Is Schwarzschild spacetime parallelizable?

[TrickyDicky]

I was wondering since it is usually foliated into 2-spheres and these are not themselves parallelizable(only the n-spheres S1, S3 and S7 are).

I know the timelike Killing vector field is not global, but is the a global basis of vector fields in Schwarzschild spacetime? I mean a basis in a Lorentzian manifold needs to include a timelike vector field and three spacelike vector fields?

 

[PeterDonis]

I know the timelike Killing vector field is not global

More precisely, the KVF usually denoted $\partial / \partial t$ is globally defined, but it is only timelike outside the horizon (on the horizon it is null, inside the horizon it is spacelike). So it can't serve to define a timelike basis vector everywhere.

is the a global basis of vector fields in Schwarzschild spacetime? I mean a basis in a Lorentzian manifold needs to include a timelike vector field and three spacelike vector fields?

Yes, there are frame fields (meaning, sets of one timelike and three spacelike unit vectors) that are defined everywhere in Schwarzschild spacetime. One obvious one is the frame field defined using Kruskal coordinates, since that chart has one timelike and three spacelike coordinates everywhere. Another is the "Painleve basis", i.e., the frame field defined by Painleve observers (i.e., the timelike basis vector is the 4-velocity of the Painleve observer at each event; the 3 spacelike vectors are the radial and tangential unit vectors orthogonal to the 4-velocity). Technically, the latter is not defined "globally" on the maximally extended Schwarzschild spacetime; it's only defined in the exterior and future interior regions (i.e., inside the black hole). But for practical purposes that's good enough.

 

[TrickyDicky]

Yes, there are frame fields (meaning, sets of one timelike and three spacelike unit vectors) that are defined everywhere in Schwarzschild spacetime. One obvious one is the frame field defined using Kruskal coordinates, since that chart has one timelike and three spacelike coordinates everywhere.

Are you sure you can define frame fields from the Kruskal coordinate basis?
Can the V coordinate be considered of timelike character globally, including the event horizon null hypersurface and the region inside?

 

[PeterDonis]

Are you sure you can define frame fields from the Kruskal coordinate basis?
Can the V coordinate be considered of timelike character globally, including the event horizon null hypersurface and the region inside?

Yes, certainly. That's obvious from the line element:

$$
ds^2 = \frac{32 M^3}{r} e^{- r / 2M} \left( - dV^2 + dU^2 \right) + r^2 \left( d \theta^2 + \sin^2 \theta d \phi^2 \right)
$$

A curve where only $dV$ is nonzero is obviously timelike. Equivalently, the vector field $\partial / \partial V$ is obviously timelike everywhere.

 

[TrickyDicky]

Yes, certainly. That's obvious from the line element:

$$
ds^2 = \frac{32 M^3}{r} e^{- r / 2M} \left( - dV^2 + dU^2 \right) + r^2 \left( d \theta^2 + \sin^2 \theta d \phi^2 \right)
$$

A curve where only $dV$ is nonzero is obviously timelike. Equivalently, the vector field $\partial / \partial V$ is obviously timelike everywhere.

Yes, and also all thru the event horizon hypersurface which is null, U=V so there it is not clear for me how a timelike and a spacelike vector fields can exist there.
As I said in my first post each point in the Kruskal chart is a 2-sphere, which can't be covered just with one ø, θ chart, so in four dimensions to each point of the Kruskal chart it corresponds a point where you can't have global basis with U,V,ø,θ. That's why I tend to think the spacetime is not parallelizable.

 

[PeterDonis]

all thru the event horizon hypersurface which is null, U=V so there it is not clear for me how a timelike and a spacelike vector fields can exist there.

Huh? The horizon is just the line $U = V$. There is no requirement that all vector fields at that line point along the line. Along that line, $\partial / \partial V$ is timelike, and $\partial / \partial U$ is spacelike, just like everywhere else. Again, that's obvious from looking at the line element; along the horizon $r = 2M$, but all that does is fix the constant in front of $\left( - dV^2 + dU^2 \right)$; it doesn't change the timelike or spacelike character of $V$ or $U$. (The tangent vector field to the horizon is $\partial / \partial V + \partial / \partial U$, which is indeed null; but that's not the same vector field as either $\partial / \partial V$ or $\partial / \partial U$.)

As I said in my first post each point in the Kruskal chart is a 2-sphere, which can't be covered just with one ø, θ chart, so in four dimensions to each point of the Kruskal chart it corresponds a point where you can't have global basis with U,V,ø,θ. That's why I tend to think the spacetime is not parallelizable.

Yes, you're right, if you want to be precise, as the coordinates are usually defined, $\partial / \partial \phi$ is not defined on the "axis" points of each 2-sphere; this reflects the fact that, as you say, you cannot define an everywhere non-vanishing vector field on $S^2$.

However, this doesn't prevent you from defining a timelike vector field that covers the entirety of each 2-sphere, because the 2-spheres are all spacelike, so a timelike vector field can be nonzero everywhere on each 2-sphere as long as it is orthogonal to the 2-sphere. The vector field $\partial / \partial V$ in Kruskal coordinates meets that requirement. (So does $\partial / \partial U$, which is therefore spacelike and nonvanishing everywhere on each 2-sphere. In other words, what you can't define is a spacelike vector field that is everywhere nonvanishing and *tangent* to a 2-sphere.)

So you're technically correct that the Kruskal chart can't define a full 4-D frame field on Schwarzschild spacetime; it can only define a 2-D frame field in the $U - V$ plane. The 4-D frame field it defines will have one undefined spacelike vector at two points on each 2-sphere. (This is true for all the standard charts on Schwarzschild spacetime, and nobody seems to be bothered by it.) But the timelike vector of the frame field will be well-defined everywhere.

 

[TrickyDicky]

Thanks, Peter, you are totally right.
I just wanted to confirm this on the Schwarzschild spacetime, and do it for the right reasons.

I was trying to relate this with the assertion frequently made that an observer near the event horizon doesn't have any local physical observation to discern what side of the event horizon he is at.