I Global simultaneity surfaces - how to adjust proper time?

[cianfa72]

TL;DR Summary

Adjust proper time for proper time synchronizable congruence of worldlines.

Hi,

searching on PF I found this old post Global simultaneity surfaces. I read the book "General Relativity for Mathematicians"- Sachs and Wu section 2.3 - Reference frames (see the page attached).

They define a congruence of worldlines as 'proper time synchronizable' iff there exist a function $t$ of spacetime such that $d\omega= - dt$. The one-form $\omega$ is defined from $g_{ab}Q^a$ where $Q$ is the vector field the worldlines of the congruence are orbits of.

Said $\gamma : I \rightarrow M$ the generic worldline in the congruence we get $du=\gamma^{*}dt$ where $\gamma^{*}$ is the pullback of the map $\gamma$ -- from my understanding $du$ is such that $du(\frac {\partial} {\partial u}) = 1$.

Then since $du=\gamma^{*}dt$ they claimed $t \circ \gamma (u) = u + c$. I've some problem to grasp the reason behind it, though.

Can you help me ? thank you.

 

[PeterDonis]

I've some problem to grasp the reason behind it, though.

I'm not sure what your question is. Can you give more detail about what you are having problems with?

 

[cianfa72]

I have not a clear understanding why from $du = \gamma^{*}dt$ follows that $t \circ \gamma (u) = u + c$ where $u$ should be the proper time used as parameter along each worldline in the congruence.

p.s. said $Q$ the vector field tangents to worldlines $\gamma$ the following condition holds $g_{ab}Q^aQ^b = g(Q,Q) = -1$ (see the attached page in the post #1).

 

[PeterDonis]

I have not a clear understanding why from $du = \gamma^{*}dt$ follows that $t \circ \gamma (u) = u + c$ where $u$ should be the proper time used as parameter along each worldline in the congruence.

I don't see this claim being made anywhere in the page that you posted. I think you need to give more context from the book (I don't have a copy of Sachs & Wu so I can't check the reference directly).

 

[cianfa72]

At the end of the page they say that if $h \circ \gamma$ is not identically equal to 1, $t \circ \gamma$ no longer equals $u$ up to an additive constant.

AFAIU the coordinate on the real interval domain $I$ is $u$ so the curve $\gamma: I \rightarrow M$ is actually a function of $u$.

So I think in our case in which $t \circ \gamma=1$ should be $t \circ \gamma (u) = u + c$

In other words the proper time $u$ elapsed along each worldline between two given hypersurfaces $t=a$ and $t=b$ is exactly $b-a$.

 

[PeterDonis]

At the end of the page they say that if $h \circ \gamma$ is not identically equal to 1, $t \circ \gamma$ no longer equals $u$ up to an additive constant.

Ah, ok. So by implication if $h \circ \gamma$ is equal to $1$, then $t \circ \gamma$ does equal $u$ up to an additive constant.

AFAIU the coordinate on the real interval domain $I$ is $u$ so the curve $\gamma: I \rightarrow M$ is actually a function of $u$.

It's a mapping from real numbers (values of $u$) to points in spacetime.

So I think in our case in which $t \circ \gamma=1$ should be $t \circ \gamma (u) = u + c$

Not quite. $t$ is a mapping from points in spacetime to real numbers. So $t \circ \gamma$ is a mapping from real numbers to real numbers. This mapping maps the real number $u$ to the real number $u + c$ (where $c$ is some real constant). So if we wanted to include $u$ in the notation, it would be something like $\left( t \circ \gamma \right) (u) = u + c$.

 

[cianfa72]

Ah, ok. So by implication if $h \circ \gamma$ is equal to $1$, then $t \circ \gamma$ does equal $u$ up to an additive constant.

Maybe it is a simple/stupid question: why that implication is true ?

 

[PeterDonis]

Maybe it is a simple/stupid question: why that implication is true ?

I'm just saying it's an implication of the way the book phrased the statement you quoted. You're the one who said I should look at that phrase when I said I didn't see the claim $t \circ \gamma = u + c$ being made anywhere in what you referenced.

