# Assignment 1 (Written): A First Look at Exterior Algebra and Exterior Calculus

The written portion of your first assignment 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 in mind a few things:

• 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.
• If the $\wedge$ and $\star$ symbols look alien to you, don’t sweat: this is not something you should know already! We’ll be talking about these objects in our lectures next week. Until then, you can (and should) get a jump on the lectures by reading the first few sections of Chapter 3 in our course notes.

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 first assignment, so that you can enjoy all the adventures yet to come…

Here is a zip file with the $\LaTeX$ source.

## 30 thoughts on “Assignment 1 (Written): A First Look at Exterior Algebra and Exterior Calculus”

1. ahagerty says:

Exercise 10 appears to contain typos- The prompt defines only vector fields $U,V$ but the questions refer to vector fields $U,V,W$.

2. Leslie says:

Could you please also release the Latex file of Assignments?
Thanks!

1. Josh says:

The source files had gotten out of sync with the official PDF, so they had been taken down. The up-to-date zip file has been added back to the above post.

3. aremondt says:

Quick questions about a few typos (maybe?):

In exercise 17, is the orientation of the edge $(A,B)$ correct? It seems to be inconsistent with the orientation of the face $(A,D,B)$, but I may have misunderstood how orientations of edges and faces are supposed to behave with respect to one another.

In exercise 18, part (c), should the second wedge product be $\wedge_{1,0}$ (and not $\wedge_{2,0}$)?

In exercise 19, should $g$ be a 0-form, or should $\hat{g}$ be a discrete 1-form perhaps? Otherwise I’m not too sure how to evaluate a 1-form at vertices.

Thank you!

1. Josh says:

Those are good questions.

17) The orientations of the faces and edges do not need to be consistent (it’s a coincidence that only one of the edges is flipped the other way).

18) Great catch! It should be $\wedge_{1, 0}$ as written. There should be an extra $d$.

19) Good eye! 🙂 I meant for $g$ to be a $0$-form.

4. peiya_wang says:

In class today you mentioned a possible extension on this assignment, so when exactly is this assignment going to be due now?

1. Keenan says:

It’s already been updated on the calendar and the written assignment PDF; in the end we only added an extra day, because otherwise it becomes too hard to get the *next* assignment done in a reasonable amount of time. But we’ll continue to adjust as needed.

5. yangxio says:

Just wanna squeeze out an extra credit from your preplanted landmines. There is no big deal.

In V4-1, Exercise 14 Exactness, if $f(x)=x^3$ then it SHOULD have the property that $\frac{d^2}{d x^2} f = 6x$ instead of the written one with $2x$.

6. plav says:

I am lost here. I’ll write my current thoughts and please tell me where I am getting something wrong here.

1) A cross product in R3 is a vector normal to the inputs with length equals area.
2) A cross product in R2 is just the number, that we interpret as form (or as an oriented area)

3a) The wedge product between vectors is the cross product.

or

3b) The wedge product is similar to cross product but substitutes the internal multiplication of the components by wedge products

4) In both 3a or 3b, there is a precise way to calculate a wedge between two vectors. However, how do you determine the wedge product between a form and a vector. I do not get if is there any coherent method for calculation or we should interpret the results case by case?

1. Josh says:

Thanks for writing down your thought process.

1) Yes, this is accurate.

2) You can think of a 2D cross product as a “number” or (0-vector), but this is not quite the same thing as a form.

3) Between (3a) and (3b), (3b) is the more accurate of the two. Note that the wedge product of two 1-vectors is a 2-vector, but the cross product in $\mathbb R^3$ takes two 1-vectors and makes another 1-vector. Therefore they are not the same operation but are highly related. Can you write down such a relationship using exterior calculus? (Hint: what relates 2-vectors and 1-vectors in $\mathbb R^3$?)

4) A form and vector are different types of objects, so you cannot take the wedge product of a form and a vector. Instead, you can apply a form to a vector in order to get a number. How to do this will be discussed in the next lecture (or you can read the posted slides).

1. plav says:

a) Please, can you show me a a concrete example comparing a wedge product and cross product?

b) Besides, the notion of a k-form is confusing for me. The text refers to the 1-form as different things:
1) the unit orthogonal 1-form $\alpha$ (kind of a vector)
2) a function $\alpha()$ 1-form that projects a given vector onto (1)
3) the result $\alpha(u)$ of the projection made by the function (2) onto (1).

