By Hitchin N.

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**Example text**

3 the cohomology class [α0 ] is non-zero. e. that α = cα0 + dγ. 64 Given α use the partition of unity to write α= ϕi α then by linearity it is sufficient to prove the result for each ϕi α, so we may assume that the support of α lies in a coordinate neighbourhood Um . e. renumbering, we have p ∈ U1 , Ui ∩ Ui+1 = ∅, q ∈ Um . Now for 1 ≤ i ≤ m − 1 choose an n-form αi with support in Ui ∩ Ui+1 and integral 1. 1, there is a form β0 with support in U1 such that α0 − α1 = dβ1 . Continuing, we get α0 − α1 α1 − α2 ...

The unit ball {x ∈ Rn : x ≤ 1} is a manifold with boundary S n−1 . 3. The M¨obius band is a 2-dimensional manifold with boundary the circle: 4. The cylinder I × S 1 is a 2-dimensional manifold with boundary the union of two circles – a manifold with two components. 59 We can define differential forms etc. on manifolds with boundary in a straightforward way. Locally, they are just the restrictions of smooth forms on some open set in Rn to (Rn )+ . A form on M restricts to a form on its boundary.

46 Proof: Consider the right hand side RX (α) = d(i(X)α) + i(X)dα. Now i(X) reduces the degree p by 1 but d increases it by 1, so RX maps p-forms to p-forms. Also, d(d(i(X)α) + i(X)dα) = di(X)dα = (di(X) + i(X)d)dα because d2 = 0, so RX commutes with d. Finally, because i(X)(α ∧ β) = i(X)α ∧ β + (−1)p α ∧ i(X)β d(α ∧ β) = dα ∧ β + (−1)p α ∧ dβ we have RX (α ∧ β) = (RX α) ∧ β + α ∧ RX (β). On the other hand ϕ∗t (dα) = d(ϕ∗t α) so differentiating at t = 0, we get LX dα = d(LX α) and ϕ∗t (α ∧ β) = ϕ∗t α ∧ ϕ∗t β and differentiating this, we have LX (α ∧ β) = LX α ∧ β + α ∧ LX β.