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

In this case we have

$$A_0 = \begin{pmatrix} a_{11} & 1 & 0 & 0 \\ a_{21} & 0 & (a_{11}^*)^{-1} & 1 \end{pmatrix}, \quad A_1 = \begin{pmatrix} b_{11} & b_{11}(a_{11})^{-1} & 0 & 0 \\ b_{21} & b_{22} & 0 & 0 \end{pmatrix} := (B \ 0),$$

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 $b_{22} - b_{21}a_{11}^{-1} = i\beta$ with $\alpha, \beta \in \mathbb{R}$.
We get $g(k_x, \kappa) = g_\infty(k_x, \kappa) + o(1)$ with

$$g_\infty(k_x, \kappa) = -\det(B)k_x^2 - \frac{i\kappa_{ev}}{a_{11}^*} + k_x \left( \frac{b_{11}\kappa_{ev}}{|a_{11}|^2} - \frac{ia_{21}b_{11}}{a_{11}} + ia_{11}b_{22} + b_{11}(-\kappa) - ib_{21} \right) + ia_{11}\kappa - a_{21} \\ - \frac{(\kappa_{ev} - \kappa)(\kappa_{ev} - ik_x)}{2k^2\epsilon} + \frac{(\kappa_{ev} - \kappa)(\kappa + ik_x)(a_{11} + ib_{11}k_x)}{2a_{11}^*k^2\epsilon} - \frac{ib_{11}(\kappa_{ev} - \kappa)k_x(\kappa_{ev} - ik_x)}{2a_{11}k^2\epsilon} \\ + \frac{b_{11}k_x(k_x - i\kappa)(k_x + i\kappa_{ev})(b_{22}k_x + \kappa_{ev})}{4k^4\epsilon^2} - \frac{b_{11}k_x(k_x - i\kappa)(k_x + i\kappa_{ev})(b_{21}a_{11}^*k_x + \kappa)}{4|a_{11}|^2k^4\epsilon^2}. \quad (5.21)$$

Passing to dual variables via (4.13) we get

$$\begin{aligned} g_0(\lambda_x, \delta) = & \frac{1}{4|a_{11}|^2\epsilon^2(\delta^2 + \lambda_x^2)^2} \left( 2\epsilon|a_{11}|^2\delta^4 - 2\epsilon a_{11}^2\delta^4 - 4\epsilon^2|a_{11}|^2a_{21}\delta^4 + 2i\epsilon\sqrt{\delta^2 + 2\lambda_x^2}a_{11}^2\delta^3 \right. \\ & \left. + 2i\epsilon|a_{11}|^2\sqrt{\delta^2 + 2\lambda_x^2}\delta^3 + 4i\epsilon^2|a_{11}|^2a_{11}\delta^3 + 2i\epsilon|a_{11}|^2\lambda_x\delta^3 + 2i\epsilon a_{11}^2\lambda_x\delta^3 + |a_{11}|^2b_{11}\lambda_x\delta^3 \right. \\ & \left. + 6\epsilon|a_{11}|^2\lambda_x^2\delta^2 - 2\epsilon a_{11}^2\lambda_x^2\delta^2 - 8\epsilon^2|a_{11}|^2a_{21}\lambda_x^2\delta^2 - i|a_{11}|^2b_{11}\lambda_x^2\delta^2 + ib_{11}\lambda_x^2\delta^2 \right. \\ & \left. + 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}\lambda_x\delta^2 + 2\epsilon|a_{11}|^2\sqrt{\delta^2 + 2\lambda_x^2}\lambda_x\delta^2 \right. \\ & \left. + i\sqrt{\delta^2 + 2\lambda_x^2}b_{11}\lambda_x\delta^2 + 2i\epsilon a_{11}b_{11}\lambda_x\delta^2 + 4i\epsilon^2|a_{11}|^2b_{21}\lambda_x\delta^2 - 2i\epsilon b_{11}a_{11}^*\lambda_x\delta^2 \right. \\ & \left. + 4i\epsilon^2 a_{21}b_{11}a_{11}^*\lambda_x\delta^2 - 4i\epsilon^2 a_{11}^2b_{22}a_{11}^*\lambda_x\delta^2 + 2i\epsilon|a_{11}|^2\lambda_x^3\delta + 2i\epsilon a_{11}^2\lambda_x^3\delta + 2|a_{11}|^2b_{11}\lambda_x^3\delta \right. \\ & \left. + b_{11}\lambda_x^3\delta + i|a_{11}|^2b_{11}b_{22}\lambda_x^3\delta - ib_{11}b_{21}a_{11}^*\lambda_x^3\delta + 2i\epsilon\sqrt{\delta^2 + 2\lambda_x^2}a_{11}^2\lambda_x^2\delta \right. \\ & \left. + 2i\epsilon|a_{11}|^2\sqrt{\delta^2 + 2\lambda_x^2}\lambda_x^2\delta + |a_{11}|^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 \right. \\ & \left. + i|a_{11}|^2\sqrt{\delta^2 + 2\lambda_x^2}b_{11}b_{22}\lambda_x^2\delta + 4i\epsilon^2|a_{11}|^2a_{11}\lambda_x^2\delta + 2\epsilon b_{11}a_{11}^*\lambda_x^2\delta - ib_{11}b_{21}a_{11}^*\lambda_x^2\sqrt{\delta^2 + 2\lambda_x^2}\delta \right. \\ & \left. + 4\epsilon^2|a_{11}|^2b_{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|a_{11}|^2\lambda_x^4 \right. \\ & \left. - 4\epsilon^2|a_{11}|^2a_{21}\lambda_x^4 - 2i|a_{11}|^2b_{11}\lambda_x^4 + |a_{11}|^2b_{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 \right. \\ & \left. + 2\epsilon|a_{11}|^2\sqrt{\delta^2 + 2\lambda_x^2}\lambda_x^3 + 4i\epsilon^2|a_{11}|^2b_{21}\lambda_x^3 + |a_{11}|^2\sqrt{\delta^2 + 2\lambda_x^2}b_{11}b_{22}\lambda_x^3 - 4i\epsilon b_{11}a_{11}^*\lambda_x^3 \right. \\ & \left. + 4i\epsilon^2 a_{21}b_{11}a_{11}^*\lambda_x^3 - 4i\epsilon^2|a_{11}|b_{22}a_{11}\lambda_x^3 - 4\det(B)\epsilon^2|a_{11}|^2\lambda_x^2 + 4\epsilon^2\sqrt{\delta^2 + 2\lambda_x^2}a_{11}\lambda_x^2 \right. \\ & \left. - i|a_{11}|^2b_{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} - 2i\epsilon a_{11}b_{11}\lambda_x^2\sqrt{\delta^2 + 2\lambda_x^2} \right. \\ & \left. - 2i\epsilon b_{11}a_{11}^*\lambda_x^2\sqrt{\delta^2 + 2\lambda_x^2} - 4i\epsilon^2 b_{11}\lambda_x\sqrt{\delta^2 + 2\lambda_x^2} \right). \end{aligned} \quad (5.22)$$

The denominator does not contribute to $w_\infty$ so we replace $g_0$ by $2\epsilon|a_{11}|^2(\delta^2 + \lambda_x^2)^2g_0$. 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

$$\delta^{-2}g_0(u\delta, \pm\delta) = G_{\pm}(u) + R_{\pm}(u)$$
5.4 Class $\mathfrak{A}_{2,4}$

In this case we have

$$A_0 = \begin{pmatrix} a_{11} & 1 & 0 & 0 \\ a_{21} & 0 & (a_{11}^*)^{-1} & 1 \end{pmatrix}, \quad A_1 = \begin{pmatrix} b_{11} & b_{11}(a_{11})^{-1} & 0 & 0 \\ b_{21} & b_{22} & 0 & 0 \end{pmatrix} := (B \ 0),$$

