Killing Vector on S^2: Solving the Killing Equation

[negru]

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

ds^2={d\theta}^2+sin^2 {\theta} {d\phi}^2.

are:

R=\frac{d}{d\phi}}
S=cos {\phi} \frac{d}{d\theta}}-cot{\theta} sin {\phi} \frac{d}{d\phi}}
T=-sin {\phi} \frac{d}{d\theta}}-cot{\theta} cos {\phi} \frac{d}{d\phi}}

I'm not sure how to use the Killing equation, basically because I am confused by R=\frac{d}{d\phi}} 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 cot{\theta} and sin{\theta}cos{\theta} and others zero, but next what are \frac{dR_a}{dx^b}? And {\Gamma}^k_a_b{R_k} for that matter.
Is \frac{dR_1}{dx^2} just equal to \frac{d^2}{d\phi^2} ?

 

[nicksauce]

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.

 

[dx]

d/dtheta and d/dphi are the basis tangent vectors, not one-forms.

 

[nicksauce]

d/dtheta and d/dphi are the basis tangent vectors, not one-forms.

Yeah... that makes more sense.

 

[negru]

Ok, so now what? Is \frac{dR_1}{dx^2}=\frac{d^2}{d\phi^2} correct then? What do I do with something that looks like
\frac{d^2}{d\phi^2}+cot{\theta}\frac{d}{d\phi} ? 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?

 

[Fredrik]

I am confused by R=\frac{d}{d\phi}} not being a vector?

If you think that d/d\theta isn't a vector, you might want to look up the definition of "tangent space".

The components of V\in T_pM in a coordinate system x:M\rightarrow\mathbb R^n are V^i=V(x^i).

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

 

[negru]

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

 

[Fredrik]

x would be the function that takes a point p on the sphere to the pair (\theta(p),\phi(p)), so we can write x^1=\theta as you did, and

R^1=R(x^1)=\frac{\partial}{\partial\phi}(\theta)=\frac{\partial}{\partial x^2}(\theta)=(\theta\circ x^{-1}),_2=0

Note that \theta\circ x^{-1} is the map that takes (\theta(p),\phi(p))=x(p) to \theta(p). That means it's just the map (x,y)\mapsto x (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.

 

[negru]

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?
R^1=R(x^1)=R({\theta})=\frac{d}{d\phi}{\theta}=0 ?
S^1=cos {\phi} \frac{d}{d\theta}}{\theta}-cot{\theta} sin {\phi} \frac{d}{d\phi}}{\theta}=cos{\phi}
S^2=cos {\phi} \frac{d}{d\theta}}{\phi}-cot{\theta} sin {\phi} \frac{d}{d\phi}}{\phi}=-cot{\theta}sin{\phi}

Is this right then?

 

[Fredrik]

It looks right to me.

 

[negru]

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

 

[Fredrik]

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:

\frac{\partial}{\partial x^i}\bigg|_p f=(f\circ x^{-1}),_i(x(p))

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.