For your next assignment, **due on Friday, March 4th**, you will simply complete the exercises you previously skipped over that required use of exterior calculus. These exercises can all be found in the class notes, and include:

- Chapter 5, Exercise 13 (
*Hint: read very carefully the end of Section 3.2*) - Chapter 5, Exercise 16
- Chapter 6, Exercise 20
- Chapter 6, Exercise 26

You should also carefully read Section 6.3 (which leads up to Exercise 26), which provides an alternative perspective on the Laplacian to the finite element approach you derived in your last homework, and will be very useful to understand for upcoming assignments. For the exercises in Chapter 5, it may also be helpful to review your reading of Chapter 2 (on the geometry of surfaces), re-thinking these concepts in terms of differential forms. (For example, the Weingarten map \(dN\) and differential \(df\) of the immersion can both be viewed as vector-valued 1-forms.) These exercises are a bit more challenging than the “warm up” exercises from your readings, so please post comments if you have any trouble.

Sharing a book about Exterior Calculus which has some introductory stuff in Chapter 2

http://link.springer.com/book/10.1007%2F978-1-84996-290-2

For exercise 16, it may help to go review some of the discussion of curvature from Sections 2.3 and 2.4, especially the discussion of how derivatives of tangent vectors are related to curvature, and what principal curvatures are / how they’re related to mean curvature.

What time is it due on Friday? The usual of 11 PM?

Correct.

What does $df$ mean in exercise 16? In the readings, there always is a unit direction vector $X$ on a point on $M$ so that we can define the one-form directional differential as $df(X)$. In the problem, it it not obvious how $X$ is defined on $M$. Do I misunderstand anything here? Thanks.

Sorry, it’s exercise 13

It’s not defined on $M$, it’s defined on $\partial M$. We have an integral over the surface $M$, and then somehow using Stoke’s theorem, we can instead talk about an integral along some curve on the boundary of that surface ($\partial M$). $df(X)$ is then the infinitesimal length of the steps we take along that curve, if I understand correctly.

If $\alpha$ is a 1-form, then $\alpha \wedge \alpha$ should equal 0 by anti-symmetry. On the other hand, if $\alpha$ is vector valued, and we use the cross product, then $\alpha \wedge \alpha (u,v) = 2 \alpha (u) \times \alpha (v)$ which may not equal zero. How can this discrepancy be explained?

With a 1 form, you’re only measuring one direction, so it can only possibly be 0. With a 2 form, you’re measuring the cross product of two different directions $u, v$ (after passing through $\alpha$), which might not be equal to zero.

The antisymmetry property relates to the commutativity of the codomain. Most of our language pertains to scalar-valued forms, which we implicitly call forms, but here we’re working with vector-valued forms (equipped with the cross product), which is not commutative, so $\alpha \wedge \alpha$ is not necessarily $0$.