De Rham Cohomology

Recall the exterior derivative d: \Omega^k(M) \rightarrow \Omega^{k+1}(M) that took k-forms on M to (k+1)-forms on M such that

  1. d is an \mathbb{R}-linear satisfying d(\omega \wedge \tau) = d\omega \wedge \tau + (-1)^k\omega \wedge d\tau
  2. d^2=0
  3. f \in \Omega^0(M) (a C^\infty function on M), X a C^\infty-vector field on M then (df)(X)=Xf, that is, on 0-forms f the exterior derivative behaves like the differential df.

Recall also Stokes’ theorem and a useful corollary.

Theorem. (Stokes’ theorem) Let M be an oriented n-manifold with boundary \partial M given the induced boundary orientation. Then for any (n-1)-form \omega with compact support on M we have

\displaystyle \int_M d\omega = \int_{\partial M}\omega

Theorem. (Fundamental theorem for line integrals) Let C be a curve in \mathbb{R}^3 parametrised by r(t)=(x(t), y(t), z(t)) for t \in [a,b], and let F be a vector field on \mathbb{R}^3. If F = grad(f) is the gradient of some scalar function f then

\displaystyle \int_C F\cdot dr = f(r(b))-f(r(a))

Motivated then by the fundamental theorem for line integrals we ask, when is a vector field F on X the gradient of a scalar function f?

Theorem. (Poincaré’s lemma, first attempt) On \mathbb{R}^3, a vector field F is the gradient of a scalar function if and only if curl(F)=0

This is somehow the best we could have hoped for. Indeed, if F = grad(f) then curl(F)=curl(grad(f))=0 because curl \circ grad =0.

Is something like Poincaré’s lemma always true? Recall the following example we saw in class.

Example. Let X = \mathbb{R}^3 \backslash \{z-\mbox{axis}\} and let \displaystyle F = \left\langle \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}, 0 \right\rangle

Exercise. Show curl(F)=0.

Question. Is F the gradient of a scalar function f on X? If something like Poincaré’s lemma held on X 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 C would be 0. That is,

\displaystyle \int_C \frac{-y}{x^2+y^2}dx + \frac{x}{x^2+y^2}dy =0

However, choose the unit circle C: x^2+y^2=1 and compute,

\displaystyle \int_C -ydx+xdy= \int_0^{2\pi}-(\sin t)d(\cos t) + (\cos t)d(\sin t)=2\pi \neq 0

de Rham cohomology

Fix a manifold M. We want a way to know when something like Poincaré’s lemma holds on a space M, and when it fails. Further, we want to measure the extent to which it fails. Let’s recall some language.

Definition. A differential form \omega on M is said to be

  1. Closed if d\omega=0
  2. Exact if \omega = d\tau for some differential form \tau of degree one less then \omega.

We define two vector spaces

Z^k(M) = the vector space of all closed k-forms on M

B^k(M) = the vector space of all exact k-forms on M

Exercise. Show that these are vector spaces over \mathbb{R}.

So in this language our question becomes how similar/different are the vector spaces Z^k(M) and B^k(M) for different M? As a first attempt to compare them we make the observation that B^k(M) \subseteq Z^k(M) which follows from the relation d^2=0. Indeed, suppose \omega \in B^k(M), that is \omega = d\tau some \tau \in \Omega^{k-1}(M). Then d\omega=d^2\tau=0.

The idea of de Rham cohomology is to take the quotient vector space

Definition. The de Rham cohomology in degree k is

\displaystyle H^k_{dR}(M)=Z^k(M)/B^k(M)

Let’s apply the definition of a quotient vector space given in class to H^k_{dR} for clarity. We define an equivalence relation \sim on Z^k(M) as follows

\omega \sim \omega' \Leftrightarrow \omega-\omega' \in B^k(M)

and the quotient is defined to be the vector space of equivalence classes of \sim.

Exercise. Show that \sim is an equivalence relation, and think about what it means to be equivalent to 0.

In this language we can restate

Theorem. (Poincaré’s lemma) For every k \geq 1, every closed k form is exact, that is, H^k_{dR}(\mathbb{R}^n)=0

Example. Let’s see this in action in dimension 1. First observe that there are no 2-forms on \mathbb{R}, since \mathbb{R} is 1-dimensional. So every 1-form is closed (indeed for any 1-form \omega we have d\omega \in \Omega^2(\mathbb{R})=0 so d\omega=0). A general 1-form looks like f(x)dx and it is exact if there exists a smooth function g(x) on \mathbb{R} such that f(x)dx=dg=g'(x)dx. Pick

\displaystyle g(x) = \int_0^x f(t)dt

Thus every 1-form is exact. Consider \omega \in Z^1(\mathbb{R})=\Omega^1(\mathbb{R}). Since \omega is exact, that is

\omega - 0 \in B^1(\mathbb{R})

the equivalence relation tells us \omega (and hence any 1-form) is the zero equivalence class (i.e., [\omega]=[0]) and so H^1_{dR}(\mathbb{R})=0.

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