The exterior derivitive 'd'

[r16]

I am attempting to teach myself differential geometry (being bereft of creditied educational institutions in my area) and I am being irked by that fact that I do not have a good physical/geometrical view of the exterior derivitive 'd'; which is a necessity of being a visual learner.

Does anyone have one ?

 

[Tantoblin]

For the purposes of visualisation, it is easier to look at the coboundary operator in algebraic topology. In the book "Algebraic Topology" by Allen Hatcher (freely available from http://www.math.cornell.edu/~hatcher/AT/ATpage.html) there is an excellent introduction to the coboundary operator (look at chapter 3, in particular the first paragraph, "the idea of cohomology").

The advantage of Hatcher's treatment is that it requires only simple tools, whereas the exterior differential is really quite advanced. However, all of his statements have a natural counterpart in the theory of differential forms.

Basically, what he is doing is to consider a number of simple cases, which is always a good thing to do when trying to visualise differential forms. For example, if one has a manifold M on which a function f is defined, how should one interpret the differential df?

Well, look at the level surfaces of f (the points where f takes a constant value). The whole of all level surfaces is again M, and the exterior differential df is related to the rate of change of f as one goes from one surface to the next. There is a natural way to interpret df as an integrable distribution of hyperplanes on M, and if you're interested I can explain.

(edit: please feel free to ask further question because my reply is rather incoherent).

 

[r16]

I downloaded and read the passage out of that book, however i feel i comprehended very little of it except the analogy to the trails and the circuts. I have practically no experience in topology, group theory, or lie algebra, so I was quite confused.

Several texts have good explanations / visualizations of the fundamental theorm of exterior calculus:

$$\int_R d \alpha = \int_{\partial R} \alpha$$

with examination of the special case

$$\int^b_a df = f(b) - f(a)$$

I was attempting to backfit that into a definition for 'd', but ran into a couple condrumns.

In the case of a 0-form f, the action of 'd' on f creates a series of perpidicular lines (the geometrical picture of a 1-form) on f so that df and f are always perpindicular. The "density" of df; how close the perpindicular lines of the 1-form are spaced, could be determined by examining the function; but the actual mechanism of how what values f maps to the reals determines how the 1-form elements are spaced is a mystery to me.

This definition would give a good physical interpertation on the integration of a 1-form. Each 1-form line would carry a constant value(how would this be determinded from the fuction?), and the sum of each 1-form perpinducular to f in the interval from b to a would give you the value of the integral. This is consistant with the definition of the fundamental theory of exterior calculus, where the value in the interavl would be the difference of the values of the function at the endpoints.

The operation of 'd' would be the determinitation of the perpindicular hyperplanes, their density, and their constant value to the path which f takes from a to b.

I do not know about the validity of the previous statement, but this is what i came up with after brainstorming and conjecturing on this topic. I have not tried to comprehend how this model would work in higher dimensions or how it would work on a p-form where p != 0.

As for the explanation of the hyperplane model of 1-forms, it always helps to hear another explanation of a topic. Mabey some light could be shed on where the idea to use perpindicular hyperplanes was concieved.

 

[Oxymoron]

It's hard to get a geometrical "feel" for the exterior derivative because in Euclidean spaces where we have good, intuitive notions of geometry the exterior derivative acts on scalars and the result is much like the regular partial derivative. However, unlike the partial derivative, the exterior derivative can be extended to an operator on all differential forms (not just scalars).

The Exterior Derivative (which I will denote by $\mbox{d}$) of a function, $\mbox{d}f$ is a 1-form such that:

$$\mbox{d}f(X) = Xf$$

The Exterior derivative maps p-forms to p+1-forms:

$$\mbox{d}\,:\,\Omega^p \rightarrow \Omega^{p+1}$$

The Exterior derivative on a wedge product obeys the anti-derivation law:

$$\mbox{d}(\omega\wedge\phi) = \mbox{d}\omega\wedge\phi + (-1)^p\omega\wedge\mbox{d}\phi$$

The Exterior derivative of a 0-form vanishes:

$$\mbox{d}^2 f = 0$$

So basically, the Exterior derivative is just another operator to put with all your other operators. It is kind of like the partial derivative, except that it has more uses.