1. Keenan says:

a) Suppose we have two vectors in $R^3$: $u := e_1 + e_2 – 2e_3$ and $v := 3e_1 + e_3$. Then $u \times v = e_1 – 7e_2 – 3e_3$ and $u \wedge v = (e_1+e_2-2e_3) \wedge (3e_1 + e_3)$ which when expanded becomes $e_1 \wedge e_3 + 3 e_2 \wedge e_1 + e_2 \wedge e_3 – 6e_3 \wedge e_1$. Simplifying this expression we get the final answer $u \wedge v = e_2 \wedge e_3 – 7e_3 \wedge e_1 – 3e_1 \wedge e_2$.

b) A 1-form is a linear map from a vector to a scalar. So, in your three examples, $\alpha$ by itself is the 1-form; $\alpha(u)$ is the result of applying the 1-form to a vector (producing a scalar). (I’m not sure what you mean by (1).)

Hope that helps! 🙂

1. plav says:

$a_1$) Now I understand the operation. However, I still have few questions here:
is $u \wedge v$ should happen for a certain basis, right? In this case, should the operation you computed be represented $e_1 \wedge e_2 \wedge e_3 (u, v)$
$a_2$) Just to confirm. While we understand the result of the cross product in $R^3$ as a normal vector, the result of $u \wedge v$ in $R^3$ should add the components to represent the area of the plane, right?
$a_3$) In my point (1) I was referring to “we represent this plane via a pair of unit orthogonal 1-forms $\alpha$ and $\beta$”. Notice that they are treated as unit vectors representing the base of the space for the projection – not as the “projection” map (point 2: the right definition) or as the result of the mapping (my point 3).

2. plav says:

$a_4$) Following your example, the anti-symmetry property holds for the wedge product of the components $e_i$ or for the wedge product of vectors in $R^3$?

Well, that is all. Thanks again.

3. Keenan says:

Hi Plav,

(a) I’m not sure what you mean about a basis. You do not *need* a basis in order to talk about wedge products. However, to do concrete calculations it is often convenient to write things in a basis. To make an analogy, you can understand what I mean if I say “add the vectors u and v”, even if I don’t explicitly expand u and v in a basis, i.e., I don’t need to write $u = u_1 e_1 + u_2 e_2$ and $v = v_1 e_1 + v_2 e_2$ in order for the expression “u+v” to make sense. Likewise, I don’t need to write u and v in components in order for the expression “u ^ v” to make sense. Geometrically, “u ^ v” means, “please give me a little patch of in space whose area is equal to the area of the parallelogram spanned by u and v, and which ‘points’ in the same direction as this parallelogram” (at least intuitively).

Also, you asked, “don’t you need to write it as $e_1 \wedge e_2 \wedge e_3 (u,v,w)$. Here it’s important to understand the difference between exterior algebra (which we discussed in the previous lecture), and exterior calculus (which we will discuss in the next lecture). In exterior algebra, you *only* have wedge products, like “u ^ v”. You do not apply this wedge product to vectors. In exterior calculus, on the other hand, you can apply a wedge product to vectors, e.g., “$\alpha \wedge \beta(u,v)$”. In the former case, the wedge just represents a “little volume.” In the latter case, the wedge represents a “little volume *measurement*”.

If this is all a bit confusing, I wouldn’t worry—it takes a while for it to ‘sink in,’ and we haven’t even covered most of it in class! :-). But you’re asking the right kinds of questions.

(a2) Yes, the cross product $u \times v$ and the wedge product $u \wedge v$ are indeed different types of quantities. (I think this is a homework exercise, so I do not want to completely give away the solution! :-)). But also please be careful: you say that we should “add” the components of $u$ and $v$ to get the wedge product $u \wedge v$. This is not quite true: we are not “adding”, but rather performing a completely different operation called the “wedge product.” For instance, if $u$ and $v$ are vectors in $\mathbb{R}^2$ expressed as $u := a e_1 + b e_2$ and $v := c e_1 + d e_2$, then their wedge product is $u \wedge v = (a e_1 + b e_2) \wedge (c e_1 + d e_2)$ and distributing the wedge over the product we get $(a e_1 + b e_2) \wedge (c e_1) + (a e_1 + b e_2) \wedge (d e_2) = ac \underbrace{e_1 \wedge e_1}_{=0} + bc \underbrace{e_2 \wedge e_1}_{=-e_1 \wedge e_2} + ad e_1 \wedge e_2 + bd \underbrace{e_2 \wedge e_2}_{=0}$ which when simplified is just $(ad-bc) e_1 \wedge e_2$.

