Lecture 18 - De Rham Cohomology

#math

We are picking off from the last class in our discussion of closed exact 1 forms.

Motivation: How a Euclidean geometry describing $2$ D subspace of $R^{3}$ via $\hat{n}, \hat{u}, \hat{v}$ . We have the notion of the cross product (that $u \times v = - v \times u$ ). How do we extend this notion to higher dimensional manifolds?

We want to define a skew symmetric bilinear product $B (v_{i}, v_{j}) = - B (v_{j}, v_{i})$ . In other words we want to work exterior algebras.

Definition

The wedge product of two vectors $u$ and $v$ measures the noncommutativity of their tensor product. Thus, the wedge product $u \land v$ is the square matrix defined by:

u \land v = u \otimes v - v \otimes u

Equivalently,

(u \land v)_{i j} = (u_{i} v_{j} - v_{i} u_{j})

Now recall the following:

$Z^{k} (M) =$ vector space of all closed $k -$ forms.
$B^{k} (M) =$ vector space of all exact $k -$ forms.
$H^{k} (M) =$ $Z^{k} (M) / B^{K} (M)$ measures the extent to which closed $k -$ forms fail to be exact.

(Write about differential forms here)
https://en.wikipedia.org/wiki/Closed_and_exact_differential_forms

Proposition

For any connected smooth manifold $M$ of dimension n.

H^{0} (M) ≅ R

Proof

Let $f$ be a closed $0 -$ form. That is $d f = 0$ this implies $f$ is constant on $U$ . This holds for every coordinated neibourhood on $M$ since $M$ is connected.
Since the function are $0 -$ forms we cannot have exact forms thus $B^{0} (M) = 0$ thus:

H^{0} (M) = {f \in C^{\infty} | f : M \to R}

which is isomorphic to $R$ .

Example

(De Rham cohomology of $R^{1}$ ) We know that:

H^{0} (R) = R

Let us compute $H^{1} (R)$ .
Let $α$ be a one form, that is $α = f (x) d x$ , then let us define $g (x) = \int^{x} f (u) d u$ then its clear we obtain that:

d g = f (x) d x

Thus every 1 form is exact. Thus:

H^{1} (R) = 0

(Write about De Rham cohomology of spheres)

Let $ω$ be 1-form on $S^{1}$ :

ω = - y d x + x d x

Then $ω (X)$ is nowhere vanishing for every vector field $X$ .

Proposition

Let $G : M \to N$ is a smooth map between manifolds. Let $G^{*}$ be the pullback map then it takes closed forms to closed forms and also takes exact forms to exact forms. Thus:

\begin{aligned} G^{*} : H^{*} (N) \to H^{*} (M) \\ G^{*} [ω] & = [G^{*} ω] \end{aligned}

Proof

Let $ω$ be an exact form that is $ω = d η$ then $G^{*} ω = G^{*} d η = d (G^{*} η)$ .
If $ω$ is closed then $d (G^{*} ω) = G^{*} d ω = 0$
Thus if $[ω] = [ω^{'}]$ then $G^{*} [ω] = G^{*} [ω^{'}]$
Thus $G^{*}$ decends to a linear map on the cohomoology.

Lemma

(Poincare Lemma) If $U$ is any contractible subset of $R^{n}$ then $H^{n} (U) = 0$ for any $n \geq 1$ .

Let $M$ be a smooth manifold and let $Λ^{n} (M)$ with $d : Λ^{p} (M) \to Λ^{p + 1} (M)$ .
Let $ω$ be smooth $p -$ form on a smooth manifold $M$ of dimension $n$ . Then:

d ω (u_{1}, . . ., u_{p + 1}) = \sum_{i = 1}^{p + 1} (- 1)^{i + 1} u_{i} (u_{1}, . . ., {\hat{u}}_{i}, . . ., u_{p + 1}) + \sum_{1 \leq i < j \leq p + 1} (- 1)^{i + j} ω ([u_{i}, u_{j}, u_{1}], . . ., {\hat{u}}_{i}, . ., {\hat{u}}_{j}, . . ., u_{p + 1})

A smooth autonomous dynamical system on a 2D manifold

We have a local chart $(U, x, y)$ .

\begin{aligned} \frac{d x}{d t} & = f (x, y) \\ \frac{d y}{d t} & = g (x, y) \end{aligned}

We have the vector field $X = f (x, y) \partial x + g (x, y) \partial y$ . We have the projection map $τ_{M} : T M \to M$ .

$R \times T M ≅ J^{1} (R, M)$ is called the Jet bundle.

A smooth curve, $γ : t \to γ (t)$ defines an integral curve if the contact forms (one forms) vanish on $γ$ . The contact forms are:

{\begin{cases} ω^{1} = d t \\ ω^{2} = d x - f d t \\ ω^{3} = d y - g d t \end{cases}

Notice the following:

\begin{aligned} d ω^{1} & = 0 \\ d ω^{2} & = - f_{x} d x \land d t - f_{y} d y \land d t \\ = - f_{x} (d x - f d t) \land d t - f_{y} (d y - g d t) \land d t \\ = - f_{x} ω^{2} \land ω^{1} - f_{y} ω^{3} \land ω^{1} \\ d ω^{3} & = - g_{x} d_{x} \land d t - g_{y} d y \land d t \\ = g_{x} ω^{1} \land ω^{2} + g_{y} ω^{1} \land ω^{3} \end{aligned}

Thus

\begin{aligned} ω^{2} \land d ω^{3} + ω^{3} \land d ω^{2} & = 0 \\ ω^{2} \land d ω^{2} - ω^{3} \land d ω^{3} & = 0 \end{aligned}

Thus if we allow $ω = ω^{2} + i ω^{3}$ this gives us $ω \land d ω = 0$ . This also gives us the Cauchy Riemann Equations:

\begin{aligned} \frac{\partial f}{\partial y} & = - \frac{\partial g}{\partial x} \\ \frac{\partial f}{\partial x} & = \frac{\partial g}{\partial y} \end{aligned}

If we have an almost complex structure such that where we have $- J^{2} = - I$ then $J X = f (x, y) \partial y - g (x, y) \partial x$ . If CR relations holds then $[X, J X] = 0$ .

i : Σ ↪ J^{1} (R, M)

g_{Σ} = \sum ω^{i} \times ω^{i} d ω^{1} = 0 d ω^{2} = f_{x} ω^{1} \land ω^{2} + f_{y} ω^{1} \land ω^{3} d ω^{3} = - f_{y} ω^{1} \land ω^{2} + f_{x} ω^{1} \land ω^{3}

That is if we let:

d ω^{i} = - \sum θ_{j}^{i} \land ω^{j}

Curvature 2-form (Wald Second Structure equation)