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

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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

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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 $\mathbb{R}$.

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$^1$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 (2,860 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

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