Diffeomorphic Invariance implies Poincare Invariance?

[FedEx]

But what was written was {set of riemannian metrics on M} / {set of diffeomorphisms on M}. How is the word quotient defined in this context? It certainly isn't the definition of quotient from algebra or topology. What you have written is completely different in that what you have stated is [M] = {all manifolds diffeomorphic to M} as the equivalence class.

For instance the classes are, as you would already know, [M,g], [M,\Omega^2 g] etc.. To me that's the essence with which Beger is talking. And i don't think that we require anything else.

 

[micromass]

My guess is the following. Take a smooth manifold $M$. We say that two metrics $g$ and $h$ are equivalent if and only if there exists a diffeomorphism $\varphi:M\rightarrow M$ such that $\varphi^*g=h$.

So if $X$ is the set of all metrics on $M$, then the above relation is an equivalence relation. The weird notation $RM/Diffeo(M)$ would then be the quotient wrt the equivalence relation.

Anyway, I don't think it's a good idea to quote Berger his book. Berger is a nice, informal book. But it shouldn't be used as a reference because it is informal. I would suggest sticking to more formal books.

 

[FedEx]

I think [url="http://www.math.umn.edu/~xuxxx225/docs/A%20Panoramic%20View%20of%20Riemannian%20Geometry.asp"Berger[/url] is using "quotient" is in the same sense as its used in algebra. I guess it all comes down to whether f*g is counts as equivalence under diffeomorphisms or not. Since f* is naturally induced by f, then it would seem yes.

The confusing thing is Berger also defines Isom(M), so one wonders why doesn't he write:

RS(M)=RM(M)/Isom(M) ?

Cause the certainly larger set of RS(M)=RM(M)/Diff(M) corresponds to the same physical situation.

 

[NanakiXIII]

What? Are you talking about the pushforward? It certainly isn't an isometry in general (the term isometry won't even make sense without a Riemannian structure; of course I mean isometry in the diff geo sense and not the sense of metric spaces) so you must be talking about something else.

It induces a map on your metric.

<br /> \phi^\ast g(x,y) = g(\phi_\ast x, \phi_\ast y)<br />

This makes it an isometry. I do believe this is the kind of mappings we usually deal with.

 

[FedEx]

It induces a map on your metric.

<br /> \phi^\ast g(x,y) = g(\phi_\ast x, \phi_\ast y)<br />

This makes it an isomorphism. I do believe this is the kind of mappings we usually deal with.

<br /> \phi^\ast g(x,y) = g(\phi_\ast x, \phi_\ast y) ≠ g(x,y)<br />

Hence it is not an isometry necessarily.

 

[WannabeNewton]

For instance the classes are, as you would already know, [M,g], [M,\Omega^2 g] etc.. To me that's the essence with which Beger is talking. And i don't think that we require anything else.

It's fine, my problem is with the notation RM / diffeo(M) so it isn't a problem with what you said. Anyways, back to the discussion at hand xD.

 

[WannabeNewton]

It induces a map on your metric.

<br /> \phi^\ast g(x,y) = g(\phi_\ast x, \phi_\ast y)<br />

This makes it an isomorphism. I do believe this is the kind of mappings we usually deal with.

This is false. A pullback induced by an arbitrary diffeomorphism is certainly not an isometry of the metric tensor as FedEx already noted. A trivial example is a conformal isometry which is a diffeomorphism \phi:M \rightarrow M such that \phi ^{*}g_{ab} = \Omega ^{2}g_{ab}.

EDIT: I just read that you said isomorphism here. But before you said isometry, which is what I and FedEx were objecting to. Which do you mean? An isomorphism is a very general term from category theory.

 

[atyy]

But what was written was {set of riemannian metrics on M} / {set of diffeomorphisms on M}. What you have written is completely different in that what you have stated is [M] = {all manifolds diffeomorphic to M} as the equivalence class. What I'm asking for is: what is the equivalence relation for the quotient given by Berger?

My guess is the following. Take a smooth manifold $M$. We say that two metrics $g$ and $h$ are equivalent if and only if there exists a diffeomorphism $\varphi:M\rightarrow M$ such that $\varphi^*g=h$.

So if $X$ is the set of all metrics on $M$, then the above relation is an equivalence relation. The weird notation $RM/Diffeo(M)$ would then be the quotient wrt the equivalence relation.