 

[r16]

i guess i will have to live with the fact that the exteror derivitive is kind of vauge for the time being. Its not with a normal derivitive as measuring the instant rate of change of a function or the tangent line but some other action. Not knowing what it does physically doesn't diminsh the algebric significance, but it would give me a warm and fuzzy feeling if i could visually comprehend it.

 

[Oxymoron]

I guess you would have to be able to visualise differentiable forms before you can visualise what the exterior derivative is. Unfortunately I don't even know what a Diff. form looks like (yet), so I won't be much help there.

 

[Hurkyl]

(Disclaimer: I certainly cannot be called an expert in this stuff)

I don't think that forms are something that's meant to be pictured; forms are functions of the geometrical objects.

A zero-form is something that measures points.
A one-form is something that measures paths.
A two-form is something that measures surfaces.
An k-form is something that measures k-dimensional surfaces.

So what you really want to be doing is not trying to picture the forms themselves, but try to picture what they measure, or how they measure it.

If w is an k-form, then the picture of dw's action is that dw is the form that takes a (k+1)-surface, and measures w around its boundary.


Well, I guess you can picture k-forms as sections of the bundle of k-forms on your manifold, but that picture doesn't really connect with their algebraic properties.


Well, I suppose there are ways of recasting this picture. For example, if you have a metric, you can stop thinking about one-forms, and think instead about taking the dot product of vector fields... and whatever the higher dimensional analog of this looks like.

Also, vector fields can be turned into differential forms: for example, in three space, the vector field <f, g, h> can be turned into the 2-form

f dy dz - g dx dz + h dx dy

or in general, we can go from $X^i$ to the n-1 form $\epsilon_{i_1 i_2 \cdots i_n}X^{i_1} dx^{i_2} \wedge \cdots \wedge dx^{i_n}$, and this looks like it would work for any k-vector field of rank less than n. I'm not really sure what this gives us, though.

 

[mathwonk]

come on guys. everyone has known the meaning of these objects for years, decades, centuries.

on functions d is the "gradient" or direction of greatest increase, on one forms d is the "curl" of a vector field or its tendency to rotate at a point, on 2 forms, d is the "divergence" of avector field, or the edxtent to which it expands out from a point or to which that point is a "source".

read the intro to maxwells electricity and magnetism.

this is an example of the loss of understanding that comes with modern definitions.

we are all physicists here right?

 

[Hurkyl]

I think my biggest stumbling block in learning differential geometry is understanding just what everything is. Euclidean space encourages confusing the notions of vector and covector, and I feel that is harmful when passing to the general case!

I count myself lucky that I figured out early on (still in the Euclidean setting) that the gradient was different from a vector -- in coordinates, the gradient is a row, while a vector is a column... and I have since felt that this distinction has frequently helped me work through situations where I was confused.

Of course, maybe my experience is just highly unusual.


Now, I do think that in the case of curl and divergence, my description of d is the same as yours; I had even meant to give it as an example!

If w is the 2-form that measures the flux of a vector field V across a surface, then my description says that dw measures a region of space by measuring the flux through its boundary.

If we take that region of space to be a tiny ball around a point, then the flux through its surface is nothing more than how much the vector field is pointing away from that point!

And note that it's not trivial to say that "d" is simply the divergence or curl of a vector field! You have to know to go over to that 2-form as an intermediate step. Three months ago, before I was introduced to this idea, I would still have known that $d (\vec{F} \cdot d\vec{A}) = \mathop{\text{div}} \vec{F} \, dV$ had to be right because I know the divergence theorem is supposed to be a special case of the generalizaed Stoke's theorem... but I think it would have taken me much work to figure out why it's true. (or even if I could have)

Geometric intuition isn't much help if you don't actually know how the geometry connects to the algebra!



In the Euclidean space, any two form can be viewed as measuring the flux of a vector field through a surface, but that's very specific to a 3-dimensional space with a metric... what if we're looking at 2-forms in a 4-dimensional space? Or a space where we don't have a metric?


