Building Intuition for Differential forms, exterior derivative, wedge [duplicate]
- 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 fully anti-symmetric tensor (think determinant) of order 2. A 2-form is a thing that takes two vectors and returns a scalar. If, for example, we plug a vector into a 1-form we get a number. We also know that forms are 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 takes 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 taken by the 2-form to 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.
- Why does it make sense that $d(d\alpha)=0$ for every differential form $\alpha$
We know that the exterior derivative $d$ takes 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.
- 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 $\phi$?