# Killing vector on S^2

1. Feb 14, 2009

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

2. Feb 14, 2009

### 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.

3. Feb 15, 2009

### dx

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

4. Feb 15, 2009

### nicksauce

Yeah... that makes more sense.

5. Feb 15, 2009

### 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?

6. Feb 15, 2009

### Fredrik

Staff Emeritus
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 $V^i=V(x^i)$.

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

7. Feb 15, 2009

### 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.

8. Feb 15, 2009

### Fredrik

Staff Emeritus
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.

Last edited: Feb 15, 2009
9. Feb 15, 2009

### 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?

10. Feb 15, 2009

### Fredrik

Staff Emeritus
It looks right to me.

11. Feb 15, 2009

### 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

Last edited: Feb 15, 2009
12. Feb 15, 2009

### Fredrik

Staff Emeritus
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.