Second order asymptotics for Krein indefinite multipliers with multiplicity two

TL;DR

This paper derives second order asymptotic expansions for eigenvalues bifurcating from non-real Krein indefinite eigenvalues with multiplicity two in linear Hamiltonian systems in four dimensions.

Contribution

It provides a detailed second order asymptotic analysis of eigenvalues near non-real Krein indefinite eigenvalues with multiplicity two.

Findings

Second order asymptotics for eigenvalues bifurcating from non-real Krein indefinite eigenvalues.

Analysis applies to 4x4 Hamiltonian systems with symmetric, continuously varying matrices.

Results enhance understanding of spectral bifurcations in Hamiltonian dynamics.

Abstract

We consider linear Hamiltonian equations in $\mathbb{R}^{4}$ of the following type \begin{equation} \frac{\mathrm{d}\gamma}{\mathrm{d}t}(t)=J_{4}A(t)\gamma(t), \gamma(0)\in\operatorname{Sp}(4,\mathbb{R}), \end{equation} where $J=J_{4}\overset{\text{def}}{=}\begin{bmatrix}0 & \operatorname{Id}_2\\-\operatorname{Id}_2 & 0\end{bmatrix}$ and $A:t\mapsto A(t)$ is a $C^1$-continuous curve in the space of $4\times 4$ real matrices which are symmetric. We obtain second order asymptotics for the eigenvalues bifurcated from non-real Krein indefinite eigenvalues with multiplicity two.

Equations

$$\frac{\mathrm{d}\gamma }{\mathrm{d}t}(t)=J_{4}A(t)\gamma (t),\gamma (0)\in \mathrm{Sp}(4,\mathbb{R}),$$

$$\frac{\partial \gamma }{\partial t}(t,\epsilon )=J_{4}A(t,\epsilon )\gamma (t,\epsilon ),\gamma (0,\epsilon )={\mathrm{Id}}_4,$$

$$\frac{\partial }{\partial \epsilon }\gamma (T,\epsilon )=J_{4}B(T,\epsilon )\gamma (T,\epsilon ),$$

$$B(T,\epsilon )=$$

$$=$$

$$\gamma (0)({\eta }_1,{\eta }_2,{\bar{\eta }}_1,{\bar{\eta }}_2)=({\eta }_1,{\eta }_2,{\bar{\eta }}_1,{\bar{\eta }}_2)\left[\begin{array}{cccc}{\lambda }_0& {\lambda }_0& 0& 0\\ 0& {\lambda }_0& 0& 0\\ 0& 0& {\bar{\lambda }}_0& {\bar{\lambda }}_0\\ 0& 0& 0& {\bar{\lambda }}_0\end{array}\right]$$

$${\lambda }_j(t)=$$

$$+\frac{{\lambda }_0}{2}\left(\frac{⟨A(0){\eta }_1,{\eta }_1⟩}{⟨{\eta }_2,J_4{\eta }_1⟩}+\frac{⟨A(0){\eta }_1,{\eta }_2⟩}{⟨{\eta }_1,J_4{\eta }_2⟩}+\frac{⟨A(0){\eta }_2,{\eta }_1⟩}{⟨{\eta }_2,J_4{\eta }_1⟩}-\frac{⟨A(0){\eta }_1,{\eta }_1⟩⟨{\eta }_2,J_4{\eta }_2⟩}{⟨{\eta }_2,J_4{\eta }_1⟩⟨{\eta }_1,J_4{\eta }_2⟩}\right)t+o(t).$$

$$\mathrm{Re}\left({\bar{\lambda }}_0\frac{\mathrm{d}}{\mathrm{d}t}({\lambda }_1(t)+{\lambda }_2(t)){|}_{t=0}\right)=\frac{⟨A(0){\eta }_1,{\eta }_1⟩}{⟨{\eta }_2,J_4{\eta }_1⟩}.$$

$$\gamma (T,0)({\eta }_1,{\eta }_2,{\bar{\eta }}_1,{\bar{\eta }}_2)=({\eta }_1,{\eta }_2,{\bar{\eta }}_1,{\bar{\eta }}_2)\left[\begin{array}{cccc}{\lambda }_0& {\lambda }_0& 0& 0\\ 0& {\lambda }_0& 0& 0\\ 0& 0& {\bar{\lambda }}_0& {\bar{\lambda }}_0\\ 0& 0& 0& {\bar{\lambda }}_0\end{array}\right].$$

