I About global inertial frame in GR - revisited

[cianfa72]

In almost all such cases, there is only one such KVF. The only example that comes to mind where there is more than one is flat Minkowski spacetime.

You mean in almost such cases (i.e. spacetime types) in which there are more than one timelike KVF, only one of them is hypersurface orthogonal.

 

[PeterDonis]

You mean in almost such cases (i.e. spacetime types) in which there are more than one timelike KVF, only one of them is hypersurface orthogonal.

No. I mean that in almost all cases where there is a timelike KVF at all in the spacetime, there is only one such KVF. And in most cases where there is a timelike KVF, it is not hypersurface orthogonal--i.e., most stationary spacetimes are not static.

 

[PeterDonis]

Sorry for my poor english: what do you mean with "where there is a timelike KVF at all in the spacetime" ?

If a timelike KVF exists.

 

[cianfa72]

No. I mean that in almost all cases where there is a timelike KVF at all in the spacetime, there is only one such KVF.

So, in almost all cases when a timelike KVF exists, there is actually just/only one. Then it may or may not be hypersurface orthogonal.

 

[PeterDonis]

So, in almost all cases when a timelike KVF exists, there is actually just/only one. Then it may or may not be hypersurface orthogonal.

Yes.

 

[cianfa72]

Another point related to the Frobenius condition in the covector field/differential one-form formulation.

Wald in appendix B.3 claims:

Let $T^*$ a smooth specification of 1-dimensional subspace of one-forms. Then the associated $n-1$ dimensional subspace $W$ of the tangent space at each point admits integral submanifolds if and only if for all $\omega \in T^*$ we have $d\omega = \sum_{\alpha} \mu^{\alpha} \wedge v^{\alpha}$, where each $\mu^{\alpha} \in T^*$

Now each $\mu^{\alpha}$ and $v^{\alpha}$ should be covector field/one-forms. So why they are written like vectors -- i.e. with an upper index instead of a lower index ?

 

[Orodruin]

Another point related to the Frobenius condition in the covector field/differential one-form formulation.

Wald in appendix B.3 claims:Now each $\mu^{\alpha}$ and $v^{\alpha}$ should be covector field/one-forms. So why they are written like vectors -- i.e. with an upper index instead of a lower index ?

$\alpha$ enumerates the one-form. It is not a one-form index.

 

[cianfa72]

$\alpha$ enumerates the one-form. It is not a one-form index.

So in that specific case we've just one one-form, therefore the Frobenius's condition becomes $d\omega=\omega \wedge v$ for a one-form $v$. This condition is equivalent to $\omega \wedge d\omega = 0$.