example of de Rham cohomology
If is a differential form on a smooth manifold , then it is always true that if is exact ( for some other differential form ), then is closed (). On some manifolds, the opposite is also the case: all closed forms of degree at least 1 are exact. However, in general this is not true. The idea of de Rham cohomology is to measure the extent to which closed differential forms are not exact in terms of real vector spaces.
The simplest example of a differential manifold (apart from the empty manifold) is the zero-dimensional manifold consisting of a single point. Here the only differential forms are those of degree 0; actually, if is a single point. Applying the definition of the de Rham cohomology gives .
Next, we use the fact that the de Rham cohomology is a homotopy invariant functor to show that for any the de Rham cohomology groups of are
The reason for this is that is contractible (homotopy equivalent to a point), and so has the same de Rham cohomology. More generally, any contractible manifold has the de Rham cohomology of a point; this is essentially the statement of the Poincaré lemma.
The first example of a non-trivial for is the circle . In fact, we have
where is any 1-form on with . The standard volume form on , which it inherits from if we view as the unit circle, is such a form. The notation is somewhat misleading since it is not the differential of a global function ; this is exactly the reason it appears in . (However, by the Poincaré lemma, it can locally be viewed as the differential of a function.)
For arbitrary , the dimensions of the de Rham cohomology groups of are given by for or , and otherwise. A couple of methods exist for calculating the de Rham cohomology groups for and other, more complicated, manifolds. The Mayer-Vietoris sequence is an example of such a tool.
|Title||example of de Rham cohomology|
|Date of creation||2013-03-22 14:25:01|
|Last modified on||2013-03-22 14:25:01|
|Last modified by||pbruin (1001)|