$${\lambda }_j(\epsilon )=$$

$$+\frac{{\lambda }_0}{2}(-{\int }_0^T\frac{⟨\frac{\partial }{\partial \epsilon }A(t,0){\eta }_1(t),{\eta }_1(t)⟩}{⟨{\eta }_2,J_4{\eta }_1⟩}\mathrm{d}t+{\int }_0^T\frac{⟨\frac{\partial }{\partial \epsilon }A(t,0){\eta }_1(t),{\eta }_2(t)⟩}{⟨{\eta }_1,J_4{\eta }_2⟩}\mathrm{d}t$$

$$+{\int }_0^T\frac{⟨\frac{\partial }{\partial \epsilon }A(t,0){\eta }_2(t),{\eta }_1(t)⟩}{⟨{\eta }_2,J_4{\eta }_1⟩}\mathrm{d}t-{\int }_0^T\frac{⟨\frac{\partial }{\partial \epsilon }A(t,0){\eta }_1(t),{\eta }_1(t)⟩⟨{\eta }_2,J_4{\eta }_2⟩}{⟨{\eta }_2,J_4{\eta }_1⟩⟨{\eta }_1,J_4{\eta }_2⟩}\mathrm{d}t)\epsilon +o(\epsilon ).$$

$$\mathrm{Re}\left({\bar{\lambda }}_0\frac{\mathrm{d}}{\mathrm{d}\epsilon }({\lambda }_1(\epsilon )+{\lambda }_2(\epsilon )){|}_{\epsilon =0}\right)={\int }_0^T\frac{⟨\frac{\partial }{\partial \epsilon }A(t,0){\eta }_1(t),{\eta }_1(t)⟩}{⟨{\eta }_2,J_4{\eta }_1⟩}\mathrm{d}t.$$

$$⟨x,y⟩=\sum _{j=1}^m{x}_j{\bar{y}}_j.$$

$$p(\lambda ,t)=\det (\lambda \cdot \mathrm{Id}-\gamma (t)).$$

$$\bigwedge (n,k,A)(v_1\wedge \cdots \wedge v_n)\stackrel{\mathrm{def}}{=}\sum _{\sigma \in \{0,1{\}}^n:{\sum }_i{\sigma }_i=k}{\wedge }_{i=1}^n({\sigma }_i\cdot A{v}_i+(1-{\sigma }_i)\cdot v_i).$$

$$\begin{gathered} \bigwedge (n,k_1,k_2,A_1,A_2)(v_1\wedge \cdots \wedge v_n) \\ \stackrel{\mathrm{def}}{=}\sum _{\sigma \in \{0,1,2{\}}^n:{\sum }_i{1}_{{\sigma }_i=1}=k_1,{\sum }_i{1}_{{\sigma }_i=2}=k_2}{\wedge }_{i=1}^n(1_{{\sigma }_i=1}\cdot A_1{v}_i+1_{{\sigma }_i=2}\cdot A_2{v}_i+1_{{\sigma }_i=0}\cdot v_i), \end{gathered}$$

$$\begin{gathered} p(\lambda ,t)=\det (\lambda \cdot {\mathrm{Id}}_{2n}-\gamma (t))=\det ((\lambda -{\lambda }_0)\cdot {\mathrm{Id}}_{2n}+({\lambda }_0\cdot {\mathrm{Id}}_{2n}-\gamma (0))-(\gamma (t)-\gamma (0))) \\ =\sum _{k=0}^{2n}(\lambda -{\lambda }_0{)}^k\sum _{k_1+k_2=2n-k,k_1\ge 0,k_2\ge 0}(-1{)}^{k_2}\cdot \bigwedge (2n,k_1,k_2,{\lambda }_0\cdot {\mathrm{Id}}_{2n}-\gamma (0),\gamma (t)-\gamma (0)). \end{gathered}$$

$$p(\lambda ,t)=\sum _{k=0}^4{c}_k(t)(\lambda -{\lambda }_0{)}^k.$$

$$c_k(t)=\sum _{k_2=0}^{4-k}(-t{)}^{k_2}{c}_{4-k-k_2,k_2}(t),$$

$$c_{k_1,k_2}(t)=\bigwedge (4,k_1,k_2,{\lambda }_0\cdot \mathrm{Id}-\gamma (0),\frac{1}{t}(\gamma (t)-\gamma (0))).$$

$$c_k(t)=c_{4-k,0}-t{c}_{3-k,1}(0)+O(t^2).$$

$$c_{4,0}=c_0(0)=0,c_{3,0}=c_1(0)=0,$$

$$c_{2,0}=c_2(0)=({\lambda }_0-{\bar{\lambda }}_0{)}^2,$$

$$c_{1,0}=c_3(0)=2({\lambda }_0-{\bar{\lambda }}_0).$$

$$c_{2,0}{a}^2=c_{3,1}(0).$$

$$c_4(t)(a{t}^{\frac{1}{2}}+\mu t{)}^4=O(t^2),$$

$$c_3(t)(a{t}^{\frac{1}{2}}+\mu t{)}^3=(c_{1,0}+O(t))(a^3{t}^{\frac{3}{2}}+O(t^2))=c_{1,0}{a}^3{t}^{\frac{3}{2}}+O(t^2),$$

