# Analytic proof that Lyapunov exponents in Hamiltonian systems pairwise sum to zero

+ 4 like - 0 dislike
711 views

I have read that in Hamiltonian systems, Lyapunov exponents come in pairs $(\lambda_i, \lambda_{2N-i+1})$ such that their sum is equal to zero.

Is there a way of proving this analytically?

EDIT: Saw this here.

In symplectic systems, LEs come in pairs $(\lambda_i, \lambda_{2N-i+1})$ such that their sum is equal to zero. This means that the Lyapunov spectrum is symmetric. It is a way of emphasizing the invariance of Hamiltonian dynamics under change of the time arrow.

This post imported from StackExchange Physics at 2015-12-07 13:22 (UTC), posted by SE-user Cheeku

edited Dec 7, 2015
Ah, I see. Still, you should a) give a reference for where you read that and b) show some research effort. Just typing "lyapunov exponents pair sum zero" into Google gives me this paper as a result, which even shows a generalization to non-Hamiltonian systems.

This post imported from StackExchange Physics at 2015-12-07 13:22 (UTC), posted by SE-user ACuriousMind
@ACuriousMind I agree. Since my own search is filled with stuff in context, I stepped over the need to provide one while asking on forum. Edited

This post imported from StackExchange Physics at 2015-12-07 13:22 (UTC), posted by SE-user Cheeku

+ 6 like - 0 dislike
1. We are considering a discrete time evolution $$x_{n}~=~f(x_{n-1})~=~f^{n\circ}(x_0), \qquad n~\in~\mathbb{N},$$ in a $2N$-dimensional symplectic manifold $(M,\omega)$, where $f$ is a symplectomorphism.

2. Let us for simplicity work in local coordinates. Define the Jacobian matrix as $$\tag{1} A(x,n)^{i}{}_{j}~:=~\frac{\partial (f^{n\circ} (x))^i}{\partial x^j}.$$

3. In local Darboux coordinates, the Jacobian matrix (1) is a symplectic matrix $$\tag{2} A^T\Omega A~=~ \Omega, \qquad \Omega ~:=~\begin{bmatrix} 0_N & -I_N \cr I_N & 0_N \end{bmatrix}.$$

4. Note that the transposed $A^T$ is also a symplectic matrix. Note that $A^TA$ is a positive definite symplectic matrix.

5. Symplectic quartet mechanism: For a diagonalizable$^1$ symplectic matrix, the eigenvalues form quartets $$\tag{3} \{\lambda,\bar{\lambda}, \lambda^{-1},\bar{\lambda}^{-1}\}$$ in the complex plane $\mathbb{C}$. A quartet becomes a doublet on the real axis and on the unit circle.

6. Define the Lyapunov exponents $$\tag{4} \left\{\lambda_1(x,n), \ldots, \lambda_{2N}(x,n)\right\}~\subset~\mathbb{R}$$ as the eigenvalues of the Hermitian matrix $$\tag{5} \Lambda(x,n)~:=~\frac{1}{2n}\ln \left(A(x,n)^TA(x,n)\right).$$

7. It follows from the symplectic doublet mechanism (3), that the eigenvalues (4) are distributed symmetrically around 0 on the real axis $\mathbb{R}$.

--

$^1$Not all symplectic matrices are diagonalizable. 2D Counterexample: $$\tag{6} A~=~\begin{bmatrix} 1 & 1 \cr 0 & 1 \end{bmatrix}.$$

This post imported from StackExchange Physics at 2015-12-07 13:22 (UTC), posted by SE-user Qmechanic
answered Oct 23, 2015 by (3,120 points)
+1, Just a brief comment, since any continuous-time-translation is a symplectic transformation, then the continuous-time case is a trivial generalization. I also believe that the matrix $A^T A$ is positive semi-definite and thus will have exclusively $\lambda=\bar{\lambda}$ degenerate quartets.

This post imported from StackExchange Physics at 2015-12-07 13:22 (UTC), posted by SE-user Void
@Void: I agree.

This post imported from StackExchange Physics at 2015-12-07 13:22 (UTC), posted by SE-user Qmechanic

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsO$\varnothing$erflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.