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 $b_{22} - b_{21}a_{11}^{-1} = i\beta$ with $\alpha, \beta \in \mathbb{R}$.
We get $g(k_x, \kappa) = g_\infty(k_x, \kappa) + o(1)$ with

$$g_{\infty}(k_x, \kappa) = -\det(B)k_x^2 - \frac{i\kappa_{ev}}{a_{11}^*} + k_x \left( \frac{b_{11}\kappa_{ev}}{|a_{11}|^2} - \frac{ia_{21}b_{11}}{a_{11}} + ia_{11}b_{22} + b_{11}(-\kappa) - ib_{21} \right) + ia_{11}\kappa - a_{21} \\ - \frac{(\kappa_{ev} - \kappa)(\kappa_{ev} - ik_x)}{2k^2\epsilon} + \frac{(\kappa_{ev} - \kappa)(\kappa + ik_x)(a_{11} + ib_{11}k_x)}{2a_{11}^*k^2\epsilon} - \frac{ib_{11}(\kappa_{ev} - \kappa)k_x(\kappa_{ev} - ik_x)}{2a_{11}k^2\epsilon} \\ + \frac{b_{11}k_x(k_x - i\kappa)(k_x + i\kappa_{ev})(b_{22}k_x + \kappa_{ev})}{4k^4\epsilon^2} - \frac{b_{11}k_x(k_x - i\kappa)(k_x + i\kappa_{ev})(b_{21}a_{11}^*k_x + \kappa)}{4|a_{11}|^2k^4\epsilon^2}. \quad (5.21)$$

Passing to dual variables via (4.13) we get

$$\begin{aligned} g_0(\lambda_x, \delta) = & \frac{1}{4|a_{11}|^2\epsilon^2(\delta^2 + \lambda_x^2)^2} \left( 2\epsilon|a_{11}|^2\delta^4 - 2\epsilon a_{11}^2\delta^4 - 4\epsilon^2|a_{11}|^2a_{21}\delta^4 + 2i\epsilon\sqrt{\delta^2 + 2\lambda_x^2}a_{11}^2\delta^3 \right. \\ & \left. + 2i\epsilon|a_{11}|^2\sqrt{\delta^2 + 2\lambda_x^2}\delta^3 + 4i\epsilon^2|a_{11}|^2a_{11}\delta^3 + 2i\epsilon|a_{11}|^2\lambda_x\delta^3 + 2i\epsilon a_{11}^2\lambda_x\delta^3 + |a_{11}|^2b_{11}\lambda_x\delta^3 \right. \\ & \left. + 6\epsilon|a_{11}|^2\lambda_x^2\delta^2 - 2\epsilon a_{11}^2\lambda_x^2\delta^2 - 8\epsilon^2|a_{11}|^2a_{21}\lambda_x^2\delta^2 - i|a_{11}|^2b_{11}\lambda_x^2\delta^2 + ib_{11}\lambda_x^2\delta^2 \right. \\ & \left. + 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}\lambda_x\delta^2 + 2\epsilon|a_{11}|^2\sqrt{\delta^2 + 2\lambda_x^2}\lambda_x\delta^2 \right. \\ & \left. + i\sqrt{\delta^2 + 2\lambda_x^2}b_{11}\lambda_x\delta^2 + 2i\epsilon a_{11}b_{11}\lambda_x\delta^2 + 4i\epsilon^2|a_{11}|^2b_{21}\lambda_x\delta^2 - 2i\epsilon b_{11}a_{11}^*\lambda_x\delta^2 \right. \\ & \left. + 4i\epsilon^2 a_{21}b_{11}a_{11}^*\lambda_x\delta^2 - 4i\epsilon^2 a_{11}^2b_{22}a_{11}^*\lambda_x\delta^2 + 2i\epsilon|a_{11}|^2\lambda_x^3\delta + 2i\epsilon a_{11}^2\lambda_x^3\delta + 2|a_{11}|^2b_{11}\lambda_x^3\delta \right. \\ & \left. + b_{11}\lambda_x^3\delta + i|a_{11}|^2b_{11}b_{22}\lambda_x^3\delta - ib_{11}b_{21}a_{11}^*\lambda_x^3\delta + 2i\epsilon\sqrt{\delta^2 + 2\lambda_x^2}a_{11}^2\lambda_x^2\delta \right. \\ & \left. + 2i\epsilon|a_{11}|^2\sqrt{\delta^2 + 2\lambda_x^2}\lambda_x^2\delta + |a_{11}|^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 \right. \\ & \left. + i|a_{11}|^2\sqrt{\delta^2 + 2\lambda_x^2}b_{11}b_{22}\lambda_x^2\delta + 4i\epsilon^2|a_{11}|^2a_{11}\lambda_x^2\delta + 2\epsilon b_{11}a_{11}^*\lambda_x^2\delta - ib_{11}b_{21}a_{11}^*\lambda_x^2\sqrt{\delta^2 + 2\lambda_x^2}\delta \right. \\ & \left. + 4\epsilon^2|a_{11}|^2b_{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|a_{11}|^2\lambda_x^4 \right. \\ & \left. - 4\epsilon^2|a_{11}|^2a_{21}\lambda_x^4 - 2i|a_{11}|^2b_{11}\lambda_x^4 + |a_{11}|^2b_{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 \right. \\ & \left. + 2\epsilon|a_{11}|^2\sqrt{\delta^2 + 2\lambda_x^2}\lambda_x^3 + 4i\epsilon^2|a_{11}|^2b_{21}\lambda_x^3 + |a_{11}|^2\sqrt{\delta^2 + 2\lambda_x^2}b_{11}b_{22}\lambda_x^3 - 4i\epsilon b_{11}a_{11}^*\lambda_x^3 \right. \\ & \left. + 4i\epsilon^2 a_{21}b_{11}a_{11}^*\lambda_x^3 - 4i\epsilon^2|a_{11}|b_{22}a_{11}\lambda_x^3 - 4\det(B)\epsilon^2|a_{11}|^2\lambda_x^2 + 4\epsilon^2\sqrt{\delta^2 + 2\lambda_x^2}a_{11}\lambda_x^2 \right. \\ & \left. - i|a_{11}|^2b_{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} - 2i\epsilon a_{11}b_{11}\lambda_x^2\sqrt{\delta^2 + 2\lambda_x^2} \right. \\ & \left. - 2i\epsilon b_{11}a_{11}^*\lambda_x^2\sqrt{\delta^2 + 2\lambda_x^2} - 4i\epsilon^2 b_{11}\lambda_x\sqrt{\delta^2 + 2\lambda_x^2} \right). \end{aligned} \quad (5.22)$$

The denominator does not contribute to $w_\infty$ so we replace $g_0$ by $2\epsilon|a_{11}|^2(\delta^2 + \lambda_x^2)^2g_0$. 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

$$
\begin{align}
G_+(u) &= -2 \det(B) \epsilon |a_{11}|^2 u^2 - 2i \epsilon b_{11} u \sqrt{1+2u^2} + 2 \epsilon b_{11} |a_{11}|^2 u \nonumber \\
G_-(u) &= -2 \det(B) \epsilon |a_{11}|^2 u^2 - 2i \epsilon b_{11} u \sqrt{1+2u^2} - 2 \epsilon b_{11} |a_{11}|^2 u. \tag{5.23}
\end{align}
$$

