Linearly Independent Killing Fields in n-D Manifold

[Cexy]

A question on a General Relativity exam that I have asks how many linearly independent Killing fields there can be in an n-dimensional manifold. I'm sure I've seen this question before and I think that the answer is n(n+1)/2, but I can't remember why!

Any help?

 

[George Jones]

A question on a General Relativity exam that I have asks how many linearly independent Killing fields there can be in an n-dimensional manifold. I'm sure I've seen this question before and I think that the answer is n(n+1)/2, but I can't remember why!

Any help?

The reason might might be something like this.

Killing's equation is anti-symmetric in 2 indices, which gives

<br /> \left(<br /> \begin{array}{cc}<br /> n\\<br /> 2<br /> \end{array}<br /> \right)<br />

degrees of freedom, plus n initial degrees of freedom, one for each component of a killing vector.

Regards,
George

 

[Timbuqtu]

Another way of understanding this number is considering the symmetries belonging to the Killing fields.

The simplest example of a maximally symmetric (n dim.) manifold is just Euclidian \mathbb{R}^n (or Minkowski space in case of Lorentzian metric). Independent symmetries are given by the translations and rotations. There are n independent translations and you can convince yourself that there are n(n-1)/2 independent rotations (the dimension of SO(4)). So the total number of independent symmetries is n+n(n-1)/2=n(n+1)/2.

So any (Riemannian) manifold can have at most n(n+1)/2 symmetries, and at most the same number of independent Killing fields.

-- Timbuqtu

 

[selfAdjoint]

Another way of understanding this number is considering the symmetries belonging to the Killing fields.

The simplest example of a maximally symmetric (n dim.) manifold is just Euclidian \mathbb{R}^n (or Minkowski space in case of Lorentzian metric). Independent symmetries are given by the translations and rotations. There are n independent translations and you can convince yourself that there are n(n-1)/2 independent rotations (the dimension of SO(4)). So the total number of independent symmetries is n+n(n-1)/2=n(n+1)/2.

So any (Riemannian) manifold can have at most n(n+1)/2 symmetries, and at most the same number of independent Killing fields.

-- Timbuqtu

This is excellent. My only gripe is using the term "rotation" for an element of SO(n). Isometry would be better.

 

[Cexy]

Okay, thanks for the help guys. I suppose the question to ask now is: Why do Killing vectors have a one-to-one correspondence to spacetime symmetries?

 

[Timbuqtu]

The killing equation \nabla_{(\mu}k_{\nu)}=0 is equivalent to the vanishing of the Lie-derivative of the metric along k, \mathcal{L}_k g_{\mu\nu}=0. This last equation actually just tells you that the metric does not change if you pull it along the vector field k, so k represents a symmetry of the metric (and therefore a symmetry of the space time).

-- Timbuqtu

 

[Cexy]

Ah, fantastic. Cheers. :)