20] but classify them all.

The first result of this paper is to classify all self-adjoint boundary
conditions for the massive and regularized Dirac Hamiltonian. We identify them
with stable subspaces of $2\times 4$ matrices of maximal rank. Up to
equivalence, we get seven classes based on the Schubert cell decomposition of
the Grassmannian $\mathrm{Gr}(2,4)$. The main result of this paper is then to
compute the asymptotic winding number, denoted $w_{\infty}\in\mathbb{Z}$
below, for any boundary condition among theses classes. Remarkably, the
classification can be done exhaustively. One consequence is that anomalies are
actually ubiquitous and very stable: this is not a fine-tuning effect.
Nevertheless, many boundary conditions (e.g. Dirichlet) are also anomaly-free
and stable as well.

To do so, we develop a formalism to extract precisely the dominant scale of
the scattering amplitude in the asymptotic part of the spectrum, in order to
compute the winding number $w_{\infty}$. Then we manage to apply this strategy
to any boundary condition, class by class. In some cases the expressions
contain up to 60 terms, but using a symbolic computation software
–Mathematica– our approach reduces them to small and tractable complex curves.
Surprisingly, among the many parameters in each class, only one or two drive
the anomaly, so that $w_{\infty}$ can be always computed exactly and moreover
the whole classification can be summarized in a rather compact way. Finally,
the classification also reveals a family of anomalous boundary conditions
where the ghost charge is $2$ instead of $1$, which was unexpected and never
noticed before.

The general message of this paper is that one should always consider with
extra care apparent topological edge modes of unbounded operators, as they
might be anomalous. However, there are several way to circumvent anomalous
boundary conditions. Considering a smooth interface or domain wall instead of
a sharp boundary usually removes the anomaly and restores the bulk-edge
correspondence [34, 32]. Moreover, there are situations where an interface
index can be properly defined even though there is no bulk index [13, 12, 3,
30]. Furthermore it is also possible to avoid the asymptotic part of the
spectrum even with a sharp boundary condition, for example via a spectral flow
of boundary conditions [36] or using an external scalar field [27]. Finally,
we also mention a related anomalous problem for the magnetic Dirac
Hamiltionian in graphene [37], in a slightly different context.

The rest of the paper is organized as follows: In Section 2 we describe the
model, explain the scattering theory to define $w_{\infty}$ and present the
main results: Theorem 2 and 11, associated with Table 1 and 2, respectively.
Section 3 proves the boundary condition classification. Section 4 develops the
main tools to expand the scattering amplitude near infinity, and Section 5
applies these tools to classify all the anomalies.

##  2 Setting and main results

###  2.1 Dirac Hamiltonian and its topological index

The massive and regularized Dirac Hamiltonian is a self-adjoint operator on
$L^{2}(\mathbb{R}^{2})$ with domain $H^{1}(\mathbb{R}^{2})\oplus
H^{1}(\mathbb{R}^{2})$ given by

$\mathcal{H}=\begin{pmatrix}m+\epsilon(\partial_{x}^{2}+\partial_{y}^{2})&{\mathrm{i}}\partial_{x}+\partial_{y}\\\
{\mathrm{i}}\partial_{x}-\partial_{y}&-m-\epsilon(\partial_{x}^{2}+\partial_{y}^{2})\end{pmatrix}\,.$
(2.1)

In the following we assume $m>0$, $\epsilon>0$ and $\epsilon<1/2m$. The mass
parameter $m$ opens a gap in the spectrum, see (2.4) below, and the term
$\epsilon(\partial_{x}^{2}+\partial_{y}^{2})$ regularizes the problem at high
energy. It also has a physical interpretation in the context of superfluid
Helium [38]. By translation invariance in space, the stationary solutions to
the Schrödinger equation ${\mathrm{i}}\partial_{t}\psi=\mathcal{H}\psi$ are
given by normal modes

$\psi:=\widehat{\psi}(k_{x},k_{y},\omega){\mathrm{e}}^{{\mathrm{i}}(k_{x}x+k_{y}x-\omega
t)}\,.$ (2.2)

They have momentum $\boldsymbol{k}=(k_{x},k_{y})\in\mathbb{R}^{2}$, energy
$\omega\in\mathbb{R}$ and correspond to the eigenvalue problem:

$H\widehat{\psi}=\omega\widehat{\psi}\,,\quad\widehat{\psi}=\begin{pmatrix}\widehat{\psi}_{1}\\\
\widehat{\psi}_{2}\end{pmatrix}\,,\quad
H(\boldsymbol{k})=\begin{pmatrix}m-\epsilon\boldsymbol{k}^{2}&-k_{x}+{\mathrm{i}}k_{y}\\\
-k_{x}-{\mathrm{i}}k_{y}&-m+\epsilon\boldsymbol{k}^{2}\end{pmatrix}\,,$ (2.3)

with $\boldsymbol{k}^{2}=k_{x}^{2}+k_{y}^{2}$ and $H(\boldsymbol{k})$ a
Hermitian matrix. The system admits two energy bands

$\omega_{\pm}=\pm\sqrt{\boldsymbol{k}^{2}+\left(m-\epsilon\boldsymbol{k}^{2}\right)^{2}}$
(2.4)

separated by a gap of size $m$. We denote by $P_{\pm}(\boldsymbol{k})$ the
corresponding eigenprojections. Since $\epsilon\neq 0$, these projections are
single-valued at $\boldsymbol{k}\to\infty$, which allows to consider the
associated fibre bundle over the compactified plane
$\mathbb{R}^{2}\cup\\{\infty\\}\cong S^{2}$. See [20] or Appendix A for more
details. Consequently, each projection $P$ has an associated fibre bundle
index –the Chern number– given by

$C(P)=\frac{1}{2\pi{\mathrm{i}}}\int_{S^{2}}dk_{x}dk_{y}\operatorname{tr}\left(P\left[\partial_{k_{x}}P,\partial_{k_{y}}P\right]\right)\,.$
(2.5)

A short computation shows that

$C_{\pm}:=C(P_{\pm})=\pm\dfrac{\operatorname{sign}{m}+\operatorname{sign}{\epsilon}}{2}=\pm
1.$

Thus, each energy band has an associated non-trivial bulk topological index.

###  2.2 Self-adjoint boundary conditions

We restrict the spatial domain to $\\{(x,y)\,|\,y\geq
0\\}\subset\mathbb{R}^{2}$ and study the restriction of $\mathcal{H}$ to it,
named $\mathcal{H}^{\sharp}$. This domain preserves translation invariance in
$x$-direction, so we focus on normal modes of the form
$\psi:=\widetilde{\psi}(y;k_{x},\omega){\mathrm{e}}^{{\mathrm{i}}(k_{x}x-\omega
t)}$. The Schrödinger equation
${\mathrm{i}}\partial_{t}\psi=\mathcal{H}^{\sharp}\psi$ is reduced to the
study of a one-parameter family of operators on $L^{2}(\mathbb{R}^{+})$:

$H^{\sharp}\widetilde{\psi}=\omega\widetilde{\psi},\qquad
H^{\sharp}(k_{x})=\begin{pmatrix}m-\epsilon
k_{x}^{2}+\epsilon\partial_{y}^{2}&-k_{x}+\partial_{y}\\\
-k_{x}-\partial_{y}&-m+\epsilon
k_{x}^{2}-\epsilon\partial_{y}\end{pmatrix}\,,$ (2.6)

for $k_{x},\omega\in\mathbb{R}$. We consider any local boundary conditions for
this problem:

$\left.(B_{0}+ik_{x}B_{1})\widetilde{\psi}+B_{2}\widetilde{\psi}^{\prime}\right|_{y=0}=0$
(2.7)

where $B_{0},B_{1},B_{2}\in\mathrm{M}_{2}(\mathbb{C})$ are $k_{x}$ and
$y$-independent and

$\widetilde{\psi}=\begin{pmatrix}\psi_{1}\\\
\psi_{2}\end{pmatrix},\qquad\widetilde{\psi}^{\prime}=\begin{pmatrix}\partial_{y}\psi_{1}\\\
\partial_{y}\psi_{2}\end{pmatrix}.$ (2.8)

For example, Dirichlet’s boundary condition corresponds to $B_{0}=I_{2}$ and
$B_{1}=B_{2}=0$. However, not all matrices in (2.7) define a self-adjoint
operator associated to (2.6). The first result of this paper is to classify
all local self-adjoint boundary conditions. It appears convenient to regroup
the $B_{i}$ matrices in $2\times 4$-matrices
$A,A_{0},A_{1}\in\mathrm{M}_{2,4}(\mathbb{C})$:

$A:=A_{0}+{\mathrm{i}}k_{x}A_{1},\qquad A_{0}:=[B_{0}|B_{2}],\qquad
A_{1}:=[B_{1}|0],$ (2.9)

so that (2.7) is equivalent to

$\left.A\Psi\right|_{y=0}=0,\qquad\Psi=\begin{pmatrix}\widetilde{\psi}\\\
\widetilde{\psi}^{\prime}\end{pmatrix}\in\mathbb{C}^{4}.$ (2.10)

######  Definition 1.

Two boundary conditions given by $A$ and $\widetilde{A}$ are equivalent if
$\widetilde{A}=BA$ for some $B\in\mathrm{GL}_{2}(\mathbb{C})$.

Equivalent boundary conditions preserve self-adjointness as well as anomalies,
see below.

######  Theorem 2.

For almost every $k_{x}\in\mathbb{R}$ the operator $H^{\sharp}(k_{x})$
together with boundary condition (2.10) is self-adjoint on
$L^{2}(\mathbb{R}_{+})$ if and only if $A\in\mathrm{M}_{2,4}(\mathbb{C})$ is a
rank-2 matrix and is equivalent to one of the seven classes
$\mathfrak{A}_{i,j},\mathfrak{B}$ and $\mathfrak{C}$ given in Table 1.

Class |  $A_{0}$ |  $A_{1}$
---|---|---
$\mathfrak{A}_{1,2}$ |  $\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\end{pmatrix}$ |  $\begin{pmatrix}b_{11}&b_{12}&0&0\\\ b_{21}&b_{22}&0&0\end{pmatrix}$
$\mathfrak{A}_{1,4}$ |  $\begin{pmatrix}1&0&0&0\\\ 0&\alpha&0&1\end{pmatrix}$ |  $\begin{pmatrix}b_{11}&0&0&0\\\ b_{21}&{\mathrm{i}}\beta&0&0\end{pmatrix}$
|  $\alpha\in\mathbb{R}$ |  $\beta\in\mathbb{R}$
$\mathfrak{A}_{2,3}$ |  $\begin{pmatrix}0&1&0&0\\\ \alpha&0&1&0\end{pmatrix}$ |  $\begin{pmatrix}0&b_{12}&0&0\\\ {\mathrm{i}}\beta&b_{22}&0&0\end{pmatrix}$
|  $\alpha\in\mathbb{R}$ |  $\beta\in\mathbb{R}$
$\mathfrak{A}_{2,4}$ |  $\begin{pmatrix}a_{11}&1&0&0\\\ a_{21}&0&(a_{11}^{*})^{-1}&1\end{pmatrix}$ |  $\begin{pmatrix}b_{11}&b_{11}(a_{11})^{-1}&0&0\\\ b_{21}&b_{22}&0&0\end{pmatrix}$
|  $\scriptstyle a_{11}\in\mathbb{C}\setminus\\{0\\},\quad a_{21}=\alpha a_{11}+\epsilon^{-1},\quad\alpha\in\mathbb{R}$ |  $\scriptstyle b_{22}-b_{21}a_{11}^{-1}={\mathrm{i}}\beta,\quad\beta\in\mathbb{R}$
$\mathfrak{A}_{3,4}$ |  $\begin{pmatrix}\alpha_{1}&a_{12}&1&0\\\ {\epsilon^{-1}}-a_{12}^{*}&\alpha_{2}&0&1\end{pmatrix}$ |  $\begin{pmatrix}{\mathrm{i}}\beta_{1}&b_{12}&0&0\\\ b_{12}^{*}&{\mathrm{i}}\beta_{2}&0&0\end{pmatrix}$
|  $\alpha_{1},\alpha_{2}\in\mathbb{R}$ |  $\beta_{1},\beta_{2}\in\mathbb{R}$
$\mathfrak{B}$ |  $\begin{pmatrix}a_{1}&a_{2}&{\mathrm{i}}\alpha&-{\mathrm{i}}\mu^{*}\alpha\\\ \mu a_{1}&\mu a_{2}&{\mathrm{i}}\mu\alpha&-{\mathrm{i}}|\mu|^{2}\alpha\end{pmatrix}$ |  $\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\end{pmatrix}$
|  $\scriptstyle\alpha\in\mathbb{R}$, $\scriptstyle\alpha\big{(}\alpha\operatorname{Im}(\mu)-\epsilon\operatorname{Re}(a_{1}-a_{2}\mu)\big{)}=0$ |
$\mathfrak{C}$ |  $\begin{pmatrix}a_{1}&a_{2}&0&a_{4}\\\ \mu a_{1}&\mu a_{2}&0&\mu a_{4}\end{pmatrix}$ |  $\begin{pmatrix}1&0&0&0\\\ 0&0&0&0\end{pmatrix}$
|  $\scriptstyle(a_{2},a_{4})\neq 0,\,\mu\in\mathbb{C}\setminus\\{0\\},\,\operatorname{Im}\left(a_{2}a_{4}^{*}\right)=0$ |
Table 1: Classification of local self-adjoint boundary conditions from Theorem
2, up to $\mathrm{GL}_{2}$-invariance. All unconstrained parameters are
arbitrary complex numbers: $a_{ij},b_{ij},a_{i}\in\mathbb{C}$.

The proof of this theorem can be found in Section 3. We relate self-adjoint
extensions of $H^{\sharp}$ with stable subspaces generated by $\ker A$, then
use the $\mathrm{GL}_{2}$-invariance to identify $A$ with elements of the
Grassmanian $\mathrm{Gr}(4,2)$. The subclasses $\mathfrak{A}_{i,j}$ correspond
to a Schubert cell decomposition which is compatible with the self-adjoint
constraint.

######  Remark 3.

$A_{0}$ and $A_{1}$ are $k_{x}$-independent, so each couple $(A_{0},A_{1})$
provides a self-adjoint operator $\mathcal{H}^{\sharp}$ on
$L^{2}(\mathbb{R}\times\mathbb{R}^{+})$. It may happen exceptionally that
$A=A_{0}+{\mathrm{i}}k_{x,0}A_{1}$ is not of rank $2$ for some
$k_{x,0}\in\mathbb{R}$ (e.g. $k_{x,0}=0$ in class $\mathfrak{B}$) so that that
$H^{\sharp}(k_{x,0})$ is not self-adjoint. However this would happen only for
a finite number of $k_{x}$ so that $\mathcal{H}^{\sharp}$ remains globally
self-adjoint.

###  2.3 Edge modes

For each boundary condition, equation (2.6) is a system of two coupled ODEs in
$y$ which is exactly solvable for every $k_{x},\,\omega$. We consider the
spectrum of $H^{\sharp}$, namely solutions to (2.6) which are bounded in $y$,
and plot it in the $(k_{x},\omega)$-plane. We distinguish two kinds of
solutions. For $|\omega|\geq\omega_{+}(k_{x},0)$, see (2.4), the solutions are
oscillatory modes delocalized in the upper half-plane, they correspond to the
essential spectrum of $\mathcal{H}$. We call them bulk modes. On the other
hand, for $|\omega|\leq\omega_{+}(k_{x},0)$, there may exist solutions which
decay exponentially when $y\to\infty$. We call them edge modes. As $k_{x}$
varies, edge modes draw continuous curves that we count as follows.

######  Definition 4.

The number $n_{\mathrm{b}}$ of edge modes below a bulk band is the signed
number of edge mode branches emerging $(+)$ or disappearing $(-)$ at the lower
band limit, as $k_{x}\in\mathbb{R}$ increases. The number $n_{\mathrm{a}}$ of
edge modes above a band is counted likewise up to a global sign change.

######  Example 5.

To illustrate the diversity of situations we plot three examples of the
spectrum of $H^{\sharp}$ in Figure 1 for the following boundary condition:

  * •

Dirichlet boundary condition: class $\mathfrak{A}_{12}$ with $A_{0}=[I_{2}|0]$
and $A_{1}=[0|0]$. This corresponds to $\psi_{1}=\psi_{2}=0$ at $y=0$.

  * •

Condition a: class $\mathfrak{A}_{1,4}$ with $A_{0}=\begin{pmatrix}1&0&0&0\\\
0&0&0&1\end{pmatrix}$ and $A_{1}=\begin{pmatrix}0&0&0&0\\\
0&-{\mathrm{i}}&0&0\end{pmatrix}$ which corresponds to $\psi_{1}=0$ and
$\psi_{2}^{\prime}+k_{x}\psi_{2}=0$ at $y=0$.

  * •

Condition b: class $\mathfrak{A}_{3,4}$ with $A_{0}=\begin{pmatrix}1&1&1&0\\\
9&1&0&1\\\ \end{pmatrix}$ and
$A_{1}=\begin{pmatrix}4{\mathrm{i}}&-{\mathrm{i}}&0&0\\\
{\mathrm{i}}&-4{\mathrm{i}}&0&0\\\ \end{pmatrix}$.

Figure 1: Edge spectrum of $H^{\sharp}$ for $m=1$, $\epsilon=0.1$ and various
boundary conditions from Example 5. Blue regions correspond to bulk modes and
yellow curves to edge modes. Their number below the upper band is respectively
$n_{\mathrm{b}}=1,2$ and $3$.

We focus on the edge modes below the upper band. For Dirichlet boundary
condition, we get $n_{\mathrm{b}}=1=C_{+}$, in agreement with the bulk-edge
correspondence. However, such a correspondence is violated for boundary
conditions (a) and (b), for which we get $n_{\mathrm{b}}=2$ and $3$,
respectively. Changing $-{\mathrm{i}}$ to ${\mathrm{i}}$ in $A_{1}$ of
Condition (a) leads to a spectrum with no edge modes, so that
$n_{\mathrm{b}}=0$. As we shall see, for this model all the possible values
are $n_{\mathrm{b}}=-1,0,1,2,3$, see Remark 12 below.

Similarly to [20], a boundary condition with $C_{+}\neq n_{\mathrm{b}}$ is
said to be anomalous. Notice that the mismatch cannot be cured by looking at
higher –but finite– values of $k_{x}$ and $\omega$, possibly revealing extra
edge modes that are not visible in Figure 1. The anomaly is actually rooted in
a singularity at $k_{x}\to\infty$, and can be precisely detected for any
boundary condition via the scattering amplitude formalism.

###  2.4 Scattering amplitude

For the rest of the paper we focus on the upper part of the spectrum
$\omega>0$. A similar analysis could be performed on the lower part111In
general, a boundary condition does not preserve the symmetry of the bulk, so
the nature –anomalous or not– of the upper band may be different than the
lower one. However the analysis of the lower band would be very similar.. Let
$k_{x}\in\mathbb{R}$ and fix
$\omega>\omega_{+}(k_{x},0)=\sqrt{k_{x}^{2}+(m-\epsilon k_{x}^{2})^{2}}$. The
bulk band equation

$k_{x}^{2}+k_{y}^{2}+(m-\epsilon(k_{x}^{2}+k_{y}^{2}))^{2}=\omega^{2}$ (2.11)

has four solutions in $k_{y}$, two are opposite and real and two are opposite
and purely imaginary. Let $\kappa=\kappa(k_{x},\omega)$ be the real positive
one. Then the other solutions are: $-\kappa<0$ and

$\kappa_{\mathrm{ev}/\mathrm{div}}=\pm{\mathrm{i}}\sqrt{\kappa^{2}+2k_{x}^{2}+\frac{1-2\epsilon
m}{\epsilon^{2}}}\in{\mathrm{i}}\mathbb{R}_{+}\,.$ (2.12)

Reciprocally, for fixed $k_{x}\in\mathbb{R}$ and $\kappa>0$,
$k_{y}=\kappa,-\kappa,\kappa_{\mathrm{ev}}$ and $\kappa_{\mathrm{div}}$ are
all solution to (2.11) for the same $\omega=\omega_{+}(k_{x},\kappa)$. Let
$\widehat{\psi}_{\mathrm{in}}$, $\widehat{\psi}_{\mathrm{out}}$ and
$\widehat{\psi}_{\mathrm{ev}}$ be bulk eigenstates, namely solutions of
$H\widehat{\psi}=\omega\widehat{\psi}$, see (2.3), with momentum $k_{x}$ and
$k_{y}=-\kappa,\kappa,\kappa_{\mathrm{ev}}$, respectively. The associated
normal modes are not solutions of the half-space problem, but a linear
combination can be. We define the scattering state as

$\psi_{\mathrm{s}}:=\widetilde{\psi}_{\mathrm{s}}{\mathrm{e}}^{{\mathrm{i}}(k_{x}x-\omega
t)},\qquad\widetilde{\psi}_{\mathrm{s}}:=\widehat{\psi}_{\mathrm{in}}{\mathrm{e}}^{-{\mathrm{i}}\kappa
y}+S\widehat{\psi}_{\mathrm{out}}{\mathrm{e}}^{{\mathrm{i}}\kappa
y}+T\widehat{\psi}_{\mathrm{ev}}{\mathrm{e}}^{{\mathrm{i}}\kappa_{\mathrm{ev}}y}.$
(2.13)

Such a state has well-defined momentum $k_{x}$ and energy
$\omega=\omega_{+}(k_{x},\kappa)$. Each term is respectively interpreted as an
incoming, outgoing and evanescent state with respect to the boundary $y=0$.
The coefficients $S$ and $T$ are tuned so that $\widetilde{\psi}_{\mathrm{s}}$
satisfies the boundary condition (2.10), and $S$ is called the scattering
amplitude222In this framework, $S$ is actually a pure reflection amplitude,
since the incoming wave cannot be transmitted across the sharp boundary..

We are interested in the properties of $S$ as $(k_{x},\kappa)$ varies. Let
$U_{\mathrm{out}}\subset\mathbb{R}^{2}$ be an open subset and let
$U_{\mathrm{in}}$ and $U_{\mathrm{ev}}$ be the images of it by the maps
$(k_{x},\kappa)\mapsto(k_{x},-\kappa)$ and
$(k_{x},\kappa)\mapsto(k_{x},\kappa_{\mathrm{ev}})$. We assume
$\widehat{\psi}_{\mathrm{in}}$, $\widehat{\psi}_{\mathrm{out}}$ and
$\widehat{\psi}_{\mathrm{ev}}$ to be regular and non-vanishing on their
respective domains, and also of amplitude one. Namely
$\lVert\widehat{\psi}_{\mathrm{out}}(k_{x},\kappa)\rVert=1$ for
$k_{x},\kappa\in U_{\mathrm{out}}$, and similarly for the others. In that
case, the scattering state (2.13) is uniquely defined and moreover
$S(k_{x},\kappa)\in U(1)$.

