What are the de Sitter Killing vectors?

Question

I'm trying to find the Killing vectors for de Sitter space using global coordinates, i.e. the spherical foliation. In the end, I want to know how to perform a boost on the 1+1 and 3+1 de Sitter manifolds so that I can make two points lie at the same spatial position (the spatial origin). I know how to perform a Lorentz transformation in Minkowski, but not for curved spacetimes.

Two things I've tried are re-deriving the Lorentz transformations for a curved space using the steps on the Wikipedia page for Lorentz transformations, as well as performing a regular Lorentz transformation on the hyperboloid embedded in a higher dimension.

Any hints?

Answer 1

To assist with your problem, I will describe the general approach, when we don't necessarily know what we're looking for. As you are aware, finding the Killing vectors $X^\mu$ requires solving,

$$\nabla_\mu X_\nu + \nabla_\nu X_\mu = 0$$

which is an over-determined system of differential equations. As you know, these Killing vectors are precisely those for which $\mathcal L_X g = 0$ and are the infinitesimal generators of isometries of the manifold being described. If you split them into a linearly independent set, $\{\xi_i\}$ you can work out the Lie algebra they generate and sometimes may be able to identify what they are physically.

Alternatively, one can explicitly find the finite form of the transformations they generate by solving the equation for the integral curves of the fields, that is,

$$\frac{d x^\mu}{d\lambda} = X^\mu(x)$$

where $x^\mu(\lambda)$ parametrises the integral curve, that is, you can think of it as an embedding function and $X^\mu(x)$ is the Killing field, seen as a function of the components $x^\mu$.

Choosing the right coordinates often makes the calculation simpler; it is certainly desirable to have the metric be diagonal, and well-behaved at as many points as possible.