Recalling that $\det(B) = b_{11}(b_{22} - b_{21}/a_{11})$ and that $b_{22} - b_{21}/a_{11} = i\beta$ 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

$$
S(u) = \frac{G_-(u)}{G_+(u)} = \frac{\beta |a_{11}|^2 u + \sqrt{1+2u^2} - i|a_{11}|^2}{\beta |a_{11}|^2 u + \sqrt{1+2u^2} + i|a_{11}|^2} := \frac{G(u)}{G^*(u)} \quad (5.24)
$$

and we are left with twice the winding of the numerator $G(u) = \beta|a_{11}|^2 u + \sqrt{1+2u^2} - i|a_{11}|^2$ 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

$$
\begin{align}
G_+(u) &= 2a_{11} (u^2 + 1) \epsilon \left( \sqrt{2u^2 + 1} + i(|a_{11}|^2 + u(a_{11}^* b_{21} - |a_{11}|^2 b_{22})) \right) \nonumber \\
G_-(u) &= 2a_{11} (u^2 + 1) \epsilon \left( -\sqrt{2u^2 + 1} + i((|a_{11}|^2 - u(a_{11}^* b_{21} - |a_{11}|^2 b_{22})) \right) \tag{5.25}
\end{align}
$$

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

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

We are back to the same numerator and denominator from the case $b_{11} = 0$, from which we imme-
diately 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.
with

$$
\begin{align}
G_+(u) &= -2 \det(B) \epsilon |a_{11}|^2 u^2 - 2i \epsilon b_{11} u \sqrt{1+2u^2} + 2 \epsilon b_{11} |a_{11}|^2 u \nonumber \\
G_-(u) &= -2 \det(B) \epsilon |a_{11}|^2 u^2 - 2i \epsilon b_{11} u \sqrt{1+2u^2} - 2 \epsilon b_{11} |a_{11}|^2 u. \tag{5.23}
\end{align}
$$

Recalling that $\det(B) = b_{11}(b_{22} - b_{21}/a_{11})$ and that $b_{22} - b_{21}/a_{11} = i\beta$ 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

$$
S(u) = \frac{G_-(u)}{G_+(u)} = \frac{\beta |a_{11}|^2 u + \sqrt{1+2u^2} - i|a_{11}|^2}{\beta |a_{11}|^2 u + \sqrt{1+2u^2} + i|a_{11}|^2} := \frac{G(u)}{G^*(u)} \quad (5.24)
$$

and we are left with twice the winding of the numerator $G(u) = \beta|a_{11}|^2 u + \sqrt{1+2u^2} - i|a_{11}|^2$ 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

$$
\begin{align}
G_+(u) &= 2a_{11} (u^2 + 1) \epsilon \left( \sqrt{2u^2 + 1} + i(|a_{11}|^2 + u(a_{11}^* b_{21} - |a_{11}|^2 b_{22})) \right) \nonumber \\
G_-(u) &= 2a_{11} (u^2 + 1) \epsilon \left( -\sqrt{2u^2 + 1} + i((|a_{11}|^2 - u(a_{11}^* b_{21} - |a_{11}|^2 b_{22})) \right) \tag{5.25}
\end{align}
$$

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

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

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

$$A_0 = \begin{pmatrix} \alpha_1 & a_{12} & 1 & 0 \\ \epsilon^{-1} - a_{12}^* & \alpha_2 & 0 & 1 \end{pmatrix}, \quad A_1 = \begin{pmatrix} i\beta_1 & b_{12} & 0 & 0 \\ b_{12}^* & i\beta_2 & 0 & 0 \end{pmatrix} := (B \ 0),$$

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

$$g_\infty(k_x, \kappa) = -\det(B)k_x^2 + i\alpha_2\kappa_{ev} - \kappa\kappa_{ev} + i\alpha_1\kappa + \det(A) \\ + ik_x \left( i\alpha_2\beta_1 + i\alpha_1\beta_2 - a_{12}b_{12}^* + b_{12}a_{12}^* - \beta_1\kappa - \frac{b_{12}}{\epsilon} - \beta_2\kappa_{ev} \right) \\ \frac{(\kappa_{ev} - \kappa)(a_{12}\epsilon(-\kappa_{ev} + ik_x) + b_{12}\epsilon k_x(-k_x - i\kappa_{ev}) + (k_x - i\kappa)(i\alpha_{12}^*\epsilon + b_{12}^*\epsilon k_x - i))}{2k^2\epsilon^2} \\ + \frac{(k_x - i\kappa)(-\kappa_{ev} + ik_x)(-i\det(B)k_x^2 - i\kappa\kappa_{ev} + \beta_1\kappa_{ev}k_x + \beta_2\kappa k_x)}{4k^4\epsilon^2}. \quad (5.26)$$

Passing to dual variables via (4.13) we get

$$g_0(\lambda_x, \delta) = \frac{1}{4\epsilon^2 (\delta^2 + \lambda_x^2)^2} \left( \begin{aligned} & 4 \det(A) \epsilon^2 \delta^4 + 2 \epsilon a_{12} \delta^4 - 2 \epsilon a_{12}^* \delta^4 + \delta^4 - 2i \sqrt{\delta^2 + 2\lambda_x^2} \delta^3 + 2i \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12} \delta^3 \\ & + 2i \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12} \delta^3 + 4i \epsilon^2 \alpha_1 \delta^3 - i \lambda_x \delta^3 + 2i \epsilon a_{12} \lambda_x \delta^3 + 2i \epsilon a_{12}^* \lambda_x \delta^3 + i \beta_1 \lambda_x \delta^3 + 8 \det(A) \epsilon^2 \lambda_x^2 \delta^2 \\ & + 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 \\ & + 2 \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12} \lambda_x \delta^2 + 2i \epsilon b_{12} \lambda_x \delta^2 + 2 \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12}^* \lambda_x \delta^2 - 4i \epsilon^2 b_{12} a_{12}^* \lambda_x \delta^2 - 2i \epsilon b_{12}^* \lambda_x \delta^2 \\ & + 4i \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) i \lambda_x^3 \delta + 2i \epsilon a_{12} \lambda_x^3 \delta \\ & + 2i \epsilon a_{12}^* \lambda_x^3 \delta + 2i \beta_1 \lambda_x^3 \delta - i \beta_2 \lambda_x^3 \delta + \det(B) i \sqrt{\delta^2 + 2\lambda_x^2} \lambda_x^2 \delta + 2i \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12} \lambda_x^2 \delta + 2 \epsilon b_{12} \lambda_x^2 \delta \\ & + 2i \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12}^* \lambda_x^2 \delta - 2 \epsilon b_{12}^* \lambda_x^2 \delta + 4i \epsilon^2 \alpha_1 \lambda_x^2 \delta + i \sqrt{\delta^2 + 2\lambda_x^2} \beta_1 \lambda_x^2 \delta - 4i \epsilon^2 \sqrt{\delta^2 + 2\lambda_x^2} \delta \\ & - i \lambda_x^2 \sqrt{\delta^2 + 2\lambda_x^2} \delta - 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 + 4i \epsilon^2 \beta_1 \lambda_x \delta \\ & + 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 \\ & + 2 \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12}^* \lambda_x^3 - 4i \epsilon^2 b_{12} a_{12}^* \lambda_x^3 + 4i \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 \\ & - 4 \det(B) \epsilon^2 \lambda_x^2 + 2i \epsilon \sqrt{\delta^2 + 2\lambda_x^2} b_{12}^* \lambda_x^2 - 2 \lambda_x^3 \sqrt{\delta^2 + 2\lambda_x^2} - 2i \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} \\ & - 4 \epsilon^2 \beta_2 \lambda_x \sqrt{\delta^2 + 2\lambda_x^2} \right). \quad (5.27)$$

