>Green's theorem only applies to "simply-connected" regions
So when I take the line integral of a the unit circle according to vector F, I get 2pi. Is this because the curl leaks out of the origin where F is not defined?
>>8169650
You'll learn about that when you get to know the generalized Stokes theorem and complex analysis. It's got something to do with poles and residues. You can imagine it like some "source" or "charge" being located at that singularity, so that some "force field" goes through the region.
>>8169650
The vector field in the pic is Hodge dual to (a representative of) the generator of the first de Rham cohomology group of the punctured plane. Since this group is non-zero, Poincare duality for compactly supported cohomology easily implies that the cap product with the generator of the first homology of the same region is non-zero.
>>8169731
Why do people do this ?
>>8169731
I honestly wish I could strangle people like you.
>>8169731
>babby undergrad is proud he learned about deRham cohomology for the first time
>>8169769
OP here. I'm pretty sure he's just trying to disguise the fact that 11th-dimensional space pirates are injecting extra curl through the origin. This is why Green's Theorem requires working around holes in the region, because undefined points open portals to space pirates.
>>8169731
>didn't explaine whether or not curl is leaking out of the origin.
Wew lad
>>8169650
most of that notation is retarded, done on purpose to confuse students
>>8169836
>most of that notation is retarded, done on purpose to confuse students
Welp it's working.
>>8169731
kek'd
>>8169650
It depends entirely on the residue of the undefined point in F, but sure, if that's how your mind wants to think about it.
>>8169731
Oh shit, I actually understand this. I remember now. With that function F, the point (0,0) fails Green's theorem and you need to make a circle in the middle that is clockwise and radius approaching 0. So the circulation isn't the same as if it was one complete region with no break points.
>>8169783
The deRham Complex, [math] 0 \to {\Omega ^0}\left( M \right) \to {\Omega ^1}\left( M \right) \to {\Omega ^2}\left( M \right) \to {\Omega ^3}\left( M \right) \to 0[/math] ,
is equivalent to the classical vector interpretation
[math]0 \to {C^\infty }\left( M \right)\mathop \to \limits^{\operatorname{grad} } \mathcal{V}\left( M \right)\mathop \to \limits^{\operatorname{curl} } \mathcal{V}\left( M \right)\mathop \to \limits^{\operatorname{div} } {C^\infty }\left( M \right) \to 0[/math]
.
So the first deRham Cohomology [math] H_{\operatorname{d} R}^1\left( M \right) = \frac{{\ker \left( {\operatorname{d} :{\Omega ^1}M \to {\Omega ^2}M} \right)}}{{\operatorname{im} \left( {\operatorname{d} :{\Omega ^0}M \to {\Omega ^1}M} \right)}}[/math]
Is analogous to [math] H_{\operatorname{d} R}^1\left( M \right) = \frac{{\left\{ {{\text{fields}}\;{\text{with}}\;{\text{zero}}\;\operatorname{curl} } \right\}}}{{\left\{ {{\text{gradient}}\;{\text{fields}}} \right\}}}[/math]
So let [math]M = {\mathbb{R}^2}\backslash \left\{ 0 \right\}[/math].
The field [math] {\mathbf{F}} = \left( {\frac{{ - y}}{{{x^2} + {y^2}}},\frac{x}{{{x^2} + {y^2}}}} \right) [/math] has zero curl but is not a gradient field.
So it generates the group [math] H_{\operatorname{d} R}^1\left( {{\mathbb{R}^2}\backslash \left\{ 0 \right\}} \right)[/math] .