 

[cianfa72]

I'm just saying it's an implication of the way the book phrased the statement you quoted.

Maybe I was unclear there. I think the point is that since each curve in the congruence has the property that its tangent vector field $Q$ fulfills $g(Q,Q)=-1$ then their spacetime lenghts (i.e. proper time along each of them) between $u=a$ and $u=b$ is exactly $b-a$.

In other words the parameter $u$ is actually the proper time along each curve in the congruence.

 

[PeterDonis]

the parameter $u$ is actually the proper time along each curve in the congruence.

Yes, this is correct. But that is always true for any timelike curve provided your curve parameter is properly normalized. So from this fact alone you can't deduce anything useful about particular congruences of timelike curves that have particular properties of interest, such as being synchronizable or proper time synchronizable.

Again, I'm not really clear about where you are having a problem with understanding.

 

[cianfa72]

Ok, section 2.3 defines 'proper time synchronizable' a congruence for which there exist a function $t$ such that $\omega = - dt$.

Now integrating $\omega$ along a curve we get its lenght. On the other hand since $\omega$ is exact then its integral along a curve between two hypersurface $t=a$ and $t=b$ is exactly $b-a$.

Is that correct ?

 

[PeterDonis]

Ok, section 2.3 defines 'proper time synchronizable' a congruence for which there exist a function $t$ such that $\omega = - dt$.

Yes.

Now integrating $\omega$ along a curve we get its lenght. On the other hand since $\omega$ is exact then its integral along a curve between two hypersurface $t=a$ and $t=b$ is exactly $b-a$.

Is that correct ?

Yes.

 

[cianfa72]

ok, now makes sense to me $(t \circ \gamma) (u) = u + c$. Section 2.3 then claims each observer (i.e. each worldline in the proper time synchronizable congruence) can adjust/reset his clock so that its proper time is 0 when his worldline intersects a given $t=a$ hypersurface.

 

[cianfa72]

Just to double check my understanding.

We have a smooth curve $\gamma: I \rightarrow M$. It defines a tangent vector field $Q$ on its points in $M$. Since the manifold $M$ is endowed of a metric tensor $g$ we can naturally associate to $Q$ the one-form $\omega$ defined as $\omega = g(Q, \_ )$.

In a given coordinate chart $\{x^{\mu}\}$ on $M$ we get $\omega= g_{\mu \nu}q^{\mu}dx^{\nu}$ where $q^{\mu}$ are the components of $Q$ in the coordinate basis associated to $\{x^{\mu}\}$.

Given a one-form $\omega$ defined on a n-dimensional smooth manifold (e.g. 4D spacetime) we can always integrate it along a one-dimensional curve (in our case the curve $\gamma$). The condition $\omega (Q)=-1$ along the curve $\gamma$ implies that the integral of $\omega$ along it is actually its length according to the metric $g$.

 

[PeterDonis]

Just to double check my understanding.

Nothing you say here appears to be relevant to the actual question at issue. See below.

We have a smooth curve $\gamma: I \rightarrow M$. It defines a tangent vector field $Q$ on its points in $M$.

You're doing it backwards from the way your reference does it. Your reference defines a "reference frame" $Q$ on spacetime (the term "reference frame" is not defined in the excerpt you posted from your reference, but I assume it means something like a frame field), and then derives a set of curves which are the integral curves of this reference frame. $\gamma$ is just the name given to one of those curves (which one is immaterial). Note that when referring to the tangent vectors to the curves, which are the timelike unit vectors of the reference frame at each point, your reference uses $u$, not $Q$.

Since the manifold $M$ is endowed of a metric tensor $g$ we can naturally associate to $Q$ the one-form $\omega$ defined as $\omega = g(Q, \_ )$.

As noted above, in the notation your reference is using, this would be $\omega = g(u, \_)$, since $u$ is the timelike unit vector of the reference frame.