Anyway, I don't think it's a good idea to quote Berger his book. Berger is a nice, informal book. But it shouldn't be used as a reference because it is informal. I would suggest sticking to more formal books.

Yes, that's the relation. Berger gives "Equivalence under diffeomorphisms: (M,f*g) is the same as (M,g)"

That's the same as Hawking and Ellis's definition:"Two models (M,g) and (M',g') will be taken to be equivalent if they are isometric, that is if there is a diffeomorphism θ: M → M', which carries the metric g into the metric g', ie. θ*g = g'."

Anyway, the question remaining is whether a statement eg. in a relatively "formal" physics GR book like Wald would be correct by mathematician's standards "Thus, the diffeomorphisms comprise the gauge freedom of any theory formulated in terms of tensor fields on a spacetime manifold. In particular, diffeomorphisms comprise the gauge freedom of general relativity. (Wald, p438)" (BTW, by this point, Wald has already stated the same definitions as micromass, Berger and Hawking & Ellis - so the question is whether he is going from more formal language to more informal physics language.)

 

[WannabeNewton]

I'm not sure what the trouble is in that statement exactly atyy. Isn't it akin to the usual statement from EM that if the 4 - potential A^{a} solves the usual \partial ^{a}\partial _{[a}A_{b]} = -4\pi j_b then so will A^{a} + \partial ^{a}\varphi. Here instead of adding gradients of a smooth scalar function we are talking about diffeomorphisms of a space - time manifold. Of course in the case of EM we could write maxwell's equations using the field strength tensor as \partial^{a}F_{ab} = -4\pi j_{b} and never know about said gauge invariance (I think you mentioned this earlier correct?) but I don't know if such a blindness is possible in GR in the context of what Wald was talking about.

 

[atyy]

I'm not sure what the trouble is in that statement exactly atyy. Isn't it akin to the usual statement from EM that if the 4 - potential A^{a} solves the usual \partial ^{a}\partial _{[a}A_{b]} = -4\pi j_b then so will A^{a} + \partial ^{a}\varphi. Here instead of adding gradients of a smooth scalar function we are talking about diffeomorphisms of a space - time manifold. Of course in the case of EM we could write maxwell's equations using the field strength tensor as \partial^{a}F_{ab} = -4\pi j_{b} and never know about said gauge invariance (I think you mentioned this earlier correct?) but I don't know if such a blindness is possible in GR in the context of what Wald was talking about.

Yes, it's exactly the same. I guess the question is whether it's formal or informal terminology to say that diffeomorphisms are the gauge group of GR. micromass did find Berger's RM(M)/Diff(M) weird, but that naively seems to be the notational counterpart for Wald's statement. So the question is what is what would a mathematician quotient by?

 

[sheaf]

Yes, it's exactly the same. I guess the question is whether it's formal or informal terminology to say that diffeomorphisms are the gauge group of GR. micromass did find Berger's RM(M)/Diff(M) weird, but that naively seems to be the notational counterpart for Wald's statement. So the question is what is what would a mathematician quotient by?

This quotient appears also in the treatment of http://arxiv.org/abs/gr-qc/0403081http:// , where they quotient by the (passive) diffeos which have a natural action on the space of Riemannian metrics. (There is also the interesting statement there that the passive diffeos don't form the largest group of dynamical symmetries. That title goes to the Bergmann Komar group...)

 

[NanakiXIII]

This is false. A pullback induced by an arbitrary diffeomorphism is certainly not an isometry of the metric tensor as FedEx already noted. A trivial example is a conformal isometry which is a diffeomorphism \phi:M \rightarrow M such that \phi ^{*}g_{ab} = \Omega ^{2}g_{ab}.

EDIT: I just read that you said isomorphism here. But before you said isometry, which is what I and FedEx were objecting to. Which do you mean? An isomorphism is a very general term from category theory.

Sorry, I meant isometry. And it is, since two vectors (x,y) are mapped to (\phi_\ast x, \phi_\ast y) and the inner product g&#039;(\phi_\ast x, \phi_\ast y) is pulled back to \phi^\ast g&#039;(x,y) = g(x,y). Therefore, the inner product is preserved.

