# Symplectic forms and sign of eigenvalues

This question has come out while reading J. Moser "New Aspects in the Theory of Stability of Hamiltonian Systems". I'm particularly interested to the Appendix, where one investigates the stability of elliptic fixed points of Hamiltonian dynamical systems, in the time independent case. I start presenting the framework.

Let us consider the Hamiltonian dynamical system $$\dot{x}_\nu = H_{y_\nu}(x,y), \qquad \dot{y}_\nu=-H_{x_\nu}(x,y) \qquad \qquad (1)$$

where $\nu=1,2,\dots,n$. Hamiltonian $H$ does not depend on time $t$ and it is assumed to be a real analytic function of $x_\nu,\,y_\nu$, with $\nu=1,2,\dots,n$ in the neighboorhood of $x=y=0$, the expansion of which starts with quadratic terms. Then $x=y=0$ is an equilibrium solution.

One can construct a fundamental system of solutions of exponential form $$w^{(\nu)}=e^{\gamma_\nu t}p^{(\nu)}$$ where $p^{(\nu)}$ are constant vectors or, in the case of multiple eigenvalues $\gamma_{\nu}$, possibly polinomials in $t$. The numbers $\gamma_\nu$ are obtained as the eigenvalues of the matrix determined by the linear terms of the right-hand side of (1). Suppose that all eigenvalues are distinct and purely imaginary, i.e. of the type $\gamma_\nu=i\beta_\nu$, with $\beta_\nu$ real. So the spectrum has the form $$\pm i \beta_1, \quad \pm i \beta_2, \,\dots,\, \pm i \beta_n.$$ So one obtains a collection of distinct numbers $\beta_\nu$, $-\beta_\nu$ with $\nu=1,2,\dots,n$

So far, so good.

For later purposes, one needs to define the sign of $\beta_\nu$. The Author says that the sign of $\beta_\nu$ is taken in such a way that $$\mathcal{Im}\left[w^{(\nu)},\overline{w^{(\nu)}}\right]<0.$$ Square brackets are defined as Lagrange Brackets (an outdated therminology, nowadays called symplectic form [see comments]). More precisely, given any two $2n$-dimensional vectors, $x$ and $\tilde{x}$, with components $x_\nu$ and $\tilde{x}_\nu$, their Lagrange Bracket (read: symplectic form) is defined as $$[x,\tilde{x}]=\sum_{\nu=1}^n (x_\nu\tilde{x}_{\nu+n}-x_{\nu+n}\tilde{x}_\nu)$$ In passing, one has to remember that an Hamiltonian sysytem is marked by the fact that, for any two solutions $x$ and $\tilde{x}$, the Lagrange bracket (read: symplectic form) $[x,\tilde{x}]$ is $t$-independent.

1) How to explicitly compute:

$$[w^{(\nu)},w^{(\mu)}] = ?$$

$$[w^{(\nu)},\overline{w^{(\mu)}}] = ?$$

2) How to prove that $$[w^{(\nu)},\overline{w^{(\nu)}}] = [p^{(\nu)},\overline{p^{(\nu)}}]$$

$$[p^{(\nu)},\overline{p^{(\nu)}}] \text{ is purely imaginary}$$ 3) How to define the sign $\beta_{\nu}$ in such a way that $$\mathcal{Im}\left[w^{(\nu)},\overline{w^{(\nu)}}\right]<0.$$ I am from the Physics community, so I kindly ask to display all important passages.

• The terminology is outdated. Moser's term "Lagrange bracket" is nowadays called a "symplectic form". – Ben McKay Jun 23 '18 at 16:21
• Thanks a lot for the info. Indeed this was something that was confusing me, because looking around for Lagrange Brackets, I found different definitions, including partial derivatives. So, can you please help me in computing those symplectic forms? – AndreaPaco Jun 23 '18 at 16:45
• By the way, I’ve modified the title and the text of my question accordingly. – AndreaPaco Jun 23 '18 at 16:49

I have found the solution myself.

I write the system of linear ordinary differential equations in matrix form:

$$\begin{pmatrix} \dot{\vec{x}}(t)\\ \dot{\vec{y}}(t) \end{pmatrix} =\mathbb{J} \begin{pmatrix} \vec{x}(t)\\ \vec{y}(t) \end{pmatrix}$$ where $\mathbb{J}$ is the Jacobian matrix associated to the linear system and $\vec{x}$, $\vec{y}$ are $n$-dimensional vectors.

What Moser calls "fundamental system of solutions" is simply a set of $2n$ functions $w^{(j)}(t)$ having the form $$\vec{w}^{(j)}(t)=e^{i\beta_jt} \begin{pmatrix} P_{1,j}\\ P_{2,j}\\ \dots \\ P_{2n,j} \end{pmatrix}$$ for $j=1,2,...,n$. The remaining $n$ fundamental functions are simply the complex conjugates: $$\vec{w}^{(j+n)}(t)=\overline{\vec{w}^{(j)}}(t)=e^{-i\beta_jt} \begin{pmatrix} P_{1,j}^*\\ P_{2,j}^*\\ \dots \\ P_{2n,j}^* \end{pmatrix}$$ for $j=1,2,\dots,n$. Notice that numbers $i\beta_j$ are the eigenvalues of $\mathbb{J}$ while $\vec{P}_j$ are the corresponding eigenvectors.

1)Applying the definition of symplectic form given by Moser, one can write that: $$\left[\vec{w}^{(\nu)}(t),\vec{w}^{(\mu)}(t)\right]=e^{it(\nu+\mu)} \left[\vec{P}_\nu,\vec{P}_{\mu} \right]$$ From the Theory of Hamiltonian system, it is known that the symplectic form of two solutions must be $t$-independent. This implies that $$\left[\vec{P}_\nu,\vec{P}_{\mu} \right]=0$$ (Probably there exist a proof which exploits the property of eigenvectors $\vec{P}_j$'s but I was not able to find it.)
Similarly, one can prove that $$\left[\vec{w}^{(\nu)}(t),\overline{\vec{w}^{(\mu)}}(t)\right]=e^{it(\nu-\mu)} \left[\vec{P}_\nu,\vec{P}_{\mu}^* \right]=0$$
2) If one computes the symplectic form between two complex conjugate solutions, one discovers that the exponential term cancel: $$\left[\vec{w}^{(\nu)}(t),\overline{\vec{w}^{(\nu)}}(t)\right]=e^{it(\nu-\nu)} \left[\vec{P}_\nu,\vec{P}_{\nu}^* \right]= \left[\vec{P}_\nu,\vec{P}_{\nu}^* \right]$$ One can see that $\left[\vec{P}_\nu,\vec{P}_{\nu}^* \right]$ is purely imaginary, in fact: $$\left[\vec{P}_\nu,\vec{P}_{\nu}^* \right]=\sum_{k=1}^n \left[P_{k,\nu}P_{k+n,\nu}^* -P_{k+n,\nu}P_{k,\nu}^*\right] = \sum_{k=1}^n 2\mathcal{I}\{P_{k,\nu}P_{k+n,\nu}^* \}$$
3) First of all notice that: $$\left[\vec{P}_\nu,\vec{P}_{\nu}^* \right]=-\left[\vec{P}_\nu^*,\vec{P}_{\nu} \right]$$ Notice also that, if eigenvalue $+i\beta_\nu$ is associated to eigenvector $\vec{P}_\nu$, eigenvalue $-i\beta_\nu$ is associated to eigenvector $\vec{P}_\nu^*$. So if the computation of the imaginary part of symplectic form gives a positive result, one just needs to exchange the roles of $+i\beta_\nu$ and $-i\beta_\nu$.