$$c_2(t)(a{t}^{\frac{1}{2}}+\mu t{)}^2=(c_{2,0}+O(t))(a^{2}t+2a\mu t^{\frac{3}{2}}+O(t^2))=c_{2,0}{a}^{2}t+2c_{2,0}a\mu t^{\frac{3}{2}}+O(t^2),$$

$$c_1(t)(a{t}^{\frac{1}{2}}+\mu t)=(-t{c}_{2,1}(0)+O(t^2))(a{t}^{\frac{1}{2}}+O(t))=-c_{2,1}(0)a{t}^{\frac{3}{2}}+O(t^2),$$

Full text

[TABLE]

where $J=J_{4}\overset{\text{def}}{=}\begin{bmatrix}0&\operatorname{Id}_{2}\\ -\operatorname{Id}_{2}&0\end{bmatrix}$ and $A:t\mapsto A(t)$ is a $C^{1}$-continuous curve in the space of $4\times 4$ real matrices which are symmetric. We obtain second order asymptotics for the eigenvalues bifurcated from non-real Krein indefinite eigenvalues with multiplicity two.

1 Introduction

We consider linear Hamiltonian equations in $\mathbb{R}^{4}$ of the following type

[TABLE]

where $J=J_{4}\overset{\text{def}}{=}\begin{bmatrix}0&\operatorname{Id}_{2}\\ -\operatorname{Id}_{2}&0\end{bmatrix}$ and $A:t\mapsto A(t)$ is a $C^{1}$-continuous curve in the space of $4\times 4$ real matrices which are symmetric.

The system (2) arises naturally from perturbations of linearized Hamiltonian equations. Indeed, let $\varepsilon\in\mathbb{R}$ be a real perturbation parameter. Consider

[TABLE]

where $\frac{\partial A}{\partial\varepsilon}(t,\varepsilon)$ is jointly continuous . Then, for fixed $T$, as $\varepsilon$ varies, the endpoint matrix $\gamma(T,\varepsilon)$ is a $C^{2}$-curve satisfying (2). More precisely,

[TABLE]

where

[TABLE]

Please see [CLW19, Eq. (5) and Appendix A] for a proof.

Many people have studied the system (2) or (3), see Ekeland [Eke90], Yakubovich and Starzhinskii[YS75], and the references there. We are interested in the bifurcation of a Krein indefinite eigenvalue with multiplicity two. The first order asymptotics and qualitative behavior of the bifurcated eigenvalues were firstly discovered by Krein and Lyubarskii in [KL62] for the end-point matrix $\gamma(T,\varepsilon)$ of perturbed linear Hamiltonian equations in $\mathbb{R}^{2n}$ under certain positivity and linearity assumption on $A(t,\varepsilon)$. Recently, a general version of the Krein-Lyubarskii theorem was obtained by Chang, Long and Wang in [CLW19] for $C^{1}$ paths of symplectic matrices (corresponding to the solution of (2)). In the present paper, we study the second order asymptotics and the derivative of the sum of bifurcated Krein multipliers by adapting the argument in [CLW19]. Our main results are the following two theorems.

Theorem 1.1**.**

Consider the solution of (2). Suppose $\lambda_{0}=e^{\sqrt{-1}\theta_{0}}\in U$ is an eigenvalue of $\gamma(0)$. Suppose that $\lambda_{0}\neq\pm 1$. Assume that

[TABLE]

and that $\langle A(0)\eta_{1},\eta_{1}\rangle\neq 0$.111The existence of $\eta_{1}$ and $\eta_{2}$ satisfying (6) is equivalent to the assumption that the geometric multiplicity of $\lambda_{0}$ is $1$. Denote by $\lambda_{1}(t)$ and $\lambda_{2}(t)$ the eigenvalues of $\gamma(t)$ bifurcated from $\lambda_{0}$. Then, for $j=1,2$, we have that

[TABLE]

Consequently, we have that

[TABLE]

Theorem 1.2**.**

Consider the solution of (3). Suppose $\lambda_{0}=e^{\sqrt{-1}\theta_{0}}\in U$ is an eigenvalue of $\gamma(T,0)$. Suppose that $\lambda_{0}\neq\pm 1$. Assume that

[TABLE]

