Timeline for Exact 1- and 2-forms in $R^n$
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 27, 2012 at 8:03 | comment | added | Buschi Sergio | THe space $\mathbb{R}\setminus${$0$} isnt simply connected en.wikipedia.org/wiki/Simply_connected_space. FOr closed 1-form the proof is the usual, you define the potential function $P$ by a line integral from a fixed point $P_0$ to $P$ (and this integral depend only from the $P$). ABout the 2-form, I think that your claim is false, need a n-connected condiction | |
| May 27, 2012 at 4:35 | answer | added | Igor Rivin | timeline score: 1 | |
| May 27, 2012 at 1:10 | comment | added | Misha | If your intention is to teach this to undergraduates, just tell them that the primitive of a closed 1-form $\omega$ is obtained by integrating $\omega$ along paths starting from some point $p_0\in D$. Then explain why integral is independent of the choice of the path connecting $p_0$ to $p$. This, in view of simple connectivity assumption on $D$ that you are making, amounts to verifying path-independence when $D$ is the square, which, I presume, you already know how to do. This is how one proves that irrotational vector fields are conservative in a vector calculus class. | |
| May 26, 2012 at 19:51 | comment | added | Lee Mosher | Well, what I would say is any book on de Rham cohomology --- I like Bott-Tu --- which might not be what you have in mind. However, if you already have a proof in mind for the case of open sets in $R^2$ and $R^3$, the exact same proof should work in higher dimensions. | |
| May 26, 2012 at 19:21 | comment | added | Alan Macdonald | Thank you. Do you have a reference? | |
| May 26, 2012 at 19:15 | comment | added | Lee Mosher | "Simply connected" still implies that closed 1-forms are exact, independent of dimension. | |
| May 26, 2012 at 19:10 | comment | added | Alan Macdonald | I am aware of de Rham cohomology. But I asked for something "suitable for the elementary vector calculus course". Thanks for the tip about math.stackexchange.com. I'll try there if I don't get an answer here within a couple of days. AM. | |
| May 26, 2012 at 18:42 | comment | added | Jyrki Lahtonen | Search for more information on de Rham cohomology. I'm not sure, but it may also be that your question would be more on-topic at math.stackexchange.com ? | |
| May 26, 2012 at 18:42 | answer | added | Charles Matthews | timeline score: 1 | |
| May 26, 2012 at 18:31 | history | asked | Alan Macdonald | CC BY-SA 3.0 |