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.

First Hour

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 which was the "k forms on M" which equaled


which is

1) Multilinear

2) Alternating

We had proved that :

1) is a vector space

2) there was a wedge product via that is

a) bilinear

b) associative

c) supercommutative, i.e.,

From these definitions we can define for and that with the same properties as above.


If is a basis of then is a basis of and

If and a basis of then any can be written as where are smooth.

The equivalence of these is left as an exercise.


Let us investigate (the * just means "anything").

Now, where and so


Now, and hence we get a basis.

So, {vector fields on }

where the are smooth.

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.

{functions} where again the k is just a smooth function from to .

{vector fields}


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 then via

The pullback preserves all the properties discussed above and is well defined. In particular, it is compatible with the wedge product via


Given M,  ! linear map satisfies

1) If then for

2) . I.e. if and then .


Second Hour

Some notes about the above definition:

1) When we restrict our d to functions we just get the old meaning for d.

2) Philosophically, there is a duality between differential forms and manifolds and that duality is given by integration. In this duality, d is the adjoint of the boundary operator on manifolds. For manifolds, the boundary of the boundary is empty and hence it is reasonable that on differential forms.

3) To remember the formula in 3 given above and others like it, it helps to keep in mind what objects are "odd" and what are "even" and thus when commuting such operators we will get the signs as you would expect from multiplying objects that are either odd or even.


Let us aim for a formula for d on .

Lets compute where and


The last term vanishes because of (2) in the theorem (proving uniqueness!)

Now, as an aside, we claim that for

Indeed, we know

However, which is the same.

Returning, we thus get

Thus our d takes functions to vector fields by

This is just the grad operator from calculus and we can see that the d operator appropriately takes things from to .

Now let us compute + 6 more terms representing the 3 partials of each of the last 2 terms.

As each term vanishes we are left with just,

I.e., d takes

this is just the div operator from calculus and appropriately takes vector fields to functions and represents the d from to .

We are left with computing d from to


I.e., we just have the curl operator.

Note that the well known calculus laws that curl grad = 0 and div curl = 0 are just the expression that .

To provide some physical insight to the meanings of these operators:

1) The gradient represents the direction of maximum descent. I.e. if you had a function on the plane the graph would look like the surface of a mountain range and the direction that water would run would be the gradient.

2) In a say compressible fluid, the divergence corresponds to the difference between in the inflow and outflow of fluid in some small epsilon box around a point.

3) The curl corresponds to the rotation vector for a ball. Ie consider a ball (of equal density to the liquid about it) going down a river. In the , plane the tenancy for it to rotate clockwise would be given by