 

[WannabeNewton]

\phi^\ast g&#039;(x,y) = g(x,y)

This is true only if the map is an isometry by definition. I gave you an easy counter example in the above post. A diffeomorphism between riemannian manifolds is not in general an isometry!

 

[micromass]

Sorry, I meant isometry. And it is, since two vectors (x,y) are mapped to (\phi_\ast x, \phi_\ast y) and the inner product g&#039;(\phi_\ast x, \phi_\ast y) is pulled back to \phi^\ast g&#039;(x,y) = g(x,y). Therefore, the inner product is preserved.

Are you claiming that all diffeomorphisms are isomtries?? I think there are many counterexamples for this statement. Wbn gave one already.

 

[TrickyDicky]

I believe Nanaki( and not only him, this thread seems to be going in circles because some distinctions are being overlooked) is falling into two of the mistakes I warned against in a previous post, conflating local diffeomorphisms with local isometries and also local diffeomorphisms with diffeomorphisms.
When you have a diffeomorphism that preserves the metric, you have an isometry, this has been sufficiently stressed by WN and micromass, the problem is that in GR, as commented already by haushofer, atyy and me, the diffeomorphisms alluded by the term "diffeomorphism invariance", have to do with "no prior geometry" and are related to gauge invariance so can't be promoted to isometries.
So if we want to talk about geometry we must restrict ourselves to the local geometry, that is local isometries, these are not bijective but are injective and preserve curvature which is important for a physics theory that identifies gravity with curvature.
Now local isometries are just local diffeomorphisms that pullback the metric tensor and therefore preserve only infinitesimal distances. Maybe some of the confusion of Nanaki comes from the fact that local diffeomorphisms induce by the inverse function theorem a linear isomorphism(thus this one is bijective) at each point of the manifold.

 

[NanakiXIII]

This is true only if the map is an isometry by definition. I gave you an easy counter example in the above post. A diffeomorphism between riemannian manifolds is not in general an isometry!

Granted, this is what this construction does.

Are you claiming that all diffeomorphisms are isomtries?? I think there are many counterexamples for this statement. Wbn gave one already.

I believe Nanaki( and not only him, this thread seems to be going in circles because some distinctions are being overlooked) is falling into two of the mistakes I warned against in a previous post, conflating local diffeomorphisms with local isometries and also local diffeomorphisms with diffeomorphisms.
When you have a diffeomorphism that preserves the metric, you have an isometry, this has been sufficiently stressed by WN and micromass, the problem is that in GR, as commented already by haushofer, atyy and me, the diffeomorphisms alluded by the term "diffeomorphism invariance", have to do with "no prior geometry" and are related to gauge invariance so can't be promoted to isometries.
So if we want to talk about geometry we must restrict ourselves to the local geometry, that is local isometries, these are not bijective but are injective and preserve curvature which is important for a physics theory that identifies gravity with curvature.
Now local isometries are just local diffeomorphisms that pullback the metric tensor. Maybe some of the confusion of Nanaki comes from the fact that local diffeomorphisms induce by the inverse function theorem a linear isomorphism(thus this one is bijective) at each point of the manifold.

I really don't know what your definition of a diffeomorphism is then. Mine makes no mention of geometry at all. It's a pure mapping between manifolds. See e.g. Spivak. Isn't this the type of diffeomorphism GR is invariant under? If you go to Riemannian manifolds and you include your type of transformations, where you deviate from using the pullback metric, then obviously you're going to violate things.

 

[TrickyDicky]

I really don't know what your definition of a diffeomorphism is then. Mine makes no mention of geometry at all. It's a pure mapping between manifolds. See e.g. Spivak. Isn't this the type of diffeomorphism GR is invariant under?

Sure, great then. I inferred by your posts and the replies that they were getting that you thought all diffeomorphisms were isometries.

If you go to Riemannian manifolds and you include your type of transformations, where you deviate from using the pullback metric, then obviously you're going to violate things.

I don't understand what you mean here. What things am I going to violate and why?

The other problem that I saw here is that not all(most but not all) that is true about Riemannian manifolds carries over exactly the same to spacetimes(pseudoriemannian manifolds), that is where the Einstein's hole argument came from basically.

 

[WannabeNewton]

Granted, this is what this construction does.

I'm not sure what you mean by this. Are you still trying to assert that all diffeomorphisms between Riemannian manifolds are isometries?