Incidentally, isn't it rather important about forms that they are things you can integrate? I sort of feel that learning about forms in terms of their local description as being something that at each point takes a value in a tensor product of the cotangent space has caused me to miss out on the big picture. (ha ha)

 

[mathwonk]

listen to hurkyl, as he knows both physics and math.

 

[r16]

i really like the flux analogy: i have been quite familiar with it since high school. The fact that it is an action on something and not trying actually visualize 'd' as you would a derivitave in calculus.

I stumbled upon all of the examples you pointed out when i proved maxwell's equations from the maxwell field tensor and it's hodge dual. As it was said, making the connection between the geometry and algebra is 1/2 the battle.

All of the little tricks i would use to find specifically curl a divergance make sense, taking hodge dual of a 1-form (to get a 2-form in $$R^3$$ for curl and taking the exterior derivitave of the hodge dual of a 1-form for divergance (which makes a top-form). It all lines up now.

 

[nrqed]

come on guys. everyone has known the meaning of these objects for years, decades, centuries.

on functions d is the "gradient" or direction of greatest increase...

This is where I get confused... I really liked the following post by quetzalcoat. It made sense to me...but it contradicts saying that df is the gradient.

the gradient, $$\vec{\nabla f}$$ is a vector. it has vector components $$\nabla f ^i = g^{ij} \frac{\partial f}{\partial x^i}$$

whereas $$df$$ is a one-form with covector components $$\frac {\partial f}{\partial x^i}$$

df is a one-form and so it resides in the dual space with the basis $dx^i$:

$$df = \frac{\partial f}{\partial x^i} dx^i$$

the gradient is a vector in the tangent space with the basis $\frac{\partial}{\partial x^i}$, defined as:

$$df(\vec{v}) = <{\nabla f}, \vec{v}>$$

so the gradient vector is the vector whose inner product with $\vec{v}$ maps to the same real value as acting the one-form $df$ on the vector $\vec{v}$

it is an important distinction because they live in entirely different spaces, being related by the metric tensor.

 

[robphy]

Some references:
http://homepage.mac.com/sigfpe/Mathematics/forms.pdf
http://physics.syr.edu/courses/vrml/electromagnetism/references.html
http://www.av8n.com/physics/thermo-forms.htm

 

[mathwonk]

there is some algebra involved in going back and forth between vectors and covectors but in this case to me that is not the main point. I try to distinguish them for global purposes, and functorial purposes, but here we are trying to visualize something, and something local. I agree that we agree, but I thought you were making it seem harder than necessary.

I think starting from definitions is a mistake in trying to understand the geometry, and that here the old physicists were right, and it is ok to use a metric since the question of what is being measured by a derivative is a local matter.

so the old physics books and partial diff eq books, like L. Hopf, explain it the way hurkyl did at the end, and this is what I was alluding to.

Namely to understand d of a one form, you use the green's theorem. That theorem says what is being measured at a point, is the rate of change with respect to area of the integral of the one form around smaller and smaller curves encircling the point.

Now if you want to visualize a covector, it seems to me prudent to introduce a metric and visualize it as dotting with a vector. Then the fact that the integral of a one form around a small closed curve is non zero says that vector is to some extent tangent to that curve as it goes around, i.e. it curls around.

But even if you do not introduce a metric, you are still measuring the integral of your one form around a curve and the extent to which it is non zero for a small closed curve around your point. So suppose you are imagining your one form as a family of lines in the tangent spaces at points of your curve, i.e. the lines of tangent vectors at each point, where your one form has value zero. Then you are meaeuring the extent to which these lines are transverse to your curve, which means again they look as if they are swirling around your point as you go around.to understand what d means, a basic thing to ask is "what does it mean for d to be zero?" for a function this is a measure of whether it is constant. For a one form it is a measure of whether it is locally a gradient.

Hence part of visualizing "d" is being able to visualize a gradient covector field. Now even visualizing a function requires some choice of coordinates, to be able to graph it or to see distances in the domain space.

But suppose we have a plane region as our domain and a real valued function defined on it. To some extent this function is determined by its level curves, i.e. the subsets where f = constant.

