Diffeomorphic Invariance implies Poincare Invariance?

[PeterDonis]

First just to respond to your previous post, I meant the full text. I vaguely recall from some other thread that you said you had the text?

No, unfortunately I don't, I only have the online lecture notes.

That isn't what he means although his wording is terrible. What he is saying is that if M and N are diffeomorphic and any properties regarding the smooth structure of M hold true for M then they also hold true for N.

This makes sense, but then what's the difference between M and N? He puts all the additional structure (metric, fields, etc.) into other objects, not M or N.

The concept of integral curves do not require any kind of Riemannian structure.

Agreed. I wasn't really thinking about that when I read the notes, but you're right, everything he says in this passage is valid without any metric.

And for that final point, yes the derivation of the local conservation of energy comes as a consequence of the invariance of the matter field action under diffeomorphisms. It again uses the concept of infinitesimal diffeomorphisms and the lie derivative of the metric tensor under the associated flows generated. The same argument can be used on the Hilbert action to derive the contracted Bianchi identity independent of the field equations.

Ok, so basically, the LHS of the EFE has zero covariant divergence because of the Bianchi identities; the RHS of the EFE has zero covariant divergence because of diffeomorphism invariance; and these items serve as a sanity check on the EFE itself.

 

[PeterDonis]

In your example you have matter.

Not really; Schwarzschild spacetime is a vacuum solution. Changing the M parameter doesn't "move" any matter in this particular case. (In a physically realistic solution, of course, changing the M parameter *would* require changing something about the non-vacuum region occupied by the object that collapsed to form the black hole; but we're talking here about the maximally extended Schwarzschild spacetime as a vacuum solution of the EFE, regardless of whether it's physically realistic or not.)

I agree with your more general points about having to move "active" degrees of freedom; but in the particular example I gave there actually aren't any other than the metric itself. (M is a parameter in the metric, so "moving" the metric entails "moving" M.)

Carroll explicitly says he does not require the principle of equivalence, but http://arxiv.org/abs/0805.1726 (p42, just after Eq 241) seems to indicate otherwise: "Since the matter is not minimally coupled to R, such theories will not lead to energy conservation and will generically exhibit a violation of the Equivalence Principle"

I think Carroll is restricting himself to the standard Einstein-Hilbert action for gravity, with no extra couplings between R and the matter fields. The paper you link to is talking about more general theories with extra couplings. But if Carroll only means his derivation to apply to the standard Einstein-Hilbert action, that does make me wonder why he says he doesn't require the principle of equivalence, since the standard E-H action leads directly to the vacuum EFE, which I thought entailed the principle of equivalence. Maybe he just means he doesn't need the principle of equivalence as an extra assumption.

 

[atyy]

Not really; Schwarzschild spacetime is a vacuum solution. Changing the M parameter doesn't "move" any matter in this particular case. (In a physically realistic solution, of course, changing the M parameter *would* require changing something about the non-vacuum region occupied by the object that collapsed to form the black hole; but we're talking here about the maximally extended Schwarzschild spacetime as a vacuum solution of the EFE, regardless of whether it's physically realistic or not.)

I agree with your more general points about having to move "active" degrees of freedom; but in the particular example I gave there actually aren't any other than the metric itself. (M is a parameter in the metric, so "moving" the metric entails "moving" M.)

OK, let me try this for the vacuum solution. In your example, you put test particles on the manifold, which constitute matter. (I'm not actually sure what's going in your example, I'm just playing around with ideas until maybe one works:)

I think Carroll is restricting himself to the standard Einstein-Hilbert action for gravity, with no extra couplings between R and the matter fields. The paper you link to is talking about more general theories with extra couplings. But if Carroll only means his derivation to apply to the standard Einstein-Hilbert action, that does make me wonder why he says he doesn't require the principle of equivalence, since the standard E-H action leads directly to the vacuum EFE, which I thought entailed the principle of equivalence. Maybe he just means he doesn't need the principle of equivalence as an extra assumption.

Yes, either way, it seems one needs a stronger principle that diff invariance - one needs either the EP or EFE, so I'm quite mystified by that claim in Carroll's notes.

