On structures of normal forms of complex points of small ${\mathcal {C}}^{2}$-perturbations of real 4-manifolds embedded in a complex 3-manifold

Abstract

We extend our previous result on the behaviour of the quadratic part of a complex points of a small ${\mathcal {C}}^{2}$-perturbation of a real 4-manifold embedded in a complex 3-manifold. We describe the change of the structure of the quadratic normal form of a complex point. It is an immediate consequence of a theorem clarifying how small perturbations can change the bundle of a pair of one arbitrary and one symmetric $2\times 2$ matrix with respect to an action of a certain linear group.

1 Introduction

Let M be a smooth real 2n-submanifold in ${\mathbb {C}}^{n+1}$. A point $p\in M$ is called complex when $T_pM$ is a complex subspace in $T_p {\mathbb {C}}^{n+1}$; its complex dimension is equal to n. Locally, near a complex point $p\in M$ we can see M as a graph (see e.g. [23]):

$$\begin{aligned} w=\overline{z}^TAz+\textrm{Re}(z^TBz)+o(|z|^2), \quad (z(p),w(p))=(0,0),\,\,\, A\in {\mathbb {C}}^{n\times n}, B\in {\mathbb {C}}^{n\times n}_S, \nonumber \\ \end{aligned}$$

(1.1)

in which $(z,w)=(z_1,z_2,\ldots ,z_n,w)$ are suitable local coordinates on ${\mathbb {C}}^{n+1}$, and ${\mathbb {C}}^{n\times n}$, ${\mathbb {C}}^{n\times n}_S$ are sets of all $n\times n$ matrices and $n\times n$ symmetric matrices, respectively. A complex point p is quadratically flat, if the quadratic part of (1.1) is real valued.

When $n=1$ complex points are well understood; see papers of Bishop [3], Kenig and Webster [18], Moser and Webster [19], Huang and Yin [16], Huang and Krantz [17], Bedford and Klingenberg [2] and Forstnerič [13]. They are quadratically flat and given locally by $w=z\overline{z}+\frac{\gamma }{2} (z^2+\overline{z}^2)+o(|z|^2)$, $0\le \gamma$, or $w=z^2+\overline{z}^2+o(|z|^2)$. For $n=2$ a relatively simple description of complex points up to quadratic terms was obtained by Coffman [7]; it includes two generic normal forms given by equations

$$\begin{aligned} w&=\tau z_1\overline{z}_2+z_2\overline{z}_1+\tfrac{a}{2}z_1^{2} +\tfrac{\overline{a}}{2}\overline{z}_1^{2}+b(z_1z_2 +\overline{z}_1\overline{z}_2)+\tfrac{d}{2}z_2^{2} +\tfrac{\overline{d}}{2}\overline{z}_2^{2}+o(|z|^{2}), \end{aligned}$$

(1.2)

$$\begin{aligned} w&=z_1\overline{z}_1+e^{i\theta }z_2\overline{z}_2 +\tfrac{a'}{2}(z_1^{2}+\overline{z}_1^{2})+b'z_1z_2 +\overline{b'} \overline{z}_1\overline{z}_2+\tfrac{d'}{2} (z_2^{2}+\overline{z}_2^{2})+o(|z|^{2}), \end{aligned}$$

(1.3)

in which $|a|=1$, $b>0$, $d\in {\mathbb {C}}$, $\tau \in (0,1)$ and $\theta \in (0,\pi )$, $a',d'>0$, $b'\in {\mathbb {C}}^{*}$ (apply $(A,B)= \big (\begin{bmatrix} 0 &{} 1 \\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} a &{} b \\ b &{} d \end{bmatrix}\big )$ and $(A,B)=\big (\begin{bmatrix} 1 &{} 0 \\ 0 &{} e^{i\theta } \end{bmatrix}, \begin{bmatrix} a' &{} b' \\ b' &{} d' \end{bmatrix}\big )$ to (1.1) for $n=2$, respectively). If $n>2$ quadratically flat complex points were studied by Slapar and the author [22]. We refer to the papers of Dolbeault, Tomassini and Zaitsev [9] and Fang and Huang [12] for results on holomorphic flattenability of CR-nonminimal real analytic submanifolds near complex points. Formal normal forms of CR-singularities were considered by Burcea [6] and Gong and Stolovitch [15], among others.

In this paper we continue a research started in the paper [23], in which we explained when the quadratic part of a complex point of a real 4-manifold embedded in a complex 3-manifold can be transformed under small ${\mathcal {C}}^{2}$-perturbations to the quadratic part of another complex point. For instance, [23, Corollary 3.8] implies that no sufficiently small ${\mathcal {C}}^{2}$-deformation of $w=z_1\overline{z}_1+\frac{1}{2}z_2^{2}+\frac{1}{2} \overline{z}_2^{2}+o(|z|^{2})$ near (0, 0) (with $(A,B)=\big (\begin{bmatrix} 1 &{} 0 \\ 0 &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} 0 \\ 0 &{} 1 \end{bmatrix}\big )$ in (1.1) for $n=2$) can lead to $w=\frac{1}{2} \overline{z}_1z_2+z_1\overline{z}_2+z_1^{2} +\overline{z}_1^{2}+z_1z_2+\overline{z}_1\overline{z}_2+z_2^{2} +\overline{z}_2^{2}+o(|z|^{2})$ (with $(A,B)= \big (\begin{bmatrix} 0 &{} 1 \\ \frac{1}{2} &{} 0 \end{bmatrix}, \begin{bmatrix} 2 &{} 1 \\ 1 &{} 2 \end{bmatrix}\big )$ in (1.1) for $n=2$).

We now focus on the change of the type of a complex point, i.e. on the structure of (A, B) in (1.1). In particular, we provide the following result describing possible arbitrarily small ${\mathcal {C}}^{2}$-deformations to generic normal forms.

Theorem 1.1

Let M be a real 4-manifold in ${\mathbb {C}}^{3}$ and let $p\in M$ be a complex point given locally by (1.1) with $n=2$ and $A\in {\mathbb {C}}^{2\times 2}$, $B\in {\mathbb {C}}^{2\times 2}_S$. It follows that $A\ne \begin{bmatrix} 1 &{} 0 \\ 0 &{} e^{i\theta _0} \end{bmatrix}$ with $\theta _0 \in [0,\pi )$ (or $A\ne \begin{bmatrix} 0 &{} 1 \\ \tau _0 &{} 0 \end{bmatrix}$ with $\tau _0\in [0,1)$) if and only if there exists an arbitrarily small ${\mathcal {C}}^{2}$-perturbation $M'$ of M, and such that $M'$ has a complex point $p'$, arbitrarily close to p, and $p'$ is locally given by the equation of the form (1.2) (of the form (1.3)).

A more general situation is considered in Theorem 3.4 and Corollary 3.6. Due to technical reasons, these results are stated in Sec. 3. A substantial difference in comparison to [23] is that our problem now reduces to a system of nonlinear equations with larger set of parameters. It makes the analysis considerably more involved. However, the number of cases to be considered now is smaller. We add that Theorem 3.4 is of independent interest in matrix analysis since it clarifies how small perturbations can change the bundle of a pair $(A,B)\in {\mathbb {C}}^{2\times 2}\times {\mathbb {C}}^{2\times 2}_S$ with respect to transformations $(cP^{*}AP,P^{T}BP)$ with nonsingular matrix P and $|c|=1$.

2 Normal forms in dimension 2

Any holomorphic change of coordinates that preserves (1.1) for $n=2$ transforms (1.1) into the equation that can by a slight abuse of notation be written as

$$\begin{aligned} w=\overline{z}^T\left( cP^{*}AP\right) z+\textrm{Re}\left( z^T (P^TBP) z\right) +o(|z|^2), \qquad P\in GL_2({\mathbb {C}}),\quad c\in S^{1}, \end{aligned}$$

where $S^1$ and $GL_2({\mathbb {C}})$ are a unit circle and the group of invertible $2\times 2$ matrices, respectively. Studying the quadratic part of a complex point thus means examining the action of $S^1\times GL_2({\mathbb {C}})$ on ${\mathbb {C}}^{2\times 2}\times {\mathbb {C}}^{2\times 2}_S$ (see also [7] and [23, Sec. 3]):

$$\begin{aligned} \Psi :\bigl ((c,P),(A,B)\bigr )\mapsto (cP^{*}AP,P^TBP), \qquad P\in GL_2({\mathbb {C}}),\quad c\in S^{1}. \end{aligned}$$

(2.1)

An orbit at $(A,B)\in {\mathbb {C}}^{2\times 2}\times {\mathbb {C}}^{2\times 2}_S$ with respect to (2.1) is denoted by $\textrm{Orb}_{\Psi }(A,B)$.

For some applications it is useful to have a stratification into bundles of matrices, i.e. sets of all matrices having similar properties. This notion was first introduced by Arnold [1, Section 30] for the action of similarity; two matrices are in the same bundle under similarity precisely when their Jordan canonical forms have the same structure (with bijection between the sets of distinct eigenvalues). For instance, matrices with all distinct eigenvalues form the generic bundle.

Three bundles with respect to the action (2.1) can be formed according to the sign of $\det \begin{bmatrix} A &{} \overline{B}\\ B &{} \overline{A} \end{bmatrix}$ for $(A,B)\in {\mathbb {C}}^{2\times 2}\times {\mathbb {C}}^{2\times 2}_S$ (see [7, Sec. 4]). Slapar [20] (see also [21]) proved that the bundles with nonvanishing determinant are connected components of ${\mathbb {C}}^{2\times 2}\times {\mathbb {C}}^{2\times 2}_S$ and showed that up to smooth isotopy complex points of a real 4-submanifold in ${\mathbb {C}}^{3}$ are locally given either by $w=z_1\overline{z}_1+z_2\overline{z}_2$ or $w=z_1\overline{z}_1+\overline{z}_2^{2}$.

Our goal is to understand the change of normal forms of (2.1) under small perturbations, thus we use the list [7, Sec. 7,Table 1] (see also [23, Lemma 2.2]) of normal forms for orbits under (2.1), to form bundles so that they contain normal forms of similar structure. To be more precise, each such set of normal forms is parameterized by smooth maps $\Lambda \rightarrow {\mathbb {C}}^{2\times 2}$, $\lambda \mapsto A(\lambda )$ and $\Lambda \rightarrow {\mathbb {C}}^{2\times 2}_S$, $\lambda \mapsto B(\lambda )$, in which $\Lambda \subset {\mathbb {R}}^{k}$ is a parameter set, and we define the bundle of $(A_0,B_0)=(A(\lambda _0),B(\lambda _0))$ for $\lambda _0\in \Lambda$ with respect to the action $\Psi$ in (2.1) as:

$$\begin{aligned} \textrm{Bun}_{\Psi }(A_0,B_0):=\bigcup _{\lambda \in \Lambda } \textrm{Orb}_{\Psi }(A(\lambda ),B(\lambda )). \end{aligned}$$

(2.2)

Elements of a bundle must behave similarly under small perturbations (Sect. 3).

To simplify the notation, $a\oplus d$ denotes the diagonal matrix with a, d on the diagonal, while the $2\times 2$ identity-matrix and the $2\times 2$ zero-matrix are $I_2$ and $0_2$.

Lemma 2.1

Bundles of the action (2.1), represented by pairs (A, B) given in Table 1, are immersed submanifolds in ${\mathbb {C}}^{2\times 2}\times {\mathbb {C}}^{2\times 2}_S$ with dimensions noted in the first column.

Table 1 Bundles of the action (2.1). Here $0<\tau <1$, $0<\theta <\pi$, $a,b,d>0$, $\zeta \in {\mathbb {C}}$, $\varphi \in {\mathbb {R}}$, $\zeta ^{*}\in {\mathbb {C}}^{*}$ are the parameters

Note that we arranged orbits $\textrm{Orb}_{\Psi }(1\oplus \sigma ,d_0\oplus d)$ for $\sigma \in \{1,-1\}$, $d>0$, $d_0\in \{0,d\}$ into bundles $\textrm{Bun}_{\Psi }(1\oplus \sigma ,0\oplus d)=\cup _{d>0}\textrm{Orb}_{\Psi }(1\oplus \sigma ,0\oplus d)$ and $\textrm{Bun}_{\Psi }(1\oplus \sigma ,dI_2)=\cup _{d>0}\textrm{Orb}_{\Psi }(1\oplus \sigma , d I_2)$, $\sigma \in \{1,-1\}$. Next, $\textrm{Orb}_{\Psi }\big (\begin{bmatrix} 0 &{} 1 \\ 0 &{} 0 \end{bmatrix}, \begin{bmatrix} \zeta &{} b \\ b &{} 1 \end{bmatrix}\big )$ for $\zeta \in {\mathbb {C}}$, $b>0$ are split into bundles with representatives $\big (\begin{bmatrix} 0 &{} 1 \\ 0 &{} 0 \end{bmatrix}, \begin{bmatrix} \zeta ^{*} &{} b \\ b &{} 1 \end{bmatrix}\big )$ and $\big (\begin{bmatrix} 0 &{} 1 \\ 0 &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} b \\ b &{} 1 \end{bmatrix}\big )$ for $\zeta ^{*}\in {\mathbb {C}}^{*},b>0$.

Sketch of the proof of Lemma 2.1

Fix $(A_0,B_0)\in {\mathbb {C}}^{2\times 2}\times {\mathbb {C}}_S^{2\times 2}$ from Table 1 and define

$$\begin{aligned} \Psi _{\Lambda }:S^1\times GL_2({\mathbb {C}}) \times \Lambda \rightarrow {\mathbb {C}}^{2\times 2} \times {\mathbb {C}}^{2\times 2}_S, \quad (c,P,\lambda ) \mapsto \Psi \big (c,P,A(\lambda ),B(\lambda )\big ), \end{aligned}$$

(2.3)

where $\Psi _{\Lambda }(1,I_2,\lambda _0)=(A_0,B_0)$. For every $g\in S^1\times GL_2({\mathbb {C}})$ the maps $\Psi ^g :(A,B) \mapsto \Psi (g,(A,B))$ and $R_g:h\mapsto hg$ are automorphisms of ${\mathbb {C}}^{2\times 2}\times {\mathbb {C}}^{2\times 2}_S$ and $S^1\times GL_2({\mathbb {C}})$, respectively, and we have $\Psi ^g\circ \Psi _{\Lambda }=\Psi _{\Lambda } \circ (R_g\times \textrm{id}_{\Lambda })$. Thus the rank of $d \Psi _{\Lambda }$ does not depend on $\lambda \in \Lambda$, $g\in S^1\times GL_2({\mathbb {C}})$ and by the constant rank theorem (e.g. [5, Theorem IV.5.8]) the bundle $\textrm{Bun}_{\Psi }(A_0,B_0)\subset {\mathbb {C}}^{2\times 2}\times {\mathbb {C}}_S^{2\times 2}$ is an immersed manifold.

In a similar manner as tangent spaces of orbits in [23, Lemma 2.2] are computed, tangent spaces of bundles are obtained. We choose paths (A(t), B(t)) in $\textrm{Bun}_{\Psi }(A_0, B_0)$ and

$$\begin{aligned} \gamma :(-\delta ,\delta )\rightarrow S^{1}\times GL_2({\mathbb {C}}), \quad \gamma (t)=(e^{i\alpha t},I+tX), \qquad \alpha \in {\mathbb {R}}, X\in {\mathbb {C}}^{2\times 2}, \delta >0, \end{aligned}$$

and calculate:

$$\begin{aligned}&\frac{d}{dt}\Big |_{t=0}e^{i\alpha t}\big ((I+tX)^*A(t)(I+tX)\big ) =i\alpha A_0+\tfrac{d}{dt}\big |_{t=0}A(t)+ (X^*A_0+A_0X),\\&\frac{d}{dt}\Big |_{t=0}\big ((I+tX)^TB(t)(I+tX)\big ) =\tfrac{d}{dt}\big |_{t=0}B(t)+(X^TB_0+B_0X). \end{aligned}$$

Writing $X=\sum _{j,k=1}^2(x_{jk}+iy_{jk})E_{jk}$, where $E_{jk}$ is the elementary matrix with one in the j-th row and k-th column and zeros otherwise, we deduce that

$$\begin{aligned} X^*A_0+A_0X&= \sum _{j,k=1}^2(x_{jk}-iy_{jk})E_{kj}A_0 +\sum _{j,k=1}^2(x_{jk}+iy_{jk})A_0E_{jk}\\&=\sum _{j,k=1}^2x_{jk}(E_{kj}A_0+A_0E_{jk}) +\sum _{j,k=1}^2y_{jk}i(-E_{kj}A_0+A_0E_{jk}),\\ \tfrac{d}{dt}\big |_{t=0}A(t)&=\beta _{21}E_{21} +\beta _{22}E_{22}, \beta _{22}=\left\{ \begin{array}{ll} \beta ie^{i\theta }, &{} A=1\oplus e^{i\theta }, 0<\theta<\pi \\ 0, &{} \textrm{otherwise} \end{array}\right. ,\\ \beta _{21}&=\left\{ \begin{array}{ll} \beta , &{} A=\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, 0<\tau <1\\ 0, &{} \textrm{otherwise} \end{array}\right. , \quad \beta \in {\mathbb {R}}. \end{aligned}$$

In a similar fashion we conclude that

$$\begin{aligned}&X^TB_0+B_0X = \sum _{j,k=1}^2x_{jk}(E_{kj}B_0+B_0E_{jk}) +\sum _{j,k=1}^2y_{jk}i(E_{kj}B_0+B_0E_{jk}),\\&\tfrac{d}{dt}\big |_{t=0}B(t)=\sum _{j,k=1}^2\gamma _{jk}E_{jk}, \qquad \gamma _{jk}=\left\{ \begin{array}{ll} z_{jk}, &{} B_{jk}(t)=(B_0)_{jk}+z_{jk}t, z_{jk}\in {\mathbb {C}}\\ i(B_0)_{jk}\omega _{jk}, &{} B_{jk}(t)=(B_0)_{jk}e^{i\omega _{jk}t}, \omega _{jk}\in {\mathbb {R}}\\ 0, &{} \textrm{otherwise} \end{array}\right. . \end{aligned}$$

Note that if $A_{jk}(t)$ (or $B_{jk}(t)$) is constant, then $\beta _{jk}=0$ ($\gamma _{jk}=0$).

In view of the identification ${\mathbb {R}}^{8} \times {\mathbb {R}}^{6}\approx {\mathbb {C}}^{2\times 2}\times {\mathbb {C}}^{2\times 2}_S$ we denote ($j,k\in \{1,2\}$):

$$\begin{aligned}&\widetilde{u}_{jk}\approx (0,E_{jk}), \qquad \qquad \widetilde{v}_{jk}\approx (0,iE_{jk}), \quad \qquad j\le k \\&u_{jk}\approx (E_{kj}A_0+ A_0E_{jk},E_{kj}B_0+B_0E_{jk}), \quad v_{jk}\approx i(-E_{kj}A_0+A_0E_{jk},E_{kj}B_0+B_0E_{jk}),\\&w_1\approx \left\{ \begin{array}{ll} (ie^{i\theta }E_{22},0), &{} A=1\oplus e^{i\theta }, 0<\theta<\pi \\ (E_{21},0), &{} A=\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, 0<\tau <1\\ 0, &{} \textrm{otherwise} \end{array} \right. , \qquad \begin{array}{l} w_2\approx (iA,0),\\ w_{3} \approx (0,i(B_0)_{11}E_{11}),\\ w_{4} \approx (0,i(B_0)_{22}E_{22}). \end{array} \end{aligned}$$

The tangent space of $\textrm{Bun}_{\Psi }(A_0,B_0)$ can be seen as a linear space spanned by vectors $\{w_1,w_2\}\cup \{u_{jk},v_{jk}\}_{j,k \in \{1,2\} }$ and a subset of vectors $\{w_{3},w_{4}\} \cup \{\widetilde{u}_{jk},\widetilde{v}_{jk}\}_{j,k\in \{1,2\},j\le k}$. If $B_{jj}(t)=(B_0)_{jj}(\lambda _0)e^{i\omega _{jj}t}$ for $j\in \{1,2\}$, then $w_{j+2}$ is in the span, while for $B_{jk}(t)=(B_0)_{jk}+z_{jk}t$, $z_{jk}\ne 0$ vectors $\widetilde{u}_{jk},\widetilde{v}_{jk}$ are in the span. It is straightforward to compute the dimensions; see [23, Lemma 2.2] for the details in the case of orbits. $\square$

3 Change of the normal form under small perturbations

In this section we study how small deformations of a pair of one arbitrary and one symmetric matrix can change its bundle under the action (2.1). For the sake of clarity the notion closure graph for bundles for an action is introduced; compare it with the closure graph for orbits in [23]. Given an action $\Phi$, vertices of its closure graph are pairwise disjoint bundles of orbits with respect to $\Phi$, and there is a path from a vertex $\widetilde{{\mathcal {V}}}$ to a vertex ${\mathcal {V}}$ precisely when $\widetilde{{\mathcal {V}}}$ lies in the closure of ${\mathcal {V}}$. The path from $\widetilde{{\mathcal {V}}}$ to ${\mathcal {V}}$ is denoted by $\widetilde{{\mathcal {V}}}\rightarrow {\mathcal {V}}$. To simplify the notation we usually write $\widetilde{V}\rightarrow V$ for $\widetilde{V}\in \widetilde{{\mathcal {V}}}$, $V\in {\mathcal {V}}$ (instead of $\widetilde{{\mathcal {V}}}\rightarrow {\mathcal {V}}$). We also require that if $\widetilde{V}\in {\mathcal {\widetilde{V}}}$ (hence $\textrm{Orb}_{\Phi }(\widetilde{V})$) is contained in the closure of ${\mathcal {V}}$, then whole bundle $\widetilde{{\mathcal {V}}}$ must lie in the closure of ${\mathcal {V}}$; it does not hold in general. Closure graphs are reflexive and transitive.

When $\widetilde{V}\not \rightarrow V$ it is useful to know the distance from $\widetilde{V}$ to the bundle ${\mathcal {V}}\ni V$. It suffices to consider the distance from the normal form of $\widetilde{V}$ (see e.g. [23, Remark 3.2]). We use the max norm $\Vert X\Vert =\max _{j,k\in \{1, 2\}}|x_{j,k}|$, $X=[x_{j,k}]_{j,k=1}^{2}\in {\mathbb {C}}^{2\times 2}$ to measure the distance between matrices.

To emphasize the difference between the closure graphs for orbits and bundles we take look at the action of similarity on ${\mathbb {C}}^{2\times 2}$. Given $\lambda ,\mu \in {\mathbb {C}}$ with $\lambda \ne \mu$ we have $\lambda \oplus \lambda \not \rightarrow \lambda \oplus \mu$ in the closure graph for orbits (eigenvalues depend continuously on the entries of the matrix), but $\lambda \oplus \lambda \rightarrow \lambda \oplus \mu$ in the closure graph for bundles (the bundle of $\lambda \oplus \mu$ is dense in ${\mathbb {C}}^{2\times 2}$). For a comprehensive theory on closure hierarchy of matrices under similarity we refer to [10] and [11].

The action (2.1) is closely related to the following two actions:

$$\begin{aligned}&\Psi _1:(c,P,A)\big )\mapsto cP^{*}AP, \quad P\in GL_2({\mathbb {C}}), \,c\in S^{1}, \,A\in {\mathbb {C}}^{2\times 2} \end{aligned}$$

(3.1)

$$\begin{aligned}&\Psi _2:(P,B)\mapsto P^TBP, \quad P\in GL_2({\mathbb {C}}),\, B \in {\mathbb {C}}_S^{2\times 2}. \end{aligned}$$

(3.2)

Bundles under these actions are defined the same way as bundles for $\Psi$ in (2.2).

The closure graph for (3.2) with trivial bundles (hence orbits) is simple (see [23, Lemma 3.2]); we add a few necessary conditions on its paths and prove them in Sec. 4. For closure graphs of all $2\times 2$ or $3\times 3$ matrices see [8].

Lemma 3.1

The closure graph for the action (3.2) is

$$\begin{aligned} 0_2 \rightarrow 1\oplus 0\rightarrow I_2, \end{aligned}$$

(3.3)

in which $1\oplus 0$ and $I_2$ correspond to bundles of symmetric matrices of rank 1 and 2. Furthermore, let $B=\begin{bmatrix} a &{} b\\ b &{} d \end{bmatrix}\in {\mathbb {C}}^{2\times 2}_S$, $\widetilde{B}=\begin{bmatrix} \widetilde{a} &{} \widetilde{b} \\ \widetilde{b} &{} \widetilde{d} \end{bmatrix}\in {\mathbb {C}}^{2\times 2}_S$, $P=\begin{bmatrix} x &{} y\\ u &{} v \end{bmatrix}\in GL_2({\mathbb {C}})$ and $F=\begin{bmatrix} \epsilon _1 &{} \epsilon _2\\ \epsilon _2 &{} \epsilon _4 \end{bmatrix}\in {\mathbb {C}}^{2\times 2}_S$ be such that $P^{T} BP=\widetilde{B}+F$. Then the following statements hold:

• (1) If $\widetilde{B}$, B are normal forms in (3.3) and such that $\widetilde{B}\not \rightarrow B$, then $\Vert F\Vert \ge 1$.

• (2) If $\widetilde{B}\rightarrow B$, then there exist $\epsilon _2',\epsilon _2''\in {\mathbb {C}}$, $|\epsilon _2'|,|\epsilon _2''|\le \left\{ \begin{array}{ll} \tfrac{\Vert F\Vert (4\Vert \widetilde{B}\Vert +2+\sqrt{|\det \widetilde{B}|})}{\sqrt{|\det \widetilde{B}|}}, &{} \det \widetilde{B}\ne 0\\ \sqrt{\Vert F\Vert (4\Vert \widetilde{B}\Vert +3)}, &{} \det \widetilde{B}= 0 \end{array}\right.$, so that equations listed in the third column (and in the line corresponding to B) of Table 2 are valid.

Table 2 Necessary conditions on B and P (given that $P^{T} BP=\widetilde{B}+F$)

By making a more detailed analysis than in [23, Lemma 3.4] (see also [14, Theorem 2.2]) we get the closure graph for bundles under the action (3.1) along with necessary conditions related to its paths; the proof is given in Sect. 4.

Lemma 3.2

The closure graph for bundles under the action (3.1) is drawn in Fig. 1. It contains six vertices corresponding to bundles (orbits) with normal forms $0_2$, $1\oplus 0$, $I_2$, $1\oplus -1$, $\begin{bmatrix} 0 &{} 1\\ 0 &{} 0 \end{bmatrix}$, $\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}$, and two vertices for bundles with normal forms $1\oplus e^{i\theta }$ for $\theta \in (0,\pi )$ and $\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}$ for $\tau \in (0,1)$.

Fig. 1

The closure graph for the action (3.1)

Furthermore, let $\widetilde{A}$, A be normal forms in Fig. 1, and let $E=cP^{*} AP-\widetilde{A}$ for some $c\in S^{1}$, $P=\begin{bmatrix} x &{} y\\ u &{} v \end{bmatrix}\in GL_2({\mathbb {C}})$, $E\in {\mathbb {C}}^{2\times 2}$ with $\Vert E\Vert <1$. Then the following statements hold:

• (1) If $\widetilde{A}\not \rightarrow A$, then there exists a constant $\mu >0$ which does not depend on c, P and such that $\Vert E\Vert \ge \mu$.

• (2) If $\widetilde{A}\rightarrow A$, then there is a constant $\nu >0$ such that the moduli of expressions listed in the fourth column (and in the line corresponding to $\widetilde{A}$, A) of Table 3 are bounded by $\nu \sqrt{\Vert E\Vert }$. (If $\widetilde{A}\in GL_2({\mathbb {C}})$ then also $\Vert E\Vert \le \tfrac{|\det \widetilde{A}|}{8\Vert \widetilde{A}\Vert +4}$ is assumed.)

Table 3 Necessary conditions on A, P, c (given that $cP^{*} AP=\widetilde{A}+E$)

Remark 3.3

For calculations of $\mu$, $\nu$ in Lemma 3.2 see the proof of the lemma.

We are ready to state the main results of the paper. The proof is given in Sect. 5.

