- #1

- 4

- 0

- Thread starter SomeGuy
- Start date

- #1

- 4

- 0

- #2

mathwonk

Science Advisor

Homework Helper

- 11,041

- 1,232

If k is less than the dimension of V, and the dimension is n > k, then there are exactly the binomial coefficient "n choose k" linearly independent such functions, formed as follows. Take k vectors from V and place them in order in a k by n matrix. Then choose k of the n columns, and take the determinant of the resulting square k by k subdeterminant.

There are exactly "n choose k" ways to do this. Moreover one can form linear combinations of these basic functions and get thus a vector space of such functions of dimension "n choose k". Thus there is an "n choose k" dimensional vector space of such k dimensional oriented volume measures on an n dimensional vector space, dependeing on which set of k axes one projects ones k dimensional parallepiped onto before taking the determinant.

Of course in the case of length, there is a formula known as pythgoras theorem that lest one calculate the usual length of a segment from the lengths of its projections noto the avrious axes, and this generalizes to the case of area of parallelograms in 3 space. So perhaps there is also a way of compouting k dimensional volume of a k block from the various projections onto different choices of sets of k axes in n space.

Anyway, differential k forms on n space are a computations way of emasuring orienteds violume of k blocks in that space, generalizing determinants, and in fact are precisely computed as linear combinations of k fold subdeterminants of k by n matrices.

now if we have a family of n dimensional vector spaces, say VxU where U is an open set in R^n, then we can have a linear combination of k forms with varying coefficients. i.e. the "n choose k" coefficients expressing our k form as a linear combination of the basic subdeterminants, can vary over the open set U, in a continuous or smooth way.

This is called a differential k form on the tangent bundle VxU, of the opens et U. I.e. a smooth k form on a manifold, is locally a linear combination of k diomensional subdeterminants, with smoothly varying coefficients.

so it is a way of measuring oriented k dimensional volume in the smoothly varying family of tangent spaces to the manifold.

hows them apples?:grumpy:

- #3

mathwonk

Science Advisor

Homework Helper

- 11,041

- 1,232

consequently, if we have a parametrized k dimensional surface in ouir manifold, by looking at the images in each tangent space, of the unit k block under the derivative of our parameter map, and evaluating our oriented volume measure on each image block, we get a numerical function back on the unit block which we can integrate. I.e. we can integrate k forms over parametrized k diml surfaces.

i.e. you approximate a parametrized k surface at each point by a parametrized k block, and your k form measures its size. this approximates the stretching factor of your parameter map on the unit block, i.e. the size of the image surface.

thus since determinants are the basic way of measuring volumes, differential forms are the basic way of expressing integrals over parametrized surfaces. hence they are the natural language for formulating theiorems about integrals, such as stokes, greens, gauss, divergence, etc...

as to why those theorems are true, that relies on the FTC and fubinis theorem, i.e.repeated integration, which reduces them all to the fundamental theorem of calculus.:tongue:

i.e. you approximate a parametrized k surface at each point by a parametrized k block, and your k form measures its size. this approximates the stretching factor of your parameter map on the unit block, i.e. the size of the image surface.

thus since determinants are the basic way of measuring volumes, differential forms are the basic way of expressing integrals over parametrized surfaces. hence they are the natural language for formulating theiorems about integrals, such as stokes, greens, gauss, divergence, etc...

as to why those theorems are true, that relies on the FTC and fubinis theorem, i.e.repeated integration, which reduces them all to the fundamental theorem of calculus.:tongue:

Last edited:

- #4

mathwonk

Science Advisor

Homework Helper

- 11,041

- 1,232

wedge product is just alternating multiplication, i.e. you want something that acts like multiplication in each variable separately, but that changes sign when you change the order of x and y, so instead of the dot product x1y1 + x2y2, you take the wedge product x1y2-y1x2. then when you change order you get y1x2-x1y2.

this formula should look familiar from 2 by 2 determinants. this lets you make higher order determinants out of lower order ones, i.e. lets you get LaGranges inductive formula for expanding determinants.

this formula should look familiar from 2 by 2 determinants. this lets you make higher order determinants out of lower order ones, i.e. lets you get LaGranges inductive formula for expanding determinants.