 

[WannabeNewton]

I'll take a picture of the page I was referring to then. M and N can be completely different sets even if they are diffeomorphic. It's the same for homeomorphic topological manifolds as well-they can be different sets. Similarly, a single set can fail to be homeomorphic to another copy of itself but with a different topology e.g. one copy of Euclidean space with the Euclidean topolog and another copy with the discrete topology - they will fail to be homeomorphic. The point is that while M and N may be different sets, if they are diffeomorphic then their smooth structures are, for all practical purposes, identical.

Personally I find these things too abstract for the physics of GR. I'm perfectly content with the view that two different observers in some neighborhood of space time can have different labelings for events in this neighborhood, the labelings being represented by their respective coordinates but that the representation of measured values of tensor components in the respective coordinates are related by the tensor transformation rules, in particular the way scalar fields work. It makes the physics more intuitive for me in terms of measurements.

 

[atyy]

Not really; Schwarzschild spacetime is a vacuum solution. Changing the M parameter doesn't "move" any matter in this particular case. (In a physically realistic solution, of course, changing the M parameter *would* require changing something about the non-vacuum region occupied by the object that collapsed to form the black hole; but we're talking here about the maximally extended Schwarzschild spacetime as a vacuum solution of the EFE, regardless of whether it's physically realistic or not.)

I agree with your more general points about having to move "active" degrees of freedom; but in the particular example I gave there actually aren't any other than the metric itself. (M is a parameter in the metric, so "moving" the metric entails "moving" M.)

I looked up Hawking and Ellis's statement. They say:

"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'."

So perhaps by this definition, two spacetimes are equivalent as long as there is an isometry between them, even though there are other diffeomorphisms between them that are not isometries?

 

[TrickyDicky]

I looked up Hawking and Ellis's statement. They say:

"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'."

So perhaps by this definition, two spacetimes are equivalent as long as there is an isometry between them,

This is exactly what the definition means.

even though there are other diffeomorphisms between them that are not isometries?

If one is concerned with the equivalence between spacetimes, like in GR where manifolds are (M,g) pairs , isometries are the only diffeomorphisms of interest. I guess I can't see where this question is coming from unless you are talking about background independence, see my next post.

 

[TrickyDicky]

I'll take a picture of the page I was referring to then.

Hey WN, I'm trying to locate my copy of Carroll, but IIRC from when I read that appendix , Carroll was rather sloppy with the terminology, he indeed was most of the time using the term diffeomorphism for what were isometries, which is not exactly wrong since all isometries are diffeomorphisms but since the converse is not true it might be misleading.
Furthermore I believe he didn't made clear either the distinction between isometries and local isometries, the former preserve the metric but the latter only preserve curvature and are not diffeomorphisms of the manifold, only of open sets of the manifold(local diffeomorphisms).
As mentioned by Haushofer in a previous post this has to do with the hole argument, and the way I think Einstein solved that issue was acknowledging that general covariance in GR was restricted to preserving the local geometry, that is, curvature. He of course didn't use these words he said : "All our spacetime verifications invariably amount to a determination of spacetime coincidences. If, for example, events consisted merely in the motion of material points, then ultimately nothing would be observable but the meeting of two or more of these points." (Einstein, 1916)
This is nowadays called "no prior (global) geometry" or "background independence". Background-independence is the requirement that the theory be formulated in a way that it only depends on a bare differentiable manifold, but not on any prior geometry. Only with this premise(that is, forgetting about preserving the metric) can one talk about diffeomorphism invariance in GR. (Diffeomorphisms as gauge.)

 

[atyy]

Not really; Schwarzschild spacetime is a vacuum solution. Changing the M parameter doesn't "move" any matter in this particular case. (In a physically realistic solution, of course, changing the M parameter *would* require changing something about the non-vacuum region occupied by the object that collapsed to form the black hole; but we're talking here about the maximally extended Schwarzschild spacetime as a vacuum solution of the EFE, regardless of whether it's physically realistic or not.)

So I take back my statement about matter, test or non-test, in the vacuum solution (post #73). I think it has to be along the lines Hawking and Ellis's definition (post #75). I found another comment about equivalence up to isometry in Berger (p202), and the nice thing is his notation:

"We are interested only in the geometric “structure” of a Riemannian manifold, which is to say in a Riemannian metric up to isometries. What we will call a Riemannian structure on a given manifold M is an element of the quotient of of the set of all possible Riemannian metrics on M by the group of all diffeomorphisms of M. Let Diff (M) be the group of diffeomorphisms. For the total sets we will use the notations RM(M) and RS (M):

RM(M) = {Riemannian metrics on M}
RS(M) = {Riemannian structures on M} = RM(M) / Diff (M)

 

[WannabeNewton]

Hey WN, I'm trying to locate my copy of Carroll, but IIRC from when I read that appendix , Carroll was rather sloppy with the terminology, he indeed was most of the time using the term diffeomorphism for what were isometries...

