0708-1300/Class notes for Tuesday, November 13

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Typed Notes

The notes below are by the students and for the students. Hopefully they are useful, but they come with no guarantee of any kind.

We begin with a review of last class. Since no one has typed up the notes for last class yet, I will do the review here.

Recall had an association $M\rightarrow \Omega ^{k}(M)$ which was the "k forms on M" which equaled $\{w:\ p\in M\rightarrow A^{k}(T_{p}M)\}$

where

$A^{k}(V):=\{w:\underbrace {V\times \ldots \times V} _{k\ times}\rightarrow \mathbb {R} \}$ which is

1) Multilinear

2) Alternating

We had proved that :

1) $A^{k}(V)$ is a vector space

2) there was a wedge product $\wedge :A^{k}(V)\times A^{l}(V)\rightarrow A^{k+l}(V)$ via $\omega \wedge \lambda (v_{1},\ldots ,v_{k+l})=$${\frac {1}{k!l!}}\sum _{\sigma \in S_{k+l}}(-1)^{\sigma }\omega (v_{\sigma (1)},\ldots ,v_{\sigma (k)})\lambda (v_{\sigma (k+1)},\ldots ,v_{\sigma (k+l)})$ that is

a) bilinear

b) associative

c) supercommutative, i.e., $\omega \wedge \lambda =(-1)^{deg(w)deg(\lambda )}\lambda \wedge \omega$

From these definitions we can define for $\omega \in \Omega ^{k}(M)$ and $\lambda \in \Omega ^{l}(M)$ that $\omega \wedge \lambda \in \Omega ^{k+l}(M)$ with the same properties as above.

Claim

If $\omega _{1},\ldots ,\omega _{n}$ is a basis of $A^{1}(V)=V^{*}$ then $\{\omega _{I}=w_{i_{1}}\wedge \ldots \wedge \omega _{i_{k}}\ :\ I=(i_{1},\ldots ,i_{k})\ with\ 1\leq i_{1}<\ldots <i_{k}\leq n\}$ is a basis of $A^{k}(V)$ and $dimA^{k}(V)=nCk$

If $\omega _{1},\ldots ,\omega _{n}\in \Omega ^{1}(M)$ and $w_{1}|_{p},\ldots ,\omega _{n}|_{p}$ a basis of $(T_{p}M)^{*}\ \forall p\in M$ then any $\lambda$ can be written as $\lambda =\sum _{I}a_{I}(p)\omega _{I}$ where $a_{I}:M\rightarrow \mathbb {R}$ are smooth.

The equivalence of these is left as an exercise.

Example

Let us investigate $\Omega ^{*}(\mathbb {R} ^{3})$ (the * just means "anything").

Now, $(T_{p}(\mathbb {R} ^{3}))^{*}=<dx_{1},dx_{2},dx_{3}>$ where $x_{i}:\mathbb {R} ^{3}\rightarrow \mathbb {R}$ and so $dx_{i}|_{p}:T_{p}\mathbb {R} ^{3}\rightarrow T_{x_{i}(p)}\mathbb {R} =\mathbb {R}$

Hence, $dx_{i}|_{p}\in (T_{p}\mathbb {R} ^{3})^{*}$

Now, $dx_{2}({\frac {\partial }{\partial x}})={\frac {\partial }{\partial x_{i}}}x_{2}=\delta _{i2}$ and hence we get a basis.

So, $\Omega ^{1}(\mathbb {R} ^{3})=\{g_{1}dx_{1}+g_{2}dx_{2}+g_{3}dx_{3}\}\approx \{g_{1},g_{2},g_{3}\}\approx$ {vector fields on $\mathbb {R} ^{3}$}

where the $g_{i}:\mathbb {R} ^{3}\rightarrow \mathbb {R}$ are smooth.

$\Omega ^{0}(\mathbb {R} ^{3})=\{f:\mathbb {R} ^{3}\rightarrow \mathbb {R} \}$

This is because to each point p we associate something that takes zero copies of the tangent space into the real numbers. Thus to each p we associate a number.

$\Omega ^{3}(\mathbb {R} ^{3})=\{kdx_{1}\wedge dx_{2}\wedge dx_{3}\}\approx$ {functions} where again the k is just a smooth function from $\mathbb {R} ^{3}$ to $\mathbb {R}$.

$\Omega ^{2}(\mathbb {R} ^{3})=\{h_{1}dx_{2}\wedge dx_{3}+h_{2}dx_{3}\wedge dx_{1}+h_{3}dx_{1}\wedge dx_{2}\}\approx \{h_{1},h_{2},h_{3}\}\approx$ {vector fields}

Aside

Recall our earlier discussion of how points and things like points (curves, equivalence classes of curves) pushfoward while things dual to points (functions) pullback and that things dual to functions (such as derivations) push forward. See earlier for the precise definitions.

Now differential forms pull back, i.e., for $\phi :M\rightarrow N$ then $\phi ^{*}(\lambda )\in \Omega ^{k}(M)\leftarrow \lambda \in \Omega ^{k}(N)$ via $\phi ^{*}(\lambda )(v_{1},\ldots ,v_{k})=\lambda (\phi _{*}v_{1},\ldots \phi _{*}v_{k})$

The pullback preserves all the properties discussed above and is well defined. In particular, it is compatible with the wedge product via $\phi ^{*}(\omega \wedge \lambda )=\phi ^{*}(\omega )\wedge \phi ^{*}(\lambda )$

Theorem-Defintion

Given M, $\exists$ ! linear map $d:\Omega ^{*}(M)\rightarrow \Omega ^{k*+1}(M)$ satisfies

1) If $f\in \Omega ^{0}(M)$ then $df(X)=X(f)$ for $X\in TM$

2) $d^{2}=0$. I.e. if $d_{k}:\Omega ^{k}\rightarrow \Omega ^{k+1}$ and $d_{k+1}:\Omega ^{k+1}\rightarrow \Omega ^{k+2}$ then $d_{k+1}\circ d_{k}=0$.

3) $d(\omega \wedge \lambda )=d\omega \wedge \lambda +(-1)^{deg\omega }\omega \wedge d\lambda$