## Killing vector on S^2

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 $$\frac{d^2}{d\phi^2}$$ ?

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 Recognitions: Homework Help Science Advisor 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.
 Blog Entries: 1 Recognitions: Gold Member Homework Help d/dtheta and d/dphi are the basis tangent vectors, not one-forms.

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## Killing vector on S^2

 Quote by dx d/dtheta and d/dphi are the basis tangent vectors, not one-forms.
Yeah... that makes more sense.

 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?

Mentor
 Quote by negru 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 $V^i=V(x^i)$.

$$V(x^i)=V^j\frac{\partial}{\partial x^j}\bigg|_p x^i=V^j(x^i\circ x^{-1}),_j(x(p))=V^j\delta^i_j=V^i$$

 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.
 Mentor 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 $$R^1=R(x^1)=\frac{\partial}{\partial\phi}(\theta)=\frac{\partial}{\parti al x^2}(\theta)=(\theta\circ x^{-1}),_2=0$$ 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.
 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?
 Mentor It looks right to me.
 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
 Mentor 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.

 Tags killing vector, two-sphere