And the gradient of the function, i.e. df, is a family of covectors that vanish on lines tangent to these curves, hence are visualized either as a family of lines tangent to these curves, or equivalently by vectors perpendicular to these curves.

now ask yourself, what is it about this family of vectors that is peculiar to their being orthogonal to level curves of a function?

perhaps you will see that it is their total absence of any points where they wind around. i.e. they always either glide past or go into or out of a point, or do both in a hyperbolic sort of shape, inwards in one direction, outwards in another.

so you never get a non zereo integral around any small closed curve.

i.e. a one form w has dw = 0 if and only if locally it defines a "conservative" covector field, one such that integration around small closed curves is always zero.

this is the explanation given by greens theorem, that the integral around a small closed curve equals the integral of dw over the interior of the curve.
now imagine a two form A, and ask what dA means. by the divergence theorem this measures the rate of change of the integral of A over small spheres centered at the point, i.e. if you represent the 2 form as taking the cross product of a piece of surface with a vector, the extent to which that vector is perpendicular to the surface, or the tendency of it to measure flow or "flux" through the surface.

again dA = 0, means the 2 form A has integral zero over small closed surfaces centered at your point.

In general, as alluded to above in the realm of algebraic topology, thinking of a k form Z as something that acts on pieces of k dimensional surface via integration, having dZ = 0, means, by stokes theorem, that the integral of Z is always zero over a piece of k dimensional surface which is itself the boundary of a chunk of k+1 dimesnional space.

so if you think of integration of Z over a surface S as measuring some kind of flow or flux of a substance through or along S, then dZ = 0, means there is no such flow through the boundary of any small k+1 dimensional ball.

of course whether the geometry isflow or circulation is somewhat up to you. I.e. a family of vectors tangent to a closed plane curve can be replaced by their orthocomplements using a metric, and turned into a family of vectors pointing out of the circle. this changes measuring rotation nito measuring divergence or flux.

similarly a family of covectors annihilating tangent planes to a sphere, may be represented by that family of oriented planes tangent to the sphere, hence representing rotationalmotion on the shperes surface, or by the family of orthocomplements to thsoe tangent planes, giving a family of vector pointing out of the sphere, hence can also be thought of as repesenting divergence or flux through the sphere.

 

[mathwonk]

in topology this duality is also mirrored via the poincare duality theorem as intersection theory of cycles. for instance suppsoe we look at a torus, i.e. the surface of a doughnut, and we draw a closed loop all the way around one hole in the doughnut. this is a closed curve that cannot be shrunk to a point, and does not form the boundary of any piece of surface on the torus.now there is a vector field parallel to this curve on the doughnut, i.e. there is an ordinary differential equation for which this curve is a solution.

but we can also look for a differential one form represented by this curve, a covector field represented by vectors orthogonal to this curve without choosing any metrics. Just form a little local collar around the curve, thickening it slightly, and define a local function with level curves parallel to this curve. then take d of that function, getting zero at a small distance from the curve, and extend by zero globally.

Then this one form will have integral zero over this curve, but will have non zero integral over any closed curve transverse to this one and intersecting it once, i.e. over any curve which goes around the "other" hole in the doughnut.

then the one form w defined by the first curve, has integral which represents "intersection with" the first curve". this one form w will also have dw equal to zero, because it vanishes on any closed curve that bounds a piece of surface on the doughnut.

so we have used the same curve to represent both a vector field and a covector field, all without choosing any metrics.

 

[mathwonk]

After re reading all the posts, I still think that the formalism of differential geometry, worrying about vectors and covectors and all the silly notation, has hampered most people here from becoming comfortable with the concepts the way the old guys were.

tensors, covectors, manifolds, bundles, these are all nonsense as far as understanding a local phenomenon goes, in my opinion.

i mean how do you see a force field? you put a particle in there and watch it accelerate right? that's what integration means, letting it act over time on something.

these mathematical constructs were introduced by physicists trying to write down phenomena of electricity and magnetism, and developed further to treat gravitation, so it makes little sense to me to try to interpret the mathematics geometrically without going back to those ideas, of the action of magnetic and other fields of force.

