12
$\begingroup$

Barrett O'Neill's Differential Geometry book says that Classical vector analysis avoids the use of differential forms on $\mathbb{R}^3$ by converting 1-forms and 2-forms into vector fields via the following 1-1 correspondences:

$$\begin{array}{*5c}\text{(1-form)} &\leftrightarrow &\text{Vector field}& \leftrightarrow& \text{(2-form)}\\ \sum f_idx_i &\leftrightarrow &\sum f_iU_i &\leftrightarrow &f_1dx_2dx_3 -f_2dx_1dx_3+f_3dx_1dx_2\end{array} $$

where $U_i$ denote the natural coordinate vector fields that constitute an orthonormal frame.

I superfically grasp the first correspondence although the motivation behind that escapes me as I have always thought of a 1-form as the dual of a vector field in that at each point of the manifold; the form associates a linear functional on a tangent vector at that point while the Vector field actually gives the tangent vector.

As for the second one , I am still not able to visualise 2-forms as anything but functionals on a pair of tangent vectors and the correspondence is unnatural.

O'Neill uses this to frame the Curl and Gradient and Divergence definitions in the language of forms. Could someone explain the need for this as well as good way to visualise two forms and higher degree ones in the same way as 1-forms???

$\endgroup$
  • $\begingroup$ Look at T. Frankel "The Geometry of Physics" book. $\endgroup$ – Roberto Mendes Jan 8 '14 at 12:47
8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks a lot. So this is akin to identifying the dual and the vector space in the first case?? I didnt get what you said about bidual of functions being functions. Do you mean something like $V^{**} \cong V$ ?? V here is just a vector space. $\endgroup$ – Vishesh Oct 8 '13 at 9:27
  • 1
    $\begingroup$ Yes, for both of your questions. $\endgroup$ – martini Oct 8 '13 at 9:38
  • $\begingroup$ @martini 'This corresponcence - applied pointwise - assiociates to a vector field....'. I didn't get this could you please explain? $\endgroup$ – RagingBull Oct 9 '15 at 17:01
10
$\begingroup$

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 ${\mathbb R}^3$.

$\endgroup$
  • $\begingroup$ Thank you very much Sir, for your neat example. I need a slight clarification. Would it be okay to look at $\mathbf{x}$ as a coordinate patch in the 2nd case?? $\endgroup$ – Vishesh Oct 8 '13 at 10:33
  • $\begingroup$ Thanks, the edit has made things clear. I really liked the explanation. $\endgroup$ – Vishesh Oct 8 '13 at 15:25
  • $\begingroup$ So is anything in basic physics genuinely a vector field, as opposed to a $1$-form or a $2$-form? $\endgroup$ – goblin Oct 30 '17 at 14:53
5
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ Thanks. But $\mathbf{i}$ is quite unlike the exterior derivative operator on forms?? $\endgroup$ – Vishesh Oct 14 '13 at 14:16
  • 1
    $\begingroup$ Multiplication by $i$ is related to the Hodge star, $*$ or $\star$. For instance, if $A$ is a 1-form, then you would take the divergence of $A$ via $*d(*A)$. In geometric calculus, this would be written $i^{-1} \nabla \wedge (iA)$. But as an algebraic object, $i$ can be moved through wedges at the cost of turning them into dots. So we get $i^{-1} (i) \nabla \cdot A = \nabla \cdot A$, exactly the way you would write the divergence in vector calculus. $\endgroup$ – Muphrid Oct 14 '13 at 14:21
  • $\begingroup$ What a beautiful, didactic, answer. In particular: "$i$ can be moved through wedges at the cost of turning them into dots". Where can I learn this stuff? $\endgroup$ – étale-cohomology Sep 15 '17 at 12:55

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.