The relevance of the scattering amplitude comes from the following result:

######  Theorem 6.

[20, Thm. 9] Let $\Gamma$ be any smooth, counter-clockwise and not self-
intersecting loop in the half-plane
$\\{(k_{x},\kappa)\in\mathbb{R}^{2},\kappa>0\\}$. Then

$\dfrac{1}{2\pi{\mathrm{i}}}\int_{\Gamma}S^{-1}{\mathrm{d}}S=C_{+}$ (2.14)

Moreover,

$\lim_{\kappa\to
0}\int_{\mathbb{R}}{\mathrm{d}}k_{x}(S^{-1}\partial_{k_{x}}S)(k_{x},\kappa)=n_{\mathrm{b}}$
(2.15)

The scattering amplitude appears as a pivotal point to understand the bulk-
edge correspondence. On the one hand, $S$ can be interpreted as a transition
function between two bulk sections, and its winding is naturally related to
the Chern number. Eq. (2.14) is sometimes called the bulk-scattering
correspondence. On the other hand, $\omega(k_{x},\kappa)\to\omega(k_{x},0)$ as
$\kappa\to 0$ so that Eq. (2.15) is the change of argument of $S$ near the
bottom of the bulk spectrum. This quantity detects merging events of edge mode
branches with the bulk band, thank to a relative version of Levinson’s
theorem, originally developped in [21].

###  2.5 Detecting anomalies: winding number

One sees from Theorem 6 above that in order for the bulk-edge correspondence
to hold, the scattering amplitude needs to be regular and not winding at
infinity. In that case, we can deform the loop $\Gamma$ to the real line
$\kappa=0$ and recover $C_{+}=n_{\mathrm{b}}$. In order to investigate the
infinite part of the spectrum, we need more on the scattering matrix. For a
bulk section $\widehat{\psi}(k_{x},k_{y})$, solution of
$H\widehat{\psi}=\omega\widehat{\psi}$, see (2.3), we denote by

$\widehat{\Psi}(k_{x},k_{y})=\begin{pmatrix}\widehat{\psi}(k_{x},k_{y})\\\
{\mathrm{i}}k_{y}\widehat{\psi}(k_{x},k_{y})\end{pmatrix}\in\mathbb{C}^{4}$
(2.16)

and notice that $A\widehat{\Psi}$ is a column vector of size two for
$A\in\mathrm{M}_{2,4}(\mathbb{C})$.

######  Proposition 7.

Let $\widehat{\psi}^{0}$ a bulk section that is regular and non-vanishing in
$\mathbb{R}^{2}\setminus\\{0\\}\cup\\{\infty\\}$ and let
$\widehat{\psi}^{\infty}$ be a bulk section that is regular and non-vanishing
in $\mathbb{R}^{2}$. Let $A\in\mathrm{M}_{2,4}(\mathbb{C})$ of rank 2 be a
boundary condition as in (2.10). The scattering amplitude reads:

$S(k_{x},\kappa)=-\dfrac{g(k_{x},-\kappa)}{g(k_{x},\kappa)},\qquad
g(k_{x},\kappa)=\det([A\widehat{\Psi}^{0}(k_{x},\kappa),A\widehat{\Psi}^{\infty}(k_{x},\kappa_{\mathrm{ev}})]),$
(2.17)

and is regular in the limit $k_{x}^{2}+\kappa^{2}\to\infty$. In particular,
equivalent boundary conditions have the same scattering amplitude $S$.

This proposition is proved in Section 4.1. The last sentence is obvious from
the expression of $g$ above and ensures that the boundary condition
classification from Theorem 2 is relevant for the investigation of the
anomalous bulk-edge correspondence. To study the winding phase of $S$ near
infinity, we consider the following path for $\lambda>0$ and $\delta>0$:

$\Gamma_{\delta}(\lambda)=\Big{\\{}\Big{(}k_{x}=\dfrac{-\lambda_{x}}{\lambda_{x}^{2}+\delta^{2}},\kappa=\dfrac{\delta}{\lambda_{x}^{2}+\delta^{2}}\Big{)},\lambda_{x}\in[-\lambda,\lambda]\Big{\\}}$
(2.18)

where $\lambda_{x},\delta$ are called the dual variables to $k_{x},\kappa$.

######  Definition 8.

We say that $S$ winds at $\infty$ if $w_{\infty}\neq 0$ with

$w_{\infty}:=\lim_{\lambda\to 0}\lim_{\delta\to
0}\int_{\Gamma_{\delta}(\lambda)}S^{-1}{\mathrm{d}}S$

For small $\delta$, $\Gamma_{\delta}(\lambda)$ explores the region near
$k_{x}=0$, $\kappa=\infty$. The limit $\lambda\to 0$ ensures that we are not
counting other edge modes at finite but large $k_{x}$.

######  Lemma 9.

$C_{+}=n_{\mathrm{b}}+w_{\infty}$.

###### Proof.

According to Theorem 6 one has

$C_{+}=\dfrac{1}{2\pi{\mathrm{i}}}\int_{\Gamma}S^{-1}{\mathrm{d}}S$

for any simple and counterclockwise loop $\Gamma$. Fix $\delta>0$ and consider

$\mathcal{C}_{\delta}=\Big{\\{}\Big{(}k_{x}=\dfrac{-\lambda_{x}}{\lambda_{x}^{2}+\delta^{2}},\kappa=\dfrac{\delta}{\lambda_{x}^{2}+\delta^{2}}\Big{)},\lambda_{x}\in\mathbb{R}\Big{\\}}$

This loop naturally splits into
$\mathcal{C}_{\delta}=\Gamma_{\delta}(\lambda)\cup(\Gamma_{\delta}(\lambda))^{c}$
for any $\lambda>0$, with

$(\Gamma_{\delta}(\lambda))^{c}=\Big{\\{}\Big{(}k_{x}=\dfrac{-\lambda_{x}}{\lambda_{x}^{2}+\delta^{2}},\kappa=\dfrac{\delta}{\lambda_{x}^{2}+\delta^{2}}\Big{)},\lambda_{x}\in]\lambda,+\infty[\cup]-\infty,\lambda[\Big{\\}}$

In particular, $(\Gamma_{\delta}(\lambda))^{c}$ becomes the real line
$\kappa=0$, $k_{x}\in\mathbb{R}$ as $\delta\to 0$ and $\lambda\to 0$, and $S$
is regular in the upper half-plane $\kappa>0$ so that

$\lim_{\lambda\to 0}\lim_{\delta\to
0}\int_{(\Gamma_{\delta}(\lambda))^{c}}S^{-1}{\mathrm{d}}S=\lim_{\kappa\to
0}\int_{\mathbb{R}}{\mathrm{d}}k_{x}(S^{-1}\partial_{k_{x}}S)(k_{x},\kappa)=n_{\mathrm{b}}$

The remaining part of the integral along $\Gamma_{\delta}$ leads to
$w_{\infty}$. ∎

Consequently, $w_{\infty}$ detects when a boundary condition is anomalous at
infinity, but also predicts the number of edge modes below the upper bulk band
by $n_{\mathrm{b}}=C_{+}-w_{\infty}$ since $C_{+}=1$ is fixed. The main idea
to compute $w_{\infty}$ is to expand $S$ and $g$ from Proposition 7 near
$k_{x}^{2}+\kappa^{2}\to\infty$, then extract the leading term $\mathcal{S}$
along the path $\Gamma_{\delta}(\lambda)$. Ultimately, the winding number is
reduced to

$w_{\infty}=\int_{-\infty}^{+\infty}\mathcal{S}^{-1}(u)\mathcal{S}^{\prime}(u){\mathrm{d}}u,\qquad\mathcal{S}(u):=\dfrac{G_{-}(u)}{G_{+}(u)}.$

with $G_{\pm}:\mathbb{R}\to\mathbb{C}^{*}$ two complex curves whose winding
phases can be computed explicitly.

######  Example 10.

We continue Example 5.

  * •

Dirichlet boundary condition: one has $G_{+}=G_{-}=1$ so that $w_{\infty}=0$.

  * •

Condition a: one has $G_{-}(u)=-u-{\mathrm{i}}$ and $G_{+}(u)=G_{-}^{*}(u)$.
As $u$ goes from $-\infty$ to $+\infty$, $G_{-}$ goes from
$+\infty-{\mathrm{i}}$ to $-\infty-{\mathrm{i}}$ so that its change of
argument is $-\pi$. Consequently $G_{+}$ argument change is $\pi$, so that $S$
winds by $-2\pi$ and $w_{\infty}=-1$. We recover $n_{\mathrm{b}}=1-(-1)=2$.

  * •

Condition b: one has

$G_{-}(u)=-15u^{2}+4u\sqrt{2u^{2}+1}+{\mathrm{i}}\left(\sqrt{2u^{2}+1}-4u\right),\qquad
G_{+}(u)=G^{*}_{-}(u).$

A detailed analysis of this complex curve (see Section 5.5) shows that it is
dominated by its real part as $u\to\pm\infty$. $\operatorname{Re}(G_{-})$ goes
from $-\infty$ (with $\operatorname{Im}(G_{-})>0$) to $-\infty$ (with
$\operatorname{Im}(G_{-})<0$). Its imaginary part is strictly decreasing and
vanishes at $u_{0}$ with $\operatorname{Re}(G_{-}(u_{0}))>0$. Thus $G_{-}$
winds once clockwise around zero as $u\in\mathbb{R}$, and $G_{+}=G_{-}^{*}$
winds once anti-clockwise. Thus $w_{\infty}=-2$, and we recover
$n_{\mathrm{b}}=3$.

What is remarkable for this problem is that $G_{\pm}$, $\mathcal{S}$ and
$w_{\infty}$ can be computed not only for specific examples, but actually for
any boundary condition from Table 1. Moreover, among the many free parameters
in each class, only a few of them survive near infinity. The anomalous nature
(and the value of $w_{\infty}$) is driven by one or two parameters only, so
that an exhaustive classification can be formulated in a rather compact way.
This is the central result of this paper.

######  Theorem 11.

$w_{\infty}$ can be systematically computed for almost every self-adjoint
boundary condition. Its value is given in Table 2. Each expression for
$\mathcal{S}$ can be found in Section 5.

Class |  $w_{\infty}$ |  Condition
---|---|---
$\mathfrak{A}_{1,2}$ |  $0$ |  None
$\mathfrak{A}_{1,4}$ |  $\operatorname{sign}_{0}(\beta)$ |  None
|  $-\operatorname{sign}(\beta)$ |  $\beta\neq 0$
$\mathfrak{A}_{2,3}$ |  $1$ |  $\beta=0$ and $|\alpha-\epsilon^{-1}|<1/\sqrt{2}$
|  $0$ |  $\beta=0$ and $|\alpha-\epsilon^{-1}|>1/\sqrt{2}$
$\mathfrak{A}_{2,4}$ |  $\operatorname{sign}(\beta)$ |  $\beta|a_{11}|^{2}>\sqrt{2}$ or $\beta|a_{11}|^{2}<-\sqrt{2}$
|  $0$ |  $-\sqrt{2}<\beta|a_{11}|^{2}<\sqrt{2}$
|  $\operatorname{sign}(B_{+})$ |  $b_{12}\neq 0$ and $B_{+}B_{-}<0$
$\mathfrak{A}_{3,4}$ |  $0$ |  $b_{12}\neq 0$ and $B_{\pm}>0$
|  $2\times\operatorname{sign}(\sqrt{2}-\beta_{1})$ |  $b_{12}\neq 0$ and $B_{\pm}<0$
|  $\operatorname{sign}_{0}(\beta)$ |  $b_{12}=0$ and $\beta_{1}^{2}\neq 2$
$\mathfrak{B}$ |  $0$ |  None
$\mathfrak{C}$ |  $1$ |  $a_{2}=0$
|  $0$ |  $a_{2}\neq 0$
Table 2: Anomaly classification for any self-adjoint boundary condition. The
class and parameters refer to Table 1. For class $\mathfrak{A}_{3,4}$ we also
define $B_{\pm}:=\beta_{2}(\beta_{1}\pm\sqrt{2})+|b_{12}|^{2}$. We distinguish
the sign function $\operatorname{sign}:\mathbb{R}\setminus\\{0\\}\to\\{\pm
1\\}$ from its extended version $\operatorname{sign}_{0}:\mathbb{R}\to\\{0,\pm
1\\}$ which sends $0$ to $0$.

######  Remark 12.

Beyond the classification in itself, several important consequences can be
inferred:

  * •

Class $\mathfrak{A}_{1,2}$ and $\mathfrak{B}$ are never anomalous:
$w_{\infty}=0$, whereas all other classes are of both nature with
$w_{\infty}\in\\{0,\pm 1,\pm 2\\}$. By Lemma 9, we deduce that
$n_{\mathrm{b}}\in\\{-1,0,1,2,3\\}$.

  * •

Among the many parameters in each class, only a few (up to three) actually
drive the value of $w_{\infty}$. Thus each phase is rather stable: in both
cases (anomalous or not), several free parameters can vary without changing
$w_{\infty}$. In particular, anomalies are not fine-tuned effects among the
boundary condition, but are as stable as the non-anomalous regime.

  * •

The possibility that $w_{\infty}=\pm 2$ was unexpected from previous works and
could be revealed only through the full classification.

  * •

There is another possibility for the upper band to be anomalous, with
$w_{\infty}=0$ and an edge mode branch merging exactly at infinity, see [20]
for an example. The detection of asymptotic edge modes is another distinct
classification that we postpone to future work.

  * •

Class $\mathfrak{A}_{1,2},\mathfrak{A}_{1,4},\mathfrak{B}$ and $\mathfrak{C}$
are completely classified. Some threshold cases remain: in
$\mathfrak{A}_{2,3}$ ($\alpha-\epsilon^{-1}=\pm 1/\sqrt{2}$),
$\mathfrak{U}_{2,4}$ ($\beta|a_{11}|^{2}=\pm\sqrt{2}$) and
$\mathfrak{U}_{3,4}$ ($B_{+}=0$ or $B_{-}=0$), where we suspect $w_{\infty}$
to be ill-defined. In that sense, the classification is almost exhaustive.

The rest of the paper is organized as follows. In Section 3 we prove Theorem 2
and establish Table 1 of self-adjoint boundary conditions. In Section 4 we
prove Proposition 7 and expand the scattering amplitude $S$ near infinity.
Then we develop a general formalism to extract the dominant scale
$\mathcal{S}$ of $S$. There, the main technical results are Proposition 17 and
19. Finally, we apply this strategy to each class in Section 5. Due to the
large number of parameters in some classes, some expression for $S$ are very
large and tedious to manipulate (up to 60 terms, see e.g. (5.42)). Thus, some
computation where assisted with a symbolic computation software (Mathematica),
but without any numerical approximation. We also give a detailed example in
Section 4.4 where all computations can be done by hand to facilitate the
reading.

##  3 Self-adjoint boundary conditions

In this section we prove Theorem 2 and establish Table 1. We adapt the method
from Appendix B of [20] for shallow-water wave operator to the Dirac
Hamiltonian. We study the edge operator $H^{\sharp}$ from (2.6) associated to
states $\tilde{\psi}(y;k_{x})$ from (2.8). In this section, we drop the
$k_{x}$ dependence as well as $\sharp$ and $\tilde{}$ notations, so that we
work instead with $H$ and $\psi(y)$.

######  Lemma 13.

Self-adjoint realizations of the Hamiltonian H correspond to subspaces
$M\subset\mathbb{C}^{4}$ with

$\Omega M=M^{\perp}\,,$ (3.1)

where

$\Omega=\begin{pmatrix}0&1&\epsilon&0\\\ -1&0&0&-\epsilon\\\
-\epsilon&0&0&0\\\ 0&\epsilon&0&0\end{pmatrix}\,.$ (3.2)

The domain $\mathcal{D}(H)$ is a subspace of the Sobolev space $H^{2}\oplus
H^{2}$ which is characterized by

$\mathcal{D}(H):=\left\\{\psi\in H^{2}\oplus
H^{2}\,|\,\Psi=\begin{pmatrix}\psi(0)\\\ \psi^{\prime}(0)\end{pmatrix}\in
M\right\\}.$ (3.3)

###### Proof.

Let $\psi,\phi\in H^{2}\oplus H^{2}$. A partial integration of
$\langle\phi,H\psi\rangle$ yields a boundary term at $y=0$:

$\displaystyle\langle\phi,H\psi\rangle-\langle H\phi,\psi\rangle$
$\displaystyle=\phi_{1}^{*}\psi_{2}-\phi_{2}^{*}\psi_{1}+\epsilon\left(\phi_{1}^{*}\psi_{1}^{\prime}-\phi_{2}^{*}\psi_{2}^{\prime}-\psi_{1}(\phi_{1}^{*})^{\prime}+\psi_{2}(\phi_{2}^{*})^{\prime}\right)$
(3.4) $\displaystyle=\Phi^{*}\Omega\Psi\,,$ (3.5)

with $\Psi$ and $\Phi$ as in (3.3), $\Omega$ as in (3.2). In order for $H$ to
be self-adjoint, we require $\Psi\in M\subset\mathbb{C}^{4}$ with the
following properties

  1. i)

$\Phi^{*}\Omega\Psi=0$, $\Psi\in M$ $\Rightarrow$ $\Phi\in M$,

  2. ii)

$\Psi\in M$ $\Rightarrow$ $\Phi^{*}\Omega\Psi=0$ for all $\Phi\in M$.

By (3.5), the properties (i) and (ii) imply $H^{*}\subset H$ and $H\subset
H^{*}$ respectively, and thus $H=H^{*}$.

Moreover these two properties imply $(\Omega M)^{\perp}=M$. Indeed, let
$\Phi\in(\Omega M)^{\perp}$ and $\Psi\in M$. One has $\Phi^{*}\Omega\Psi=0$ so
that $\Phi\in M$ by (i). Thus $(\Omega M)^{\perp}\subset M$. Conversely, let
$\Psi\in M$. For any $\Phi\in M$ one has $\Phi^{*}\Omega\Psi=0$ by (ii), which
implies $\Psi^{*}\Omega\Phi=0$ since $\Omega^{*}=-\Omega$. Thus
$\Psi\in(\Omega M)^{\perp}$ and $M\subset(\Omega M)^{\perp}$.

Reciprocally, it is not hard to check that $(\Omega M)^{\perp}=M$ implies (i)
and (ii), so that $H$ is self-adjoint if and only if $(\Omega M)^{\perp}=M$,
namely $\Omega M=M^{\perp}$. ∎

######  Remark 14.

The matrix $\Omega$ defines a symplectic form on $\mathbb{C}^{4}$ and the
subspace $M$ is sometimes called a Lagrangian space, a general structure that
naturally appears for self-adjoint extensions of elliptic operators, see e.g.
[18].

######  Proposition 15.

  1. i)

Subspaces $M\subset\mathbb{C}^{4}$ with $\dim M=2$ are determined by $2\times
4$ matrices $A$ of maximal rank, i.e. $\operatorname{rk}A=2$ by means of

$M=\\{\Psi\in\mathbb{C}^{4}\lvert A\Psi=0\\}=\ker A\,,$ (3.6)

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

  2. ii)

Self-adjoint boundary conditions as in Lemma 13 are characterized by matrices
$A$ as in i) with

$A\Omega^{-1}A^{*}=0\,.$ (3.7)

###### Proof.

  1. i)

The first part is clear as $A\Psi=0$ gives two conditions if $A$ is of
$\operatorname{rk}$ 2\. In other words $\Omega$ is invertible and thus
$\ker\Omega=\\{0\\}$. For the second part (the last sentence) suppose that $A$
and $\widetilde{A}$ determine the same subspace, then there is a map
$B:\mathbb{C}^{2}\to\mathbb{C}^{2}$, which is well-defined by
$Av\mapsto\widetilde{A}v$, $v\in\mathbb{C}^{4}$ as $\ker A=\ker\widetilde{A}$.
If the kernel is to be preserved then $B\in\mathrm{GL}(2)$.

  2. ii)

Self-adjoint boundary conditions correspond to subspaces
$M\subset\mathbb{C}^{4}$ with $\Omega M=M^{\perp}$. Moreover, $\Omega$ is
invertible so that $\Omega M=M^{\perp}$ is equivalent to
$\Omega^{-1}M^{\perp}\subset M$ and $\dim M=2$. Thus, by (i), $M=\ker A$ for
some $A\in\mathrm{M}_{2,4}(\mathbb{C})$ of rank 2. Moreover one has
$M^{\perp}=\operatorname{ran}A^{*}=\\{A^{*}v\lvert v\in\mathbb{C}^{2}\\}$. Let
$v_{1},v_{2}\in\mathbb{C}^{2}$. One has

$\langle v_{1},A\Omega^{-1}A^{*}v_{2}\rangle=\langle
A^{*}v_{1},\Omega^{-1}A^{*}v_{2}\rangle=0$

since $A^{*}v_{1}\in M^{\perp}$ and
$\Omega^{-1}A^{*}v_{2}\in\Omega^{-1}M^{\perp}\subset M$.

Reciprocally, if $M=\ker A$ with $A\in\mathrm{M}_{2,4}(\mathbb{C})$ of rank 2
then $\dim M=2$ and one can check that
$\Omega^{-1}M^{\perp}\subset(M^{\perp})^{\perp}=M$.

∎

###  3.1 Grassmannian

Self-adjoint boundary conditions are given by $2\times 4$ matrices of rank 2
up to $\mathrm{GL}_{2}$-invariance. This set of matrices corresponds to the
Grassmannian $\mathrm{Gr}(4,2)$, which parametrizes $2$-dimensional subspaces
in the $4$-dimensional vector space $\mathbb{C}^{4}$:

$\mathrm{Gr}(4,2)\cong\frac{\\{A\in\mathrm{M}_{2,4}(\mathbb{C})\,|\,\operatorname{rk}(A)=2\\}}{\mathrm{GL}_{2}(\mathbb{C})}\,.$
(3.8)

Grassmannians are complex manifolds for which a lot is known, see e.g. [22].
We shall use the following facts: since $\operatorname{rk}(A)=2$ there are at
least 2 columns among the 4 of $A\in\mathrm{M}_{2,4}(\mathbb{C})$ which are
linearly independent vectors in $\mathbb{C}^{2}$. Then one can use the
$\mathrm{GL}_{2}(\mathbb{C})$ invariance to set the first one to
$\begin{pmatrix}1\\\ 0\end{pmatrix}$ and the second to $\begin{pmatrix}0\\\
1\end{pmatrix}$ and reduce the matrix $A$ to its row echelon form (from the
right). This provides the following partition

$\mathrm{Gr}(4,2)=\bigsqcup_{\boldsymbol{j}}\mathcal{C}_{\boldsymbol{j}}\,.$
(3.9)

where $\boldsymbol{j}=\\{1,2\\}$, $\\{1,3\\}$, $\\{1,4\\}$, $\\{2,3\\}$,
$\\{2,4\\}$ and $\\{3,4\\}$, with

$\displaystyle\mathcal{C}_{\\{1,2\\}}:\begin{pmatrix}1&0&0&0\\\
0&1&0&0\end{pmatrix}\,,$
$\displaystyle\mathcal{C}_{\\{1,3\\}}:\begin{pmatrix}1&0&0&0\\\
0&\star&1&0\end{pmatrix}\,,\quad\mathcal{C}_{\\{1,4\\}}:\begin{pmatrix}1&0&0&0\\\
0&\star&\star&1\end{pmatrix}\,,$ (3.10)
$\displaystyle\mathcal{C}_{\\{2,3\\}}:\begin{pmatrix}\star&1&0&0\\\
\star&0&1&0\end{pmatrix}\,,$
$\displaystyle\mathcal{C}_{\\{2,4\\}}:\begin{pmatrix}\star&1&0&0\\\
\star&0&\star&1\end{pmatrix}\,,\quad\mathcal{C}_{\\{3,4\\}}:\begin{pmatrix}\star&\star&1&0\\\
\star&\star&0&1\end{pmatrix}\,.$

The $\mathcal{C}_{\boldsymbol{j}}$ are called the Schubert cells of
$\mathrm{Gr}(4,2)$. Other Schubert cells of $\mathrm{Gr}(4,2)$ exist but the
six above are enough to describe each element uniquely. Notice that
$\dim(\mathcal{C}_{\boldsymbol{j}})=\sum_{i=1}^{2}j_{i}-i$ and
$\mathcal{C}_{\boldsymbol{j}}\cong\mathbb{C}^{\dim(\mathcal{C})}$ so that the
Schubert cells above have complex dimensions $0,1,2,2,3$ and $4$,
respectively.