 

[mathwonk]

basically i was just struck by the feeling that since "d" was developed to generalize and formalize the operations of div, grad, curl, that it really seemed absurd to take the abstract definition of d, and try to understand it intuitively without looking back at the meaning of the earlier concepts.

And I did not really notice that being emphasized in the first few posts, but perhaps I was not perceptive enough.

 

[mathwonk]

let me make another point about the intrinsic nature of duality in a global form. in the example above we began with one curve on a torus and passed to another curve that intersected the first one exactly once. this other curve then has a field of tangent vectors which although not necessarily perpendicular to those of the first curve are nonetheless transverse to them. this notion of transversality, i.e. the opposite of tangency, is intrinsic and independent of coordinates whereas perpendicularity is not.

nonetheless, integrating the differential form associated to the fiurst curve over the second curve yields the same answer, for any two homotopic curves, hence it is the transversality which is detected by the integral, and perpendicularity is irrelevant.

i.e. we do not need a metric to choose a transversal representativem for the second curve, and any choice gives us a geometric representative of the vector field perpendicular to the tangent field to the original curve.by the way the poster said he was trying to learn differential geometry, not differential topology, and differential geometry is by definition the study of differential manifolds with a given riemannian metric. (otherwise curvature would not make sense.)

 

[ObsessiveMathsFreak]

these mathematical constructs were introduced by physicists trying to write down phenomena of electricity and magnetism, and developed further to treat gravitation, so it makes little sense to me to try to interpret the mathematics geometrically without going back to those ideas, of the action of magnetic and other fields of force.

I agree with you 100%. My own view is that the best way to teach a mathematical method is to return to the problem in which it first arose. Let the student see the context of why the method is useful, and why it developed the way it did. For example, the best way to teach Fourier analysis is through an examination of the original problem faced by Fourier when solving his heat and string equations.

Defining mathematics via axioms and theorems alone is fundamentally flawed, like learning the grammar rules of a language without ever having seen a single sentance.

 

[mathwonk]

first there is a problem, then a solution, then a mathematical theory.

 

[mathwonk]

to understand a mathematical construct, ask yourself what it is supposed to measure? any specific tool in mathematics is a way to render a particular intuitive concept calculable.


div is short for divergence, i.e. how much stuff is flowing out from a certain region in a certain amiount of time. so to emasure it you need to measure some volume. hence it aint no shock that p forms are certain vloume measures. and the volume of a flat block squished into a plane should be zero, so naturallky a block spanned by three dependent vectors should be zero. hence the form should be alternating.

instead of trying to "learn" it, i.e. memorize the definitions, try to understand why they were made. the only way to understand math is to practice doing it.

 

[mathwonk]

i.e. rookies in math say things like " that's not the definition i learned." more experienced heads say things like, "well that may be the definition in the book, but that is not what a good definition really should say, and here is why".

 

[ObsessiveMathsFreak]

It's the difference between knowing what something is, and knowing what it really is.

Unfortunately in my case, I'll be dealing with 4-forms, so falling back on the old curl and divergence operators won't be an option. Practice makes perfect.

 

[mathwonk]

just as with determinants, 4 diml ones are defineable inductively in terms of lower dimensioinal ones so in fact the same pattern holds for 4 forms as 2 and 3 forms. i.e. they measure 4 diml "volumes".

think about what it should mean. ask what it is trying to measure.

 

[ObsessiveMathsFreak]

I've been grappling with the exterior derivative for about a week now working mostly from David Bachman's book. I unfortunately do not have a good geometrical interpretation, but I think I might be closer to some kind of understanding.

$$d\omega$$ seems to be a very strange quantity. If $$\omega$$ is a zero-form, we have $$d\omega$$ as the gradient. If one-form, we have a curl. If a three-form, we have a divergence. The only thing these three quantities seem to have in common is that they obey the generalised Stokes Equation.

