G.R. Belitskii and A.Ya. Kopanskii, "Equivariant Sternberg theorem", preprint mp-arc 00-54; "Sternberg theorem for equivariant hamiltonian vector fields", preprint mp-arc 00-449 Home/Search Document Details and Download
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Poincaré normal forms and simple compact Lie groups - Gaeta (Correct)
....to its linear part. ffl Sternberg theorem [3, 4] if the linear part A of the system is hyperbolic, then formally conjugacy to the normal form implies C (but in general not analytic) conjugacy. The theorem was proved by Sternberg and Chen [12, 36] and recently extended to symmetric systems [6]. ffl BMW C theory: if the linear part of the system satisfies a very general arithmetic condition known as condition (introduced by Bruno [10] building on work by Siegel and Pliss [3, 33] and the normal form satisfies condition A , i.e. can be written as x = 1 ff(x) Ax, where ff(x) is ....
G.R. Belitskii and A.Ya. Kopanskii, "Equivariant Sternberg theorem", preprint mp-arc 00-54; "Sternberg theorem for equivariant hamiltonian vector fields", preprint mp-arc 00-449
Poincar'e Renormalized Forms and Regular Singular Points of Vector .. - Gaeta (Correct)
....and [5, 34] 15 of the Poincar e criterion. If 1 = 2 is irrational, the NF reduces to the linear form, and no further normalization is needed; moreover, convergence can be guaranteed on the basis of Pliss theorem [41] If 1 = 2 = p=q 2 Q, there can be resonances; in this case Sternberg theorem [2, 6, 7, 18, 44] guarantees that the NF is smoothly (but in general, not analytically) equivalent to the original system. The PRF for this case has not been studied so far, and we study it later on in section 8. In case N1 the eigenvalues are in a Poincar e domain, as 6= 0; moreover there are no resonances, ....
....( Delta ) 0 (both eigenvalues have the same sign) we are in a Poincar e domain, so the convergence of the transformation to NF is guaranteed; on the other hand, if 0 (i.e. we have an hyperbolic saddle point in the origin) we are not in a Poincar e domain. However, the ChenSternberg theorem [2, 6, 7, 18, 44] guarantees the system is C conjugated to its normal form; as for the analytic conjugacy in this case, this is guaranteed if j=j is irrational, due to Pliss theorem [41] We also noticed that if = is irrational, there are no resonances, i.e. the NF is linear; in this case we do not need (nor ....
G.R. Belitskii and A.Ya. Kopanskii, "Equivariant Sternberg theorem", preprint mp-arc 00-54
Sternberg Theorem for Equivariant Hamiltonian Vector Fields - Belitskii Ya Kopanskii Self-citation (Belitskii Kopanskii) (Correct)
.... see [2] This activity faces, in particular, a similar equivariant problem: could one provide a conjugacy (smooth, formal) of two symmetric or anti symmetric vector fields via a transformation keeping the property The linear and the formal aspects of the problem were considered in [3, 4] In [5] the authors proved a related version of Smooth Conjugacy Sternberg Theorem. A similar question arised for hamiltonian systems: is it possible to conjugate two hamiltonian vector fields via a canonical coordinate change The affirmative answer was given in [6, 7, 8] These two results are in the ....
....we have to solve the cohomology equation (3) then find the corresponding vector field and integrate the system x = x) 1. 4 G equivariant conjugacy First of all, note that solvability of equation (3) under the assumption of Theorem 2. 1 was proved in [8] for k 1 ) and [5] (for k = 1 ) Prove that the solution can be choosed (G; oe) equivariant. Let be a (G; oe) equivariant symplectic form. Take a matrix U 2 G, U = A B C D ; where A; B; C and D are d Theta d matrices. Since is in the standard form then (see, for example, 12] Proposition 5.5.6) A t ....
BELITSKII G. & KOPANSKII A., Equivariant Sternberg theorem, Phisica D, in press.
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