Last edited:

- #5

mathwonk

Science Advisor

Homework Helper

- 11,041

- 1,232

and it behaves like a volume, i.e. if the k block has one side stretched by the factor c, then the number spit out is multiplied by c.

and it is oriented, so if we interchange the ordering of two sides of the k block, the number spit out chjanges sign. thats all.

but because we want to express them cimoputationally in terms of a bsis of them, it becoems operationally complicated. i.e. already the formula for the area of a parallelogram in terms of the 4 heights and bases is complicated ad-bc.

the area of a parallelogram in space in terms of its porojections on the three pairs of axes is complicated, and the volume of an n block in terms of the various coordinates fo the spanning vectors is a determinant, hence coimp[licated.

so k forms are intuitively easy, just oriented volume measures, but operationally complex. since the simplest ones are determinants, they cannot be made simpler than determinants.

david bachmans little book on them studied here some time back is highly recommended for an intuitive grasp of them. the little technical flaws I argued about in the presentation at the time can safely be ignored.

- #6

- 117

- 0

I'm been trying to answer this question for little while now.

For what its worth, at least so far, I don't see much value in JUST differential forms/wedge products. As Mathwonk points out, it is basically just different notation for the volume invariant change of coordinates formula (i.e. the determinant of the jacobian).

So it doesn't really give you much additional power compared to the usual vector calculus scale factors.

The true power of comes when you add in all the other machinery that calculus on manifolds and differential topology brings to the table.

I'm working thru:

Bamberg, Sternberg: A course in mathemathics for students of physics

Arnold: Mathematical methods of classical mechanics

Lee: Introduction to smooth manifolds

I was always unsatisfied with the apparently ad hoc manner of scale factors in usual vector calculus, and I wanted a better understanding.

These 3 books have greatly increased my understanding. I still have a long way to go.

brian

For what its worth, at least so far, I don't see much value in JUST differential forms/wedge products. As Mathwonk points out, it is basically just different notation for the volume invariant change of coordinates formula (i.e. the determinant of the jacobian).

So it doesn't really give you much additional power compared to the usual vector calculus scale factors.

The true power of comes when you add in all the other machinery that calculus on manifolds and differential topology brings to the table.

I'm working thru:

Bamberg, Sternberg: A course in mathemathics for students of physics

Arnold: Mathematical methods of classical mechanics

Lee: Introduction to smooth manifolds

I was always unsatisfied with the apparently ad hoc manner of scale factors in usual vector calculus, and I wanted a better understanding.

These 3 books have greatly increased my understanding. I still have a long way to go.

brian

Last edited:

- #7

- 406

- 6

You may have seen the visual interpreatation of a one-form as a set of parallel gradient lines. The visual interpretation of a two form is two overlapping set of such lines creating a gird of parallelograms. Now, here's where forms get messy.

Take a two form in two diemtsions. This form takes two vectors and then, careful now, returns the area of the parallelogram spanned by those vectors divided by the area of the parallelograms formed by the overlapping one form lines.

Now, that second part about dividing by the area of the parallelogram may seem a little odd. That's because it is It's a big problem with forms, because what a form is giving you is not the area of the parallelogram spanned by two vectors, but the area of the parallelogram spanned by the vectors

The situation gets even worse when you move to differentiation, with the omnipresent exterior derivative overshadowing all.

dw. What is it? No idea. It's usually attempted to be passed off as the curl in 2d or divergence in 3d, but this is simply wrong. The curl is an operator on one vector field to produce another, and the divergence is an operator on a vector field to produce a scalar field. The exterior derivative of a one form produced a two form, and of a two form produces a three form. Of course, the coefficients of these produced two and three forms happen to coincide with the coeffients for the curl and divergence, but this isn't really very helpful.

And the classic argument for forms is that they compress Maxwell's equations from four to three.

dF=0

d*F=J

dJ=0

Which means......? I have no idea. dw meant little to begin with and d*F=J means even less. At least in the vector form of Maxwells equations you could see what was going on. Differential forms tends to strip all semantic meaning from familiar equations.