Fig. 2

Paths not mentioned in Theorem 3.4 (1), (2), (3), (4); $a,b,d>0$, $\zeta ^{*}\in {\mathbb {C}}^{*}$, $\tau \in (0,1)$, $\theta \in (0,\pi )$, $\varphi \in [0,\pi )$

Fig. 3

Paths not mentioned in Theorem 3.4 (1), (2), (3), (4); $a,b,d>0$, $\zeta \in {\mathbb {C}}$, $\tau \in (0,1)$, $\theta \in (0,\pi )$

Theorem 3.4

Let bundles with normal forms of types from Lemma 2.1 be vertices in the closure graph for the action $\Psi$ in (2.1). The graph contains precisely the paths described by the following statements:

• (1) There is a path from $(0_2,0_2)$ to any bundle. There exist paths from $\textrm{Bun}_{\Psi }(1\oplus 0,0_2)$ to all bundles, except to $\textrm{Bun}_{\Psi }(0_2,B)$ for $B\in {\mathbb {C}}_S^{2\times 2}$. Furthermore, there are paths from $(1\oplus 0,\widetilde{a}\oplus 0)$ with $\widetilde{a}>0$ to all bundles, except to $(0_2,B)$ for $B\in {\mathbb {C}}_S^{2\times 2}$ and $(A,0_2)$ for $A\in {\mathbb {C}}^{2\times 2}$.

• (2) There exist paths from $\textrm{Bun}_{\Psi }(0_2,1\oplus 0)$ to all bundles, except to $\textrm{Bun}_{\Psi }(A,0_2)$ for $A\in {\mathbb {C}}^{2\times 2}$.

• (3) From every bundle, except $\textrm{Bun}_{\Psi }(1\oplus e^{i\theta },B)$ for $0\le \theta <\pi$, $B\in {\mathbb {C}}_S^{2\times 2}$, there exists a path to the bundle $\textrm{Bun}_{\Psi }\big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} e^{i\varphi } &{} b\\ b &{} \zeta \end{bmatrix}\big )$ with $0\le \varphi <\pi$, $0<b$, $\zeta \in {\mathbb {C}}$.

• (4) From every bundle, except $\textrm{Bun}_{\Psi }\big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix},B\big )$ for $0\le \tau < 1$, $B\in {\mathbb {C}}_S^{2\times 2}$, there exists a path to the bundle $\textrm{Bun}_{\psi }\big (1\oplus e^{i\theta },\begin{bmatrix} a &{} \zeta ^{*}\\ \zeta ^{*} &{} d \end{bmatrix}\big )$ with $0\le \theta <\pi$, $\zeta ^{*}\in {\mathbb {C}}^{*}$ and $a,d>0$.

• (5) All other paths that are not mentioned in (1), (2), (3), (4) are noted in Figs. 2 and 3. (Dimensions of bundles are indicated on the right.)

Remark 3.5

We prove $(\widetilde{A},\widetilde{B})\rightarrow (A,B)$ by finding $(A(s),B(s))\in \textrm{Bun}(A,B)$, $c(s)\in S^{1}$, $P(s)\in GL_2({\mathbb {C}})$ such that $c(s) (P(s))^*A(s)P(s)\rightarrow \widetilde{A}$ and $(P(s))^TB(s)P(s)\rightarrow \widetilde{B}$ as $s\rightarrow 0$. It often includes tedious calculations and intriguing estimates; but since these do not seem to be of any special interest we omit them and thus shorten the proof significantly. When $(\widetilde{A}, \widetilde{B})\not \rightarrow (A,B)$, then a lower bound for the distance from $(\widetilde{A},\widetilde{B})$ to $\textrm{Bun}_{\Psi }(A,B)$ will be provided as part of the proof of Theorem 3.4. Note that if $\textrm{dim}\textrm{Bun}_{\Psi }(A,B)\le \textrm{dim}\textrm{Orb}_{\Psi }(\widetilde{A}, \widetilde{B})$, then it implies $(\widetilde{A},\widetilde{B}) \not \rightarrow (A,B)$ ([4, Propositions 2.8.13,2.8.14]), but it gives no estimate on the distance from $(\widetilde{A},\widetilde{B})$ to $\textrm{Bun}_{\Psi }(A,B)$.

The following result is an immediate consequence of Theorem 3.4 (see [23, Corollary 3.8] for an analogous result in the case of the closure graph for orbits).

Corollary 3.6

Let M be a compact real 4-manifold embedded ${\mathcal {C}}^2$-smoothly in a complex 3-manifold X and let $p_1,\ldots ,p_k\in M$ be precisely (all) its complex points with the corresponding normal forms up to quadratic terms $(A_1,B_1),\ldots ,(A_k,B_k)\in {\mathbb {C}}^{n\times n}\times {\mathbb {C}}_S^{n\times n}$. Assume that $M'$ is a deformation of M obtained by a smooth isotopy of M, and let $p\in M'$ be a complex point with the corresponding quadratic normal form (A, B). If the isotopy is sufficiently ${\mathcal {C}}^2$-small then p is arbitrarily close to some $p_{j_0}$, $j_{0}\in \{1,\ldots ,k\}$, and $(A_{j_{0}},B_{j_{0}})\rightarrow (A,B)$ is a path in the closure graph for bundles for the action (2.1).

Remark 3.7

The lower bounds for the distances from normal forms to other bundles give the estimate how small the isotopy in the corollary needs to be.

4 Proof of Lemmas 3.1 and  3.2

In this section we prove Lemma 3.1 and Lemma 3.2. We start with a technical lemma which is an adaptation of [23, Lemma 4.1] to the case of bundles.

Lemma 4.1

Suppose $P\in GL_2({\mathbb {C}})$, $\widetilde{A},A,E,\widetilde{B},B,F\in {\mathbb {C}}^{2\times 2}$, $c\in S^{1}$.

• (1) If $cP^*AP=\widetilde{A}+E$, $\Vert E\Vert \le \left\{ \begin{array}{ll} \min \{\tfrac{|\det \widetilde{A}|}{8\Vert \widetilde{A}\Vert +4},1\}, &{} \det \widetilde{A}\ne 0\\ 1 &{} \det \widetilde{A}=0 \end{array}\right. ,$ it then follows that

  $$\begin{aligned}&\big |\sqrt{\det A}\big |\,|\det P|=\bigl |\sqrt{\det \widetilde{A}} \bigr |+r, \qquad |r|\le \left\{ \begin{array}{ll} \tfrac{\Vert E\Vert (4\Vert \widetilde{A}\Vert +2)}{\sqrt{|\det \widetilde{A}|}}, &{} \det \widetilde{A}\ne 0\\ \sqrt{\Vert E\Vert (4\Vert \widetilde{A}\Vert +2)}, &{} \det \widetilde{A}= 0 \end{array}\right. . \nonumber \\ \end{aligned}$$

  (4.1)

  Moreover, if $A,\widetilde{A}\in GL_2({\mathbb {C}})$ and $\Delta :=\arg \bigl (\tfrac{\det \widetilde{A}}{\det A}\bigr )$ we have

  $$\begin{aligned}&c= (-1)^{k}e^{ \frac{i\Delta }{2}}+g, \quad c^{-1}= (-1)^{k} e^{-\frac{i\Delta }{2}}+\overline{g}, \qquad k\in {\mathbb {Z}},\quad |g|\le \tfrac{\Vert E\Vert (8\Vert \widetilde{A}\Vert +4)}{|\det \widetilde{A}|}. \nonumber \\ \end{aligned}$$

  (4.2)

• (2) If $P^TBP=\widetilde{B}+F$, $\Vert F\Vert \le \min \{\tfrac{|\det \widetilde{B}|}{4\Vert \widetilde{B}\Vert +2},1\}$, then

  $$\begin{aligned} \sqrt{\det B} \,\det P=\sqrt{\det \widetilde{B}}+r, \qquad |r|\le \left\{ \begin{array}{ll} \tfrac{\Vert F\Vert (4\Vert \widetilde{B}\Vert +2)}{\sqrt{|\det \widetilde{B}|}}, &{} \det \widetilde{B}\ne 0\\ \sqrt{\Vert F\Vert (4\Vert \widetilde{B}\Vert +2)}, &{} \det \widetilde{B}= 0 \end{array}\right. . \end{aligned}$$

• (3) Let further $A,\widetilde{A}\in GL_2({\mathbb {C}})$, $\Vert E\Vert \le \min \{1,||\widetilde{A}^{-1}||^{-1},\frac{|\det \widetilde{A}|}{8\Vert \widetilde{A}\Vert +4}\}$ and $cP^*AP =\widetilde{A}+E$, $P^TBP=\widetilde{B}+F$. It then implies that

  $$\begin{aligned}&|\det \widetilde{A}\det B|=|\det \widetilde{B}\det A|+r,\\&|r|\le \max \{\Vert E\Vert ,\Vert F\Vert \}\tfrac{|\det A|}{|\det \widetilde{A}|} \big (4\max \{\Vert \widetilde{A}\Vert ,\Vert \widetilde{B}\Vert ,|\det \widetilde{A}|,|\det \widetilde{B}|\}+2\big )^{2}. \end{aligned}$$

  Moreover, if in addition $B,\widetilde{B}$ are nonsingular and $|\det A|=|\det \widetilde{A}|=\Vert \widetilde{A}\Vert =1$, $\Vert E\Vert ,\Vert F\Vert \le \frac{|\det \widetilde{B}|}{4(4\max \{1,\Vert \widetilde{B}\Vert ,\det \widetilde{B}\}+2)^{2}}$, $\Gamma :=\arg \big (\frac{\det \widetilde{B}}{\det B}\big )$, then we have

  $$\begin{aligned} \det P= (-1)^{l}e^{i\frac{\Gamma }{2}}+p,\qquad l \in {\mathbb {Z}}, \quad |p|\le \Vert F\Vert \tfrac{8\Vert \widetilde{B}\Vert +4}{\sqrt{3}|\det \widetilde{B}|}. \end{aligned}$$

Proof

For $\xi ,h\in {\mathbb {C}}$, $\zeta \in {\mathbb {C}}^{*}$ we have $\xi \zeta ^{-1}=1+\frac{h}{\zeta }=|1+\frac{h}{\zeta }|e^{i \psi }$ with $|\frac{h}{\zeta }|\le \frac{1}{2}$, hence $\psi \in (-\tfrac{\pi }{2},\tfrac{\pi }{2})$ and $|\sin \psi |=\bigl |\textrm{Im}\big (\tfrac{1+\frac{h}{\zeta }}{|1+\frac{h}{\zeta }|}\big )\bigr | \le \tfrac{|\textrm{Im}\frac{h}{\zeta }|}{|1+\frac{h}{\zeta }|} \le \tfrac{|\frac{h}{\zeta }|}{1-|\frac{h}{\zeta }|}\le \tfrac{2|h|}{|\zeta |}$. Thus

$$\begin{aligned} \xi =\zeta +h, \,|h|\le \tfrac{|\zeta |}{2}\ne 0 \quad \textrm{implies}\quad \arg (\xi )-\arg (\zeta ) =\psi \in \left( -\tfrac{\pi }{2},\tfrac{\pi }{2}\right) ,\,\,| \sin \psi |\le 2\left| \tfrac{h}{\zeta }\right| . \end{aligned}$$

(4.3)

Estimating the absolute values of the entries of the matrices by the max norm of the matrices, and by slightly simplifying, we obtain that for any $X,D\in {\mathbb {C}}^{2\times 2}$:

$$\begin{aligned} \big ||\det (X+D)|-|\det X|\big |&\le \big |\det (X+D) -\det X \big |\le \Vert D\Vert \big (4\Vert X\Vert +2\Vert D\Vert \big ). \end{aligned}$$

(4.4)

Furthermore, we apply the determinant to $cP^{*}AP=\widetilde{A}+E$, $Q^{T}BQ=\widetilde{B}+F$ to get

$$\begin{aligned} c^2|\det P|^2\det A= \det (\widetilde{A}+E), \qquad (\det Q)^2 \det B= \det (\widetilde{B}+F). \end{aligned}$$

(4.5)

Assuming $\Vert E\Vert ,\Vert F\Vert \le 1$ and using (4.4) for $X=\widetilde{A}$, $D=E$ and $X=\widetilde{B}$, $D=F$ gives

$$\begin{aligned}&|\det A|\,|\det P|^{2}=|\det \widetilde{A}|+p, \qquad |p| \le \Vert E\Vert (4\Vert \widetilde{A}\Vert +2),\nonumber \\&\det B(\det Q)^{2}=\det \widetilde{B}+q, \qquad |q| \le \Vert E\Vert (4\Vert \widetilde{B}\Vert +2), \end{aligned}$$

(4.6)

respectively. Equations (4.6) immediately imply statements (4.1) for $\det \widetilde{A}=0$ and (2) for $\det \widetilde{B}=0$. Next, we observe another simple fact. If $|s|\le 1$ then there exists $s'$ so that

$$\begin{aligned} \sqrt{1+s}=(-1)^{l}(1+s'), \qquad l\in {\mathbb {Z}}, \,\,\textrm{Re}(s')\ge -1 , \,\, |s'|\le |s|. \end{aligned}$$

(4.7)

We apply (4.7) to (4.6) with $\Vert E\Vert \le \tfrac{|\det \widetilde{A}|}{4\Vert \widetilde{A}\Vert +2}$ and $\Vert F\Vert \le \tfrac{|\det \widetilde{B}|}{4\Vert \widetilde{B}\Vert +2}$ to obtain (4.1) for $\det \widetilde{A}=0$ and (2) for $\det \widetilde{B}\ne 0$.

The right-hand side of (4.4) for X nonsingular and D with $\Vert D\Vert \le 1$ leads to

$$\begin{aligned} \left| \tfrac{\det (X+D)}{\det (X)}-1\right| \le \tfrac{\Vert D\Vert (4\Vert X\Vert +2)}{|\det X|}. \end{aligned}$$

(4.8)

By assuming $\Vert E\Vert \le \tfrac{|\det \widetilde{A}|}{8\Vert \widetilde{A}\Vert +4}$ and applying (4.3) to (4.8) for $X=\widetilde{A}$, $D=E$ we obtain

$$\begin{aligned} \psi =\arg \big (\tfrac{\det (\widetilde{A}+E)}{\det \widetilde{A}}\big ) \in \left( -\tfrac{\pi }{2},\tfrac{\pi }{2}\right) ,\qquad |\sin \psi | \le \tfrac{\Vert E\Vert (8\Vert \widetilde{A}\Vert +4)}{|\det \widetilde{A}|}. \end{aligned}$$

(4.9)

From (4.5) we get

$$\begin{aligned} c^2|\det P|^2=\tfrac{\det (\widetilde{A}+E)}{\det A} =\tfrac{\det (\widetilde{A}+E)}{\det \widetilde{A}} \tfrac{\det \widetilde{A}}{\det A}, \end{aligned}$$

(4.10)

and it follows that $c=(-1)^{k}e^{i(\frac{\Delta }{2}+\frac{\psi }{2})}$ with $k\in {\mathbb {Z}}$, $\Delta =\arg \bigl (\frac{\det \widetilde{A}}{\det A}\bigr )$. Using the identity $e^{i\frac{\psi }{2}}=1+2i(\sin \frac{\psi }{4})e^{i\frac{\psi }{4}}$ and the inequality $2|\sin \frac{\psi }{4}|\le |\frac{\psi }{2}|\le |\sin \psi |$ for $\psi \in (-\tfrac{\pi }{2},\tfrac{\pi }{2})$, we deduce (4.2).

We multiply (4.5) for $P=Q$ by $\det \widetilde{B}$ and $\det \widetilde{A}$. By comparing the moduli of the expressions, and assuming $\Vert E\Vert \le ||\widetilde{A}^{-1}||^{-1}$ (hence $\det (\widetilde{A}+E)\ne 0$), we get

$$\begin{aligned} |\det B||\det \widetilde{A}|=|\det A|\tfrac{|\det \widetilde{A}|| \det (\widetilde{B}+F)|}{|\det (\widetilde{A}+E)|}. \end{aligned}$$

(4.11)

Set $d_{X,D}:=|\det (X+D)|-|\det (X)|$ and apply (4.4) for $X=\widetilde{A}$, $D=E$ and $X=\widetilde{B}$, $D=F$:

$$\begin{aligned} \big |\tfrac{|\det \widetilde{A}||\det (\widetilde{B}+F)|}{|\det (\widetilde{A}+E)|}-|\det \widetilde{B}|\big | =\big |\tfrac{d_{\widetilde{B},F}|\det \widetilde{A}| -d_{\widetilde{A},E}|\det \widetilde{B}|}{d_{\widetilde{A},E} +|\det (\widetilde{A})|} \big |\le \tfrac{|\det \widetilde{B}| \Vert E\Vert \big (4\Vert \widetilde{A}\Vert +2\big )+|\det \widetilde{A}| \Vert F\Vert \big (4\Vert \widetilde{B}\Vert +2\big )}{|\det \widetilde{A}| -\Vert E\Vert \big (4\Vert \widetilde{A}\Vert +2\big )}, \end{aligned}$$

provided that $\Vert E\Vert \le \min \big \{||\widetilde{A}^{-1}||^{-1}, \tfrac{|\det \widetilde{A}|}{8\Vert \widetilde{A}\Vert +4}\big \}$. We combine it with (4.11):

$$\begin{aligned}&\big ||\det \widetilde{A}\det B|-|\det \widetilde{B}\det A|\big | =|\det A| \big | \tfrac{|\det \widetilde{A}||\det (\widetilde{B}+F)|}{|\det (\widetilde{A}+E)|}-| \det \widetilde{B}| \big | \nonumber \\ \quad&\le \tfrac{|\det A|}{|\det \widetilde{A}|} \max \{\Vert E\Vert ,\Vert F\Vert \}4\max \big \{|\det \widetilde{A}|,| \det \widetilde{B}|\big \}\big (4\max \{\Vert \widetilde{A}\Vert , \Vert \widetilde{B}\Vert \} +2\big ). \end{aligned}$$

(4.12)

Further, let $B,\widetilde{B}$ be nonsingular and $|\det A|=|\det \widetilde{A}|=\Vert \widetilde{A}\Vert =1$, $\Vert F\Vert \le \{\tfrac{|\det \widetilde{B}|}{4\Vert \widetilde{B}\Vert +2},1\}$, $r:=|\det B|-|\det \widetilde{B}|$. Applying (4.8) for $X=\widetilde{B}$, $D=F$ and (4.5) for $Q=P$ yields

$$\begin{aligned}&(\det P)^{2}=\tfrac{\det (\widetilde{B}+F)}{\det \widetilde{B}} \tfrac{\det \widetilde{B}}{\det B}= e^{i\Gamma } \big (1-\tfrac{r}{|\det \widetilde{B}|+r}\big )(1+\epsilon '),\\&\quad \Gamma =\arg \big (\tfrac{\det {\widetilde{B}}}{\det B}\big ), \, |\epsilon '|\le \Vert F\Vert \tfrac{4\Vert \widetilde{B}\Vert +2}{|\det \widetilde{B}|}. \end{aligned}$$

Provided that $\Vert E\Vert ,\Vert F\Vert \le \frac{|\det \widetilde{B}|}{4(4\max \{1,\Vert \widetilde{B}\Vert ,\det \widetilde{B}\}+2)^{2}}$ we use (4.12) to assure $|r|\le \frac{|\det \widetilde{B}|}{4}$ (hence $|1-\frac{r}{|\det \widetilde{B}|+r}|\le \frac{4}{3}$). By applying (4.7) we complete the proof of (3). $\square$

We proceed with a simple proof of Lemma 3.1.

Proof of Lemma 3.1

The closure graph for $2\times 2$ symmetric matrices is obtained by an easy and straightforward calculation.

We write the matrix equation $P^{T} BP=F+\widetilde{B}$ for $B=\begin{bmatrix} 0 &{} b\\ b &{} d \end{bmatrix}$ componentwise:

$$\begin{aligned}&2bux+du^2=\widetilde{a}+\epsilon _1\nonumber \\&bvx+buy+duv=\widetilde{b}+\epsilon _2\nonumber \\&2byv+dv^2=\widetilde{d}+\epsilon _4. \end{aligned}$$

(4.13)

By adding and subtracting $b\det P=b(vx-uy)$ from the second equation yields

$$\begin{aligned} 2bvx+duv=b\det P+\widetilde{b}+\epsilon _2, \qquad 2buy+duv=\widetilde{b}+\epsilon _2-b\det P. \end{aligned}$$

(4.14)

We multiply the first (the second) equation of (4.14) by u (by v) and compare it with the first (the last) equation of (4.13), multiplied by v (by u):

$$\begin{aligned} u(b\det P+\widetilde{b}+\epsilon _2)=v(\widetilde{a}+\epsilon _4), \qquad v(-b\det P+\widetilde{b}+\epsilon _2)=u(\widetilde{d}+\epsilon _4). \end{aligned}$$

(4.15)

For $b=0$ we obtain (D4). Since $\det B=-b^{2}$ we deduce from Lemma 4.1 (2) that

$$\begin{aligned} b\det P=i(-1)^{l}\sqrt{\det \widetilde{B}}+r, \qquad l\in {\mathbb {Z}},\,\,\, |r|\le \left\{ \begin{array}{ll} \tfrac{\Vert F\Vert (4\Vert \widetilde{B}\Vert +2)}{\sqrt{|\det \widetilde{B}|}}, &{} \det \widetilde{B}\ne 0\\ \sqrt{\Vert F\Vert (4\Vert \widetilde{B}\Vert +2)}, &{} \det \widetilde{B}= 0 \end{array}\right. . \end{aligned}$$

(4.16)

Together with (4.14) for $d=0$ and (4.15) this gives (D3) and (D1).

Next, the equation $P^{T} AP=F+\widetilde{B}$ for $B=\begin{bmatrix} a &{} b\\ b &{} 0 \end{bmatrix}$ yields

$$\begin{aligned}&ax^2+2bux=\widetilde{a}+\epsilon _1\nonumber \\&axy+bvx+buy=\widetilde{b}+\epsilon _2\nonumber \\&ay^2+2byv=\widetilde{d}+\epsilon _4. \end{aligned}$$

(4.17)

We add and subtract $b\det P=b(vx-uy)$ from the second equation of (4.17):

$$\begin{aligned} 2bvx+axy=b\det P+\widetilde{b}+\epsilon _2, \qquad 2buy+axy=\widetilde{b}+\epsilon _2-b\det P. \end{aligned}$$

By multiplying the first (the second) equation by y (by x) and comparing it with the last (the first) equation of (4.17), multiplied by x (by y), gives

$$\begin{aligned} y(b\det P+\widetilde{b}+\epsilon _2)=x(\widetilde{d}+\epsilon _4), \qquad x(-b\det P+\widetilde{b}+\epsilon _2)=y(\widetilde{a}+\epsilon _1). \end{aligned}$$

(4.18)

For $b=0$ we get (D5), while using (4.16) and (4.18) we obtain (D2). $\square$

Proof of Lemma 3.2

For actions $\Psi$, $\Psi _1$ (see (2.1) and (3.1)), it follows that $(A',B')\in \textrm{Orb}_{\Psi }(A,0)$ if and only if $B'=0$ and $A'\in \textrm{Orb}_{\Psi _1}(A)$. Hence $\textrm{dim}\bigl (\textrm{Orb}_{\Psi _1}(A)\bigr )=\textrm{dim}\bigl (\textrm{Orb}_{\Psi }(A,0)\bigr )$, where dimensions of orbits of $\Psi$ are obtained from Lemma 2.1.

To prove $\widetilde{A}\rightarrow A$ it suffices to find $c(s)\in S^{1}$, $P(s)\in GL_2({\mathbb {C}})$, $A(s)\in \textrm{Bun}(A)$ so that

$$\begin{aligned} c(s)(P(s))^*A(s)P(s)-\widetilde{A}\rightarrow 0\ \textrm{as}\ s \rightarrow 0. \end{aligned}$$

(4.19)

Trivially $0_2\rightarrow 1\oplus 0$, $\begin{bmatrix} 0 &{} 1 \\ 0 &{} 0 \end{bmatrix} \rightarrow \begin{bmatrix} 0 &{} 1 \\ \tau &{} 0 \end{bmatrix}$, $\begin{bmatrix} 0 &{} 1 \\ 1 &{} 0 \end{bmatrix} \rightarrow \begin{bmatrix} 0 &{} 1 \\ \tau &{} 0 \end{bmatrix}$ for $0<\tau <1$ and $1\oplus e^{i\widetilde{\theta }} \rightarrow 1\oplus e^{i\theta }$ for $\widetilde{\theta }\in \{0,\pi \}$, $0< \theta < \pi$. It is not too difficult to show $1\oplus 0 \rightarrow 1\oplus \lambda$, $1\oplus 0 \rightarrow \begin{bmatrix} 0 &{} 1 \\ \tau &{} 0 \end{bmatrix}$ for $0\le \tau \le 1$, $1\oplus -1 \rightarrow \begin{bmatrix} 0 &{} 1 \\ 1 &{} i \end{bmatrix}$ and $\begin{bmatrix} 0 &{} 1 \\ 1 &{} i \end{bmatrix} \rightarrow \begin{bmatrix} 0 &{} 1 \\ \tau &{} 0 \end{bmatrix}$ for $0<\tau <1$, we take $P(s)=1\oplus s$, $P(s)=\frac{1}{\sqrt{1+\tau }}\begin{bmatrix} 1 &{} 0 \\ 1 &{} s \end{bmatrix}$, $P(s)=\frac{1}{\sqrt{2}}\begin{bmatrix} s^{-1} &{} s^{-1} \\ s &{} -s \end{bmatrix}$ and $P(s)=\frac{1}{2\sqrt{s}}\begin{bmatrix} s &{} -2i \\ -is &{} 2 \end{bmatrix}$ with $\tau (s)=1-s$ in (4.19), respectively; in all cases $c(s)=1$. Finally, $A(s)=1\oplus e^{i\theta (s)}$ with $\cos (\frac{\theta (s)}{2})=\frac{s}{2}$, $c(s)=1$, $P(s)=\sqrt{s} \begin{bmatrix} i &{} is^{-1}\\ 0 &{} -is^{-1} \end{bmatrix}$ proves $\begin{bmatrix} 0 &{} 1 \\ 1 &{} i \end{bmatrix} \rightarrow 1\oplus e^{i\theta }$ for $0<\theta <1$.

It is left to find necessary conditions for the existence of these paths, i.e. given $\widetilde{A}$, E, we must find out how c, P, A depend on E, $\widetilde{A}$, if the following is satisfied:

$$\begin{aligned} cP^*AP=\widetilde{A}+E, \qquad c\in S^{1}, P\in GL_2({\mathbb {C}}). \end{aligned}$$

(4.20)

On the other hand, if (4.20) fails for every sufficiently small E, it gives $\widetilde{A}\not \rightarrow A$. In such cases the lower estimates for $\Vert E\Vert$ will be provided. These easily follow for $\widetilde{A}\ne 0$, $A=0$ and $\det \widetilde{A}\ne 0$, $\det A=0$ (Lemma 4.1 (1)).

Throughout the rest of the proof we denote

$$\begin{aligned} \widetilde{A}=\begin{bmatrix} \alpha &{} \beta \\ \gamma &{} \omega \end{bmatrix}, \qquad E=\begin{bmatrix} \epsilon _1 &{} \epsilon _2\\ \epsilon _3 &{} \epsilon _4 \end{bmatrix}, \qquad P=\begin{bmatrix} x &{} y\\ u &{} v \end{bmatrix}. \end{aligned}$$

(4.21)

Case I.:

$A=\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}$ ($\textrm{Bun}_{\Psi _1}(A)=\textrm{Orb}_{\Psi _1}(A)$)

This case coincides with [23, Lemma 3.4. Case I]; see (C6), (C8).