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

$$G_+(u) = -\det(B)u^2 - \beta_2 u \sqrt{2u^2+1} - i\sqrt{2u^2+1} + i\beta_1 u \\ G_-(u) = -\det(B)u^2 - \beta_2 u \sqrt{2u^2+1} + i\sqrt{2u^2+1} - i\beta_1 u \quad (5.28)$$
## 5.5 Class $\mathfrak{A}_{3,4}$

In this case we have

$$A_0 = \begin{pmatrix} \alpha_1 & a_{12} & 1 & 0 \\ \epsilon^{-1} - a_{12}^* & \alpha_2 & 0 & 1 \end{pmatrix}, \quad A_1 = \begin{pmatrix} i\beta_1 & b_{12} & 0 & 0 \\ b_{12}^* & i\beta_2 & 0 & 0 \end{pmatrix} := (B \ 0),$$

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

$$g_\infty(k_x, \kappa) = -\det(B)k_x^2 + i\alpha_2\kappa_{ev} - \kappa\kappa_{ev} + i\alpha_1\kappa + \det(A) \\ + ik_x \left( i\alpha_2\beta_1 + i\alpha_1\beta_2 - a_{12}b_{12}^* + b_{12}a_{12}^* - \beta_1\kappa - \frac{b_{12}}{\epsilon} - \beta_2\kappa_{ev} \right) \\ \frac{(\kappa_{ev} - \kappa)(a_{12}\epsilon(-\kappa_{ev} + ik_x) + b_{12}\epsilon k_x(-k_x - i\kappa_{ev}) + (k_x - i\kappa)(i\alpha_{12}^*\epsilon + b_{12}^*\epsilon k_x - i))}{2k^2\epsilon^2} \\ + \frac{(k_x - i\kappa)(-\kappa_{ev} + ik_x)(-i\det(B)k_x^2 - i\kappa\kappa_{ev} + \beta_1\kappa_{ev}k_x + \beta_2\kappa k_x)}{4k^4\epsilon^2}. \quad (5.26)$$

Passing to dual variables via (4.13) we get

$$g_0(\lambda_x, \delta) = \frac{1}{4\epsilon^2 (\delta^2 + \lambda_x^2)^2} \left( \begin{aligned} & 4 \det(A) \epsilon^2 \delta^4 + 2 \epsilon a_{12} \delta^4 - 2 \epsilon a_{12}^* \delta^4 + \delta^4 - 2i \sqrt{\delta^2 + 2\lambda_x^2} \delta^3 + 2i \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12} \delta^3 \\ & + 2i \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12} \delta^3 + 4i \epsilon^2 \alpha_1 \delta^3 - i \lambda_x \delta^3 + 2i \epsilon a_{12} \lambda_x \delta^3 + 2i \epsilon a_{12}^* \lambda_x \delta^3 + i \beta_1 \lambda_x \delta^3 + 8 \det(A) \epsilon^2 \lambda_x^2 \delta^2 \\ & + 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 \\ & + 2 \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12} \lambda_x \delta^2 + 2i \epsilon b_{12} \lambda_x \delta^2 + 2 \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12}^* \lambda_x \delta^2 - 4i \epsilon^2 b_{12} a_{12}^* \lambda_x \delta^2 - 2i \epsilon b_{12}^* \lambda_x \delta^2 \\ & + 4i \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) i \lambda_x^3 \delta + 2i \epsilon a_{12} \lambda_x^3 \delta \\ & + 2i \epsilon a_{12}^* \lambda_x^3 \delta + 2i \beta_1 \lambda_x^3 \delta - i \beta_2 \lambda_x^3 \delta + \det(B) i \sqrt{\delta^2 + 2\lambda_x^2} \lambda_x^2 \delta + 2i \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12} \lambda_x^2 \delta + 2 \epsilon b_{12} \lambda_x^2 \delta \\ & + 2i \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12}^* \lambda_x^2 \delta - 2 \epsilon b_{12}^* \lambda_x^2 \delta + 4i \epsilon^2 \alpha_1 \lambda_x^2 \delta + i \sqrt{\delta^2 + 2\lambda_x^2} \beta_1 \lambda_x^2 \delta - 4i \epsilon^2 \sqrt{\delta^2 + 2\lambda_x^2} \delta \\ & - i \lambda_x^2 \sqrt{\delta^2 + 2\lambda_x^2} \delta - 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 + 4i \epsilon^2 \beta_1 \lambda_x \delta \\ & + 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 \\ & + 2 \epsilon \sqrt{\delta^2 + 2\lambda_x^2} a_{12}^* \lambda_x^3 - 4i \epsilon^2 b_{12} a_{12}^* \lambda_x^3 + 4i \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 \\ & - 4 \det(B) \epsilon^2 \lambda_x^2 + 2i \epsilon \sqrt{\delta^2 + 2\lambda_x^2} b_{12}^* \lambda_x^2 - 2 \lambda_x^3 \sqrt{\delta^2 + 2\lambda_x^2} - 2i \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} \\ & - 4 \epsilon^2 \beta_2 \lambda_x \sqrt{\delta^2 + 2\lambda_x^2} \right). \quad (5.27)$$

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

$$G_+(u) = -\det(B)u^2 - \beta_2 u \sqrt{2u^2+1} - i\sqrt{2u^2+1} + i\beta_1 u \\ G_-(u) = -\det(B)u^2 - \beta_2 u \sqrt{2u^2+1} + i\sqrt{2u^2+1} - i\beta_1 u \quad (5.28)$$
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

$$ G(u) = -G_{-}(u) = \det(B)u^{2} + \beta_{2}u\sqrt{2u^{2} + 1} + i(\beta_{1}u - \sqrt{2u^{2} + 1}) := G_{r}(u) + iG_{i}(u). \quad (5.29) $$

**Case 1:** $\Delta < 0$. In that case one can check that $G'_r(u)$ is nowhere vanishing so that $G_r$ is strictly monotonic. Moreover, the real part dominates the behavior of $G$ asymptotically:

$$ G(u) \underset{+\infty}{\sim} (\det(B) + \sqrt{2}\beta_2)u^2, \quad G(u) \underset{-\infty}{\sim} (\det(B) - \sqrt{2}\beta_2)u^2. $$

In particular

$$ \lim_{u \to \infty} \frac{G(u)}{G(-u)} = \frac{\Delta}{(\det(B) - \sqrt{2}\beta_2)^2} < 0 $$

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

$$ G_i(u) \underset{+\infty}{\sim} (\beta_1 - \sqrt{2})u, \quad G_i(u) \underset{-\infty}{\sim} (\beta_1 + \sqrt{2})u. $$

Moreover $G_i(u)$ vanishes only if $\beta_1^2 > 2$, in which case $G_i(u_0) = 0$ with

$$ u_0 = \frac{\operatorname{sign}(\beta_1)}{\sqrt{\beta_1^2 - 2}}. \quad (5.30) $$

and

$$ G_r(u_0) = -\frac{|b_{12}|^2}{\beta_1^2 - 2} < 0 $$

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 $+\pi$.
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

$$ G(u) = -G_{-}(u) = \det(B)u^{2} + \beta_{2}u\sqrt{2u^{2} + 1} + i(\beta_{1}u - \sqrt{2u^{2} + 1}) := G_{r}(u) + iG_{i}(u). \quad (5.29) $$

**Case 1:** $\Delta < 0$. In that case one can check that $G'_r(u)$ is nowhere vanishing so that $G_r$ is strictly monotonic. Moreover, the real part dominates the behavior of $G$ asymptotically:

$$ G(u) \underset{+\infty}{\sim} (\det(B) + \sqrt{2}\beta_2)u^2, \quad G(u) \underset{-\infty}{\sim} (\det(B) - \sqrt{2}\beta_2)u^2. $$