In a given coordinate chart $\{x^{\mu}\}$ on $M$ we get $\omega= g_{\mu \nu}q^{\mu}dx^{\nu}$ where $q^{\mu}$ are the components of $Q$ in the coordinate basis associated to $\{x^{\mu}\}$.

Yes, as long as you understand the pitfalls in the notation you have adopted (and again, in the notation your reference is using, it would be $u^\mu$, not $q^\mu$). But your reference is not reasoning this way; nothing it says has anything to do with coordinate charts, and nothing it says requires choosing a coordinate chart to prove.

Given a one-form $\omega$ defined on a n-dimensional smooth manifold (e.g. 4D spacetime) we can always integrate it along a one-dimensional curve (in our case the curve $\gamma$).

Yes, you can, but that's not what your reference is doing.

The condition $\omega (Q)=-1$ along the curve $\gamma$

This condition is not just true along the curve $\gamma$. It is true at every point of spacetime, from the definition of $\omega$ and the fact that $Q$ (or more properly $u$) is a unit timelike vector field. And, as I have remarked more than once, this condition tells you nothing at all useful since it's just restating the fact that $Q$ (or more properly $u$) is a unit timelike vector field.

implies that the integral of $\omega$ along it is actually its length according to the metric $g$.

Again, this is just restating the fact that $Q$ (or more properly $u$) is a unit timelike vector field, so it is telling you nothing at all useful.

 

[PeterDonis]

Nothing you say here appears to be relevant to the actual question at issue.

An obvious reason for this is that your post doesn't even mention $t$, the time function on spacetime, so obviously you can't be answering the question of why $\left( t \circ \gamma \right) (u) = u + c$.

 

[PeterDonis]

since $du=\gamma^{*}dt$ they claimed $t \circ \gamma (u) = u + c$. I've some problem to grasp the reason behind it, though.

What do you get if you just integrate both sides of the equation $du = \gamma^* dt$?

 

[ergospherical]

Note that when referring to the tangent vectors to the curves, which are the timelike unit vectors of the reference frame at each point, your reference uses $u$, not $Q$.

This actually isn’t right - cianfa had the right idea. Sachs and Wu use the letter $u$ as the worldline parameter, and denote the tangent vector to $\gamma$ as $\gamma_* u = (\gamma_* [d/du])(u)$. Meanwhile $Q$ is the vector field which generates the integral curves $\gamma$, and $\omega$ is its metric dual.

 

[PeterDonis]

Sachs and Wu use the letter $u$ as the worldline parameter

Sort of. They define the 1-form $du = - \gamma^* \omega$, but a 1-form is not the exterior derivative of a worldline parameter, it's the exterior derivative of a function on spacetime. So $u$ must be a function on spacetime that gives the worldline parameter at a given point in spacetime belonging to the integral curve of $Q$ that passes through that point. You're right that I misspoke when I said $u$ itself was the tangent vector; see below.

and denote the tangent vector to $\gamma$ as $\gamma_* u = (\gamma_* [d/du])(u)$.

Near the top of the excerpt the OP posted, we see the equation $\left( \gamma^* \omega \right) \left( d / du \right) = \omega(\gamma_*) = g(Q, \gamma_*) = g(\gamma_*, \gamma_*) = -1$. Here it looks like they are using just $\gamma_*$ to denote the tangent vector.

 

[ergospherical]

So $u$ must be a function on spacetime that gives the worldline parameter at a given point in spacetime belonging to the integral curve of $Q$ that passes through that point.

It’s not a function on space time, it’s the inclusion function (which is on the interval $\mathscr{E}$)

Near the top of the excerpt the OP posted, we see the equation $\left( \gamma^* \omega \right) \left( d / du \right) = \omega(\gamma_*) = g(Q, \gamma_*) = g(\gamma_*, \gamma_*) = -1$. Here it looks like they are using just $\gamma_*$ to denote the tangent vector.