Case II.:

$A=1\oplus \lambda$, $|\lambda |\in \{1,0\}$

The equation (4.20) multiplied by $c^{-1}$, written componentwise and rearranged is:

$$\begin{aligned}&|x|^2+\lambda |u|^2 -c^{-1}\alpha =c^{-1}\epsilon _1, \quad \overline{x}y+\lambda \overline{u}v-c^{-1} \beta =c^{-1}\epsilon _2,\nonumber \\&\overline{y}x+\lambda \overline{v}u-c^{-1}\gamma =c^{-1}\epsilon _3, \quad |y|^2+\lambda |v|^2-c^{-1}\omega =c^{-1}\epsilon _4. \end{aligned}$$

(4.22)

Subtracting the second complex-conjugated equation (and multiplied by $\lambda$) from the third equation (and multiplied by $\overline{\lambda }$) for $\beta , \gamma \in {\mathbb {R}}$ gives

$$\begin{aligned}&2\textrm{Im}(\lambda )\overline{v}u-c^{-1}\gamma +\overline{c}^{-1} \beta = c^{-1}\epsilon _3- \overline{c}^{-1}\overline{\epsilon }_2,\nonumber \\&-2\textrm{Im}(\lambda )\overline{y}x-c^{-1}\overline{\lambda }\gamma +\overline{c}^{-1}\lambda \beta =c^{-1}\overline{\lambda }\epsilon _3 -\overline{c}^{-1}\lambda \overline{\epsilon }_2. \end{aligned}$$

(4.23)

• (a) $\lambda =e^{i\theta }$,    $0\le \theta \le \pi$

  From (4.23) for $\beta =\gamma =0$, $\textrm{Im}(\lambda )=\sin \theta$ we get

  $$\begin{aligned} \big |(\sin \theta )\overline{v}u\big |\le \Vert E\Vert , \qquad \big |(\sin \theta )\overline{x}y\big |\le \Vert E\Vert . \end{aligned}$$

  (4.24)

  We take the (real) imaginary parts of the (last) first equation of (4.22) for $\lambda =e^{i\theta }$:

  $$\begin{aligned}&(\sin \theta ) |u|^2=\textrm{Im}(c^{-1}\alpha +c^{-1}\epsilon _1), \qquad |x|^2+(\cos \theta )|u|^2 =\textrm{Re}(c^{-1}\alpha +c^{-1}\epsilon _1),\nonumber \\&(\sin \theta ) |v|^2=\textrm{Im}(c^{-1}\omega +c^{-1}\epsilon _4), \qquad |y|^2+(\cos \theta )|v|^2=\textrm{Re}(c^{-1}\omega +c^{-1}\epsilon _4). \nonumber \\ \end{aligned}$$

  (4.25)

  If $\alpha =0$ we further have:

  $$\begin{aligned}&(\sin \theta )|u|^2\le \Vert E\Vert , \quad (\sin \theta )|x|^2 \le \Vert E\Vert (\sin \theta +|\cos \theta |),\nonumber \\&\big |(\sin \theta )|v|^2-\textrm{Im}(c^{-1}\omega )\big |\le \Vert E\Vert , \quad \big |(\sin \theta )|y|^2-\textrm{Re}(c^{-1}\omega )\big | \le \Vert E\Vert (\sin \theta +|\cos \theta |). \end{aligned}$$

  (4.26)

  • (i) $\widetilde{A}=\begin{bmatrix} 0 &{} 1\\ \widetilde{\tau } &{} 0 \end{bmatrix}$, $0\le \widetilde{\tau }\le 1$

    If $1\le \widetilde{\tau }< 1$, then by applying the triangle inequality to the first equation of (4.23) for $\beta =1$, $\gamma =\widetilde{\tau }$, $\textrm{Im}(\lambda )=\sin \theta$ and using the first estimates of (4.26) for $\omega =0$ we obtain $2\Vert E\Vert \ge 2(\sin \theta )| uv|\ge 1-\widetilde{\tau }-2\Vert E\Vert$, which fails for $\Vert E\Vert <\frac{1-\widetilde{\tau }}{4}$.

  • (ii) $\widetilde{A}=\begin{bmatrix} 0 &{} 1\\ 1 &{} \omega \end{bmatrix}$,    $\omega \in \{0,i\}$

    By applying the triangle inequality to the second equation of (4.22), and using (4.26) with $|c^{-1}\omega |\le 1$ leads to the inequality:

    $$\begin{aligned} (\sin \theta )(1-\Vert E\Vert )\le \sin \theta |\overline{x}y +\lambda \overline{u}v|\le \sqrt{\Vert E\Vert (1+\Vert E\Vert )}+\sqrt{2\Vert E\Vert (1+2\Vert E\Vert )}. \end{aligned}$$

    If $\Vert E\Vert \le \frac{1}{12}$ then we deduce $\sin \theta \le 3\sqrt{\Vert E\Vert }$ and $\cos ^{2} \theta \ge 1-9\Vert E\Vert$. If $\theta$ is close to 0 then the second and the last equation of (4.25) for $\alpha =0$, $|c^{-1}\omega |\le 1$ imply that $|x|^{2},|u|^{2}\le \frac{\Vert E\Vert }{\sqrt{1-9\Vert E\Vert }}$ and $|y|^{2},|v|^{2}\le \frac{1+\Vert E\Vert }{\sqrt{1-9\Vert E\Vert }}$, respectively. For $\Vert E\Vert$ so small that $1>2\frac{\sqrt{\Vert E\Vert (1+\Vert E\Vert )}}{\sqrt{1-9\Vert E\Vert }} +\Vert E\Vert$, the second equation of (4.22) for $\beta =1$ fails. Next, when $\theta$ is close to $\pi$, we deduce that $\frac{1+\cos \theta }{\sin \theta }=\cot \frac{\theta }{2}$ is close to 0 and $\pi -\theta \in (0,\frac{\pi }{2})$, hence

    $$\begin{aligned}&\left| \cos \tfrac{\theta }{2}\right| =\left| \sin \left( \tfrac{\pi - \theta }{2}\right) \right| \le \sin (\pi -\theta )=\sin \theta ,\quad \left| \cos \left( \tfrac{\theta +\pi }{4}\right) \right| =\left| \sin \left( \tfrac{\pi - \theta }{4}\right) \right| \le \sin \theta ,\nonumber \\&1+\cos \theta = \tfrac{\cos \frac{\theta }{2}\sin \theta }{\sin \frac{\theta }{2}}\le \tfrac{\sin \theta }{\sqrt{1-\sin ^{2}\theta }}\sin \theta , \quad 1-\sin \tfrac{\theta }{2}=\tfrac{\cos ^{2} \frac{\theta }{2}}{1+\sin \frac{\theta }{2}}\le \sin \theta . \end{aligned}$$

    (4.27)

    We have $c^{-1}=-i(-1)^{k}e^{i\frac{\theta }{2}}+\overline{g}$, $|g|\le 12\Vert E\Vert$ with $\Vert E\Vert \le \frac{1}{12}$ (Lemma 4.1 (4.2)), thus $\big |\textrm{Re}(c^{-1}i)\big |=|\cos \frac{\theta }{2}+i\overline{g}|\le 3\sqrt{\Vert E\Vert }+12\Vert E\Vert$ (since $\sin \theta \le 3\sqrt{\Vert E\Vert }$). Using the second (fourth) equation of (4.25) and (4.26) for $\alpha =0$, $\omega \in \{0,i\}$ with (4.27) further implies

    $$\begin{aligned} \big ||x|^{2}-|u|^{2}\big |-\Vert E\Vert&\le \big ||x|^2-|u|^{2} +(1+\cos \theta )|u|^2\big |\nonumber \\&=\big ||x|^{2}+(\cos \theta )|u|^{2}\big |\le \Vert E\Vert , \nonumber \\ \big ||y|^{2}-|v|^{2}\big |-\tfrac{3\sqrt{\Vert E\Vert }(1+\Vert E\Vert )}{\sqrt{1-9\Vert E\Vert }}&\le \big ||y|^{2}-|v|^{2}+(1+\cos \theta )|v|^{2}\big |=\big ||y|^{2} +(\cos \theta )|v|^{2}\big |\nonumber \\&\le 3\sqrt{\Vert E\Vert }+13\Vert E\Vert . \end{aligned}$$

    (4.28)

    Using the second equation of (4.22) and (4.26), (4.27) (for $\alpha =0$, $\omega \in \{0,i\}$) we get:

    $$\begin{aligned} 14\Vert E\Vert&\ge |\overline{x}y+e^{i\theta }\overline{u}v+i(-1)^{k} e^{i\frac{\theta }{2}}|=\nonumber \\&= \big |(\overline{x}y-\overline{u}v-(-1)^{k})+2 \left( \cos \tfrac{\theta }{2}\right) e^{i\frac{\theta }{2}} \overline{u}v+2(-1)^{k}\left( \cos \tfrac{\theta +\pi }{4}\right) e^{i\frac{\theta +\pi }{4}} \big |\nonumber \\&\ge \big |\overline{x}y-\overline{u}v-(-1)^{k}\big | -2\sqrt{\Vert E\Vert (1+\Vert E\Vert )}+6\sqrt{\Vert E\Vert }. \end{aligned}$$

    (4.29)

    For $\omega =i$ we have $\textrm{Im}(c^{-1}i)=\sin \frac{\theta }{2}+\textrm{Im}(i\overline{g})$, $|g|\le 12\Vert E\Vert$, therefore (4.26) yields

    $$\begin{aligned} 13\Vert E\Vert \ge \big |(\sin \theta ) |v|^{2}-(-1)^k-(-1)^{k} \left( \sin \tfrac{\theta }{2}-1\right) \big |\ge \big |(\sin \theta ) |v|^{2}-(-1)^k\big |-3\sqrt{\Vert E\Vert }. \end{aligned}$$

    Together with (4.28) and (4.29) it proves (C2). Note that the third equation of (4.25) for $\theta =\pi$, $\omega =i$ fails for $\Vert E\Vert <\frac{1}{13}$.

  • (iii) $\widetilde{A}=\alpha \oplus 0$, $\alpha \in \{0,1\}$

    If $\theta \in \{0,\pi \}$, then (4.22) for $e^{i\theta }=\sigma$ yields (C10).

    By (4.24) and the second equation of (4.22) we have

    $$\begin{aligned} \big |\overline{x}y+(\cos \theta )\overline{u}v\big |\le 2\Vert E\Vert . \end{aligned}$$

    (4.30)

    If $0<\theta \le \pi$, then (4.22), (4.24), (4.30) for $\omega =\beta =\gamma =0$, $\lambda =e^{i\theta }$ give (C1).

  • (iv) $\widetilde{A}=1\oplus e^{i\widetilde{\theta }}$, $0\le \widetilde{\theta } \le \pi$.

    By Lemma 4.1 (4.2) we have $c^{-1}=(-1)^k e^{i\frac{\theta -\widetilde{\theta }}{2}}+\overline{g}$, $|\overline{g}|\le 12\Vert E\Vert$, assuming that $\Vert E\Vert \le \frac{1}{12}$. Thus the first and the last equation of (4.22) for $\alpha =1$, $\lambda =e^{i\widetilde{\theta }}$ are of the form:

    $$\begin{aligned}&|x|^2+e^{i\theta }|u|^2 =(-1)^k e^{i\frac{\theta -\widetilde{\theta }}{2}} +(\overline{g}+c^{-1}\epsilon _1),\nonumber \\&|y|^2+e^{i\theta }|v|^2=(-1)^k e^{i\frac{\widetilde{\theta }+\theta }{2}} +(\overline{g}e^{i\widetilde{\theta }}+c^{-1}\epsilon _4). \end{aligned}$$

    (4.31)

    We take the imaginary parts of (4.31) and apply the triangle inequality:

    $$\begin{aligned}&\bigl | |u|^2\sin \theta -(-1)^k\sin \left( \tfrac{\theta -\widetilde{\theta }}{2}\right) \bigr | \le \bigl | \textrm{Im}(\overline{g}) +\textrm{Im}(\epsilon _1)\bigr |\le 13\Vert E\Vert ,\nonumber \\&\bigl | |v|^2\sin \theta -(-1)^k\sin \left( \tfrac{\widetilde{\theta }+\theta }{2}\right) \bigr | \le \bigl | \textrm{Im}(\overline{g}e^{i\widetilde{\theta }} +\overline{c}^{-1}\epsilon _4)\bigr |\le 13\Vert E\Vert . \end{aligned}$$

    (4.32)

    In particular we have

    $$\begin{aligned} |u|^2\sin \theta \ge \left| \sin \left( \tfrac{\widetilde{\theta }-\theta }{2}\right) \right| -13\Vert E\Vert , \quad |v|^2\sin \theta \ge |\sin \left( \tfrac{\widetilde{\theta } +\theta }{2}\right) | -13\Vert E\Vert . \end{aligned}$$

    By multiplying these inequalities and using the triangle inequality we deduce

    $$\begin{aligned} (\sin ^2\theta )|uv|^2&\ge \big |\sin \left( \tfrac{\widetilde{\theta }-\theta }{2}\right) \sin \left( \tfrac{\widetilde{\theta }+\theta }{2}\right) \big |\\&\quad -13\Vert E\Vert \bigl (\big |\sin \left( \tfrac{\widetilde{\theta } -\theta }{2}\right) \big |+\big |\sin \left( \tfrac{\widetilde{\theta }+\theta }{2}\right) \big |\bigr ) -169\Vert E\Vert ^2 . \end{aligned}$$

    By combining it with (4.24) and rearranging the terms we obtain

    $$\begin{aligned} \tfrac{1}{2}|\cos \widetilde{\theta }-\cos \theta |=\big |\sin \left( \tfrac{\widetilde{\theta }-\theta }{2}\right) \sin \left( \tfrac{\widetilde{\theta }+\theta }{2}\right) \big | \le 170 \Vert E\Vert ^{2} + 26\Vert E\Vert \le 196 \Vert E\Vert . \end{aligned}$$

    (4.33)

    If $\theta \in \{0,\pi \}$ with $\widetilde{\theta }\ne \theta$ then (4.33) fails for $\Vert E\Vert <\frac{1-|\cos \widetilde{\theta }|}{392}$.

    We take the real parts in the first equation of (4.31), multiply them by $\sin \theta$, then rearrange the terms and apply (4.32):

    $$\begin{aligned}&(\sin \theta ) \big ||x|^{2}-(-1)^{k} \cos \left( \tfrac{\widetilde{\theta }-\theta }{2}\right) \big |\nonumber \\&\quad =\big |-\sin \theta \cos \theta |u|^{2}+(\sin \theta ) \textrm{Re}(\overline{g}+c^{-1}\epsilon _1)\big |,\nonumber \\&(\sin \theta ) \big ||x|^{2}-(-1)^{k}\big |-(\sin \theta ) \big |\cos \left( \tfrac{\widetilde{\theta }-\theta }{2}\right) -1\big | \le \big |\sin \left( \tfrac{\theta -\widetilde{\theta }}{2}\right) \big | +13\Vert E\Vert +13\Vert E\Vert . \end{aligned}$$

    (4.34)

    Next, let $0<\widetilde{\theta },\theta <\pi$. Thus $\frac{\theta -\widetilde{\theta }}{2}\in (-\frac{\pi }{2},\frac{\pi }{2})$ and $\frac{\theta +\widetilde{\theta }}{2}\in (\frac{\widetilde{\theta }}{2},\frac{\widetilde{\theta }+\pi }{2})\subset (0,\pi )$ with $\sin (\frac{\widetilde{\theta }+\theta }{2})\ge \min \{\sin \frac{\widetilde{\theta }}{2},\cos \frac{\widetilde{\theta }}{2}\}$. We apply (4.33) and make a trivial estimate:

    $$\begin{aligned} \tfrac{196\Vert E\Vert }{\min \{\sin \widetilde{\theta }, \cos \widetilde{\theta }\}}\ge \big |\sin \left( \tfrac{\widetilde{\theta }-\theta }{2}\right) \big |\ge \big |\sin \left( \tfrac{\widetilde{\theta }-\theta }{4}\right) \big |=\tfrac{1}{\sqrt{2}}\big |\cos \left( \tfrac{\widetilde{\theta }-\theta }{2}\right) -1\big |^{\frac{1}{2}}. \end{aligned}$$

    (4.35)

    By combining (4.34) and (4.35) it is straightforward to get a constant $C>0$ so that

    $$\begin{aligned} (\sin \theta )\big ||x|^{2}-(-1)^{k}\big | \le \tfrac{196\Vert E\Vert }{\min \{\sin \widetilde{\theta }, \cos \widetilde{\theta }\}}+2\big (\tfrac{196\Vert E\Vert }{\min \{\sin \widetilde{\theta },\cos \widetilde{\theta }\}}\big )^{2} +26\Vert E\Vert \le C \Vert E\Vert . \end{aligned}$$

    (4.36)

    We multiply the second equation of (4.31) by $e^{-i\theta }$. Then we take the imaginary parts or only rearrange the terms; in both cases we also use (4.35):

    $$\begin{aligned} (\sin \theta )|y|^{2}&\le \big | \sin \left( \tfrac{\widetilde{\theta }-\theta }{2}\right) \big | +14\Vert E\Vert \le C'\Vert E\Vert , \quad C':=\tfrac{196}{\min \{\sin \widetilde{\theta },\cos \widetilde{\theta }\}}+14,\nonumber \\ \big ||v|^{2}-(-1)^{k}\big |&\le |e^{i\frac{\widetilde{\theta } -\theta }{2}}-1|+ |y|^{2}+|\overline{g}e^{i\widetilde{\theta }} +c^{-1}\epsilon _4|\nonumber \\&\le 2\big (\tfrac{196\Vert E\Vert }{\min \{\sin \widetilde{\theta }, \cos \widetilde{\theta }\}}\big )^{2} +13\Vert E\Vert +|y|^{2}. \end{aligned}$$

    (4.37)

    From the first estimate in (4.32) we similarly obtain $(\sin \theta )|u|^{2}\le C'\Vert E\Vert$. If $\sin \theta \le \max \{\sqrt{C},\sqrt{C'}\}\sqrt{\Vert E\Vert }$, then (4.33) yields a contradiction for sufficiently small $\Vert E\Vert$. Otherwise $|u|^{2}\le \sqrt{C'\Vert E\Vert }$ and (4.36), (4.37) imply $\big ||x|^{2}-(-1)^{k}\big |\le \sqrt{C\Vert E\Vert }$, $|y|^{2}\le \sqrt{C'\Vert E\Vert }$, respectively. The last estimate in (4.37) concludes the proof of (C7).

    Finally, suppose $0<\theta <\pi$ and $\widetilde{\theta }\in \{0,\pi \}$; hence $\frac{\widetilde{\theta }-\theta }{2}\in (-\frac{\pi }{2},\frac{\pi }{2})$. We apply (4.33) and use (4.32) for $\widetilde{\theta }=0$ or $\widetilde{\theta }=\pi$ to deduce

    $$\begin{aligned} 14 \sqrt{\Vert E\Vert }\ge \big |\sin \left( \tfrac{\widetilde{\theta }-\theta }{2}\right) \big | \ge \big |\sin \left( \tfrac{\widetilde{\theta }-\theta }{4}\right) \big |, \quad |u|^2\sin \theta ,|v|^2\sin \theta \le 13 \Vert E\Vert +14\sqrt{\Vert E\Vert }. \end{aligned}$$

    (4.38)

    Assume now that $\sqrt{\Vert E\Vert }\le \frac{\sqrt{2}}{28}$. If $\widetilde{\theta }=0$, then $|\cos \frac{\theta }{2}|\ge \frac{\sqrt{2}}{2}$, therefore $1-\cos \theta = (\sin \theta )|\tan \frac{\theta }{2}|\le \sqrt{2}\sin \theta$. Similarly, for $\widetilde{\theta }=\pi$ we have $|\sin \frac{\theta }{2}|\ge \frac{\sqrt{2}}{2}$ and so $1+\cos \theta = (\sin \theta )|\cot \frac{\theta }{2}|\le \sqrt{2}\sin \theta$. We take the real parts of the first equation (4.31) for $\sigma =e^{i\widetilde{\theta }}$ with $\widetilde{\theta }\in \{0,\pi \}$, rearrange the terms, and apply the triangle inequality:

    $$\begin{aligned} 13\Vert E\Vert&\ge \big | |x|^{2}+ \sigma |u|^{2}-(-1)^{k}+(-1)^{k} \left( 1-\cos \left( \tfrac{\theta -\widetilde{\theta }}{2}\right) \right) -|u|^{2}(\sigma -\cos \theta ) \big | \nonumber \\&\ge \big | |x|^{2}+\sigma |u|^{2}-(-1)^{k}\big |-392\Vert E\Vert -\sqrt{2}(13 \Vert E\Vert +14\sqrt{\Vert E\Vert }). \end{aligned}$$

    (4.39)

    The same proof applies if we replace $x,u,(-1)^{k}$ by $y,v,\sigma (-1)^{k}$, respectively. The second equation (4.22) for $\beta =0$, $\lambda =e^{i\theta }$ and (4.24) finally yield

    $$\begin{aligned} \Vert E\Vert&\ge \big |\overline{x}y+e^{i\theta }\overline{u}v\big | =\big |\overline{x}y+\sigma \overline{u}v-(\sigma -\cos \theta ) \overline{u}v+i(\sin \theta )\overline{u}v \big | \nonumber \\&\ge \big |\overline{x}y+\sigma \overline{u}v\big | -(1+\sqrt{2})\big (13 \Vert E\Vert +14\sqrt{\Vert E\Vert }\big ). \end{aligned}$$

    Thus (C11) follows.

• (b) $\lambda =0$    (hence $\det \widetilde{A}=0$.)

  If $\widetilde{A}=\alpha \oplus 0$ for $\alpha \in \{0,1\}$, then (C12) follows from (4.22) for $\omega =\lambda =0$. Applying (4.3) for $\Vert E\Vert \le \frac{1}{2}$ to the first equation of (4.22) for $\alpha =1$, $\lambda =0$ (multiplied by c), yields $c=e^{i\psi }=1+2i(\sin \frac{\psi }{2})e^{i\frac{\psi }{2}}$ with $|\sin \frac{\psi }{2}|\le 2\Vert E\Vert$. If $\widetilde{A}=\begin{bmatrix} 0 &{} 1\\ 0 &{} 0 \end{bmatrix}$, then (4.22) for $\lambda =\alpha =\omega =0$ yields $|x|^{2},|y|^{2}\le \Vert E\Vert$, thus (4.22) fails for $\lambda =\gamma =0$, $\Vert E\Vert <\frac{1}{2}$.

Case III.:

$A=\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}$, $0\le \tau \le 1$

From (4.20) multiplied by $c^{-1}$ we obtain

$$\begin{aligned} \overline{x}u+\tau \overline{u}x -c^{-1}\alpha =c^{-1}\epsilon _1, \qquad \overline{x}v+\tau \overline{u}y-c^{-1}\beta =c^{-1}\epsilon _2,\nonumber \\ \tau \overline{v}x+\overline{y}u-c^{-1}\gamma =c^{-1}\epsilon _3, \qquad \overline{y}v+\tau \overline{v}y-c^{-1}\omega =c^{-1}\epsilon _4. \end{aligned}$$

(4.40)

Rearranging the terms of the first and the last equation immediately yields

$$\begin{aligned} (1+\tau )\textrm{Re}(\overline{x}u)+i(1-\tau )\textrm{Im}(\overline{x}u) =c^{-1}\alpha +c^{-1}\epsilon _1,\nonumber \\ (1+\tau )\textrm{Re}(\overline{y}v)+i(1-\tau )\textrm{Im}(\overline{y}v) =c^{-1}\omega +c^{-1}\epsilon _4, \end{aligned}$$

(4.41)

while multiplying the third (second) complex-conjugated equation with $\tau$, subtracting it from the second (third) equation, and rearranging the terms, give

$$\begin{aligned} (1-\tau ^2)\overline{x}v=&c^{-1}(\beta +\epsilon _2) -\tau \overline{c}^{-1}(\overline{\gamma }+\overline{\epsilon _3}) =(c^{-1}\beta -\tau \overline{c}^{-1}\overline{\gamma })+(c^{-1} \epsilon _2-\tau \overline{c}^{-1}\overline{\epsilon _3})\nonumber \\ (1-\tau ^{2})\overline{y}u=&c^{-1}(\gamma +\epsilon _3) -\tau \overline{c}^{-1}(\overline{\beta }+\overline{\epsilon _2}) =(c^{-1}\gamma -\tau \overline{c}^{-1}\overline{\beta })+(c^{-1} \epsilon _3-\tau \overline{c}^{-1}\overline{\epsilon _2}). \end{aligned}$$

(4.42)

For the existence of paths to $\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}$ ($*$-congruent to $1\oplus -1$) see Case II

Using (4.40) we obtain that

$$\begin{aligned} (1+\tau )|\overline{x}u|\ge |\alpha +\epsilon _1| \ge (1-\tau )|\overline{x}u|,\qquad (1+\tau )|\overline{y}v| \ge |\omega +\epsilon _4|\ge (1-\tau )|\overline{y}v|. \end{aligned}$$

(4.43)

By multiplying the left-hand and the right-hand sides of these inequalities we get

$$\begin{aligned}&(1+\tau )^{2}|\overline{x}u\overline{y}v|\ge |\alpha \omega | -\bigl (|\alpha |+|\omega |\bigr )\Vert E\Vert -\Vert E\Vert ^2, \end{aligned}$$

(4.44)

$$\begin{aligned}&|\alpha \omega |+\bigl (|\alpha |+|\omega |\bigr )\Vert E\Vert +\Vert E\Vert ^2 \ge (1-\tau )^{2}|\overline{x}u\overline{y}v|. \end{aligned}$$

(4.45)

• (a) $\widetilde{A}=\begin{bmatrix} 0 &{} 1\\ \gamma &{} \omega \end{bmatrix}$, either $0 \le \gamma \le 1$, $\omega =0$ or $\gamma =1$, $\omega =i$

  Equations (4.42) for $\beta =1$, $0\le \gamma \le 1$ imply

  $$\begin{aligned} (1-\tau ^2)|\overline{x}v|\ge |\tau \gamma -1|-(\tau +1)\Vert E\Vert , \quad (1-\tau ^{2})|\overline{u}y|\ge |\gamma -\tau |-(1+\tau )\Vert E\Vert . \end{aligned}$$

  By combining these inequalities and making some trivial estimates we deduce

  $$\begin{aligned} (1-\tau ^{2})^2|\overline{y}u\overline{x}v|\ge |\tau \gamma -1|\,| \gamma -\tau |-(1+\tau )\bigl (\tau \gamma +1+\gamma +\tau \bigr )\Vert E\Vert -(1+\tau )^2\Vert E\Vert ^2. \end{aligned}$$

  Together with (4.45) for $\alpha =0$ and using $\Vert E\Vert \ge \Vert E\Vert ^{2}$ we get

  $$\begin{aligned}&(1+\tau )^2(1+|\omega |)\Vert E\Vert \ge |\tau \gamma -1|\,|\gamma -\tau | -(1+\tau )^{2}(\gamma +1)\Vert E\Vert -(1+\tau )^2\Vert E\Vert ,\nonumber \\&(1+\tau )^2\big (3+|\omega |+\gamma \big )\Vert E\Vert \ge |\tau \gamma -1| \,|\gamma -\tau |\ge |1-\gamma |\,|\gamma -\tau |. \end{aligned}$$

  (4.46)

  If $0\le \gamma <1$ (if $\gamma =1$) then the right-hand (the left-hand) side of (4.46) implies

  $$\begin{aligned} |\gamma -\tau |\le \left\{ \begin{array}{ll} \frac{(1+\tau )^{2}}{1-\gamma }\big (4+|\omega |\big )\Vert E\Vert , &{} 0\le \gamma <1\\ (1+\tau )\sqrt{(4+|\omega |)\Vert E\Vert }, &{} \gamma =1 \end{array}.\right. \end{aligned}$$

  (4.47)

  When either $\tau =0$, $\gamma > 0$ or $\tau =1$, $\gamma < 1$ (and $\Vert E\Vert$ is small enough), then (4.47) fails. If $0\le \gamma <1$ and $\Vert E\Vert \le \frac{(1-\gamma )^{2}}{2(1+\tau )^{2} (4+|\omega |)}$ (hence $1-\tau \ge |1-\gamma |-|\gamma -\tau |\ge \frac{1-\gamma }{2}$), then (4.43) for $\alpha =0$ (for $\omega =0$) yields $|\overline{x}u|\le \frac{2}{1-\gamma }\Vert E\Vert$ (and $|\overline{y}v|\le \frac{2}{1-\gamma }\Vert E\Vert$). Next, (4.42), (4.47) for $\beta =1$, $\gamma =0$, imply $|\overline{y}u|\le C \Vert E\Vert$ and $|\overline{x}v-c^{-1}|\le C \Vert E\Vert$ for some constant $C>0$ (see (C9) for $\widetilde{\tau }=0$, $0\le \tau <1$).