Define $\eta_{1}(t)=\gamma(t,0)\eta_{1}$ and $\eta_{2}(t)=\gamma(t,0)\eta_{2}$ and that $\int_{0}^{T}\langle\frac{\partial}{\partial\varepsilon}A(t,0)\eta_{1}(t),\eta_{1}(t)\rangle\,\mathrm{d}t\neq 0$. Let $\varepsilon\geq 0$. Denote by $\lambda_{1}(\varepsilon)$ and $\lambda_{2}(\varepsilon)$ the eigenvalues of $\gamma(T,\varepsilon)$ bifurcated from $\lambda_{0}$ as $\varepsilon$ varies. Then, for $j=1,2$, we have that

[TABLE]

Consequently, we have that

[TABLE]

Our study on the second order asymptotics of bifurcated Krein multipliers and the derivative of the sum of bifurcated Krein multipliers was motivated by [KY06] of Kuwamura and Yanagida. For perturbed linear Hamiltonian equations as (3) but with a general form of $J_{4}$, they point out that the sum of the bifurcated eigenvalues is differentiable, though neither of them is differentiable. However, their expression (1.7) of the derivative of the sum of the eigenvalues seems to be incorrect.

Note that (8) is useful for studying the strong stability of $\gamma(t)$. See Remark 3.1.

By using [CLW19, Lemma B.1], Theorem 1.1 and 1.2 could be generalized to the equations on $\mathbb{R}^{2n}$ if the algebraic multiplicity of $\lambda_{0}$ is two and the geometric multiplicity is one.

The organization of this paper is as follows. We collect definitions and notations, prepare some useful properties in Section 2. We prove Theorem 1.1 in Section 3 and Theorem 1.2 in Section 4.

2 Preliminaries

2.1 Notations and definitions

• •

For a matrix $M$, we denote by $M^{T}$ the transpose of $M$. For a complex matrix $M$, we denote by $M^{*}$ the conjugate transpose of $M$.

• •

For $n\geq 1$, we denote by $\operatorname{Id}_{n}$ the $n\times n$ identity matrix and define $J_{2n}\overset{\text{def}}{=}\begin{bmatrix}0&\operatorname{Id}_{n}\\ -\operatorname{Id}_{n}&0\end{bmatrix}$. Then, $J_{2n}^{*}=J_{2n}^{T}=-J_{2n}$ and $J_{2n}^{2}=-\operatorname{Id}_{n}$.

• •

For vectors $v_{1},\ldots,v_{n}$ in a vector space $V$, we denote by $\wedge_{j=1}^{n}v_{j}$ the exterior product $v_{1}\wedge v_{2}\wedge\cdots\wedge v_{n}$. (Note that $\wedge$ is associative.) We denote by $\Lambda^{n}(V)$ the space of the linear span of all such $\wedge_{j=1}^{n}v_{j}$.

• •

For $m\geq 1$, the inner product $\langle\cdot,\cdot\rangle$ on $\mathbb{C}^{m}$ is defined by

[TABLE]

• •

Denote by $p(\lambda,t)$ the characteristic polynomial of the matrix $\gamma(t)$, i.e.,

[TABLE]

2.2 Exterior powers of linear maps

We recall exterior powers of a linear map $A$ and its relation with its determinant $\det(A)$.

Definition 2.1**.**

Let $A:V\to V$ be a linear map on an $n$-dimensional vector space $V$. For $k=0,\ldots,n$, we define the exterior powers $\bigwedge(n,k,A):\Lambda^{n}(V)\to\Lambda^{n}(V)$ as a linear map as follows:

[TABLE]

Similarly, for two linear maps $A_{1},A_{2}:V\to V$, for two integers $k_{1},k_{2}=0,\ldots,n$, we define the linear map $\bigwedge(n,k_{1},k_{2},A_{1},A_{2}):\Lambda^{n}(V)\to\Lambda^{n}(V)$ as follows:

[TABLE]

Since $\Lambda^{n}(V)$ is $1$-dimensional, we identify the $\bigwedge(n,k,A)$ (or $\bigwedge(n,k_{1},k_{2},A_{1},A_{2})$) with the unique scaling factor, which is also denoted by $\bigwedge(n,k,A)$ (or $\bigwedge(n,k_{1},k_{2},A_{1},A_{2})$).

In the above definition, for each vector $v_{i}$, we choose one from the three linear maps $\operatorname{Id}$, $A_{1}$ and $A_{2}$ and apply it to $v_{i}$. For the assignment of linear maps to the linear basis, the only constraint is that the map $A_{1}$ occurs $k_{1}$ many times and the map $A_{2}$ occurs exactly $k_{2}$ many times. All these assignments have equal weight.