I've recently discovered Geometric Algebra. I'm not going to argue that it's some kind of miracle cure, as it has some serious, often glaring issues. But frankly, it's better. In Geometric Algebra notation, you set [tex]F=\mathbf{E}+i\mathbf{B}[/tex] and pretty much immediately get

[tex]

(\frac{1}{c}\partial_t +\nabla)F = \rho -\frac{1}{c}\mathbf{J}

[/tex]

Yeah. That's right. Maxwell's equations become a single line, first order wave equation. Semantic or what. I was pretty much sold on GA at this point, but YMMV.

So basically, IMHO, forms work, but not very well. There are alternatives out there, and if forms are giving you pains in your back teeth, I would suggest looking them up.

- #8

- 289

- 0

Perhaps I'm missing something subtle or just there's a matter of less than rigorous notation, but isn't that equation wrong on the grounds that the LHS is a matrix and the RHS is a vector, if that! Even [tex] \rho -\frac{1}{c}\mathbf{J}[/tex] seems to be questionable, you're adding a scalar and a 3-vector. Even if you meant to say [tex](\rho , -\frac{1}{c}\mathbf{J})[/tex] you're still mixing matrices and vectors.In Geometric Algebra notation, you set [tex]F=\mathbf{E}+i\mathbf{B}[/tex] and pretty much immediately get

[tex]

(\frac{1}{c}\partial_t +\nabla)F = \rho -\frac{1}{c}\mathbf{J}

[/tex]

Yeah. That's right. Maxwell's equations become a single line, first order wave equation. Semantic or what. I was pretty much sold on GA at this point, but YMMV.

Or am I missing something?

BTW, Mathwonk, excellent posts :)

- #9

- 406

- 6

Elements in Geometric algebra exist in a 2^N dimensional "multivector" space, with the usual N dimensional vectors as a subset of that space. For example in 2d, you have a 4 dimensional extended space with 1, i, j and i^j as basis "vectors". As you can see, we have i and j as the usual vector basis, but now we append scalars and directed area elements, or "bivectors" (i^j) to them to form a more complete space.Or am I missing something?

In this fashion, expressions like 7+4i-6j+3i^j are valid if one simply considers them as a "multivector" which componentwise would be (7,4,-6,3). Addition and subtraction work as expected. Multiplication however does not, with multiplication by another multivector potentially swapping elements about in a strange, but well defined way.

Technically in GA, if I remember correctly, E is a regular vector as is J. Charge (rho) is a scalar and B is a "bivector", or directed area segement. This is analagous to the expression in differential forms notation of E as a one form, and B as a two form. In both cases the fact that E has units of V/m and B has units of Wb/m^2 is expressed clearly.

GA takes a bit of getting used to, but you do get used to it. Something I can't readily say about differential forms.

- #10

- 289

- 0

Is that not just another way of talking about [tex]\Omega^{\ast} = \Omega^{0} \oplus \Omega^{1} \oplus \Omega^{2}[/tex] ? Or is that kind of the point, it's an analogous way of describing the space of p-forms but without some of the complications?Elements in Geometric algebra exist in a 2^N dimensional "multivector" space, with the usual N dimensional vectors as a subset of that space. For example in 2d, you have a 4 dimensional extended space with 1, i, j and i^j as basis "vectors". As you can see, we have i and j as the usual vector basis, but now we append scalars and directed area elements, or "bivectors" (i^j) to them to form a more complete space.

I'm only just getting into p-forms myself and I agree it can be somewhat perplexing at times. I didn't get the whole "It gives div, grad, curl and laplacian without coordinates" thing for quite a while but I don't agree that it's not geometrically intuitive. Once things begin to click into place, I think it does give a very useful way of describing geometric properties.

Nakahara - Geometry, Topology and Physics gets onto differential forms about 1/3 the way through and it builds up to them by moving through a number of geometric systems, putting them in context. An excellent book for more mathematical physicists.

- #11

- 406

- 6

No. Multivectors exists in the tangent space, by themselves. Forms exists in the dual space and you need to apply vectors to them to evaluate them to a number.Is that not just another way of talking about [tex]\Omega^{\ast} = \Omega^{0} \oplus \Omega^{1} \oplus \Omega^{2}[/tex] ? Or is that kind of the point, it's an analogous way of describing the space of p-forms but without some of the complications?

