Recall the exterior derivative that took -forms on to -forms on such that
- is an -linear satisfying
- (a function on ), a -vector field on then , that is, on -forms the exterior derivative behaves like the differential .
Recall also Stokes’ theorem and a useful corollary.
Theorem. (Stokes’ theorem) Let be an oriented -manifold with boundary given the induced boundary orientation. Then for any -form with compact support on we have
Theorem. (Fundamental theorem for line integrals) Let be a curve in parametrised by for , and let be a vector field on . If is the gradient of some scalar function then
Motivated then by the fundamental theorem for line integrals we ask, when is a vector field on the gradient of a scalar function ?
Theorem. (Poincaré’s lemma, first attempt) On , a vector field is the gradient of a scalar function if and only if
This is somehow the best we could have hoped for. Indeed, if then because .
Is something like Poincaré’s lemma always true? Recall the following example we saw in class.
Example. Let and let
Exercise. Show .
Question. Is the gradient of a scalar function on ? If something like Poincaré’s lemma held on then the answer would be yes.
Answer. No! Suppose it were true. Then we could apply the fundamental theorem for line integrals, which would say that the integral around any closed curve would be 0. That is,
However, choose the unit circle and compute,
de Rham cohomology
Fix a manifold . We want a way to know when something like Poincaré’s lemma holds on a space , and when it fails. Further, we want to measure the extent to which it fails. Let’s recall some language.
Definition. A differential form on is said to be
- Closed if
- Exact if for some differential form of degree one less then .
We define two vector spaces
the vector space of all closed -forms on
the vector space of all exact -forms on
Exercise. Show that these are vector spaces over .
So in this language our question becomes how similar/different are the vector spaces and for different ? As a first attempt to compare them we make the observation that which follows from the relation . Indeed, suppose , that is some . Then .
The idea of de Rham cohomology is to take the quotient vector space
Definition. The de Rham cohomology in degree is
Let’s apply the definition of a quotient vector space given in class to for clarity. We define an equivalence relation on as follows
and the quotient is defined to be the vector space of equivalence classes of .
Exercise. Show that is an equivalence relation, and think about what it means to be equivalent to .
In this language we can restate
Theorem. (Poincaré’s lemma) For every , every closed form is exact, that is,
Example. Let’s see this in action in dimension 1. First observe that there are no 2-forms on , since is 1-dimensional. So every 1-form is closed (indeed for any 1-form we have so ). A general 1-form looks like and it is exact if there exists a smooth function on such that . Pick
Thus every 1-form is exact. Consider . Since is exact, that is
the equivalence relation tells us (and hence any 1-form) is the zero equivalence class (i.e., ) and so .