Killing fields commute if they are asymptotically coordinate?

[ergospherical]

Some axially symmetric star has two independent KVFs, $T$ and $\Phi$. We don't know the expressions for these at all points -- the only thing we know is that as $r^2 + z^2 \rightarrow \infty$, that $T \rightarrow \partial/\partial t$ and $\Phi \rightarrow \partial/\partial \phi$. The question is to show whether $T$ and $\Phi$ commute, everywhere.

We can show that the commutator of two KVFs is itself a KVF. So let $X = [T, \Phi]$, and clearly $X = 0$ at infinity. We can also show that KVFs satisfy $\nabla_a \nabla_b X_c = -R_{bcad} \xi^d$, from which we know that if you specify $\xi_a$ and $\nabla_a \xi_b$ at any point, then this specifies $\xi_a$ and $\nabla_a \xi_b$ everywhere (i.e. we know $u^a \nabla_a \xi_b$ and also $u^a \nabla_a (\nabla_b \xi_c) = -R_{bcad} \xi^d u^a$).

So the last bit is just to show that $\nabla_a X_b = 0$ at infinity. Maybe I'm being thick - I would expect that if $X \rightarrow 0$ asymptotically then also $\nabla X \rightarrow 0$. Are there some constraints on this?

 

[PeterDonis]

Can you find a chart in which $T = \partial / \partial t$ and $\Phi = \partial / \partial \phi$ everywhere? If the two KVFs commute in that chart, they would have to commute in any chart, since any valid coordinate transformation will have to preserve commutators of vector fields.

 

[ergospherical]

No, or at least it cannot be assumed. We only know that they take this form in the asymptotic limit.

 

[PeterDonis]

it cannot be assumed

I'm not suggesting that it be assumed. Isn't it already known to be true in, for example, the Boyer-Lindquist chart?

 

[PeterDonis]

We only know that they take this form in the asymptotic limit.

The relevant properties in the asymptotic limit are that $T$ has unit norm in that limit and $\Phi$ is orthogonal to $T$ in that limit. That is different from them being coordinate basis vector fields only in that limit.

 

[ergospherical]

It could be something like a rotating star with some small perturbations. So you cannot come up with exact expressions for the KVF at finite distance. But you know the asymptotic behaviour.

The question is, can you figure out whether the commutator is zero at finite distance, using just this information…

 

[PeterDonis]

you cannot come up with exact expressions for the KVF at finite distance

True, but you have specified that the spacetime is stationary and axisymmetric. You have also implicitly specified that it is asymptotically flat. Wald, in section 7.1, defines "stationary and axisymmetric" to include the property that $\Phi$ commutes with $T$. (Wald then goes on to show how coordinates adapted to the two commuting KVFs can be derived--what he ends up with, if further restricted to the asymptotically flat case, is basically Boyer-Lindquist coordinates.) So really I think the question you are asking is whether it is possible to have an asymptotically flat spacetime that has two KVFs $T$ and $\Phi$ with the appropriate properties that do not commute everywhere; by Wald's definition such a spacetime would not be properly called stationary and axisymmetric.

 

[PeterDonis]

the last bit is just to show that $\nabla_a X_b = 0$ at infinity. Maybe I'm being thick - I would expect that if $X \rightarrow 0$ asymptotically then also $\nabla X \rightarrow 0$.

I don't know if this is true for all KVFs, but we know $X$ is a commutator of two KVFs whose asymptotic behavior is known, which makes things simpler:

$$
\nabla X = (\nabla T) \Phi - (\nabla \Phi) T
$$

So if $\nabla T \to 0$ and $\nabla \Phi \to 0$ in the asymptotic limit, that would be sufficient to show that $\nabla X \to 0$.