  By Lemma 4.1 (4.2) for $1\ge \tau > 0$, $\widetilde{A}=\begin{bmatrix} 0 &{} 1\\ \gamma &{} \omega \end{bmatrix}$ with $1\ge \gamma > 0$ and $\Vert E\Vert \le \frac{\gamma }{12}\le \frac{1}{12}$, we have $c^{-1}=(-1)^{k}+\overline{g}$, $k\in {\mathbb {Z}}$, $|g|\le \frac{12}{\gamma }\Vert E\Vert$, thus (4.42) for $\beta =1$ (and $\gamma \in {\mathbb {R}}$) gives

  $$\begin{aligned}&(1-\tau ^2)\overline{x}v=\big ((-1)^{k}(1-\tau \gamma ) -g\tau \gamma +\overline{g}\big )+(c^{-1}\epsilon _2 -\tau \overline{c}^{-1}\overline{\epsilon _3})\\&(1-\tau ^{2})\overline{y}u=(-1)^{k}(\gamma -\tau ) +\gamma \overline{g}-\tau g+(c^{-1}\epsilon _3 -\tau \overline{c}^{-1}\overline{\epsilon _2}). \end{aligned}$$

  We further obtain

  $$\begin{aligned} (1-\tau ^{2})|\overline{y}u|&\le (\gamma -\tau ) +(\tau \gamma +1)\tfrac{12}{\gamma }\Vert E\Vert +(1+\tau )\Vert E\Vert , \nonumber \\ (1-\tau ^2)\big |\overline{x}v-(-1)^{k}\big |&\le \tau (\gamma -\tau )+\tfrac{12(\tau \gamma +1)}{\gamma }\Vert E\Vert +(\tau +1)\Vert E\Vert . \end{aligned}$$

  (4.48)

  Using (4.47) for $0<\gamma <1$ we deduce that the left-hand sides of (4.48) are bounded by $D \Vert E\Vert$, where $D:=\frac{4(4+|\omega |)}{1-\gamma }+\frac{12(\gamma +1)}{\gamma }+2$. Thus either $1-\tau ^{2}\le \sqrt{D}\sqrt{\Vert E\Vert }$ and

  $$\begin{aligned} |1-\gamma |\le |\tau -\gamma |+|1-\tau | \le \tfrac{(1+\tau )^{2}}{1-\gamma }(2+|\omega |)\Vert E\Vert +\tfrac{\sqrt{D}\sqrt{\Vert E\Vert }}{2} \end{aligned}$$

  fails for small $\Vert E\Vert$, or we have $|\overline{y}u|,\big |\overline{x}v-(-1)^{k}\big |\le \sqrt{D}\sqrt{\Vert E\Vert }$ (see (C9) for $0<\tau _0,\tau <1$). The second equation of (4.40) with $\beta =1$, $c^{-1}=(-1)^{k}+\overline{g}$, $k\in {\mathbb {Z}}$, $|g|\le 12\Vert E\Vert$ gives

  $$\begin{aligned} \big |\overline{x}v+\overline{u}y-(-1)^{k}\big |-(1-\tau )| \overline{u}y|\le \big |\overline{x}v+\tau \overline{u}y-(-1)^{k}\big |\le 12\Vert E\Vert +\Vert E\Vert . \end{aligned}$$

  (4.49)

  From (4.40), (4.41), (4.47), (4.48), (4.49) for $\alpha =0$, $\omega \in \{0,i\}$, $\gamma =1$ we deduce (C4). If $\omega =i$, $\tau =1$ and $\Vert E\Vert <\frac{1}{13}$, then the second equality of (4.41) fails.

• (b) $\widetilde{A}=\alpha \oplus \omega$

  From (4.42) for $\beta =\gamma =0$ it follows that

  $$\begin{aligned} (1-\tau ^{2})|\overline{x}v|\le (1+\tau )\Vert E\Vert , \quad (1-\tau ^{2})|\overline{u}y|\le (1+\tau )\Vert E\Vert , \quad (1-\tau )^{2}|\overline{x}v\overline{u}y|\le \Vert E\Vert ^2. \end{aligned}$$

  (4.50)

  Next, (4.50) yields either $(1-\tau )|\overline{x}u|\le \Vert E\Vert$ or $(1-\tau )|\overline{y}v|\le \Vert E\Vert$.

  By Lemma 4.1 (4.2) for $0<\tau \le 1$, $\widetilde{A}=1\oplus e^{i\widetilde{\theta }}$, we have $c^{-1}=(-1)^{k}e^{-i\frac{\widetilde{\theta }+\pi }{2}}+\overline{g}$, $k\in {\mathbb {Z}}$, $|g|\le 12\Vert E\Vert$. We take the imaginary parts of (4.41) with $\alpha =1$, $\omega =e^{i\widetilde{\theta }}$, $0<\tau <1$ to deduce $|\cos \frac{\widetilde{\theta }}{{2}}|\le 14\Vert E\Vert$, which fails for $0\le \widetilde{\theta } <\pi$ and small $\Vert E\Vert$.

  By combining (4.50) with (4.44) for $|\alpha |=|\omega |=1$ and using $\Vert E\Vert \le \frac{1}{4}$, we get

  $$\begin{aligned} \tfrac{1}{4}(1-\tau )^{2}\le (1-\tau )^{2}(1-2\Vert E\Vert -\Vert E\Vert ^2) \le (1-\tau ^{2})^{2}|\overline{x}v\overline{u}y| \le (1+\tau )^{2}\Vert E\Vert ^{2}. \end{aligned}$$

  Thus $1-\tau \le 4\Vert E\Vert$. (In particular, we obtain a contradiction for $\tau =0$, $|\alpha |=|\omega |=1$.) When $\widetilde{\theta }=\pi$ (i.e. $\widetilde{A}=1\oplus -1$, $c^{-1}=(-1)^{k+1}+\overline{g}$, $k\in {\mathbb {Z}}$, $|g|\le 12\Vert E\Vert$), we use (4.40), (4.41) for $\beta =0$, $\alpha =-\omega =1$ to get $(1-\tau ) \textrm{Im}(\overline{x}u),(1-\tau ) \textrm{Im}(\overline{x}u)\le 13\Vert E\Vert$ and

  $$\begin{aligned}&|\overline{x}v+\overline{u}y|-2\Vert E\Vert \le |\overline{x}v +\overline{u}y|-(1-\tau )|\overline{u}y|\le |\overline{x}v +\tau \overline{u}y|\le \Vert E\Vert ,\nonumber \\&\bigl |2\textrm{Re}(\overline{x}u)-(-1)^{k+1}\bigr | =\tfrac{2}{1+\tau }\bigl |(1+\tau )\textrm{Re}(\overline{x}u) -(-1)^{k+1}+(-1)^{k+1}\tfrac{1-\tau }{2}\bigr |\le 30 \Vert E\Vert ,\nonumber \\&\bigl |2\textrm{Re}(\overline{y}v)-(-1)^{k}\bigr |=\tfrac{2}{1+\tau } \bigl |(1+\tau )\textrm{Re}(\overline{y}v)-(-1)^{k}+(-1)^{k} \tfrac{1-\tau }{2}\bigr |\le 30 \Vert E\Vert . \end{aligned}$$

  (4.51)

  It gives (C5). The first line of (4.51) is valid also for $\alpha \in \{0,1\}$, $\beta =\omega =0$ (see (4.50)). If $\alpha =1$, then (4.41) for $\tau =1$ yields $2c\textrm{Re}(\overline{x}u)=1+\epsilon _1$. By applying (4.3) for $\Vert E\Vert \le \frac{1}{2}$ we get $c=(-1)^{k}e^{i\psi }$, $k\in {\mathbb {Z}}$, $\psi \in (-\frac{\pi }{2},\frac{\pi }{2})$, $|\sin \psi |\le 2\Vert E\Vert$. Moreover, $\big |c-(-1)^{k}\big |=2|\sin \frac{\psi }{2}|\le 4\Vert E\Vert$. To conclude, (4.41), (4.43), (4.50) provides (C3).

This completes the proof of the lemma. $\square$

5 Proof of Theorem 3.4

To prove the nonexistence of some paths in the closure graph for bundles under (2.1), the proof of [23, Theorem 3.6] (the closure graph for orbits) applies mutatis mutandis; we shall not rewrite the proof in these cases, instead we refer to [23] for the proof. However, we reprove the existence of paths for bundles consisting of one orbit, since short and plausible arguments can be given (see e.g. (5.2)).

Proof of Theorem 3.4

Given normal forms $(\widetilde{A},\widetilde{B})$, (A, B) from Lemma 2.1 the existence of a path $(\widetilde{A},\widetilde{B})\rightarrow (A,B)$ in the closure graph for bundles for the action (2.1) immediately implies $\widetilde{A}\rightarrow A$, $\widetilde{B}\rightarrow B$. When this is not fulfilled, then $(\widetilde{A},\widetilde{B})\not \rightarrow (A,B)$ and we already have a lower estimate on the distance from $(\widetilde{A},\widetilde{B})$ to the bundle of (A, B) (see Lemmas 3.1,  3.2). Further, $(\widetilde{A},0_2)\rightarrow (A,0_2)$ (or $(0_2,\widetilde{B})\rightarrow (0_2,B)$) if and only if $\widetilde{A}\rightarrow A$ (or $\widetilde{B}\rightarrow B$), and trivially $(A,B)\rightarrow (A,B)$ for any A, B.

From now on suppose $\widetilde{A}\rightarrow A$, $\widetilde{B}\rightarrow B\ne 0$ with $(\widetilde{A},\widetilde{B})\not \in \textrm{Bun}_{\Psi }(A,B)$ and let

$$\begin{aligned} cP^{*}AP=\widetilde{A}+E, \quad P^{T}BP=\widetilde{B}+F, \qquad c\in S^{1},\,P\in GL_2({\mathbb {C}}),\quad E,F \in {\mathbb {C}}^{2\times 2}. \nonumber \\ \end{aligned}$$

(5.1)

Due to Lemmas 3.2 and 4.1 (1) the first equation of (5.1) yields restrictions on P, c, A imposed by $\Vert E\Vert$, $\widetilde{A}$. The trick of the proof is to use these to analyse the second equation of (5.1). We now work with equations with larger set of parameters than in [23, Theorem 3.6], and it usually makes the analysis more involved. If it eventually leads to an inequality that fails for any sufficiently small E and F, it will prove $(\widetilde{A},\widetilde{B})\not \rightarrow (A,B)$; it is then straightforward to estimate how small E, F should be, thus we omit this calculation. Otherwise, to prove $(\widetilde{A},\widetilde{B})\rightarrow (A,B)$, we find $c(s)\in S^{1}$, $P(s)\in GL_2({\mathbb {C}})$, $(A(s),B(s))\in \textrm{Bun}(A,B)$ such that

$$\begin{aligned} c(s) \bigl (P(s)\bigr )^*A(s)P(s)-\widetilde{A}=:E(s) {\mathop {\longrightarrow }\limits ^{s\rightarrow 0}} 0, \quad \bigl (P(s)\bigr )^T B(s)P(s)-\widetilde{B}=:F(s){\mathop {\longrightarrow }\limits ^{s\rightarrow 0}} 0. \nonumber \\ \end{aligned}$$

(5.2)

When we can arrange the parameter s so that $A(s)\rightarrow \widetilde{A}$ and $B(s)\rightarrow \widetilde{B}$, this is trivial.

Throughout the proof we denote $\delta =\nu \sqrt{\Vert E\Vert }$ for $\nu >0$ (Lemma 3.2 (2)), $\epsilon =\Vert F\Vert$,

$$\begin{aligned} B=\begin{bmatrix} a &{} b\\ b &{} d \end{bmatrix}, \quad \widetilde{B}=\begin{bmatrix} \widetilde{a} &{} \widetilde{b}\\ \widetilde{b} &{} \widetilde{d} \end{bmatrix}, \qquad F=\begin{bmatrix} \epsilon _1 &{} \epsilon _2\\ \epsilon _2 &{} \epsilon _4 \end{bmatrix}, \qquad P=\begin{bmatrix} x &{} y\\ u &{} v \end{bmatrix}, \end{aligned}$$

where sometimes polar coordinates for x, y, u, v in P might be preferred:

$$\begin{aligned} x=|x|e^{i\phi },\quad y=|y|e^{i\varphi },\quad u=|u|e^{i\eta }, \quad v=|v|e^{i\kappa }, \qquad \phi ,\varphi ,\eta ,\kappa \in {\mathbb {R}}. \end{aligned}$$

(5.3)

The second matrix equation of (5.1) can thus be written componentwise as:

$$\begin{aligned}&ax^2+2bux+du^2=\widetilde{a}+\epsilon _1,\nonumber \\&axy+buy+bvx+duv=\widetilde{b}+\epsilon _2,\nonumber \\&ay^2+2bvy+dv^2=\widetilde{d}+\epsilon _4. \end{aligned}$$

(5.4)

For the sake of simplicity some estimates in the proof are crude, and it is always assumed $\epsilon ,\delta \le \frac{1}{2}$. Since we shall often apply Lemma 4.1, we take for granted that $(\frac{\delta }{\nu })^{2}=\Vert E\Vert \le \left\{ \begin{array}{ll} \min \{\tfrac{|\det \widetilde{A}|}{8\Vert \widetilde{A}\Vert +4},1\}, &{} \det \widetilde{A}\ne 0\\ 1 &{} \det \widetilde{A}=0 \end{array}\right.$, $\epsilon =\Vert F\Vert \le \left\{ \begin{array}{ll} \min \{\tfrac{|\det \widetilde{B}|}{8\Vert \widetilde{B}\Vert +4},1\}, &{} \det \widetilde{B}\ne 0\\ 1 &{} \det \widetilde{B}=0 \end{array}\right.$. If $A,\widetilde{A}$ are nonsingular we also assume $\Vert E\Vert \le \Vert \widetilde{A}^{-1}\Vert ^{-1}$, while for $B,\widetilde{B}$ nonsingular with $1=|\det A|=|\det \widetilde{A}|=\Vert \widetilde{A}\Vert$ it is assumed $\Vert E\Vert ,\Vert F\Vert \le \frac{|\det \widetilde{B}|}{4(4\max \{1,\Vert \widetilde{B}\Vert ,|\det \widetilde{B}|\}+2)^{2}}$.

We split our analysis into several cases (see Lemma 2.1 for normal forms). The notation $(\widetilde{A},\widetilde{B}) \dashrightarrow (A,B)$ is used when the existence of a path is yet to be considered.

Case I.:

$(1\oplus e^{i\widetilde{\theta }},\widetilde{B})\dashrightarrow (1\oplus e^{i\theta },B)$, $0\le \widetilde{\theta }\le \pi$, $0\le \theta \le \pi$

• (a) $0<\widetilde{\theta },\theta <\pi$

  From Lemma 3.2 (2) for (C7) we get

  $$\begin{aligned} |y|^{2},|u|^{2}\le \delta , \qquad \bigl ||v|^{2}-1 \bigr |,\, \bigl ||x|^{2}-1\bigr |\le \delta . \end{aligned}$$

  (5.5)

  • (i) $B=\begin{bmatrix} a &{} b\\ b &{} 0 \end{bmatrix}$, $b,a\ge 0$

    Using (5.5) and Lemma 3.1 (D2) we get a contradiction for small $\epsilon ,\delta$ and $\widetilde{d}\ne 0$. Next, Lemma 4.1 (3) for $\widetilde{d}=0$ gives $b^{2}=|\widetilde{b}|^{2}+\delta _5$, $|\delta _5|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\} \big (4\max \{1,|\widetilde{b}|^{2},|\widetilde{b}|\}+2\big )^{2}$. It fails for $b=0$, $\widetilde{b}\ne 0$ and $\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}<\widetilde{b}^{2} \big (4\max \{1,|\widetilde{b}|^{2},|\widetilde{b}|\}+2\big )^{-2}$, while the case $b=\widetilde{b}=0$ is trivial. For $a=0$, $\widetilde{a}\ne 0$ then the first equation of (5.4) for $a=d=0$ and (5.5) yields an inequality that fails for $\epsilon ,\delta$ small enough:

    $$\begin{aligned}&|\widetilde{a}|=|\epsilon _1-2bux|\le \epsilon +2(\widetilde{b}+\delta _5)\sqrt{\delta (1+\delta )}. \end{aligned}$$

  • (ii) $B=\begin{bmatrix} 0 &{} b\\ b &{} d \end{bmatrix}$, $b\ge 0$, $d\ne 0$

    Due to a symmetry we deal with this case similarly as with Case I (a) (i).

  • (iii) $B=a\oplus d$,    $a,d>0$

    From (5.4) for $b=0$ we obtain

    $$\begin{aligned}&ax^2+du^2=\widetilde{a}+\epsilon _1, \nonumber \\&axy+duv=\widetilde{b}+\epsilon _2, \nonumber \\&ay^2+dv^2=\widetilde{d}+\epsilon _4. \end{aligned}$$

    (5.6)

    By multiplying the first and the last equation of (5.6) by $\delta _6=\frac{y}{x}$ and $\delta _5=\frac{u}{v}$, respectively, and by slightly simplifying them, we get

    $$\begin{aligned} axy+duv\delta _5\delta _6=\delta _6(\widetilde{a}+\epsilon _1), \qquad axy\delta _5\delta _6+duv=\delta _5(\widetilde{d}+\epsilon _4). \end{aligned}$$

    Adding these two equations and using the second equation of (5.6) we deduce

    $$\begin{aligned} (\widetilde{b}+\epsilon _2)(1+\delta _5\delta _6)=\delta _5 (\widetilde{d}+\epsilon _4)+\delta _6(\widetilde{a}+\epsilon _1), \end{aligned}$$

    which fails for $\widetilde{b}\ne 0$ and sufficiently small $\epsilon ,\delta$ (by (5.5) we have $|\delta _5|,|\delta _6|\le \frac{\delta }{1-\delta }$ ).

• (b) $\widetilde{\theta }\in \{0,\pi \}$

  Set $\sigma =e^{i\widetilde{\theta }}\in \{1,-1\}$. Lemma 3.2 (C11) yields

  $$\begin{aligned} |x|^2+\sigma |u|^2=(-1)^{k}+\delta _1, \quad \overline{x}y+\sigma \overline{u} v=\delta _2, \quad |y|^2+\sigma |v|^2=\sigma (-1)^{k}+\delta _4, \nonumber \\ \end{aligned}$$

  (5.7)

  where $|\delta _1|,|\delta _2|,|\delta _4|\le \delta$. Next, for $v\ne 0$, $\big (|x|-|u|\big )^{2}\le \big ||x|^2-|u|^2\big |=: 1+\delta _1'$ we deduce

  $$\begin{aligned} |\overline{x}y+\sigma \overline{u} v|&\ge \big ||\overline{x}y|-|\overline{x}v|+|\overline{x}v| -|\overline{u} v|\big |\ge |v|\big |\overline{x}|-| \overline{u} |\big |-|\overline{x}|\big ||y|-|v|\big | \nonumber \\&\ge \tfrac{\big |\overline{x}|^{2}-|\overline{u} |^{2} \big |}{\frac{1}{|v|}(|\overline{x}|+|\overline{u} |)} -\tfrac{|x|\big |\overline{y}|^{2}-|\overline{v} |^{2} \big |}{|\overline{y}|+|\overline{v} |} \ge \tfrac{1-|\delta _1'|}{2\frac{|u|}{|v|} +\frac{\sqrt{1+|\delta _1'|}}{|v|}} -\bigg (\tfrac{|u|}{|v |}+\tfrac{\sqrt{1+|\delta _1'|}}{|v|}\bigg ) \big ||\overline{y}|^{2}-|\overline{v} |^{2}\big |. \end{aligned}$$

  (5.8)

  • (i) $B=\begin{bmatrix} a &{} b\\ b &{} d \end{bmatrix}$, $a,d,b\ge 0$, $a+d\ne 0$

    Let first $B=a\oplus d$. Using the notation (5.3) the following calculation is trivial:

    $$\begin{aligned}&ax^{2}+du^{2}= ae^{2i\phi }\big (|x|^{2}+\sigma |u|^{2}\big ) -u^{2} (\sigma a e^{2i(\phi -\eta )}-d), \qquad \sigma \in \{-1,1\},\nonumber \\&ay^{2}+dv^{2}=a e^{2i\varphi }\big (|y|^{2}+\sigma |v|^{2}\big ) -v^{2} (\sigma ae^{2i(\varphi -\kappa )}-d),\nonumber \\&ay^{2}+dv^{2}=d\sigma e^{2i\kappa }\big (|y|^{2}+\sigma |v|^{2}\big ) -y^{2} (\sigma de^{2i(\kappa -\varphi )}-a). \end{aligned}$$

    (5.9)

    Furthermore, one easily writes:

    $$\begin{aligned} a xy+d uv&=ae^{2i\phi }(\overline{x}y+\sigma \overline{u}v) -uv(\sigma a e^{2i(\phi -\eta )}-d), \qquad \sigma \in \{-1,1\},\nonumber \\ a xy+d uv&=ae^{2i\varphi }(x\overline{y}+\sigma u\overline{v}) -uv(\sigma a e^{2i(\varphi -\kappa )}-d),\nonumber \\ a xy+d uv&=d\sigma e^{2i\kappa }(x\overline{y}+\sigma u \overline{v})-xy(d\sigma e^{2i(\kappa -\varphi )}-a). \end{aligned}$$

    (5.10)

    Rearranging the terms in (5.9), (5.10) and using (5.6), (5.7) yields for $\sigma \in \{-1,1\}$:

    $$\begin{aligned}&u^{2}(\sigma a e^{2i(\phi -\eta )}-d)= ae^{2i\phi } ((-1)^{k}+ \delta _1)-\widetilde{a}-\epsilon _1,\\&uv(\sigma a e^{2i(\phi -\eta )}-d)=ae^{2i\phi } \delta _2 -\widetilde{b}- \epsilon _2, \\&v^{2}(\sigma a e^{2i(\varphi -\kappa )}-d)= a e^{2i\varphi }(\sigma (-1)^{k}+ \delta _4)-\widetilde{d}-\epsilon _4,\\&uv(\sigma a e^{2i(\varphi -\kappa )}-d)= a e^{2i\varphi }\overline{\delta }_2-\widetilde{b}-\epsilon _2,\\&y^{2}(\sigma d e^{2i(\kappa -\varphi )}-a) = d\sigma e^{2i\kappa }(\sigma (-1)^{k}+ \delta _4) -\widetilde{d}-\epsilon _4,\\&xy(d\sigma e^{2i(\kappa -\varphi )}-a)= d\sigma \overline{\delta }_2-\widetilde{b}-\epsilon _2. \end{aligned}$$

    By dividing the equations in each line we get

    $$\begin{aligned} \tfrac{u}{v}=\tfrac{ae^{2i\phi }((-1)^{k}+ \delta _1) -\widetilde{a}-\epsilon _1}{ae^{2i\phi } \delta _2-\widetilde{b}- \epsilon _2} =\tfrac{ ae^{2i\varphi }\overline{\delta }_2-\widetilde{b} -\epsilon _2}{ae^{2i\varphi }(\sigma (-1)^{k}+ \delta _4) -\widetilde{d}-\epsilon _4},\qquad \tfrac{x}{y} =\tfrac{d\sigma \overline{\delta }_2-\widetilde{b} -\epsilon _2}{d\sigma e^{2i\kappa }(\sigma (-1)^{k} + \delta _4)-\widetilde{d}-\epsilon _4}. \end{aligned}$$

    (5.11)

    If $\widetilde{B}=\begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 0 \end{bmatrix}$ for $\widetilde{b}>0$ (hence $\sigma =-1$, $d\ge a> 0$), then Lemma 4.1 (3) implies $a^{2}\le ad=\widetilde{b}^{2}+\epsilon '$, $|\epsilon '|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\}\big (4\max \{1,| \widetilde{b}|,|\widetilde{b}|^{2}\}+2\big )^{2}$. From the first equation of (5.11) for $\widetilde{d}=\widetilde{a}=0$, $\sigma =1$ we now obtain a contradiction for small $\epsilon ,\delta$. Similarly, it follows from Lemma 4.1 (3) for $\widetilde{B}=\widetilde{a}\oplus \widetilde{d}$ and $B=aI_2$ that $a^{2}=\widetilde{a}\widetilde{d}+\epsilon '$, $|\epsilon '|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\}(4\max \{1,|\widetilde{d}|,| \widetilde{d}\widetilde{a}|\}+2)^{2}$. If $\widetilde{d}>\widetilde{a}> 0$, then the first equation of (5.11) (with $\sigma \in \{-1,1\}$, $\widetilde{b}=0$) fails as well. Next, when $\widetilde{a}=0$ we have $a^{2}=\epsilon '$. Hence (5.11) for $\sigma \in \{-1,1\}$, $\widetilde{a}=\widetilde{b}=0$ yields $|\frac{u}{v}|,|\frac{x}{y}|\le \frac{\epsilon +\sqrt{\epsilon '}\delta }{\widetilde{d} -\epsilon -\sqrt{\epsilon '}(1+\delta )}$. Further, the third equation of (5.9) with (5.6), (5.7) for $a=d=\sqrt{\epsilon '}$, $\widetilde{b}=0$ gives $\frac{1}{|v|^{2}}\le \frac{2\epsilon '}{\widetilde{d}-\epsilon -\epsilon '(1+\delta )}$. We apply this and (5.7) to (5.8) to deduce a contradiction for small $\epsilon ,\delta$ and $\widetilde{d}\ne 0$.

    Take $P(s)=\frac{1}{\sqrt{\widetilde{d}+\sigma \widetilde{a}}} \begin{bmatrix} -i\sqrt{\widetilde{d}} &{} \sqrt{\widetilde{a}}\\ i\sqrt{\widetilde{a}} &{} \sigma \sqrt{\widetilde{d}} \end{bmatrix}$, $B(s)=\begin{bmatrix} 0 &{} \sqrt{\widetilde{a}\widetilde{d}}+s\\ \sqrt{\widetilde{a}\widetilde{d}}+s &{} \widetilde{d}-\sigma \widetilde{a}+s \end{bmatrix}$, $c(s)=1$, $e^{i\theta }\rightarrow \sigma$ in (5.2) to see $\big (1\oplus \sigma , \widetilde{a}\oplus \widetilde{d}\big ) \rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} 0 &{} b\\ b &{} d \end{bmatrix}\big )$, and $P(s) =\frac{1}{\sqrt{\widetilde{d}+\sigma \widetilde{a}}} \begin{bmatrix} i\sqrt{\widetilde{a}} &{} \sigma \sqrt{\widetilde{d}}\\ -i\sqrt{\widetilde{d}} &{} \sqrt{\widetilde{a}} \end{bmatrix}$, $B(s)=\begin{bmatrix} \widetilde{d}-\sigma \widetilde{a}+s &{} \sqrt{\widetilde{a}\widetilde{d}}+s\\ \sqrt{\widetilde{a}\widetilde{d}}+s &{} 0 \end{bmatrix}$, $c(s)=\sigma$, $e^{i\theta }\rightarrow \sigma$ to show$\big (1\oplus \sigma , \widetilde{a}\oplus \widetilde{d}\big ) \rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} a &{} b\\ b &{} 0 \end{bmatrix}\big )$, $0<\theta <\pi$.

  • (ii) $B=\begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}$, $b>0$

    From (5.4) for $a=d=0$ we obtain

    $$\begin{aligned}&2bux=\widetilde{a}+\epsilon _1, \nonumber \\&buy+bvx=\widetilde{b}+\epsilon _2,\nonumber \\&2bvy=\widetilde{d}+\epsilon _4. \end{aligned}$$

    (5.12)

    It suffices to take $0\le \widetilde{a}\le \widetilde{d}$, $\widetilde{d}> 0$, $\widetilde{b}=0$. By Lemma 4.1 (3) and (4.7) we have $b=\sqrt{\widetilde{a}\widetilde{d}}+\delta _5>0$, $|\delta _5|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\}(4\max \{1,\widetilde{d}, \widetilde{a}\widetilde{d}\}+2)^{2}$. Thus (5.12) and (5.7) give:

    $$\begin{aligned} |v|^{2},|y|^{2}\le \tfrac{\widetilde{d}+\epsilon }{2(\sqrt{\widetilde{a}\widetilde{d}}+\delta _5)}+1+\delta ,\quad |u|^{2},|x|^{2}\le \tfrac{\widetilde{a}+\epsilon }{2(\sqrt{\widetilde{a}\widetilde{d}}+\delta _5)}+1+\delta . \end{aligned}$$

    (5.13)

    Using Lemma 3.1 (D1), (D2) for $\det \widetilde{B}=\widetilde{a}\widetilde{d}$ we get

    $$\begin{aligned} \begin{array}{ll} u(\widetilde{d}+\epsilon _4)=v\big (-i(-1)^{l} \sqrt{\widetilde{a}\widetilde{d}}+\epsilon _2''\big )\\ x(\widetilde{d}+\epsilon _4)=y\big (i(-1)^{l} \sqrt{\widetilde{a}\widetilde{d}}+\epsilon _2'\big ) \end{array}, |\epsilon _2'|,|\epsilon _2''| \le \left\{ \begin{array}{ll} \tfrac{\epsilon (4\max \{\widetilde{d},\widetilde{a}\}+2 +\widetilde{d}\widetilde{a})}{\widetilde{d}\widetilde{a}}, &{} \widetilde{a}\widetilde{d}\ne 0\\ \scriptstyle {\sqrt{\epsilon } (4\max \{\widetilde{d}, \widetilde{a}\}+3)^{\frac{1}{2}}}, &{} \widetilde{a}\widetilde{d}= 0 \end{array}\right. . \end{aligned}$$

    (5.14)

    By further applying the first and the third equality of (5.7) we deduce

    $$\begin{aligned} (-1)^{k}+\delta _1&=|x|^2+\sigma |u|^2=\tfrac{|(-1)^{l} \sqrt{\widetilde{a}\widetilde{d}}+\epsilon _2'|^{2}}{|\widetilde{d}+\epsilon _4|^{2}}|y|^{2}+\sigma \tfrac{|-(-1)^{l}\sqrt{\widetilde{a}\widetilde{d}} +\epsilon _2''|^{2}}{|\widetilde{d}+\epsilon _4|^{2}}|v|^{2}\nonumber \\&=\tfrac{|-(-1)^{l}\sqrt{\widetilde{a}\widetilde{d}} +\epsilon _2'|^{2}}{|\widetilde{d}+\epsilon _4|^{2}} \big (\sigma (-1)^{k}+\delta _4\big )+\delta '|v|^{2} \end{aligned}$$

    (5.15)

    with $|\delta '|\le C\max \{\epsilon ,\delta \}$, where $C>0$ is a constant that can be computed easily.

