10
$\begingroup$

I think I understood 1-forms fairly well with the help of these two sources. They are dual to vectors, so they measure them which can be visualized with planes the vectors pierce.

But I struggle with the explanations for higher order forms.

The goal is to answer and understand these questions with drawings:

  1. How can I visualize the wedge between two 1-forms $\alpha\wedge\beta$? I think I understood the wedge between two vectors, as the parallelogram created by the two in a "area sense". The determinant comes in to make it only about the area which is why $v\wedge w = \frac{1}{2}v\wedge 2w$ since stretching the parallelogram by two in the w direction is compensated by squishing it in the v direction, so the area stays constant. So the wedge between two vectors is the area it spans with its vectors. But how does that translate to the dual space? Where 1-forms measure the length of the component of its dual vector. And can be visualized as planes the vectors pierce through. What is the visualization between two of these 1-forms as a wedge?

Gravitation has this picture:

2-forms Dan-Piponi drew it like this:

2-forms dan

Now these pictures make some sense as they are generated by intersecting the 1-forms. But I am not quite getting how the result is evaluated. The result (2-form) should map two vectors as input to a number. And I don't see how these intersections do that. While for 1-forms you count the numbers of planes a vector pierced.

  1. Why does it make sense that $d(d\alpha)=0$ for every differential form $\alpha$
  2. What does Dan Piponi mean by saying: "exterior derivative is none other than finding the boundary of the picture" (4 Exterior Derivatives)
  3. Understand part 5 about Stokes' theorem from Dan Piponi's paper

Note: I should maybe add that I have no background in physics, so I didn't understand a lot of the stuff in Gravitation. I just tried to read it after I couldn't quite understand other source since it was cited there.

Similar questions:

$\endgroup$
  • 2
    $\begingroup$ Hi, I drew some pictures to visualize differential forms before, with explanation for $dd\alpha=0$ (but not for wedge products, and different from Dan Piponi's illustrations). I hope these pictures are also helpful to you. math.stackexchange.com/a/2750165/266571 $\endgroup$ – autodavid Jan 20 '19 at 18:00
  • $\begingroup$ @autodavid thank you, that explained a lot of my questions, I am just still not getting how to interpret the pictures I posted but I realized (because of your suggestion to view differential forms as functions) that since $<A\wedge B , v\otimes w>= <A\otimes B, v\wedge w>$ I can now view 2-forms as a linear function evaluating parallelograms, so I think you deserve 50 reputation for that. $\endgroup$ – Felix B. Jan 21 '19 at 20:09
  • $\begingroup$ I'm glad that it helps. But I have to say that my explanation for exterior derivative is just an intuition and not a way to calculate $d\alpha$. So the answer is modified to warn this. (It is still a helpful intuition, which comes from singular cohomology). Viewing differential forms as "alternating linear functions" pointwisely is still no problem because it is just the definition of forms. $\endgroup$ – autodavid Jan 22 '19 at 19:21
  • $\begingroup$ Hmm... Is it really true that $\langle A\wedge B , v\otimes w\rangle = \langle A\otimes B, v\wedge w \rangle$ ? I think it should be $\langle A\wedge B , v\otimes w\rangle = \langle A\wedge B, v\wedge w \rangle$. $\endgroup$ – autodavid Jan 22 '19 at 19:26
  • $\begingroup$ $A\wedge B = A\otimes B - B\otimes A $ thus $\langle A\wedge B, v\otimes w\rangle = \langle A\otimes B, v\otimes w\rangle -\langle B\otimes A, v\otimes w\rangle = A(v)B(w)-B(v)A(w) = A(v)B(w) + A(-w)B(v) = A\otimes B (v\otimes w - v\otimes w) = \langle A \otimes B, v\wedge w\rangle $ $\endgroup$ – Felix B. Jan 22 '19 at 21:49
0
$\begingroup$
  1. How can I visualize the wedge between two 1-forms $α∧β$?

First we need to understand what the wedge actually does. In your case it creates a new full anti-symmetric tensor (think determinant) of order 2. A 2-form is a thing that take two vectors and return a scalar. If, for example, we plug a vector into a 1-form we get a number. We also know that forms are a multi-linear, therefore we can pull out all the factors and apply the form on each basis-vector individually. So in practice a 1-form just projects a vector and measures the length of that projection. Following this, a 2-form can be visualized as a thing that first take a vector and becomes a 1-form.

But how does that translate to the dual space? What is the visualization between two of these 1-forms as a wedge?

Let's say we are in $\Lambda(\mathbb{R}^3)$, your 2-vector $a \wedge b$ can be thought of as something with the magnitude of the enclosed parallelogram of $a,b$ and an additional other property path orientation. These two properties are computed by the 2-form and return a number. You might get the impression that this looks very similar to integration, and you would be right. The way they scale and how they compute vectors is exactly how integration works.

  1. Why does it make sense that $d(d\alpha)=0$ for every differential form $\alpha$

We know that the exterior derivative $d$ take a (n-1)-form to an n-form. Since a form is also full anti-symmetric we combine that with the Schwarz integrability condition, that the second derivatives are symmetric and arrive at statements like $ \operatorname{div}(\operatorname{rot}a)= 0$ or $\operatorname{rot}(\operatorname{grad}\phi)= 0$. This is just the general statement.

  1. What does Dan Piponi mean by saying: "exterior derivative is none other than finding the boundary of the picture" (4 Exterior Derivatives)

Let us operate in $\Omega (\mathbb{R^3})$ and look at the object \begin{eqnarray} \phi &=& x_1+x_2+x_3\\ d\phi &=& \frac{\partial \phi}{\partial x_1}dx^1+\frac{\partial \phi}{\partial x_2}dx^2+\frac{\partial \phi}{\partial x_3}dx^3\\ d\phi &=& dx^1+dx^2+dx^3\\ \end{eqnarray} Over what boundary do you need to integrate $d\phi$ to retrieve the information of $\phi$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.