- #71
cianfa72
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ok, the latter part of post #69 is in fact consistent to assume the function ##h## to be ##h>0## everywhere (as done from Sachs and Wu in their definition of synchronizable congruence).
Yes, for 1-forms this local equivalence holds.cianfa72 said:BTW I was thinking that their definition of locally proper time synchronizable (i.e. ##d\xi=0##) is actually the same as the existence *locally* of a function f such that ##\xi= df## -- in force of Poincaré lemma.
No, they can't. The poles are not "level sets"; they're single points. Single points won't work as "level sets" since each point in a "level set" has to have an open neighborhood that is also in the "level set".cianfa72 said:Consider the set of latitude circles. They can be described as the level sets of a global function ##g## defined on the sphere
The one you defined does not work, as above. I think the sphere in general does not work for this kind of global construction because it's a compact manifold.cianfa72 said:I believe the problem is that such globally defined function ##g## actually does not exist.
Yes, this is another reason why the sphere would not work for this kind of global construction. One can still do it on an open subset of the sphere, as @PAllen said in the thread you linked to.cianfa72 said:If there was then ##\omega=dg## would define a smooth unit vector field ##Q## such that ##\omega(Q)=-1## not vanishing at any point. We know it cannot exist -- see Hair ball theorem.
Ah ok, so this is not actually a counterexample. The claim that if exists a smooth global function ##g## such that ##\omega = dg## (i.e. a unit vector field ##Q## defined from ##\omega(Q)=-1##) then there is a *global* coordinate chart such that ##g_{00}=1## and ##g_{0\alpha}=0## holds true.PeterDonis said:No, they can't. The poles are not "level sets"; they're single points. Single points won't work as "level sets" since each point in a "level set" has to have an open neighborhood that is also in the "level set".
Correct.cianfa72 said:this is not actually a counterexample.
I went back and looked at this again, and I don't think my previous comment, quoted above, was correct.PeterDonis said:Formally, ##d \alpha = \beta \wedge \omega## is another possibility to make the first term vanish, yes. However, I believe it is ruled out because continuing to apply ##dd = 0## leads to an infinite regress.
ok, I believe it is locally the expected result.PeterDonis said:In short, there is no way to have ##d\alpha = \beta \wedge \omega \neq 0## without leading to a contradiction.
Any closed 1-form is locally exact, yes. This result does not generalize to higher rank forms, though.cianfa72 said:By Poincarè lemma locally any closed form is even exact hence *locally*
This only follows from the Poincare lemma if we can prove that ##d \alpha = 0##, where ##d \omega = \alpha \wedge \omega##. To do that, we have to rule out the possibility that ##d \alpha = \beta \wedge \omega \neq 0## for some ##\beta##. That's what my post #78 is about.cianfa72 said:(i.e. in any open neighborhood) ##\omega \wedge d\omega=0## is really equivalent to the existence of functions ##f## and ##g## such that ##\omega=fdg##.
Sorry, not sure to understand. You said there is no way to have ##d \alpha = \beta \wedge \omega \neq 0## without leading to a contradiction. Hence it has to be ##d \alpha=0## I believe...PeterDonis said:This only follows from the Poincare lemma if we can prove that ##d \alpha = 0##, where ##d \omega = \alpha \wedge \omega##. To do that, we have to rule out the possibility that ##d \alpha = \beta \wedge \omega \neq 0## for some ##\beta##. That's what my post #78 is about.
Yes, that was the point of my post #78.cianfa72 said:You said there is no way to have ##d \alpha = \beta \wedge \omega \neq 0## without leading to a contradiction.
Once we have correctly shown that the other possibility leads to a contradiction, yes. As I noted in post #78, I don't think my previous claim about the other possibility was correct; that's why I posted post #78, to correctly show why ##d \alpha = \beta \wedge \omega## does not work.cianfa72 said:Hence it has to be ##d \alpha=0##