No, $\gamma_*$ is the tangent vector field (which here coincides with $Q$), whilst $\gamma_* u$ is the tangent vector at the point on $\gamma$ at worldline parameter $u$

 

[PeterDonis]

It’s not a function on space time, it’s the inclusion function (which is on the interval $\mathscr{E}$)

I don't see how that can be true if $du$ is a 1-form, but perhaps $du$ means something else. As I said in an earlier post, I don't have this reference, and the notation they are using is not one I have a lot of experience with.

 

[ergospherical]

I don't see how that can be true if $du$ is a 1-form, but perhaps $du$ means something else. As I said in an earlier post, I don't have this reference, and the notation they are using is not one I have a lot of experience with.

The differential of a function is a 1-form

 

[cianfa72]

It’s not a function on space time, it’s the inclusion function (which is on the interval $\mathscr{E}$)

Yes, it is the inclusion of $\mathscr{E}$ in $\mathbb R$, so basically it assigns to $\mathscr{E}$ the coordinate $u$. So $du$ is the differential of coordinate $u$.

No, $\gamma_*$ is the tangent vector field (which here coincides with $Q$), whilst $\gamma_* u$ is the tangent vector at the point on $\gamma$ at worldline parameter $u$

Yes, that’s right. The one-form $du$ is the pullback of $\omega$ via the pullback $\gamma^*$

 

[PeterDonis]

The differential of a function is a 1-form

The exterior derivative (which is what the $d$ operator appears to mean in the reference the OP gave) of a function on spacetime is a 1-form on spacetime. But you're saying $u$ isn't a function on spacetime, which means you can't apply the exterior derivative operator to it. That's why I'm confused about what $du$ is supposed to mean in the reference the OP gave. If the $d$ in $du$ just means "take the differential", which would make $du$ a "1-form" on $\mathscr{E}$, not spacetime, then it seems like their notation is confusing, since they also use the same $d$ to mean "take the exterior derivative".

 

[PeterDonis]

the interval $\mathscr{E}$

Isn't $\mathscr{E}$ just the reals?

 

[ergospherical]

Nobody says that $d$ is only defined on functions on spacetime…

 

[PeterDonis]

Nobody says that $d$ is only defined on functions on spacetime…

The exterior derivative operator on spacetime is a different operator from the "take the differential" operator on the space of functions over a manifold such as $\mathscr{E}$.

I've been lazy in typing up to now, but I'll stop. What I've been typing as $d$ so far is actually $\mathbf{d}$ in the reference, and $\mathbf{d}$ is a standard notation for the exterior derivative operator. Whereas an ordinary $d$, not boldface, is a common notation for "take the differential". So I'm confused why $du$ isn't just written that way in the reference, instead of $\mathbf{d} u$, as it's actually written in the reference. $\mathbf{d} u$ is naturally interpreted as "take the exterior derivative of the function $u$ on spacetime". Whereas $du$ (no boldface) is naturally interpreted as "take the differential of the function $u$ on $\mathscr{E}$", which is what you are saying it actually means.

 

[ergospherical]

You’re making a distinction where there is none

 

[PeterDonis]

The one-form $du$ is the pullback of $\omega$ via the pullback $\gamma^*$

But the reference you gave actually writes this as $\mathbf{d} u = - \gamma^* \mathbf{\omega}$, which, per my previous post, doesn't really make sense. If $u$ is a function on $\mathscr{E}$, and $\gamma^*$ is the pullback from spacetime to $\mathscr{E}$ (which makes sense), then it would seem more natural to write $du = - \gamma^* \mathbf{\omega}$, to make clear that $du$ is the differential of a function on $\mathscr{E}$, not the exterior derivative of a function on spacetime.

 

[PeterDonis]

You’re making a distinction where there is none

Nonsense. There is most certainly a distinction between spacetime and $\mathscr{E}$, since they're different manifolds, and therefore there is also most certainly a distinction between functions on spacetime and functions on $\mathscr{E}$, and operators over such functions. There are of course mappings between these things, but saying there is a mapping between two things is not at all the same as saying they're identical.