In particular

$$ \lim_{u \to \infty} \frac{G(u)}{G(-u)} = \frac{\Delta}{(\det(B) - \sqrt{2}\beta_2)^2} < 0 $$

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

$$ G_i(u) \underset{+\infty}{\sim} (\beta_1 - \sqrt{2})u, \quad G_i(u) \underset{-\infty}{\sim} (\beta_1 + \sqrt{2})u. $$

Moreover $G_i(u)$ vanishes only if $\beta_1^2 > 2$, in which case $G_i(u_0) = 0$ with

$$ u_0 = \frac{\operatorname{sign}(\beta_1)}{\sqrt{\beta_1^2 - 2}}. \quad (5.30) $$

and

$$ G_r(u_0) = -\frac{|b_{12}|^2}{\beta_1^2 - 2} < 0 $$

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

$$ \int_{-\infty}^{\infty} G^{-1}(u) \partial_u G(u) du = \operatorname{sign}(B_+) \pi $$

so that

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

**Case 2:** $\Delta > 0$. In that case one can check that $G'_r(u)$ vanishes exactly once. Moreover, similarly to case 1, we compute

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

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 (5.30), 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:

$$ \begin{array}{|c|c|c|} \hline B_{+} & B_{-} & w_{\infty} \\ \hline > 0 & < 0 & 1 \\ \hline < 0 & > 0 & -1 \\ \hline > 0 & > 0 & 0 \\ \hline < 0 & < 0 & \pm 2 \\ \hline \end{array} \qquad (5.31) $$

with $\pm 2 = 2\operatorname{sign}(\sqrt{2} - \beta_1)$.
**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

$$ \int_{-\infty}^{\infty} G^{-1}(u) \partial_u G(u) du = \operatorname{sign}(B_+) \pi $$

so that

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

**Case 2:** $\Delta > 0$. In that case one can check that $G'_r(u)$ vanishes exactly once. Moreover, similarly to case 1, we compute

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

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 (5.30), 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:

$$ \begin{array}{|c|c|c|} \hline B_{+} & B_{-} & w_{\infty} \\ \hline > 0 & < 0 & 1 \\ \hline < 0 & > 0 & -1 \\ \hline > 0 & > 0 & 0 \\ \hline < 0 & < 0 & \pm 2 \\ \hline \end{array} \qquad (5.31) $$

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

$$G(u) = (\beta_1 u - \sqrt{2u^2 + 1})(-\beta_2 u + i).$$

The first factor vanishes at $u_0 = \text{sign} \, \beta_1(\beta_1^2 - 2)^{-1/2}$, 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

$$S(u) = \frac{G(u)}{G^*(u)} = \frac{-\beta_2 u + i}{-\beta_2 u - i}$$

so that, as in previous classes, we infer

$$w_{\infty} = \text{sign}(\beta_2).$$

### 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

$$G_+(u) = (\alpha_2 u^2 + i) (\beta_1 u - \sqrt{2\sqrt{u^2}}), \quad G_-(u) = G^*(u).$$

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

$$S(u) = \frac{G_-(u)}{G_+(u)} = \frac{\alpha_2 u^2 - i}{\alpha_2 u^2 + i},$$

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

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

$$\Delta = ((\beta_1 + \sqrt{2})\beta_2 + |b_{12}|^2)((\beta_1 - \sqrt{2})\beta_2 + |b_{12}|^2)$$

except if $\beta_2 = b_{12} = 0$, which is treated in the last case above. The four remaining cases would be

1. $\beta_2 \neq 0, b_{12} = 0$ and $\beta_1 = \pm \sqrt{2}$,

2. $\beta_1^2 \neq 2, b_{12} \neq 0$ and

$$\beta_2 = - \frac{|b_{12}|^2}{\beta_1 \pm \sqrt{2}}.$$

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.
### 5.5.2 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

$$G(u) = (\beta_1 u - \sqrt{2u^2 + 1})(-\beta_2 u + i).$$

The first factor vanishes at $u_0 = \text{sign} \, \beta_1(\beta_1^2 - 2)^{-1/2}$, 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

$$S(u) = \frac{G(u)}{G^*(u)} = \frac{-\beta_2 u + i}{-\beta_2 u - i}$$

so that, as in previous classes, we infer

$$w_{\infty} = \text{sign}(\beta_2).$$

### 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

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

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

$$S(u) = \frac{G_{-}(u)}{G_{+}(u)} = \frac{\alpha_{2}u^{2} - i}{\alpha_{2}u^{2} + i}$$

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

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

$$\Delta = ((\beta_1 + \sqrt{2})\beta_2 + |b_{12}|^2)((\beta_1 - \sqrt{2})\beta_2 + |b_{12}|^2)$$

except if $\beta_2 = b_{12} = 0$, which is treated in the last case above. The four remaining cases would be

$$1. \beta_2 \neq 0, b_{12} = 0 \text{ and } \beta_1 = \pm \sqrt{2},$$

$$2. \beta_1^2 \neq 2, b_{12} \neq 0 \text{ and}$$

$$\beta_2 = - \frac{|b_{12}|^2}{\beta_1 \pm \sqrt{2}}.$$

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 B

In this case we have

$$A_0 = \begin{pmatrix} a_1 & a_2 & i\alpha & -i\mu^*\alpha \\ \mu a_1 & \mu a_2 & i\mu\alpha & -i|\mu|^2\alpha \end{pmatrix}, \quad A_1 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix},$$

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

$$g_\infty(k_x, \kappa) = -k_x^2 + ia_2\mu k_x + ia_1 k_x + \frac{i\kappa_{ev}k_x^3 + \kappa\kappa_{ev}k_x^2 - i\kappa k_x^3 + k_x^4}{4k^4\epsilon^2}. \quad (5.32)$$

Passing to dual variables via (4.13) we get

$$g_0(\lambda_x, \delta) = \frac{1}{4\epsilon^2 (\delta^2 + \lambda_x^2)^2} \left( -4ia_2\delta^2\mu\epsilon^2\lambda_x - 4ia_1\delta^2\epsilon^2\lambda_x - 4ia_2\mu\epsilon^2\lambda_x^3 - 4ia_1\epsilon^2\lambda_x^3 + \lambda_x^3\sqrt{\delta^2 + 2\lambda_x^2} \right. \\ \left. + i\delta\lambda_x^2\sqrt{\delta^2 + 2\lambda_x^2} + i\delta\lambda_x^3 + \lambda_x^4 - 4\epsilon^2\lambda_x^2 \right). \quad (5.33)$$

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

$$G_+(u) = -ia_2\mu - ia_1 - u, \quad G_-(u) = G_+(u). \quad (5.34)$$

$G_+$ and $G_-$ do not vanish if $a_2\mu + a_1 \notin i\mathbb{R}$, and we can check that Proposition 17 applies, from which we immediately infer $S = 1$ and $w_\infty = 0$. If $a_2\mu + a_1 \in i\mathbb{R}$, since $G_+ = G_-$ then we can replace $g_0$ by $g_0 + i\eta$ 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 $G_+ + i\eta = G_- + i\eta$. Again, we get $w_\infty = 0$.