    By combining (5.15) and (5.13) we obtain a contradiction for $0<\widetilde{a}< \widetilde{d}$ and sufficiently small $\epsilon , \delta$. Next, let $\widetilde{a}=0$, $\widetilde{d}>0$. From (5.14) it follows $|\frac{u}{v}|\le \frac{|\epsilon _2'|}{|\widetilde{d}|-\epsilon }$, $|\frac{x}{y}|\le \frac{|\epsilon _2''|}{|\widetilde{d}|-\epsilon }$ ($y=0$ or $v=0$ would contradict (5.12) for $|\widetilde{d}|> \epsilon$). By applying this with (5.7) and (5.15) (hence |v| is large) to (5.8), we obtain a contradiction for small $\epsilon , \delta$.

    We take $P(s)=\frac{1}{\sqrt{2}} \begin{bmatrix} 1 &{} i\\ 1 &{} -i \end{bmatrix}$, $B(s)=(\widetilde{d}+s) \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}$ and $c(s)=1$, $e^{i\theta }\rightarrow 1$ in (5.2) to prove $\big (I_2, \widetilde{d}I_2\big ) \rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}\big )$, $b>0$, $0\le \widetilde{d}$, $0<\theta <\pi$. Using (5.7) for $\sigma =-1$ leads to

    $$\begin{aligned} \delta \ge |\overline{x}y-\overline{u}v|\ge \big ||x|^{2}| \tfrac{y}{x}|-|u|^{2}|\tfrac{v}{u}|\big |\ge \big ||x|^{2} -|u|^{2}\big |-|x|^{2}\big (1-|\tfrac{y}{x}|\big )-|u|^{2} \big (1-|\tfrac{v}{u}|\big ). \end{aligned}$$

    (5.16)

    From (5.14) we get that $|\frac{y}{x}|,|\frac{v}{u}|$ are close to 1, and (5.13) implies that $|u|^{2},|x|^{2}$ are bounded. Thus the last two terms on the right-hand side of (5.16) are small, while the first one is close to 1 (see (5.7) for $\sigma =-1$). For small $\epsilon ,\delta$ we get a contradiction.

Case II.:

$\bigl (\begin{bmatrix} 0 &{} 1\\ \widetilde{\tau } &{} 0 \end{bmatrix}, \begin{bmatrix} \widetilde{a} &{} \widetilde{b}\\ \widetilde{b} &{} \widetilde{d} \end{bmatrix}\bigr )\dashrightarrow \bigl (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} a &{} b\\ b &{} d \end{bmatrix}\bigr )$, $\widetilde{b},b \ge 0$, $(\widetilde{\tau },\tau )\in \big ([0,1)\times (0,1)\big )\cup \{(0,0)\}$

By Lemma 3.2 (2) for (C9) we have

$$\begin{aligned} |xu|,|yu|,|vy|\le \delta , \qquad \big ||vx|-1\big |\le \delta . \end{aligned}$$

(5.17)

It yields $\delta _6=\frac{y}{x}=\frac{yv}{xv}$ with $|\delta _6|\le \frac{\delta }{1-\delta }\le 2\delta$, $\delta _5=\frac{u}{v}=\frac{ux}{xv}$ with $|\delta _5|\le \frac{\delta }{1-\delta }\le 2\delta$ and $\delta _7=\frac{uy}{vx}$ with $|\delta _7|\le 2\delta$ (note $\delta \le \frac{1}{2}$).

• (a) $B=\begin{bmatrix} 0 &{} b\\ b &{} d \end{bmatrix}$, $b\ge 0$, $|d|\in \{0,1\}$, $|b|+|d|\ne 0$

  By multiplying the last two equations of (4.13) by $\delta _5=\frac{u}{v}$ and using $\delta _7=\frac{uy}{vx}$ we get

  $$\begin{aligned} du^{2}+(1+\delta _7)bux=(\widetilde{b}+\epsilon _2)\delta _5, \qquad 2\delta _7 bvx+dvu=(\widetilde{d}+\epsilon _4)\delta _5. \end{aligned}$$

  (5.18)

  Subtracting the first and the second equation of (5.18) from the first and the second equation of (4.13) (in the form $duv+b(1+\delta _7)vx=\widetilde{b}+\epsilon _2$), we deduce

  $$\begin{aligned} (1-\delta _7)bux=\widetilde{a}+\epsilon _1-(\widetilde{b}+\epsilon _2) \delta _5,\quad (1-\delta _7)bvx=\widetilde{b}+\epsilon _2 -(\widetilde{d}+\epsilon _4)\delta _5. \end{aligned}$$

  (5.19)

  It is clear that the first (the second) equality in (5.19) fails for $\widetilde{a}\ne 0$ (for $\widetilde{b}\ne 0$) and $b=0$, provided that $\epsilon ,\delta$ are sufficiently small. Next, from the second equation of (5.19) and using $vx=e^{i\vartheta }-\delta _0$ with $|\delta _0|\le \delta$, $\vartheta \in {\mathbb {R}}$ (see (5.17)) we obtain

  $$\begin{aligned} b=\tfrac{\widetilde{b}+\epsilon _2-(\widetilde{d}+\epsilon _4) \delta _5}{(1-\delta _7)(e^{i\vartheta }-\delta _0)}=e^{-i\vartheta } \widetilde{b}+\tfrac{e^{-i\vartheta }\widetilde{b}(\delta _7 +e^{i\vartheta }\delta _0-\delta _0\delta _7)+\epsilon _2 -(\widetilde{d}+\epsilon _4)\delta _5}{(1-\delta _7) (e^{i\vartheta }-\delta _0)} \end{aligned}$$

  (5.20)

  From (5.20) and $|ux|\le \delta$ (and $|yv|\le \delta$) we get that the first equation of (5.19) fails for $\widetilde{a}\ne 0$ (the last equation of (4.13) fails for $\widetilde{d}\ne 0$, $d=0$), and $\epsilon ,\delta$ small enough.

  Finally, it is easy to check that $P(s)= \begin{bmatrix} s^{-1} &{} 0\\ s^2 &{} s \end{bmatrix}$, $B(s)= \begin{bmatrix} 0 &{} b(s)\\ b(s) &{} d \end{bmatrix}$ with $b(s)\rightarrow \widetilde{b}$, $A(s)= \begin{bmatrix} 0 &{} 1\\ \widetilde{\tau }+s &{} 0 \end{bmatrix}$, $c(s)=1$ in (5.2) proves $\big (\begin{bmatrix} 0 &{} 1\\ \widetilde{\tau } &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 0 \end{bmatrix}\big ) \rightarrow \big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} b\\ b &{} d \end{bmatrix}\big )$, $b\ge \widetilde{b}\ge 0$.

• (b) $B=\begin{bmatrix} 1 &{} b\\ b &{} 0 \end{bmatrix}$, $b\ge 0$, $\tau =0$

  We argue similarly as in Case II (a). We have equations (4.17); by multiplying the first two equations by $\delta _6=\frac{y}{x}$ and using $\delta _7=\frac{uy}{vx}$ we obtain

  $$\begin{aligned} ay^{2}+(1+\delta _7)bvy=(\widetilde{b}+\epsilon _2)\delta _6, \qquad 2\delta _7 bvx+axy=(\widetilde{a}+\epsilon _1)\delta _6. \end{aligned}$$

  (5.21)

  Subtracting the first and the second equation of (5.21) from the last and the second equation of (4.13) (written as $axy+b(1+\delta _7)vx =\widetilde{b}+\epsilon _2$), respectively, we get

  $$\begin{aligned} (1-\delta _7)bvy=\widetilde{d}+\epsilon _4-(\widetilde{b}+\epsilon _2) \delta _5,\quad (1-\delta _7)bvx=\widetilde{b}+\epsilon _2 -(\widetilde{a}+\epsilon _1) \delta _6. \end{aligned}$$

  (5.22)

  The first (the second) equality in (5.22) fails for $\widetilde{d}\ne 0$ (for $\widetilde{b}\ne 0$) and $b=0$, provided that $\epsilon ,\delta$ are sufficiently small. We obtain a similar expression for b as in (5.20). It yields a contradiction for $b=0$, $\widetilde{b}\ne 0$ and $\delta ,\epsilon$ small enough, while by combining it with $|yv|\le \delta$ (and $|ux|\le \delta$) we contradict the first equation of (5.22) for $\widetilde{d}\ne 0$ (or (4.17) for $\widetilde{a}\ne 0$, $a=0$), provided that $\epsilon ,\delta$ are small. Take $P(s)= \begin{bmatrix} s &{} s^2\\ 0 &{} s^{-1} \end{bmatrix}$, $B(s)=\begin{bmatrix} 1 &{} b(s)\\ b(s) &{} 0 \end{bmatrix}$, $b(s)\rightarrow \widetilde{b}$, $c(s)=1$ in (5.2) to prove $\big (\begin{bmatrix} 0 &{} 1\\ 0 &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 0 \end{bmatrix}\big ) \rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 0 &{} 0 \end{bmatrix}, \begin{bmatrix} 1 &{} b\\ b &{} 0 \end{bmatrix}\big )$, $b\ge \widetilde{b}$.

• (c) $B=1\oplus d$, $d\in {\mathbb {C}}$ ($0<\tau <1$) or $B=a\oplus 1$, $a>0$ ($\tau =0$)

  Since $|\frac{y}{x}|\le \frac{\delta }{1-\delta }$ and $|\frac{u}{v}|\le \frac{\delta }{1-\delta }$ the same proof as in Case I (a) (iii) applies.

  From (5.2) for $P(s)=\begin{bmatrix} s &{} s^2\\ 0 &{} s^{-1} \end{bmatrix}$, $B(s)=1\oplus s^{2}\widetilde{d}$ and $P(s)= \sqrt{\widetilde{a}}\oplus \frac{1}{\sqrt{\widetilde{a}}}$, $B(s)= 1\oplus \widetilde{a}\widetilde{d}$ with $\tau \rightarrow \widetilde{\tau }$, $c(s)=1$, in (5.2) we obtain $\big (\begin{bmatrix} 0 &{} 1\\ \widetilde{\tau } &{} 0 \end{bmatrix}, \widetilde{a}\oplus \widetilde{d}\big ) \rightarrow \big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, 1\oplus d\big )$ (with $0<\tau <1$) for $\widetilde{a}=0$ and $\widetilde{a}>0$, respectively. Finally, $P(s)=\begin{bmatrix} s^{-1} &{} 1\\ s^2 &{} s \end{bmatrix}$, $B(s)=(\widetilde{d}s^{2}+s^{3})\oplus 1$ with $c(s)=1$ gives $\big (\begin{bmatrix} 0 &{} 1\\ 0 &{} 0 \end{bmatrix}, \widetilde{d}\oplus 0\big ) \rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 0 &{} 0 \end{bmatrix}, a\oplus 1\big )$, $a>0$, $\widetilde{d}\in \{0,1\}$.

• (d) $B=\begin{bmatrix} e^{i\varphi } &{} b\\ b &{} \zeta \end{bmatrix}$, $\zeta \in {\mathbb {C}}$, $\varphi \in [0,\pi )$, $\tau \in (0,1)$ or $B=\begin{bmatrix} \zeta ^{*} &{} b\\ b &{} 1 \end{bmatrix}$, $\zeta ^{*}\in {\mathbb {C}}^{*}$, $\tau =0$; $b>0$

  Let $B=\begin{bmatrix} e^{i\varphi } &{} b\\ b &{} \zeta \end{bmatrix}$, $\zeta \in {\mathbb {C}}$, $0\le \varphi <\pi$. If $\widetilde{B}$ is either $\begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} \widetilde{d} \end{bmatrix}$ or $\begin{bmatrix} \widetilde{\zeta } &{} \widetilde{b}\\ \widetilde{b} &{} 1 \end{bmatrix}$ with $\widetilde{\zeta }\ne 0$ we take $P(s)=\begin{bmatrix} s &{} s^{2}\\ 1 &{} s^{-1} \end{bmatrix}$, $B(s)=\begin{bmatrix} e^{i\varphi } &{} \widetilde{b}+s\\ \widetilde{b}+s &{} \widetilde{d}s^{2} \end{bmatrix}$ or $P(s)=|\widetilde{\zeta }| e^{i\frac{k\pi }{2}}\oplus \frac{1}{|\widetilde{\zeta }|} e^{i\frac{k\pi }{2}}$, $c(s)=(-1)^{k}$, $B(s)= \begin{bmatrix} e^{i\varphi } &{} \widetilde{b}+s\\ \widetilde{b}+s &{} (-1)^{k}|\widetilde{\zeta }|^{2} \end{bmatrix}$ with $\arg \widetilde{\zeta }=\arg (\varphi +k\pi )$ in (5.2) to get a path. Next, $B(s)= \begin{bmatrix} \widetilde{a}s^{2}+s^{3} &{} \widetilde{b}+s\\ \widetilde{b}+s &{} 1 \end{bmatrix}$, $c(s)=1$, $P(s)=\begin{bmatrix} s^{-1} &{} 1\\ s^2 &{} s \end{bmatrix}$ shows $\big (\begin{bmatrix} 0 &{} 1\\ 0 &{} 0 \end{bmatrix}, \begin{bmatrix} \widetilde{a} &{} \widetilde{b}\\ \widetilde{b} &{} 0 \end{bmatrix}\big ) \rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 0 &{} 0 \end{bmatrix}, \begin{bmatrix} a &{} b\\ b &{} 1 \end{bmatrix}\big )$, $\widetilde{b}\ge 0$, $\widetilde{a}\in \{0,1\}$.

Case III.:

$(1\oplus -1, \widetilde{B}) \dashrightarrow \big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, B\big )$, $0<\tau \le 1$

Lemma 3.2 (2) with (C5) for $\alpha =-\omega =1$, $\beta =0$ gives ($|\delta _1|,|\delta _2|,|\delta _4|\le \delta$):

$$\begin{aligned} 2\textrm{Re}(\overline{x}u)=(-1)^{k}+\delta _1, \quad 2 \textrm{Re}(\overline{y}v)=-(-1)^{k}+\delta _2, \quad \overline{x}v+\overline{u}y=\delta _4, \quad 1-\tau , \quad k\in {\mathbb {Z}}, \end{aligned}$$

(5.23)

Observe that $u,v\ne 0$, otherwise (5.23) fails. We compute

$$\begin{aligned} \overline{x}v+\overline{u}y&=e^{-2i\phi }(xv-yu) +2\cos (\phi -\eta )e^{-i(\phi +\eta )}uy =e^{-2i\phi }\det P +\textrm{Re}(\overline{x}u)\tfrac{y}{x}, \end{aligned}$$

(5.24)

$$\begin{aligned} \overline{x}v+\overline{u}y&=-e^{-2i\eta }\det P +2\cos (\phi -\eta )e^{-i(\phi +\eta )}vx =-e^{-2i\eta }\det P +\textrm{Re}(\overline{x}u)\tfrac{v}{u}.\nonumber \\ \end{aligned}$$

(5.25)

Therefore, by combining (5.24) and (5.25) with (5.23) we obtain

$$\begin{aligned} \tfrac{y}{x}=\tfrac{\delta _4-e^{-2i\phi }\det P}{(-1)^{k}+\delta _1}, \qquad \tfrac{v}{u}=\tfrac{\delta _4+e^{-2i\eta }\det P}{(-1)^{k}+\delta _1}. \end{aligned}$$

(5.26)

• (a) $B=a\oplus d$,    $a\ge 0$

  Equations (5.6) and (5.26) yield

  $$\begin{aligned} \widetilde{b}+\epsilon _2&= axy+duv= \tfrac{1}{(-1)^{k} +\delta _1}\big (ax^{2}(\delta _4-e^{-2i\phi }\det P) +du^{2}(\delta _4+e^{-2i\eta }\det P) \big )\nonumber \\&= \tfrac{1}{(-1)^{k}+\delta _1}\big (\delta _4(ax^{2}+du^{2}) +\det P(-ax^{2}e^{-2i\phi }+du^{2}e^{-2i\eta }) \big )\nonumber \\&= \tfrac{1}{(-1)^{k}+\delta _1}\big (\delta _4(\widetilde{a}+\epsilon _1) +\det P(-a|x|^{2}+d|u|^{2}) \big ), \end{aligned}$$

  (5.27)

  and further for $a,\widetilde{a}\in {\mathbb {R}}$:

  $$\begin{aligned} \widetilde{d}+\epsilon _4&= ay^{2}+dv^{2}=\tfrac{1}{((-1)^{k} +\delta _1)^{2}}\big (ax^{2}(\delta _4-e^{-2i\phi }\det P)^{2} +du^{2}(\delta _4+e^{-2i\eta }\det P)^{2} \big )\nonumber \\&= \tfrac{1}{((-1)^{k}+\delta _1)^{2}}\big ( \delta _4^{2}(ax^{2}+du^{2}) +2\delta _4 \det P(-a|x|^{2}+d|u|^{2}) +(\det P)^{2}(a\overline{x}^{2} +d\overline{u}^{2}) \big )\nonumber \\&= \tfrac{1}{((-1)^{k}+\delta _1)^{2}}\Big (\delta _4^{2} (\widetilde{a}+\epsilon _1)+2\delta _4 \big ((\widetilde{b} +\epsilon _2)((-1)^{k}+\delta _1)-\delta _4 (\widetilde{a}+\epsilon _1)\big )\nonumber \\&\quad +(\det P)^{2}(\widetilde{a}+\overline{\epsilon }_1 +2i\textrm{Im}(d)\overline{u}^{2}) \Big ). \end{aligned}$$

  (5.28)

  The equation (5.27) gives ($a\in {\mathbb {R}}$):

  $$\begin{aligned} \textrm{Im}(d)|u|^{2}=\textrm{Im}\left( \tfrac{1}{\det P} \big ((\widetilde{b}+\epsilon _2)((-1)^{k}+\delta _1) -\delta _4(\widetilde{a}+\epsilon _1)\big )\right) . \end{aligned}$$

  (5.29)

  Lemma 4.1 (1) yields $|\det P|\ge \frac{1-\frac{6\delta ^{2}}{\nu ^{2}}}{\sqrt{1-\delta }}$ (note $1-\tau \le \delta$ by (5.23). It follows for $\widetilde{b}=0$ that $|\textrm{Im}(d)u^{2}|\le \nu ^{2}\frac{\epsilon (1+\delta )^{\frac{3}{2}}+\delta (\widetilde{a}+\epsilon ) \sqrt{1+\delta }}{\nu ^{2}-6\delta ^{2}}$. It contradicts (5.28) for $\widetilde{a}<\widetilde{d}$, $\widetilde{b}=0$ and $\epsilon ,\delta$ small enough. Next, $c(s)=1$, $P(s)=\sqrt{\frac{\widetilde{b}}{2}}\begin{bmatrix} 1 &{} 1\\ \widetilde{b}^{-1} &{} -\widetilde{b}^{-1} \end{bmatrix}$, $B(s)=1\oplus \widetilde{b}^{2}e^{is}$ (or $B(s)=1\oplus -\widetilde{b}^{2}e^{-is}$) yields a path from $(1\oplus -1, \widetilde{b}I_2)$, $\widetilde{b}> 0$ (from $\big (1\oplus -1, \begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 0 \end{bmatrix}\big )$, $\widetilde{b}> 0$) to $\big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix},1 \oplus d\big )$, $\textrm{Im}(d)>0$. For $P(s)=\frac{1}{2}\begin{bmatrix} 2s &{} s^{-1}\\ 2s &{} -s^{-1} \end{bmatrix}$ we get $(1\oplus -1,0_2)\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix},1 \oplus 0\big )$.

• (b) $B=\begin{bmatrix} a &{} b\\ b &{} d \end{bmatrix}$, $b>0$

  Let $\widetilde{B}=\begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 0 \end{bmatrix}$, $\widetilde{b}>0$. For $a=0$ we have $b^{2}= \widetilde{b}^{2}-(1-\tau )\widetilde{b}^{2}+\epsilon '$ with $1-\tau \le \delta$, $|\epsilon '|\le \max \{\epsilon , \frac{\delta ^{2}}{\nu ^{2}}\}(4\max \{\widetilde{b}, \widetilde{b}^{2},1\}+2)^{2}$ (Lemma 4.1 (3)). If $d=e^{i\varphi }$ with $\varphi <\pi$, the proof in [22, Theorem 3.6, Case VII. (b) (i)] applies, while for $d=0$ the first equation of (5.12) for $\widetilde{a}=0$ and (5.23) yield $b(1-\delta )\le 2b|ux|\le \epsilon$, which fails for small $\epsilon ,\delta$.

  Suppose $\widetilde{B}=\widetilde{a}\oplus \widetilde{d}$ for $0\le \widetilde{a}\le \widetilde{d}$. If $d=e^{i\varphi }$ the proof in [22, Theorem 3.6, Case VII. (b) (ii)] for $\widetilde{a}\ne \widetilde{d}$ applies almost mutatis mutandis, we only replace $\frac{|\det \widetilde{A}|}{|\det A|}=\frac{|\det \widetilde{B}|}{|\det B|}=|\frac{\widetilde{a} \widetilde{d}}{b^{2}}|=1$ with $b^{2}=\widetilde{a} \widetilde{d}-(1-\tau )\widetilde{a}\widetilde{d}+\epsilon '$, $1-\tau \le \delta$, $|\epsilon '|\le \max \{\epsilon , \frac{\delta ^{2}}{\nu ^{2}}\}(4\max \{\widetilde{d}, \widetilde{a}\widetilde{d},1\}+2)^{2}$ (Lemma 4.1 (3)). If $d=0$, the first equation of (5.12) for $\widetilde{a}=0$ and (5.23) give $(\widetilde{a}\widetilde{d}-\delta \widetilde{a} \widetilde{d}-|\epsilon '|)(1-\delta )^{2}\le 4b^{2}|ux|^{2}\le |\widetilde{a}+\epsilon |^{2}$, which fails for small $\epsilon ,\delta$. Note, $c(s)=1$, $P(s)=\begin{bmatrix} \frac{1}{2s} &{} -\frac{i}{2s}\\ s &{} is \end{bmatrix}$, $B(s)=\begin{bmatrix} 0 &{} \widetilde{d}+s\\ \widetilde{d}+s &{} 1 \end{bmatrix}$ in (5.2) implies $(1\oplus -1,\widetilde{d}I_2)\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} b\\ b &{} 1 \end{bmatrix}\big )$ for $\widetilde{d}\ge 0$. By conjugating with $\frac{1}{2}\begin{bmatrix} 1 &{} -2\\ -1 &{} -2 \end{bmatrix}$ and $r\oplus \frac{1}{r}$ for $r>0$, we get a path

  $$\begin{aligned} (1\oplus -1,\widetilde{a}\oplus \widetilde{d})&\approx \left( \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, \tfrac{1}{4}\begin{bmatrix} \widetilde{a}+\widetilde{d} &{} 2(\widetilde{d}-\widetilde{a})\\ 2(\widetilde{d}-\widetilde{a}) &{} 4(\widetilde{a}+\widetilde{d}) \end{bmatrix}\right) \rightarrow \left( \begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} r^{2}e^{i\varphi } &{} b\\ b &{} {r^{-2}\zeta } \end{bmatrix}\right) \\&\approx \left( \begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} e^{i\varphi } &{} b\\ b &{} \zeta \end{bmatrix}\right) . \end{aligned}$$

Case IV.:

$\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, \widetilde{B}\big )\dashrightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, B\big )$

Lemma 3.2 (2) with (C8) for, $\beta =1$, $\omega =i$, $\alpha =k=0$ (since $||v|^{2}-(-1)^{k}|<\delta$) gives

$$\begin{aligned} \big |\overline{x}v+\overline{u}y-1\big |\le \delta , \quad |u|^2\le \delta , \quad \big ||v|^{2}-1\big | \le \delta , \quad |\textrm{Re}(\overline{x}u)|,|\textrm{Re}(\overline{y}v)|\le \delta . \end{aligned}$$

(5.30)

• (a) $B=a\oplus d$, $a\ge 0$, $d\in {\mathbb {C}}$

  It is not difficult to check that $B(s)= s \oplus \frac{\widetilde{b}^{2}}{s}$, $c(s)=1$, $P(s)=e^{-i\frac{\pi }{4}} \begin{bmatrix} 1 &{} i\widetilde{b}s^{-1}\\ s^2e^{i\frac{\pi }{4}} &{} 1 \end{bmatrix}$ in (5.2) proves $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, \begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 0 \end{bmatrix}\big ) \rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, a\oplus d\big )$, $d\in {\mathbb {C}}$, $a>0$, $\widetilde{b}\ge 0$.

  Next, let $B=0\oplus d$, $d>0$, $\widetilde{B}=\widetilde{a}\oplus \widetilde{d}$, $\widetilde{a}>0$. Using (5.6) for $a=0$ and $|u|^2\le \delta$ we get

  $$\begin{aligned} \widetilde{a}+\epsilon \le |du^{2}|\le d\delta , \qquad d(1-\delta )\le |dv^{2}|\le \epsilon +|\widetilde{d}|. \end{aligned}$$

  Hence $\frac{|\widetilde{d}|+\epsilon }{1-\delta }\delta \ge \widetilde{a}+\epsilon$, which fails for sufficiently small $\epsilon ,\delta$.

• (b) $B=\begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}$, $b>0$, (hence $\widetilde{B}=\widetilde{a} \oplus \widetilde{d}$ by Lemma 2.1)

  The proof in [22, Theorem 3.6, Case V. (b)] applies mutatis mutandis. Note, $B(s)= \frac{\widetilde{d}s}{2}\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}$, $P(s)=e^{-i\frac{\pi }{4}} \begin{bmatrix} e^{i\frac{\pi }{4}}s &{} s^{-1}\\ s &{} i \end{bmatrix}$ in (5.2) implies $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, 0\oplus \widetilde{d}\big ) \rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, \begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}\big )$, $\widetilde{d}>0$.

Case V.:

$\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, \widetilde{B}\big ) \dashrightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix},B\big )$

Lemma 3.2 (2) with (C8) for $\alpha =\omega =0$, $\beta =0$ yields

$$\begin{aligned} |u|^{2},|v|^{2}\le \delta ,\quad \big |2 \textrm{Re}(\overline{y}v)\big |\le \delta , \quad \big |2 \textrm{Re}(\overline{x}u)\big |\le \delta , \quad \big | \overline{x}v+\overline{u}y-(-1)^{k}\big | \le \delta ,\,k\in {\mathbb {Z}}. \end{aligned}$$

(5.31)

• (a) $B=a\oplus d$, $a\ge 0$

  Taking $c(s)=1$, $P(s)=\begin{bmatrix} 1 &{} s^{-1}\\ s &{} 0 \end{bmatrix}$, $B(s)=0\oplus \frac{1}{s}$ in (5.2) proves $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix},\widetilde{a}\oplus 0\big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix},0\oplus d \big )$, $\widetilde{a}\in \{0,1\}$, $d>0$. Next, $c(s)=1$, $P(s)=e^{i\frac{1}{2}\widetilde{\vartheta }}\begin{bmatrix} 1 &{} s^{-1}\\ s &{} 0 \end{bmatrix}$, $B(s)=(|\widetilde{d}|+s)s^{2}\oplus \frac{1}{s^{2}}e^{-i\widetilde{\theta }}$ yields $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, 1\oplus \widetilde{d}\big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, a\oplus d \big )$, $\widetilde{d} =|\widetilde{d}|e^{i\widetilde{\vartheta }}$, $a>0$, $d\in {\mathbb {C}}$.

