Reading 8 — Exterior Calculus Part I

A1

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.

  1. 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.
  2. Using the same values of \(u\) and \(\alpha\), what is \(\alpha(u)\)?
  3. 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).
  4. 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!)
  5. 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:

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

1 thought on “Reading 8 — Exterior Calculus Part I”

  1. Here are the solutions:

    1. In coordinates \(x,y,z\), the column vector u can be expressed as a vector \(u = \tfrac{\partial}{\partial x} + 2\tfrac{\partial}{\partial y} + 3\tfrac{\partial}{\partial z}\). The row vector can be expressed as a 1-form \(\alpha = 4 dx + 5 dy + 6 dz\). We could have also used coordinates \(x^1,x^2,x^3\). The distinction is largely superficial, though one advantage of using indices instead of letters is that it becomes easier to write sums (i.e., by summing over an index). For calculations in low dimensions, it is often “prettier” to just use a few letters (\(x,y,z\) or \(u,v\) or \(s,t\), etc.).
    2. \(\alpha(u) = 4dx(u) + 5dy(u) + 6dz(u) = 4 \cdot 1 + 5 \cdot 2 + 6 \cdot 3 = 4 + 10 + 18 = 32.\)
    3. To determine the vector \(\alpha^\sharp\), we consider that for all vectors \(v\) we should have \(\alpha(v) = g(\alpha^\sharp,v) = 2\langle \alpha^\sharp, v \rangle = \langle 2\alpha^\sharp, v \rangle\). In other words, \(\alpha^\sharp\) should be a vector whose components are simply half the 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 \(u^\flat = 2 dx + 4 dy + 6 dz\).
    4. \(\alpha \wedge \beta = (2 dx + 3 dy + 4 dz) \wedge (5 dx + 6 dy + 7 dz)\). Three like terms cancel (e.g., \(2 dx \wedge 5 dx = 10 dx \wedge dx = 0\), and hence we are left with the six terms \(12 dx \wedge dy + 14 dx \wedge dz + 15 dy \wedge dx + 21 dy \wedge dz + 20 dz \wedge dx + 24 dz \wedge dy\). Taking advantage of antisymmetry of the (real!) wedge product, we can simplify this expression to \(-3 dx \wedge dy – 3 dy \wedge dz + 6 dz \wedge dx\).
    5. \((8dx + 9 dy) \wedge (8dx + 9 dy) = 64 \underbrace{dx \wedge dx}_{=0} + 72 dx \wedge dy + 72 dy \wedge dx + 81 \underbrace{dz \wedge dz}_{=0} = 72 dx \wedge dy – 72 dx \wedge dy = 0.\)

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