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Killing vector on S^2 
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#1
Feb1409, 08:40 PM

P: 308

Hi, I'm trying to understand isometries, for example between S^2 (two sphere) and SO(3).
For this I need to show that the killing vectors for S^2 [tex]ds^2={d\theta}^2+sin^2 {\theta} {d\phi}^2.[/tex] are: [tex]R=\frac{d}{d\phi}}[/tex] [tex]S=cos {\phi} \frac{d}{d\theta}}cot{\theta} sin {\phi} \frac{d}{d\phi}} [/tex] [tex]T=sin {\phi} \frac{d}{d\theta}}cot{\theta} cos {\phi} \frac{d}{d\phi}} [/tex] I'm not sure how to use the Killing equation, basically because I am confused by [tex]R=\frac{d}{d\phi}}[/tex] not being a vector? How do I calculate the comma derivative of R then? I suppose I could convert to cartesian coordinates or something, but there has to be a direct way. I can get that some components of the Christoffel symbol are [tex]cot{\theta}[/tex] and [tex]sin{\theta}cos{\theta}[/tex] and others zero, but next what are [tex]\frac{dR_a}{dx^b}[/tex]? And [tex]{\Gamma}^k_a_b{R_k}[/tex] for that matter. Is [tex]\frac{dR_1}{dx^2}[/tex] just equal to [tex]\frac{d^2}{d\phi^2}[/tex] ? 


#2
Feb1409, 11:06 PM

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My guess might be that d/d theta and d/d phi are meant to be the basis oneforms, but i could very well be wrong.



#3
Feb1509, 06:38 AM

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PF Gold
P: 1,961

d/dtheta and d/dphi are the basis tangent vectors, not oneforms.



#4
Feb1509, 09:42 AM

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Killing vector on S^2



#5
Feb1509, 09:46 AM

P: 308

Ok, so now what? Is [tex]\frac{dR_1}{dx^2}=\frac{d^2}{d\phi^2}[/tex] correct then? What do I do with something that looks like
[tex]\frac{d^2}{d\phi^2}+cot{\theta}\frac{d}{d\phi}[/tex] ? Probably not solve it, since I'm supposed to check that those are killing vectors, not find the geodesic? Or is that the only way? 


#6
Feb1509, 10:24 AM

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PF Gold
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The components of [itex]V\in T_pM[/itex] in a coordinate system [itex]x:M\rightarrow\mathbb R^n[/itex] are [itex]V^i=V(x^i)[/itex]. [tex]V(x^i)=V^j\frac{\partial}{\partial x^j}\bigg_p x^i=V^j(x^i\circ x^{1}),_j(x(p))=V^j\delta^i_j=V^i[/tex] 


#7
Feb1509, 10:58 AM

P: 308

I'm not sure I got what you mean. What is [itex]R(x^1)=R({\theta})[/itex] then? I guess my main problem is that I don't have clear definitions of all these notations.



#8
Feb1509, 01:22 PM

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PF Gold
P: 9,226

x would be the function that takes a point p on the sphere to the pair [itex](\theta(p),\phi(p))[/itex], so we can write [itex]x^1=\theta[/itex] as you did, and
[tex]R^1=R(x^1)=\frac{\partial}{\partial\phi}(\theta)=\frac{\partial}{\parti al x^2}(\theta)=(\theta\circ x^{1}),_2=0[/tex] Note that [itex]\theta\circ x^{1}[/itex] is the map that takes [itex](\theta(p),\phi(p))=x(p)[/itex] to [itex]\theta(p)[/itex]. That means it's just the map [itex](x,y)\mapsto x[/itex] (here x is just a number, not a coordinate system), and the partial derivative of that with respect to the second variable is of course 0. 


#9
Feb1509, 01:31 PM

P: 308

But R looks like an operator, which acts on something, not like a function of something. Is the argument of an operator just the function you apply it to?
[tex]R^1=R(x^1)=R({\theta})=\frac{d}{d\phi}{\theta}=0 ?[/tex] [tex]S^1=cos {\phi} \frac{d}{d\theta}}{\theta}cot{\theta} sin {\phi} \frac{d}{d\phi}}{\theta}=cos{\phi} [/tex] [tex]S^2=cos {\phi} \frac{d}{d\theta}}{\phi}cot{\theta} sin {\phi} \frac{d}{d\phi}}{\phi}=cot{\theta}sin{\phi} [/tex] Is this right then? 


#11
Feb1509, 01:42 PM

P: 308

Ok, I got it now, thanks! Interesting, so the derivative basically acts like a unit vector, and applying it to the coordinates is like doing a dot product. Never thought of it that way



#12
Feb1509, 01:57 PM

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PF Gold
P: 9,226

The tangent space can be defined as the vector space that's spanned by the partial derivative operators on the manifold, which are defined using a coordinate system:
[tex]\frac{\partial}{\partial x^i}\bigg_p f=(f\circ x^{1}),_i(x(p))[/tex] It can also be defined in a coordinate independent way (see e.g. the GR book by Wald), but then you can prove that these operators are basis vectors of that space. You get a different basis for each coordinate system of course, but you can calculate the relationship between them using the chain rule. 


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