But again like I said, Wald talks about active and passive BEFORE even definining isometries. I see nothing in his definition of active vs passive that would even care about if they were isometries or not (appendix C).

 

[FedEx]

So I take back my statement about matter, test or non-test, in the vacuum solution (post #73). I think it has to be along the lines Hawking and Ellis's definition (post #75). I found another comment about equivalence up to isometry in Berger (p202), and the nice thing is his notation:

"We are interested only in the geometric “structure” of a Riemannian manifold, which is to say in a Riemannian metric up to isometries. What we will call a Riemannian structure on a given manifold M is an element of the quotient of of the set of all possible Riemannian metrics on M by the group of all diffeomorphisms of M. Let Diff (M) be the group of diffeomorphisms. For the total sets we will use the notations RM(M) and RS (M):

RM(M) = {Riemannian metrics on M}
RS(M) = {Riemannian structures on M} = RM(M) / Diff (M)

I don't understand. Are you quotient-ing with the Diffeos that are Isometries or arbritary Diffeos?

Or is it just the way he defines it?

 

[NanakiXIII]

If that's true, then the statement that GR is diffeomorphism invariant is either trivial or meaningless. The whole point of a coordinate transformation is to leave the underlying geometry invariant, not just the underlying topological manifold. So it seems to me that GR physicists, at least, must mean something more by "diffeomorphism" than just a transformation of the underlying manifold; by that definition all GR coordinate transformations are just the identity diffeomorphism.

It is somewhat trivial, in that in GR, active coordinate transformations (as I defined them) are no longer distinct from passive ones. However, it is non-trivial in the sense that other physical theories (where the background geometry is fixed) do not have this property.

Example: taking Schwarzschild spacetime with mass M1, to Schwarzschild spacetime with a different mass M2, using the same coordinate chart, say Schwarzschild coordinates. Formally, the expression for the metric changes, and so do the components of the Einstein Field Equation (though not much, since only one parameter changes). But again, both metrics express a formally valid vacuum solution to the EFE. However, now scalar invariants are changed.

It may well be that "diffeomorphism" is not a good term for these transformations, mathematically speaking. But it seems to be often used by physicists in the way I've used it here.

Agreed, terminology may be a part of the issue here. But let me explain why I would not call your example a diffeomorphism. A diffeomorphism is defined (whether you ask a physicist or a mathematician) as a map

$$\phi : M \to N$$

such that it is bijective, differentiable, etc.

Such a diffeomorphism induces a natural (mathematicians may be less inclined to use this word) map on your tangent bundle

$$\phi_{\ast} : TM \to TN$$

that will do exactly what you want in GR; the diffeomorphism is also an isometry. (I forget whether the asterisk goes in sub- or superscript.)

Now, I cannot think of any diffeomorphism that naturally induces the map $M_1 \to M_2$ in the Schwarzschild metric.

Of course, you're free to drop this last bit. What a map naturally does or doesn't induce is a matter of taste. However, in my opinion you should either 1) use this natural consequence which most people seem to agree upon, 2) define your own type of induced map on the tangent bundle or 3) drop any reference to any induced map altogether. What you're doing is taking a diffeomorphism (the identity map) and gluing a map on your metric to it as you please. This map you're gluing onto it doesn't even act on your tangent bundle but only on this one-parameter dependence of the metric. This is ugly and not very general.

So I grant you, if you want to allow this kind of map and call it a diffeomorphism, then no, GR is not diffeomorphism invariant but behaves highly nontrivial under these mappings (since you can just define anything you want.) However, this is not a very meaningful statement at all.

 

[atyy]

I don't understand. Are you quotient-ing with the Diffeos that are Isometries or arbritary Diffeos?

Or is it just the way he defines it?

Let me report what Berger seems to be doing first, then let's figure out if it's right.

I looked at Berger's text again, and he does seem to be quotienting with arbitrary diffeos.