###  3.2 Local boundary conditions

According to (2.7) we are interested in local boundary condition of the form
$A\Psi=0$ with
$A=A_{0}+{\mathrm{i}}k_{x}A_{1}\in\mathrm{M}_{2,4}(\mathbb{C})$. We shall
start with

$A_{0}=\begin{pmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\\
a_{21}&a_{22}&a_{23}&a_{24}\end{pmatrix}=[A_{1}\quad A_{2}],\qquad
A_{1}=\begin{pmatrix}b_{11}&b_{12}&0&0\\\
b_{21}&b_{22}&0&0\end{pmatrix}=[B\quad 0]$ (3.11)

with $A_{1},A_{2}$ and $B\in\mathrm{M}_{2}(\mathbb{C})$. According to
Proposition 15, $A$ has to be a rank-2 matrix. Thus we consider the various
cases with respect to the rank of $A_{0}$ and $A_{1}$. Depending on their
rank, such matrices can be simplified further using the
$\mathrm{GL}_{2}(\mathbb{C})$-invariance.

Moreover, $A$ satisfies

$A\Omega^{-1}A^{*}=0$

with

$\Omega^{-1}=\epsilon^{-2}\begin{pmatrix}0&0&-\epsilon&0\\\ 0&0&0&\epsilon\\\
\epsilon&0&0&-1\\\
0&-\epsilon&1&0\end{pmatrix}:=\epsilon^{-2}\begin{pmatrix}0&-\Omega_{1}\\\
\Omega_{1}&\Omega_{2}\end{pmatrix}\,.$ (3.12)

with $\Omega_{1},\Omega_{2}\in\mathrm{M}_{2}(\mathbb{C})$ and
$\Omega_{1}^{*}=\Omega_{1}$ and $\Omega_{2}^{*}=\Omega_{2}$. The condition
$A\Omega^{-1}A^{*}=0$ becomes

$-A_{1}\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}A_{1}^{*}+A_{2}\Omega_{2}A_{2}^{*}-{\mathrm{i}}k_{x}(B\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}B^{*})=0.$

This relation has to be valid for every $k_{x}\in\mathbb{R}$ so we infer

$\displaystyle-
A_{1}\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}A_{1}^{*}+A_{2}\Omega_{2}A_{2}^{*}=0,\qquad
B\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}B^{*}=0.$ (3.13)

###  3.3 Class $\mathfrak{A}$: $\operatorname{rk}(A_{0})=2$

In this section we assume $\operatorname{rk}(A_{0})=2$ and $A_{1}$ arbitrary:

$A_{1}=\begin{pmatrix}b_{11}&b_{12}&0&0\\\
b_{21}&b_{22}&0&0\end{pmatrix}=[B\quad 0]$

The $\mathrm{GL}_{2}(\mathbb{C})$-invariance allows to reduce $A_{0}$ to one
of the six Schubert cells from (3.10). For each of them we investigate (3.13)
and possibly restrict some parameters.

##### Class $\mathfrak{A}_{1,2}$.

In that case $A_{0}\in\mathcal{C}_{\\{1,2\\}}$, namely

$A_{0}=\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\end{pmatrix}$

One can check that (3.13) is satisfied for any
$B\in\mathbb{\mathrm{missing}}{M}_{2}(\mathbb{C})$.

##### Class $\mathfrak{A}_{1,3}$.

In that case $A_{0}\in\mathcal{C}_{\\{1,3\\}}$, namely

$A_{0}=\begin{pmatrix}1&0&1&0\\\ 0&a_{22}&0&0\end{pmatrix}.$

In that case we have

$-A_{1}\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}A_{1}^{*}+A_{2}\Omega_{2}A_{2}^{*}=\begin{pmatrix}0&-\epsilon\\\
\epsilon&0\end{pmatrix}$

which never vanishes for $\epsilon>0$ so that (3.13) is never satisfied. Thus
there is no self-adjoint boundary condition in this class, this is why it is
not appearing in Table 1.

##### Class $\mathfrak{A}_{1,4}$.

In that case $A_{0}\in\mathcal{C}_{\\{1,4\\}}$, namely

$A_{0}=\begin{pmatrix}1&0&0&0\\\ 0&a_{22}&a_{23}&1\end{pmatrix}.$

In that case we have

$-A_{1}\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}A_{1}^{*}+A_{2}\Omega_{2}A_{2}^{*}=\left(\begin{array}[]{cc}0&-\epsilon\left(a_{23}\right){}^{*}\\\
a_{23}\epsilon&-\epsilon\left(a_{22}\right){}^{*}+\left(a_{23}\right){}^{*}+a_{22}\epsilon-
a_{23}\\\ \end{array}\right),$

from which we infer $a_{23}=0$ and then $a_{22}\in\mathbb{R}$. Moreover, with
that knowledge, we compute

$B\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}B^{*}=\left(\begin{array}[]{cc}0&-\epsilon
b_{12}\\\
-\epsilon\left(b_{12}\right){}^{*}&-\epsilon(b_{22}+\left(b_{22}\right){}^{*})\\\
\end{array}\right),$

from which we infer $b_{12}=0$ and $b_{22}\in{\mathrm{i}}\mathbb{R}$. Denoting
$a_{22}=\alpha$ and $b_{22}={\mathrm{i}}\beta$ we deduce the self-adjoint
matrices in that case

$A_{0}=\begin{pmatrix}1&0&0&0\\\ 0&\alpha&0&1\end{pmatrix},\qquad
A_{1}=\begin{pmatrix}b_{11}&0&0&0\\\
b_{21}&{\mathrm{i}}\beta&0&0\end{pmatrix},$

with $\alpha,\beta\in\mathbb{R}$ and $b_{11},b_{21}\in\mathbb{C}$.

##### Class $\mathfrak{A}_{2,3}$.

In that case $A_{0}\in\mathcal{C}_{\\{2,3\\}}$, namely

$A_{0}=\begin{pmatrix}a_{11}&1&0&0\\\ a_{21}&0&1&0\end{pmatrix}.$

In that case we have

$-A_{1}\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}A_{1}^{*}+A_{2}\Omega_{2}A_{2}^{*}=\left(\begin{array}[]{cc}0&-\epsilon
a_{11}\\\
\epsilon\left(a_{11}\right){}^{*}&\epsilon\left(a_{21}\right){}^{*}-a_{21}\epsilon\\\
\end{array}\right),$

from which we infer $a_{11}=0$ and $a_{21}=\alpha\in\mathbb{R}$. Moreover one
has

$B\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}B^{*}=\left(\begin{array}[]{cc}0&b_{11}\epsilon\\\
\epsilon\left(b_{11}\right){}^{*}&\epsilon\left(b_{21}\right){}^{*}+b_{21}\epsilon\\\
\end{array}\right)$

from which we infer $b_{11}=0$ and
$b_{21}={\mathrm{i}}\beta\in{\mathrm{i}}\mathbb{R}$. The self-adjoint matrices
in that case are

$A_{0}=\begin{pmatrix}0&1&0&0\\\ \alpha&0&1&0\end{pmatrix},\qquad
A_{1}=\begin{pmatrix}0&b_{12}&0&0\\\
{\mathrm{i}}\beta&b_{22}&0&0\end{pmatrix},$

with $\alpha,\beta\in\mathbb{R}$ and $b_{12},b_{22}\in\mathbb{C}$.

##### Class $\mathfrak{A}_{2,4}$.

In that case $A_{0}\in\mathcal{C}_{\\{2,4\\}}$, namely

$A_{0}=\begin{pmatrix}a_{11}&1&0&0\\\ a_{21}&0&a_{23}&1\end{pmatrix}.$

In that case we have

$-A_{1}\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}A_{1}^{*}+A_{2}\Omega_{2}A_{2}^{*}=\left(\begin{array}[]{cc}0&\epsilon(1-a_{11}\left(a_{23}\right){}^{*})\\\
-\epsilon(1-a_{23}\left(a_{11}\right){}^{*})&\epsilon(a_{23}\left(a_{21}\right){}^{*}-a_{21}\left(a_{23}\right){}^{*})+\left(a_{23}\right){}^{*}-a_{23}\\\
\end{array}\right)$

from which we infer $1-a_{11}(a_{23})^{*}=0$, that can be rewritten
$a_{11}\neq 0$ and $a_{23}=(a_{11}^{-1})^{*}$. The lower right coefficient of
the matrix above then implies

$\epsilon\left(\left(\dfrac{a_{21}}{a_{11}}\right)^{*}-\dfrac{a_{21}}{a_{11}}\right)+\dfrac{1}{a_{11}}-\dfrac{1}{(a_{11})^{*}}=0$

which can be rephrased as

$\operatorname{Im}\left(\dfrac{a_{21}-\epsilon^{-1}}{a_{11}}\right)=0.$

Thus we have $a_{21}=\alpha a_{11}+\epsilon^{-1}$ with $\alpha\in\mathbb{R}$.

The second part of (3.13) reads

$B\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}B^{*}=\left(\begin{array}[]{cc}0&\epsilon(b_{11}a_{11}^{-1}-b_{12})\\\
\epsilon(b_{11}a_{11}^{-1}-b_{12})^{*}&\epsilon\left(\left(b_{21}a_{11}^{-1}\right)^{*}+b_{21}a_{11}^{-1}-\left(b_{22}\right){}^{*}-b_{22}\right)\\\
\end{array}\right)$

which implies $b_{12}=b_{11}a_{11}^{-1}$ and

$\operatorname{Re}\left(b_{21}a_{11}^{-1}-b_{22}\right)=0.$

Thus we have $b_{21}a_{11}^{-1}-b_{22}={\mathrm{i}}\beta$ with
$\beta\in\mathbb{R}$.

The self-adjoint matrices in that case are

$A_{0}=\begin{pmatrix}a_{11}&1&0&0\\\
a_{21}&0&(a_{11}^{-1})^{*}&1\end{pmatrix},\qquad
A_{1}=\begin{pmatrix}b_{11}&b_{11}a_{11}^{-1}&0&0\\\
b_{21}&b_{22}&0&0\end{pmatrix},$

with $a_{11}\in\mathbb{C}\setminus\\{0\\}$,
$a_{21},b_{11},b_{21},b_{22}\in\mathbb{C}$ and $a_{21}=\alpha
a_{11}+\epsilon^{-1}$, $\alpha\in\mathbb{R}$ as well as
$b_{21}a_{11}^{-1}-b_{22}={\mathrm{i}}\beta$, $\beta\in\mathbb{R}$.

##### Class $\mathfrak{A}_{3,4}$.

In that case $A_{0}\in\mathcal{C}_{\\{3,4\\}}$, namely

$A_{0}=\begin{pmatrix}a_{11}&a_{12}&1&0\\\ a_{21}&a_{22}&0&1\end{pmatrix}.$

In that case we have

$0=-A_{1}\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}A_{1}^{*}+A_{2}\Omega_{2}A_{2}^{*}=\left(\begin{array}[]{cc}\epsilon\left(a_{11}\right){}^{*}-a_{11}\epsilon&\epsilon\left(a_{21}\right){}^{*}+a_{12}\epsilon-1\\\
-\epsilon\left(a_{12}\right){}^{*}-a_{21}\epsilon+1&a_{22}\epsilon-\epsilon\left(a_{22}\right){}^{*}\\\
\end{array}\right),$

from which we infer $a_{11}=\alpha_{1}\in\mathbb{R}$,
$a_{22}=\alpha_{2}\in\mathbb{R}$ and $a_{21}=\epsilon^{-1}-(a_{12})^{*}$.
Moreover, we have

$0=B\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}B^{*}=\left(\begin{array}[]{cc}\epsilon\left(b_{11}\right){}^{*}+b_{11}\epsilon&\epsilon\left(b_{21}\right){}^{*}-b_{12}\epsilon\\\
b_{21}\epsilon-\epsilon\left(b_{12}\right){}^{*}&-\epsilon\left(b_{22}\right){}^{*}-b_{22}\epsilon\\\
\end{array}\right),$

from which we infer $b_{11}={\mathrm{i}}\beta_{1}\in{\mathrm{i}}\mathbb{R}$,
$b_{22}={\mathrm{i}}\beta_{2}\in{\mathrm{i}}\mathbb{R}$ and
$b_{21}=(b_{12})^{*}$.

The self-adjoint matrices in that case are

$A_{0}=\begin{pmatrix}\alpha_{1}&a_{12}&1&0\\\
\epsilon^{-1}-(a_{12})^{*}&\alpha_{2}&0&1\end{pmatrix},\qquad
A_{1}=\begin{pmatrix}{\mathrm{i}}\beta_{1}&b_{12}&0&0\\\
(b_{12})^{*}&{\mathrm{i}}\beta_{2}&0&0\end{pmatrix},$

with $\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}\in\mathbb{R}$ and
$a_{12},b_{12}\in\mathbb{C}$.

###  3.4 Class $\mathfrak{B}$: $\operatorname{rk}(A_{1})=2$

In this section we assume $\operatorname{rk}(A_{1})=2$. Since
$A_{1}=[B\,|\,0]$ this means that $\operatorname{rk}(B)=2$ and thus by the
$\mathrm{GL}_{2}(\mathbb{C})$-invariance we can reduce the study to

$A_{1}=\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\end{pmatrix}$

To avoid any overlap with class $\mathfrak{A}$ it is sufficient to consider
the case where $\operatorname{rk}(A_{0})\leq 1$. Thus we write

$A_{0}=\begin{pmatrix}a_{1}&a_{2}&a_{3}&a_{4}\\\ \mu a_{1}&\mu a_{2}&\mu
a_{3}&\mu a_{4}\end{pmatrix}$

with $a_{1},a_{2},a_{3},a_{4},\mu\in\mathbb{C}$.

In that case we have

$0=B\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}B^{*}=\left(\begin{array}[]{cc}\epsilon\left(a_{3}\right){}^{*}+a_{3}\epsilon&\epsilon\left(\mu
a_{3}\right){}^{*}-a_{4}\epsilon\\\
a_{3}\mu\epsilon-\epsilon\left(a_{4}\right){}^{*}&-\epsilon\left(\mu
a_{4}\right){}^{*}-a_{4}\mu\epsilon\\\ \end{array}\right),$

from which we infer $a_{3}={\mathrm{i}}\alpha\in{\mathrm{i}}\mathbb{R}$ and
$a_{4}={\mathrm{i}}\alpha\mu^{*}$ (in particular $\mu
a_{4}\in{\mathrm{i}}\mathbb{R}$). Moreover, we have

$\displaystyle 0$
$\displaystyle=-A_{1}\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}A_{1}^{*}+A_{2}\Omega_{2}A_{2}^{*}$
$\displaystyle=\alpha\left(\alpha(\mu^{*}-\mu)+{\mathrm{i}}\epsilon\alpha(a_{1}^{*}+a_{1}-a_{2}\mu-
a_{2}^{*}\mu^{*})\right)\begin{pmatrix}1&\mu^{*}\\\
\mu&|\mu|^{2}\end{pmatrix},$

from which we infer

$\alpha\Big{(}\alpha\operatorname{Im}(\mu)-\epsilon\operatorname{Re}(a_{1}-a_{2}\mu)\Big{)}=0.$

The self-adjoint matrices in that case are

$A_{0}=\begin{pmatrix}a_{1}&a_{2}&\alpha&\alpha\mu^{*}\\\ \mu a_{1}&\mu
a_{2}&\alpha\mu&\alpha|\mu|^{2}\end{pmatrix},\qquad
A_{1}=\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\end{pmatrix},$

with $a_{1},a_{2},\mu\in\mathbb{C}$, $\alpha\in\mathbb{R}$ and
$\alpha\big{(}\alpha\operatorname{Im}(\mu)-\epsilon\operatorname{Re}(a_{1}-a_{2}\mu)\big{)}=0$.
This last condition is equivalent to one of the three cases:

  1. 1.

$\alpha=0$

  2. 2.

$\alpha\neq
0,\,\mu\in\mathbb{R},\,a_{1}=a_{2}\mu+{\mathrm{i}}\beta,\,\beta\in\mathbb{R}$

  3. 3.

$\mu\in\mathbb{C}\setminus\mathbb{R}$ and
$\alpha=\tfrac{\epsilon}{\operatorname{Im}(\mu)}\operatorname{Re}(a_{1}-a_{2}\mu)\in\mathbb{R}$

but we shall not use them explicitly below so we keep the general constraint
instead.

###  3.5 Class $\mathfrak{C}$:
$\operatorname{rk}(A_{0})=\operatorname{rk}(A_{1})=1$

In this section we consider the case where $A_{0}$ and $A_{1}$ are exactly of
rank 1, with $\operatorname{rk}(A_{0}+{\mathrm{i}}k_{x}A_{1})=2$. Since
$A_{1}=[B\,|\,0]$ this means that $\operatorname{rk}(B)=1$ and thus by the
$\mathrm{GL}_{2}(\mathbb{C})$-invariance we can reduce the study to

$A_{1}=\begin{pmatrix}1&0&0&0\\\ 0&0&0&0\end{pmatrix}$

Then, since $\operatorname{rk}(A_{0})=1$ we write

$A_{0}=\begin{pmatrix}a_{1}&a_{2}&a_{3}&a_{4}\\\ \mu a_{1}&\mu a_{2}&\mu
a_{3}&\mu a_{4}\end{pmatrix}$

with $a_{1},a_{2},a_{3},a_{4},\mu\in\mathbb{C}$. Moreover one has $\mu\neq 0$
and $(a_{2},a_{3},a_{4})\neq 0$, otherwise
$\operatorname{rk}(A_{0}+{\mathrm{i}}k_{x}A_{1})=1<2$.

In that case we have

$0=B\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}B^{*}=\left(\begin{array}[]{cc}\epsilon\left(a_{3}\right){}^{*}+a_{3}\epsilon&\epsilon\left(\mu
a_{3}\right){}^{*}\\\ a_{3}\mu\epsilon&0\\\ \end{array}\right)$

from which we infer $a_{3}=0$. Moreover, we have

$\displaystyle 0$
$\displaystyle=-A_{1}\Omega_{1}A_{2}^{*}+A_{2}\Omega_{1}A_{1}^{*}+A_{2}\Omega_{2}A_{2}^{*}=\epsilon\left(a_{2}\left(a_{4}\right){}^{*}-a_{4}\left(a_{2}\right){}^{*}\right)\left(\begin{array}[]{cc}1&\mu^{*}\\\
\mu&|\mu|^{2}\\\ \end{array}\right)$

from which we infer $\operatorname{Im}(a_{2}a_{4}^{*})=0$.

The self-adjoint matrices in that case are

$A_{0}=\begin{pmatrix}a_{1}&a_{2}&0&a_{4}\\\ \mu a_{1}&\mu a_{2}&0&\mu
a_{4}\end{pmatrix},\qquad A_{1}=\begin{pmatrix}1&0&0&0\\\
0&0&0&0\end{pmatrix},$

with $a_{1},a_{2},a_{4},\mu\in\mathbb{C}$ such that $\mu\neq 0$,
$(a_{2},a_{4})\neq 0$ and $\operatorname{Im}(a_{2}a_{4}^{*})=0$.

