Time to get busy with exterior calculus. Very roughly speaking, exterior calculus is a “souped up” version of vector calculus that will let us do lots of interesting calculations on surfaces, both mathematically and computationally. This material is *critical* to the assignments we’ll do during the rest of the course; it is well-worth your time to understand it well! Your reading assignment for **Thursday** (*not* Tuesday) is to read the first half of Chapter 3 from our course notes, namely Sections 3.1–3.3 (pp. 23-35). We have also given you four mini-exercises below, which you should submit answers to either by email, or in-class if you prefer to do them on paper. These shouldn’t take more than a few minutes, and are mostly to confirm your own understanding; if you’re having trouble, please don’t hesitate to ask questions either here on the website or via email.

- Write the column vector \(u = [\ 1\ 2\ 3\ ]^T\) and the row vector \(\alpha = [\ 4\ 5\ 6\ ]\) using the notation described for vectors and 1-forms (respectively) in Section 3.1.
- Using the same values of \(u\) and \(\alpha\), what is \(\alpha(u)\)?
- Bonus question: suppose that for vectors \(u,v \in \mathbb{R}^n\), we define the metric as \(g(u,v) := 2\langle u, v \rangle\) where \(\langle \cdot, \cdot \rangle\) is the usual Euclidean inner product, i.e., our metric \(g\) is simply a conformal rescaling of the Euclidean metric by a factor 2. Based on the comment immediately preceding Section 3.1.1., what do you think the 1-form \(u^\flat\) and the vector \(\alpha^\sharp\) should be (using the same values for \(u\) and \(\alpha\) as in the previous question)? You can work out the answer using matrices if you like, but try writing the final answer in terms of the basis vectors and 1-forms (as in the previous question).
- Letting \(\alpha := 2 dx + 3 dy + 4 dz\) and \(\beta := 5 dx + 6 dy + 7 dz\), calculate \(\alpha \wedge \beta\). (This time, no matrices allowed!)
- Letting \(\alpha := 8 dx \wedge 9 dy\), write out \(d\alpha \wedge d\alpha\), showing explicitly why it’s equal to zero.

To give a couple concrete examples of what these calculations look like: if \(u := \tfrac{\partial}{\partial x} – 2 \tfrac{\partial}{\partial y}\), \(\alpha = dx – dy\), and \(\beta = 2 dx + dy\), then

\[

\alpha(u) = dx(u) – dy(u) = dx(\tfrac{\partial}{\partial x} – 2 \tfrac{\partial}{\partial y}) – dy(\tfrac{\partial}{\partial x} – 2 \tfrac{\partial}{\partial y}) = 1 – (-2) = 3,

\]

and

\[

\alpha \wedge \beta = (dx-dy) \wedge (2dx + dy) = 2 dx \wedge dx + dx \wedge dy – 2 dy \wedge dx – dy \wedge dy = 0 + dx \wedge dy + 2 dx \wedge dy – 0 = 3 dx \wedge dy.

\]

**Submission:** As usual, please send an email to kmcrane@cs.cmu.edu and nsharp@cs.cmu.edu no later than **10:00 AM on Thursday, February 24th** including the string **DDGSpring2016 **in your subject line. Your email for readings should always include:

- a short (2-3 sentence) summary of what you read, and
- at least one question about something you found confusing / interesting / incomplete / not addressed.

Here are the solutions:

halfthe corresponding component values from \(\alpha\), i.e., \(\alpha^\sharp = 2 \tfrac{\partial}{\partial x} + \tfrac{5}{2} \tfrac{\partial}{\partial y} + 3 \tfrac{\partial}{\partial z}\). A similar argument shows that \(u^\flat = 2 dx + 4 dy + 6 dz\).