non-degeneracy condition for hamiltonian

Question

Let $ H : M^{2n} \longrightarrow \mathbb{C}$ an analytic Hamiltonian on symplectic manifold $M$. In KAM theory we ask that $H$ is non-degenerate, that is $$\frac{\partial ^2 H}{\partial x_i \partial x_j} \neq 0.$$ I don' t understand why we need this condition for proving KAM theorem, which says that provided the Diophantine condition on the frequency vector, the invariant tori of $H$ are preserved under a small perturbation.

Answer 1

There are a few proofs of KAM and I'm only familiar with one of them. Please read for yourself, I only summarize it here.

One starts with a perturbed hamiltonian $H(p,q) = h(p) + \epsilon f_*(p,q)$ real analytic on $D \times \mathbb{T}^n$ for some domain $D \in \mathbb{R}^n$. For the case of $\epsilon = 0$ the system is integrable with standard orbits $(p(t),q(t))$, $$ p(t) = p_0,\qquad q(t) = q_0 + \omega(p_0) t,\qquad \omega(p_0) = h_p(p_0), $$ where subscript denotes partial derivation. One says that the hamiltonian is nondegenerate if $\det(h_{pp}) \neq 0$. Notice that when one talks about a nondegenerate hamiltonian, one refers to a condition on the integrable part. In particular, the map $$ h_p : D \to \Omega \subset \mathbb{R}^n,\qquad p \mapsto \omega(p) $$ is a diffeomorphism. The idea of the proof (the one I'm considering) is to consider a family of hamiltonians parametrized by $\omega \in \Omega$ instead of the single hamiltonian that we started with. One then proves that each parametrized hamiltonian $H_\omega$ has one invariant torus. This works if each $p \in D$ can be uniquely represented by frequencies $\omega \in \Omega$, and this is the case if we require nondegeneracy.