##  4 Asymptotic expansion of the scattering amplitude

###  4.1 Bulk eigensections

Here we prove Proposition 7 and also provide explicit expression for bulk
sections which are used further below. Boundary condition (2.10) reads

$A\Psi|_{y=0}=0\,,\quad\Psi=\begin{pmatrix}\widetilde{\psi}\\\
\widetilde{\psi}^{\prime}\end{pmatrix},$ (4.1)

for $A\in\mathrm{M}_{2,4}(\mathbb{C})$. We apply it to the scattering state
from (2.13) at $y=0$,

$\displaystyle\widetilde{\psi}_{s}=\widehat{\psi}_{\mathrm{in}}+S\widehat{\psi}_{\mathrm{out}}+T\widehat{\psi}_{\mathrm{ev}}$
(4.2)
$\displaystyle\widetilde{\psi}_{s}^{\prime}=-{\mathrm{i}}\kappa\widehat{\psi}_{\mathrm{in}}+{\mathrm{i}}\kappa
S\widehat{\psi}_{\mathrm{out}}+{\mathrm{i}}\kappa_{\mathrm{ev}}T\widehat{\psi}_{\mathrm{ev}}$
(4.3)

where $\widehat{\psi}_{\mathrm{in/out/ev}}$ are bulk solution at momentum
$k_{x}$ and $k_{y}=-\kappa,\kappa,\kappa_{\mathrm{ev}}$ respectively. Thus at
$y=0$ we have

$\Psi_{s}=\widehat{\Psi}_{\mathrm{in}}+S\widehat{\Psi}_{\mathrm{out}}+T\widehat{\Psi}_{\mathrm{ev}}$

as in (2.16). This amounts to

$A\Psi_{s}=\begin{pmatrix}-\,a_{1}\,-\\\
-\,a_{2}\,-\end{pmatrix}\Psi_{s}=\begin{pmatrix}a_{1}\cdot\Psi_{s}\\\
a_{2}\cdot\Psi_{s}\end{pmatrix}=\begin{pmatrix}0\\\ 0\end{pmatrix}\,,$ (4.4)

where $a_{i}\in M_{1,4}(\mathbb{C})$ is the $i$-th row of $A$ for $i=1,2$.
This means that we get a $2\times 2$ linear system of equations to be solved
for $S$ and $T$

$\displaystyle
a_{1}\cdot(\widehat{\Psi}_{\mathrm{in}}+S\widehat{\Psi}_{\mathrm{out}}+T\widehat{\Psi}_{\mathrm{ev}})$
$\displaystyle=0\,,$ (4.5) $\displaystyle
a_{2}\cdot(\widehat{\Psi}_{\mathrm{in}}+S\widehat{\Psi}_{\mathrm{out}}+T\widehat{\Psi}_{\mathrm{ev}})$
$\displaystyle=0\,.$ (4.6)

Its solution for $S$ is

$S=-\frac{\begin{vmatrix}a_{1}\cdot\widehat{\Psi}_{\mathrm{in}}&a_{1}\cdot\widehat{\Psi}_{\mathrm{ev}}\\\
a_{2}\cdot\widehat{\Psi}_{\mathrm{in}}&a_{2}\cdot\widehat{\Psi}_{\mathrm{ev}}\end{vmatrix}}{\begin{vmatrix}a_{1}\cdot\widehat{\Psi}_{\mathrm{out}}&a_{1}\cdot\widehat{\Psi}_{\mathrm{ev}}\\\
a_{2}\cdot\widehat{\Psi}_{\mathrm{out}}&a_{2}\cdot\widehat{\Psi}_{\mathrm{ev}}\end{vmatrix}}=-\frac{\begin{vmatrix}A\widehat{\Psi}_{\mathrm{in}}&A\widehat{\Psi}_{\mathrm{ev}}\end{vmatrix}}{\begin{vmatrix}A\widehat{\Psi}_{\mathrm{out}}&A\widehat{\Psi}_{\mathrm{ev}}\end{vmatrix}}$
(4.7)

Let $\widehat{\psi}^{0}$ a bulk section that is regular and non-vanishing in
$\mathbb{R}^{2}\setminus\\{0\\}\cup\\{\infty\\}$ and let
$\widehat{\psi}^{\infty}$ be a bulk section that is regular and non-vanishing
in $\mathbb{R}^{2}$. Then for $k_{x}\in\mathbb{R}$ and $\kappa>0$ we set
$\widehat{\psi}_{\mathrm{in}}(k_{x},-\kappa)=\widehat{\psi}_{0}(k_{x},-\kappa)$,
$\widehat{\psi}_{\mathrm{out}}(k_{x},\kappa)=\widehat{\psi}_{0}(k_{x},\kappa)$
and
$\widehat{\psi}_{\mathrm{ev}}(k_{x},\kappa_{\mathrm{ev}})=\widehat{\psi}_{\infty}(k_{x},\kappa_{\mathrm{ev}})$,
which leads to the expression (2.17) for $S$ and $g$ from Proposition 7.

To expand $S$ near $\infty$ we provide explicit expressions for the sections
$\widehat{\psi}_{0}$ and $\widehat{\psi}_{\infty}$. We concentrate on the
upper part of the spectrum and only discuss the eigensections corresponding to
$\omega_{+}(k)=+\sqrt{k^{2}+(m-\epsilon k^{2})^{2}}$. Assuming $m$,
$\epsilon>0$ the corresponding Chern number is $C_{+}=1$, and therefore it is
impossible to find a global bulk eigensection $\widehat{\psi}$ regular for all
points on the compactified $k$-plane, $\mathbb{C}\cup\\{\infty\\}\cong S^{2}$.
To cover the sphere we need at least two distinct ones overlapping in a region
and regular where they are defined. Explicitly, one section satisfying
$H\widehat{\psi}=\omega_{+}\widehat{\psi}$ is given by

$\widehat{\psi}_{0}(k_{x},k_{y})=\frac{1}{\sqrt{2}\sqrt{1-q(\boldsymbol{k})}}\begin{pmatrix}\frac{k_{x}-{\mathrm{i}}k_{y}}{\omega_{+}(k)}\\\
q(\boldsymbol{k})-1\end{pmatrix}\,,\quad
q(\boldsymbol{k}):=\frac{m-\epsilon\boldsymbol{k}^{2}}{\omega_{+}}\,.$ (4.8)

One can check that $\left\lVert\widehat{\psi}_{0}(k_{x},k_{y})\right\rVert=1$.
Notice that $\omega_{+}(k)\sim\epsilon k^{2}$ and
$q\to-\epsilon/\left\lvert\epsilon\right\rvert=-1$ as $k\to\infty$. On the
other hand, $\omega_{+}(k)\to m$ and

$q(k)=1-\dfrac{k^{2}}{2m^{2}}+o(k^{2})$

as $k\to 0$. Thus, writing
$k_{x}+{\mathrm{i}}k_{y}=k{\mathrm{e}}^{{\mathrm{i}}\phi}$,

$\lim_{k\to
0}\widehat{\psi}_{0}(k_{x},k_{y})=\begin{pmatrix}{\mathrm{e}}^{-{\mathrm{i}}\phi}\\\
0\end{pmatrix},\qquad\lim_{k\to\infty}\widehat{\psi}_{0}(k_{x},k_{y})=\begin{pmatrix}0\\\
-1\end{pmatrix}\,.$ (4.9)

Therefore, $\psi_{0}$ is regular at $k=\infty$ but not at $k=0$. There, it has
a removable phase singularity up to a gauge transformation. We define
$\psi_{\infty}=\lambda\psi_{0}$ with
$\lambda={\mathrm{e}}^{{\mathrm{i}}\phi}=k^{-1}(k_{x}+{\mathrm{i}}k_{y})$, or
explicitly

$\widehat{\psi}_{\infty}(k_{x},k_{y})=\frac{1}{k\sqrt{2}\sqrt{1-q(\boldsymbol{k})}}\begin{pmatrix}\frac{k^{2}}{\omega_{+}(k)}\\\
(k_{x}+{\mathrm{i}}k_{y})(q(\boldsymbol{k})-1)\end{pmatrix}\,.$ (4.10)

In particular, $\psi_{\infty}$ is regular at $k=0$ but not at $k=\infty$:

$\lim_{k\to 0}\widehat{\psi}_{\infty}(k_{x},k_{y})=\begin{pmatrix}1\\\
0\end{pmatrix},\qquad\lim_{k\to\infty}\widehat{\psi}_{\infty}(k_{x},k_{y})=\begin{pmatrix}0\\\
-{\mathrm{e}}^{{\mathrm{i}}\phi}\end{pmatrix}\,.$

Now for $k_{x}\in\mathbb{R}$ and $\kappa>0$ consider
$k_{y}=\kappa_{\mathrm{ev}}(k_{x},\kappa)\in{\mathrm{i}}\mathbb{R}$ , see
(2.12). Define

$\tilde{k}^{2}=k_{x}^{2}+\kappa_{\mathrm{ev}}^{2}=-(k_{x}^{2}+\kappa^{2})-\dfrac{1-2\epsilon
m}{\epsilon^{2}}<0,$

as well as
$\tilde{\omega}_{+}=\sqrt{\tilde{k}^{2}+(m-\epsilon\tilde{k}^{2})^{2}}$ and
$\tilde{q}=(m-\epsilon\tilde{k}^{2})\tilde{\omega}_{+}^{-1}$. One can check
that $\tilde{q}>1$ so that $\tilde{k}^{2}(1-\tilde{q})>0$ and thus
$\tilde{k}\sqrt{1-\tilde{q}}$ is well defined, and so is
$\widehat{\psi}_{\infty}(k_{x},\kappa_{\mathrm{ev}}(k_{x},\kappa))$. In
particular

$\lim_{(k_{x},\kappa)\to\infty}\tilde{q}=1,\qquad\lim_{(k_{x},\kappa)\to\infty}\tilde{k}\sqrt{1-\tilde{q}}=\dfrac{1}{\sqrt{2}\epsilon},$

so that

$\lim_{(k_{x},\kappa)\to\infty}\widehat{\psi}_{\infty}(k_{x},\kappa_{\mathrm{ev}}(k_{x},\kappa))=\begin{pmatrix}1\\\
0\end{pmatrix}\,.$

Notice that $\kappa_{\mathrm{ev}}\to{\mathrm{i}}\epsilon^{-1}\sqrt{1-2\epsilon
m}\neq 0$ as $(k_{x},\kappa)\to 0$ so that
$\widehat{\psi}_{\infty}(k_{x},\kappa_{\mathrm{ev}}(k_{x},\kappa))$ is defined
for all $\kappa>0$ and $k_{x}\in\mathbb{R}$, including at $\infty$. We then
use
$\widehat{\psi}_{\mathrm{in}}(k_{x},-\kappa)=\widehat{\psi}_{0}(k_{x},-\kappa)$,
$\widehat{\psi}_{\mathrm{out}}(k_{x},\kappa)=\widehat{\psi}_{0}(k_{x},\kappa)$
and
$\widehat{\psi}_{\mathrm{ev}}(k_{x},\kappa_{\mathrm{ev}})=\widehat{\psi}_{\infty}(k_{x},\kappa_{\mathrm{ev}}(k_{x},\kappa))$
in the definition (2.13) of the scattering state. Then recall that

$\widehat{\Psi}_{0}(k_{x},\pm\kappa)=\begin{pmatrix}\widehat{\psi}_{0}(k_{x},\pm\kappa)\\\
\pm{\mathrm{i}}\kappa\widehat{\psi}_{0}(k_{x},\pm\kappa)\end{pmatrix},\qquad\widehat{\Psi}_{\infty}(k_{x},\kappa_{\mathrm{ev}})=\begin{pmatrix}\widehat{\psi}_{\infty}(k_{x},\kappa_{\mathrm{ev}})\\\
{\mathrm{i}}\kappa_{\mathrm{ev}}\widehat{\psi}_{\infty}(k_{x},\kappa_{\mathrm{ev}})\end{pmatrix}\,,$
(4.11)

which provides a fully explicit expression for the scattering amplitude $S$.
In particular, it is regular near infinity.

###  4.2 Expansion near $\infty$

In order to expand the scattering amplitude near $\infty$, we focus on the
expansion of $g$. The latter requires to expand
$\widehat{\psi}_{0}(k_{x},\kappa)$ and
$\widehat{\psi}_{\infty}(k_{x},\kappa_{\mathrm{ev}})$ near $\infty$. Such an
expansion is a bit tricky. Even though the limits are single-valued, the next
terms are not, as they are $k_{x},k_{y}$ dependent. Some of these terms are
relevant in the expansion of $g$ and have to be properly included. We shall
then distinguish in each section the radial part, which depends only on the
variable $k=\sqrt{k_{x}^{2}+\kappa^{2}}$ and can be simply expanded, from the
directional part which depends on $k_{x},\kappa$ and $\kappa_{\mathrm{ev}}$.
Thus we get

$\widehat{\psi}_{0}(k_{x},\kappa)=\begin{pmatrix}(k_{x}-{\mathrm{i}}\kappa)\left(\frac{1}{2\epsilon
k^{2}}+o\left(\frac{1}{k^{3}}\right)\right)\\\ -1+\frac{1}{8\epsilon
k^{2}}+o\left(\frac{1}{k^{3}}\right)\end{pmatrix}$

For the expansion of $\widehat{\psi}_{\infty}$, we first notice that
$\tilde{k}$, $\tilde{q}$ and $\tilde{\omega}_{\pm}$ are all functions of
$k^{2}=k_{x}^{2}+\kappa^{2}$ only, so that they can be expanded as well as a
radial part. The remaining terms depends on $k_{x}$ and
$\kappa_{\mathrm{ev}}$. Explicitly we get

$\widehat{\psi}_{\infty}(k_{x},\kappa_{\mathrm{ev}})=\begin{pmatrix}1+\frac{1}{8\epsilon
k^{2}}+o\left(\frac{1}{k^{3}}\right)\\\
(k_{x}+{\mathrm{i}}\kappa_{\mathrm{ev}})\left(-\frac{1}{2\epsilon
k^{2}}+o\left(\frac{1}{k^{3}}\right)\right)\\\ \end{pmatrix}$

Notice that $\kappa_{\mathrm{ev}}\sim\sqrt{2k_{x}^{2}+\kappa^{2}}$ as
$k_{x},\kappa\to\infty$. This way one has for example

$\dfrac{k_{x}-{\mathrm{i}}\kappa}{k^{2}}\to
0,\qquad\dfrac{k_{x}+{\mathrm{i}}\kappa_{\mathrm{ev}}}{k^{2}}\to 0$

as $k_{x},\kappa\to\infty$, independent from the direction towards infinity,
whereas expressions like

$k_{x}(k_{x}-{\mathrm{i}}\kappa)k^{-2}$ have a limit which depends on the
$(k_{x},\kappa)$ direction. Such expressions naturally appear in the
scattering amplitude through $\widehat{\Psi}_{0}$ and
$\widehat{\Psi}_{\infty}$, see (4.11), which involve
$\kappa\widehat{\psi}_{0}$ and $\kappa_{\mathrm{ev}}\widehat{\psi}_{\infty}$,
and through $A\widehat{\Psi}_{0}$ and $A\widehat{\Psi}_{\infty}$, which
involve $k_{x}\widehat{\psi}_{0}$ and $k_{x}\widehat{\psi}_{\infty}$.
Consequently, an expansion of the radial part of $\widehat{\psi}_{0}$ and
$\widehat{\psi}_{\infty}$ up to order $k^{-3}$ is required to get and
expansion of $g$ up to order 1 in the limit $k_{x},\kappa\to\infty$. The non
radial part of the expansion will be first kept as it is and specified later
when passing to dual variables. Thus we will use

$\widehat{\psi}_{0}(k_{x},\kappa)=\begin{pmatrix}\dfrac{k_{x}-{\mathrm{i}}\kappa}{2\epsilon
k^{2}}\\\ -1+\dfrac{1}{8\epsilon
k^{2}}\end{pmatrix}+o\left(\dfrac{1}{k^{2}}\right),\qquad\widehat{\psi}_{\infty}(k_{x},\kappa_{\mathrm{ev}})=\begin{pmatrix}1+\dfrac{1}{8\epsilon
k^{2}}\\\ -\dfrac{k_{x}+{\mathrm{i}}\kappa_{\mathrm{ev}}}{2\epsilon k^{2}}\\\
\end{pmatrix}+o\left(\dfrac{1}{k^{2}}\right)$ (4.12)

###  4.3 Detecting anomalies

From the explicit expression of $g$ from Proposition 7 we compute its
expansion near $\infty$ using the expansion of $\widehat{\psi}_{0}$ and
$\widehat{\psi}_{\infty}$ from the previous section and keeping all the terms
that do not vanish in the limit:

$g(k_{x},\kappa)=g_{\infty}(k_{x},k)+o(1),$ (4.13)

with $o(1)$ in the limit $k_{x}^{2}+\kappa^{2}\to\infty$, independent from the
direction. The exact expression for $g_{\infty}$ depends on the boundary
condition. The winding number of $S$ near infinity is thus reduced to the
study the winding phase of $g_{\infty}$ computed along the curve
$\Gamma_{\delta}(\lambda)$ with dual variables
$\lambda_{x}\in\mathbb{[}-\lambda,\lambda]$ and $\delta>0$, see (2.18). Then
consider

$g_{0}(\lambda_{x},\delta)=g_{\infty}\Big{(}-\dfrac{\lambda_{x}}{\lambda_{x}^{2}+\delta^{2}},\dfrac{\delta}{\lambda_{x}^{2}+\delta^{2}}\Big{)}.$
(4.14)

######  Lemma 16.

Assume the existence of $M>0$ as well as $\lambda_{0}>0$ and $\delta_{0}>0$
sufficiently small such that $|g_{0}|>M$ and $g_{0}$ is continuously
differentiable on $(0,\delta_{0}]\times[-\lambda_{0},\lambda_{0}]$. Then

$\lim_{\lambda\to 0}\lim_{\delta\to
0}\int_{\Gamma_{\delta}(\lambda)}g^{-1}{\mathrm{d}}g=\lim_{\lambda\to
0}\lim_{\delta\to
0}\int_{-\lambda}^{\lambda}g_{0}^{-1}\partial_{\lambda_{x}}g_{0}{\mathrm{d}}\lambda_{x}$

###### Proof.

The winding of $g$ reads

$\int_{\Gamma_{\delta}(\lambda)}g^{-1}{\mathrm{d}}g=\int_{-\lambda}^{\lambda}\tilde{g}^{-1}\partial_{\lambda_{x}}\tilde{g}{\mathrm{d}}\lambda_{x},\qquad\tilde{g}(\lambda_{x},\delta)=g\Big{(}-\dfrac{\lambda_{x}}{\lambda_{x}^{2}+\delta^{2}},\dfrac{\delta}{\lambda_{x}^{2}+\delta^{2}}\Big{)}.$

Since $k_{x}^{2}+\kappa^{2}=(\lambda_{x}^{2}+\delta^{2})^{-1}$, one has
$\tilde{g}(\lambda_{x},\delta)=g_{0}(\lambda_{x},\delta)+o(1)$ with $o(1)$ in
the limit $\lambda_{x}^{2}+\delta^{2}\to 0$, independent from the direction.
Thus we decompose

$\tilde{g}=g_{0}+R=g_{0}\left(1+\dfrac{R}{g_{0}}\right).$

with $R(\lambda_{x},\delta)\to 0$ as $\lambda_{x}^{2}+\delta^{2}\to 0$. Since
$|g_{0}|>M$ on $(0,\delta_{0}]\times[-\lambda,\lambda]$ then $\tfrac{R}{G}\to
0$ as $\lambda_{x}^{2}+\delta^{2}\to 0$.

Moreover, the multiplicative property of winding phase gives

$\int_{-\lambda}^{\lambda}\tilde{g}^{-1}\partial_{\lambda_{x}}\tilde{g}{\mathrm{d}}\lambda_{x}=\int_{-\lambda}^{\lambda}g_{0}^{-1}\partial_{\lambda_{x}}g_{0}{\mathrm{d}}\lambda_{x}+\int_{-\lambda}^{\lambda}\left(1+\dfrac{R}{g_{0}}\right)^{-1}\partial_{\lambda_{x}}\left(1+\dfrac{R}{g_{0}}\right){\mathrm{d}}\lambda_{x}.$

Let $\varepsilon>0$ and $\delta_{1},\lambda_{1}$ such that
$\left|\tfrac{R}{g_{0}}\right|<\varepsilon$ on
$(0,\delta_{1}]\times[-\lambda_{1},\lambda_{1}]$.

Consequently, for $|\lambda|<\min(\lambda_{0},\lambda_{1})$ and
$\delta<\min(\delta_{0},\delta_{1})$ one has

$\left|\arg\left(1+\dfrac{R}{g_{0}}\right)\right|\leq\dfrac{\varepsilon}{1-\varepsilon}$

so that

$\left|\int_{-\lambda}^{\lambda}\left(1+\dfrac{R}{g_{0}}\right)^{-1}\partial_{\lambda_{x}}\left(1+\dfrac{R}{g_{0}}\right){\mathrm{d}}\lambda_{x}\right|\leq\dfrac{2\varepsilon}{1-\varepsilon}$

Thus the winding phase of $\left(1+\tfrac{R}{g_{0}}\right)$ goes to $0$ as
$\delta\to 0$ and $\lambda\to 0$. ∎

One can check that the assumption of the lemma above is satisfied for all
explicit $g_{0}$ below by checking that the limits of $|g_{0}|$ as $\delta\to
0$ or $\lambda\to 0$ are either finite and non vanishing or infinite. Thus, to
compute $w_{\infty}$ we can simply replace $S$ by $S_{0}$ with

$S_{0}(\lambda_{x},\delta)=\dfrac{g_{0}(\lambda_{x},-\delta)}{g_{0}(\lambda_{x},\delta)}.$

The next step is to extract the dominant scale from $g_{0}$ in the limit
$\delta\to 0$ and $\lambda\to 0$. First, due to the expression of $S_{0}$ we
can replace $g_{0}$ by any function that is even in $\delta$, typically
$g_{0}\mapsto g_{0}f$ with
$f(\delta,\lambda)=(\lambda_{x}^{2}+\delta^{2})^{2}$ or
$(\lambda_{x}^{2}+\delta^{2})^{2}$, see other examples below. We shall keep
the same symbol for the simplified expression of $g_{0}$.

Finally, we use the following proposition to extract the leading contribution
from $g_{0}$.

######  Proposition 17.

Let $g_{0}:\mathbb{R}\times\mathbb{R}_{+}^{*}\to\mathbb{C}^{*}$ continuously
differentiable. Let $s>0$, $u=\frac{\lambda_{x}}{\delta^{s}}$ and $r>0$.
Decompose

$\delta^{-r}g_{0}(u\delta^{s},\delta)=G(u)+R(u,\delta)$

with $G:\mathbb{R}\to\mathbb{C}^{*}$ and
$R:\mathbb{R}\times\mathbb{R}_{+}^{*}\to\mathbb{C}$ continuously
differentiable and such that

  1. 1.

For any $u\in\mathbb{R}$, $\displaystyle\lim_{\delta\to 0}R(u,\delta)=0$,

  2. 2.

For $\delta>0$,

$\displaystyle\dfrac{R(u,\delta)}{G(u)}\sim Cu^{\gamma}\delta^{s\gamma}$

as $u\to+\infty$, with $C\in\mathbb{C}^{*}$ independent from $\delta$, and
$\gamma>0$.

  3. 3.

Assumption 2 also holds as $u\to-\infty$ with possibly distinct constants
$\widetilde{C}$ and $\widetilde{\gamma}$.

Then,

$\lim_{\lambda\to 0}\lim_{\delta\to
0}\int_{-\lambda}^{\lambda}g_{0}^{-1}\partial_{\lambda_{x}}g_{0}{\mathrm{d}}\lambda_{x}=\int_{-\infty}^{\infty}G^{-1}(u)G^{\prime}(u){\mathrm{d}}u.$
(4.15)

###### Proof.

First notice that for $\delta>0$ the winding phase of $g_{0}$ and
$\delta^{-r}g_{0}$ are the same. We change the integration variable
$\lambda_{x}$ to $u=\tfrac{\lambda_{x}}{\delta^{s}}$ and replace $g_{0}$ by
$g_{0}\delta^{-r}=G+R$. Then we rewrite $G+R=G(1+\tfrac{R}{G})$ and use the
additive property of winding phase:

$\displaystyle\int_{-\lambda}^{\lambda}g_{0}^{-1}\partial_{\lambda_{x}}g_{0}{\mathrm{d}}\lambda_{x}$
$\displaystyle=\int_{-\frac{\lambda}{\delta^{s}}}^{\frac{\lambda}{\delta^{s}}}g_{0}^{-1}\partial_{u}g_{0}{\mathrm{d}}u$
(4.16)
$\displaystyle=\int_{-\frac{\lambda}{\delta^{s}}}^{\frac{\lambda}{\delta^{s}}}G^{-1}(u)G^{\prime}(u){\mathrm{d}}u+\int_{-\frac{\lambda}{\delta^{s}}}^{\frac{\lambda}{\delta^{s}}}\Big{(}1+\frac{R}{G}\Big{)}^{-1}(u,\delta)\partial_{u}\Big{(}1+\frac{R}{G}\Big{)}(u,\delta){\mathrm{d}}u.$
(4.17)

The integrand of the first term does not depend on $\delta$, so in the limit
$\delta\to 0$ we get the right hand side of (4.15).

