Ivancevic_Applied-Diff-Geom

Applied Manifold Geometry 1853.6.3.3 De Rham Complex and Homotopy OperatorsAfter an intuitive introduction of (co)homology ideas, we now turn to theirproper definitions. Given a smooth manifold M, let Ω p (M) denote thespace of all smooth p−forms on M. The differential d, mapping p−formsto (p + 1)−forms, serves to define the de Rham complex on M0 → Ω 0 (M)d 0 ✲ Ω 1 (M)d 1 ✲ ...d n−1 ✲ Ω n (M) → 0. (3.42)Recall (from section 6.200 above) that in general, a complex is definedas a sequence of vector spaces, and linear maps between successive spaces,with the property that the composition of any pair of successive maps isidentically 0. In the case of the de Rham complex (3.42), this requirement isa restatement of the closure property for the exterior differential: d ◦ d = 0.In particular, for n = 3, the de Rham complex on a manifold M reads0 → Ω 0 (M)d 0 ✲ Ω 1 (M)If ω ≡ f(x, y, z) ∈ Ω 0 (M), thend 0 ω ≡ d 0 f = ∂f∂xdx +∂f∂yd 1 ✲ Ω 2 (M)If ω ≡ fdx + gdy + hdz ∈ Ω 1 (M), then( ∂gd 1 ω ≡∂x − ∂f ) ( ∂hdx∧dy+∂y∂y − ∂g )dy∧dz+∂z∂fdy + dz = grad ω.∂zd 2 ✲ Ω 3 (M) → 0.(3.43)( ∂f∂z − ∂h∂xIf ω ≡ F dy ∧ dz + Gdz ∧ dx + Hdx ∧ dy ∈ Ω 2 (M), thend 2 ω ≡ ∂F∂x + ∂G∂y + ∂H∂z= div ω.Therefore, the de Rham complex (3.43) can be written as)dz∧dx = curl ω.0 → Ω 0 (M)grad ✲ →Ω 1 (M)curl✲ Ω 2 (M)div ✲ Ω 3 (M) → 0.Using the closure property for the exterior differential, d ◦ d = 0, we get thestandard identities from vector calculuscurl · grad = 0 and div · curl = 0.The definition of the complex requires that the kernel of one of thelinear maps contains the image of the preceding map. The complex is