  Proceed with $\widetilde{b}=\begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 1 \end{bmatrix}$, $\widetilde{b}>0$; we conjugate the first pair with $\frac{1}{2} \begin{bmatrix} 2 &{} -2\\ 1 &{} 1 \end{bmatrix}$:

  $$\begin{aligned}{} & {} \left( \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 1 \end{bmatrix}\right) \approx \nonumber \\{} & {} \qquad \left( 1\oplus -1, \tfrac{1}{4}\begin{bmatrix} 4\widetilde{b}+1 &{} 1\\ 1 &{} -4\widetilde{b}+1 \end{bmatrix}\right) \dashrightarrow \left( \begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, a\oplus d\right) , \quad d\in {\mathbb {C}}, a>0.\nonumber \\ \end{aligned}$$

  (5.32)

  Using ideas from Case III (a) we find $c(s)=-1$, $P(s)=\frac{e^{i\frac{\pi }{4}}}{\sqrt{2}} \begin{bmatrix} \frac{i}{s^{2}}e^{i\alpha (s)} &{} \frac{i}{s^{2}}e^{-i\alpha (s)}\\ se^{-i\alpha (s)} &{} se^{i\alpha (s)} \end{bmatrix}$ with $\sin (2 \alpha (s))=s$, $B(s)=\widetilde{b}s^{3}\oplus (\frac{\widetilde{b}}{s^{3}}-\frac{i}{2\,s^{2}})$ (see (5.2)), which proves the existence of (5.32).

• (b) $B=\begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}$, $b>0$

  If $\widetilde{B}=\begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 1 \end{bmatrix}$, $\widetilde{b}>0$, the proof in [22, Theorem 3.6, Case VI. (b) (i)] applies mutatis mutandis, we only use $b^{2}=\widetilde{b}^{2}+\epsilon '$, $|\epsilon '|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\} (4\max \{\widetilde{b},\widetilde{b}^{2},1\}+2)^{2}$ (see Lemma 4.1 (3)) instead of $1=\frac{|\det \widetilde{A}|}{|\det A|}=\frac{|\det \widetilde{B}|}{|\det B|}=\frac{\widetilde{b}^{2}}{b^{2}}$. For $\widetilde{B}=1\oplus \widetilde{d}$, $\widetilde{d}\ne 0$ we apply [22, Theorem 3.6, Case VI. (b) (ii)], we only replace $\frac{|\det \widetilde{A}|}{|\det A|}=\frac{|\det \widetilde{B}|}{|\det B|}=|\frac{\widetilde{d}}{b^{2}}|=1$ with $b^{2}=|\widetilde{d}|+\epsilon '$, $|\epsilon '|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\} (4\max \{|\widetilde{d}|,1\}+2)^{2}$ (Lemma 4.1 (3)).

Case VI.:

$(1\oplus -1,\widetilde{B}) \dashrightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, B\big )$

Lemma 3.2 (2) with (C8) for $-\omega =\alpha =1$, $\beta =0$ yields ($|\delta _1|,|\delta _2|,|\delta _4|<\delta$, $k\in {\mathbb {Z}}$):

$$\begin{aligned} 2\textrm{Re}(\overline{x}u)=(-1)^{k}+ \delta _1, \quad 2\textrm{Re}(\overline{y}v)=-(-1)^k+\delta _2, \,\, |u|^{2},|v|^2\le \delta , \quad \overline{x}v+\overline{u}y= \delta _4. \end{aligned}$$

(5.33)

• (a) $B=\begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}$, $b>0$

  The proof in [22, Theorem 3.6, Case V. (b) (i)] applies mutatis mutandis for $\widetilde{B}=\begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 0 \end{bmatrix}$, $\widetilde{b}> 0$; recall $b^{2}=\widetilde{b}^{2}+\epsilon '$, $|\epsilon '|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\}(2\max \{1,\widetilde{b},\widetilde{b}^{2}\}+1)^{2}$ (Lemma 4.1 (3)).

  Let $\widetilde{B}=\widetilde{a}\oplus \widetilde{d}$, $\widetilde{d}\ge \widetilde{a}\ge 0$. If $\widetilde{d}>\widetilde{a}>0$ the proof in [22, Theorem 3.6, Case V. (b) (ii)] applies for $b^{2}=\widetilde{a}\widetilde{d}+\epsilon '$, $|\epsilon '|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\}(4\max \{1, \widetilde{d},\widetilde{a}\widetilde{d}\}+2)^{2}$ (Lemma 4.1 (3)). For $c(s)=-1$, $P(s)=\frac{1}{\sqrt{2}}\begin{bmatrix} i s^{-1} &{} s^{-1}\\ -i s &{} s \end{bmatrix}$, $B(s)=(\widetilde{d}+s)\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}$ in (5.2) we get $(1\oplus -1,\widetilde{d}I_2)\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, \begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}\big )$, $\widetilde{d}\ge 0$. If $\widetilde{a}=0$, $\widetilde{d}> 0$ then Lemma 3.1 (D1) yields $\widetilde{d}-\epsilon \le |\frac{v}{u}|\sqrt{\epsilon }(4\widetilde{d}+2)^{\frac{1}{2}}$, and Lemma 4.1 (1) gives $|\det P|\le 1 +\frac{6\delta ^{2}}{\nu ^{2}}$. By applying this and (5.33) to (5.25) implies $(1-\delta )(\widetilde{d}-\epsilon )\le \sqrt{\epsilon }(4\widetilde{d}+2)^{\frac{1}{2}} (\delta +1+\frac{6\delta ^{2}}{\nu ^{2}})$, which fails for small $\epsilon ,\delta$.

• (b) $B=a\oplus d$,    $a\ge 0$, $d\in {\mathbb {C}}$

  If $\widetilde{b}=0$, $0\le \widetilde{a}<\widetilde{d}$ the same proof as in Case III (a) applies (see (5.23) and (5.33)).

Case VII.:

$\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, \widetilde{B}\big )\dashrightarrow (1\oplus -1,B)$

Lemma 3.2 (2) with (C2) for $\omega =0$, $\theta =\pi$ gives

$$\begin{aligned} |x|^2-|u|^2= \delta _1,\;\; \overline{x}y-\overline{u}v-(-1)^{k} = \delta _2,\;\; |y|^2-|v|^2= \delta _4,\;\; |\delta _1|,| \delta _2|,|\delta _4|\le \delta ,\,k\in {\mathbb {Z}}. \end{aligned}$$

(5.34)

• (a) $B=a\oplus d$,    $0\le a\le d$, $d>0$

  • (i) $\widetilde{B}=\begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 1 \end{bmatrix}$, $\widetilde{b}>0$

    First, $c(s)=-1$, $P(s)=\begin{bmatrix} \frac{-i}{2}s &{} is^{-1}\\ \frac{s}{2} &{} s^{-1} \end{bmatrix}$, $B(s)=\widetilde{b}\oplus (\widetilde{b}+s^2)$ in (5.2) gives $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 1 \end{bmatrix}\big )\rightarrow (1\oplus -1,a\oplus d)$, $a<d$. For $a=d$ we apply the proof of [22, Theorem 3.6, Case VIII (a) (ii)], but replace $\begin{bmatrix} 0 &{} d\\ d &{} 1 \end{bmatrix}$ with $\begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 1 \end{bmatrix}$; and use $d^{2}=|\widetilde{b}|^{2}+\epsilon '$, $|\epsilon '|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\} \big (4\max \{|\widetilde{b}|,|\widetilde{b}|^{2},1\}+2\big )^{2}$ (Lemma 4.1 (3)) at the end of the proof.

  • (ii) $\widetilde{B}=1\oplus \widetilde{d}$, $\widetilde{d}\in {\mathbb {C}}$

    We prove $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix},1 \oplus 0 \big )\rightarrow (1\oplus -1,0\oplus d)$ with $P(s)=\begin{bmatrix} s^{-1} &{} \frac{1}{2}s\\ s^{-1} &{} -\frac{1}{2}s \end{bmatrix}$, $c(s)=1$, $B(s)=0\oplus s^{2}$.

    Proceed with $B=a\oplus d$, $0<a\le d$. We have equations (5.6) for $\widetilde{a}=1$, $\textrm{Im}\widetilde{d}>0$, $\widetilde{b}=0$. By combining them with (5.9), (5.10) for $\sigma =-1$ and with (5.34) we get

    $$\begin{aligned}&\epsilon _1+1- ae^{2i\phi }\delta _1 = u^{2} (ae^{2i(\phi -\eta )}+d),\nonumber \\&a\big ((-1)^{k}+ \delta _2\big )-e^{-2i\phi } \epsilon _2=-e^{-2i\phi }\big (uv(ae^{2i(\phi -\eta )}+d)\big ),\nonumber \\&a\big ((-1)^{k}+ \delta _2\big )-e^{-2i\varphi } \epsilon _2=-e^{-2i\varphi }\big (uv (ae^{2i(\varphi -\kappa )}+d)\big ),\nonumber \\&\epsilon _4+\widetilde{d}-ae^{2i\varphi }\delta _4= v^{2} (ae^{2i(\varphi -\kappa )}+d),\nonumber \\&d\big ((-1)^{k}+ \delta _2\big )+e^{-2i\kappa }\epsilon _2 =e^{-2i\kappa }\big (xy(de^{2i(\kappa -\varphi )}+a)\big ), \end{aligned}$$

    (5.35)

    We have $ad=|\widetilde{d}|+\delta '$, $|\delta '|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\}\big (4 \max \{1,|\widetilde{d}|\}+2\big )^{2}$ (Lemma 4.1), hence $a\le \sqrt{|\widetilde{d}|}+1$, provided that $\max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\}\le \frac{1}{(4\max \{1,|\widetilde{d}|\}+2)^{2}}$. Next, we divide the first and the second (the third and the fourth) two equations of (5.35) to get

    $$\begin{aligned} \tfrac{u}{v}=\tfrac{\epsilon _1+1- ae^{2i\phi }\delta _1}{a\big ((-1)^{k} + \delta _2\big )-e^{-2i\phi }\epsilon _2}(-e^{-2i\phi }) =\tfrac{a\big ((-1)^{k}+ \delta _2\big )-e^{-2i\varphi } \epsilon _2}{\epsilon _4+\widetilde{d}-ae^{2i\varphi } \delta _4}(-e^{2i\varphi }). \end{aligned}$$

    The second equality yields that there is a (computable) constant $D>0$ so that

    $$\begin{aligned} a^{2}=\widetilde{d}e^{-2i(\phi +\varphi )}+\delta _5, \quad d^{2}=\tfrac{(|\widetilde{d}|+\delta ')^{2}}{\widetilde{d} e^{2i(\phi -\varphi )}+\delta _5},\qquad |\delta _5| \le D \max \{\epsilon ,\delta \}, \end{aligned}$$

    (5.36)

    Furthermore, we divide the third and the fifth equation of (5.35) to conclude:

    $$\begin{aligned} \tfrac{xy}{uv}=\tfrac{(d((-1)^{k}+ \delta _2) +e^{-2i\kappa }\epsilon _2)}{(a((-1)^{k}+ \delta _2) e^{2i\varphi }\epsilon _2)}=1+\delta _6, \qquad |\delta _6| \le C\max \{\epsilon ,\delta \}, \end{aligned}$$

    (5.37)

    while the first four equations of (5.35) yield

    $$\begin{aligned} \tfrac{1}{\widetilde{d}}+\delta _0= & {} \tfrac{\big (1+\epsilon _1 -ae^{2i\phi }\delta _1\big )\big (a(-1)^{k}+ a\delta _2-e^{-2i\phi } \epsilon _2\big )}{\big (\widetilde{d}+\epsilon _1- ae^{2i\varphi } \delta _4\big )\big (a(-1)^{k}+ a\delta _2-e^{-2i\varphi } \epsilon _2\big )}\\= & {} e^{i(2\eta -2\kappa -2\phi +2\varphi )} \tfrac{|u|^{2}}{|v|^{2}},\quad |\delta _0|\le K \max \{\epsilon ,\delta \}, \end{aligned}$$

    where constants $C,K>0$ can be computed. By applying (4.7) for $\widetilde{d}=|\widetilde{d}| e^{i\widetilde{\vartheta }}$ we get $2\eta -2\kappa -2\phi +2\varphi +\widetilde{\vartheta }=\psi$ with $|e^{i\frac{\psi }{2}}-1|=|\sin \frac{\psi }{4}|\le |\sin \psi |\le \delta _0$. Using (5.37) we get

    $$\begin{aligned} \big |\tfrac{\overline{x}y}{\overline{u}v} -1\big |&=\big |\tfrac{|xy|}{|uv|}e^{i(\phi -\varphi -\kappa +\eta )} -1\big | =\big ||1+\delta _6|e^{i\left( -\frac{\widetilde{\vartheta }}{2} +\frac{\psi }{2}\right) }-1\big |\\&=\big |e^{i\frac{\psi }{2}} \left( e^{-i\frac{\widetilde{\vartheta }}{2}}+1\right) -\left( e^{i\frac{\psi }{2}}-1\right) +(|1-\delta _6|-1) e^{i\left( -\frac{\widetilde{\vartheta }}{2}+\frac{\psi }{2}\right) }\big |\\&\ge |e^{-i\frac{\widetilde{\vartheta }}{2}}+1|-|\delta _0| -|\delta _6| \ge \cos \tfrac{\widetilde{\vartheta }}{4}, \end{aligned}$$

    provided that $\epsilon ,\delta$ are such that $\tfrac{1}{4}|e^{-i\frac{\widetilde{\vartheta }}{2}}+1|=\frac{1}{2}\cos \frac{\widetilde{\vartheta }}{4}\ge |\delta _0|,|\delta _6|$ with $0<\widetilde{\vartheta }<\pi$. Thus:

    $$\begin{aligned} 2&\ge 1+\delta \ge |\overline{x}y-\overline{u}v|= |\overline{u}v|\big |\tfrac{\overline{x}y}{\overline{u}v}-1\big | \ge \tfrac{1}{2}|uv|\cos \tfrac{\widetilde{\vartheta }}{4},\quad |u|^{2}=\tfrac{|u|}{|v|}|uv|\le 4\tfrac{|\widetilde{d}|^{-1} +|\delta _0|}{\cos \frac{\widetilde{\vartheta }}{4}}. \end{aligned}$$

    We simplify the first and the third equation of (5.35) and rearrange the terms:

    $$\begin{aligned}&2au^{2}\cos (\phi -\eta )e^{i(\phi -\eta )}=1+\epsilon _1-ae^{2i\phi } \delta _1-(d-a)u^{2}, \nonumber \\&-2auv\cos (\varphi -\kappa )e^{-i(\varphi +\kappa )}=a(-1)^{k} +a\delta _2-e^{-2i\varphi }\epsilon _2+(d-a)uve^{-2i\varphi }. \end{aligned}$$

    (5.38)

    By applying (4.3) with $\widetilde{d}=|\widetilde{d}|e^{i\widetilde{\vartheta }}$ to (5.36) and (5.38) we deduce ($L>0$ is a constant):

    $$\begin{aligned}&\psi _0=\widetilde{\theta }-2(\phi +\varphi ), \qquad |\sin \psi | \le \tfrac{2|\delta _5|}{|\widetilde{d}|}, \\&\psi _1=(\phi +\eta )-\pi l_1,\qquad |\sin \psi _1|\le L \max \{\epsilon ,\delta \},\quad l_1\in {\mathbb {Z}}, \\&\psi _2=(\eta -\varphi )-\pi (k+l_2),\qquad |\sin \psi _2| \le L\max \{\epsilon ,\delta \},\quad l_2\in {\mathbb {Z}}. \end{aligned}$$

    Thus $\big |\sin (\psi _0+2\psi _1-2\psi _2)\big |=|\sin \widetilde{\vartheta }|\le \tfrac{2|\delta _5|}{|\widetilde{d}|} +4\,L \max \{\epsilon ,\delta \}$ and it fails for small $\epsilon ,\delta$.

• (b) $B=\begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}$, $b>0$

  If $\widetilde{B}=\begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 1 \end{bmatrix}$ for $\widetilde{b}>0$ we can apply the proof of [22, Theorem 3.6, Case VIII (b) (i)], recall $b^{2}=\widetilde{b}^{2}+\epsilon '$, $|\epsilon '|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\}(4 \max \{|\widetilde{b}|,|\widetilde{b}|^{2},1\}+2)^{2}$ (Lemma 4.1 (3)).

  Let $\widetilde{B}=1\oplus \widetilde{d}$, $\widetilde{d}\in {\mathbb {C}}$. To get $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix},1 \oplus 0 \big )\rightarrow \big (1\oplus -1,\begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}\big )$, we take $P(s)=\frac{1}{2}\begin{bmatrix} 2s^{-1} &{} s\\ 2s^{-1} &{} -s \end{bmatrix}$, $c(s)=1$, $B(s)=\frac{s^{2}}{2}\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}$ in (5.2). If $\widetilde{d}=|\widetilde{d}|e^{i\widetilde{\vartheta }}$, $0<\widetilde{\vartheta }<\pi$ Lemma 3.1 (D3) implies

  $$\begin{aligned} bvx=\tfrac{1}{2}\left( \epsilon _2'+(-1)^{l}i\sqrt{|\widetilde{d}|} e^{i\tfrac{\widetilde{\vartheta }}{2}}\right) ,\qquad buy=\tfrac{1}{2}\left( \epsilon _2''-(-1)^{l}i\sqrt{|\widetilde{d}|} e^{i\tfrac{\widetilde{\vartheta }}{2}}\right) , \end{aligned}$$

  where $|\epsilon _2'|,|\epsilon _2''| \le \tfrac{\epsilon (4\max \{1,|\widetilde{d}|\}+2 +|\widetilde{d}|)}{|\widetilde{d}|}$. By applying (4.3) to these two equations and to the first equality of (5.12) we get $\psi _1,\psi _2,\psi _3\in (-\frac{\pi }{2},\frac{\pi }{2})$ such that:

  $$\begin{aligned}&\psi _1=\phi +\kappa -\tfrac{\pi }{2}-\tfrac{\widetilde{\vartheta }}{2} -l\pi +2\pi l_3, \quad |\sin \psi _1|\le \tfrac{2|\epsilon _2'|}{\sqrt{|\widetilde{d}|}}, \\&\psi _2=\varphi +\eta -\tfrac{\pi }{2}-\tfrac{\widetilde{\vartheta }}{2} -(l+1)\pi +2\pi l_4, \quad |\sin \psi _2|\le \tfrac{2|\epsilon _2''|}{\sqrt{|\widetilde{d}|}},\\&\psi _3=\phi +\eta +2\pi l_1, \quad |\sin \psi _3|\le \epsilon . \end{aligned}$$

  Therefore

  $$\begin{aligned}&(-1)^{k}+\delta _2\\&=\overline{x}y-\overline{u}v=|xy| e^{i(\varphi -\phi )}-|uv|e^{i(\kappa -\eta )} =e^{-i(\phi +\eta )}\big (|xy|e^{i(\varphi +\eta )} -|uv|e^{i(\kappa +\phi )}\big )\\&=e^{i(-\psi _3+2\pi l_1)}\big (|xy|e^{i(\psi _2 -2\pi l_4+\frac{\pi }{2}+\frac{\widetilde{\vartheta }}{2} +(l+1)\pi )}-|uv|e^{i(\frac{\pi }{2}+\frac{\widetilde{\vartheta }}{2} +l\pi -2\pi l_3+\psi _1)}\big )\\&=e^{i(\psi _2-\psi _3+\frac{\widetilde{\vartheta }}{2}+(l+1)\pi +\frac{\pi }{2})}\big (|xy|+|uv|e^{i(\psi _1-\psi _2)}\big ). \end{aligned}$$

  Since $\psi _1,\psi _2,\psi _3\in (-\frac{\pi }{2},\frac{\pi }{2})$ are close to 0, the argument of the second factor is close to 0, too. Using (4.3) again we obtain a contradiction for $\epsilon ,\delta$ small enough:

  $$\begin{aligned}&\psi = k\pi -\big (\psi _2-\psi _3+\tfrac{\widetilde{\vartheta }}{2} +(l+1)\pi +\tfrac{\pi }{2}\big )-(\psi _1-\psi _2), \qquad |\sin \psi |\le 2\delta ,\\&0\ne |\cos \tfrac{\widetilde{\vartheta }}{2}| =\big |\sin \left( \tfrac{\widetilde{\vartheta }}{2} +\tfrac{\pi }{2}\right) \big | \le \big |\sin (\psi _3+\psi _1)\big |\le 2\epsilon _2'+2\delta . \end{aligned}$$

Case VIII.:

$\bigl (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, \widetilde{B}\bigr ) \dashrightarrow \bigl (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix},B\bigr )$

• (a) $B=1\oplus d$, $\textrm{Im}d> 0$, $\widetilde{B}=\begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 1 \end{bmatrix}$, $\widetilde{b}>0$

  We can apply the proof of [22, Theorem 3.6, Case IX (b)], and use $|d|=|\widetilde{b}|^{2}+\epsilon '$, $|\epsilon '|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\} (4\max \{|\widetilde{b}|,|\widetilde{b}|^{2},1\}+2)^{2}$ (Lemma 4.1 (3)).

• (b) $B=\begin{bmatrix} 0 &{} b\\ b &{} 1 \end{bmatrix}$, $b>0$

  For $P(s)=\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}$, $B=\begin{bmatrix} 0 &{} s\\ s &{} 1 \end{bmatrix}$, $c(s)=1$ in (5.2) we get $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix},1\oplus 0\big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix},\begin{bmatrix} 0 &{} b\\ b &{} 1 \end{bmatrix}\big )$, $b>0$. For $\widetilde{B}=1\oplus \widetilde{d}$, $\widetilde{d}\ne 0$ we use the proof of [22, Theorem 3.6, Case IX (c)], but replace $\frac{|\det \widetilde{B}|}{|\det B|}=\frac{\widetilde{a}\widetilde{d}}{b^{2}}$ with $b^{2}=|\widetilde{d}|+\epsilon '$, $|\epsilon '|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\} (4\max \{|\widetilde{b}|,|\widetilde{b}|^{2},1\}+2)^{2}$ (Lemma 4.1 (3)).

Case IX.:

$\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} \omega \end{bmatrix}, \widetilde{B}\big )\dashrightarrow (1\oplus e^{i\theta },B)$, $0< \theta <\pi$, $\omega \in \{0,i\}$

From Lemma 3.2 (C2) we get

$$\begin{aligned}&\big ||u|^2-|x|^2\big | \le \delta , \quad \big ||v|^2-|y|^2\big | \le \delta , \quad \big |\overline{x}y-\overline{u}v-(-1)^{k}\big | \le \delta ,\quad k\in {\mathbb {Z}},\,\,\sin \theta \le \delta ; \nonumber \\&\textrm{if}\quad \omega =i,\quad \textrm{then} \quad (\sin \theta ) |v|^2=1 +\delta _2,\,\, (\sin \theta )|u|^{2}=\delta _3, \,\, |\delta _2|,|\delta _3|\le \delta . \end{aligned}$$

(5.39)

For $\omega =i$ we further deduce

$$\begin{aligned} \quad \big |(\sin \theta )|y|^{2}- 1\big | \le \delta +\delta ^{2}, \quad (\sin \theta )|x|^{2}\le \delta +\delta ^{2}. \end{aligned}$$

(5.40)

• (a) $B=\begin{bmatrix} 0 &{} b\\ b &{} d \end{bmatrix}$, $b\ge 0$, $d> 0$

  Lemma 3.1 (D1) for $\widetilde{B}=\widetilde{a}\oplus \widetilde{d}$, $\widetilde{a}\ne 0$ and (5.39) for $\omega =i$ (hence $(1+\delta _2)|u|^{2}=\delta _3|v|^{2}$) yield a contradiction for small $\epsilon ,\delta$. Next, $c(s)=1$, $P(s)=i\sqrt{\widetilde{d}+s}\begin{bmatrix} \frac{s}{\widetilde{d}+s} &{} s^{-1}\\ 0 &{} -s^{-1} \end{bmatrix}$, $\cos (\frac{\theta }{2})=\left\{ \begin{array}{ll} \frac{s^{2}}{2(\widetilde{d}+s)}, &{} \omega =i\\ s^{3}, &{} \omega =0 \end{array}\right.$, $B(s)=\begin{bmatrix} 0 &{} \widetilde{b} \\ \widetilde{b} &{} 2\widetilde{b}-s^2 \end{bmatrix}$ in (5.2) proves $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} \omega \end{bmatrix}, \begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} \widetilde{d} \end{bmatrix}\big )\rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} 0 &{} b\\ b &{} d \end{bmatrix}\big )$, $\widetilde{b}> 0$, either $\omega =0$, $\widetilde{d}=1$ or $\omega =i$, $\widetilde{d}=0$. Taking $c(s)=1$, $P(s)=\sqrt{\widetilde{d}+s}\begin{bmatrix} \frac{s}{\widetilde{d}+s} &{} s^{-1}\\ 0 &{} -s^{-1} \end{bmatrix}$, $\cos (\frac{\theta }{2}) =\frac{s^{2}}{2(\widetilde{d}+s)}$, $B(s)=0\oplus s^2$ shows $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, 0\oplus \widetilde{d} \big )\rightarrow (1\oplus e^{i\theta }, 0\oplus d)$. Finally, $c(s)=-ie^{i\frac{\widetilde{\vartheta }}{2}}$, $\cos (\frac{\theta (s)}{2})=s^{3}$, $P(s)=\frac{1}{s} e^{-i\frac{\pi }{4}}\begin{bmatrix} e^{i\alpha (s)} &{} ie^{-i\alpha (s)}\sqrt{\widetilde{d}+s}\\ -e^{-i\alpha (s)} &{} -ie^{i\alpha (s)}\sqrt{\widetilde{d}+s} \end{bmatrix}$, $\sin (2\alpha (s))=\frac{s^{2}}{2| \sqrt{\widetilde{d}+s}|}$, $B(s)=\big |\sqrt{\widetilde{d}+s}\big |\begin{bmatrix} 0 &{} 1 \\ 1 &{} 2 \end{bmatrix}$ in (5.2) proves $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, 1\oplus \widetilde{d}\big )\rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} 0 &{} b\\ b &{} d \end{bmatrix}\big )$, $b > 0$, $\textrm{Im}(\widetilde{d})> 0$.

• (b) $B=\begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}$, $b>0$

  Let $\widetilde{B}=\begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 0 \end{bmatrix}$, $\widetilde{b}> 0$ and $\omega =i$. It follows from Lemma 4.1 (3) that $b^{2}=\widetilde{b}^{2}+\epsilon '$, $|\epsilon '|\le \max \{\epsilon ,\frac{\delta ^{2}}{\nu ^{2}}\}\big (4 \max \{1,|\widetilde{b}|,|\widetilde{b}|^{2}\}+2\big )^{2}$, so the third equation of (5.12) for $\widetilde{d}=0$ yields $(yv)^{2}=\frac{\epsilon _4^{2}}{4(\widetilde{b}^{2}+\epsilon ')}$. By combining it with (5.39) and (5.40) we deduce

  $$\begin{aligned} \big (1- \delta (1+\delta )\big )(1-\delta )\le (\sin \theta )^{2}|yv|^{2} =\tfrac{\delta ^{2}\epsilon ^{2}}{4|\widetilde{b}^{2}+\epsilon '|}, \end{aligned}$$

  which fails for $\epsilon ,\delta$ small enough. Next, $c(s)=1$, $\cos (\frac{\theta (s)}{2})=s^{2}$, $P(s)=\frac{i}{\sqrt{2}} \begin{bmatrix} s &{} s^{-1}\\ s &{} -s^{-1} \end{bmatrix}$, $B(s)=(\widetilde{d}+s)s^{2}\begin{bmatrix} 0 &{} 1 \\ 1 &{} 0 \end{bmatrix}$ in (5.2) gives $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, 0\oplus \widetilde{d}\big )\rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}\big )$, $\widetilde{d}\ge 0$.