Bachman's book, along with just about every other source, doesn't really seem to go into detail about what the exterior derivative of a form actually is, in and of itself, other than to say, or imply, that the exterior derivative is just something that will make Stokes Equation work. To this end, quite complicated definitions of how to find the exterior derivative are needed. I was never a big fan of long definitions, and the definitions of the exterior derivative are a bit cumbersome. Plus the fact that they change for every type of form, getting longer and longer as you go, doesn't help in conceptual understanding of what makes all these exterior derivatives "the same" in some way.

Anyway, after fighting with this for a while, I think I've managed to come to terms with "why" all these things called exterior derivatives are the same underlying relation or operation. But I'm not certain if it is actually correct. Here's my current thinking.

Taking the case of three dimensional vectors, if you do something, a little risque, and define the following "one-form"
$$\nabla = \frac{\partial}{\partial x} dx +\frac{\partial}{\partial y} dy +\frac{\partial}{\partial z} dz$$

Then, at least for zero, one and two forms, because they're the only ones I've checked
$$d\omega \equiv \nabla\wedge\omega$$

Or in a more descriptive form.
$$d\omega(V_1,\ldots,V_{n+1}) \equiv \nabla\wedge\omega(V_1,\ldots,V_{n+1})$$
where
$$\omega(V_1,\ldots,V_n)$$
is an n-form.

Now I haven't actually checked this for the general case, i.e. of all n-dimensional vectors and k-forms, but I've got a feeling it should be OK, and hopefully it can be extended. The reason I'm posting is to ask if this interpretation is correct? If so, can anyone give it the thumbs up? And if I'm on the wrong track, please let me know!

 

[mathwonk]

isnt that the definition of dw? i.e. if w = f dx^dy, the dw = (df)^dx^dy.

 

[ObsessiveMathsFreak]

I had simply viewed that as a consequence of the exterior derivative, rather than its definition. Anyway, the $$\nabla$$ kind of makes it clearer for me at least, as to what's going on.

 

[Hurkyl]

Or another way to say the same thing, just compute a formal derivative (complete with product rule), with the relation that d(dx) = 0 for any x.

 

[mathwonk]

theres only two ways to go, either define it in corordinates, and prove it is the adjoint of the boundary operator [i.e. satisfies the stokes theorem), or define it as the adjoint of the boundary operator and compute it in coordinates.

what is your definition of dw?

 

[ObsessiveMathsFreak]

My definition up to this point has been Bachman's. Namely;

$$d\omega(V^1, \ldots,V^{n+1}) = \sum_{i=1}^{n+1} (-1)^{i+1} \nabla_{V^i} \omega(V^1, \ldots, V^{i-1},V^{i+1}, \ldots ,V^{n+1})$$

Which wasn't very helpful. I didn't find "d" very helpful either as it didn't really make clear that the order of the form was being increased, as well as the fact that this "d" means something completely different to those in "dx" and "dy". With $$d\omega = \nabla\wedge\omega$$ you can see where the additional wedge product is coming from in things like $$\nabla\wedge \omega (f dx) \equiv d(f dx) = df \wedge dx$$

Actually, I don't think it will be too tricky to prove that the definitions are equivilant. In fact, I think it's pretty trivial.

If
$$\omega(V^1,\ldots,V^n) = \left | \begin{array}{cccc} \omega_1(V^1) & \omega_1(V^2) & \ldots & \omega_1(V^n) \\ \omega_2(V^1) & \omega_2(V^2) & \ldots & \omega_2(V^n) \\ \vdots & \vdots & & \vdots \\ \omega_n(V^1) & \omega_n(V^2) & \ldots & \omega_n(V^n) \\ \end{array} \right|$$

and
$$\nabla(V) = \sum_{i=1}^{N} \frac{\partial}{\partial x_i} dx_i , N=dim(V)$$
then
$$\nabla \wedge \omega(V^1,\ldots,V^{n+1}) = \left | \begin{array}{ccccc} \nabla(V^1) & \nabla(V^2) & \ldots & \nabla (V^n) & \nabla(V^{n+1}) \\ \omega_1(V^1) & \omega_1(V^2) & \ldots & \omega_1(V^{n}) &\omega_1(V^{n+1}) \\ \omega_2(V^1) & \omega_2(V^2) & \ldots & \omega_2(V^{n}) &\omega_2(V^{n+1}) \\ \vdots & \vdots & & \vdots & \vdots \\ \omega_n(V^1) & \omega_n(V^2) & \ldots & \omega_n(V^{n}) & \omega_n(V^{n+1}) \\ \end{array} \right|$$

