The written portion of assignment 1 is now available (below), which covers some of the fundamental tools we’ll be using in our class. Initially this assignment may look a bit intimidating, but keep a few things in mind:

- The homework is not as long as it might seem: all the text in the big gray blocks contains supplementary, formal definitions that you do not need to know in order to complete the assignments.
- Moreover, note that you are required to complete
**only three problems from each section**.

Finally, don’t be shy about asking us questions here in the comments, via email, or during office hours. *We want to help you succeed* on this assignment, so that you can enjoy all the adventures yet to come…

This assignment is due on Thursday, February 25.

Typo: on page 2 section 1.1.1 the course notes “3.1.3 The Hodge Star” should be “4.1.3 The Hodge Star” in this year’s notes.

Do $e_1, e_2, \dots, e_n$ always represent the canonical orthonormal vectors in $\mathbb{R}^n$?

Usually although there are a couple instances where e_i represents edges in a mesh. But this should be clear from the context. When talking about bases e_1,…,e_n can be assumed to be the canonical one for R^n.

Hi. How does one take wedge product of two 2-vectors? Does the associativity rule demonstrated using a 2-vector and a one vector hold true in this case as well? Also, what does the wedge product of two 2-vectors mean geometrically?

In notation, my question is how to do: $(\alpha \wedge \beta) \wedge (\gamma \wedge \tau)$ and what is means geometrically. I think I can do this: $((\alpha \wedge \beta) \wedge \gamma) \wedge ((\alpha \wedge \beta) \wedge \tau)$, but I am not sure of the geometric meaning.

Pretty sure the wedge product does not distribute over itself (in the same way that multiplication does not distribute over itself i.e. a * (b * c) does not equal (a * b) * (a * c)). To see this, note that the second thing you’ve written is actually a 6-form, while the first thing is a 4-form (if $\alpha$, $\beta$, … are 1-forms)!

The wedge product is associative though! So you should be able to re-distribute the parantheses however you see fit.

Yep! That’s it. Because of associativity, the wedge product you wrote should exactly match your intuition for any 4-form geometrically. In other words the 4-form measures the projection of a volume spanned by 4 vectors onto the 4-form which itself can be visualized as a volume spanned by 4 1-forms. It’s helpful to think of lower dimensional analogues of measuring the shadow of a vector onto a 1-form or the shadow of a plane onto another plane.

And to assist with the computation, note that the wedge product is associative and distributive (but only when the forms you are distributing over are the same degree).

What happens when the discrete differential for a triangle mesh is evaluated for an edge that doesn’t exist? Is it 0? How about if it’s an edge that’s the opposite orientation from an existing one?

Typo: exercise 12 says ‘as discussed in section 4.4 of the course notes,’ but the discussion is in section 4.6.

Should the identifier be DDG20A1 or DDG_A1, or does it not matter?

Please use DDG20A1.