Then we claim that for all $\varepsilon>0$ there exist $\delta_{0}>0$ and
$\lambda_{0}>0$ such that

$\forall
u\in\left[-\frac{\lambda}{\delta^{s}},\frac{\lambda}{\delta^{s}}\right],\quad\forall\delta<\delta_{0},\quad\forall\lambda<\lambda_{0},\qquad\left|\dfrac{R(u,\delta)}{G(u)}\right|\leq\varepsilon\,,$

which implies, similarly to the proof of the Lemma above, that the winding
phase of $u\mapsto 1+\tfrac{R}{G}$ is arbitrarily small for
$u\in\left[-\tfrac{\lambda}{\delta^{s}},\tfrac{\lambda}{\delta^{s}}\right]$ if
$\delta$, $\lambda$ are sufficiently small. Consequently, the second term in
the right hand side of (4.16) vanishes in the limit $\delta\to 0$, $\lambda\to
0$, and we get (4.15).

Let $\varepsilon>0$. In order to prove the claim above, we split the study of
the ratio by into the asymptotic and finite regions. First, by Assumption 2,
there exists a $u_{1}>0$ such that,

$\forall
u>u_{1},\qquad\left|\dfrac{R(u,\delta)}{G(u)}\right|\leq(1+\varepsilon)Cu^{\gamma}\delta^{s\gamma}$

Let $\delta_{1}>0$ such that $\tfrac{\lambda}{\delta_{1}^{\alpha}}>u_{1}$. For
$\delta<\delta_{1}$ and $u$ such that $u_{1}<u\leq\tfrac{\lambda}{\delta^{s}}$
one has

$\left|\dfrac{R(u,\delta)}{G(u)}\right|\leq(1+\varepsilon)C\lambda^{\gamma},$

so that for $\lambda<\lambda_{1}$ sufficiently small the right hand side is
smaller than $\varepsilon$.

Similarly, at $-\infty$ there exists $u_{2}>0$ and $\delta_{2}>0$ such that
for $\delta<\delta_{2}$ and $u$ such that $-\tfrac{\lambda}{\delta^{s}}\leq
u<-u_{2}$ one has

$\left|\dfrac{R(u,\delta)}{G(u)}\right|\leq(1+\varepsilon)\widetilde{C}\lambda^{\widetilde{\gamma}}\,.$

So that for $\lambda<\lambda_{2}$ sufficiently small the right hand side is
smaller than $\varepsilon$.

Finally, $G$ and $R$ are continuous on $[-u_{2},u_{1}]$ and $G(u)\neq 0$. Let
$M>0$ such that $|G(u)|>M$ for $u\in[-u_{2},u_{1}]$. Moreover, by Assumption 1
one has $R(u,\delta)\to 0$ as $\delta\to 0$. For $u\in[-u_{2},u_{1}]$, this
limit is uniform in $\delta$. So there exists a $\delta_{3}>0$ such that

$\forall 0<\delta<\delta_{3},\forall
u\in[-u_{2},u_{1}],\qquad|R(u,\delta)|<\dfrac{\varepsilon}{M}\,,$

which implies

$\forall 0<\delta<\delta_{3},\forall
u\in[-u_{2},u_{1}],\qquad\left|\dfrac{R(u,\delta)}{G(u)}\right|<\varepsilon\,.$

Putting all together, we get the claim if
$\delta<\min(\delta_{1},\delta_{2},\delta_{3})$ and
$\lambda<\min(\lambda_{1},\lambda_{2})$. ∎

This lemma drastically simplifies the computation of the winding number since
the leading contribution $G$ usually consists of only a few terms of the
numerous ones of $g_{0}$.

######  Corollary 18.

Assume

$\displaystyle\delta^{-r}g_{0}(u\delta^{s},\delta)=G_{+}(u)+R_{+}(u,\delta),\qquad\delta^{-r}g_{0}(u\delta^{s},-\delta)=G_{-}(u)+R_{-}(u,\delta)$

with $G_{\pm},R_{\pm}$ satisfying the assumption of Proposition 17. Then

$w_{\infty}=\int_{-\infty}^{\infty}\mathcal{S}^{-1}(u)\mathcal{S}^{\prime}(u){\mathrm{d}}u,\qquad\mathcal{S}(u)=\dfrac{G_{-}(u)}{G_{+}(u)}\,.$

A detailed application of this result with a specific boundary condition is
given in Section 4.4 below. Before to move on with the full classification, we
need to anticipate one possible issue: It may occasionally happen for specific
boundary conditions below that $G_{+}(u_{0})=0=G_{-}(u_{0})$ for some
$u_{0}\in\mathbb{R}$. In that case Proposition 17 does not apply for two
reasons:

  1. 1.

The winding phases of $G_{+}$ and $G_{-}$ are ill-defined

  2. 2.

$\tfrac{R_{+}}{G_{+}}$ and $\tfrac{R_{-}}{G_{-}}$ may be very large near
$u_{0}$.

However, if $G_{+}$ and $G_{-}$ are such that the ratio
$\mathcal{S}=\tfrac{G_{-}}{G_{+}}$ has a well defined the winding phase then
point 1 is actually not an issue to compute $w_{\infty}$. Thus we only need to
check that $R_{-}$ and $R_{+}$ do not contribute to the winding phase of $S$
in that case. The proposition below provides another criterion and generalizes
Proposition 17.

######  Proposition 19.

Let $g_{0}:\mathbb{R}\times\mathbb{R}_{+}^{*}\to\mathbb{C}^{*}$ continuously
differentiable. Let $s>0$, $u=\tfrac{\lambda_{x}}{\delta^{s}}$ and $r>0$.
Decompose

$\delta^{-r}g_{0}(u\delta^{s},\delta)=G_{+}(u)+R_{+}(u,\delta),\qquad\delta^{-r}g_{0}(u\delta^{s},-\delta)=G_{-}(u)+R_{-}(u,\delta)$

with $G_{+},G_{-}:\mathbb{R}\to\mathbb{C}$ and
$R_{+},R_{-}:\mathbb{R}\times\mathbb{R}_{+}^{*}\to\mathbb{C}$ continuously
differentiables and such that Assumptions 1, 2 and 3 of Proposition 17 hold.
Assume moreover that $G_{+}(u_{0})=0=G_{-}(u_{0})$ for some
$u_{0}\in\mathbb{R}$, but such that the ratios

$\mathcal{S}(u)=\dfrac{G_{-}(u)}{G_{+}(u)},\qquad
F(u,\delta)=\dfrac{1+\frac{R_{-}(u,\delta)}{G_{-}(u)}}{1+\frac{R_{+}(u,\delta)}{G_{+}(u)}}$

remain continuously differentiables near $u_{0}$ and that

$\lim_{\delta\to 0}\lim_{u\to u_{0}}F(u,\delta)=1.$

Then,

$\lim_{\lambda\to 0}\lim_{\delta\to
0}\int_{-\lambda}^{\lambda}S_{0}^{-1}\partial_{\lambda_{x}}S_{0}{\mathrm{d}}\lambda_{x}=\int_{-\infty}^{\infty}\mathcal{S}^{-1}(u)\mathcal{S}^{\prime}(u){\mathrm{d}}u.$
(4.18)

###### Proof.

We proceed as in the proof of Proposition 17: multiply numerator and
denominator by $\delta^{-r}$, change the variable and split the integral:

$\displaystyle\int_{-\lambda}^{\lambda}S_{0}^{-1}\partial_{\lambda_{x}}S_{0}{\mathrm{d}}\lambda_{x}$
$\displaystyle=\int_{-\frac{\lambda}{\delta^{s}}}^{\frac{\lambda}{\delta^{s}}}S_{0}^{-1}\partial_{u}S_{0}{\mathrm{d}}u$
(4.19)
$\displaystyle=\int_{-\frac{\lambda}{\delta^{s}}}^{\frac{\lambda}{\delta^{s}}}S^{-1}(u)S^{\prime}(u){\mathrm{d}}u+\int_{-\frac{\lambda}{\delta^{s}}}^{\frac{\lambda}{\delta^{s}}}F^{-1}(u,\delta)\partial_{u}F(u,\delta){\mathrm{d}}u.$
(4.20)

The first term gives the expected result and we show that the second has no
winding phase in the limit. We split as before between small and large $u$

$\displaystyle\int_{-\frac{\lambda}{\delta^{s}}}^{\frac{\lambda}{\delta^{s}}}F^{-1}(u,\delta)\partial_{u}F(u,\delta){\mathrm{d}}u$
(4.21) $\displaystyle=$
$\displaystyle\int_{M}^{\frac{\lambda}{\delta^{s}}}\Big{(}1+\frac{R_{+}}{G_{+}}\Big{)}^{-1}\partial_{u}\Big{(}1+\frac{R_{+}}{G_{+}}\Big{)}{\mathrm{d}}u+\int_{M}^{\frac{\lambda}{\delta^{s}}}\Big{(}1+\frac{R_{-}}{G_{-}}\Big{)}^{-1}\partial_{u}\Big{(}1+\frac{R_{-}}{G_{-}}\Big{)}{\mathrm{d}}u$
(4.22)
$\displaystyle+\int_{-\frac{\lambda}{\delta^{s}}}^{M}\Big{(}1+\frac{R_{+}}{G_{+}}\Big{)}^{-1}\partial_{u}\Big{(}1+\frac{R_{+}}{G_{+}}\Big{)}{\mathrm{d}}u+\int_{-\frac{\lambda}{\delta^{s}}}^{M}\Big{(}1+\frac{R_{-}}{G_{-}}\Big{)}^{-1}\partial_{u}\Big{(}1+\frac{R_{-}}{G_{-}}\Big{)}{\mathrm{d}}u$
(4.23) $\displaystyle+\int_{-M}^{M}F^{-1}(u,\delta)\partial_{u}F(u,\delta).$
(4.24)

We treat the four first terms as before: the asymptotic properties are the
same, so similarly to the proof of Proposition 17, we show that for any
$\varepsilon>0$ there exist $\delta$ and $\lambda$ sufficiently small and some
$M>|u_{0}|$ such that

$|u|>M\quad\Rightarrow\quad\left|\dfrac{R_{\pm}(u,\delta)}{G_{\pm}(u)}\right|\leq\varepsilon$

so that in these regions the associated winding phase of
$1+\tfrac{R_{\pm}}{G_{\pm}}$ vanish as $\delta\to 0$ and $\lambda\to 0$.

Finally, for the fifth term, we use the fact that $F$ is continuous for all
$u\in[-M,M]$ and that $F(u,\delta)\to 1$ and $\delta\to 0$. Moreover, the
limit can be taken uniformly on $u\in[-M,M]$ so that
$|F(u,\delta)-1|\leq\epsilon$ for all $u\in[-M,M]$ and $\delta,\lambda$
sufficiently small. Again, $F$ has no winding phase in that region in the
limit. ∎

###  4.4 Case study

######  Example 20.

We illustrate Lemma 16, Proposition 17 and Corollary 18 above in the following
simple case of the classification where the boundary condition is

$A_{0}=\begin{pmatrix}1&0&0&0\\\ 0&\alpha&0&1\end{pmatrix},\qquad
A_{1}=\begin{pmatrix}0&0&0&0\\\
b_{21}&{\mathrm{i}}\beta&0&0\end{pmatrix}:=\begin{pmatrix}B&0\end{pmatrix},$
(4.25)

with $b_{12}\in\mathbb{C}$ and $\alpha,\beta\in\mathbb{R}$, which is part of
class $\mathfrak{A}_{1,4}$. We further assume $\beta\neq 0$ here. Notice that
for $b_{12}=\alpha=0$ and $\beta=-1$ we recover condition (b) from Examples 5
and 10. Inserting (4.12) and (4.25) into (2.17) we get

$\displaystyle g(k_{x},\kappa)=\frac{1}{64k^{4}\epsilon^{4}}\Big{(}$
$\displaystyle\,-\alpha-{\mathrm{i}}\kappa+\beta
k_{x}+64k^{4}\epsilon^{4}\left(\alpha+{\mathrm{i}}\kappa-\beta k_{x}\right)$
(4.26)
$\displaystyle+16\epsilon^{2}\left(\kappa+{\mathrm{i}}k_{x}\right)\left(\kappa_{\text{ev}}-{\mathrm{i}}k_{x}\right)\left(\alpha+{\mathrm{i}}\kappa_{\text{ev}}-\beta
k_{x}\right)\Big{)}+o(1)$ (4.27)

Many terms vanish in the limit $k\to\infty$ so this further simplifies to
$g(k_{x},\kappa)=g_{\infty}(k_{x},\kappa)+o(1)$ with

$g_{\infty}(k_{x},\kappa)=\alpha+{\mathrm{i}}\kappa-{\mathrm{i}}k_{x}\beta.$

Passing to dual variables we get

$g_{0}(\lambda_{x},\delta)=g_{\infty}\Big{(}-\dfrac{\lambda_{x}}{\lambda_{x}^{2}+\delta^{2}},\dfrac{\delta}{\lambda_{x}^{2}+\delta^{2}}\Big{)}=\frac{\alpha\lambda_{x}^{2}+\beta\lambda_{x}+\alpha\delta^{2}+{\mathrm{i}}\delta}{\delta^{2}+\lambda_{x}^{2}},$

and one can check that Lemma 16 applies. Then as discussed above, the
denominator does not contribute to the winding phase so we replace $g_{0}$ by
$(\delta^{2}+\lambda_{x}^{2})g_{0}$ (and keep the same symbol for $g_{0}$),
leading to

$g_{0}(\lambda_{x},\delta)=\alpha\lambda_{x}^{2}+\beta\lambda_{x}+\alpha\delta^{2}+{\mathrm{i}}\delta$
(4.28)

Now, the dominant scale is given by $s=r=1$, namely

$\delta^{-1}g(u\delta,\pm\delta)=G_{\pm}(u)+R_{\pm}(u,\delta)$

with

$G_{+}(u)={\mathrm{i}}+\beta u,\qquad G_{-}(u)=-{\mathrm{i}}+\beta u.\qquad
R_{+}(u)=R_{-}(u)=\alpha\delta(u^{2}+1)$ (4.29)

One can check that Proposition 17 applies. In particular, since we assume
$\beta\neq 0$

$\dfrac{R_{\pm}}{G_{\pm}}\mathop{\sim}_{u\to+\infty}\dfrac{\alpha}{\beta}u\delta,$

namely $\gamma=1$, and similarly as $u\to-\infty$. Thus we are left with the
winding number of

$\mathcal{S}=\dfrac{G_{-}(u)}{G_{+}(u)}=\dfrac{-{\mathrm{i}}+\beta
u}{{\mathrm{i}}+\beta u}$

which can be computed as twice the winding phase of $G_{-}(u)$ since
$G_{+}(u)=G^{*}_{-}(u)$. The real part of $G_{-}$ is constant negative so that
the winding phase is simply computed by the limits of $\arg(G_{-}(u))$ as
$u\to\pm\infty$. For $\beta>0$, the winding phase of $G_{-}$ is $\pi$ and for
$\beta<0$ it is $-\pi$. Thus in that example we infer
$w_{\infty}=\operatorname{sign}(\beta)$, regardless of the values of $\alpha$
and $b_{21}$. The relevant quantities of this example are illustrated in
Figure 2 below.

Figure 2: Argument of the scattering amplitude near $\lambda_{x}=0$ for
$\alpha=2$, $\beta=-1$ and $b_{12}=0$. Moreover, $m=1,\epsilon=0.1$ and
$\delta=0.001$. (Left) Argument of $S$ with respect to $\lambda_{x}$. The blue
curve corresponds to the exact $S$, which winds at $\lambda_{x}=0$ but also at
near $-0.5$ and $0.1$, detecting edge modes at finite $k_{x}\approx 2$ and
$-10$ (compare with Figure 1, middle). The red curve corresponds to $S_{0}$,
computed with $g_{0}$, which approximates $g$ well near $0$ but also keeps
track (poorly) of unwanted finite edge modes. The green curve corresponds to
$\mathcal{S}$ computed with $G_{\pm}$ (actually $\delta
G_{\pm}(\lambda_{x}/\delta)$ here, to compare with other curves), which
exactly captures only the winding of $S$ near infinity. (Right) Complex curve
$G_{-}(u)=-{\mathrm{i}}+\beta u$ for $\beta<0$. Its argument changes by $-\pi$
as $u$ goes from $-\infty$ to $+\infty$.

##  5 Anomaly classification

##### General strategy.

We apply the statements from Section 4.3 to each class of self-adjoint
boundary conditions from Table 1, going from $g$ to its leading contribution
$G$, and inferring $w_{\infty}$. More precisely:

  1. 1.

Pick a matrix $A=A_{0}+{\mathrm{i}}k_{x}A_{1}$ from Table 1 and insert it in
expression (2.17) for $g$, together with the asymptotic expansion of the bulk
eigensections (4.12). This leads to
$g(k_{x},\kappa)=g_{\infty}(k_{x},k)+o(1)$. We shall not detail this step
below and give only the final expression.

  2. 2.

Pass from $(k_{x},\kappa)$ to dual variables via (4.14), compute
$g_{0}(\lambda_{x},\delta)$ and simplify it.

  3. 3.

Find the dominant scale
$\delta^{-r}g_{0}(u\delta^{s},\pm\delta)=G_{\pm}(u)+R_{\pm}(u)$ and check that
$G_{\pm}$ and $R_{\pm}$ satisfy either the assumptions of Proposition 17 or
that of Proposition 19. We shall not always give the explicit expressions of
$R_{\pm}$ but they can be inferred from $g_{0}$ and $G_{\pm}$.

  4. 4.

Study the complex curves $G_{\pm}$ and compute their winding phases, from
which we deduce $w_{\infty}$.

  5. 5.

For specific values of the parameters in $A$, Assumption 2 of Proposition 17
may fail to hold so that another scale has to be found instead. This leads to
various subcases for which we repeat steps 4 and 5.

###  5.1 Class $\mathfrak{A}_{1,2}$

In that case we have

$A_{0}=\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\end{pmatrix},\qquad
A_{1}=\begin{pmatrix}b_{11}&b_{12}&0&0\\\
b_{21}&b_{22}&0&0\end{pmatrix}:=\begin{pmatrix}B&0\end{pmatrix},$

where $B\in\mathrm{M}_{2}(\mathbb{C})$ is arbitrary. The expansion of $g$ near
$\infty$ leads to $g(k_{x},\kappa)=g_{\infty}(k_{x},k)+o(1)$ where $o(1)$ in
the limit $k_{x}^{2}+\kappa^{2}\to\infty$, independent from the direction, and
with

$g_{\infty}(k_{x},\kappa)=-\det(B)k_{x}^{2}+{\mathrm{i}}\operatorname{tr}(B)k_{x}+\det(B)\dfrac{k_{x}^{2}\big{(}k_{x}^{2}+\kappa\kappa_{\mathrm{ev}}+{\mathrm{i}}k_{x}(\kappa_{\mathrm{ev}}-\kappa)\big{)}}{4\epsilon^{2}k^{4}}+1.$
(5.1)

We then move to the dual variables $\lambda_{x},\delta$ via (4.14), leading to

$\displaystyle g_{0}(\lambda_{x},\delta)=$
$\displaystyle\,\frac{1}{4\epsilon^{2}\left(\delta^{2}+\lambda_{x}^{2}\right){}^{2}}\Big{(}4\delta^{4}\epsilon^{2}+{\mathrm{i}}\delta\det(B)\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}+\det(B)\lambda_{x}^{3}\sqrt{\delta^{2}+2\lambda_{x}^{2}}$
(5.2)
$\displaystyle+{\mathrm{i}}\delta\det(B)\lambda_{x}^{3}+\det(B)\lambda_{x}^{4}-4\det(B)\epsilon^{2}\lambda_{x}^{2}-4{\mathrm{i}}\delta^{2}\operatorname{tr}(B)\epsilon^{2}\lambda_{x}$
(5.3)
$\displaystyle-4{\mathrm{i}}\operatorname{tr}(B)\epsilon^{2}\lambda_{x}^{3}+8\delta^{2}\epsilon^{2}\lambda_{x}^{2}+4\epsilon^{2}\lambda_{x}^{4}\Big{)}$
(5.4)

##### Case $\det B=0$.

In that case one has $g(\lambda_{x},\delta)=g_{0}(\lambda_{x},-\delta)$ so
that $S_{0}=1$ and $w_{\infty}=0$.

##### Case $\det B\neq 0$.

The denominator does not contribute to $w_{\infty}$ so we replace $g_{0}$ by
$(\delta^{2}+\lambda_{x}^{2})^{2}g_{0}$, and we we extract the dominant scale
by considering

$\delta^{-4}g_{0}(u\delta^{2},\delta)=G_{+}(u)+R_{+}(u,\delta)$

with

$\displaystyle G_{+}(u)=$
$\displaystyle\,1-{\mathrm{i}}\operatorname{tr}(B)u-\det(B)u^{2},$ (5.5)
$\displaystyle R_{+}(u,\delta)=$
$\displaystyle\,\frac{\delta^{4}\det(B)u^{4}}{4\epsilon^{2}}+\frac{{\mathrm{i}}\delta^{3}\det(B)u^{3}}{4\epsilon^{2}}+\frac{\delta^{2}\det(B)u^{3}\sqrt{\delta^{2}+2\delta^{4}u^{2}}}{4\epsilon^{2}}$
(5.6)
$\displaystyle+\frac{{\mathrm{i}}\delta\det(B)u^{2}\sqrt{\delta^{2}+2\delta^{4}u^{2}}}{4\epsilon^{2}}-{\mathrm{i}}\delta^{2}\operatorname{tr}(B)u^{3}+\delta^{4}u^{4}+2\delta^{2}u^{2}.$
(5.7)

One has $R_{+}(u,0)=0$ and

$\dfrac{R(u,\delta)}{G(u)}\sim-\dfrac{(1\pm\sqrt{2})\det(B)+4\epsilon^{2}}{4\det(B)\epsilon^{2}}u^{2}\delta^{4}$

as $u\to\pm\infty$, so that Proposition 17 applies333One can check that
Proposition 17 also applies in the fine-tuned case where
$\det(B)=-\tfrac{1\pm\sqrt{2}}{4\epsilon^{2}}$, then a lower order computation
leads to
$\dfrac{R(u,\delta)}{G(u)}\sim-\frac{{\mathrm{i}}(\sqrt{2}+1)\delta^{2}u(\delta+\operatorname{tr}(B))}{4\epsilon^{2}}$
when $G_{+}$ has no zero.