Note that $\det(A)$ is identified with the linear map $\bigwedge(n,n,A)$ on the $1$-dimensional vector space $\Lambda^{n}(V)$. In particular, for an eigenvalue $\lambda_{0}$ of the $2n\times 2n$ matrix $\gamma(0)$, we have that

[TABLE]

In the above calculation, we express the determinant by wedge powers of the sum of linear maps $(\lambda-\lambda_{0})\cdot\operatorname{Id}_{2n}$, $\lambda_{0}\cdot\operatorname{Id}_{2n}-\gamma(0)$ and $\gamma(0)-\gamma(t)$, expand it according to distributive law and collect the terms with the same times of occurrence, where $k_{1}$ is the time of occurrence of $\lambda_{0}\cdot\operatorname{Id}_{2n}-\gamma(0)$ and $k_{2}$ is the time of occurrence of $\gamma(0)-\gamma(t)$.

2.3 Continuity of roots of polynomials

We will need the following lemma on the continuity of the roots of polynomials as the coefficients vary.

Lemma 2.1**.**

[CLW19, Lemma 2.1]** Let $W$ be a neighborhood of [math]. Let $P_{t}(z)=\sum_{j=0}^{n}c_{j}(t)z^{j}$, where $c_{j}(t)\in\mathbb{C}$ and $t\in W$. Suppose that $t\mapsto c_{j}(t)$ is continuous for $j=0,\ldots,n$ and $t\in W$. Denote by $d(t)$ the degree of the polynomial $P_{t}$. Suppose that $d(t)=n$ for $t\in W\setminus\{0\}$ and $d(0)=m\leq n$. Then, there exist $m$ continuous complex valued functions $z_{1},\ldots,z_{m}$ on $W$ and $n-m$ continuous complex valued functions $z_{m+1},\ldots,z_{n}$ on $W\setminus\{0\}$ such that

• •

for $t\neq 0$, $z_{1}(t),\ldots,z_{n}(t)$ are roots of $P_{t}$,

• •

for $t=0$, $z_{1}(0),\ldots,z_{m}(0)$ are roots of $P_{0}$,

• •

for $i=m+1,\ldots,n$, we have that $\lim_{t\to 0}z_{i}(t)=\infty$.

3 Proof of Theorem 1.1

We assume $t\geq 0$ in the following proof. The proof for $t\leq 0$ is similar and we left it to the reader. Recall Definition 2.1 and (13). We expand the characteristic polynomial $p(\lambda,t)=\det(\lambda\cdot\operatorname{Id}_{4}-\gamma(t))$ at $\lambda_{0}=e^{\sqrt{-1}\theta_{0}}$:

[TABLE]

Since $p(\lambda,t)$ is a monic polynomial in $\lambda$, we have that $c_{4}(t)\equiv 1$. For $k=0,\ldots,4$,

[TABLE]

where

[TABLE]

Note that $c_{k,0}(t)$ doesn’t depend on $t$ for $k=0,1,\ldots,4$. For simplicity, we denote it by $c_{k,0}$. Since $t\mapsto\gamma(t)$ is $C^{2}$, we have that $\frac{1}{t}(\gamma(t)-\gamma(0))=\dot{\gamma}(0)+O(t)$ as $t\to 0$. Hence, for $k=0,\ldots,3$,

[TABLE]

In particular, $c_{k}(0)=c_{4-k,0}$. Since the four eigenvalues of $\gamma(0)$ are $\lambda_{0}$, $\lambda_{0}$, $\bar{\lambda_{0}}$, $\bar{\lambda}_{0}$, $p(\lambda,0)=\det(\lambda\cdot\operatorname{Id}_{4}-\gamma(0))=(\lambda-\lambda_{0})^{2}(\lambda-\bar{\lambda}_{0})^{2}=(\lambda-\lambda_{0})^{4}+2(\lambda_{0}-\bar{\lambda}_{0})(\lambda-\lambda_{0})^{3}+(\lambda_{0}-\bar{\lambda}_{0})^{2}(\lambda-\lambda_{0})^{2}$. Hence,

[TABLE]

[TABLE]

[TABLE]

Define $q(\mu,t)=p(\lambda_{0}+at^{1/2}+\mu t,t)=\sum_{k=0}^{4}c_{k}(t)(at^{\frac{1}{2}}+\mu t)^{k}$, where $a$ is chosen such that

[TABLE]

