By Michael Spivak

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**Extra resources for A Comprehensive Introduction to Differential Geometry, Vol. 4, 3rd Edition **

**Example text**

0 be the volume form on R2n . 47) Ω = dH ∧ α on U. 48) µ = j∗α on S is clearly a volume form on S. 47). Indeed, if Ω = dH ∧ α = dH ∧ β then dH ∧ (α − β) = 0 so that α − β = dH ∧ γ for a (2n − 2)-form γ on U . In view of j ∗ dH = 0 we ﬁnd j ∗ α = j ∗ β + j ∗ (dH ∧ γ) = j ∗ β as claimed. 49) (ϕt )∗ µ = µ. 20 Chapter 1 Introduction This follows from (ϕt )∗ Ω = Ω and (ϕt )∗ dH = dH. 47) we ﬁrst ﬁnd Ω = dH ∧ (ϕt )∗ α, and using the uniqueness we conclude j ∗ (ϕt )∗ α = j ∗ α. 49). 50) = m(A) , A ⊂ S.

The polynomial in the parenthesis is homogeneous of degree m and the bracket {·, ·} stands for the Poisson bracket deﬁned for two functions F and G as follows: {F, G} = ω(XG , XF ). We see that we can modify H by functions of the form {H2 , P }. Denote by Vm the vector space of homogeneous polynomials of degree m and let L : Vm −→ Vm be the linear operator deﬁned by L(P ) = {H2 , P } for P ∈ Vm . 44 Chapter 1 Introduction Lemma 1. The kernel K = K(L) and the range R = R(L) of the linear operator L are complementary: Vm = K + R K ∩ R = {0} .

Pm−1 + Pm + . . + Ps with Pj ∈ N for 3 ≤ j ≤ m − 1, and show that also Pm ∈ N . In order to determine the terms of order m, we use the formula H ◦ exp XG = H + {H, G} + {{H, G}, G} + . . for two functions H, G. This follows from the Taylor expansion of H ◦ exp tXG in t at t = 0. If now Fˆ = H2 + Fˆ3 + . . + Fˆs and similarly, F = H2 + F3 + . . + Fs we ﬁnd Fˆm + {H2 , Pm } + Am = Fm , where, by the induction assumption, Am is a sum of Poisson brackets of elements in N . Since N is a subalgebra we have Am ⊂ N , and therefore, {H2 , Pm } ∈ N .