Similarly, replacing $\delta$ by $-\delta$ in (5.2) one gets
$\delta^{-4}g_{0}(u\delta^{2},-\delta)=G_{-}(u)+R_{-}(u,\delta)$ with

$\displaystyle G_{-}(u)=$ $\displaystyle\,G_{+}(u),$ (5.8) $\displaystyle
R_{-}(u,\delta)=$
$\displaystyle\,\frac{\delta^{4}\det(B)u^{4}}{4\epsilon^{2}}-\frac{{\mathrm{i}}\delta^{3}\det(B)u^{3}}{4\epsilon^{2}}+\frac{\delta^{2}\det(B)u^{3}\sqrt{\delta^{2}+2\delta^{4}u^{2}}}{4\epsilon^{2}}$
(5.9)
$\displaystyle-\frac{{\mathrm{i}}\delta\det(B)u^{2}\sqrt{\delta^{2}+2\delta^{4}u^{2}}}{4\epsilon^{2}}-{\mathrm{i}}\delta^{2}\operatorname{tr}(B)u^{3}+\delta^{4}u^{4}+2\delta^{2}u^{2}.$
(5.10)

and one can check that Proposition 17 also applies. In particular

$\mathcal{S}(u)=\dfrac{G_{-}(u)}{G_{+}(u)}=1$

so that $w_{\infty}=0$.

######  Remark 21 (About vanishing cases).

$G_{+}$ and $G_{-}$ may occasionally vanish at some $u_{0}\in\mathbb{R}$ for
exceptional values of $B$, but $w_{\infty}$ remains the same. Let
$\det(B)=d_{r}+{\mathrm{i}}d_{i}$ and
$\operatorname{tr}(B)=t_{r}+{\mathrm{i}}t_{i}$ with $d_{r},d_{i},t_{r}$ and
$t_{i}\in\mathbb{R}$. One has

$\operatorname{Re}(G_{+}(u))=1+t_{i}u-d_{r}u^{2},\qquad\operatorname{Im}(G_{+}(u))=-u(t_{r}+ud_{i}).$

The function $G_{+}$ vanishes in the following three cases :

  1. 1.

If $d_{r}=0,t_{i}\neq 0,d_{i}\neq 0$ and $d_{i}=-t_{r}t_{i}$ at
$u_{0}=-\tfrac{1}{t_{i}}$

  2. 2.

If $d_{r}\neq 0$, $\Delta=t_{i}^{2}+4d_{r}>0$ and

    1. (a)

$t_{r}=d_{i}=0$ at $u_{0\pm}=\tfrac{t_{i}\pm\sqrt{\Delta}}{2d_{r}}$,

    2. (b)

$d_{i}\neq 0$ with $2t_{r}d_{r}=-d_{i}(t_{i}+\sqrt{\Delta})$ or
$2t_{r}d_{r}=-d_{i}(t_{i}-\sqrt{\Delta})$, at $u_{0}=-\tfrac{t_{r}}{d_{i}}$.

In all cases above one could check that Proposition 19 applies. However, one
still has $G_{+}=G_{-}$, so we can instead replace $g_{0}$ by $g_{0}+z_{0}$
for some $z_{0}\in\mathbb{C}^{*}$ with $|z_{0}|$ sufficiently small so that it
does not change the winding phase of $g_{0}$. The computation above leads to

$g_{0}(\lambda_{x},\delta)+z_{0}=G_{+}(u)+z_{0}+R_{+}(u,\delta),\qquad
g_{0}(\lambda_{x},-\delta)+z_{0}=G_{-}(u)+z_{0}+R_{-}(u,\delta)$

with again $G_{-}(u)=G_{+}(u)$. Moreover, $z_{0}$ can be chosen such that
$G_{+}(u)+z_{0}\in\mathbb{C}^{*}$ for all $u\in\mathbb{R}$ and, consequently,
$G_{-}(u)+z_{0}\in\mathbb{C}^{*}$ . The asymptotic properties are preserved as
$u\to\pm\infty$ so that Proposition 17 now applies to $(G_{+}+z_{0},\,R_{+})$
and $(G_{-}+z_{0},\,R_{-})$. Thus
$\mathcal{S}=\tfrac{G_{-}+z_{0}}{G_{+}+z_{0}}=1$ and we get $w_{\infty}=0$
also in that case.

##### Summarizing.

Thus $S$ does not wind and $w_{\infty}=0$ for any boundary condition in this
class.

###  5.2 Class $\mathfrak{A}_{1,4}$

In this case we have

$A_{0}=\begin{pmatrix}1&0&0&0\\\ 0&\alpha&0&1\end{pmatrix},\qquad
A_{1}=\begin{pmatrix}b_{11}&0&0&0\\\
b_{21}&{\mathrm{i}}\beta&0&0\end{pmatrix}:=\begin{pmatrix}B&0\end{pmatrix},$

with $b_{11},b_{21}\in\mathbb{C}$ and $\alpha,\beta\in\mathbb{R}$. We get
$g(k_{x},\kappa)=g_{\infty}(k_{x},\kappa)+o(1)$ with

$\displaystyle g_{\infty}(k_{x},\kappa)=$
$\displaystyle-\det(B)k_{x}^{2}-b_{11}k_{x}\kappa+{\mathrm{i}}\kappa+{\mathrm{i}}k_{x}\left(\alpha
b_{11}+{\mathrm{i}}\beta\right)+\alpha$ (5.11)
$\displaystyle+\det(B)\dfrac{k_{x}^{2}\big{(}k_{x}^{2}+\kappa\kappa_{\mathrm{ev}}+{\mathrm{i}}k_{x}(\kappa_{\mathrm{ev}}-\kappa)\big{)}}{4\epsilon^{2}k^{4}}+b_{11}\frac{k_{x}\kappa_{\mathrm{ev}}\big{(}k_{x}^{2}+\kappa\kappa_{\mathrm{ev}}+{\mathrm{i}}k_{x}(\kappa_{\mathrm{ev}}-\kappa)\big{)}}{4k^{4}\epsilon^{2}}\,.$
(5.12)

Passing to dual variables via (4.14) we get

$\displaystyle g_{0}(\lambda_{x},\delta)=$
$\displaystyle\,\dfrac{1}{4\epsilon^{2}(\lambda_{x}^{2}+\delta^{2})^{2}}\Big{(}4\alpha\delta^{4}\epsilon^{2}-4{\mathrm{i}}\alpha
b_{11}\delta^{2}\epsilon^{2}\lambda_{x}-4{\mathrm{i}}\alpha
b_{11}\epsilon^{2}\lambda_{x}^{3}+b_{11}\delta^{3}\lambda_{x}-{\mathrm{i}}b_{11}\delta^{2}\lambda_{x}^{2}$
(5.13)
$\displaystyle+b_{11}\delta\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}-{\mathrm{i}}b_{11}\lambda_{x}^{3}\sqrt{\delta^{2}+2\lambda_{x}^{2}}+2b_{11}\delta\lambda_{x}^{3}+4b_{11}\delta\epsilon^{2}\lambda_{x}-2{\mathrm{i}}b_{11}\lambda_{x}^{4}$
(5.14)
$\displaystyle+4{\mathrm{i}}\delta^{3}\epsilon^{2}+{\mathrm{i}}\delta\det(B)\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}+\det(B)\lambda_{x}^{3}\sqrt{\delta^{2}+2\lambda_{x}^{2}}+{\mathrm{i}}\delta\det(B)\lambda_{x}^{3}+\det(B)\lambda_{x}^{4}$
(5.15)
$\displaystyle-4\det(B)\epsilon^{2}\lambda_{x}^{2}+8\alpha\delta^{2}\epsilon^{2}\lambda_{x}^{2}+4\alpha\epsilon^{2}\lambda_{x}^{4}+4\beta\delta^{2}\epsilon^{2}\lambda_{x}+4\beta\epsilon^{2}\lambda_{x}^{3}+4{\mathrm{i}}\delta\epsilon^{2}\lambda_{x}^{2}\Big{)}\,.$
(5.16)

The denominator does not contribute to $w_{\infty}$ so we replace $g_{0}$ by
$(\delta^{2}+\lambda_{x}^{2})^{2}g_{0}$. Then the dominant scale of $g_{0}$
depends on some parameter vanishing or not.

##### Case $\det(B)\neq 0$.

Since $\det(B)={\mathrm{i}}\beta b_{11}$ one has $b_{11}\neq 0$ and $\beta\neq
0$. The dominant scale is

$\delta^{-2}g_{0}(u\delta,\pm\delta)=G_{\pm}(u)+R_{\pm}(u)$

with

$G_{+}(u)=ub_{11}(1-{\mathrm{i}}\beta u),\qquad
G_{-}(u)=-ub_{11}(1+{\mathrm{i}}\beta u)\,.$ (5.17)

$G_{+}$ and $G_{-}$ vanish both at $u=0$ but one can check that Proposition 19
applies, in particular that the ratio $F(u,\delta)\to 1$ as $u\to 0$ and
$\delta\to 0$. Thus we are left with the winding number of

$\mathcal{S}(u)=\dfrac{G_{-}(u)}{G_{+}(u)}=-\dfrac{1+{\mathrm{i}}\beta
u}{1-{\mathrm{i}}\beta u}=\dfrac{{\mathrm{i}}-\beta u}{{\mathrm{i}}+\beta u}$
(5.18)

Numerator and denominator are conjugated so $w_{\infty}$ is twice the winding
phase of the numerator, which has constant positive imaginary part. Its
winding phase is the difference of limit of argument between $+\infty$ and
$-\infty$. If $\beta>0$ this winding phase is $\pi$. If $\beta<0$ it is
$-\pi$. Thus we infer $w_{\infty}=\operatorname{sign}(\beta)$.

##### Case $b_{11}=0$ and $\beta\neq 0$.

This case has been completely treated already in Example 20, and also leads to
$w_{\infty}=\operatorname{sign}(\beta)$.

##### Case $b_{11}\neq 0$ and $\beta=0$.

In that case $\det(B)=0$ and we go back to expression (5.13) for $g_{0}$ and
replace it again by $(\delta^{2}+\lambda_{x}^{2})^{2}g_{0}$. The dominant
scale is

$\delta^{-3/2}g_{0}(u\delta^{1/2},\pm\delta)=G_{\pm}(u)+R_{\pm}(u)$

with

$G_{+}(u)=b_{11}u\left(1-{\mathrm{i}}\alpha u^{2}\right),\qquad
G_{-}(u)=b_{11}u\left(-1-{\mathrm{i}}\alpha u^{2}\right)=-G^{*}_{+}(u)$ (5.19)

$G_{+}$ and $G_{-}$ vanish both at $u=0$ but one can check that Proposition 19
applies, in particular that the ratio $F(u,\delta)\to 1$ as $u\to 0$ and
$\delta\to 0$. Thus we are left with the winding number of

$\mathcal{S}(u)=\dfrac{G_{-}(u)}{G_{+}(u)}=\dfrac{1-{\mathrm{i}}\alpha
u^{2}}{1+{\mathrm{i}}\alpha u^{2}}.$

Notice that $\operatorname{Re}(1-{\mathrm{i}}\alpha u^{2})=1>0$ so that its
argument goes continuously from $\pi/2$ to $\pi/2$ as $u$ goes from $-\infty$
to $\infty$. Consequently, its winding phase vanishes, and similarly for
$1+{\mathrm{i}}\alpha u^{2}$. Thus $w_{\infty}=0$.

##### Case $b_{11}=0$ and $\beta=0$

In that case we go back to (5.11) which simplifies to
$g_{\infty}(k_{x},\kappa)=\alpha+{\mathrm{i}}\kappa$ leading to

$g_{0}(\lambda_{x},\delta)=\frac{\alpha\lambda_{x}^{2}+\alpha\delta^{2}+{\mathrm{i}}\delta}{\delta^{2}+\lambda_{x}^{2}},$

Replacing $g_{0}$ by $g_{0}(\delta^{2}+\lambda_{x}^{2})$, the dominant scale
is

$\delta^{-1}g(u\delta^{1/2},\pm\delta)=G_{\pm}(u)+R_{\pm}(u,\delta)$

with $G_{\pm}(u)={\mathrm{i}}\pm u^{2}\alpha$ and
$R_{\pm}(u,\delta)=\pm\alpha\delta$. Notice that Proposition 17 does not apply
here but the situation is simpler since $R_{\pm}$ do not depend on $u$ and
$G_{\pm}$ do not vanish hence one has
$\left|\tfrac{R_{\pm}}{G_{\pm}}\right|\to 0$ as $\delta\to 0$, for any
$\lambda>0$. Thus we are left with the winding number of

$\mathcal{S}(u)=\dfrac{G_{-}(u)}{G_{+}(u)}=\dfrac{{\mathrm{i}}-\alpha
u^{2}}{{\mathrm{i}}+\alpha u^{2}}.$

which is 0 as discussed in the case above. Consequently $w_{\infty}=0$.

##### Summarizing.

In class $\mathfrak{A}_{1,4}$ one has $w_{\infty}=\operatorname{sign}(\beta)$
if $\beta\neq 0$ and $w_{\infty}=0$ if $\beta=0$.

###  5.3 Class $\mathfrak{A}_{2,3}$

In this case we have

$A_{0}=\begin{pmatrix}0&1&0&0\\\ \alpha&0&0&1\end{pmatrix},\qquad
A_{1}=\begin{pmatrix}0&b_{12}&0&0\\\
{\mathrm{i}}\beta&b_{22}&0&0\end{pmatrix}:=\begin{pmatrix}B&0\end{pmatrix},$

with $b_{12},b_{22}\in\mathbb{C}$ and $\alpha,\beta\in\mathbb{R}$. We get
$g(k_{x},\kappa)=g_{\infty}(k_{x},\kappa)+o(1)$ with

$\displaystyle g_{\infty}(k_{x},\kappa)=$
$\displaystyle-\alpha+k_{x}\left(\beta-{\mathrm{i}}\alpha
b_{12}\right)-\det(B)k_{x}^{2}$ (5.20)
$\displaystyle+{\mathrm{i}}\det(B)\frac{k_{x}^{2}\left(k_{x}-{\mathrm{i}}\kappa\right)\left(\kappa_{\text{ev}}-{\mathrm{i}}k_{x}\right)}{4k^{4}\epsilon^{2}}-\frac{{\mathrm{i}}\left(\kappa_{\text{ev}}-\kappa\right)\left(b_{12}k_{x}-{\mathrm{i}}\right)\left(\kappa_{\text{ev}}-{\mathrm{i}}k_{x}\right)}{2k^{2}\epsilon}$
(5.21)

Passing to dual variables via (4.14) we get

$\displaystyle g_{0}(\lambda_{x},\delta)=$
$\displaystyle\,\dfrac{1}{4\epsilon^{2}(\lambda_{x}^{2}+\delta^{2})^{2}}\Big{(}-4\alpha\delta^{4}\epsilon^{2}+4{\mathrm{i}}\alpha
b_{12}\delta^{2}\epsilon^{2}\lambda_{x}+4{\mathrm{i}}\alpha
b_{12}\epsilon^{2}\lambda_{x}^{3}-2{\mathrm{i}}b_{12}\delta^{2}\epsilon\lambda_{x}$
(5.22)
$\displaystyle+2b_{12}\delta\epsilon\lambda_{x}\sqrt{\delta^{2}+2\lambda_{x}^{2}}-2{\mathrm{i}}b_{12}\epsilon\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}+2b_{12}\delta\epsilon\lambda_{x}^{2}-4{\mathrm{i}}b_{12}\epsilon\lambda_{x}^{3}+2\delta^{4}\epsilon$
(5.23)
$\displaystyle+{\mathrm{i}}\delta\det(B)\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}+\det(B)\lambda_{x}^{3}\sqrt{\delta^{2}+2\lambda_{x}^{2}}+{\mathrm{i}}\delta\det(B)\lambda_{x}^{3}+\det(B)\lambda_{x}^{4}-8\alpha\delta^{2}\epsilon^{2}\lambda_{x}^{2}$
(5.24)
$\displaystyle-4\det(B)\epsilon^{2}\lambda_{x}^{2}-4\alpha\epsilon^{2}\lambda_{x}^{4}-4\beta\delta^{2}\epsilon^{2}\lambda_{x}-4\beta\epsilon^{2}\lambda_{x}^{3}+2{\mathrm{i}}\delta^{3}\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}+2{\mathrm{i}}\delta^{3}\epsilon\lambda_{x}+6\delta^{2}\epsilon\lambda_{x}^{2}$
(5.25)
$\displaystyle+2\delta^{2}\epsilon\lambda_{x}\sqrt{\delta^{2}+2\lambda_{x}^{2}}+2{\mathrm{i}}\delta\epsilon\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}+2\epsilon\lambda_{x}^{3}\sqrt{\delta^{2}+2\lambda_{x}^{2}}+2{\mathrm{i}}\delta\epsilon\lambda_{x}^{3}+4\epsilon\lambda_{x}^{4}\Big{)}$
(5.26)

The denominator does not contribute to $w_{\infty}$ so we replace $g_{0}$ by
$2(\delta^{2}+\lambda_{x}^{2})^{2}g_{0}$. Then the dominant scale of $g_{0}$
depends on some parameter vanishing or not.

##### Case $\det(B)\neq 0$.

Since $\det(B)=-{\mathrm{i}}\beta b_{12}$ then $\beta\neq 0$ and $b_{12}\neq
0$. The dominant scale is

$\delta^{-4}g_{0}(u\delta^{2},\pm\delta)=G_{\pm}(u)+R_{\pm}(u)$

with

$\displaystyle
G_{+}(u)=(1+{\mathrm{i}})-2\alpha\epsilon+\left(2{\mathrm{i}}\alpha
b_{12}\epsilon-2\beta\epsilon+(1-{\mathrm{i}})b_{12}\right)u-2\epsilon\det(B)u^{2},$
(5.27) $\displaystyle
G_{-}(u)=(1-{\mathrm{i}})-2\alpha\epsilon+\left(2{\mathrm{i}}\alpha
b_{12}\epsilon-2\beta\epsilon-(1+{\mathrm{i}})b_{12}\right)u-2\epsilon\det(B)u^{2}.$
(5.28)

The two expressions above factorize to

$\displaystyle
G_{+}(u)=\big{(}(1+{\mathrm{i}})b_{12}u+(-1+{\mathrm{i}})\big{)}\big{(}(1+i)\alpha\epsilon+(1+{\mathrm{i}})\beta
u\epsilon-1\big{)},$ (5.29) $\displaystyle
G_{-}(u)=\big{(}(1+{\mathrm{i}})b_{12}u+(-1+{\mathrm{i}})\big{)}\big{(}(1+i)\alpha\epsilon+(1+{\mathrm{i}})\beta
u\epsilon-{\mathrm{i}}\big{)}.$ (5.30)

The first factor is common to $G_{+}$ and $G_{-}$ and vanishes at
$u_{0}=-{\mathrm{i}}/b_{12}$. The second factors do not vanish for
$u\in\mathbb{R}$. Thus if $b_{12}\notin{\mathrm{i}}\mathbb{R}$ then
Proposition 17 applies and otherwise Proposition 19 applies, as long as
$\det(B)\neq 0$. In both cases the ratio simplifies to

$\dfrac{G_{-}(u)}{G_{+}(u)}=\dfrac{2\epsilon\beta
u+2\alpha\epsilon-1+{\mathrm{i}}}{2\epsilon\beta
u+2\alpha\epsilon-1-{\mathrm{i}}}\,.$ (5.31)

Finally, the map $u\to 2\epsilon\beta u+2\alpha\epsilon-1+{\mathrm{i}}$ winds
by $-\operatorname{sign}(\beta)\pi$ as $y$ goes from $-\infty$ to $+\infty$
whereas $u\to 2\epsilon\beta u+2\alpha\epsilon-1-{\mathrm{i}}$ winds by
$\operatorname{sign}(\beta)\pi$, so that

$w_{\infty}=-\operatorname{sign}(\beta).$

##### Case $\beta=0$ and $b_{12}\neq 0$.

The dominant scale is

$\delta^{-3}g_{0}(u\delta,\pm\delta)=G_{\pm}(u)+R_{\pm}(u)$

with

$\displaystyle
G_{+}(u)=b_{12}u(u+{\mathrm{i}})\left(2\alpha\epsilon-1-{\mathrm{i}}(2u(1-\alpha\epsilon)-\sqrt{1+2u^{2}})\right)$
(5.32) $\displaystyle
G_{-}(u)=b_{12}u(u-{\mathrm{i}})\left(2\alpha\epsilon-1+{\mathrm{i}}(2u(1-\alpha\epsilon)-\sqrt{1+2u^{2}})\right)=G^{*}_{+}(u)\,.$
(5.33)

These two functions vanish at $u=0$, and we can check that Proposition 19
applies when $b_{12}\neq 0$. Thus we are left with studying the winding phase
of

$G(u)=(u-{\mathrm{i}})\left(2\alpha\epsilon-1+{\mathrm{i}}(2u(1-\alpha\epsilon)-\sqrt{1+2u^{2}})\right):=(u-{\mathrm{i}})\widetilde{G}(u).$

The first factor always winds by $\pi$. As for the second factor
$\widetilde{G}(u)$, its real part is constant. Moreover, one has

$\widetilde{G}(u)\mathop{\sim}_{+\infty}{\mathrm{i}}(2(1-\alpha\epsilon)+\sqrt{2})u,\qquad\widetilde{G}(u)\mathop{\sim}_{-\infty}{\mathrm{i}}(2(1-\alpha\epsilon)-\sqrt{2})u\,.$

Thus if $\alpha<\tfrac{1}{\epsilon}\left(1-\tfrac{1}{\sqrt{2}}\right)$ or
$\alpha>\tfrac{1}{\epsilon}\left(1+\tfrac{1}{\sqrt{2}}\right)$ one can check
that $\widetilde{G}$ winds by $-\pi$, so that $G$ does not wind. Otherwise, if

$\dfrac{1}{\epsilon}\left(1-\dfrac{1}{\sqrt{2}}\right)<\alpha<\dfrac{1}{\epsilon}\left(1+\dfrac{1}{\sqrt{2}}\right)$

then $\widetilde{G}$ does not wind, so that $G$ winds by $\pi$. To summarize
one has

$w_{\infty}=\begin{cases}1&\text{if}\qquad\frac{1}{\epsilon}\left(1-\frac{1}{\sqrt{2}}\right)<\alpha<\frac{1}{\epsilon}\left(1+\frac{1}{\sqrt{2}}\right),\\\
0&\text{if}\qquad\alpha<\frac{1}{\epsilon}\left(1-\frac{1}{\sqrt{2}}\right)\quad\text{or}\quad\alpha>\frac{1}{\epsilon}\left(1+\frac{1}{\sqrt{2}}\right).\end{cases}$

##### Case $\beta=0$ and $b_{12}=0$.

The dominant scale is

$\delta^{-4}g_{0}(u\delta,\pm\delta)=G_{\pm}(u)+R_{\pm}(u)$

with

$\displaystyle G_{-}(u)=$ $\displaystyle 1-2\alpha\epsilon-2\alpha
u^{4}\epsilon-4\alpha
u^{2}\epsilon+2u^{4}+u^{3}\sqrt{2u^{2}+1}+3u^{2}+u\sqrt{2u^{2}+1}$ (5.34)
$\displaystyle+{\mathrm{i}}\left(-u^{3}-u^{2}\sqrt{2u^{2}+1}-\sqrt{2u^{2}+1}-u\right)$
(5.35)

and $G_{+}(u)=G^{*}_{-}(u)$. It is worth noticing here that

$R_{+}=R-=0$

so that Proposition 17 is not even required. $G_{-}(u)$ never vanishes and its
imaginary part is always strictly negative. Moreover,

