Difference between revisions of "07081300/Class notes for Tuesday, November 20"
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<math>= (v_1 w)(v_2,v_3)  (v_2 w)(v_1,v_3) + (v_3 w)(v_1,v_2)</math>  <math>= (v_1 w)(v_2,v_3)  (v_2 w)(v_1,v_3) + (v_3 w)(v_1,v_2)</math>  
−  Now, <math>v_1 f = lim\frac{(p+\epsilon v_1)  f(p)}{\epsilon}</math>, or, loosely <math>f(p+v_1)  f(p)</math>  +  Now, <math>v_1 f = lim\frac{f(p+\epsilon v_1)  f(p)}{\epsilon}</math>, or, loosely <math>f(p+v_1)  f(p)</math> 
So this corresponds to the difference between <math>\omega</math> calculated on each of the two faces parallel to the <math>v_3,v_2</math> plane.  So this corresponds to the difference between <math>\omega</math> calculated on each of the two faces parallel to the <math>v_3,v_2</math> plane. 
Revision as of 14:47, 9 February 2008

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 .
Recall we had that
Now we want to compute 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 .
Feeding in the parallelepiped we get
Now, , or, loosely
So this corresponds to the difference between calculated on each of the two faces parallel to the plane.
Hence, our 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, 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 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 and is unique.
Define
Proof
We need to check that this satisfies the properties 1)  3) from last lecture:
1) and so satisfies property 1)
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) 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 and then
so
Via assignment 3 this is unique.
Q.E.D.
Now, we can extend this definition on manifolds by using coordinate charts.
Claim
Properties 13 imply that on any M, d is local. That is, if then
Proof: Exercise.
Definition
For the
Then has compact support if the supp is compact.
Define := the compactly supported
Definition
For , we define by
I.e.,
Second Hour
In general if we have a diffeomorphism then the normal integral
is not equal to
However we claim that this IS true for differential forms. I.e.,
Claim
as forms
This is very important because it essentially means we can integrate in whatever charts we like and get the same thing.
Proof
Now by chain rule and so we extend to
Hence,
but the wedge product is zero unless and in that case yields
Hence we get,
where is the determinant of the Jacobian matrix.
Hence,
Q.E.D
If we restrict our attention to just the orientation preserving 's so that then we will always get the + in the end.
Definition
An orientation of M is an assignment of the charts to ,
Then if the domains on and overlap then
By definition, M is orientable if we can find an orientation.
Examples
1) . We declare the identity positive and then all other designations follow from this.
2) The finite cylinder
We can put two charts on the cylinder by considering two rectangles which overlap by a little bit that cover the whole cylinder. If we denote one of these as positive, the overlap makes the other positive. We compare any other chart to these two.
3) Consider the mobius strip and the same attempted charts as for the cylinder. If we label one section positive, the other section must be positive due to the overlap on one side, but must be negative due to the overlap on the other side. Thus the mobius strip is not orientable.
Definition
An orientation of a vector space V is an equivalence class of ordered bases of V, where if the determinant of the transition matrix between these two bases is positive.
Definition
An orientation of M is a continuous choice of orientations for for any p. We haven't technically defined what it means to be continuous in this sense but the meaning is clear.
Definition
let be an oriented manifold and let . Let be a collection of positive charts that cover M. Let be a partition of unity subordinate to this cover then
Note all the intermediate steps were merely properties we would LIKE the integral to have, the actual definition is the equality of the left most and right most expressions.
Theorem
is independent of the choices.