6
votes
2answers
112 views

Relation between exterior (second) derivative $d^2=0$ and second derivative in multi-variable calculus.

What does an exterior (second) derivative such as in $d^2=0$ have to do with second derivatives as in single- or multi-variable calculus? Is this a correct start: Calculus derivatives are good for ...
0
votes
1answer
82 views

Wedge product and change of variables

The question is: Let $\phi : \mathbb{R}^n \to \mathbb{R}^n$ be a $C^1$ map and let $y = \phi(x)$ be the change of variables. Show that $$dy_1\wedge\dots\wedge dy_n = ...
4
votes
2answers
234 views

algebraic manipulation of differential form

Suppose $\phi_1, \phi_2, \dots, \phi_k \in (\mathbb{R}^n)^*$, and $\mathbf{v}_1, \dots, \mathbf{v}_k \in \mathbb{R}^n$ $(\mathbb{R}^n)^*$ stands for the space of all linear transformations that goes ...
2
votes
1answer
254 views

Taking the exterior derivative of a 0-form

I'm attempting to show that $dg(\vec{x})=\alpha$, where $\alpha=\Sigma^n_{i=1}f_idx_i$ and $g(\vec x)={1\over{p+1}}\Sigma_{i=1}^nx_if_i(\vec x)$...and $d$ is the exterior derivative. The $f_i$ are ...
2
votes
1answer
427 views

“And” symbol? Wedge product in a surface integral? — Is this a typo, or did I miss an important lecture?

This is the question I got on my final assignment (Calculus III): Evaluate the surface integral \begin{equation} \int \int_S xy \; \; dy\wedge dz - yz \; \; dz\wedge dx + xz \; \; dx\wedge ...