Killing vector on S^2

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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] ?
 

Answers and Replies

  • #2
nicksauce
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My guess might be that d/d theta and d/d phi are meant to be the basis one-forms, but i could very well be wrong.
 
  • #3
dx
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d/dtheta and d/dphi are the basis tangent vectors, not one-forms.
 
  • #4
nicksauce
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d/dtheta and d/dphi are the basis tangent vectors, not one-forms.

Yeah... that makes more sense.
 
  • #5
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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
Fredrik
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I am confused by [tex]R=\frac{d}{d\phi}}[/tex] not being a vector?
If you think that [itex]d/d\theta[/itex] isn't a vector, you might want to look up the definition of "tangent space". :smile:

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
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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
Fredrik
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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}{\partial 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.
 
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  • #9
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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?
 
  • #10
Fredrik
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It looks right to me.
 
  • #11
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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
 
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  • #12
Fredrik
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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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