$G(u)\mathop{\sim}_{+\infty}(2(1-\alpha\epsilon)+\sqrt{2})u^{4},\qquad
G(u)\mathop{\sim}_{-\infty}(2(1-\alpha\epsilon)-\sqrt{2})u^{4}$

A similar analysis may be performed as in the previous case, and we end up
again with

$w_{\infty}=\begin{cases}1&\text{if}\qquad\frac{1}{\epsilon}\left(1-\frac{1}{\sqrt{2}}\right)<\alpha<\frac{1}{\epsilon}\left(1+\frac{1}{\sqrt{2}}\right),\\\
0&\text{if}\qquad\alpha<\frac{1}{\epsilon}\left(1-\frac{1}{\sqrt{2}}\right)\quad\text{or}\quad\alpha>\frac{1}{\epsilon}\left(1+\frac{1}{\sqrt{2}}\right).\end{cases}$

##### Case $b_{12}=0$ and $\beta\neq 0$.

In that case we go back to

$\displaystyle
g_{\infty}(k_{x},\kappa)=-\alpha+k_{x}\beta-\frac{\left(\kappa_{\text{ev}}-\kappa\right)\left(\kappa_{\text{ev}}-{\mathrm{i}}k_{x}\right)}{2k^{2}\epsilon}$
(5.36)

Passing to dual variables via (4.14) we get

$\displaystyle g_{0}(\lambda_{x},\delta)=$
$\displaystyle\,\dfrac{-2\alpha\delta^{2}\epsilon+\delta^{2}-2\alpha\epsilon\lambda_{x}^{2}-2\beta\epsilon\lambda_{x}+{\mathrm{i}}\delta\sqrt{\delta^{2}+2\lambda_{x}^{2}}+\lambda_{x}\sqrt{\delta^{2}+2\lambda_{x}^{2}}+{\mathrm{i}}\delta\lambda_{x}+2\lambda_{x}^{2}}{2\epsilon(\lambda_{x}^{2}+\delta^{2})}\,.$
(5.37)

The denominator does not contribute to $w_{\infty}$ so we replace $g_{0}$ by
$2(\delta^{2}+\lambda_{x}^{2})\epsilon g_{0}$. The dominant scale is

$\delta^{-2}g_{0}(u\delta^{2},\pm\delta)=G_{\pm}(u)+R_{\pm}(u)$

with

$G_{+}(u)=-2\alpha\epsilon-2\beta u\epsilon+1+{\mathrm{i}},\qquad
G_{-}(u)=G^{*}_{+}(u).$

Proposition 17 applies and we infer that $G_{-}$ has a winding phase of
$-\pi\operatorname{sign}(\beta)$ so that

$w_{\infty}=-\operatorname{sign}(\beta).$

##### Summarizing.

If $\beta\neq 0$ then $w_{\infty}=-\operatorname{sign}(\beta)$. If $\beta=0$
then

$w_{\infty}=\begin{cases}1&\text{if}\qquad\frac{1}{\epsilon}\left(1-\frac{1}{\sqrt{2}}\right)<\alpha<\frac{1}{\epsilon}\left(1+\frac{1}{\sqrt{2}}\right),\\\
0&\text{if}\qquad\alpha<\frac{1}{\epsilon}\left(1-\frac{1}{\sqrt{2}}\right)\quad\text{or}\quad\alpha>\frac{1}{\epsilon}\left(1+\frac{1}{\sqrt{2}}\right).\end{cases}$

######  Remark 22.

At the threshold cases $\beta=0$ and
$\alpha=\tfrac{1}{\epsilon}\big{(}1\pm\tfrac{1}{\sqrt{2}}\big{)}$, one can
check that the winding phase of $G_{+}$ is not a multiple of $\pi$ so that
$w_{\infty}\notin\mathbb{Z}$. We suspect some collapse of edge mode branch
occurring at infinity, similarly to the case $a=\pm\sqrt{2}$ in [20]. We do
not investigate further this fine-tuned case and consider it out of the
$w_{\infty}$-classification.

###  5.4 Class $\mathfrak{A}_{2,4}$

In this case we have

$A_{0}=\begin{pmatrix}a_{11}&1&0&0\\\
a_{21}&0&(a_{11}^{*})^{-1}&1\end{pmatrix},\qquad
A_{1}=\begin{pmatrix}b_{11}&b_{11}(a_{11})^{-1}&0&0\\\
b_{21}&b_{22}&0&0\end{pmatrix}:=\begin{pmatrix}B&0\end{pmatrix},$

with $a_{11}\in\mathbb{C}^{*}$, $b_{11},b_{21},b_{22}\in\mathbb{C}$, such that
$a_{21}=\alpha a_{11}+\epsilon^{-1}$ and
$b_{22}-b_{21}a_{11}^{-1}={\mathrm{i}}\beta$ with $\alpha,\beta\in\mathbb{R}$.
We get $g(k_{x},\kappa)=g_{\infty}(k_{x},\kappa)+o(1)$ with

$\displaystyle g_{\infty}(k_{x},\kappa)=$ (5.38)
$\displaystyle-\det(B)k_{x}^{2}-\frac{{\mathrm{i}}\kappa_{\text{ev}}}{a_{11}^{*}}+k_{x}\left(\frac{b_{11}\kappa_{\text{ev}}}{|a_{11}|^{2}}-\frac{{\mathrm{i}}a_{21}b_{11}}{a_{11}}+{\mathrm{i}}a_{11}b_{22}+b_{11}(-\kappa)-{\mathrm{i}}b_{21}\right)+{\mathrm{i}}a_{11}\kappa-
a_{21}$ (5.39)
$\displaystyle-\frac{\left(\kappa_{\text{ev}}-\kappa\right)\left(\kappa_{\text{ev}}-{\mathrm{i}}k_{x}\right)}{2k^{2}\epsilon}+\frac{\left(\kappa_{\text{ev}}-\kappa\right)\left(\kappa+{\mathrm{i}}k_{x}\right)\left(a_{11}+{\mathrm{i}}b_{11}k_{x}\right)}{2a_{11}^{*}k^{2}\epsilon}-\frac{{\mathrm{i}}b_{11}\left(\kappa_{\text{ev}}-\kappa\right)k_{x}\left(\kappa_{\text{ev}}-{\mathrm{i}}k_{x}\right)}{2a_{11}k^{2}\epsilon}$
(5.40)
$\displaystyle+\frac{b_{11}k_{x}\left(k_{x}-{\mathrm{i}}\kappa\right)\left(k_{x}+{\mathrm{i}}\kappa_{\text{ev}}\right)\left(b_{22}k_{x}+\kappa_{\text{ev}}\right)}{4k^{4}\epsilon^{2}}-\frac{b_{11}k_{x}\left(k_{x}-{\mathrm{i}}\kappa\right)\left(k_{x}+{\mathrm{i}}\kappa_{\text{ev}}\right)\left(b_{21}a_{11}^{*}k_{x}+\kappa\right)}{4|a_{11}|^{2}k^{4}\epsilon^{2}}\,.$
(5.41)

Passing to dual variables via (4.14) we get

$\displaystyle g_{0}(\lambda_{x},\delta)=$ (5.42)
$\displaystyle\,\dfrac{1}{4|a_{11}|^{2}\epsilon^{2}\left(\delta^{2}+\lambda_{x}^{2}\right){}^{2}}\Big{(}2\epsilon\left|a_{11}\right|{}^{2}\delta^{4}-2\epsilon
a_{11}^{2}\delta^{4}-4\epsilon^{2}\left|a_{11}\right|{}^{2}a_{21}\delta^{4}+2{\mathrm{i}}\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{11}^{2}\delta^{3}$
(5.43)
$\displaystyle+2{\mathrm{i}}\epsilon\left|a_{11}\right|{}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\delta^{3}+4{\mathrm{i}}\epsilon^{2}|a_{11}|^{2}a_{11}\delta^{3}+2{\mathrm{i}}\epsilon\left|a_{11}\right|{}^{2}\lambda_{x}\delta^{3}+2{\mathrm{i}}\epsilon
a_{11}^{2}\lambda_{x}\delta^{3}+\left|a_{11}\right|{}^{2}b_{11}\lambda_{x}\delta^{3}$
(5.44)
$\displaystyle+6\epsilon\left|a_{11}\right|{}^{2}\lambda_{x}^{2}\delta^{2}-2\epsilon
a_{11}^{2}\lambda_{x}^{2}\delta^{2}-8\epsilon^{2}\left|a_{11}\right|{}^{2}a_{21}\lambda_{x}^{2}\delta^{2}-{\mathrm{i}}\left|a_{11}\right|{}^{2}b_{11}\lambda_{x}^{2}\delta^{2}+{\mathrm{i}}b_{11}\lambda_{x}^{2}\delta^{2}$
(5.45)
$\displaystyle+4\epsilon^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{11}\delta^{2}+2\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{11}^{2}\lambda_{x}\delta^{2}+2\epsilon\left|a_{11}\right|{}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\lambda_{x}\delta^{2}$
(5.46)
$\displaystyle+{\mathrm{i}}\sqrt{\delta^{2}+2\lambda_{x}^{2}}b_{11}\lambda_{x}\delta^{2}+2{\mathrm{i}}\epsilon
a_{11}b_{11}\lambda_{x}\delta^{2}+4{\mathrm{i}}\epsilon^{2}\left|a_{11}\right|{}^{2}b_{21}\lambda_{x}\delta^{2}-2{\mathrm{i}}\epsilon
b_{11}a_{11}^{*}\lambda_{x}\delta^{2}$ (5.47)
$\displaystyle+4{\mathrm{i}}\epsilon^{2}a_{21}b_{11}a_{11}^{*}\lambda_{x}\delta^{2}-4{\mathrm{i}}\epsilon^{2}a_{11}^{2}b_{22}a_{11}^{*}\lambda_{x}\delta^{2}+2{\mathrm{i}}\epsilon\left|a_{11}\right|{}^{2}\lambda_{x}^{3}\delta+2{\mathrm{i}}\epsilon
a_{11}^{2}\lambda_{x}^{3}\delta+2\left|a_{11}\right|{}^{2}b_{11}\lambda_{x}^{3}\delta$
(5.48)
$\displaystyle+b_{11}\lambda_{x}^{3}\delta+{\mathrm{i}}\left|a_{11}\right|{}^{2}b_{11}b_{22}\lambda_{x}^{3}\delta-{\mathrm{i}}b_{11}b_{21}a_{11}^{*}\lambda_{x}^{3}\delta+2{\mathrm{i}}\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{11}^{2}\lambda_{x}^{2}\delta$
(5.49)
$\displaystyle+2{\mathrm{i}}\epsilon\left|a_{11}\right|{}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\lambda_{x}^{2}\delta+\left|a_{11}\right|{}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}b_{11}\lambda_{x}^{2}\delta+\sqrt{\delta^{2}+2\lambda_{x}^{2}}b_{11}\lambda_{x}^{2}\delta+2\epsilon
a_{11}b_{11}\lambda_{x}^{2}\delta$ (5.50)
$\displaystyle+{\mathrm{i}}\left|a_{11}\right|{}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}b_{11}b_{22}\lambda_{x}^{2}\delta+4{\mathrm{i}}\epsilon^{2}|a_{11}|^{2}a_{11}\lambda_{x}^{2}\delta+2\epsilon
b_{11}a_{11}^{*}\lambda_{x}^{2}\delta-{\mathrm{i}}b_{11}b_{21}a_{11}^{*}\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\delta$
(5.51)
$\displaystyle+4\epsilon^{2}\left|a_{11}\right|{}^{2}b_{11}\lambda_{x}\delta+2\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{11}b_{11}\lambda_{x}\delta+2\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}b_{11}a_{11}^{*}\lambda_{x}\delta+4\epsilon\left|a_{11}\right|{}^{2}\lambda_{x}^{4}$
(5.52)
$\displaystyle-4\epsilon^{2}\left|a_{11}\right|{}^{2}a_{21}\lambda_{x}^{4}-2{\mathrm{i}}\left|a_{11}\right|{}^{2}b_{11}\lambda_{x}^{4}+\left|a_{11}\right|{}^{2}b_{11}b_{22}\lambda_{x}^{4}-b_{11}b_{21}a_{11}^{*}\lambda_{x}^{4}+2\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{11}^{2}\lambda_{x}^{3}$
(5.53)
$\displaystyle+2\epsilon\left|a_{11}\right|{}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\lambda_{x}^{3}+4{\mathrm{i}}\epsilon^{2}\left|a_{11}\right|{}^{2}b_{21}\lambda_{x}^{3}+\left|a_{11}\right|{}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}b_{11}b_{22}\lambda_{x}^{3}-4{\mathrm{i}}\epsilon
b_{11}a_{11}^{*}\lambda_{x}^{3}$ (5.54)
$\displaystyle+4{\mathrm{i}}\epsilon^{2}a_{21}b_{11}a_{11}^{*}\lambda_{x}^{3}-4{\mathrm{i}}\epsilon^{2}|a_{11}|b_{22}a_{11}\lambda_{x}^{3}-4\det(B)\epsilon^{2}\left|a_{11}\right|{}^{2}\lambda_{x}^{2}+4\epsilon^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{11}\lambda_{x}^{2}$
(5.55)
$\displaystyle-{\mathrm{i}}\left|a_{11}\right|{}^{2}b_{11}\lambda_{x}^{3}\sqrt{\delta^{2}+2\lambda_{x}^{2}}-b_{11}b_{21}a_{11}^{*}\lambda_{x}^{3}\sqrt{\delta^{2}+2\lambda_{x}^{2}}-2{\mathrm{i}}\epsilon
a_{11}b_{11}\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}$ (5.56)
$\displaystyle-2{\mathrm{i}}\epsilon
b_{11}a_{11}^{*}\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}-4{\mathrm{i}}\epsilon^{2}b_{11}\lambda_{x}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\Big{)}.$
(5.57)

The denominator does not contribute to $w_{\infty}$ so we replace $g_{0}$ by
$2\epsilon|a_{11}|^{2}(\delta^{2}+\lambda_{x}^{2})^{2}g_{0}$. Then the
dominant scale of $g_{0}$ depends on some parameter vanishing or not.

##### Case $b_{11}\neq 0$.

If $b_{11}\neq 0$, the dominant scale is

$\delta^{-2}g_{0}(u\delta,\pm\delta)=G_{\pm}(u)+R_{\pm}(u)$

with

$\displaystyle
G_{+}(u)=-2\det(B)\epsilon|a_{11}|^{2}u^{2}-2{\mathrm{i}}\epsilon
b_{11}u\sqrt{1+2u^{2}}+2\epsilon b_{11}|a_{11}|^{2}u$ (5.58) $\displaystyle
G_{-}(u)=-2\det(B)\epsilon|a_{11}|^{2}u^{2}-2{\mathrm{i}}\epsilon
b_{11}u\sqrt{1+2u^{2}}-2\epsilon b_{11}|a_{11}|^{2}u\,.$ (5.59)

Recalling that $\det(B)=b_{11}(b_{22}-b_{21}/a_{11})$ and that
$b_{22}-b_{21}/a_{11}={\mathrm{i}}\beta$ with $\beta\in\mathbb{R}$. $G_{+}$
and $G_{-}$ vanish at $u=0$ and we can check that Proposition 19 applies when
$b_{11}\neq 0$. We are left with the ratio

$\mathcal{S}(u)=\dfrac{G_{-}(u)}{G_{+}(u)}=\dfrac{\beta|a_{11}|^{2}u+\sqrt{1+2u^{2}}-{\mathrm{i}}|a_{11}|^{2}}{\beta|a_{11}|^{2}u+\sqrt{1+2u^{2}}+{\mathrm{i}}|a_{11}|^{2}}:=\dfrac{G(u)}{G^{*}(u)}$
(5.60)

and we are left with twice the winding of the numerator
$G(u)=\beta|a_{11}|^{2}u+\sqrt{1+2u^{2}}-{\mathrm{i}}|a_{11}|^{2}$ which has
constant negative imaginary part so its winding phase is the difference of
limits of argument between $+\infty$ and $-\infty$. If
$\beta|a_{11}|^{2}>\sqrt{2}$ then

$\lim_{u\to+\infty}\frac{G(u)}{G(-u)}=-1$

and $G$ winds by $\pi$ so that $w_{\infty}=1$. Similarly, if
$\beta|a_{11}|^{2}<-\sqrt{2}$ then $G$ winds by $-\pi$ so that
$w_{\infty}=-1$. Otherwise, if $|\beta||a_{11}|^{2}<\sqrt{2}$ then

$\lim_{u\to+\infty}\frac{G(u)}{G(-u)}=1$

and $G$ does not wind, so that $w_{\infty}=0$.

##### Case $b_{11}=0$.

If $b_{11}=0$, the dominant scale is

$\delta^{-3}g_{0}(u\delta,\pm\delta)=G_{\pm}(u)+R_{\pm}(u)$

with

$\displaystyle
G_{+}(u)=2a_{11}\left(u^{2}+1\right)\epsilon\left(\sqrt{2u^{2}+1}+{\mathrm{i}}\big{(}|a_{11}|^{2}+u(a_{11}^{*}b_{21}-|a_{11}|^{2}b_{22})\big{)}\right)$
(5.61) $\displaystyle\,,$ $\displaystyle
G_{-}(u)=2a_{11}\left(u^{2}+1\right)\epsilon\left(-\sqrt{2u^{2}+1}+{\mathrm{i}}\big{(}(|a_{11}|^{2}-u(a_{11}^{*}b_{21}-|a_{11}|^{2}b_{22})\big{)}\right)$
(5.62)

The self-adjoint condition implies $b_{22}-b_{21}/a_{11}={\mathrm{i}}\beta$
with $\beta\in\mathbb{R}$. $G_{+}$ and $G_{-}$ do not vanish and Proposition
17 applies. The ratio simplifies to:

$\mathcal{S}(u)=\dfrac{G_{-}(u)}{G_{+}(u)}=\dfrac{-\sqrt{2u^{2}+1}+{\mathrm{i}}|a_{11}|^{2}-\beta|a_{11}|^{2}u}{\sqrt{2u^{2}+1}+{\mathrm{i}}|a_{11}|^{2}+\beta|a_{11}|^{2}u}=-\dfrac{G(u)}{G^{*}(u)}\,.$

We are back to the same numerator and denominator from the case $b_{11}=0$,
from which we immediately infer

$w_{\infty}=\begin{cases}1,&\beta|a_{11}|^{2}>\sqrt{2},\\\
0,&-\sqrt{2}<\beta|a_{11}|^{2}<\sqrt{2},\\\
-1,&\beta|a_{11}|^{2}<-\sqrt{2}.\end{cases}$

######  Remark 23.

At the threshold cases $\beta|a_{11}|^{2}=\pm\sqrt{2}$, one can check that the
winding phase of $G_{+}$ is not a multiple of $\pi$ so that
$w_{\infty}\notin\mathbb{Z}$. We suspect some collapse of edge mode branch
occurring at infinity, similarly to the case $a=\pm\sqrt{2}$ in [20]. We do
not investigate further this fine-tuned case and consider it out of the
$w_{\infty}$-classification.