## 5.7 Class C

In this case we have

$$A_0 = \begin{pmatrix} a_1 & a_2 & 0 & a_4 \\ \mu a_1 & \mu a_2 & 0 & \mu a_4 \end{pmatrix}, \quad A_1 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix},$$

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 $\operatorname{Im}(a_2a_4^*) = 0$, 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

$$g_\infty(k_x, \kappa) = -a_4\mu k_x\kappa + ia_2\mu k_x + \frac{a_4\mu k_x (\kappa\kappa_{ev}^2 + \kappa_{ev}k_x^2 + i\kappa_{ev}^2k_x - i\kappa\kappa_{ev}k_x)}{4k^4\epsilon^2}. \quad (5.35)$$

Passing to dual variables via (4.13) we get

$$g_0(\lambda_x, \delta) = \frac{1}{4\epsilon^2(\lambda_x^2 + \delta^2)^2} \left( -ia_4\mu\lambda_x^3\sqrt{\delta^2 + 2\lambda_x^2} - ia_4\delta^2\mu\lambda_x^2 + a_4\delta\mu\lambda_x^2\sqrt{\delta^2 + 2\lambda_x^2} + a_4\delta^3\mu\lambda_x \right. \\ \left. - 4ia_2\delta^2\mu\epsilon^2\lambda_x + 2a_4\delta\mu\lambda_x^3 + 4a_4\delta\mu\epsilon^2\lambda_x - 2ia_4\mu\lambda_x^4 - 4ia_2\mu\epsilon^2\lambda_x^3 \right). \quad (5.36)$$

The denominator, as well as some common $\lambda_x\mu$ factor, do not contribute to $w_\infty$ so we replace $g_0$ by $\lambda_x^{-1}\mu^{-1}(\delta^2 + \lambda_x^2)^2g_0$. Then the dominant scale of $g_0$ depends on some parameter vanishing or not.
## 5.6 Class B

In this case we have

$$A_0 = \begin{pmatrix} a_1 & a_2 & i\alpha & -i\mu^*\alpha \\ \mu a_1 & \mu a_2 & i\mu\alpha & -i|\mu|^2\alpha \end{pmatrix}, \quad A_1 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix},$$

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

$$g_\infty(k_x, \kappa) = -k_x^2 + ia_2\mu k_x + ia_1 k_x + \frac{i\kappa_{ev}k_x^3 + \kappa\kappa_{ev}k_x^2 - i\kappa k_x^3 + k_x^4}{4k^4\epsilon^2}. \quad (5.32)$$

Passing to dual variables via (4.13) we get

$$g_0(\lambda_x, \delta) = \frac{1}{4\epsilon^2 (\delta^2 + \lambda_x^2)^2} \left( -4ia_2\delta^2\mu\epsilon^2\lambda_x - 4ia_1\delta^2\epsilon^2\lambda_x - 4ia_2\mu\epsilon^2\lambda_x^3 - 4ia_1\epsilon^2\lambda_x^3 + \lambda_x^3\sqrt{\delta^2 + 2\lambda_x^2} \right. \\ \left. + i\delta\lambda_x^2\sqrt{\delta^2 + 2\lambda_x^2} + i\delta\lambda_x^3 + \lambda_x^4 - 4\epsilon^2\lambda_x^2 \right). \quad (5.33)$$

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

$$G_+(u) = -ia_2\mu - ia_1 - u, \quad G_-(u) = G_+(u). \quad (5.34)$$

$G_+$ and $G_-$ do not vanish if $a_2\mu + a_1 \notin i\mathbb{R}$, and we can check that Proposition 17 applies, from which we immediately infer $S = 1$ and $w_\infty = 0$. If $a_2\mu + a_1 \in i\mathbb{R}$, since $G_+ = G_-$ then we can replace $g_0$ by $g_0 + i\eta$ 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 $G_+ + i\eta = G_- + i\eta$. Again, we get $w_\infty = 0$.

## 5.7 Class C

In this case we have

$$A_0 = \begin{pmatrix} a_1 & a_2 & 0 & a_4 \\ \mu a_1 & \mu a_2 & 0 & \mu a_4 \end{pmatrix}, \quad A_1 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix},$$

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 $\operatorname{Im}(a_2a_4^*) = 0$, 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

$$g_\infty(k_x, \kappa) = -a_4\mu k_x\kappa + ia_2\mu k_x + \frac{a_4\mu k_x (\kappa\kappa_{ev}^2 + \kappa_{ev}k_x^2 + i\kappa_{ev}^2k_x - i\kappa\kappa_{ev}k_x)}{4k^4\epsilon^2}. \quad (5.35)$$

Passing to dual variables via (4.13) we get

$$g_0(\lambda_x, \delta) = \frac{1}{4\epsilon^2(\lambda_x^2 + \delta^2)^2} \left( -ia_4\mu\lambda_x^3\sqrt{\delta^2 + 2\lambda_x^2} - ia_4\delta^2\mu\lambda_x^2 + a_4\delta\mu\lambda_x^2\sqrt{\delta^2 + 2\lambda_x^2} + a_4\delta^3\mu\lambda_x \right. \\ \left. - 4ia_2\delta^2\mu\epsilon^2\lambda_x + 2a_4\delta\mu\lambda_x^3 + 4a_4\delta\mu\epsilon^2\lambda_x - 2ia_4\mu\lambda_x^4 - 4ia_2\mu\epsilon^2\lambda_x^3 \right). \quad (5.36)$$

The denominator, as well as some common $\lambda_x\mu$ factor, do not contribute to $w_\infty$ so we replace $g_0$ by $\lambda_x^{-1}\mu^{-1}(\delta^2 + \lambda_x^2)^2g_0$. Then the dominant scale of $g_0$ depends on some parameter vanishing or not.
**Case $a_2 \neq 0$ and $a_4 = ra_2$ with $r \in \mathbb{R}^*$.** The dominant scale is

$$ \delta^{-1}g_0(u\delta^{1/2}, \pm\delta) = G_{\pm}(u) + R_{\pm}(u) $$

with

$$ \begin{aligned} G_{+}(u) &= a_{2}(1 - i r u^{2}), \\ G_{-}(u) &= a_{2}(1 + i r u^{2}) = G_{+}^{*}(u). \end{aligned} \qquad (5.37) $$

Proposition 17 applies, and $G_{-}(u)$ has no winding phase as $u$ goes from $-\infty$ to $+\infty$, so that $w_{\infty} = 0$.

**Case $a_2 = 0$ and $a_4 \neq 0$.** In that case we also divide $g_0$ by $a_4$. The dominant scale is

$$ \delta^{-1}g_0(u\delta^{1/3}, \pm\delta) = G_{\pm}(u) + R_{\pm}(u) $$

with $G_{+}(u) = 1 - i \frac{2u + \sqrt{2}|u|}{4\epsilon^2}u^2$, $G_{-}(u) = 1 + i \frac{2u + \sqrt{2}|u|}{4\epsilon^2}u^2 = G_{+}^{*}(u)$. Proposition 17 does not apply but $G_{+}$ and $G_{-}$ do not vanish and

$$ \frac{R + (u, \delta)}{G_{+}(u)} \underset{u \to \pm\infty}{\sim} \frac{i\delta^{2/3}}{u}, \qquad \frac{R - (u, \delta)}{G_{-} + (u)} \underset{u \to \pm\infty}{\sim} \frac{-i\delta^{2/3}}{u} $$

so that $|\frac{R_{\pm}}{G_{\pm}}| \to 0$ as $\delta \to 0$ and all $u \in [-\frac{\lambda}{\delta^{1/3}}, \frac{\lambda}{\delta^{1/3}}]$, and we get the same conclusion than Proposition 17. The real part of $G_{-}$ is constant and positive, and

$$ G_{-}(u) \underset{u \to \pm\infty}{\sim} \frac{2 \pm \sqrt{2}}{4\epsilon^2} u^3 $$

so that the winding phase of $G_{-}$ is $\pi$, and since $G_{+} = G_{-}^{*}$ we infer $w_{\infty} = 1$ for any $a_4 \neq 0$.

