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Unformatted text preview: Diﬀerential Geometry 2
Test 1 Let (M, ω ) be a symplectic manifold and let H ∈ C (M ) be a smooth function.
Say that F ∈ C (M ) is a constant of the H motion if and only if F is constant
along each integral curve of the Hamiltonian vector ﬁeld ξH .
(a) Prove that F is a constant of the H motion if and only if the Poisson
bracket {F, H } is zero.
(b) Prove that the set of all constants of the H motion is a subspace of
C (M ) that is closed under Poisson bracket.
(c) What conclusion can be drawn from the hypothesis that every smooth
function on M is a constant of the H motion?
(d) Can ξH have a dense integral curve?
Solutions
(a) Let γ be an integral curve of ξH and compute:
(F ◦ γ ) (t) = γ (t)F = ξH (γ (t))F = ξH F (γ (t)) = {F, H }(γ (t))
˙
so that F is constant along γ iﬀ {F, H } is zero along γ .
(b) Let CH (M ) denote the set of all constants of the H motion. That
CH (M ) is a subspace of C (M ) follows immediately from the fact that {·, ·} is
reallinear in its ﬁrst variable. That CH (M ) is closed under Poisson bracket
follows from the Jacobi identity: if F1 , F2 ∈ C (M ) then
{{F1 , F2 }, H } + {{F2 , H }, F1 } + {{H, F1 }, F2 } = 0;
if F1 , F2 ∈ CH (M ) then {F2 , H } = {H, F1 } = 0 and so {{F1 , F2 }, H } = 0.
(c) If every F ∈ C (M ) satisﬁes ξH (F ) = 0 then ξH = 0 so that dH =
−ξH ω = 0. It follows that H is locally constant; in particular, if M is
connected then H is constant.
(d) Recall that H is constant along each integral curve of ξH . If (the
image of) some integral curve is dense, it follows that H is constant on a
dense set and hence (by continuity) constant; this forces ξH to be zero and
therefore to have constant (singleton image) integral curves. Absurd. 1 ...
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 Spring '09
 Robinson

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