Differential Forms and Vector Fields correspondence

Solution 1:

For the first correspondence note, that by the inner product on $\mathbb R^3$ we have a 1-1-correspondence of linear forms $\mathbb R^3 \to \mathbb R$ and vectors in $\mathbb R^3$ where $v \in \mathbb R^3$ corresponds to $w \mapsto \left<v,w\right>$. This corresponcence - applied pointwise - assiociates to a vector field $\sum_{i} f_i U_i$ (where $U_i$ denote the constant orthogonal coordinate frame?) the form $\sum_i f_i \, dx_i$.

For the second one, we note that given a 2-form $\omega$, we have a map from 1-forms to 3-forms given by $\eta \mapsto \omega \wedge \eta$, as the 3-forms are a module of rank one over the functions, i. e. each three form is of the form $f\, dx_1\,dx_2\, dx_3$, we have a map from 1-forms to functions, that is a functional on the 1-forms, which can be represented (the bidual of the functions are the functions) by a vector field. Now condsider the 2-form $$ \omega = f_1\, dx_2 \,dx_3 - f_2\, dx_1 \, dx_3 + f_3 \, dx_1\, dx_2 $$ We have \begin{align*} \omega \wedge dx_1 &= f_1\; dx_1\, dx_2\, dx_3\\ \omega \wedge dx_2 &= -f_2\; dx_2\, dx_1\, dx_3\\ &= f_2\; dx_1\, dx_2\, dx_3\\ \omega \wedge dx_3 &= f_3\; dx_1\, dx_2 \, dx_3 \end{align*} so $\omega$ acts on the 1-forms in the same way as $\sum_i f_i\, U_i$ does.

Solution 2:

In the following I shall give some physical interpretation of the identifications alluded to in your book.

In physics we have force fields ${\bf F}$ and flow fields ${\bf v}$. In order to visualize such fields we draw in both cases little arrows (vectors) attached to the points ${\bf x}$ in the domain $\Omega\subset{\mathbb R}^3$. But mathematically there is a huge difference between the two: Force fields are integrated along curves, the result being work done, and flow fields are integrated across surfaces. In the second case the result is the amount of fluid traversing the surface per second.

A force field ${\bf F}=(F_1,F_2,F_3)$ is the same as the $1$-form $\alpha$ defined by $$\alpha({\bf X}):={\bf F}\cdot{\bf X}=F_1 X_1+F_2X_2+F_3X_3\ ,$$ and as $X_i=dx_i({\bf X})$ we can write$$\alpha=F_1dx_1+F_2dx_2+F_3dx_3\ .$$ In this way for any curve $\gamma:\ t\mapsto{\bf g}(t)$ $\>(a\leq t\leq b)$ in $\Omega$ one has $$\int\nolimits_\gamma \alpha=\int\nolimits_\gamma {\bf F}\cdot d{\bf x}=\int_a^b{\bf F}\bigl({\bf g}(t)\cdot{\bf g}'(t)\bigr)\ dt\ .$$

On the other hand, given a flow field ${\bf v}$ and an oriented surface $$S:\ (u,v)\mapsto {\bf f}(u,v)\in\Omega\qquad\bigl((u,v)\in B\bigr)\ ,$$ where $B\subset{\mathbb R}^2$ is the parameter domain, the amount (volume) of fluid flowing through $S$ per second is given by $$\int\nolimits_S{\bf v}\cdot d{\bf \omega}=\int\nolimits_B {\bf v}\bigl({\bf f}(u,v)\bigr)\cdot\bigl({\bf f}_u(u,v)\times{\bf f}_v(u,v)\bigr)\ {\rm d}(u,v)\ .$$ Here the vector field ${\bf v}$ becomes a $2$-form $\beta$ via $$\beta({\bf X},{\bf Y}):={\rm vol}({\bf v},{\bf X},{\bf Y})={\bf v}\cdot\bigl({\bf X}\times{\bf Y}\bigr)\ ,$$ where the $3$-form ${\rm vol}$ is the standard volume-form in ${\mathbb R}^3$.

Solution 3:

Let me give a perspective from geometric algebra and calculus.

Geometric algebra, or clifford algebra, imposes a "geometric" product of vectors. If $a, b, c$ are vectors, then $a(b+c) = ab + ac$, and $(ab)c = a(bc) = abc$, so it's associative and distributive (and several vectors can be involved in a series of products).

Thus, the general objects of a geometric algebra are called multivectors. Components of a multivector are often separated into blades, where a blade of grade $k$ can be written as some geometric product of $k$ orthogonal vectors. As you might expect, when the base vector space of the GA is $\mathbb R^3$, then there are only 4 grades to consider: grade-0 (scalars), grade-1 (vectors), grade-2 (dubbed "bivectors"), and grade-3 ("trivectors," or in 3d, also called "pseudoscalars").

So you can see already that there is a relationship between the grades of multivectors and 0-forms, 1-forms, 2-forms, and 3-forms. A geometric algebra usually identifies forms with vectors through the usual inner product structure: if $w$ is a 1-form and $v$ is a 1-vector, then $w(v) \equiv w \cdot v$ is just the dot product. (Yes, perhaps the $w$ on the right isn't exactly the same kind of thing as the $w$ on the left, but I'm not aware what the standard notation might be for this concept.) While $k$-vectors and $k$-forms can still be said to transform differently under coordinate transformations, they're both still considered to be elements of the same geometric algebra.

So, having done away with one layer of distinction between $k$-vectors and $k$-forms, we can focus on how traditional vector calculus gets away with using only scalar and vector fields. As has been said, the key here is duality.

In differential forms parlance, this has to do with the Hodge star. In GA parlance, this has to do with the pseudoscalar, usually called $i$. In 3d, $ii = i^2 = -1$, so the notation is suggestive. Multiplication by $i$ changed the grade of what it acts upon. If $u = u_k$ is a $k$-vector, then $iu_k$ is a $3-k$ vector. So $i$ turns vectors to bivectors, scalars to pseudoscalars. This duality makes it possible to describe $2,3$-vectors in terms of their dual $0,1$-vectors--that is, wholly in terms of vectors and scalars.

Geometrically, you can picture a 2-vector field as a field of oriented planes through space. The dual vector field is just a vector field of normals to those planes. Similarly, a 3-vector field is a field of oriented volumes, but since all volumes are scalar multiples of one another, a scalar field contains all the same information (though it is not quite the same geometrically).