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I read Arnold's book Mathematical Methods of Classical Mechanics and come across with three problems in page 229.

1.Let $\lambda$ and $\bar{\lambda}$ be simple (multiplicity 1) eigenvalues of a symplectic transformation $S$ with $|\lambda|=1$. Show that the two-dimensional invariant plane $\pi_\lambda$ corresponding to $\lambda,\bar{\lambda}$ is nonnull.

2.Let $\xi$ be a real vector of plane $\pi_\lambda$, where $Im~\lambda > 0$ and $|\lambda| = 1$. The eigenvalue $\lambda$ is called positive if $[S\xi,\xi] > 0$. Show that this definition does not depend on the choice of $\xi \neq 0$ in the plane $\pi_\lambda$.

3.Show that $S$ is strong stable if and only if all the eigenvalues $\lambda$ lie on the unit circle and are of definite sign.

In my opinion, it will be difficult to deal with these question with the knowledge in this book.

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1 Answer 1

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1.Since $[,]$ is non-degenerate, there should be such an eigenvector $\eta$ corresponding to the eigenvalue $\lambda'$ for the eigenvector $\xi$ corresponding to the eigenvalue $\lambda$ that $[\xi, \eta] \neq 0$, which is possible only if $\lambda' = \bar{\lambda}$. Thus, the two-dimensional invariant plane $\pi_\lambda$ is nonnull.

2.Since $\xi$ is a real vector, there is a number $a$ such that $\xi = a\xi_1 + \bar{a}\xi_1$, where $\xi_1$ is the eigenvector with the eigenvalue $\lambda$. Then
$$ [S(a\xi_1 + \bar{a}\bar{\xi}_1), a\xi_1 + \bar{a}\bar{\xi}_1] = (\lambda - \bar{\lambda}) |a|^2 [\xi_1, \bar{\xi}_1] = (2|a|^2 Im \lambda) i(I\xi_1, \xi_1), $$ the sign of which is independent of $a$, i.e. independent of $\xi$. Note that $i(I\xi_1, \xi_1)$ is a real number.

3.Denote $(G\xi,\xi) = i(I\xi_1, \xi_1)$, and denote $S$ by $M$. Assume $M$ is stable and all its eigenvalues are Krein-definite. If $M$ is not strongly stable, there would be $\{M_n\}$ of unstable symplectic matrices converging to $M$. Either $M_n$ has eigenvalue outside the unit circle, or $M_n$ has an eigenvalue on the unit circle which is not semi-simple. Thus, there is a $G$-isotropic unit eigenvector $x_n$: $$M_n x_n = \lambda_n x_n,~(Gx_n, x_n) = 0.$$ Since $\lambda_n$ is a root of $|M_n - z I|$, we can extract from the sequence $\lambda_n$ a subsequence converging to a root of $M - zI$, i.e. $x_n \rightarrow x, \lambda_n \rightarrow \lambda$. Then $(Gx,x) = 0$ which is impossible since all eigenvalues of $M$ is Krein-definite.

Conversely, assume $M$ is strongly stable. Then all the eigenvalues of $M$ lie on the circle and are semi-simple. For every eigenvalue $\lambda$ with positive imaginary part, we can choose in the eigenspace $Ker(M-\lambda I)$ a $G$-orthogonal basis, say $[\xi_1, \cdots, \xi_m]$ with $(G\xi_k,\xi_k) = \pm 1$. We can take $[\bar{\xi_1},\cdots,\bar{\xi_m}]$ as a basis for the conjugate eigenspace $Ker(M-\bar{\lambda} I)$. If $\pm 1$ is an eigenvalue, the corresponding eigenspace is real and even-dimensional; we can choose its $G$-orthogonal basis to be $[\xi_1,\cdots,\xi_m,\bar{\xi_1},\cdots,\bar{\xi_m}]$ with $(G\xi_k,\xi_k)=1=-(G\bar{\xi}_k,\bar{\xi}_k)$. Putting everything together, we get a $G$-orthonogonal basis of eigenvectors of $[\xi_1,\cdots,\xi_n,\bar{\xi}_1,\cdots,\bar{\xi}_n]$ for $C^{2n}$. We have $(G\xi_k,\xi_k)=-(G\bar{\xi}_k,\bar{\xi}_k)$, aand by rearranging the basis, we can assume that $(G\xi_k,\xi_k)=1$ for $1 \leq k \leq n$.

Assume there is an eigenvalue $\lambda$ which is not definite. It must have two eigenvectors with opssite $G$-norms, say $\xi_1$ and $\bar{\xi}_1$ if $\lambda = \pm 1$, and $\xi_1, \xi_2$ if $\lambda \neq \pm 1$. Define a linear transformation $M_\tau$ by setting : $$ M_\tau \xi_1 = \lambda(\xi_1 cosh\tau + \bar{\xi}_1 sinh\tau), M_\tau \bar{\xi}_1 = \lambda(\xi_1 sinh\tau + \bar{\xi}_1 cosh\tau), $$ if $\lambda = \pm 1$, and $$ M_\tau \xi_1 = \lambda(\xi_1 cosh\tau + \xi_2 sinh\tau), M_\tau \xi_2= \lambda(\xi_1 sinh\tau + \xi_2 cosh\tau), $$ if $\lambda \neq \pm 1$, and $M_\tau = M$ on the invariant subspace generated by the other $\xi_k$.

By construction, $M_\tau$ is real (i.e. $M_\tau R^{2n} \subset R^{2n}$), and symplectic as we readily check. On the other hand $\xi_1 + \bar{\xi}_1$ (if $\lambda = \pm 1$) or $\xi_1 + \xi_2$ (if $\lambda \neq \pm 1$) is an eigenvector of $M_\tau$ with the eigenvalue $\lambda e^\tau$, which is outside the unit circle if $\tau > 0$. So $M_\tau$ is not stable, and $M_\tau \rightarrow M$ when $\tau \rightarrow 0$. This contradicts the fact that $M$ is strongly stable.

Ref:

  1. Arnold's 'Mathematical methods of classical mechanics' section 42.
  2. Ivar Ekeland's 'Convexity methods in Hamiltonian mechanics' chapter 1.
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