**Case $a_2 \neq 0$ and $a_4 = 0$.** The dominant scale is

$$ \delta^{-2}g_0(u\delta, \pm\delta) = G_{\pm}(u) + R_{\pm}(u) $$

with

$$ G_{+}(u) = -ia_{2}(u^{2} + 1), \qquad G_{-}(u) = G_{+}(u) $$

and $R_{\pm} = 0$ here. Thus we immediately infer $w_{\infty} = 0$.

**Summarizing.** In that class, $w_{\infty} = 1$ if $a_2 = 0$ and $a_4 \neq 0$, and $w_{\infty} = 0$ otherwise.

## A Chern number

We rewrite the Hamiltonian in terms of Pauli matrices as

$$ H = \vec{d} \cdot \vec{\sigma}, \qquad \vec{d} = \begin{pmatrix} -k_x \\ -k_y \\ m - \epsilon \mathbf{k}^2 \end{pmatrix}, \qquad (A.1) $$

where $\vec{\sigma} = (\sigma_1, \sigma_2, \sigma_3)$ is a vector of Pauli matrices

$$ \sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. \qquad (A.2) $$
**Case $a_2 \neq 0$ and $a_4 = ra_2$ with $r \in \mathbb{R}^*$.** The dominant scale is

$$ \delta^{-1}g_0(u\delta^{1/2}, \pm\delta) = G_{\pm}(u) + R_{\pm}(u) $$

with

$$ \begin{aligned} G_{+}(u) &= a_{2}(1 - i r u^{2}), \\ G_{-}(u) &= a_{2}(1 + i r u^{2}) = G_{+}^{*}(u). \end{aligned} \qquad (5.37) $$

Proposition 17 applies, and $G_{-}(u)$ has no winding phase as $u$ goes from $-\infty$ to $+\infty$, so that $w_{\infty} = 0$.

**Case $a_2 = 0$ and $a_4 \neq 0$.** In that case we also divide $g_0$ by $a_4$. The dominant scale is

$$ \delta^{-1}g_0(u\delta^{1/3}, \pm\delta) = G_{\pm}(u) + R_{\pm}(u) $$

with $G_{+}(u) = 1 - i \frac{2u + \sqrt{2}|u|}{4\epsilon^2}u^2$, $G_{-}(u) = 1 + i \frac{2u + \sqrt{2}|u|}{4\epsilon^2}u^2 = G_{+}^{*}(u)$. Proposition 17 does not apply but $G_{+}$ and $G_{-}$ do not vanish and

$$ \frac{R + (u, \delta)}{G_{+}(u)} \underset{u \to \pm\infty}{\sim} \frac{i\delta^{2/3}}{u}, \qquad \frac{R - (u, \delta)}{G_{-} + (u)} \underset{u \to \pm\infty}{\sim} \frac{-i\delta^{2/3}}{u} $$

so that $|\frac{R_{\pm}}{G_{\pm}}| \to 0$ as $\delta \to 0$ and all $u \in [-\frac{\lambda}{\delta^{1/3}}, \frac{\lambda}{\delta^{1/3}}]$, and we get the same conclusion than Proposition 17. The real part of $G_{-}$ is constant and positive, and

$$ G_{-}(u) \underset{u \to \pm\infty}{\sim} \frac{2 \pm \sqrt{2}}{4\epsilon^2} u^3 $$

so that the winding phase of $G_{-}$ is $\pi$, and since $G_{+} = G_{-}^{*}$ we infer $w_{\infty} = 1$ for any $a_4 \neq 0$.

**Case $a_2 \neq 0$ and $a_4 = 0$.** The dominant scale is

$$ \delta^{-2}g_0(u\delta, \pm\delta) = G_{\pm}(u) + R_{\pm}(u) $$

with

$$ G_{+}(u) = -ia_{2}(u^{2} + 1), \qquad G_{-}(u) = G_{+}(u) $$

and $R_{\pm} = 0$ here. Thus we immediately infer $w_{\infty} = 0$.

**Summarizing.** In that class, $w_{\infty} = 1$ if $a_2 = 0$ and $a_4 \neq 0$, and $w_{\infty} = 0$ otherwise.

## A Chern number

We rewrite the Hamiltonian in terms of Pauli matrices as

$$ H = \vec{d} \cdot \vec{\sigma}, \qquad \vec{d} = \begin{pmatrix} -k_x \\ -k_y \\ m - \epsilon \mathbf{k}^2 \end{pmatrix}, \qquad (A.1) $$

where $\vec{\sigma} = (\sigma_1, \sigma_2, \sigma_3)$ is a vector of Pauli matrices

$$ \sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. \qquad (A.2) $$
The eigenprojections of $H$ are shared with those of the flat Hamiltonian $H' = \vec{e} \cdot \vec{\sigma}$, where $\vec{e} = \vec{d} / |\vec{d}|$. They are

$$P_{\pm} = \frac{1}{2} (1 \pm \vec{e} \cdot \vec{\sigma}), \qquad (A.3)$$

$$((\vec{a} \cdot \vec{\sigma})(\vec{b} \cdot \vec{\sigma}) = (\vec{a} \cdot \vec{b}) + i(\vec{a} \times \vec{b}) \cdot \vec{\sigma}). \text{ Note that } \vec{e} = \vec{e}(\mathbf{k}) \text{ is convergent for } k \to \infty$$

$$\vec{e} \to \begin{pmatrix} 0 \\ 0 \\ \pm 1 \end{pmatrix} \quad (k \to \infty), \quad \text{for } \epsilon \ge 0. \qquad (A.4)$$

Therefore, also the eigenprojections converge and the Chern number is a well-defined topological invariant

$$C(P) = \frac{1}{2\pi i} \int_{\mathbb{R}^2} dk_x dk_y \text{tr} (P [\partial_{k_x} P, \partial_{k_y} P]). \qquad (A.5)$$

If the regulator $\epsilon \neq 0$ we can compactify the momentum plane to the 2-sphere $S^2$ and we can compute the r.h.s. on a closed manifold with the map $\vec{e}: S^2 \to S^2$. According to Prop. 1 of [20], we get

$$C_{\pm} = \pm \frac{1}{4\pi} \int_{S^2} \vec{e} \cdot (\partial_1 \vec{e} \wedge \partial_2 \vec{e}) dx_1 dx_2,$$

which, in our case, leads to

$$C_{\pm} = \pm \frac{\text{sign } m + \text{sign } \epsilon}{2}. \qquad (A.6)$$

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[5] Bellissard, J., van Elst, A., and Schulz-Baldes, H. (1994). *The noncommutative geometry of the quantum Hall effect*. J. Math. Phys. **35**(10) 5373-5451

[6] Bourne, C., and Rennie, A. (2018) *Chern numbers, localisation and the bulk-edge correspondence for continuous models of topological phases*. Math. Phys. Anal. Geom. **21**(3) 16

[7] Neto, A. C., Guinea, F., Peres, N. M., Novoselov, K. S., and Geim, A. K. (2009) *The electronic properties of graphene*. Reviews of modern physics **81**(1) 109

[8] Combes, J. M., and Germinet, F. (2005) *Edge and impurity effects on quantization of Hall currents*. Commun. Math. Phys. **256**(1) 159-180

