Andrew Kennedy Atkins	Posted on Thursday, 13 May, 2004 - 09:01 pm: Points in a punctured plane, $R^2(0,0)$ are specified by their polar coordinates $(r,\theta )$. What are the necessary conditions for the differential $w=A(r,\theta )\mathrm{dr}+B(r,\theta )d\theta$ to be exact? Thanks
Matthew Buckley	Posted on Friday, 14 May, 2004 - 08:52 am: Andrew, Have a look at this . Matt.
Kerwin Hui	Posted on Friday, 14 May, 2004 - 08:57 am: Assume w is smooth (you can probably get away with C1 ). If you just want a necessary condition - that is easy. The exterior derivative of the 1-form w must be zero, i.e. w is closed. More explicitly, . If you want the necessary and sufficient condition, you need to work harder (check - dq is not exact. Why?). Heuristically, if w=df and g is a closed curve, then by Stoke's theorem we expect and this, together with the requirement w is exact, is the integrability criterion. There are various ways of proving this. You can try to write and prove the integrability criterion with a lot of (heavy) machineary from Fourier analysis. An alternative, which comes with a health warning [useful prerequisite: a course on algebraic topology] is the following: The fundamental group of X=2 -(0,0) is , the free group generated by the homotopy class of the usual embedding of S1 as r=1,q goes from 0 to 2p [Note that there is a smooth retract of X to the circle r=1]. Hence the first singular homology group is . Once again the generator can be thought of as the usual S1 (choose a suitable triangulation). The first de Rham cohomology group, defined as is, by de Rham's theorem, isomorphic to , the singular cohomology group with real coefficients, and the isomorphism is given by the de Rham map Thus the necessary and sufficient condition for w to be exact is that dw=0 and the integral along S1 of the 1-form w is zero (by de Rham theorem), i.e. and . Kerwin
Andrew Kennedy Atkins	Posted on Friday, 14 May, 2004 - 02:30 pm: Ok, thank you to both of you. I was wondering if it would make any difference if the plane is punctured. So, it doesn't make any difference for necessary condition, but it does for sufficient condition. Thanks a lot!