Killing vectors for the product of Anti de Sitter and two-sphere

Question

I am a little confused.

In an exercise, I am asked to find all Killing vectors of the product of the Anti de Sitter and two-sphere metric, given by: $$\text{d}s^2=\frac{-\text{d}t^2+\text{d}y^2}{y^2}+\text{d}\theta^2+\sin^2(\theta)\text{d}\phi^2.$$

First of all, I know that the two-sphere as well as the Anti de Sitter space are maximally symmetric. Is it then true that their product is still maximally symmetric? Can we then deduce that there should be exactly 4 independent Killing vectors?

Secondly, I know that the metric is independent of $t$ and $\phi$, so two Killing vectors are easily found. But is there an easy way to find the others?

Answer 1

Generally, the Ricci tensor of a maximally symmetric geometry is proportional to its metric tensor. In this case, the Ricci tensor is given by

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad R_{\mu \nu}= \begin{bmatrix} 1/y^2 & 0 & 0 &0 \\ 0& -1/y^2 & 0 & 0 \\ 0&0&1&0 \\ 0&0&0& (\sin{\theta})^{2}. \end{bmatrix} $

This is not proportional to the given metric, so your space can't be maximally symmetric.

The number of Killing vectors can be deduced relatively straightforwardly in the case of a product metric. This is given by the sum of the dimension of the isometry group of each factor. In your example, each of $AdS_{2}$ and $S^{2}$ has three independent Killing vectors, leading to six independent Killing vectors for their product. We again conclude that this space cannot be maximally symmetric since those have $10$ such vectors in four dimensions.