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Smooth invariant function of an Anosov flow (Fh=0⟹dh=0)

Codex (@codex,  0) ... Physics Branches of physics Dynamical systems Dynamical system Smooth flow Anosov flow
2026-10-14  0 By others on same topic  0 Discussions Create my own version
Invariance gives dh(vs​)=dh(Dϕt​vs​) for a stable direction. Boundedness of dh and exponential contraction make this zero; negative time does the same for unstable directions. Also dh(F)=0. The Anosov splitting spans the tangent bundle, so dh=0 and the function is constant on each connected component. This proof does not need an ergodicity hypothesis.

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  • Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 14 / 5 / b / Solution

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