Later, we will see that $a\neq 0$ when $\langle A(0)\xi_{1},\xi_{1}\rangle\neq 0$, see (31) below. By (15) and (16), uniformly for $|\mu|\leq 2$, we have that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Hence, $q(\mu,t)=at^{\frac{3}{2}}(2c_{2,0}\mu+c_{1,0}a^{2}-c_{2,1}(0))+O(t^{2})$. Define $r(\mu,t)=\frac{q(\mu,t)}{at^{\frac{3}{2}}}$. Then, $r$ is a polynomial in $\mu$ of degree $4$ for $t>0$. Define $r(\mu,0)=\lim_{t\downarrow 0}r(\mu,t)=2c_{2,0}\mu+c_{1,0}a^{2}-c_{2,1}(0)$. Then, by Lemma 2.1, there exists a unique solution $\mu(t)$ of $r(\mu,t)=0$ such that $\lim_{t\to 0}\mu(t)=\frac{c_{2,1}(0)-c_{1,0}a^{2}}{2c_{2,0}}$. Thus, among the eigenvalues of $\gamma(t)$, there exists a unique $\lambda_{1}(t)$ such that for $t\geq 0$,

[TABLE]

Similarly, there exists a unique eigenvalue $\lambda_{2}(t)$ such that

[TABLE]

To obtain (1.1), it remains to express the above quantities via $\lambda_{0}$, the generalized eigenvectors and $A(0)$. Assume that

[TABLE]

Then, we have

Lemma 3.1**.**

$\langle\eta_{1},J_{4}\eta_{1}\rangle=\langle\eta_{1},J_{4}\bar{\eta}_{1}\rangle=\langle\eta_{1},J_{4}\bar{\eta}_{2}\rangle=\langle\eta_{2},J_{4}\bar{\eta}_{1}\rangle=\langle\eta_{2},J_{4}\bar{\eta}_{2}\rangle=\langle\bar{\eta}_{1},J_{4}\bar{\eta}_{1}\rangle=0$.

Proof of Lemma 3.1.

Recall (6) and the assumption that $\lambda_{0}\in U$ and $\lambda_{0}\neq\pm 1$. Since $\gamma(0)$ is symplectic, we have that

[TABLE]

Hence, $\langle\eta_{1},J_{4}\eta_{1}\rangle=0$. Similarly, $\langle\bar{\eta}_{1},J_{4}\bar{\eta}_{1}\rangle=0$.

Since $\gamma(0)$ is symplectic, we have that

[TABLE]

Since $\lambda_{0}\neq\pm 1$, we have that

[TABLE]

Since $\gamma(0)$ is symplectic and (23), we have that

[TABLE]

Since $\lambda_{0}\neq\pm 1$, we have that

[TABLE]

Similarly, we have that

[TABLE]

Since $\gamma(0)$ is symplectic, we have that

[TABLE]

Since $\lambda_{0}\neq\pm 1$, we have that

[TABLE]

∎

We have computed $c_{1,0}$ and $c_{2,0}$ and we will compute $c_{3,1}(0)$ and $c_{2,1}(0)$. For simplicity of notation, define $K=\lambda_{0}\cdot\operatorname{Id}_{4}-\gamma(0)$. Then, $K\eta_{1}=0$, $K\eta_{2}=-\lambda_{0}\eta_{1}$, $K\bar{\eta}_{1}=(\lambda_{0}-\bar{\lambda}_{0})\bar{\eta}_{1}$, $K\bar{\eta}_{2}=(\lambda_{0}-\bar{\lambda}_{0})\bar{\eta}_{2}-\bar{\lambda}_{0}\bar{\eta}_{1}$. Recall Definition 2.1. Since $K\eta_{1}=0$, we have that

[TABLE]

Assume that $J_{4}A(0)\eta_{1}=\sum_{k=1}^{4}\alpha_{k}\eta_{k}$ where $\eta_{3}=\bar{\eta}_{1}$ and $\eta_{4}=\bar{\eta}_{2}$. By taking inner product with $J_{4}\eta_{1}$ and using Lemma 3.1, we get that

[TABLE]

Hence,

[TABLE]

[TABLE]

and

[TABLE]

Hence, by (17) and (30), we have that

[TABLE]

Similar to the deduction of (3), since $K\eta_{1}=0$ and $\eta_{1}\wedge K\eta_{2}=\eta_{1}\wedge(-\lambda_{0}\eta_{1})=0$, we have that

[TABLE]

By (29), we have that

[TABLE]

Similarly, we obtain that

[TABLE]

Note that

[TABLE]

Recall that $J_{4}A(0)\eta_{1}=\sum_{k=1}^{4}\alpha_{k}\eta_{k}$, where $\alpha_{2}=\frac{\langle A(0)\eta_{1},\eta_{1}\rangle}{\langle\eta_{2},J_{4}\eta_{1}\rangle}$. By taking inner product with $J_{4}\eta_{2}$, we obtain that

[TABLE]