But it doesn't give grad, curl and div. It just gives something that kinda sorta maybe but not really looks like grad, curl and div. dw isn't the curl. It's a two form. You have to give it vectorsI'm only just getting into p-forms myself and I agree it can be somewhat perplexing at times. I didn't get the whole "It gives div, grad, curl and laplacian without coordinates" thing for quite a while but I don't agree that it's not geometrically intuitive. Once things begin to click into place, I think it does give a very useful way of describing geometric properties.

I highly reccommend the following as an introduction to Geometric Algebra. Enjoy.

- #12

- 117

- 0

I.e. inverse(g)(df) where g is the metric maps the differential 1 form df to the usual gradient vector.

Curl and div mappings are slightly more involved.

- #13

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,916

- 19

It is terribly arrogant of you to think that your intuition is maximal amongst human beings. Your inability (or, more likely,Differwntial forms have no intuative interpretation. Seriously. Forms are operators on vectors. They exist in the dual space to vectors so their geometry is highly non-intuative.

Let's think physics in 3 dimensions for the moment. Would you agree that mass density is a pretty intuitive concept? Well, mass density

Of course, if really didn't want to use differential forms (or didn't know they existed, so scalars and vectors are your only tools), you could instead work with the hodge dual. That is, instead of working with the mass-density 3-form G, you could work with the mass-density scalar p, which is defined so that G = p dV. (where dV is the canonical volume measure)

Or worse, you could do what happened historically, and treat mass-density as a pseudoscalar, so as to maximize the technical awkwardness of the subject.

In particular, this happens because that's the only way to get the notion of a pseudoscalar (or pseudovector) to work out properly -- if you strip dV off of the mass-density 3-form to turn it into a pseudoscalar, you're going to have to reattach it if you ever want to integrate anything.Of course, areas in the pullback space aren't much use to you so nine times out of ten, the only two forms anyone talks about are dx^dy

Now, let's move up to 4 dimensions. Here, we have the current density 3-form J: it's the thing you integrate over a 3-volume to obtain the charge lying on that 3-volume. (Compare with how awkward it is to define such a thing without using higher differential forms)

Charge can be neither created nor destroyed: so the charge going into any 4-volume must equal the charge leaving the 4-volume. In other words:

[tex]\int_{\delta R} J = 0[/tex]

which, by Stoke's theorem,

Note that, if you were pretending current density is a vector, you would be using the (4-analogue of) divergence here.

And once we have

- #14

- 289

- 0

But if multivectors exist in the tangent space, do you not need to apply dual forms to them to get them to a number? A 0-form is a number or function, just like your '1', then I've got a 1-form and you've got i and j.No. Multivectors exists in the tangent space, by themselves. Forms exists in the dual space and you need to apply vectors to them to evaluate them to a number.

It would seem to me you're algebra is very much like p-forms, the same 'total space', similar formulism.

It does give the grad, curl and div. It depends what form w is. If w is a 0-form, then dw is grad. If w is a 1-form, then dw is curl. If w is a 2-form then dw is div.But it doesn't give grad, curl and div. It just gives something that kinda sorta maybe but not really looks like grad, curl and div. dw isn't the curl. It's a two form.

I was confused about it too, but once it falls into place it's very elegant.

You can look at the coefficents of the various basis terms of your p-form.You have to give it vectors and integrate for it to have any solid meaning.

For instance, suppose [tex]\omega = \omega_{x}dx + \omega_{y} dy + \omega_{z} dz[/tex]. Here w is a 1-form.

[tex]d\omega = \left( \frac{\partial \omega_{y}}{\partial x}- \frac{\partial \omega_{x}}{\partial y} \right) dx \wedge dy + \left( \frac{\partial \omega_{z}}{\partial y}- \frac{\partial \omega_{y}}{\partial z} \right) dy \wedge dz+ \left( \frac{\partial \omega_{x}}{\partial z}- \frac{\partial \omega_{z}}{\partial x} \right) dz \wedge dx[/tex]

Immediately, you've got the coefficents for [tex]\nabla \times \omega[/tex].

