Differential Form: Closed/Exact

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Homework Help Overview

The discussion revolves around a differential form represented as ##\omega = F_1 dx + F_2 dy + F_3 dz##, with a focus on determining whether this form is closed and exact. Participants are examining the implications of the domains of the functions involved and how they relate to the overall domain of the differential form.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand why the domain of ##\omega## is stated as ##\mathbb{R}^2## despite containing three variables. They question whether the function ##g## would also have the same domain. Other participants raise concerns about the interpretation of domains and suggest that the domain of a function can still be ##\mathbb{R}^3## even if not all variables are present in its expression.

Discussion Status

Participants are actively questioning the assumptions regarding the domains of the functions involved. Some guidance has been offered regarding the interpretation of domains, but there is no explicit consensus on the matter. The discussion remains open as participants explore different interpretations and implications.

Contextual Notes

The original problem requires determining if there exists a function ##g## such that ##\omega = dg## for the given differential forms. There is an emphasis on understanding the domains of the functions involved, which is central to the discussion.

Karnage1993
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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)##.

Homework 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
 
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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.
 
Can you also post the text of your problem, so we can judge what you wrote. Thanks!
 
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.
 
Karnage1993 said:
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.

Yes, I think it's implicit that the domain of all three terms is R^3.
 
Oh, okay, that makes everything easier. Thanks for the help!
 

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