# Differential Form: Closed/Exact

1. Feb 16, 2013

### Karnage1993

1 & 3
I have this differential form:

$\omega = F_1 dx + F_2 dy + F_3 dz$

And I concluded that $\omega$ is closed because I calculated the partials and found out that $\displaystyle \frac{\partial F_i}{\partial x_j} = \frac{\partial F_j}{\partial x_i}$.

Also, $F_1$ contains only $x,y$ terms, $F_2$ contains $x,y,z$ terms and $F_3$ only $y,z$ terms.

So according to an equation from class, the Domain of $\omega$ = Domains of $F_1 \cap F_2 \cap F_3 = \mathbb{R}^2 \cap \mathbb{R}^3 \cap \mathbb{R}^2 = \mathbb{R}^2$.

Here's where I'm confused. How is the domain of $\omega = \mathbb{R}^2$? The differential form contains all 3 parameters so I don't see how it can be. Also, would the $g$ also have domain $\mathbb{R}^2$? This domain problem is preventing me from concluding that $\omega$ is exact. Once I figure out it's exact, then I can carry out the computations to find $g(x,y,z)$.

2. Relevant equations
Definition of exact is:

Let $\omega$ be a first order differential form in $\mathbb{R}^n$. If $\omega = dg$, for some $g : \mathbb{R}^n \to \mathbb{R}$, then $\omega$ is said to be exact.

exact $\Rightarrow$ closed

2. Feb 16, 2013

### Dick

What's this equation from class? I suspect you are all worried about nothing. F(x,y,z)=x+z still has domain R^3 even though there is no y in it. You can't even really add things that have different domains.

3. Feb 16, 2013

### dextercioby

Can you also post the text of your problem, so we can judge what you wrote. Thanks!

4. Feb 16, 2013

### Karnage1993

So does that mean something like $F_1 dx$ = $(2x + xy^2)dx$ would mean that the domain of $F_1$ is $\mathbb{R}^3$ as well?

The equation is really from an example. It just says that the domain of $\omega$ = Domain $F_1$ intersect Domain $F_2$ intersect Domain $F_3$. I suspect this is only for $\mathbb{R}^3$.

EDIT: Here's the problem:

"For each of the following differential forms $\omega$ determine if there exists a function $g$ such that $\omega$ = $dg$."

The specific $\omega$ I'm working on is pretty long but it is exactly as I described it with the specific parameters I mentioned.

5. Feb 16, 2013

### Dick

Yes, I think it's implicit that the domain of all three terms is R^3.

6. Feb 16, 2013

### Karnage1993

Oh, okay, that makes everything easier. Thanks for the help!