  We apply the same proof as in Case VII (compare (5.34) and (5.39)) to show $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 1 \end{bmatrix}\big )\not \rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}\big )$, $\widetilde{b}>0$ and $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, 1\oplus \widetilde{d}\big )\not \rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}\big )$, $\textrm{Im}\widetilde{d}> 0$.

• (c) $\begin{bmatrix} a &{} b\\ b &{} 0 \end{bmatrix}$, $a>0$, $b\ge 0$

  We multiply the squared equation in Lemma 3.1 (D2) for $\widetilde{B}=\widetilde{a}\oplus \widetilde{d}$ with $(\sin \theta )^{2}$:

  $$\begin{aligned} (\widetilde{a}+\epsilon _1)^{2}y^{2}\sin ^{2} \theta= & {} \big (-i(-1)^{l}\sqrt{\widetilde{a}\widetilde{d}} +\epsilon _2'\big )x^{2} \sin ^{2} \theta , \,\, |\epsilon _2'|\\\le & {} \left\{ \begin{array}{ll} \tfrac{\epsilon (4\max \{\widetilde{a},\widetilde{d}\}+2 +|\widetilde{a}\widetilde{d}|)}{|\widetilde{a}\widetilde{d}|}, &{} \widetilde{a}\widetilde{d}\ne 0\\ \sqrt{\epsilon (4\max \{\widetilde{a},\widetilde{d}\}+3)}, &{} \widetilde{a}\widetilde{d}= 0 \end{array}\right. . \end{aligned}$$

  By applying (5.39) and (5.40) (for $\omega =i$) we get $|\widetilde{a}+\epsilon _1|^{2}(1-\delta )\le \big (\sqrt{| \widetilde{a}\widetilde{d}|}+|\epsilon _2'|\big )(\delta +\delta ^{2})$, which fails for $\widetilde{a}\ne 0$ and small $\epsilon ,\delta$. For $c(s)=e^{i\frac{\widetilde{\vartheta }}{2}}$, $P(s)=\frac{1}{s}e^{-i\frac{\pi }{4}}\begin{bmatrix} -e^{-i\alpha } &{} -ie^{i\alpha }\sqrt{\widetilde{d}+s} \\ e^{i\alpha } &{} ie^{-i\alpha }\sqrt{\widetilde{d}+s} \end{bmatrix}$ with $\sin (2\alpha (s)) =\frac{s^{2}}{2|\sqrt{\widetilde{d}+s}|}$, $\cos \frac{\theta }{2}=s^{3}$, $B(s)=|\sqrt{\widetilde{d}+s}|\begin{bmatrix} 2 &{} 1 \\ 1 &{} 0 \end{bmatrix}$ in (5.2), it follows $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, 1\oplus \widetilde{d}\big )\rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} a &{} b\\ b &{} 0 \end{bmatrix}\big )$, $\pi> \widetilde{\vartheta } =\arg \widetilde{d}> 0$ or $\widetilde{d}=0$. Taking $c(s)=1$, $P(s)=\frac{1}{\sqrt{\widetilde{d}+s}}\begin{bmatrix} 0 &{} \frac{1}{s}(\widetilde{d}+s) \\ s &{} -\frac{1}{s}(\widetilde{d}+s) \end{bmatrix}$, $B(s)=\begin{bmatrix} 2b(s)+s^2 &{} b(s) \\ b(s) &{} 0 \end{bmatrix}$, $b(s)\rightarrow \widetilde{b}$, $\cos (\frac{\theta }{2})=\left\{ \begin{array}{ll} \frac{s^{2}}{2(\widetilde{d}+s)}, &{} \omega =i\\ s^{3}, &{} \omega =0 \end{array}\right.$ proves $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} \omega \end{bmatrix}, \begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} \widetilde{d} \end{bmatrix}\big )\rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} a &{} b\\ b &{} 0 \end{bmatrix}\big )$, $b\ge \widetilde{b}\ge 0$, either $\omega =0$, $\widetilde{b}>0$, $\widetilde{d}=1$ or $\omega =i$, $\widetilde{d},\widetilde{b}\ge 0$.

• (d) $B=\begin{bmatrix} a &{} b\\ b &{} d \end{bmatrix}$, $a,d>0$, $b\in {\mathbb {C}}$

  First let $b=0$. We deal with the case $\omega =0$ in the same manner as in Case VII (a) (ii) (compare also (5.34) and (5.39); observe that the proof works in the case $a>d$, too). If $\omega =i$ we have $|v|^{2}\ge 1$, $\tfrac{|u|^{2}}{|v|^{2}}\le \delta \le \frac{1}{2}$ and using (5.39) we easily verify

  $$\begin{aligned} \tfrac{|x|^{2}}{|y|^{2}}=\tfrac{|u|^{2}+\delta }{|v|^{2}-\delta } =\tfrac{|u|^{2}}{|v|^{2}}+\tfrac{|v|^{2} \delta +|u|^{2}\delta }{(|v|^{2}-\delta )|v|^{2}} \le \delta +\tfrac{|v|^{2}\delta +\frac{1}{2}|v|^{2}\delta }{\frac{1}{2}|v|^{2}}\le 4\delta . \end{aligned}$$

  Multiplying the second equation of (5.6) with $\delta _5=\frac{x}{y}$ and $\delta _6=\frac{u}{v}$ yields

  $$\begin{aligned} ax^{2}+dv^{2}\delta _6\delta _5=(\widetilde{b}+\epsilon _2) \delta _5,\qquad ay^{2}\delta _6\delta _5+du^{2} =(\widetilde{b}+\epsilon _2)\delta _6. \end{aligned}$$

  By adding them and using (5.6) yields a contradiction for $\widetilde{a}\ne 0$ and small $\epsilon ,\delta$:

  $$\begin{aligned} (\widetilde{a}+\epsilon _2)+(\widetilde{d}+\epsilon _2) \delta _6\delta _5=(\widetilde{b}+\epsilon _2)(\delta _5+\delta _6). \end{aligned}$$

  It is tedious to find $c(s)=1$, $\cos (\frac{\theta (s)}{2})=s^{2}$, $B(s)=\frac{1}{2}\begin{bmatrix} \widetilde{a}s^{-2} &{} \widetilde{a}s^{-2}-2ds^2 \\ \widetilde{a}s^{-2}-2ds^2 &{} \widetilde{a}s^{-2} \end{bmatrix}$, $P(s)=\frac{1}{\sqrt{2}}\begin{bmatrix} s &{} s^{-1}\\ s &{} -s^{-1} \end{bmatrix}$ in (5.2) to prove $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, \widetilde{a}\oplus \widetilde{d}\big )\rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} a &{} b\\ b &{} d \end{bmatrix}\big )$, $a,d,\widetilde{a}> 0$, $b\in {\mathbb {C}}^{*}$, $\widetilde{d}\in {\mathbb {C}}$.

Case X.:

$\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} \omega \end{bmatrix}, \widetilde{B}\big )\dashrightarrow \big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} a &{} b\\ b &{} d \end{bmatrix}\big )$, $0< \tau <1$, $\omega \in \{0,i\}$

The following expressions are bounded by $\delta$ (Lemma 3.2 (2) (C4) for $\alpha =1$):

$$\begin{aligned} \textrm{Re}(\overline{x}u), (1-\tau )\textrm{Im}(\overline{x}u), \textrm{Re}(\overline{y}v),(1-\tau )\textrm{Im}(\overline{y}v) -(-1)^{k}|\omega |, 1-\tau , \overline{x}v+\overline{u}y-(-1)^{k}, \nonumber \\ \end{aligned}$$

(5.41)

where $k\in {\mathbb {Z}}$. If in addition $\omega =i$, it then follows that

$$\begin{aligned}&\delta _5=\tfrac{|xu|}{|yv|}=\tfrac{(1-\tau )|xu|}{(1-\tau )|yv|} \le \tfrac{\big |(1-\tau )\textrm{Re}(\overline{x}u)\big |+\big |(1-\tau ) \textrm{Im}(\overline{x}u)\big |}{\big |(1-\tau )\textrm{Im}(\overline{y}v)\big | -\big |(1-\tau )\textrm{Re}(\overline{y}v)\big |} \le \tfrac{\delta +\delta ^{2}}{1-\delta -\delta ^{2}},\nonumber \\&\delta |yv|\ge \big |(1-\tau )\textrm{Im}(\overline{y}v)\big | \ge 1-\delta , \end{aligned}$$

(5.42)

$$\begin{aligned}&(1+\delta )\tfrac{|v|}{|u|}\ge |\overline{u}y+\overline{x}v| \tfrac{|v|}{|u|}\ge |vy|-\tfrac{|xv|}{|uy|}|vy|= |vy| \big (1-\tfrac{|xu|}{|vy|}\,|\tfrac{v}{u}|^{2}\big ),\nonumber \\&(1+\delta )\tfrac{|y|}{|x|}\ge |\overline{u}y +\overline{x}v|\tfrac{|y|}{|x|}\ge |vy| -\tfrac{|uy|}{|xv|}|vy|= |vy|\big (1-\tfrac{|xu|}{|vy|}\,| \tfrac{y}{x}|^{2}\big ). \end{aligned}$$

(5.43)

• (a) $B=a\oplus d$

  Let $B=0\oplus 1$. If $\widetilde{a}\ne 0$ (hence $\widetilde{b}=\widetilde{d}=0$, $\omega =i$) then (5.4) for $a=b=\widetilde{d}=0$, $d=1$ yields $(\frac{v}{u})^{2}=\frac{\epsilon _2}{\widetilde{a}+\epsilon _1}$, thus (5.42), (5.43) give a contradiction for small $\epsilon ,\delta$. Taking $c(s)=1$, $\tau (s)=1-s$, $P(s)=\frac{1}{\sqrt{\widetilde{d}+s}}\begin{bmatrix} 1 &{} -i s^{-1} \\ 0 &{} \widetilde{d}+s \end{bmatrix}$ proves $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, 0\oplus \widetilde{d}\big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, 0\oplus 1\big )$, $\widetilde{d}\ge 0$.

  Next, $B=1\oplus d$, $d\in {\mathbb {C}}$. If either $|\frac{x}{y}|\ge 1$ (or $|\frac{u}{v}|\ge 1$), then in case $\omega =i$ the second (the first) inequality of (5.43) yields a contradiction. When $|\frac{x}{y}|,|\frac{u}{v}|\le 1$ we multiply the second equation of (5.4) for $b=0$, $a=1$ with $\frac{u}{v}$ and $\frac{x}{y}$, and simplify them:

  $$\begin{aligned} \delta _5y^{2}+du^{2}=(\widetilde{b}+\epsilon _2)\tfrac{u}{v}, \quad x^{2}+\delta _5dv^{2}=(\widetilde{b}+\epsilon _2)\tfrac{x}{y} \quad \qquad (\delta _5 \le \tfrac{\delta +\delta ^{2}}{1-\delta -\delta ^{2}}). \end{aligned}$$

  We add these equations and use (5.4) for $b=0$, $a=1$ to get $\delta _5(\widetilde{d}+\epsilon _4)+(\widetilde{a}+\epsilon _1) =(\widetilde{b}+\epsilon _2)\tfrac{u}{v}+(\widetilde{b}+\epsilon _2) \tfrac{x}{y}$. Since $|\frac{x}{y}|,|\frac{u}{v}|\le 1$, it fails for $\widetilde{a}\ne 0$, $\widetilde{b}=0$ and small $\epsilon ,\delta$. Finally, $c(s)=1$, $\tau (s)=1-s^{2}$, $P(s)=\frac{1}{\sqrt{\widetilde{b}}}e^{-i\frac{\pi }{4}}\begin{bmatrix} s^{2}e^{i\frac{\pi }{4}} &{} \widetilde{b} s^{-1} \\ s &{} s^{-1} \end{bmatrix}$, $B(s)=1\oplus \widetilde{b}^{2}$ $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, \begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 0 \end{bmatrix}\big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, 1\oplus d\big )$, while, $c(s)=1$, $\tau (s)=1-s^{3}$, $P(s)=\frac{1}{\sqrt{2}}e^{i\frac{\pi }{4}}\begin{bmatrix} s\widetilde{b}e^{-i\alpha (s)} &{} -is^{-1}e^{i\alpha (s)} \\ -is e^{i\alpha (s)} &{} (\widetilde{b}s)^{-1}e^{-i\alpha (s)} \end{bmatrix}$, $B(s)=1\oplus \widetilde{b}^{2}e^{4\alpha (s)+\beta (s)}$, $\sin (\alpha (s))=s^{3}$, $\sin (\frac{\beta (s)}{2})=-s^{2}$ gives $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 1 \end{bmatrix}\big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, 1\oplus d\big )$, $\widetilde{b} > 0$, $d\in {\mathbb {C}}$.

• (b) $B=\begin{bmatrix} 0 &{} b\\ b &{} e^{i\varphi } \end{bmatrix}$,    $0\le \varphi <\pi$, $b>0$

  Let $a=0$ and $\widetilde{B}=\widetilde{a}\oplus \widetilde{d}$. Lemma 3.1 (D1) for $\widetilde{a}\ne 0$ implies $\tfrac{v}{u}=\tfrac{i(-1)^{l} \sqrt{\widetilde{a}\widetilde{d}} +\epsilon _2'}{\widetilde{a}+\epsilon _1}=i(-1)^{l} \sqrt{\frac{\widetilde{d}}{\widetilde{a}}}+\epsilon _2''$, $|\epsilon _2'|\le \left\{ \begin{array}{ll} \tfrac{\epsilon (4|\max \{\widetilde{d},\widetilde{a}\}|+2 +|\widetilde{a}\widetilde{d}|)}{|\widetilde{a}\widetilde{d}|}, &{} \det \widetilde{B}\ne 0\\ \sqrt{\epsilon (4|\max \{\widetilde{d},\widetilde{a}\}|+3)}, &{} \det \widetilde{B}= 0 \end{array}\right.$, $|\epsilon _2''|\le \frac{2}{\widetilde{a}}(|\epsilon _2'|+\epsilon \sqrt{\frac{\widetilde{d}}{\widetilde{a}}})$, $l\in {\mathbb {Z}}$, provided that $\epsilon \le \frac{|\widetilde{a}|}{2}$. It contradicts (5.42), (5.43) for $\omega =i$. If $\widetilde{a}=1$, $\widetilde{d}=|\widetilde{d}|e^{i\widetilde{\vartheta }}\ne 0$, $0<\widetilde{\vartheta }<\pi$ we apply (4.3) to deduce $\psi =\kappa -\eta -\frac{\widetilde{\vartheta }}{2}-\frac{\pi }{2}-l\pi$ with $|\sin \psi |\le \frac{|\epsilon _2''|}{|\sqrt{\widetilde{d}}|}$. Hence

  $$\begin{aligned} \overline{x}v+\overline{u}y&=\overline{x}u\tfrac{v}{u} +y\overline{v}\tfrac{\overline{u}}{\overline{v}} =-(-1)^{l}e^{i(\frac{\widetilde{\vartheta }}{2}+\psi )} \big (\textrm{Im}(\overline{x}u)|\tfrac{v}{u}|+\textrm{Im}(y\overline{v})| \tfrac{\overline{u}}{\overline{v}}|\big )\\&\quad +\textrm{Re}(\overline{x}u)\tfrac{v}{u}+\textrm{Re}(y\overline{v}) \tfrac{\overline{u}}{\overline{v}}. \end{aligned}$$

  Using (5.41)) and $\big ||\tfrac{v}{u}|-|\sqrt{\frac{\widetilde{d}}{\widetilde{a}}}| \big |\le |\epsilon _2''|$, the above calculation and (4.3) gives

  $$\begin{aligned} \psi '=k\pi -\big (\tfrac{\widetilde{\vartheta }}{2}+\psi +(l+1)\pi \big ), \quad |\sin \psi '|\le 2\delta \left( 1+|\sqrt{\tfrac{\widetilde{d}}{\widetilde{a}}}| +|\epsilon _2''|+(|\sqrt{\tfrac{\widetilde{d}}{\widetilde{a}}}| -|\epsilon _2''|)^{-1} \right) , \end{aligned}$$

  which fails for small $\epsilon ,\delta$ (recall $|\sin \psi |\le \frac{|\epsilon _2''|}{|\sqrt{\widetilde{d}}|}$, $0<\widetilde{\vartheta }<\pi$). Next, $c(s)=-1$, $P(s)=\begin{bmatrix} -\frac{2is}{3\widetilde{b}} &{} \frac{1}{s}\\ \frac{s}{3} &{} \frac{2i\widetilde{b}}{s} \end{bmatrix}$, $B(s)=\begin{bmatrix} 0 &{} \widetilde{b} \\ \widetilde{b} &{} i \end{bmatrix}$, $\tau (s)=1-\frac{s^{2}}{2\widetilde{b}}$ implies $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, \begin{bmatrix} 0 &{} \widetilde{b}\\ \widetilde{b} &{} 0 \end{bmatrix}\big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} b\\ b &{} e^{i\varphi } \end{bmatrix}\big )$, $\widetilde{b}> 0$.

  For $\tau (s)=\left\{ \begin{array}{ll} 1-s\sqrt{\widetilde{a}+s}, &{} \omega =i\\ 1-s^{2}, &{} \omega =0 \end{array}\right.$, $P(s)=e^{i\frac{\pi }{4}}\begin{bmatrix} \sqrt{\widetilde{a}+s} &{} \frac{-i}{s}\\ s^{3}e^{-i\frac{\pi }{4}} &{} \frac{1}{\sqrt{a}+s} \end{bmatrix}$, $B(s)=\begin{bmatrix} -i &{} \frac{\sqrt{a}+s}{s} \\ \frac{\sqrt{\widetilde{a}+s}}{s} &{} -i(\widetilde{a}+s) (\widetilde{d}-\frac{1}{s^{2}}) \end{bmatrix}$, $c(s)=-1$ we get $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} \omega \end{bmatrix},\widetilde{a}\oplus \widetilde{d}, \big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} e^{i\varphi } &{} b\\ b &{} d \end{bmatrix}\big )$, $d\in {\mathbb {C}}$, $\widetilde{a}\ge 0$.

• (c) $B=\begin{bmatrix} 0 &{} b \\ b &{} 0 \end{bmatrix}$, $b>0$

  We multiply the first and the second equality of (D1) and (D2) of Lemma 3.1 for $\widetilde{a}\ne 0$ to get a contradiction with (5.42) for $\omega =i$ and small $\epsilon ,\delta$. Taking $c(s)=1$, $\tau (s)=1-s$, $P(s)=e^{-i\frac{\pi }{4}}\begin{bmatrix} se^{i\frac{\pi }{4}} &{} s^{-1} \\ s &{} i \end{bmatrix}$, $B(s)=\frac{\widetilde{d}+s}{2}s\begin{bmatrix} 0 &{} 1 \\ 1 &{} 0 \end{bmatrix}$ shows $\big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, 0\oplus \widetilde{d}\big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}\big )$.

  For $\widetilde{a}=0$, $\widetilde{b}>0$ we have $b=\widetilde{b}+\delta '$, with $|\delta '|\le \delta \widetilde{b}^{2}+\max \{\epsilon , \frac{\delta ^{2}}{\nu ^{2}}\}(4\max \{\widetilde{b}, \widetilde{b}^{2},1\}+2)^{2}$ (see Lemma 4.1 (3 and (4.7)); recall $1-\tau \le \delta$). If $\omega =i$ (hence $\widetilde{d}=0$) the last equation of (5.12) for $\widetilde{d}=0$ contradicts the second estimate of (5.42). Next, let $\omega =0$ (hence $\widetilde{d}=1$). Using $2bvy=1+\epsilon _4$ (see (5.12)) and $|\textrm{Re}(\overline{y}v)|\le \delta$ (see (5.41)), we have $|\textrm{Im}(\overline{y}v)|\ge |yv|-|\textrm{Re}(\overline{y}v)|\ge \frac{1-\epsilon }{\overline{b}+|\delta '|}-\delta$. Further Lemma 3.1 gives $2bvx=((-1)^{l+1}+1)\widetilde{b}+\epsilon _2'$, $2buy=((-1)^{l}+1)\widetilde{b}+\epsilon _2''$, $l\in {\mathbb {Z}}$, where $|\epsilon _2'|,|\epsilon _2''|\le \frac{\epsilon (4\max \{1,\widetilde{b}\}+2 +\widetilde{b}^{2})}{\widetilde{b}^{2}}$. So either $2bvx=2\widetilde{b}+\epsilon _2'$, $2buy=\epsilon _2''$ or $2buy=2\widetilde{b}+\epsilon _2''$, $2bvx=\epsilon _2'$. In the first case we also have $\overline{x}v=(-1)^{k}+\delta _2'$ with $|\delta _2'|\le \delta +\frac{|\epsilon _2''|}{2(\widetilde{b}-|\delta '|)}$ (see (5.41)). We combine all facts:

  $$\begin{aligned} |\tfrac{y}{x}|^{2}=\tfrac{2bvy \,\overline{y}v}{2bvx\,\overline{x}v} =\tfrac{(1+\epsilon _4)(i\textrm{Im}(\overline{y}v)+\delta _0)}{(2\overline{b}+\epsilon _2')((-1)^{k}+\delta _2')} \end{aligned}$$

  For sufficiently small $\epsilon ,\delta$ the right-hand (the left-hand) side is (not) real, a contradiction. The other case is treated similarly and yields a contradiction as well.

Case XI.:

$(1\oplus 0, \widetilde{B}) \dashrightarrow (1\oplus 0,B)$

If $B=\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}$, $\widetilde{B}=\widetilde{a}\oplus 1$, $\widetilde{a}\ge 0$, then [22, Theorem 3.6, Case XI (a)] applies. (Taking $c(s)=1$, $P(s)=\begin{bmatrix} 1 &{} s \\ \frac{\widetilde{a}}{2} &{} 0 \end{bmatrix}$ in (5.2) proves ${\ (1\oplus 0, \widetilde{a}\oplus 0)\rightarrow \big (1\oplus 0, \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}\big )}$.)

Next, Lemma 3.2 (2) with (C12) for $\alpha =1$ gives $\big ||x|^2-1\big |\le \delta$ and $|y|^{2}\le \delta$, hence $|\frac{y}{x}|^{2}\le \frac{\delta }{1-\delta }$. When $B=a\oplus 0$ for $a\ge 0$, then dividing the last two equalities of (5.4) for $b=d=\widetilde{b}=0$, $\widetilde{d}=1$ yields $\frac{x}{y}=\frac{\epsilon _2}{1+\epsilon _4}$, which contradicts $|\frac{y}{x}|^{2}\le \frac{\delta }{1-\delta }$ for small $\epsilon ,\delta$.

Finally, $c(s)=1$, $P(s)=\begin{bmatrix} 1 &{} 0 \\ \sqrt{\widetilde{a}-a} &{} s \end{bmatrix}$ in (5.2) proves $(1\oplus 0,\widetilde{a}\oplus 0)\rightarrow (1\oplus 0,a\oplus 1)$, $a\ge 0$, and $c(s)=1$, $P(s)=\begin{bmatrix} i &{} s^{3} \\ s^{-1} &{} s \end{bmatrix}$, $B(s)=\frac{1}{s^{2}}\oplus 1$ implies $\big (1\oplus 0, \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}\big )\rightarrow \big (1\oplus 0, a\oplus 1\big )$, $a>0$.

Case XII.:

$(1\oplus 0, \widetilde{B})\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, B\big )$

Lemma 3.2 (2) for (C5) for $\alpha =1$, $\beta =\omega =0$ yields

$$\begin{aligned} 2\textrm{Re}(\overline{x}u)=(-1)^{k}+\delta _1,\, 2\textrm{Re}(\overline{y}v)=\delta _2, \, \overline{x}v+\overline{u}y = \delta _4, \quad k\in {\mathbb {Z}}, |\delta _1|,| \delta _2|,|\delta _4|\le \delta . \end{aligned}$$

(5.44)

Next, (5.26) (compare (5.44) with (5.23)) is valid in this case as well. Since $|\det P|\le \frac{\delta \sqrt{6}}{\nu }$ by Lemma 4.1 (1), it follows from (5.26) that

$$\begin{aligned} |\tfrac{v}{u}|,|\tfrac{y}{x}|\le \delta \tfrac{\nu +\sqrt{6}}{\nu (1-\delta )}. \end{aligned}$$

(5.45)

• (a) $B=1\oplus 0$

  The bundle consists of one orbit, hence [22, Theorem 3.6, Case XV (c)] applies. (We take $c(s)=1$ and $P(s)=\begin{bmatrix} \sqrt{\widetilde{a}+s} &{} 0 \\ \frac{1}{2\sqrt{\widetilde{a}+s}}&{} s \end{bmatrix}$ to get $(1\oplus 0, \widetilde{a}\oplus 0)\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, 1\oplus 0\big )$ for $\widetilde{a}\ge 0$.)

• (b) $B=1\oplus d$,    $\textrm{Im}(d)>0$

  For $\widetilde{B}=\widetilde{a} \oplus 1$ we have (5.6) with $\widetilde{b}=0$, $a=1$. By multiplying the second equation of (5.6) for $\widetilde{b}=0$ with $\delta _4:=\frac{v}{u}$, $\delta _5:=\frac{y}{x}$ and by simplifying it we obtain

  $$\begin{aligned} ax^{2}\delta _4\delta _5+dv^{2}=\epsilon _2\delta _4, \qquad ay^{2}+dv^{2}\delta _4\delta _5=\epsilon _2\delta _5, \end{aligned}$$

  (5.46)

  respectively. We add these equalities and using the first and the last equation of (5.6) we get the equality that fails for $\widetilde{d}\ne 0$ and $\epsilon ,\delta$ small enough (recall (5.45)):

  $$\begin{aligned} \epsilon _2(\delta _4+\delta _5)=(ax^{2}+du^{2})\delta _4\delta _5 +(ay^{2}+dv^{2})=(\widetilde{a}+\epsilon _1)\delta _4\delta _5 +\widetilde{d}+\epsilon _4. \end{aligned}$$

  (5.47)

  Note that $\big (1\oplus 0, \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}\big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, 1\oplus d\big )$ will follow after we prove $\big (1\oplus 0, \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}\big )\rightarrow \big (1\oplus -1,\begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}\big )$ (see Case XIV (a)).

• (c) $B=\begin{bmatrix} 0 &{} b\\ b &{} 1 \end{bmatrix}$, $b>0$

  Let $\widetilde{B}=\widetilde{a}\oplus 1$, $\widetilde{a}\ge 0$. From Lemma 3.1 (D1) for $\widetilde{b}=0$, $\widetilde{d}=1$ we get:

  $$\begin{aligned} |u|\le \tfrac{\sqrt{\widetilde{a}}+|\epsilon _2''|}{1-\epsilon }|v|, \end{aligned}$$

  (5.48)

  which clearly contradicts (5.45) for sufficiently small $\epsilon ,\delta$.

  For $P(s)=\begin{bmatrix} -\frac{1}{2}s &{} s^{4}\\ s^{-1} &{} 2s \end{bmatrix}$, $c(s)=1$, $B(s)=\begin{bmatrix} 0 &{} s^{-2} \\ s^{-2} &{} 1 \end{bmatrix}$ we show $\big (1\oplus 0, \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}\big ) \rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} b\\ b &{} 1 \end{bmatrix}\big )$.

