# Killing Vector on S^2: Solving the Killing Equation

• negru
In summary, the conversation discusses the killing vectors for the two sphere and how to use them to show isometries between the two sphere and SO(3). The killing vectors are R, S, and T, which are calculated using the killing equation. The conversation also touches on the definition of tangent space and how it can be defined using coordinate systems. It is also mentioned that the derivative acts like a unit vector and applying it to coordinates is like doing a dot product.
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 $$\frac{d^2}{d\phi^2}$$ ?

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.

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

dx said:
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?

negru said:
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.

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.

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

Is this right then?

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

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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:

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

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.

## 1. What is a Killing vector on S^2?

A Killing vector on S^2 is a vector field that preserves the metric and thus, the geometry of the 2-dimensional sphere. In other words, it describes a symmetry of the sphere that leaves the distance between points unchanged.

## 2. What is the Killing equation?

The Killing equation is a differential equation that describes the conditions under which a vector field is a Killing vector. In the case of S^2, the Killing equation takes the form of a system of partial differential equations.

## 3. How is the Killing equation solved on S^2?

The Killing equation on S^2 can be solved using a variety of techniques, such as separation of variables or the method of characteristics. The solutions to the Killing equation form a set of vector fields that are Killing vectors on S^2.

## 4. What is the significance of Killing vectors on S^2?

Killing vectors on S^2 play an important role in the study of the geometry of the sphere. They are used to define and characterize isometries, which are transformations that preserve distances and angles on the sphere. They also have applications in physics, particularly in the theory of general relativity.

## 5. Are Killing vectors unique on S^2?

No, there are infinitely many Killing vectors on S^2. However, they are not all independent, as they form a vector space. This means that any linear combination of Killing vectors on S^2 is also a Killing vector. Additionally, there are certain special Killing vectors on S^2, such as the rotational Killing vectors, that have particular significance.

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