 

[NanakiXIII]

Sure, great then. I inferred by your posts and the replies that they were getting that you thought all diffeomorphisms were isometries.


I don't understand what you mean here. What things am I going to violate and why?

You're violating the condition that seems to me to be crucial to doing something useful within the context of GR: since the metric is a dynamical field, you cannot simply change the metric without changing the rest of your content (field, particles) as well. If you do that, of course you end up with a different physical situation. And it's not a diffeomorphism in the sense that, while you may be acting on your manifold with a diffeomorphism, you're acting on your metric separately with another function. If someone opposes this last part of my statement (which I tried to elaborately explain in a previous post) then please point out what is wrong about it. Since a diffeomorphism has nothing to do with geometry, I really don't see how a mapping defined on your geometry can be a diffeomorphism.

 

[TrickyDicky]

You're violating the condition that seems to me to be crucial to doing something useful within the context of GR: since the metric is a dynamical field, you cannot simply change the metric without changing the rest of your content (field, particles) as well. If you do that, of course you end up with a different physical situation. And it's not a diffeomorphism in the sense that, while you may be acting on your manifold with a diffeomorphism, you're acting on your metric separately with another function. If someone opposes this last part of my statement (which I tried to elaborately explain in a previous post) then please point out what is wrong about it.

I'm not changing the metric if I stick to a local patch of the manifold and only care about a neighbourhood of the point that interests me, with curvature, geodesic length and proper time and all physically meaningful observables in GR preserved, that is the key to my distinction between global and local isometries that you seem to be missing. Even though this distinction is explained in every book about differential topology/geometry usually in the first pages, I'm yet to see a physicist or a GR book that makes this distinction.

Since a diffeomorphism has nothing to do with geometry, I really don't see how a mapping defined on your geometry can be a diffeomorphism.

A diffeomorphism has nothing to do with geometry but all (global) isometries happen to be diffeomorphisms, is this what confuses you?

 

[NanakiXIII]

I'm not changing the metric if I stick to a local patch of the manifold and only care about a neighbourhood of the point that interests me, with curvature, geodesic length and proper time and all physically meaningful observables in GR preserved, that is the key to my distinction between global and local isometries that you seem to be missing. Even though this distinction is explained in every book about differential topology/geometry usually in the first pages, I'm yet to see a physicist or a GR book that makes this distinction.

Perhaps this distinction between local and global diffeomorphisms does elude me; I'm not sure I see its significance.

I also still don't understand what part of what I said you are objecting to.

A diffeomorphism has nothing to do with geometry but all (global) isometries happen to be diffeomorphisms, is this what confuses you?

No. What is confusing me is how for example WannabeNewton's conformal mapping or Peter's example of changing the mass in the Schwarzschild metric can be considered diffeomorphisms. Their transformations specifically act on the metric. A diffeomorphism only acts on the manifold. Therefore, and apparently I am wrong here, but I don't see how, therefore their transformations cannot be considered diffeomorphisms (unless there is some diffeomorphism that induces these mappings, in which case I would like to see those diffeomorphisms explicitly.)

 

[WannabeNewton]

The conformal isometry (the diffeomorphism defined above) itself acts on the manifold. It's pullback acts on the metric tensor. This is the same thing with an isometry too obviously: the isometry acts on the manifold but its pullback acts on the metric tensor. By your claim an isometry wouldn't even be a diffeomorphism because it acts on the manifold and not the metric tensor. No smooth map between manifolds acts on the tensor fields themselves, their pull backs and pushforwards (when definable) are what act on the tensor fields.

 

[TrickyDicky]

Perhaps this distinction between local and global diffeomorphisms does elude me; I'm not sure I see its significance.

I must admit I was exaggerating a bit when i claimed that GR books don't make the distinction, many do, (although maybe not stressed enough given how much confusion around thse issues seems to exist), for instance when distingusihing between isometries and infinitesimal isometries.

 

[NanakiXIII]

The conformal isometry (the diffeomorphism defined above) itself acts on the manifold. It's pullback acts on the metric tensor. This is the same thing with an isometry too obviously: the isometry acts on the manifold but its pullback acts on the metric tensor. By your claim an isometry wouldn't even be a diffeomorphism because it acts on the manifold and not the metric tensor. No smooth map between manifolds acts on the tensor fields themselves, their pull backs and pushforwards (when definable) are what act on the tensor fields.

