1. Dec 6, 2011

### TranscendArcu

1. The problem statement, all variables and given/known data
Show that exterior differentiation of a 0-form f on R3 is essentially the same as calculating the gradient of f.

3. The attempt at a solutionLet U be a differentiable 0-form on R3. I think

$$dU = \sum _{j=1} ^n \frac{δF_I}{δx_j}dx_j dx_I$$However, since U is a 0-form, I can write U = FI and drop the dxI, right? I would then have,

$$dU = \sum _{j=1} ^n \frac{δF_I}{δx_j}dx_j = \frac{U dx}{δx} + \frac{U dy}{δy} + \frac{U dz}{δz}$$This, I think, has partial derivatives in it, but looks more like $\nabla • U$ than $\nabla U$, the latter of which should be a vector.

2. Dec 6, 2011

### Dick

You probably want to use '\nabla' for your partial derivative symbol. But no, it's not like a divergence. It's not a scalar sum. dx is the dual form to the basis vector associated with x. To find rate of change along a direction v you dot the gradient vector with v. To find the rate of change using your form, you evaluate the form on the vector. You get the same thing.

3. Dec 6, 2011

### TranscendArcu

Okay. First of all, what is a dual form to the basis vector? Also, what do you mean by evaluate the form on the vector? Could you give an example?

4. Dec 6, 2011

### Dick

Take e_x=(1,0,0), e_y=(0,1,0), and e_z=(0,0,1). Those are your basis vectors. dx is a linear function on vectors, yes? And dx(e_x)=1, dx(e_y)=0, dx(e_z)=0. Etc. That's what I mean by evaluate the form on vectors. If v=a*e_x+b*e_y+c*e_z, then dx(v)=a, dy(v)=b, dz(v)=c. How did you define 'dx'?

5. Dec 6, 2011

### TranscendArcu

The only definition of dx I know is that of the differential distance. That is, taking small (I suppose linear) steps in the direction of x. Is that what you're looking for?

In any case, do I correctly sense that I have to make U into $\vec{U}$?

6. Dec 6, 2011

### Dick

If that's your definition of dx as a differential form, I think you may have skipped a chapter. Try and look back in the text and find a real definition. And no, you can't make U into a vector, it's not a vector. It's a real function on R^3, just like any 0-form.

7. Dec 6, 2011

### TranscendArcu

Well, truth be told, I only have the mathematical background up to multivariable calculus. That is, the class one might take at the conclusion of BC in high school. What I'm working on here is an extra-credit assignment given our by my instructor. My regular calculus textbook really has nothing in it on either differential k-forms. So I'm trying to piece together what these things k-forms are by googling examples and asking around here.

In other words, I don't have another calculus reference to get a real definition for dx. If you know where I might find one, I'd be very appreciative.

8. Dec 6, 2011

### Dick

I'm having a hard time finding one just now. The trouble is that you usually start talking about differential forms when you are dealing with tangent spaces to manifolds. So the language can get hard to deal with pretty quickly. Try starting with this one. http://www.sjsu.edu/faculty/watkins/difforms0.htm I only fished through the first half dozen or so links I found, so there's probably something better as well.

9. Dec 7, 2011

### TranscendArcu

Okay. So I took some time to read through the link. I didn't understand everything (even though I have done a bit with vectors, I haven't seen the term "vector space" before). Anyway, let me try a simpler application of these differential forms. Show that exterior differentiation of 0-forms on R1 is essentially the same as ordinary differentiation of smooth functions. So I have,
$$dU = \sum _{j=1} ^n \frac{δF_I}{δx_j}dx_j = \frac{U dx}{δx} + \frac{U dy}{δy} + \frac{U dz}{δz}$$Let U be a differential 0-form such that U=f(x). Thus,
$$dU = \sum _{j=1} ^1 \frac{δf(x)}{δx_j}dx_j = \frac{δf(x)}{δx} dx$$Which I think I can rewrite as

$$dU = f'(x) dx$$This is what I wanted to show, right?

10. Dec 7, 2011

### TranscendArcu

Whoops, I copied that first line of math incorrectly. I should probably just scrap it. I think I should just have:

Let U be a differential 0-form such that U=f(x). Thus,
$$dU = \sum _{j=1} ^1 \frac{δf(x)}{δx_j}dx_j = \frac{f(x)}{δx} dx$$Which I think I can rewrite as

$$dU = f'(x) dx$$That's what I wanted, I think.

11. Dec 7, 2011

### Dick

You should figure out a better way to display partial derivatives, like I said, use \nabla. But the math part looks ok to me.

12. Dec 7, 2011

### TranscendArcu

http://img338.imageshack.us/img338/8190/skjermbilde20111207kl10.png [Broken]
I was trying to recreate the loopy d's in the equation you see above. What is the code for those?

Last edited by a moderator: May 5, 2017
13. Dec 7, 2011

### Dick

Oops, I've been saying the wrong thing. The code is \partial. $$\frac{ \partial f }{ \partial x}$$

14. Dec 7, 2011

### TranscendArcu

Okay. So the next part of the question is "Show that exterior differentiation of a 0-form f on R2 is essentially the same as calculating the gradient of f."

So I have $$dU = \sum _{j=1} ^2 \frac{\partial{f(x)}}{\partial{x_j}}dx_j = \frac{f(x)}{\partial{x}} dx +\frac{f(x)}{\partial{y}} dy$$As a vector, this would be written as $<\frac{f(x)}{\partial{x}},\frac{f(x)}{\partial{y}}>$ right? And is this, then, not just the same as $\nabla U$?

15. Dec 7, 2011

### Dick

Like the question said, it is "essentially the same thing". But it's not the SAME thing. It's not a vector. It's a linear function on vectors.

16. Dec 7, 2011

### TranscendArcu

$$dU = \sum _{j=1} ^2 \frac{\partial{f(x)}}{\partial{x_j}}dx_j = \frac{\partial{f(x)}}{\partial{x}} dx +\frac{\partial{f(x)}}{\partial{y}} dy$$So, right now, the only similarity I see between this and the gradient is that they both have partial derivatives, and that both require you to calculate partial derivatives. So, yes, computationally I can see the relationship between the two. But is that really the only connection?

Also, notice that I've written $\frac{\partial{f(x)}}{\partial{y}} dy$ Is this bad notation? Would it be better to write f(x,y)?

Also, how come I can't find the edit button anywhere?

17. Dec 7, 2011

### Dick

f(x,y) would certainly be better. Go back and reread post 2 for the connection. And the Edit button has been on and off lately.