On a compact oriented Finsler manifold without boundary, every cohomology class has a unique harmonic representation. The dimension of the space of all harmonic forms of degree is the th Betti number of the manifold.
The homomorphism which, according to the snake lemma, permits construction of an exact sequence(1)from the above commutative diagram with exact rows. The homomorphism is defined by(2)for all , denotes the image, and is obtained through the following construction, based on diagram chasing.1. Exploit the surjectivity of to find such that . 2. Since because of the commutativity of the right square, belongs to , which is equal to due to the exactness of the lower row at . This allows us to find such that . While the elements and are not uniquely determined, the coset is, as can be proven by using more diagram chasing. In particular, if and are other elements fulfilling the requirements of steps (1) and (2), then and , and(3)hence because of the exactness of the upper row at . Let be such that(4)Then(5)because the left square is commutative. Since is injective, it follows that(6)and so(7)..
When two cycles have a transversal intersection on a smooth manifold , then is a cycle. Moreover, the homology class that represents depends only on the homology class of and . The sign of is determined by the orientations on , , and .For example, two curves can intersect in one pointon a surface transversally, sinceThe curves can be deformed so that they intersect three times, but two of those intersections sum to zero since two intersect positively and one intersects negatively, i.e., with the manifold orientation of the curves being the reverse orientation of the ambient space.On the torus illustrated above, the cycles intersectin one point.The binary operation of intersection makes homology on a manifold into a ring. That is, it plays the role of multiplication, which respects the grading. When and , then . In fact, intersection is the dual to the cup product in Poincaré duality. That is, if is the Poincaré dual to and is the dual..
de Rham cohomology is a formal set-up for the analytic problem: If you have a differential k-form on a manifold , is it the exterior derivative of another differential k-form ? Formally, if then . This is more commonly stated as , meaning that if is to be the exterior derivative of a differential k-form, a necessary condition that must satisfy is that its exterior derivative is zero.de Rham cohomology gives a formalism that aims to answer the question, "Are all differential -forms on a manifold with zero exterior derivative the exterior derivatives of -forms?" In particular, the th de Rham cohomology vector space is defined to be the space of all -forms with exterior derivative 0, modulo the space of all boundaries of -forms. This is the trivial vector space iff the answer to our question is yes.The fundamental result about de Rham cohomology is that it is a topological invariant of the manifold, namely: the th de Rham cohomology vector space..