As
$$\nabla(V^i) \equiv \nabla_{V^i}$$

If you simply compute the determinant you get;
$$d\omega(V^1, \ldots,V^{n+1}) = \sum_{i=1}^{n+1} (-1)^{i+1} \nabla_{V^i} \omega(V^1, \ldots, V^{i-1},V^{i+1}, \ldots ,V^{n+1})$$
Hopefully that was OK.

 

[mathwonk]

well that first definition, if you study it, is merely the adjoint of the boundary operator. i.e. to apply dw to a block spanned by three vectors say, you consider the boundary of that block, spanned by two of them at a atime, and apply w to each of those, but with a minus sign to give the right orientation of each face.

so this definition essentially forces stokes theorem to be true.

it helps if you know some algebraic topology, like boundaries and coboundaries of chains and cochains.

 

[mathwonk]

dont these people ever explain what they are doing? i would think bachman would in his geometric approach.

 

[ObsessiveMathsFreak]

I think it's important to have a definition independant of Stoke's theorem. Defining something to suit Stoke's theorem is somewhat circular. Defining the exterior derivative to be the wedge product of the gradient and the form is a little more straight forward, as the entity exists in its own right ratehr than having to be coupled with an adjoint or some such thing.

Edit:

I suppose an analogy might be for instance, how you would define a right angled triangle. You could define it to be a triangle whos sides a,b and c obey the rule a^2 +b^2 = c^2, pythagoreas' theorem, but this would be rather circular rule. One could imagine, that a student who learned to define a right triangle in this way, might never realize that one of the angles is 90 degrees. A right triangle should exist independant of ones knowladge of pythagoreas' theorem.

 

[Hurkyl]

as well as the fact that this "d" means something completely different to those in "dx" and "dy"

Nope; that's the same d! (Any scalar function, such as x, is a 0-form)

A right triangle should exist independant of ones knowladge of pythagoreas' theorem.

A right triangle should exist (in Euclidean geometry) independent of one's knowledge of right triangles. :tongue:


(I'm going to call a triangle that satisfies the Pythagorean identity a "Pythagorean triangle", to make the following easier to say)

The three are all equivalent:
(1) You are taught about Pythagorean triangles, and it is later shown that a triangle is Pythagorean iff it has a right angle.

(2) You are taught about right triangles, and it is later shown that a triangle is right iff it satisfies the Pythagorean identity.

(3) You are taught about right triangles and about Pythagorean triangles, and it is later shown that a triangle is right iff it is Pythagorean.

The only reason to prefer one of these over the other is for aesthetic reasons; maybe you think (2) will be easier for the student to follow, or maybe you think (3) will make the proofs more clear, or maybe you think that satisfying the Pythagorean identity is very important and you want to emphasize that by using (1).


IMHO there is a lot of value in defining something to have the properties you want it to have... rather than defining it by a calculation and then trying to prove the calculation has the properties you want it to have. (Despite the fact that it usually requires a theorem to prove that the thing you defined really does exist)



If you're interested in an algebraic perspective, there's something called a derivation that encapsulates the most important properties we associate with derivatives: it's a linear map that satisfies:

D(ab) = a(Db) + (Da)b

(where the multiplications involved are whatever is appropriate for the structures of interest)


In the current situation, the exterior derivative d is simply the (most general) derivation that satisfies d(dx) = 0 for all x, and has df being the ordinary differential when f is a 0-form.

 

[ObsessiveMathsFreak]

Well I suppose it's a matter of personal preference. I prefer to define things indepenently and then show how unexpected relationships emerge from simple definitions. That way, you don't really feel like you're hemming yourself in.

Edit:
On an aside, differential forms notation is terrible. Everything is just so lax!