Are you sure? I've never seen such a requirement in a book? If anything, you can have spaces where the volume form doesn't exist due to lack of orientability.As to geometric meaning, I've already discussed the major problem with forms, that is, the only R^3 geometry involved requires you to use the canonical volume forms. Which is a pretty big qualifier.

- #15

- 255

- 0

Actually, the existence of various, different 1-forms possible in 3 dimensions is what gives us three of the 3-dimensional geometries from Thurston's Geometrization Theorem (at least, according to the experts, it's a theorem now).As to geometric meaning, I've already discussed the major problem with forms, that is, the only R^3 geometry involved requires you to use the canonical volume forms. Which is a pretty big qualifier.

This dovetails nicely into another interpretation of forms:

Suppose in R^3 we have a smooth plane field of vectors. That is, at each point of space, we have a well-defined 2-dimensional subspace in the tangent space at that point. Smoothness essentially means that points nearby to each other will have plane fields that can smoothly transformed to each other.

At each point x, the plane field can be defined at the zero set of some linear functional [itex]\omega: T_x R^3 \rightarrow R[/itex]. Thus, we have a 1-form [itex]\omega[/itex] defined on all of R^3.

Properties of the plane field as a whole can be found by investigating the properties of this 1-form. For example, if it is the case that [itex]\omega \wedge d\omega=0[/itex] at all points on R^3, then the plane field is integrable, that is, at each point, one can embed a surface in R^3 that is tangent to the plane field at all of its points.

If [itex]\omega \wedge d\omega\neq 0[/itex] at all points in R^3, then the plane field is maximally non-integrable. Which is to say, that there is no surface in R^3 that is tangent to the plane field everywhere -- the points on the surface where it is completely tangent to the plane field has measure zero. This is called a contact structure and has popped up repeatedly in both mathematics and physics (e.g. quantum physics and thermodynamics).

Page 3 of this paper (http://www.math.gatech.edu/~etnyre/preprints/papers/legsur.pdf) shows a number of examples of these.

Although clearly not appropriate for a high school level geometry course (or most undergrad courses for that matter), one finds that many geometrical structures lend themselves easily to a differential form convention.

Last edited:

- #16

mathwonk

Science Advisor

Homework Helper

- 11,041

- 1,232

how silly of me, i thought i made this crystal clear to absolutely anyone.

- #17

- 255

- 0

Face it, MW, this is The Question That Will Never Go Away. Every discipline has one (or several).how silly of me, i thought i made this crystal clear to absolutely anyone.

In art history, it's "What does abstract art actually mean?"

In physics, it's "So, what's a quick way to summarize general relativity?"

Etc. etc.

Sometimes this is asked in good faith. But I suspect that most of the time the inquirer is hoping to reveal that it's all just a complicated shell game invented by a group of pinheads who want to appear smart or to get rich or to be famous...

What irks me, though, is that I personally spent 6 years of graduate school dedicated to learning about this subject, and yet some people seem to think that a few hours on the internet getting unpaid advice from complete strangers should be enough to understand the topic. And then get dismissive of it when they still "don't get it."

- #18

mathwonk

Science Advisor

Homework Helper

- 11,041

- 1,232

yup, yup and (to satisfy length constraints) yup.

- #19

- 8

- 0

Actually, you gave a very nice, some might say "intuitive" (at least "pictorial") description in the clipped material.Differwntial forms have no intuative interpretation. Seriously. [Clip...]

If you'd like to see some interesting pictorial depictions of forms, you can have a look at Misner, Thorne & Wheeler's Gravitation. For example, pages 57, 100, 104, 106, 107, 109, 116 & 117.

- #20

- 43

- 1

- Last Post

- Replies
- 3

- Views
- 3K

- Last Post

- Replies
- 8

- Views
- 3K

- Last Post

- Replies
- 8

- Views
- 5K

- Replies
- 5

- Views
- 2K

- Last Post

- Replies
- 24

- Views
- 2K

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 5

- Views
- 7K

- Last Post

- Replies
- 20

- Views
- 12K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 8

- Views
- 3K