Just before this he has introduced the notion and notation for isometries.

p201: "Isometries are of a different kind from general diffeomorphisms ... group of all isometries of a given Riemannian metric, which we will write Isom(M)"

His figure 4.22 defines f as a diffeomorphism from M to M, and his caption is

p202: "Equivalence under diffeomorphisms: (M,f^*g) is the same as (M,g)"

 

[atyy]

Such a diffeomorphism induces a natural (mathematicians may be less inclined to use this word)

http://www.mat.univie.ac.at/~michor/kmsbookh.pdf

Mathematicians are the worst sinners. Everything is fine after it's defined. So one can talk about a "soul" in mathematics!

 

[WannabeNewton]

the diffeomorphism is also an isometry. (I forget whether the asterisk goes in sub- or superscript.)

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.

 

[atyy]

Another point for discussion. Wald writes (p438) "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".

 

[FedEx]

Let me report what Berger seems to be doing first, then let's figure out if it's right.

I looked at Berger's text again, and he does seem to be quotienting with arbitrary diffeos.

Just before this he has introduced the notion and notation for isometries.

p201: "Isometries are of a different kind from general diffeomorphisms ... group of all isometries of a given Riemannian metric, which we will write Isom(M)"

His figure 4.22 defines f as a diffeomorphism from M to M, and his caption is

p202: "Equivalence under diffeomorphisms: (M,f^*g) is the same as (M,g)"

Hmm. Let us try to understand that.

An arbitrary diffeo would change the Metric. That would affect, say, the Mass distribution as well. But that again would correspond to the same physical situation much in the vein of what Nanaki was talking about.

For once I thought,(using an example of some kind of scaling diffeo) would that change the total mass? No, since the change in the mass distribution and the Jacobian cancel exactly.

Hence by quotienting with arbitrary diffeos, what we end up with, are the possible physically "distinct" solutions.

To quote from Caroll's notes : "It is possible that two purportedly distinct configurations (of matter and metric) in GR are actually “the same”, related by a diffeomorphism. In a path integral approach to quantum gravity, where we would like to sum over all possible configurations, special care must be taken not to overcount by allowing physically indistinguishable configurations to contribute more than once."

 

[WannabeNewton]

"We are interested only in the geometric “structure” of a Riemannian manifold, which is to say in a Riemannian metric up to isometries. What we will call a Riemannian structure on a given manifold M is an element of the quotient of of the set of all possible Riemannian metrics on M by the group of all diffeomorphisms of M. Let Diff (M) be the group of diffeomorphisms. For the total sets we will use the notations RM(M) and RS (M):

RM(M) = {Riemannian metrics on M}
RS(M) = {Riemannian structures on M} = RM(M) / Diff (M)

What is the definition of "quotient" as used here? A quotient in the algebraic or topological context refers to a set obtained by forming equivalence classes of another set of elements. I don't even know how to make sense of the word "quotient" as used in the quote from that book.

 

[FedEx]

What is the definition of "quotient" as used here? A quotient in the algebraic or topological context refers to a set obtained by forming equivalence classes of another set of elements. I don't even know how to make sense of how the word "quotient" is being used in the quote from that book.

The relation is "Diffeomorphism". [a] contains all x, which are related to a by a diffeomorphism. Does that make sense?

 

[WannabeNewton]

The relation is "Diffeomorphism". [a] contains all x, which are related to a by a diffeomorphism. Does that make sense?

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?

 

[atyy]

I think Berger (p202) 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) ?

 

[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.

$$\phi^\ast g(x,y) = g(\phi_\ast x, \phi_\ast y)$$

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.

$$\phi^\ast g(x,y) = g(\phi_\ast x, \phi_\ast y)$$

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

$$\phi^\ast g(x,y) = g(\phi_\ast x, \phi_\ast y) ≠ g(x,y)$$

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.

$$\phi^\ast g(x,y) = g(\phi_\ast x, \phi_\ast y)$$

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 $g'(\phi_\ast x, \phi_\ast y)$ is pulled back to $\phi^\ast g'(x,y) = g(x,y)$. Therefore, the inner product is preserved.

 

[WannabeNewton]

$\phi^\ast g'(x,y) = g(x,y)$

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 $g'(\phi_\ast x, \phi_\ast y)$ is pulled back to $\phi^\ast g'(x,y) = g(x,y)$. 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.