 

[PeterDonis]

The one-form $du$ is the pullback of $\omega$ via the pullback $\gamma^*$

So for a proper time synchronizable $Q$, where $\omega = - \mathbf{d} t$, we have that $du$ is the pullback of $\mathbf{d} t$ by $\gamma^*$. So again, if you integrate both sides of this, what do you get?

 

[cianfa72]

So for a proper time synchronizable $Q$, where $\omega = - \mathbf{d} t$, we have that $du$ is the pullback of $\mathbf{d} t$ by $\gamma^*$. So again, if you integrate both sides of this, what do you get?

$dt$ integrated between $a$ and $b$ is of course $b-a$. The integral of 1-form $\omega$ should result in the length of the curve. The integral of $du$ between $u=c$ and $u=d$ should be $d-c$ so we get the result $(t \circ \gamma)(u) = u + c$.

 

[cianfa72]

No, $\gamma_*$ is the tangent vector field (which here coincides with $Q$), whilst $\gamma_* u$ is the tangent vector at the point on $\gamma$ at worldline parameter $u$

Yes, in other words $\gamma_* u$ is the vector field $\gamma_*$ (or better $Q$) evaluated at point $\gamma(u)$.

 

[cianfa72]

Coming back the OP, I've a doubt about the difference between 'locally synchronizable' and 'synchronizable' congruence as defined by Sachs and Wu. The first is defined by Frobenius condition $\omega \wedge d\omega = 0$ whilst the latter requires two functions $h$ and $t$ such that $h>0$ and $\omega = -hdt$.

AFAIK $\omega \wedge d\omega = 0$ is equivalent (iff) to $\omega = -hdt$. Does this equivalence hold just locally or it actually extend globally ?

 

[PeterDonis]

Does this equivalence hold just locally or it actually extend globally ?

I think the issue is that, while $\omega \wedge d \omega = 0$ is equivalent to $\omega = - h d t$ at any given point, if we look at the entire spacetime, the $h$ and $t$ such that $\omega = - h dt$ might be different at different points. The synchronizable condition rules that out: $h$ and $t$ must be the same globally.

 

[cianfa72]

I think the issue is that, while $\omega \wedge d \omega = 0$ is equivalent to $\omega = - h d t$ at any given point, if we look at the entire spacetime, the $h$ and $t$ such that $\omega = - h dt$ might be different at different points. The synchronizable condition rules that out: $h$ and $t$ must be the same globally.

Functions $h$ and $t$ are actually defined in an open neighborhood $U$ of any given point of spacetime manifold. In the intersection of two overlapping neighborhoods those functions must be the same (i.e. they must coincide there). If we extend this result to all spacetime points then the two functions should be the same globally, I believe.

 

[PeterDonis]

In the intersection of two overlapping neighborhoods those functions must be the same

I'm not sure that's necessarily true. I think I've seen a discussion of this somewhere but I can't find a reference just now.

Does the textbook you referenced discuss this at all?

 

[cianfa72]

Does the textbook you referenced discuss this at all?

No, as far as I can tell no. However for the 'synchronizable' case Sachs and Wu explicitly say $h>0$. IDK if that really makes the difference.

 

[PeterDonis]

AFAIK $\omega \wedge d\omega = 0$ is equivalent (iff) to $\omega = -hdt$.

Well, let's see.

If $\omega = f dt$ (I use $f$ instead of $- h$ to avoid the pesky minus sign while writing the following; of course we can always just set $f = -h$ at the end to match what your reference gives), then $d \omega = df \wedge dt$, so $\omega \wedge d \omega = 0$ by inspection.

If $\omega \wedge d \omega = 0$, then there must be some $\alpha$ such that $d \omega = \alpha \wedge \omega$. Then we must have $d d \omega = 0 = d \alpha \wedge \omega + \alpha \wedge d \omega$. Since $\alpha \wedge d \omega = \alpha \wedge \alpha \wedge \omega = 0$, the second term in the sum vanishes identically; to make the first term vanish, we must have $d \alpha = 0$.