(a3) This is just one specific example in the textbook; it doesn’t represent the general case (i.e., 1-forms do not always have to be unit length, nor does a collection of 1-forms always have to be orthonormal).

(a4) Antisymmetry holds for all k-vectors, whether basis vectors or not. For instance, it is true that $e_i \wedge e_j = -e_j \wedge e_i$ for any pair of basis 1-vectors; it is also true that $u \wedge v = -v \wedge u$ for any arbitrary pair of 1-vectors $u,v$. (It might be useful for you to show that each of these statements implies the other, i.e., if the wedge on bases is antisymmetric then it is also antisymmetric for arbitrary vectors; likewise, if the wedge is antisymmetric for arbitrary vectors, it also is antisymmetric on basis vectors.)

7. Yifanh says:

Hi, I have a question about Hedge star. Consider an orthogonal basis {e1, e2} in R2. I wonder how did we get *e1 = e2. From the definition of Hedge star, we only have det(e1, *e1) = 1, which only says *e1(2) = 1 and is not enough to uniquely determine *e1. In class we use “e1 is orthogonal with *e1” to derive *e1(1) = 0. However, how is this “orthogonality” reflected in the computation of Hedge star? How to generalize it to k-vectors? Thanks in advance.

1. Josh says:

In the definition of the Hodge star, we always map a basis to $k$-vector (e.g., $e_1 \wedge \cdots \wedge e_k$) to $\pm 1$ times a basis $(n-k)$-vector. The $\det(u, \star u) = 1$ constraint forces that the choice of Hodge star is unique. Does that answer your question?

1. Yifanh says:

Thanks Josh. Your explanation could indeed lead to unique solution. There are also similar descriptions in course note Chapter 3.5. However, in the formal definition provided in homework handouts or on wiki (https://en.wikipedia.org/wiki/Hodge_star_operator), they did not mention that *u should be a basis when u is a basis. I don’t know why is that.

1. Josh says:

Ah, I see what you are asking. Let’s unpack what you are saying.

1) The definition in the homework should say that $\star \alpha$ is $\pm$ a basis $(n-k)$-vector. Good catch! 🙂 It’s been corrected.

2) The Wikipedia definition is a bit different. Instead of specifying how $\star$ behaves on a basis, they define every possible $\alpha \wedge \star \beta$. It turns out this is an identical definition. Do you see why? (Hint: try each basis $k$-vector for $\alpha$ to see what they say about $\star \beta$.)

8. aremondt says:

I’m getting confused when it comes to computing the discrete Hodge star, mostly because I’m a bit shaky as to how to construct the dual mesh of a triangle mesh.

It seems that the way the dual mesh is defined (in 3.3.1 in the assignment),
(1) the number of dual vertices must be the same as the number of (primal) faces,
(2) however the number of dual edges and the number of (primal) edges may not be the same, and in the same vein the number of dual faces and the number of (primal) vertices may not be the same.

Is that correct?

1. Josh says:

Yes, your observations (1) and (2) are correct. The problem is that for a flat surface (like the meshes in the coding/written homework) there is always boundary triangles, and it is rather cumbersome to formally define what dual edges and dual faces are on the boundary. For simplicity, we omit these boundary terms so there is a discrepancy between the numbers of the dual and primal components.

Does this clear things up?

1. aremondt says:

Yes, that helps a lot! Thank you for replying so quickly.

9. intrepidowl says:

In Exercise 16: Exactness, what does the part about the triangle mesh (V, E, F) have to do with the problem?

I also don’t think I’m understanding this problem properly since it seems that the differential of a scalar is 0 already, so the result of differentiating it twice would also produce 0.

Did you mean “discrete 0-form” instead of “0-form”?

1. Josh says:

The goal of the exercise is to show that the property that $d \circ d = 0$ for continuous differential forms continues to hold for discrete differential 0-forms. Since what a discrete differential is depends on the mesh, you need to show this for any (reasonable) mesh (V, E, F).

A scalar (function) is not the same thing as a constant, as the value of the scalar changes spatially. Check the discrete exterior calculus slides to see what happens in the discrete case.

Yes, “discrete differential 0-form” would be more accurate, but often people will just say “0-form” when the context is clear. Since it is slightly unclear in this case, I just updated the writeup.

10. anobani says:

does it always happen when integrating alpha(A,B) = – alpha(B,A) ? or there would be cases in some dimension that they are symmetric ?

1. Josh says:

Yes, just like how integrating from a to b in calculus always give the negative of integrating from b to a.
In higher dimensions, this same phenomenon holds (you flip the sign whenever there is an odd permutation).