6

This paper provides a geometric intuition of differential forms. On page $5$ it reads:

"Consider the case $xdy$. The number of horizontal lines is roughly proportional to $y$. In other words the number of lines that 'come to an end' in any area is roughly the same wherever we look. This is just like the example $dx ∧ dy$ above. You may have guessed what is going on here: $d(xdy) = dx∧dy$. In other words exterior derivative is none other than finding the boundary of the picture."

enter image description here

Why can the exterior derivative of a form $\omega$, defined as

$$d\omega = d\left(\sum_{i_1,\ldots,i_k}\omega_{i_1,\ldots,i_k}dx^{i_1}\wedge\ldots\wedge dx^{i_k}\right)\\ \ \ = \sum_{i_1,\ldots,i_k}d\omega_{i_1,\ldots,i_k}\wedge dx^{i_1}\wedge\ldots\wedge dx^{i_k}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \sum_{i_1,\ldots,i_k}\sum_{\alpha=1}^nD_{\alpha}(\omega_{i_1,\ldots,i_k})\ dx^{\alpha}\wedge dx^{i_1}\wedge\ldots\wedge dx^{i_k}\\ $$

be interpreted as the quantity of points at which the lines in the picture come to an end?

Sam
  • 4,734
  • 12
    Here's some meta-advice. Everybody I know who seriously work with differential forms never thinks about them as geometric objects. You'll gain familiarity with them as you work with them in algebraic formula. – Mr. Brown Apr 21 '23 at 00:54
  • 2
    Broadly, I find this series of questions interesting so I've been upvoting, but I do agree with the other commenter. I used smooth 1- and 2-forms exclusively in my thesis and i have no idea what is happening in these pictures nor in this paper. The diagrams are more confusing than descriptive to me. I look forward to an answer that clarifies any connection. – Rollen S. D'Souza Apr 21 '23 at 01:32
  • 3
    I disagree with Mr. Brown. I am writing an answer for your question about $dx\wedge dy$ and hope to get to this one in due time. In short(and just thinking locally/in the flat case), every function $L_x(v)$ linear in $v$ with $x$ a point has a natural "derivative" $L_x(\nabla)$ with respect to $x$ which can be shown to essentially be the average change in $L_x(v)$ over all directions $v$. – Nicholas Todoroff Apr 21 '23 at 19:22
  • The exterior derivative is just the derivative of $L_x(v) = v\wedge\omega$, or in other words $d\omega = \nabla\wedge\omega$. Of course we also need to geometrically interpret the wedge product here. – Nicholas Todoroff Apr 21 '23 at 19:22
  • @Mr.Brown As a geometer myself, I do think about differential forms geometrically, just not "visually". I don't know if that makes me non-serious. However, I indeed do not think about differential forms as pictured in the question. – Didier May 01 '23 at 12:15
  • 1
    I'd be happier to know that I'm wrong, than right @Didier! I'm looking forward to you and Nicholas' rebuttal. – Mr. Brown May 01 '23 at 12:21
  • 3
    @Mr.Brown I don't think this comment sections is suited for a such a debate: it would be out of the scope of the question (which is asking for a precise geometric vision which I do not share). I just wanted to emphasize that your peremptory statement wasn't shared by everyone here. I work with differential forms everyday and I have a very geometric vision of them. I'm not saying that all geometers do – Didier May 01 '23 at 12:29

0 Answers0