0708-1300/Class notes for Tuesday, November 20

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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.

First Hour

Recall we are ultimately attempting to understand and prove Stokes theorem. Currently we are investigating the meaning of $d\omega$.

Recall we had that $d(\sum f_{I}dx^{I})=\sum (df_{I})\wedge dx^{I}=\sum _{I,j}{\frac {\partial f_{I}}{\partial x^{j}}}dx^{j}\wedge dx^{I}=\sum dx^{j}{\frac {\partial \omega }{\partial x_{j}}}$

Now we want to compute $d\omega$ on the parallelepiped formed from k+1 tangent vectors. For instance let us suppose k= 2 then are interested in the parallelepiped formed from the three basis vectors $v_{1},v_{2},v_{3}$.

Feeding in the parallelepiped we get $\sum _{I}df_{I}(v_{1})dx^{I}(v_{2},v_{3})-\sum _{I}df_{I}(v_{2})dx^{I}(v_{1},v_{3})+\sum _{I}df_{I}(v_{3})dx^{I}(v_{1},v_{2})$$=\sum _{I}(v_{1}f_{I})dx^{I}(v_{2},v_{3})-\sum _{I}(v_{2}f_{I})dx^{I}(v_{1},v_{3})+\sum _{I}(v_{3}f_{I})dx^{I}(v_{1},v_{2})$

$=(v_{1}w)(v_{2},v_{3})-(v_{2}w)(v_{1},v_{3})+(v_{3}w)(v_{1},v_{2})$

Now, $v_{1}f=lim{\frac {(p+\epsilon v_{1})-f(p)}{\epsilon }}$, or, loosely $f(p+v_{1})-f(p)$

So this corresponds to the difference between $\omega$ calculated on each of the two faces parallel to the $v_{3},v_{2}$ plane.

Hence, our $d\omega$ on the parallelepiped is just the sum of parallelograms making the boundary of the parallelepiped counted with some signs.

We can see the loose idea of how the proof of stokes theorem is going to work: dividing the manifold up into little parallelepiped like this, $d\omega$ will just be the faces of the parallelepipeds and when summing over the whole manifold all of faces will cancel except those on the boundary thus just leaving the integral of $\omega$ along the boundary.

We note that this is similar to the proof of the fundamental theorem of calculus, where we take an integral and compute the value of f' at many little subintervals. But the value of f' is just the difference of f' at the boundary of each sub interval so when we add everything up everything cancels except the values of the function at the endpoint of the big interval.

Claim

d exists if $M=\mathbb {R} ^{n}$ and is unique.

Define $d(\omega )=d(\sum f_{I}dx^{I}):=\sum _{j,I}{\frac {\partial f_{I}}{\partial x_{j}}}dx^{j}\wedge dx^{I}$

Proof

We need to check that this satisfies the properties:

1) $df(\partial _{i})=\sum _{j}{\frac {\partial f}{\partial x_{j}}}dx_{j}(\partial _{i})=\sum _{j}{\frac {\partial f}{\partial x_{j}}}\delta _{ji}=\partial _{i}f$ and so satisfies

Note

We now adopt Einstein Summation Convention which means that if in a term there is an index that is repeated, once as a subscript and once as a superscript, it is meant as implicit that we are summing over this index. This just cleans up the notation so we don't have to have sums everywhere.

2) $d(df_{I}dx^{I})=d({\frac {\partial f_{I}}{\partial dx^{j}}}dx^{j}\wedge dx^{I})={\frac {\partial ^{2}f_{I}}{\partial x^{j}\partial x^{j'}}}dx^{j'}\wedge dx^{j}\wedge dx^{I}=0$ because the mixed partial is symmetric under exchange of indices but the wedge product is antisymmetric under exchange of indices. That is, each term cancels with the one where j and j' are exchanged.

3) let $\omega =f_{I}dx^{I}$ and $\lambda =g_{J}dx^{J}$ then $\omega \wedge \lambda =f_{I}g_{J}dx^{I}\wedge dx^{J}$

so $d(f_{I}g_{J}dx^{I}\wedge dx^{J})={\frac {\partial f_{I}g_{I}}{\partial x^{j}}}dx^{j}\wedge dx^{I}\wedge dx^{J}=({\frac {\partial f_{I}}{\partial x^{j}}}g_{J}+f_{I}{\frac {\partial g_{I}}{\partial x^{j}}})dx^{j}\wedge dx^{I}\wedge dx^{J}=d\omega \wedge \lambda \pm \omega \wedge d\lambda$

Via assignment 3 this is unique.

Q.E.D.

Now, we can extend this definition on manifolds by using coordinate charts.

Claim

Properties 1-3 imply that on any M, d is local. That is, if $\omega |_{U}=\lambda |_{U}$ then $d\omega |_{U}=d\lambda |_{U}$

Proof: Exercise.

Definition

For $\omega \in \Omega ^{k}(M)$ the $supp\ \omega ={\overline {\{p\in M\ :\ w|_{p}\neq 0\}}}$

Then $\omega$ has compact support if the supp $\omega$ is compact.

Define $\Omega _{c}^{*}(M)$ := the compactly support $w\in \Omega ^{*}(M)$

Definition

We define $\int _{\mathbb {R} ^{n}}\Omega _{c}^{n}(\mathbb {R} ^{n})\rightarrow \mathbb {R}$ by $\omega \in \Omega _{c}^{n}(\mathbb {R} ^{n})$ and thus $\omega =fdx^{1}\wedge ,\ldots ,\wedge dx^{n}$ then

$\int _{\mathbb {R} ^{n}}\omega :=\int _{\mathbb {R} ^{n}}f$

I.e., $\int _{\mathbb {R} ^{n}}fdx^{1}\wedge ,\ldots ,\wedge dx^{n}=\int _{\mathbb {R} ^{n}}fdx^{1}\ldots dx^{n}$

Second Hour

Coming soon to a wikipedia near you.