# Slides—Exterior Algebra

Our next lecture will cover one of the basic tools we’ll use throughout the rest of the course: exterior algebra. The basic idea is to add a couple new operations to our usual list of vector operations (dot product, cross product, etc.) that make it easy to talk about volumes rather than just vectors. If you felt ok working with things like the cross product and the determinant in your linear algebra/vector calculus courses, this shouldn’t be too big of a leap. (If not, could be a good moment for a review!)

These slides should also be helpful for those who have started on the homework. 🙂

## 11 thoughts on “Slides—Exterior Algebra”

1. yongOne says:

What happens when you wedge the 0 vector with a 1-vector (or any k-vector)? Intuitively it makes sense that this would go to 0, since the overall “area” would be 0.

I am specifically curious with regards to this case:
$$u \wedge v \wedge v$$

Would it be:
$$=u \wedge v \wedge v$$
$$=u \wedge 0$$
$$= 0$$

1. Josh says:

That’s a great question. Note that wedging a $0$-vector with a $k$-vector produces a $(k+0)=k$-vector, so the $k$-vector already contains the necessary “area” to be nonzero. In the case of $0$-vectors this “wedge product” ends up being scalar multiplication from linear algebra. Because of that, the wedge with $0$-vectors is often dropped, so in your example $u \wedge v \wedge v$ could be written as $v^2 u$, assuming $v$ is a $0$-vector.

One thing to be careful about is if $v$ is a $k$-vector, the “rule” that $v \wedge v = 0$ only applies for certain values of $k$, such as $k=1$. Can you describe for which $k$ this is the case? The answer changes depending on the dimension of the space!

1. anobani says:

Adding on the previous question of (u ^ v ^ v), since the wedge of 2 vectors in R2 would represent the area, and wedge if 3 vectors in R3 should represent volume, if (u) and (v) are in R2 then would we evaluate (u ^ v ^ v) in terms of area or volume? such if volume it will result in 0 since we didn’t break to the next dimension, and in terms of area, will we still evaluate it in terms of surface? ( i don’t know if this is a right way to think about it)

1. Keenan says:

Good question—the answer is: a 1-vector always encodes length; a 2-vector always encodes area; a 3-vector always encodes (three-dimensional) volume, and so on. So for a 3-vector you should ask yourself: “how much *volume* does this thing have?”

2. oongaang says:

How do we know that for a given k-vector, its representation w.r.t a set of basis k-vectors is well-defined? For example, in R3, we might imagine (e1^e2)+(e2^e3) as a parallelogram having some coordinates w.r.t. the orthonormal basis {(e1^e2), (e2^e3), (e2^e3)}. But if we replace that parallelogram with some equivalent 2-vector, say a circle having the same orientation and area, how do we know that its coordinates in the basis haven’t changed? And going one step further, what happens if we replace the basis 2-vectors with equivalent orthonormal circles, or really any orthonormal set of equivalent 2-vectors? Are the coordinates of (e1^e2)+(e2^e3) still the same? My intuition is to regard k-vectors as the volume determined by k-many vectors, but it seems that intuition doesn’t really hold for k>1.

1. Josh says:

That’s definitely a good question. I think part of the confusion is that a 2-vector is not one parallelogram, but is the equivalence class of all parallelograms with the same direction, orientation and magnitude (in this case, area). Thus, if the parallelograms (or non-parallelograms) you are thinking about preserve these properties, the coordinates of everything stay the same. Does that clear things up?

Note that the basis of 2-vectors is defined to be the all possible (nonzero) wedge products of your basis of 1-vectors. Thus, if you want to change your basis of 2-vectors in a consistent way, you would need to change your basis of 1-vectors. In that case, the coordinates of $e_1 \wedge e_2 + e_2 \wedge e_3$ would change.

3. yongOne says:

I have some questions about the hodge star:
1. Can constants be pulled out normally:
ie: $\alpha \star (e_1 \wedge e_2) = \star (\alpha e_1 \wedge e_2)$

2. Is it always possible to distribute the hodge star. The textbook states it is fine for 1-vectors but is it okay in general?

1. langxuan says:

The definition of Hodge star says it is a “linear isomorphism” … So it has linearity

4. apoorva says:

In the formal definition of the wedge product, should the codomain *technically* be $\Lambda^{\text{max}(k+l,n)}$? This may not be super important to the geometric interpretation of the wedge product, but I’m trying to figure out what formal ‘type’ we should assign to the wedge product of (say) a 3-vector and a 4-vector in in 5-dimensional space.

1. aremondt says:

I’m pretty sure (but I may be wrong) that $\Lambda^{k+l}$ is still correct, it’s just that whenever $k+l > n$, $\Lambda^{k+l}$ is a trivial vector space, i.e. only contains the zero $(k+l)$-vector.

2. Josh says:

Your second reply is correct. It’s reasonable to have the domain be $\Lambda^{k+l}$ even when $k + l > n$. It just happens in this case that $\Lambda^{d}$ is a $0$-dimensional vector space when $d > n$ (and thus is rarely mentioned). Formally, $\Lambda^{d}$ the span of all the possible ways to wedge $d$ basis vectors of $\mathbb R^n$, but as discussed in class, these are all equal to $0$. Like the analogy brought up in class about the box, you can’t have a non-zero volume in two-dimensional space.