###  5.5 Class $\mathfrak{A}_{3,4}$

In this case we have

$A_{0}=\begin{pmatrix}\alpha_{1}&a_{12}&1&0\\\
{\epsilon^{-1}}-a_{12}^{*}&\alpha_{2}&0&1\end{pmatrix},\qquad
A_{1}=\begin{pmatrix}{\mathrm{i}}\beta_{1}&b_{12}&0&0\\\
b_{12}^{*}&{\mathrm{i}}\beta_{2}&0&0\end{pmatrix}:=\begin{pmatrix}B&0\end{pmatrix},$

with $\alpha_{12},b_{12}\in\mathbb{C}$ and
$\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}\in\mathbb{R}$. We get
$g(k_{x},\kappa)=g_{\infty}(k_{x},\kappa)+o(1)$ with

$\displaystyle g_{\infty}(k_{x},\kappa)=$
$\displaystyle-\det(B)k_{x}^{2}+{\mathrm{i}}\alpha_{2}\kappa_{\text{ev}}-\kappa\kappa_{\text{ev}}+{\mathrm{i}}\alpha_{1}\kappa+\det(A)$
(5.63)
$\displaystyle+{\mathrm{i}}k_{x}\left({\mathrm{i}}\alpha_{2}\beta_{1}+{\mathrm{i}}\alpha_{1}\beta_{2}-a_{12}b_{12}^{*}+b_{12}a_{12}^{*}-\beta_{1}\kappa-\frac{b_{12}}{\epsilon}-\beta_{2}\kappa_{\text{ev}}\right)$
(5.64)
$\displaystyle\frac{\left(\kappa_{\text{ev}}-\kappa\right)\left(a_{12}\epsilon\left(-\kappa_{\text{ev}}+{\mathrm{i}}k_{x}\right)+b_{12}\epsilon
k_{x}\left(-k_{x}-{\mathrm{i}}\kappa_{\text{ev}}\right)+\left(k_{x}-{\mathrm{i}}\kappa\right)\left({\mathrm{i}}a_{12}^{*}\epsilon+b_{12}^{*}\epsilon
k_{x}-{\mathrm{i}}\right)\right)}{2k^{2}\epsilon^{2}}$ (5.65)
$\displaystyle+\frac{\left(k_{x}-{\mathrm{i}}\kappa\right)\left(-\kappa_{\text{ev}}+{\mathrm{i}}k_{x}\right)\left(-{\mathrm{i}}\det(B)k_{x}^{2}-{\mathrm{i}}\kappa\kappa_{\text{ev}}+\beta_{1}\kappa_{\text{ev}}k_{x}+\beta_{2}\kappa
k_{x}\right)}{4k^{4}\epsilon^{2}}\,.$ (5.66)

Passing to dual variables via (4.14) we get

$\displaystyle g_{0}(\lambda_{x},\delta)=$ (5.67)
$\displaystyle\,\dfrac{1}{4\epsilon^{2}\left(\delta^{2}+\lambda_{x}^{2}\right){}^{2}}\Big{(}4\det(A)\epsilon^{2}\delta^{4}+2\epsilon
a_{12}\delta^{4}-2\epsilon
a_{12}^{*}\delta^{4}+\delta^{4}-2{\mathrm{i}}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\delta^{3}+2{\mathrm{i}}\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{12}\delta^{3}$
(5.68)
$\displaystyle+2{\mathrm{i}}\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{12}^{*}\delta^{3}+4{\mathrm{i}}\epsilon^{2}\alpha_{1}\delta^{3}-{\mathrm{i}}\lambda_{x}\delta^{3}+2{\mathrm{i}}\epsilon
a_{12}\lambda_{x}\delta^{3}+2{\mathrm{i}}\epsilon
a_{12}^{*}\lambda_{x}\delta^{3}+{\mathrm{i}}\beta_{1}\lambda_{x}\delta^{3}+8\det(A)\epsilon^{2}\lambda_{x}^{2}\delta^{2}$
(5.69) $\displaystyle+6\epsilon a_{12}\lambda_{x}^{2}\delta^{2}-2\epsilon
a_{12}^{*}\lambda_{x}^{2}\delta^{2}+\beta_{1}\lambda_{x}^{2}\delta^{2}+\beta_{2}\lambda_{x}^{2}\delta^{2}-4\epsilon^{2}\alpha_{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\delta^{2}-3\lambda_{x}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\delta^{2}$
(5.70)
$\displaystyle+2\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{12}\lambda_{x}\delta^{2}+2{\mathrm{i}}\epsilon
b_{12}\lambda_{x}\delta^{2}+2\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{12}^{*}\lambda_{x}\delta^{2}-4{\mathrm{i}}\epsilon^{2}b_{12}a_{12}^{*}\lambda_{x}\delta^{2}-2{\mathrm{i}}\epsilon
b_{12}^{*}\lambda_{x}\delta^{2}$ (5.71)
$\displaystyle+4{\mathrm{i}}\epsilon^{2}a_{12}b_{12}^{*}\lambda_{x}\delta^{2}+4\epsilon^{2}\alpha_{2}\beta_{1}\lambda_{x}\delta^{2}+\sqrt{\delta^{2}+2\lambda_{x}^{2}}\beta_{2}\lambda_{x}\delta^{2}+4\epsilon^{2}\alpha_{1}\beta_{2}\lambda_{x}\delta^{2}+\det(B){\mathrm{i}}\lambda_{x}^{3}\delta+2{\mathrm{i}}\epsilon
a_{12}\lambda_{x}^{3}\delta$ (5.72) $\displaystyle+2{\mathrm{i}}\epsilon
a_{12}^{*}\lambda_{x}^{3}\delta+2{\mathrm{i}}\beta_{1}\lambda_{x}^{3}\delta-{\mathrm{i}}\beta_{2}\lambda_{x}^{3}\delta+\det(B){\mathrm{i}}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\lambda_{x}^{2}\delta+2{\mathrm{i}}\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{12}\lambda_{x}^{2}\delta+2\epsilon
b_{12}\lambda_{x}^{2}\delta$ (5.73)
$\displaystyle+2{\mathrm{i}}\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{12}^{*}\lambda_{x}^{2}\delta-2\epsilon
b_{12}^{*}\lambda_{x}^{2}\delta+4{\mathrm{i}}\epsilon^{2}\alpha_{1}\lambda_{x}^{2}\delta+{\mathrm{i}}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\beta_{1}\lambda_{x}^{2}\delta-4{\mathrm{i}}\epsilon^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\delta$
(5.74)
$\displaystyle-{\mathrm{i}}\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\delta-{\mathrm{i}}\beta_{2}\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\delta-2\epsilon
b_{12}^{*}\lambda_{x}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\delta+2\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}b_{12}\lambda_{x}\delta+4{\mathrm{i}}\epsilon^{2}\beta_{1}\lambda_{x}\delta$
(5.75)
$\displaystyle+4\det(A)\epsilon^{2}\lambda_{x}^{4}+\det(B)\lambda_{x}^{4}+4\epsilon
a_{12}\lambda_{x}^{4}+2\beta_{1}\lambda_{x}^{4}+\det(B)\sqrt{\delta^{2}+2\lambda_{x}^{2}}\lambda_{x}^{3}+2\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{12}\lambda_{x}^{3}$
(5.76)
$\displaystyle+2\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}a_{12}^{*}\lambda_{x}^{3}-4{\mathrm{i}}\epsilon^{2}b_{12}a_{12}^{*}\lambda_{x}^{3}+4{\mathrm{i}}\epsilon^{2}a_{12}b_{12}^{*}\lambda_{x}^{3}+\sqrt{\delta^{2}+2\lambda_{x}^{2}}\beta_{1}\lambda_{x}^{3}+4\epsilon^{2}\alpha_{2}\beta_{1}\lambda_{x}^{3}+4\epsilon^{2}\alpha_{1}\beta_{2}\lambda_{x}^{3}$
(5.77)
$\displaystyle-4\det(B)\epsilon^{2}\lambda_{x}^{2}+2{\mathrm{i}}\epsilon\sqrt{\delta^{2}+2\lambda_{x}^{2}}b_{12}^{*}\lambda_{x}^{2}-2\lambda_{x}^{3}\sqrt{\delta^{2}+2\lambda_{x}^{2}}-2{\mathrm{i}}\epsilon
b_{12}\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}-4\epsilon^{2}\alpha_{2}\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}$
(5.78)
$\displaystyle-4\epsilon^{2}\beta_{2}\lambda_{x}\sqrt{\delta^{2}+2\lambda_{x}^{2}}\Big{)}.$
(5.79)

The denominator does not contribute to $w_{\infty}$ so we replace $g_{0}$ by
$(\delta^{2}+\lambda_{x}^{2})^{2}g_{0}$. Then the dominant scale of $g_{0}$
depends on some parameter vanishing or not.

####  5.5.1 Case $b_{12}\neq 0$

If $b_{12}\neq 0$ the dominant scale is

$\delta^{-2}g_{0}(u\delta,\pm\delta)=G_{\pm}(u)+R_{\pm}(u)$

with

$\displaystyle
G_{+}(u)=-\det(B)u^{2}-\beta_{2}u\sqrt{2u^{2}+1}-{\mathrm{i}}\sqrt{2u^{2}+1}+{\mathrm{i}}\beta_{1}u$
(5.80) $\displaystyle
G_{-}(u)=-\det(B)u^{2}-\beta_{2}u\sqrt{2u^{2}+1}+{\mathrm{i}}\sqrt{2u^{2}+1}-{\mathrm{i}}\beta_{1}u$
(5.81)

Notice that

$\det(B)=-\beta_{1}\beta_{2}-|b_{12}|^{2}\in\mathbb{R}.$

Since $\beta_{1}\in\mathbb{R}$ then $G_{+}(u)=G^{*}_{-}(u)$. Moreover $G_{+}$
and $G_{-}$ do not vanish and Proposition 17 applies as long as $\Delta\neq 0$
with

$\Delta:=(\det B-\sqrt{2}\beta_{2})(\det B+\sqrt{2}\beta_{2})=B_{+}B_{-}$

with

$B_{\pm}:=\beta_{2}(\beta_{1}\pm\sqrt{2})+|b_{12}|^{2}.$

We are left with studying the argument of

$G(u)=-G_{-}(u)=\det(B)u^{2}+\beta_{2}u\sqrt{2u^{2}+1}+{\mathrm{i}}(\beta_{1}u-\sqrt{2u^{2}+1}):=G_{r}(u)+{\mathrm{i}}G_{i}(u).$
(5.82)

##### Case 1: $\Delta<0$.

In that case one can check that $G^{\prime}_{r}(u)$ is nowhere vanishing so
that $G_{r}$ is strictly monotonic. Moreover, the real part dominates the
behavior of $G$ asymptotically:

$G(u)\mathop{\sim}_{+\infty}(\det(B)+\sqrt{2}\beta_{2})u^{2},\qquad
G(u)\mathop{\sim}_{-\infty}(\det(B)-\sqrt{2}\beta_{2})u^{2}\,.$

In particular

$\lim_{u\to\infty}\dfrac{G(u)}{G(-u)}=\dfrac{\Delta}{(\det(B)-\sqrt{2}\beta_{2})^{2}}<0$

so that $G$ crosses the whole complex plane near the real line. Consequently,
its argument changes by $\pm\pi$.

The exact value of the winding sign of $G$ requires a detailed study of
$G_{i}(u)$, which behaves asymptotically as

$G_{i}(u)\mathop{\sim}_{+\infty}(\beta_{1}-\sqrt{2})u,\qquad
G_{i}(u)\mathop{\sim}_{-\infty}(\beta_{1}+\sqrt{2})u.$

Moreover $G_{i}(u)$ vanishes only if $\beta_{1}^{2}>2$, in which case
$G_{i}(u_{0})=0$ with

$u_{0}=\dfrac{\operatorname{sign}(\beta_{1})}{\sqrt{\beta_{1}^{2}-2}}.$ (5.83)

and

$G_{r}(u_{0})=-\dfrac{|b_{12}|^{2}}{\beta_{1}^{2}-2}<0$

since we assume $b_{12}\neq 0$.

##### Case 1a: $B_{+}>0$ and $B_{-}<0$.

This implies $B_{+}-B_{-}=2\sqrt{2}\beta_{2}>0$ so that $\beta_{2}>0$.
Moreover, $B_{-}<0$ implies $\beta_{2}(\beta_{1}-\sqrt{2})<-|b_{12}|^{2}<0$
and therefore $\beta_{1}-\sqrt{2}<0$, so that $G_{i}(u)<0$ near $+\infty$.
Moreover, in this case, one has $G_{r}(u)>0$ near $+\infty$ and $G_{r}(u)<0$
near $-\infty$. Finally, if $\beta_{1}<-\sqrt{2}$ then $G_{i}(u)>0$ near
$-\infty$ and $G_{i}(u)$ vanishes at $u_{0}$ given above, with
$G_{r}(u_{0})<0$. Putting all together, $G$ winds by $+\pi$. Otherwise, if
$\beta_{1}>-\sqrt{2}$ then $G_{i}(u)<0$ near $-\infty$ and $G_{i}(u)$ never
vanishes. Again, $G$ winds by $+\pi$.

##### Case 1b: $B_{+}<0$ and $B_{-}>0$.

Similarly $B_{+}-B_{-}<0$ so that $\beta_{2}<0$. Moreover $B_{+}<0$ implies
$\beta_{2}(\beta_{1}+\sqrt{2})<-|b_{12}|^{2}<0$ and therefore
$\beta_{1}+\sqrt{2}>0$ so that $G_{i}(u)<0$ near $-\infty$. Moreover, in this
case, one has $G_{r}(u)<0$ near $+\infty$ and $G_{r}(u)>0$ near $-\infty$.
Finally, if $\beta_{1}>\sqrt{2}$ then $G_{i}(u)>0$ near $+\infty$ and
$G_{i}(u)$ vanishes at $u_{0}$ given above, with $G_{r}(u_{0})<0$. Putting all
together, $G$ winds by $-\pi$. Otherwise, if $\beta_{1}<\sqrt{2}$ then
$G_{i}(u)<0$ near $+\infty$ and $G_{i}(u)$ never vanishes. Again, $G$ winds by
$-\pi$.

Summarizing cases 1a and 1b we get

$\int_{-\infty}^{\infty}G^{-1}(u)\partial_{u}G(u){\mathrm{d}}u=\operatorname{sign}(B_{+})\pi$

so that

$w_{\infty}=\operatorname{sign}(B_{+})=-\operatorname{sign}(B_{-})$

##### Case 2: $\Delta>0$.

In that case one can check that $G^{\prime}_{r}(u)$ vanishes exactly once.
Moreover, similarly to case 1, we compute

$\lim_{u\to\infty}\dfrac{G(u)}{G(-u)}=\dfrac{\Delta}{(\det(B)-\sqrt{2}\beta)^{2}}>0$

This time $G$ starts from $+\infty$ or $-\infty$ and comes back to the same
direction.

##### Case 2a: $B_{+}<0$ and $B_{-}<0$.

These inequalities imply $\beta_{2}(\beta_{1}\pm\sqrt{2})<-|b_{12}|^{2}<0$
which means that $\beta_{2}\neq 0$ and $\beta_{1}+\sqrt{2}$ and
$\beta_{1}-\sqrt{2}$ have the same sign. In particular $\beta_{1}^{2}-2>0$ and
$G_{i}(u)$ vanishes at $u_{0}$ given in (5.83), with $G_{r}(u_{0})<0$.
Moreover, in this case one has $G_{r}(u)>0$ near $+\infty$ and $-\infty$, so
that $G$ fully winds around $0$ as $u$ goes from $-\infty$ to $+\infty$.

If $\beta_{1}+\sqrt{2}>0$ and $\beta_{1}-\sqrt{2}>0$ then $G_{i}(u)>0$ near
$+\infty$ and $G_{i}(u)<0$ near $-\infty$. Thus $G$ winds by $-2\pi$ around
zero, so that $w_{\infty}=-2$. Otherwise, if $\beta_{1}+\sqrt{2}<0$ and
$\beta_{1}-\sqrt{2}<0$ then $G_{i}(u)<0$ near $+\infty$ and $G_{i}(u)>0$ near
$-\infty$. Thus $G$ winds by $2\pi$, so that $w_{\infty}=+2$. Summarizing,

$w_{\infty}=-2\operatorname{sign}(\beta_{1}+\sqrt{2})=-2\operatorname{sign}(\beta_{1}-\sqrt{2}).$

##### Case 2b: $B_{+}>0$ and $B_{-}>0$.

In this case one has $G_{r}(u)<0$ near $+\infty$ and $-\infty$. Either
$G_{i}(u)$ vanishes once at $u_{0}$, with $G_{r}(u_{0})<0$, or $G_{i}(u)$
never vanishes. In both cases, $G$ never crosses the real positive axis, so
that its change of argument is zero. Consequently

$w_{\infty}=0.$

Notice that this case is also valid when $\beta_{2}=0$, or
$\beta_{1}=\pm\sqrt{2}$.

##### Summarizing

If $b_{12}\neq 0$ and $\Delta\neq 0$ then we have the following table:

$\begin{array}[]{|c|c|c|}\hline\cr B_{+}&B_{-}&w_{\infty}\\\
\hline\cr>0&<0&1\\\ \hline\cr<0&>0&-1\\\ \hline\cr>0&>0&0\\\
\hline\cr<0&<0&\pm 2\\\ \hline\cr\end{array}$ (5.84)

with $\pm 2=2\operatorname{sign}(\sqrt{2}-\beta_{1})$.

####  5.5.2 Case $b_{12}=0$, $\beta_{2}\neq 0$ and $\beta_{1}^{2}\neq 2$.

In that case the dominant scale, $G_{\pm}$ and $R_{\pm}$ are the same as
before, except that $G$ further simplifies to

$G(u)=(\beta_{1}u-\sqrt{2u^{2}+1})(-\beta_{2}u+{\mathrm{i}}).$

The first factor vanishes at
$u_{0}=\operatorname{sign}{\beta_{1}}(\beta_{1}^{2}-2)^{-1/2}$, only if
$\beta_{1}^{2}>2$, in which case Proposition 19 applies. Otherwise Proposition
17 applies as long as $\beta_{1}^{2}<2$. In both cases we are left with

$\mathcal{S}(u)=\dfrac{G(u)}{G^{*}(u)}=\dfrac{-\beta_{2}u+{\mathrm{i}}}{-\beta_{2}u-{\mathrm{i}}}$

so that, as in previous classes, we infer

$w_{\infty}=\operatorname{sign}(\beta_{2}).$

####  5.5.3 Case $b_{12}=0$, $\beta_{2}=0$ and $\beta_{1}^{2}\neq 2$.

Notice that in that case one has $\Delta=0$. The dominant scale is

$\delta^{-3/2}g_{0}(u\delta^{1/2},\pm\delta)=G_{\pm}(u)+R_{\pm}(u)$

with

$G_{+}(u)=\left(\alpha_{2}u^{2}+{\mathrm{i}}\right)\left(\beta_{1}u-\sqrt{2}\sqrt{u^{2}}\right),\qquad
G_{-}(u)=G^{*}(u).$

The function $G_{+}$ and $G_{-}$ vanish at $u_{0}=0$. One can check that
Proposition 19 applies as long as $\beta_{1}^{2}\neq 2$, in which case we are
left with the winding of

$\mathcal{S}(u)=\dfrac{G_{-}(u)}{G_{+}(u)}=\dfrac{\alpha_{2}u^{2}-{\mathrm{i}}}{\alpha_{2}u^{2}+{\mathrm{i}}},$

from which we infer $w_{\infty}=0$.

##### Remaining cases.

We did not deal with the cases where $\Delta=0$ with

$\Delta=\big{(}(\beta_{1}+\sqrt{2})\beta_{2}+|b_{12}|^{2}\big{)}\big{(}(\beta_{1}-\sqrt{2})\beta_{2}+|b_{12}|^{2}\big{)}$

except if $\beta_{2}=b_{12}=0$, which is treated in the last case above. The
four remaining cases would be

  1. 1.

$\beta_{2}\neq 0$, $b_{12}=0$ and $\beta_{1}=\pm\sqrt{2}$,

  2. 2.

$\beta_{1}^{2}\neq 2$, $b_{12}\neq 0$ and

$\beta_{2}=-\dfrac{|b_{12}|^{2}}{\beta_{1}\pm\sqrt{2}}.$

The same issue with all this cases is that the dominant scale is correct at
$u\to+\infty$ and fails at $u\to-\infty$ (or conversely), so that neither
Proposition 17 or 19 apply. One could still compute $w_{\infty}$ by splitting
the integral into two parts (positive and negative $\lambda_{x}$), study them
separately with distinct dominant scales, then compute “half-winding” phases
and finally glue them together properly, but we do not expect
$w_{\infty}\in\mathbb{Z}$. We rather suspect some collapse of edge mode branch
occurring at infinity, similarly to the case $a=\pm\sqrt{2}$ in [20]. We do
not investigate further this fine-tuned case and consider it out of the
$w_{\infty}$-classification.

###  5.6 Class $\mathfrak{B}$

In this case we have

$A_{0}=\begin{pmatrix}a_{1}&a_{2}&{\mathrm{i}}\alpha&-{\mathrm{i}}\mu^{*}\alpha\\\
\mu a_{1}&\mu
a_{2}&{\mathrm{i}}\mu\alpha&-{\mathrm{i}}|\mu|^{2}\alpha\end{pmatrix},\qquad
A_{1}=\begin{pmatrix}1&0&0&0\\\ 0&1&0&0\end{pmatrix},$

with $a_{1},a_{2},\mu\in\mathbb{C}$ and $\alpha\in\mathbb{R}$. The self-
adjoint condition further requires $\alpha=0$ or
$\alpha\operatorname{Im}(\mu)-\epsilon\operatorname{Re}(a_{1}-a_{2}\mu)=0$ but
we shall keep it as an implicit constraint and keep general $a_{2},\,\alpha$
for the computations. We get $g(k_{x},\kappa)=g_{\infty}(k_{x},\kappa)+o(1)$
with

$\displaystyle g_{\infty}(k_{x},\kappa)=$ $\displaystyle-
k_{x}^{2}+{\mathrm{i}}a_{2}\mu
k_{x}+{\mathrm{i}}a_{1}k_{x}+\frac{{\mathrm{i}}\kappa_{\text{ev}}k_{x}^{3}+\kappa\kappa_{\text{ev}}k_{x}^{2}-{\mathrm{i}}\kappa
k_{x}^{3}+k_{x}^{4}}{4k^{4}\epsilon^{2}}\,.$ (5.85)

Passing to dual variables via (4.14) we get

$\displaystyle g_{0}(\lambda_{x},\delta)=$
$\displaystyle\,\dfrac{1}{4\epsilon^{2}\left(\delta^{2}+\lambda_{x}^{2}\right){}^{2}}\Big{(}-4{\mathrm{i}}a_{2}\delta^{2}\mu\epsilon^{2}\lambda_{x}-4{\mathrm{i}}a_{1}\delta^{2}\epsilon^{2}\lambda_{x}-4{\mathrm{i}}a_{2}\mu\epsilon^{2}\lambda_{x}^{3}-4{\mathrm{i}}a_{1}\epsilon^{2}\lambda_{x}^{3}+\lambda_{x}^{3}\sqrt{\delta^{2}+2\lambda_{x}^{2}}$
(5.86)
$\displaystyle+{\mathrm{i}}\delta\lambda_{x}^{2}\sqrt{\delta^{2}+2\lambda_{x}^{2}}+{\mathrm{i}}\delta\lambda_{x}^{3}+\lambda_{x}^{4}-4\epsilon^{2}\lambda_{x}^{2}\Big{)}.$
(5.87)

The denominator, as well as a common $\lambda_{x}$ factor, do not contribute
to $w_{\infty}$ so we replace $g_{0}$ by
$(\delta^{2}+\lambda_{x}^{2})^{2}\lambda_{x}^{-1}g_{0}$. The dominant scale is

$\delta^{-2}g_{0}(u\delta^{2},\pm\delta)=G_{\pm}(u)+R_{\pm}(u)$

with

$G_{+}(u)=-{\mathrm{i}}a_{2}\mu-{\mathrm{i}}a_{1}-u,\qquad G_{-}(u)=G_{+}(u).$
(5.88)

$G_{+}$ and $G_{-}$ do not vanish if
$a_{2}\mu+a_{1}\notin{\mathrm{i}}\mathbb{R}$, and we can check that
Proposition 17 applies, from which we immediately infer $\mathcal{S}=1$ and
$w_{\infty}=0$. If $a_{2}\mu+a_{1}\in{\mathrm{i}}\mathbb{R}$, since
$G_{+}=G_{-}$ then we can replace $g_{0}$ by $g_{0}+{\mathrm{i}}\eta$ with
$\eta\in\mathbb{R}^{*}$ sufficiently small so that the winding phases of
$g_{0}(\lambda_{x},\delta)$ and $g_{0}(\lambda_{x},-\delta)$ are unchanged.
Then Proposition 17 applies to
$G_{+}+{\mathrm{i}}\eta=G_{-}+{\mathrm{i}}\eta$. Again, we get $w_{\infty}=0$.

###  5.7 Class $\mathfrak{C}$

In this case we have

$A_{0}=\begin{pmatrix}a_{1}&a_{2}&0&a_{4}\\\ \mu a_{1}&\mu a_{2}&0&\mu
a_{4}\end{pmatrix},\qquad A_{1}=\begin{pmatrix}1&0&0&0\\\
0&0&0&0\end{pmatrix},$

with $a_{1}\in\mathbb{C}$, $a_{2},a_{4}\in\mathbb{C}^{2}\setminus\\{0\\}$ and
$\mu\in\mathbb{C}^{*}\setminus\\{0\\}$. The self-adjoint condition further
requires $\operatorname{Im}(a_{2}a_{4}^{*})=0$, which implies $a_{2}=0$, or
$a_{4}=0$ or $a_{4}=ra_{2}$ with $r\in\mathbb{R}^{*}$ and $a_{2}\neq 0$. We
get $g(k_{x},\kappa)=g_{\infty}(k_{x},\kappa)+o(1)$ with

$\displaystyle g_{\infty}(k_{x},\kappa)=$ $\displaystyle-a_{4}\mu
k_{x}\kappa+{\mathrm{i}}a_{2}\mu k_{x}+\frac{a_{4}\mu