@0celo7 I don't understand; the condition on $t^a$ requires that $g_{ab}$ exist to start with otherwise you cannot find the causal vector in the light cone that least deviates in direction from those in a nearby neighborhood so as to provide continuity

you need $g_{ab}$ to start with to define that light cone

Well if you have a Riemannian metric h_ab

@Slereah Yes but please show this

And some defined everywhere continuous vector field

@FenderLesPaul I think HE shows in Ch. 2 somewhere that you can construct a $g$ given a $k$ such that $t$ is timelike. Perhaps if we have a $k$ and a $g$ we can deduce there must be some $t$ that is timelike.

Then you can always define a Lorentzian metric by the metric

h_ab - 2/(h_ab X^a X^b) X_a X_b

@Slereah that's it

@0celo7 hmm I see

ughh I have to present a paper tomorrow

Woops

must stop being lazy

Lemme finish this chapter in my summer book and then I'll talk more

Poor formating

Much better

(3)=(4) is Poincare or something

It is the best theorem of all

(1)=(2) wat

The hairy balls theorem

@Slereah yup

(2)=(3) is what we need @FenderLesPaul

@0celo7 Careful, it doesn't say that every Lorentz metric is time-orientable ;)

"(1) ⇒ (2) (The converse is trivial.) The time orientable double covering ( ̃M, ̃g), satisfies (3) and hence (4). So, the latter is satisfied obviously by M"

@ACuriousMind I know this

@ACuriousMind creeping on GR talks?

creeping lurking

does a double covering have the same Euler char as the manifold or something?

I'll get back to this right after I finish practicing some picking techniques

@Slereah we need (2)=(3)

arxiv.org/pdf/gr-qc/0609119v3.pdf

Creeping is creepy

p. 4

@0celo7 No, but it's multiplicative, I think, so if one is zero, the other also is

@Slereah Link to abstracts, dammit!

Read this as a well meaning answer to the question is hilarious.

"Proposition 2.3. A Lorentzian manifold is time-orientable if and only if it admits a globally defined timelike vector field X (which can be chosen complete 3)."

@DanielSank What the hell is that user doing.

@ACuriousMind I believe the accepted term for this is "derping".

close for WTF

I think he wanted to delete his post but didn't know how to?

@Slereah PROOF

PLEASE

@ACuriousMind how did you revert without going through the pending edit?

20k superpowers

@DanielSank Click rollback in the revision history

If you want the proof, I think I got the original paper of it somewhere

Lemme see

@ACuriousMind That skips the edit queue?

I keep it in my "spacetime topology" folder

@DanielSank Apparently, yes

@ACuriousMind Weird.

Ah, there it is

rollback is a kinda weird functionality, anyway, but for vandalism, it's useful

"Line element fields and Lorentz structures on differentiable manifolds"

By I. Markus

The full proof is in it

It's a paper I should keep in my "Why does nobody like rewriting proofs" folder too

It's the kind of proof that is written exactly once and nobody ever does it again

They just reference everything back to it

@Slereah That sounds like a really useful folder to have

oh wait

That paper references an even earlier paper!

Oh wait!

That's actually a reference to Steenrod!!!

...

I remember now

That was why I bought that book

"As is well known [17, p. 207] the existence of a continuous line element field on M is equivalent to the existence of a differentiable, covariant, symmetric, second order tensor field on M, with signature of one at each point, that is, a Lorentz "metric" on M"

17 is Steenrod

And here's Steenrod :

Topology of Fibre Bundles

that it?

Yep

don't have it

"40.13 : THEOREM. If M is compact and dim M is odd, then M admits a quadratic form of signature 1. If dim M is even, then this holds if the Euler number of M is 0"

That guy didn't give a shit about GR

It seems to be a particular case of a more general theorem

"A compact differentiable manifold admits an everywhere, continuous, non-singular, quadratic form of signature k if and only if it admits a continuous field of tangent k-planes"

Which is itself a reformulation of...

1+0=0!

"The homotopy k' is a deformation retraction of B into B', and for each x it contracts Y_x over itself into Y'_x. Then, for any cross-section f of \mathcal{B}, k' provides a homotopy of f into a cross-section of \mathcal{B}'. Thus \mathcal{B} has a cross-section if and only if \mathcal{B}' has a cross-section"

Obviously

It's in the chapter "cohomology theory of bundles"

And I don't know shit about cohomologies

Perhaps@ACuriousMind might help!

Considering "cohomology" does not appear once in Wald, I think that book is going too far.

Although I am intrigued by it...it seems to be the canonical reference for a lot of this stuff.

Well you have a formula to turn a Riemannian metric into a Lorentzian one

Play around with it

But it is nice to know that the theorem is a special case of something cohomological :)

Ok, we have that formula.

I'll think about this later.

The vector field X will roughly correspond to the notion of a timelike vector

Like you can notice that if you input X in it, it will always be negative

But if it is orthogonal to X, it will be positive

But how do we use time-orientability here?

@Slereah Have you looked at Lemma 8.1.1 in Wald?

Well it's almost 2 AM on a monday, so no

Time orientability is related to the orientability of the line element?

It's in the paper I mentionned

I probably don't have access

jstor.org/stable/1970071

apparently

I probably still have it somewhere

I need an account there

dl.dropboxusercontent.com/u/19940612/…

There you go

I also have a paper called "Why the universe cannot be S4" if you want

@Slereah awesome

@Slereah cool, thanks

dl.dropboxusercontent.com/u/19940612/…

@Slereah oh god that's full on algebraic topology

Well it doesn't matter too much I suppose

The point is that all spacetimes of any topology can be made time orientable

By taking the double cover