Now we can see the issue that (I think) explains why locally synchronizable is not necessarily equivalent to synchronizable. $d \alpha = 0$ means that $\alpha$ is closed; but $\omega = f dt$ requires $\alpha$ to be exact, i.e., $\alpha = df$ [Edit: should be $\alpha = d \ln f$, see post #51]. So $\omega = f dt$ is only equivalent to $\omega \wedge d \omega = 0$ under conditions where a form being closed is equivalent to it being exact. Since that is not always the case, we cannot always say $\omega = f dt$ is equivalent to $\omega \wedge d \omega = 0$.

 

[cianfa72]

If $\omega \wedge d \omega = 0$, then there must be some $\alpha$ such that $d \omega = \alpha \wedge \omega$.

I tried to show $\omega \wedge d \omega = 0 \Leftrightarrow d \omega = \alpha \wedge \omega$ for some one-form $\alpha$. The implication $\Leftarrow$ is true by ispection.

For the other $\Rightarrow$ I tried as follows: just to fix ideas consider a 4D manifold and take a one-form $\omega$ defined on it. Since at each point of the 4D manifold the vector space of one-forms has dimension 4 we can get a basis completing $\omega$ with 3 independent one-forms $\delta,\beta,\gamma$.

$d\omega$ is a 2-form hence, as generic 2-form, can be always given as
$d\omega = a_1\delta\wedge \omega + a_2\beta \wedge \omega + a_3\gamma \wedge \omega + a_4\beta \wedge \delta+ a_5\gamma \wedge \delta+ a_6\gamma \wedge \beta$

$\begin{align} d\omega \wedge \omega & = (a_1\delta\wedge \omega + a_2\beta \wedge \omega + a_3\gamma \wedge \omega + a_4\beta \wedge \delta+ a_5\gamma \wedge \delta+ a_6\gamma \wedge \beta) \wedge \omega \nonumber \\ & = a_4\beta \wedge \delta\wedge \omega + a_5\gamma \wedge \delta\wedge \omega + a_6\gamma \wedge \beta \wedge \omega = 0 \nonumber \end{align}$

Since $\beta \wedge \delta\wedge \omega, \gamma \wedge \delta\wedge \omega, \gamma \wedge \beta \wedge \omega$ are independent we get $a_4=a_5=a_6=0$.

That means $d\omega = a_1\delta\wedge \omega + a_2\beta \wedge \omega + a_3\gamma \wedge \omega = (a_1\delta+ a_2\beta + a_3\gamma) \wedge \omega = \alpha \wedge \omega$ being $\alpha$ the one-form $\alpha = a_1\delta+ a_2\beta + a_3\gamma$.

Is that correct ?

 

[PeterDonis]

I tried to show $\omega \wedge d \omega = 0 \Leftrightarrow d \omega = \alpha \wedge \omega$ for some one-form $\alpha$.

I believe this is a special case of a general property of the wedge product, that if $\omega \wedge \beta = 0$, $\beta$ must be of the form $\alpha \wedge \omega$ for some $\alpha$. (Note that in the general case, it is possible that $\alpha$ is just a function, or 0-form, in which case $\alpha \wedge \omega$ is just an ordinary product of a function with a form. But this is ruled out in the special case under discussion since $d \omega$ is one rank higher than $\omega$.)

 

[cianfa72]

I believe this is a special case of a general property of the wedge product, that if $\omega \wedge \beta = 0$, $\beta$ must be of the form $\alpha \wedge \omega$ for some $\alpha$.

Yes, and I believe I have given a proof of it (at least for the wedge product of 1-form and a 2-form).

If $\omega \wedge d \omega = 0$, then there must be some $\alpha$ such that $d \omega = \alpha \wedge \omega$. Then we must have $d d \omega = 0 = d \alpha \wedge \omega + \alpha \wedge d \omega$. Since $\alpha \wedge d \omega = \alpha \wedge \alpha \wedge \omega = 0$, the second term in the sum vanishes identically; to make the first term vanish, we must have $d \alpha = 0$.

But...to make the first term vanish again $d\alpha = \beta \wedge \omega$ for some one-form $\beta$, don't you ?