Then it would seem I have completely misunderstood what the pullback and pushforward do; I was under the impression that indeed, the pullback on the metric defined an isometry. Having a look at the Wiki page, I see this. That seems to define an isometry. Is it incorrect or am I interpreting it wrong?

 

[micromass]

Then it would seem I have completely misunderstood what the pullback and pushforward do; I was under the impression that indeed, the pullback on the metric defined an isometry. Having a look at the Wiki page, I see this. That seems to define an isometry. Is it incorrect or am I interpreting it wrong?

If $\varphi:M\rightarrow M$ is a diffeomorphism and if $g$ is a metric. Then $\varphi$ defines an isometry between $(M,g)$ and $(M,\varphi^*g)$.
What we are saying is that $\varphi$ is not an isometry between $(M,g)$ and $(M,g)$.

 

[NanakiXIII]

If $\varphi:M\rightarrow M$ is a diffeomorphism and if $g$ is a metric. Then $\varphi$ defines an isometry between $(M,g)$ and $(M,\varphi^*g)$.
What we are saying is that $\varphi$ is not an isometry between $(M,g)$ and $(M,g)$.

Right. That's what I figured, but I fail to see what kind of physical significance you are trying to attach to this. Are you trying to say that there are non-isometries that still leave GR invariant?

I also think that, if you define \phi : M \to M, you cannot just promote this to a map \phi : (M,g) \to (M,g&#039;) without specifying what the action on g is, in general. As I said earlier, you should define an implied map on the geometry or specify that you're not doing anything to it. I was using the implied map defined by the pullbacks and pushforwards on your tangent bundle, because that seems to be general practice. I don't see any sense in allowing random mappings on g and then finding that it doesn't leave your system invariant. If that is your thought experiment, that's fine, but my answer remains the same: I don't think that's what diffeomorphism invariance in GR is about.

As this indeed seems to be just a matter of semantics, we can drop it.

 

[atyy]

Cause the certainly larger set of RS(M)=RM(M)/Diff(M) corresponds to the same physical situation.

If $\varphi:M\rightarrow M$ is a diffeomorphism and if $g$ is a metric. Then $\varphi$ defines an isometry between $(M,g)$ and $(M,\varphi^*g)$.
What we are saying is that $\varphi$ is not an isometry between $(M,g)$ and $(M,g)$.

I also think that, if you define \phi : M \to M, you cannot just promote this to a map \phi : (M,g) \to (M,g&#039;) without specifying what the action on g is, in general. As I said earlier, you should define an implied map on the geometry or specify that you're not doing anything to it. I was using the implied map defined by the pullbacks and pushforwards on your tangent bundle, because that seems to be general practice. I don't see any sense in allowing random mappings on g and then finding that it doesn't leave your system invariant. If that is your thought experiment, that's fine, but my answer remains the same: I don't think that's what diffeomorphism invariance in GR is about.

Is the following be correct?

We say diffeomorphisms are the gauge group of GR, since every diffeomorphism corresponds to an isometry, provided we move manifold and metric.

When we use the term isometry in GR, we usually refer to diffeomorphisms which move the manifold without moving the metric, so not every diffeomorphism is an isometry, eg. finding isometries of the Schwarzschild solution is finding symmetries via Killing vectors.

 

[TrickyDicky]

Is the following be correct?

We say diffeomorphisms are the gauge group of GR, since every diffeomorphism corresponds to an isometry, provided we move manifold and metric.

I don't think this is correct. When Diff(M) is viewed as a gauge , the manifold is seen as a bare differentiable manifold, so isometries are left out.

When we use the term isometry in GR, we usually refer to diffeomorphisms which move the manifold without moving the metric, so not every diffeomorphism is an isometry, eg. finding isometries of the Schwarzschild solution is finding symmetries via Killing vectors.

When the term isometry is used in GR (i.e. those infinitesimally generated by KV) is actually infinitesimal isometries that are meant. Otherwise Diff(M) invariance couldn't be thought of as a gauge invariance. Remember gauge symmetries and spacetime symmetries ( those determined by global isometries) are not the same thing. The EFE only fixes the local geometry (thus we only need infinitesimal isometries) not the global spacetime topology/geometry.