# 1:1:2 resonance revisited

The aim of this quick note is to review Duistermaat’s formal nonintegrability result for 1:1:2 resonance.

We will use complex coordinates and write the quadratic part of the Hamiltonian function as $H_2 =p_1q_1 +p_2q_2 + 2p_3q_3$

The monomes of degree 3 which commute with $H_2$ are: $p_3q_1^2, p_3q_1q_2, p_3q_2^2, q_3p_1^2,q_3p_1p_2, q_3p_2^2$

Thus, a generic $H_3$ in Birkhoff normal forms will have 6 terms (is a linear combination of the above monomes), with 6 independent coefficients

Similarly to the 1:2:2 resonance case, we have a gl(2,C) symmetry. But, unlike the 1:2:2 case where $H_3$ only has 4 coefficients, here we have 6.

In the 1:2:2 resonance case there is always an additiona quadratic 1st integral. But here the parameter space is too large, and only a small subset of the set of all possible $H_3$ (apparently of codimension 2) admits an additional quadratic first integrals. Those special $H_3$ are called “symmetric”.

Generic $H_3$ in the 1:1:2 resonance doesn’t admit an additional quadratic 1st integral. Duistermaat shows that all those $H_3$ (which donot admit an additional quadratic 1st integral) donot admit any additional integral (of any degree) at all, and thus 1systems with such $H_3$ are formally nonintegrable.

Integrable cases:

1) $H_3 = p_3 \phi^2 + q_3 \psi^2$

where $\phi$ is a linear function in $q_1,q_2$ and $\psi$ is a linear function in $q_1,p_2$

Then the additional quadratic integral is similar to the 1:2:2 resonance case

2) $H_3 = a p_3 q_1q_2 + b q_3 p_1p_2$

(or can be put into that form by a symplectic transformation)

then $p_1q_1 – p_2q_2$ will be an additional first integral

If $H_3$ is not of the above types (e.g. instead of $phi_2$ we have $\phi_1 phi_2$ where $\phi_1$ and $\phi_2$ are linear in $q_2,q_3$ and independent, and “generic”) then no addtional quadratic first integral. Question: why no first integral at all ?

Strategies for proving non-existence of 1st integrals ?

– analytic: infinite branching ?

– algebraic:  linear equations without solution ?

Idea: use the coordinate system $\phi_1,\phi_2, q_3, \psi_1, \psi_2,p_3$ instead of $p_1,p_2,p_3q_1,q_2,q_3$ $H_3 = p_3 \phi_1\phi_2 + q_3 \psi_1\psi_2$

with $\{\phi_i,\psi_j\} = \rho_{ij}$ constants, the matrix $(\rho_{ij})$ invertible. Find the symmetry condition  for the existence of an additional quadratic first integral? Apparently, the condition is that this matrix $(\rho_{ij})$ commutes with the matrix (1 0 // 0 -1)

Purely algebraic (nilpotent-type) verification that (independent) higher degree integrals don’t exist ? The purely algebraic verification is extremely difficult, due to the fact that higher degree integrals do exist ! (they are functions of H2 and H3) For example: H2^2, H2H3, H3^2, etc.

Back to analytic tricks: construct an independent 1st integral on some subset (near some special solutions) and show that it has infinite branching.

It was observed by Duistermaat that, in the real case (when H2 is really a resonant 1:1:2 harmonic oscillator),almost all orbits of the Hamiltonian vector field of H3 on the 5-dimensional subset $\{H_3=0\}$ are periodic !

What does it mean in the complex case ?

Look at orbits on H3 = 0

We can move any orbit to an orbit with $p_3 = 0$ by the action of $X_{H_2}$, so let’s assume that $p_3=0$ but $q_3 \neq 0$. Then it will be so along the orbit of $X_{H_3}$, and $H_3=0$ means that $\psi_1 \psi_2 = 0$

Let’s say $\psi_2 = 0$

On $\{p_3 = 0,\psi_2 = 0\}$ the vector field $X_{H_3}$ also preserves $\psi_1$

We may assume that $\{q_1,\psi_2\} = 0$. Then $q_1$ is also preserved by $X_{H_3}$ on $\{p_3 = 0,\psi_2 = 0\}$. Thus, we get invariant 2-dimensional affine planes with coordinates $q_2, q_3$.

The equation of motion (for $\{p_3 = 0\psi_2 = 0\}$ and $\psi_1, q_1$ fixed) is: $\dot{q}_2 = C q_3 \psi_1$, where C is some structural constant, and $\dot{q}_3 = - \phi_1 \phi_2$