[9] Cornean, H. D., Moscolari, M., and Sørensen, K. S. (2023). *Bulk-edge correspondence for unbounded Dirac-Landau operators*. Journal of Mathematical Physics **64**(2)
The eigenprojections of $H$ are shared with those of the flat Hamiltonian $H' = \vec{e} \cdot \vec{\sigma}$, where $\vec{e} = \vec{d}/|\vec{d}|$. They are

$$P_{\pm} = \frac{1}{2} (1 \pm \vec{e} \cdot \vec{\sigma}), \qquad (A.3)$$

$$((\vec{a} \cdot \vec{\sigma})(\vec{b} \cdot \vec{\sigma}) = (\vec{a} \cdot \vec{b}) + i(\vec{a} \times \vec{b}) \cdot \vec{\sigma}). \text{ Note that } \vec{e} = \vec{e}(\mathbf{k}) \text{ is convergent for } k \to \infty$$

$$\vec{e} \to \begin{pmatrix} 0 \\ 0 \\ \pm 1 \end{pmatrix} \quad (k \to \infty), \quad \text{for } \epsilon \ge 0. \qquad (A.4)$$

Therefore, also the eigenprojections converge and the Chern number is a well-defined topological invariant

$$C(P) = \frac{1}{2\pi i} \int_{\mathbb{R}^2} dk_x dk_y \text{tr} (P [\partial_{k_x} P, \partial_{k_y} P]). \qquad (A.5)$$

If the regulator $\epsilon \ne 0$ we can compactify the momentum plane to the 2-sphere $S^2$ and we can compute the r.h.s. on a closed manifold with the map $\vec{e}: S^2 \to S^2$. According to Prop. 1 of [20], we get

$$C_{\pm} = \pm \frac{1}{4\pi} \int_{S^2} \vec{e} \cdot (\partial_1 \vec{e} \wedge \partial_2 \vec{e}) dx_1 dx_2,$$

which, in our case, leads to

$$C_{\pm} = \pm \frac{\text{sign } m + \text{sign } \epsilon}{2}. \qquad (A.6)$$

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[13] Delplace, P., Marston, J. B., and Venaille, A. (2017) *Topological origin of equatorial waves*. Science **358**(6366) 1075-1077

[14] Drouot, A. (2019) *The bulk-edge correspondence for continuous honeycomb lattices*. Comm. Partial Differential Equations **44**(12) 1406-1430

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[16] Halperin, B. I. (1982) *Quantized Hall conductance, current-carrying edge states, and the existence of extended states in a two-dimensional disordered potential*. Phys. Rev. B **25**(4) 2185

[17] Gomi, K., and Thiang, G. C. (2019) *Crystallographic bulk-edge correspondence: glide reflections and twisted mod 2 indices*. Letters in Mathematical Physics **109** 857-904

[18] Gontier, D. (2023) *Edge states for second order elliptic operators in a channel*. Journal of Spectral Theory **12**(3) 1155-1202

[19] Elbau, P., and Graf, G. M. (2002). *Equality of bulk and edge Hall conductance revisited*. Communications in mathematical physics **229** 415-432.

[20] Graf, G. M., Jud, H., and Tauber, C. (2021) *Topology in shallow-water waves: a violation of bulk-edge correspondence*. Commun. Math. Phys. **383**(2) 731-761

[21] Graf, G. M., Porta, M. (2013) *Bulk-edge correspondence for two-dimensional topological insulators*. Commun. Math. Phys. **324**(3) 851-895

[22] Griffiths, P. and Harris, J. (2014). *Principles of algebraic geometry*. Wiley Online Library.

[23] Haldane, F. D. M. (1988) *Model for a quantum Hall effect without Landau levels: Condensed-matter realization of the "parity anomaly"*. Physical review letters **61**(18) 2015

[24] Hatsugai, Y. (1993) *Chern number and edge states in the integer quantum Hall effect*. Phys. Rev. Lett. **71**(22) 3697

[25] Kane, C. L., and Mele, E. J. (2005). *Quantum spin Hall effect in graphene*. Physical review letters **95**(22) 226801

[26] Mathai, V., and Thiang, G. C. (2016) *T-duality simplifies bulk-boundary correspondence*. Communications in Mathematical Physics **345** 675-701

[27] Onuki, Y., Venaille, A., and Delplace, P. (2023) *Bulk-edge correspondence recovered in incompressible continuous media*. arXiv preprint 2311.18249.

[28] Peri, V., Serra-Garcia, M., Ilan, R., and Huber, S. D. (2019) *Axial-field-induced chiral channels in an acoustic Weyl system*. Nat. Phys. **15**(4) 357
[29] Prodan, E., and Schulz-Baldes, H. (2016) Bulk and boundary invariants for complex topological insulators. From K-theory to physics Math. Phys. Stud., Springer

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[31] Raghu, S., and Haldane, F. D. M. (2008) Analogs of quantum-Hall-effect edge states in photonic crystals. Phys. Rev. A 78(3) 033834

[32] Rossi, S., and Tarantola, A. (2024). Topology of 2D Dirac operators with variable mass and an application to shallow-water waves. Journal of Physics A: Mathematical and Theoretical 57(6) 065201

[33] Schulz-Baldes, H., Kellendonk, J., Richter, T. (2000) Simultaneous quantization of edge and bulk Hall conductivity. J. Phys. A: Math. Gen. 33, L27

[34] Tauber, C., Delplace, P., and Venaille, A. (2019) A bulk-interface correspondence for equatorial waves. J. Fluid Mech. 868

[35] Tauber, C., Delplace, P., and Venaille, A. (2019) Anomalous bulk-edge correspondence in continuous media. Phys. Rev. Research 2(1) 013147

[36] Tauber, C., and Thiang, G. C. (2023) Topology in shallow-water waves: A spectral flow perspective. Annales Henri Poincaré 24(1) 107-132

[37] Treust, L. L., Barbaroux, J. M., Cornean, H. D., Stockmeyer, E., and Raymond, N. (2024). Magnetic Dirac systems: Violation of bulk-edge correspondence in the zigzag limit. arXiv preprint 2401.12569

[38] Volovik, G. E. (1988) Analogue of quantum Hall effect in a superfluid 3 He film. Zhurnal Ehksperimental’noj i Teoreticheskoj Fiziki 94(9) 123-137.
[29] Prodan, E., and Schulz-Baldes, H. (2016) Bulk and boundary invariants for complex topological insulators. From K-theory to physics Math. Phys. Stud., Springer

[30] Quinn, S., and Bal, G. (2022) Asymmetric transport for magnetic Dirac equations. arXiv preprint 2211.00726

[31] Raghu, S., and Haldane, F. D. M. (2008) Analogs of quantum-Hall-effect edge states in photonic crystals. Phys. Rev. A 78(3) 033834

[32] Rossi, S., and Tarantola, A. (2024). Topology of 2D Dirac operators with variable mass and an application to shallow-water waves. Journal of Physics A: Mathematical and Theoretical 57(6) 065201

[33] Schulz-Baldes, H., Kellendonk, J., Richter, T. (2000) Simultaneous quantization of edge and bulk Hall conductivity. J. Phys. A: Math. Gen. 33, L27

[34] Tauber, C., Delplace, P., and Venaille, A. (2019) A bulk-interface correspondence for equatorial waves. J. Fluid Mech. 868

[35] Tauber, C., Delplace, P., and Venaille, A. (2019) Anomalous bulk-edge correspondence in continuous media. Phys. Rev. Research 2(1) 013147

[36] Tauber, C., and Thiang, G. C. (2023) Topology in shallow-water waves: A spectral flow perspective. Annales Henri Poincaré 24(1) 107-132

[37] Treust, L. L., Barbaroux, J. M., Cornean, H. D., Stockmeyer, E., and Raymond, N. (2024). Magnetic Dirac systems: Violation of bulk-edge correspondence in the zigzag limit. arXiv preprint 2401.12569

[38] Volovik, G. E. (1988) Analogue of quantum Hall effect in a superfluid 3 He film. Zhurnal Ehksperimental’noj i Teoreticheskoj Fiziki 94(9) 123-137.