Hence, $\alpha_{1}=\frac{\langle A(0)\eta_{1},\eta_{2}\rangle}{\langle\eta_{1},J_{4}\eta_{2}\rangle}-\frac{\langle A(0)\eta_{1},\eta_{1}\rangle\langle\eta_{2},J_{4}\eta_{2}\rangle}{\langle\eta_{2},J_{4}\eta_{1}\rangle\langle\eta_{1},J_{4}\eta_{2}\rangle}$ and

[TABLE]

Note that

[TABLE]

By (29), we have that

[TABLE]

Next, we calculate $\eta_{1}\wedge J_{4}A(0)\eta_{2}\wedge\bar{\eta}_{1}\wedge\bar{\eta}_{2}$: Assume $J_{4}A(0)\eta_{2}=\sum_{k=1}^{4}\beta_{k}\eta_{k}$. By taking inner product with $J_{4}\eta_{1}$, we get that

[TABLE]

Hence, we have that $\beta_{2}=\frac{\langle A(0)\eta_{2},\eta_{1}\rangle}{\langle\eta_{2},J_{4}\eta_{1}\rangle}$,

[TABLE]

and that

[TABLE]

By (3), (3), (34), (35) and (37), we get that

[TABLE]

By (17), (18), (20), (21), (31) and (3), we get (1.1).

It remains to prove (8).

By (1.1), we have that

[TABLE]

Since $\gamma(0)$ is symplectic, we have that

[TABLE]

By Lemma 3.1, $\langle\eta_{1},J_{4}\eta_{1}\rangle=0$. Hence, we have that $\langle\eta_{1},J_{4}\eta_{2}\rangle+\langle\eta_{2},J_{4}\eta_{1}\rangle=0$. Note that $\langle\eta_{1},J_{4}\eta_{2}\rangle=-\langle J_{4}\eta_{1},\eta_{2}\rangle=-\overline{\langle\eta_{2},J_{4}\eta_{1}\rangle}$. Hence, $\langle\eta_{2},J_{4}\eta_{1}\rangle$ and $\langle\eta_{1},J_{4}\eta_{2}\rangle$ are real. Therefore, we have that

[TABLE]

which is purely imaginary.

Since $\langle\eta_{2},J_{4}\eta_{2}\rangle=-\langle J_{4}\eta_{2},\eta_{2}\rangle=-\overline{\langle\eta_{2},J_{4}\eta_{2}\rangle}$, the quantity $\langle\eta_{2},J_{4}\eta_{2}\rangle$ is purely imaginary. Since $\langle A(0)\eta_{1},\eta_{1}\rangle$, $\langle\eta_{2},J_{4}\eta_{1}\rangle$ and $\langle\eta_{1},J_{4}\eta_{2}\rangle$ are real and $\langle\eta_{2},J_{4}\eta_{2}\rangle$ is imaginary, $-\frac{\langle A(0)\eta_{1},\eta_{1}\rangle\langle\eta_{2},J_{4}\eta_{2}\rangle}{\langle\eta_{2},J_{4}\eta_{1}\rangle\langle\eta_{1},J_{4}\eta_{2}\rangle}$ is imaginary.

Since $\langle A(0)\eta_{1},\eta_{1}\rangle$ and $\langle\eta_{1},J_{4}\eta_{2}\rangle$ are real, $\frac{\langle A(0)\eta_{1},\eta_{1}\rangle}{\langle\eta_{2},J_{4}\eta_{1}\rangle}$ is real.

Therefore,

[TABLE]

Remark 3.1*.*

Note that (8) is useful for studying the strong stability of $\gamma(t)$. A symplectic matrix $\gamma$ is called stable if $\sup_{n\in\mathbb{Z}}||\gamma^{n}||<\infty$, where $\mathbb{Z}$ denotes the set of integers. It is known that $\gamma$ is stable if it is diagonalizable and all its eigenvalues stay on the unit circle $U\subset\mathbb{C}$. A symplectic matrix $\gamma$ is called strongly stable if there exists a neighborhood of $\gamma$ in the space of symplectic matrices containing only stable symplectic matrices. It is not hard to see that if the four eigenvalues of $\gamma$ are simple and lie on $U$, then $\gamma$ must be strongly stable. In general, the strong stability is related to the Krein type of the eigenvalues. Such a characterization of strong stability was firstly formulated by Krein [Kre50, Kre51], and later independently by Moser [Mos58].

If $\frac{\langle A(0)\eta_{1},\eta_{1}\rangle}{\langle\eta_{2},J_{4}\eta_{1}\rangle}>0$(resp. or $\frac{\langle A(0)\eta_{1},\eta_{1}\rangle}{\langle\eta_{2},J_{4}\eta_{1}\rangle}<0$), then there exists $\delta>0$ such that for $t\in(0,\delta)$, $\gamma(t)$ is unstable (resp. strongly stable) and for $t\in(-\delta,0)$, $\gamma(t)$ is strongly stable (resp. unstable).