Case XIII.:

$(1\oplus 0, \widetilde{B})\dashrightarrow \big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} a &{} b\\ b &{} d \end{bmatrix}\big )$, $0\le \tau <1$

From Lemma 3.2 (2) for (C3) with $\alpha =1$ we get

$$\begin{aligned}&\textrm{Re}(\overline{y}v)\le \delta , \quad (1-\tau ) \textrm{Im}(\overline{y}v)\le \delta , \quad (1-\tau )| \overline{x}v|\le \delta , \quad (1-\tau )| \overline{u}y|\le \delta ,\nonumber \\&\overline{x}v+\overline{u}y\le \delta , \qquad \big |(1+\tau ) \textrm{Re}(\overline{x}u)+i(1-\tau )\textrm{Im}(\overline{x}u)-\tfrac{1}{c}\big | \le \delta . \end{aligned}$$

(5.49)

The last estimate yields either $\big |(1+\tau )\textrm{Re}(\overline{x}u)\big |\ge \frac{1-\delta }{2}$ or $\big |(1-\tau )\textrm{Im}(\overline{x}u)\big |\ge \frac{1-\delta }{2}$, thus

$$\begin{aligned} |\overline{x}u|\ge \tfrac{1-\delta }{4}. \end{aligned}$$

(5.50)

• (a) $B=\begin{bmatrix} a &{} b\\ b &{} d \end{bmatrix}$, either $b> 0$ or $b=0$ and $ad=0$

  First, let $\widetilde{B}=\widetilde{a}\oplus 1$, $\widetilde{a}\ge 0$; we have (5.48). Using (5.49), (5.50) we thus get

  $$\begin{aligned} \delta \tfrac{\sqrt{\widetilde{a}}+|\epsilon _2''|}{|1-\epsilon |} \ge (1-\tau )|\overline{x}v||\tfrac{u}{v}|=(1-\tau )|\overline{x}u| \ge \tfrac{1}{4}(1-\tau ). \end{aligned}$$

  Similarly, when $d=0$ then Lemma 3.1 (D2) for $\widetilde{b}=0$, $\widetilde{d}=1$ and (5.49), (5.50) yield $|\frac{x}{y}|\le \frac{\sqrt{\widetilde{a}}+|\epsilon _2'|}{1-\epsilon }$ and $\tfrac{4\delta (\sqrt{\widetilde{a}}+|\epsilon _2'|)}{1-\epsilon }\ge 1-\tau$. From Lemma 4.1 (1) we obtain $\sqrt{\tau }|\det P|\le \frac{\delta \sqrt{6}}{\nu }$. By combining the above statements with (5.24), (5.25) we get $\textrm{Re}(\overline{x}u)\le C\delta$, where a constant $C>0$ can be computed. Hence $(1-\tau ) \textrm{Im}(\overline{x}u)\ge 1-\delta -C\delta$, and further

  $$\begin{aligned} \frac{|yv|}{|xu|}=\frac{(1-\tau )|yv|}{(1-\tau )|xu|} \le \tfrac{\big |(1-\tau )\textrm{Im}(\overline{y}v)\big |+\big |(1-\tau ) \textrm{Re}(\overline{y}v)\big |}{\big |(1-\tau )\textrm{Im}(\overline{x}u)\big | -\big |(1-\tau )\textrm{Re}(\overline{x}u)\big |} \le \frac{2\delta }{1-\delta -2C\delta }. \end{aligned}$$

  (5.51)

  It is also easy to validate

  $$\begin{aligned} |\overline{x}v+\overline{u}y||\tfrac{u}{v}|\ge |ux|\big |1 -\tfrac{|yv|}{|xu|}|\tfrac{u}{v}|^{2}\big |, \qquad |\overline{x}v+\overline{u}y||\tfrac{x}{y}| \ge |ux|\big |1-\tfrac{|yv|}{|xu|}|\tfrac{x}{y}|^{2}\big |. \end{aligned}$$

  (5.52)

  We apply (5.49) and the estimates on $|\frac{u}{v}|$, $|\frac{x}{y}|$, $|\frac{yv}{xu}|$ to (5.52) to get a contradiction for small $\epsilon ,\delta$. Next, $P(s)=\begin{bmatrix} -se^{i(\alpha (s)+\frac{\pi }{4})} &{} s^{3} \\ s^{-1}e^{i\frac{\pi }{4}} &{} 1 \end{bmatrix}$, $B(s)=\begin{bmatrix} s^{-4}e^{-i\alpha (s)} &{} s^{-2} \\ s^{-2} &{} 1 \end{bmatrix}$, $c(s)=-1$, $\tau (s)\rightarrow 0$, $\sin (\frac{\alpha (s)}{2})=\frac{\widetilde{a}s^{2}}{2}$ implies $(1\oplus 0, \widetilde{a}\oplus 1) \rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 0 &{} 0 \end{bmatrix}, \begin{bmatrix} \zeta ^{*} &{} b\\ b &{} 1 \end{bmatrix}\big )$, $\zeta ^{*}\in {\mathbb {C}}^{*}$.

  Let $B=\begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}$, $b>0$, $\widetilde{B}=\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}$. The first (the second) equation of (5.12) for $\widetilde{a}=0$ (for $\widetilde{b}=1$) combined with (5.50) (with (5.49) for $0\le \tau \le \frac{1}{2}$) yields $\epsilon \ge b|ux|\ge b\frac{1-\delta }{4}$ (and $1+\epsilon \ge b|vx+uy|\ge 4b\delta$), thus a contradiction for sufficiently small $\epsilon ,\delta$ and $0\le \tau \le \frac{1}{2}$. If $1\ge \tau \ge \frac{1}{2}$ then Lemma 4.1 (1), (2) leads to $|\det P|\le \frac{2\sqrt{3}\delta }{\nu }$ and $b|\det P|\ge 1- 6\epsilon$, hence $\frac{8\sqrt{3}\epsilon \delta }{\nu (1-\delta )}\ge b \frac{2\sqrt{3}\delta }{\nu }\ge 1-6\epsilon$, which fails for small $\epsilon ,\delta$. Taking $P(s)=\begin{bmatrix} -s &{} s^{4} \\ s^{-1} &{} 2s \end{bmatrix}$, $B(s)=\begin{bmatrix} \zeta &{} \frac{1}{2}s^{-2} \\ \frac{1}{2}s^{-2} &{} 1 \end{bmatrix}$ with $\frac{\zeta }{s^{2}}\rightarrow 0$ and $P(s)=\begin{bmatrix} s^{-1} &{} 2s \\ -s &{} s^{4} \end{bmatrix}$, $B(s)=\begin{bmatrix} 1 &{} \frac{1}{2}s^{-2} \\ \frac{1}{2}s^{-2} &{} 0 \end{bmatrix}$ (both with $c(s)=-1$, $\tau (s)\rightarrow 0$) in (5.2) proves $\big (1\oplus 0, \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}\big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} \zeta &{} b\\ b &{} e^{i\varphi } \end{bmatrix}\big )$, $\zeta \in {\mathbb {C}}$ and $\big (1\oplus 0, \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}\big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} e^{i\varphi } &{} b\\ b &{} 0 \end{bmatrix}\big )$ with $b>0$, $0\le \varphi <\pi$, respectively.

  Finally, to see $(1\oplus 0, \widetilde{a}\oplus 0)\rightarrow \big (\begin{bmatrix} 0 &{} 1 \\ \tau &{} 0 \end{bmatrix}, B\big )$, $0\le \tau <1$, $\widetilde{a}\ge 0$, where B is any of the matrices $\begin{bmatrix} a &{} b\\ b &{} e^{i\varphi } \end{bmatrix}$, $a,b\ge 0$ and $\begin{bmatrix} e^{i\varphi } &{} b\\ b &{} d \end{bmatrix}$, $d,b\ge 0$, we take $P(s)=\begin{bmatrix} \frac{1}{\sqrt{\widetilde{a}+s}} &{} s \\ \sqrt{\widetilde{a}+s} &{} s \end{bmatrix}$, $B(s)=\begin{bmatrix} a(s) &{} b(s) \\ b(s) &{} 1 \end{bmatrix}$ with $b(s)\rightarrow 0$, $\frac{a(s)}{\sqrt{s}}\rightarrow 0$ or $P(s)=\begin{bmatrix} \sqrt{\widetilde{a}+s} &{} s\\ \frac{1}{\sqrt{\widetilde{a}+s}} &{} s \\ \end{bmatrix}$, $B(s)=\begin{bmatrix} 1 &{} b(s) \\ b(s) &{} d(s) \end{bmatrix}$ with $b(s)\rightarrow 0$, $\frac{d(s)}{\sqrt{s}}\rightarrow 0$ in (5.2) ($c(s)=1$, $\tau (s)\rightarrow 0$ in both cases). To prove $(1\oplus 0, \widetilde{a}\oplus 0)\rightarrow \big (\begin{bmatrix} 0 &{} 1 \\ \tau &{} 0 \end{bmatrix}, \begin{bmatrix} 0 &{} b \\ b &{} 0 \end{bmatrix}\big )$, $b> 0$, we put $P(s)=\begin{bmatrix} 1 &{} s \\ 1 &{} 0 \end{bmatrix}$, $B(s)=\frac{\widetilde{a}+s}{2}\begin{bmatrix} 0 &{} 1 \\ 1 &{} 0 \end{bmatrix}$, $c(s)=1$, $\tau (s)\rightarrow 0$ in (5.2).

• (b) $B=a\oplus d$,    $a,d\ne 0$

  For $c(s)=-i$, $P(s)=\begin{bmatrix} s &{} s^{3} \\ is^{-1} &{} s^{2} \end{bmatrix}$, $B(s)=\frac{1}{s^{4}}\oplus 1$ we get $\big (1\oplus 0,\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}\big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 0 &{} 0 \end{bmatrix}, a\oplus 1\big )$.

  If $\tau \le \frac{1}{2}$ then we have $|xv|,|uy|\le 2\delta$, thus using (5.50 ) we get $|\frac{v}{u}|=|\frac{vx}{ux}|\le 8\delta$ and $|\frac{y}{x}|=|\frac{uy}{ux}|\le 8\delta$. On the other hand for $\tau \ge \frac{1}{2}$ we get $|\det P|\le \frac{2\sqrt{3}\delta }{\nu }$ (Lemma 4.1 (1)), therefore (5.24), (5.25), (5.49) imply $|\textrm{Re}(\overline{x}u)||\frac{v}{u}|,|\textrm{Re}(\overline{x}u)||\frac{v}{u}|\le 2\sqrt{3}\delta +\delta$. If $|\textrm{Re}(\overline{x}u)|\le \sqrt{2\sqrt{3}\delta +\delta }$, then $(1-\tau )|\textrm{Im}(\overline{x}u)|\ge 1-\sqrt{(2\sqrt{3}+1)\delta }$ and similarly as in (5.51) we obtain $|\frac{v}{u}|,|\frac{y}{x}|\le \frac{2\delta }{1-2\sqrt{(2\sqrt{3}+1)\delta }}$. If $\widetilde{B}=\widetilde{a}\oplus \widetilde{d}$ with $\widetilde{d}\ne 0$, then in any case we proceed mutatis mutandis as in Case XII (b) to get a contradiction for small $\epsilon ,\delta$.

Case XIV.:

$(1\oplus 0,\widetilde{B})\dashrightarrow (1\oplus e^{i\theta },B)$, $0\le \theta \le \pi$

From Lemma 3.2 (2) with (C1) for $\alpha =1$ and $0< \theta <\pi$ we have

$$\begin{aligned} \big ||x|^2+e^{i\theta }|u|^2-c^{-1}\big |\le \delta ,\, \big ||y|^2 +e^{i\theta }|v|^2\big |\le \delta , \, \sin (\theta ) |\overline{u}v| \le \delta , \, |\overline{x}y+\cos (\theta ) \overline{u}v|\le \delta . \end{aligned}$$

(5.53)

Furthermore, Lemma 3.2 (2) with (C10) for $\alpha =1$, $\sigma =-1$ yields

$$\begin{aligned} \big ||x|^2-|u|^{2}\big |=1 +\delta _1 , \quad \big ||y|^2-|v|^{2}\big |=\delta _4, \quad |\overline{x}y-\overline{u}v|= \delta _2, \quad |\delta _1|,|\delta _2|,|\delta _4|\le \delta ,\nonumber \\ \end{aligned}$$

(5.54)

while from (C10) for $\alpha =1$, $\sigma =1$ we deduce

$$\begin{aligned} |x|^2+|u|^{2}=1+ \delta _1 ,\qquad |y|^{2},|v|^{2} \le \delta ,\quad |\delta _1|\le \delta . \end{aligned}$$

(5.55)

• (a) $\widetilde{B}=\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}$

  Taking $c(s)=1$, $P(s)=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 &{} s\sqrt{2} \\ i &{} -is\sqrt{2} \end{bmatrix}$, $B(s)=\frac{\sqrt{2}}{2\,s}I_2$ gives $\big (1\oplus 0, \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}\big )\rightarrow (I_2, aI_2)$, $a> 0$. If $B=dI_2$ and $\theta =\pi$, then Lemma 4.1 (1), (2) gives $|\det P|\le \frac{\delta \sqrt{6}}{\nu }$ and $d|\det P|\ge 1-6\epsilon$. The first equation of (5.4) for $a=d$,$b=\widetilde{a}=0$ yields $\epsilon \ge \big |d(x^2+u^{2})\big |\ge |d|\big ||x|^2-|u|^{2}\big |\ge |d|(1-\delta )$ (see (5.54)). Thus $\frac{\epsilon \delta \sqrt{6}}{\nu } \ge (1-\delta )(1-6\epsilon )$, which fails for $\epsilon ,\delta \le \frac{1}{12}$.

• (b) $\widetilde{B}=\widetilde{a}\oplus 0$, $\widetilde{a}\ge 0$

  We take $c(s)=e^{-i\theta }$, $P(s)=\begin{bmatrix} s &{} s \\ 1 &{} s \end{bmatrix}$, $B(s)=\begin{bmatrix} a(s) &{} b(s) \\ b(s) &{} d(s) \end{bmatrix}$ with $d(s)\rightarrow \widetilde{a}$, $sa(s),b(s)\rightarrow 0$ to prove a path $(1\oplus 0, \widetilde{a}\oplus 0)\rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} a &{} b \\ b &{} d\end{bmatrix}\big )$ for $d> 0$, $b\ge 0$, $0\le \theta \le \pi$.

• (c) $\widetilde{B}=\widetilde{a}\oplus 1$, $\widetilde{a}\ge 0$

  • (i) $B=\begin{bmatrix} a &{} b\\ b &{} d \end{bmatrix}$, $|a|+|d|\ne 0$, $a\ne d$

    For $c(s)=1$, $P(s)=\begin{bmatrix} 1 &{} s \\ 0 &{} s \end{bmatrix}$, $B(s)=\begin{bmatrix} a(s) &{} b(s) \\ b(s) &{} s^{-2} \end{bmatrix}$, $a(s)\rightarrow \widetilde{a}$, $sb(s)\rightarrow 0$ and $c(s)=e^{-i\theta }$, $P(s)=\begin{bmatrix} 0 &{} s \\ 1 &{} s \end{bmatrix}$, $B(s)=\begin{bmatrix} s^{-2} &{} b(s) \\ b(s) &{} d(s) \end{bmatrix}$ with $d(s)\rightarrow \widetilde{a}$, $sb(s)\rightarrow 1$, we get $(1\oplus 0, \widetilde{a}\oplus 1)\rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} a &{} b\\ b &{} d \end{bmatrix}\big )$ for $b\ge 0$, $d>0$ and $b\ge 0$, $a>0$, respectively. Next, $c(s)=1$, $P(s)=\begin{bmatrix} -i &{} 0 \\ is &{} \frac{s}{\sqrt{\widetilde{a}+s}} \end{bmatrix}$, $B(s)=(\widetilde{a}+s)\begin{bmatrix} 0 &{} s^{-1} \\ s^{-1} &{} s^{-2} \end{bmatrix}$ and $c(s)=e^{-i\theta }$, $P(s)=\begin{bmatrix} is &{} \frac{s}{\sqrt{\widetilde{a}+s}}\\ -i &{} 0 \end{bmatrix}$, $B(s)=(\widetilde{a}+s)\begin{bmatrix} s^{-2} &{} s^{-1} \\ s^{-1} &{} 0 \end{bmatrix}$ in (5.2) imply $(1\oplus 0, \widetilde{a}\oplus 1)\rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} 0 &{} b\\ b &{} d \end{bmatrix}\big )$ and $(1\oplus 0, \widetilde{a}\oplus 1)\rightarrow \big (1\oplus e^{i\theta }, \begin{bmatrix} a &{} b\\ b &{} 0 \end{bmatrix}\big )$ for $b,a,d > 0$.

  • (ii) $B=\begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}$, $b> 0$    ($0<\theta \le \pi$)

    The second estimate of (5.53) gives $(\sin \theta ) |v|^{2}\le \delta$, thus either $|v|^{2}\le \sqrt{\delta }$ or $\sin \theta \le \sqrt{\delta }$ (or both). If $|v|^{2}\le \sqrt{\delta }$, then the second estimate of (5.53) (or (5.54)) implies $|y|^{2}\le \delta +\sqrt{\delta }$. Since we have (5.14) for $\widetilde{d}=1$, we further get $|u|,|x|\le (\delta +\sqrt{\delta }) \frac{|\sqrt{\widetilde{a}}|+\max \{|\epsilon _2'|,| \epsilon _2''|\}}{(1-\epsilon )}$, which contradicts the first estimate of (5.53) and (5.54).

    Let now $v,y\ne 0$ and $\sin \theta \le \sqrt{\delta }$. If $\theta \in (0,\frac{\pi }{4})$, then $(1-\cos \theta )|v|^{2}=2(\sin ^{2}\frac{\theta }{2})|v|^{2}\le 2(\sin ^{2}\theta )|v|^{2}$, hence the second estimate of (5.53) yields

    $$\begin{aligned} \delta \ge \big ||y|^{2}+\cos \theta |v|^{2}\big |\ge \big ||y|^{2} + |v|^{2}\big |-(1-\cos \theta )|v|^{2}\ge \big ||y|^{2}+|v|^{2}\big | -2\delta . \end{aligned}$$

    Hence $|y|^{2},|v|^{2}\le 3 \delta$ and it gives a contradiction again. If $\theta \in (\frac{3\pi }{4},\pi ]$, then $|\cos \frac{\theta }{2}|=|\sin \frac{\pi -\theta }{2}|\le |\sin (\pi -\theta )|$, and by combining it with the first equation in (5.14) for $\widetilde{d}=1$ and the third estimate of (5.53) we get $(\cos \frac{\theta }{2})|u|^{2}\le (\sin \theta )|uv||\frac{u}{v}|\le \delta \frac{\sqrt{\widetilde{a}}+|\epsilon _2'|}{1-\epsilon }$. Since $|x|^{2}+e^{i\theta }|u|^{2}=|x|^{2}- |u|^{2}+2(\cos \frac{\theta }{2})|u|^{2}e^{i\frac{\theta }{2}}$, the first estimate of (5.53) yields

    $$\begin{aligned} |x|^{2}-|u|^{2}=c^{-1}+\delta _5, \qquad |\delta _5|\le \delta +2\delta \tfrac{\sqrt{\widetilde{a}}+|\epsilon _2'|}{1-\epsilon } . \end{aligned}$$

    (5.56)

    Next, (5.14) for $\widetilde{d}=1$ yields $|\frac{x}{y}|,|\frac{u}{v}|\le \frac{|\sqrt{\widetilde{a}}| +\max \{|\epsilon _2'|,|\epsilon _2''|\}}{(1-\epsilon )}$. From the first estimate of (5.53) (or (5.54)) we deduce either $|x|^{2}\ge \frac{1-\delta }{2}$ or $|u|^{2}\ge \frac{1-\delta }{2}$, and the second estimate of (5.53) (or (5.54)) gives $|y|,|v|\ge \frac{(1-\epsilon )(1-\delta )}{2(|\sqrt{\widetilde{a}}|+\max \{|\epsilon _2'|,|\epsilon _2''|\})} -\sqrt{\delta }$. To conclude we use the (5.8) with (5.53), (5.54) and (5.56) to obtain an inequality that fails for small $\epsilon ,\delta$:

    $$\begin{aligned} \delta\ge & {} \big |\overline{x}y+(\cos \theta )\overline{u} v\big | \ge \big ||\overline{x}y|-|\overline{u} v|\big |-|uv||1-\cos \theta |\\\ge & {} \tfrac{1-|\delta _5|}{2\frac{|u|}{|v|} +\frac{\sqrt{1+|\delta _5|}}{|y|}}-\big (\tfrac{|u|}{|v |} +\tfrac{\sqrt{1+|\delta _5|}}{|v|}\big )\delta -2\delta . \end{aligned}$$

  • (iii) $B=aI_2$, $a>0$ (hence $A=1\oplus \sigma$, $\sigma =e^{i\theta }\in \{1, -1\}$)

    The first equation of (5.6) for $a=d$ and (5.54) yield

    $$\begin{aligned} \tfrac{\epsilon +|\widetilde{a}|}{a}\ge |x^2+u^{2}| \ge \big ||x|^2-|u|^{2}\big |. \end{aligned}$$

    (5.57)

    If $\sigma =1$, then the last equation of (5.6) for $a=d$ and the last estimates in (5.55) imply $\frac{1-\epsilon }{a}\le y^{2}+v^{2}\le 2\delta$. Hence (5.57) gives $\big ||x|^2-|u|^{2}\big |\le \delta _0:=\frac{2\delta (|\widetilde{a}|+\epsilon )}{1-\epsilon }$. The first equation of (5.6) further yields that $|x|,|u|\ge \frac{1-\delta }{2}-\delta _0$ with $\frac{|v|}{|u|},\frac{|y|}{|x|} \le \frac{2\delta }{\frac{1-\delta }{2}-2\delta _0}$. If $\widetilde{B}=\widetilde{a}\oplus 1$, we proceed mutatis mutandis as in Case XII (b) to get a contradiction.

    Let $\sigma =-1$. By Lemma 4.1 (1), (2) we have $a \frac{\delta \sqrt{6}}{\nu }\ge a|\det P|=|\sqrt{\widetilde{a}}+\delta '|$ with $\delta '\le \epsilon \frac{4\widetilde{a}+2}{\widetilde{a}}$ if $\widetilde{a}\ne 0$ (or $\delta '\le \epsilon \sqrt{4\widetilde{a}+2}$ if $\widetilde{a}= 0$). If $\widetilde{a}\ne 0$, we combine it with the first equality of (5.54) and (5.57), to obtain $\frac{\delta \sqrt{6}(\epsilon +|\widetilde{a}|)}{\nu (| \sqrt{\widetilde{a}}|-|\delta '|)}\ge \big ||x|^2-|u|^{2} \big |\ge (1-\delta )$, which fails for small $\epsilon ,\delta$. Next, if $\widetilde{a}=0$ then (5.57) and (5.54) imply $a\le \frac{\epsilon }{1-\delta }$. Using the second equation of (5.9) and (5.10) we deduce $\frac{u}{v}=\frac{ae^{2i\varphi }\delta _2 -\epsilon _4}{ae^{2i\varphi }\delta _4-1-\epsilon _4}$, while the last equation of (5.6) for $a=d$, $\widetilde{d}=1$ and the second inequality of (5.54) give $2|v^{2}|\ge \frac{1-\epsilon }{2a}-\delta$. Applying this and (5.54) to (5.8) leads to an inequality that fails for small $\epsilon ,\delta$.

Case XV.:

$(1\oplus 0, \widetilde{B}) \dashrightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix},B\big )$,

From Lemma 3.2 (2) for (C6) with $\alpha =1$, $c^{-1}=e^{i\Gamma }$ we deduce

$$\begin{aligned} |\overline{x}v+\overline{u}y|\le \delta , \quad |v|^{2},| \overline{u}v|\le \delta , \quad \big |2\textrm{Re}(\overline{y}v)\big | \le \delta , \quad \big |2\textrm{Re}(\overline{x}u)+i|u|^{2} -e^{i\Gamma }\big |\le \delta . \end{aligned}$$

(5.58)

• (a) $B=\begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}$, $b>0$

  If $\widetilde{B}=\widetilde{a}\oplus 1$ we again have (5.14) for $\widetilde{d}=1$, and by combining it with $|v|^{2}\le \delta$ (see (5.58)), we get $|u|\le \frac{(\sqrt{\widetilde{a}}+|\epsilon _2''|)\sqrt{\delta }}{1-\epsilon }$ with $\epsilon _2''$ is as in (5.14). The last estimate of (5.58) then yields $|2\textrm{Re}(\overline{x}u)| \ge 1+\delta +\tfrac{\delta (\sqrt{\widetilde{a}}+|\epsilon _2'|)^{2}}{(1-\epsilon )^{2}}$. By applying this, $\frac{\delta \sqrt{6}}{\nu }\ge |\det P|$ (Lemma 4.1 (1)) and the first estimate of (5.58) to (5.25) we get $|\frac{v}{u}|(1-\delta -\frac{\delta (\sqrt{\widetilde{a}} +|\epsilon _2''|)^{2}}{(1-\epsilon )^{2}})\le \delta -\frac{\delta \sqrt{6}}{\nu }$, which contradicts (5.14) for small $\epsilon ,\delta$.

  Taking $B(s)=\frac{1}{s}\begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}$, $P(s)=\begin{bmatrix} s^{2} &{} s \\ 1 &{} s \end{bmatrix}$, $c(s)=-i$ gives $\big (1\oplus 0, \begin{bmatrix} 0 &{} 1\\ 1 &{} 0 \end{bmatrix}\big )\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix}, \begin{bmatrix} 0 &{} b\\ b &{} 0 \end{bmatrix}\big )$.

• (b) $B=a\oplus d$,    $a\ge 0$, $d\in {\mathbb {C}}$

  First, $B(s)=\frac{1}{s^{2}}\oplus \widetilde{a}$, $P(s)= \begin{bmatrix} s^{2} &{} s \\ 1 &{} s^{2} \end{bmatrix}$, $c(s)=-i$ give $(1\oplus 0, \widetilde{a}\oplus 1)\rightarrow \big (\begin{bmatrix} 0 &{} 1\\ 1 &{} i \end{bmatrix},a\oplus d\big )$, $a>0$.

  If $\widetilde{B}=\widetilde{a}\oplus 1$, $\widetilde{a}\ge 0$ and $a=0$, then the last two equations of (5.6) for $\widetilde{d}=1$, $a=\widetilde{b}=0$ give $(1+\epsilon _4)u=\epsilon _2v$ with $|\frac{u}{v}|\le \frac{\epsilon }{1-\epsilon }$. Thus $|u|^{2}=|\frac{u}{v}||uv|\le \frac{\epsilon ^{2}\delta }{(1-\epsilon )^{2}}$ and $|2\textrm{Re}(\overline{u}x)|\ge 1-\delta -\frac{\delta \epsilon ^{2}}{(1-\epsilon )^{2}}$. By applying this, $\frac{\delta \sqrt{6}}{\nu }\ge |\det P|$ (Lemma 4.1 (1)) and (5.58) to (5.25) we get $|\frac{v}{u}|(1-\delta -\frac{\delta \epsilon ^{2}}{(1-\epsilon )^{2}})\le \delta -\frac{\delta \sqrt{6}}{\nu }$; it contradicts $|\frac{u}{v}|\le \frac{\epsilon }{1-\epsilon }$ for small $\epsilon ,\delta$.

So far we have proved Theorem 3.4 (1), (5). Furthermore, Theorem 3.4 (3), (4) can be concluded for all cases except maybe for $(0_2,1\oplus 0)$.

Case XVI.:

$(0_2,1\oplus \sigma )\dashrightarrow (A,B)$

• (a) $\sigma =1$    ($\widetilde{B}=I_2$)

  We prove $(0_2,I_2)\rightarrow \big (A,\begin{bmatrix} a &{} b\\ b &{} d \end{bmatrix}\big )$, $b>0$, $A\in {\mathbb {C}}^{2\times 2}$ by taking $P(s)=\frac{s}{\sqrt{2}}e^{i\frac{\pi }{4}}\begin{bmatrix} 1 &{} -i \\ -i &{} 1 \end{bmatrix}$, $c(s)=1$, $B(s)=\begin{bmatrix} a(s) &{} s^{-2}\\ s^{-2} &{} d(s) \end{bmatrix}$, $a(s),d(s)\le \frac{1}{s}$. Next, $P(s)=\frac{1}{\sqrt{2}}\begin{bmatrix} s &{} s\\ 1 &{} -1 \end{bmatrix}$, $B(s)=\frac{1}{s^{2}}\oplus 1$, $c(s)=1$ give $(0_2,I_2)\rightarrow (A,a\oplus 1)$ with $a>0$ and either $A=\begin{bmatrix} 0 &{} 1\\ 0 &{} 0 \end{bmatrix}$ or $A=1\oplus 0$, while $P(s)=\frac{1}{\sqrt{2}}\begin{bmatrix} s &{} s\\ s &{} -s \end{bmatrix}$, $B(s)=\frac{1}{s^{2}}\oplus (\frac{1}{s^2}+ \frac{d-a}{s})$, $c(s)=1$ yield $(0_2,I_2)\rightarrow (1\oplus \sigma ,a\oplus d)$, $d\ge a>0$.

• (b) $\sigma =0$    ($\widetilde{B}=1\oplus 0$)

  To prove $(0_2, 1\oplus 0)\rightarrow (1\oplus 0, a \oplus 0)$ for $a> 0$ we take $B(s)=\frac{1}{s^{2}}\oplus 0$, $P(s)=sI_2$, $c(s)=1$ in (5.2). From what we proved so far this implies $(0_2, 1\oplus 0)\rightarrow (A, B)$ for all $B\ne 0_2$.

This completes the proof of the theorem. $\square$