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Extra resources for Billiards

Example text

First, one may assume that f is bounded. Otherwise, consider the set S = {x : |f | ≤ c}, and replace f by χ(S)f , where χ is the indicator of the set (1 inside, 0 outside). This new function is still Ta -invariant and bounded, so the result for f will follow from that for χ(S)f as c → ∞. , rn ) is an integer vector, and the summation is over all such vectors. Since f is Ta -invariant, f (Ta x) ∼ cr e2πi e2πi = cr e2πi ∼ f (x) (mod 0). The Fourier coefficients are unique, hence cr e2πi = cr for all r.

Some higher-dimensional generalizations are available too – [B-P]. In our exposition we follow [He 1] and [Ma 2]. Theorem 2. Let T be an orientation and area-preserving diffeomorphism of an open cylinder S 1 × (−1, 1), that maps each topological end to itself. Let U be an open T -invariant subset, which contains the lower end S1 × (−1, −1 + ) and whose closure does not intersect the upper 43 end S 1 × (1 − , 1), > 0 being sufficiently small. Suppose also that the circle S 1 × {−1 + } is a deformation retract of U .

Equal to dF for a generating function F . Let x be an n-periodic point: T n x = x. Assign to this point the number n F (T i x). L(x) = i=1 This number depends on the choice of the generating function, determined up to an additive constant. One normalizes F in a natural way: say, by requiring that its average vanish, or that F vanish on a boundary component (in the case of billiards). Notice that L(x) depends on the periodic orbit, but not on a particular choice of the point: L(x) = L(T x). Finally L(x) is independent of a change of α by a coboundary: if α = α + dφ, then F = F + φ ◦ T − φ and L (x) = i F (T i x) + φ(T i+1 x) − φ(T i x) = F (T i x) = L(x).