Proof of Remark 3.1.

We give the proof for the case $\frac{\langle A(0)\eta_{1},\eta_{1}\rangle}{\langle\eta_{2},J_{4}\eta_{1}\rangle}>0$. The case $\frac{\langle A(0)\eta_{1},\eta_{1}\rangle}{\langle\eta_{2},J_{4}\eta_{1}\rangle}<0$ is quite similar.

Let’s first prove that for $t$ positive and sufficiently small, $\gamma(t)$ has four eigenvalues outside of $U$. Hence, such $\gamma(t)$ is unstable. Otherwise, there exists a sequence $\{t_{n}\}_{n\geq 1}$ strictly decreasing to [math] such that $\gamma(t_{n})$ has four semi-simple eigenvalues $\lambda_{1}(t_{n})$, $\lambda_{2}(t_{n})$, $\bar{\lambda}_{1}(t_{n})$ and $\bar{\lambda}_{2}(t_{n})$ on $U$ and that $\lim_{n\to\infty}\lambda_{1}(t_{n})=\lim_{n\to\infty}\lambda_{2}(t_{n})=\lambda_{0}$. Then, $\frac{1}{2}(\lambda_{1}(t_{n})+\lambda_{2}(t_{n}))$ stay in the unit disk $D\subset\mathbb{C}$. Hence, $\lim_{n\to\infty}\mathrm{Re}\left(\bar{\lambda}_{0}\frac{\lambda_{1}(t_{n})+\lambda_{2}(t_{n})-2\lambda_{0}}{t_{n}}\right)\leq 0$, which contradicts with the assumption $\frac{\langle A(0)\eta_{1},\eta_{1}\rangle}{\langle\eta_{2},J_{4}\eta_{1}\rangle}>0$ and (8).

Let’s show that for $t$ negative and sufficiently small, $\gamma(t)$ has four distinct eigenvalues on the unit circle $U\subset\mathbb{C}$. Hence, such $\gamma(t)$ is strongly stable. Otherwise, there exists a sequence $\{t_{n}\}_{n\geq 1}$ strictly increasing to [math] such that $\gamma(t_{n})$ has four eigenvalues $\lambda_{1}(t_{n})=r(t_{n})e^{\sqrt{-1}\theta(t_{n})}$, $\lambda_{2}(t_{n})=\frac{1}{r(t_{n})}e^{\sqrt{-1}\theta(t_{n})}$, $\bar{\lambda}_{1}(t_{n})=r(t_{n})e^{-\sqrt{-1}\theta(t_{n})}$ and $\bar{\lambda}_{2}(t_{n})=\frac{1}{r(t_{n})}e^{-\sqrt{-1}\theta(t_{n})}$ on $U$, where $\lim_{n\to\infty}\theta(t_{n})=\theta_{0}$ and $\lim_{n\to\infty}r(t_{n})=1$. (Recall that $\lambda_{0}=e^{\sqrt{-1}\theta_{0}}$.) Let $c=\mathrm{Im}\left(\bar{\lambda}_{0}\frac{\mathrm{d}}{\mathrm{d}t}(\lambda_{1}(t)+\lambda_{2}(t))|_{t=0}\right)$. Then, we have that

[TABLE]

Hence, $\lim_{n\to\infty}\frac{\theta(t_{n})-\theta_{0}}{t_{n}}=\frac{c}{2}$. Since $t_{n}<0$ and $r(t_{n})+\frac{1}{r(t_{n})}\geq 2$, we have that

[TABLE]

which contradicts with the assumption $\frac{\langle A(0)\eta_{1},\eta_{1}\rangle}{\langle\eta_{2},J_{4}\eta_{1}\rangle}>0$ and (8). ∎

4 Proof of Theorem 1.2

Recall (1), $\lambda_{0}\in U$, $\eta_{1}(t)=\gamma(t,0)\eta_{1}$ and $\eta_{2}(t)=\gamma(t,0)\eta_{2}$. Note that

[TABLE]

Note that $\gamma(T,0)\eta_{1}=\lambda_{0}\eta_{1}$ and $\gamma(T,0)\eta_{2}=\lambda_{0}\eta_{2}+\lambda_{0}\eta_{1}$. Hence, $\gamma(T,0)^{-1}\eta_{1}=\bar{\lambda}_{0}\eta_{1}$ and $\gamma(T,0)^{-1}\eta_{2}=\bar{\lambda}_{0}\eta_{2}-\bar{\lambda}_{0}\eta_{1}$. Therefore, we obtain that

[TABLE]

Similarly, we have that

[TABLE]

and that

[TABLE]

The result follows from Theorem 1.1 and (40), (41) and (42).