Slides—Differential Forms in $R^n$

Following our lecture on exterior algebra, we will start building up differential forms, which is the next step on our journey toward doing computation on meshes with discrete exterior calculus. This material may be helpful for those of you working through the second part of the written homework:

7 thoughts on “Slides—Differential Forms in $R^n$”

1. langxuan says:

For the $\frac{\partial}{\partial x}$ notation for basis of a vector field, I think the story is: when we consider a differentiable manifold $M$ with a local chart $\varphi$, for each point $p \in M$ we have a tangent space $T_p(M)$ associated to $p$, where $\frac{\partial \varphi}{\partial x_i}$ evaluated at the point $p$ turns out to be the basis of $T_p(M)$, and it does depend on the local chart $\varphi$. Right? Probably we will learn about this later in the course.

1. Keenan says:

Right, the basic idea is that one can put vector fields and derivatives of scalar functions into correspondence. For anyone who’s interested, you can read a bit about it here:

Leibniz notation for vector fields

Though even here the author admits that the notation is “extremely strange the first time one encounters it”.

In practice, even to this day, it is very rare that I ever think about the basis vector fields $\tfrac{\partial}{\partial x^i}$ as derivatives. So that’s my practical advice: if you find it fun or interesting to think about the basis vector fields as derivatives, go forth and learn about it! But when it comes to actually doing practical calculations, you may be better served by simply ignoring the connection to derivatives.

2. plav says:

1) The application of a differential 1-form $\alpha$ to a Vector field U results in a scalar function $R^n \rightarrow R$ that measures the non-normalized projection (i.e. dot product) of each vector of $U_{x, y}$ in relation to the vector $\alpha(x), \alpha(v)$. Therefore, $\alpha$ will result in different values for the angles (also influenced by magnitude) between $U_{x, y}$ and $\alpha(x), alpha(v)$:
– negative for angle $>\frac{1}{2}\pi$
– zero for angle $= 0$
– positive for inner angle $<\frac{1}{2}\pi$

2) I am confused with slide 46 and ex. 2.2. I will use an example. Let's say I have a vector field $V = y \frac{\partial}{\partial x} – x\frac{\partial}{\partial y}$ and a differential 1-form $\alpha = x dx + 2y dy$.
-2.1: If I evaluate U at (3, 5), I get the vector $V_{3, 5} = 5\frac{\partial u}{\partial x} – 3\frac{\partial u}{\partial y}$
-2.2 Can I evaluate $\alpha(.)$ for any input with coordinates or only for vector-field? If I apply $\alpha$ for regular coordinates: $\alpha_{3, 5} = 3 dx + 10 dy$. By definition, I should get a real number, but in this case, I get kind of a vector. I feel I can only apply $\alpha$ to vector fields, because dx will cancel $\frac{\partial}{\partial x}$ and dy will cancel $\frac{\partial}{\partial y}$, resulting in the real number. Is this interpretation correct?
-2.3 If I evaluate $\alpha(V)$, I foresee two options… (a) I treat (x, y) as the original coordinates (x, y) in $R^3$, and $\alpha(V) = xy – 2yx = -xy$; (b) I treat (x, y) as the coordinates of the argument, so I have $\alpha(V) = y – 2x$. I am tending to the interpretation (a).

1. Josh says:

1) If you think of $\alpha$ has a vector, then evaluating $\alpha(U)$ is like taking a dot product. This is true in the special case of $\mathbb R^n$, although it doesn’t generalize to infinite-dimensional vector spaces.

2) What you did for 2.1 is correct.

I’m not sure what you mean in 2.2 by “cancel.” Remember that $dx(\frac{\partial}{\partial x}) = 1$ and $dx(\frac{\partial}{\partial y}) = 0$, etc.

For 2.3, your perspective from (a) is the correct one. At each point, you can evaluate what $V$ and $\alpha$ are, then you can apply $\alpha$ to $V$.

3. yongOne says:

When computing projected area, we had the relationship that: $(\alpha \wedge \beta)(u,v)=\alpha(u)\beta(v) – \alpha(v)\beta(u)$ from the determinant relationship. Can we assume that a similar relationship holds for differential 1-forms (well technically I suppose differential 2-form in this case) applied to vector fields?

1. Keenan says:

Yes. The general rule is: to evaluate an operation on a differential k-form, just apply the same operation you normally would for a single k-form, but apply it at each point. So for instance if $\alpha$ and $\beta$ are differential 1-forms and $X,Y$ are vector fields, then at each point $p$ of the domain

$(\alpha \wedge \beta)(X,Y)_p := \alpha_p(X_p) \beta_p(Y_p) – \alpha_p(Y_p) \beta_p(X_p),$

i.e., the value of the resulting function at the point $p$ is obtained by grabbing the 1-form and vector values at $p$ and evaluating the usual expression. (You may start to see why we generally omit the subscript “$p$”!)