 

[PeterDonis]

to make the first term vanish again $d\alpha = \beta \wedge \omega$ for some one-form $\beta$, don't you ?

$d \alpha = 0$ will certainly make the first term vanish.

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. I suspect there is a more straightforward way of ruling out that possibility but I have not been able to find one.

 

[PeterDonis]

I believe I have given a proof of it

Your "proof" looks to me like a circular argument, since this step...

Since $\beta \wedge \delta\wedge \omega, \gamma \wedge \delta\wedge \omega, \gamma \wedge \beta \wedge \omega$ are independent we get $a_4=a_5=a_6=0$.

...appears to amount to assuming what is to be proved, namely that all parts of $d \omega$ that are linearly independent of $\omega$ must vanish.

 

[cianfa72]

...appears to amount to assuming what is to be proved, namely that all parts of $d \omega$ that are linearly independent of $\omega$ must vanish.

Why ? One-forms $\omega, \delta, \beta, \gamma$ are linear independent by construction (starting from $\omega$ we build a basis completing it with 3 independent one-forms).

By construction $\beta \wedge \delta \wedge \omega, \gamma\wedge \delta \wedge \omega, \gamma \wedge \beta \wedge \omega$ are 3 independent 3-forms (the dimension of 3-forms vector space in 4 dimension is $\binom 4 3 = 4$). The coefficients of the unique linear combination to get the null 3-form must vanish.

 

[PeterDonis]

Why ?

I'm not saying the forms you say are linearly independent aren't. I'm just saying that saying $d \omega = \alpha \wedge \omega$ is saying the same thing as "there are no non-vanishing components of $d \omega$ that are linearly independent of $\omega$". Saying the latter is not "proving" the former; it's just restating it.

 

[cianfa72]

I'm not saying the forms you say are linearly independent aren't. I'm just saying that saying $d \omega = \alpha \wedge \omega$ is saying the same thing as "there are no non-vanishing components of $d \omega$ that are linearly independent of $\omega$". Saying the latter is not "proving" the former; it's just restating it.

Sorry, not sure to get your point. To me it seems to have proved that if by hypothesis $d\omega \wedge \omega=0$ then necessarily $d\omega = \alpha \wedge \omega$ for some one-form $\alpha$.

Note that the first 3 terms involving $a_1, a_2, a_3$ actually vanish since in the wedge product there are two occurrence of $\omega$.

 

[PeterDonis]

To me it seems to have proved that if by hypothesis $d\omega \wedge \omega=0$ then necessarily $d\omega = \alpha \wedge \omega$ for some one-form $\alpha$.

If it helps you to understand what's going on, that's fine. But saying that $d \omega \wedge \omega = 0$ is already saying that $d \omega$ and $\omega$ are not linearly independent; that's what a wedge product vanishing means. Saying that $d \omega = \alpha \wedge \omega$ for some $\alpha$ is just another way of saying that $d \omega$ is not linearly independent of $\omega$. No further proof is required.

 

[cianfa72]

Saying that $d \omega = \alpha \wedge \omega$ for some $\alpha$ is just another way of saying that $d \omega$ is not linearly independent of $\omega$. No further proof is required.

Ah ok, nevertheless my proof should be correct I believe.

 

[cianfa72]

$d \alpha = 0$ means that $\alpha$ is closed; but $\omega = f dt$ requires $\alpha$ to be exact, i.e., $\alpha = df$. So $\omega = f dt$ is only equivalent to $\omega \wedge d \omega = 0$ under conditions where a form being closed is equivalent to it being exact. Since that is not always the case, we cannot always say $\omega = f dt$ is equivalent to $\omega \wedge d \omega = 0$.

Coming back to your post #39, you mean since from $\omega = f dt$ we get $d\omega = df \wedge dt$ then the latter is in the form $d\omega = \alpha \wedge \omega/f$ in which $\alpha = df$ is exact, nevertheless it is not in the form $d\omega = \alpha \wedge \omega$.