Diffeomorphism invariance of GR

[Frank Castle]

Peter, note that what Frank is talking about is a situation where the metric and any tensors on the image manifold are related to those in the initial manifold by the pushforward/pullback relations. As such, the induced metric on the target manifold is flat if the one on the original manifold is and so on. Of course, it could happen that you can construct a diffeomorphism between two manifolds that are already endowed with incompatible metrics such that $\phi^{-1*} g \neq \tilde g$, where $\tilde g$ is the metric already existing on the target manifold. However, if there is a diffeomorphism $\phi$, then the exists a metric such that the new manifold describes the same physical situation as the original manifold, namely $\phi^{-1*}g$. I think that you are talking around each other by everyone else assuming this metric and you assuming that there already exists a metric on the target manifold.

Yes, this is what I was thinking of. Although I'm not too sure on the subject.

 

[vanhees71]

If the metrics on the two manifolds are "incompatible" this just means that ther is no isomorphism between the two manifolds. Diffeomorphisms are the isomorphisms beetween differentiable manifolds but not necessarily between Riemannian ones, where in addition to being a diffeomorphism they must also keep the metric the same to be an isomorphism.

 

[Frank Castle]

they must also keep the metric the same to be an isomorphism.

What confuses me about this is doesn't this imply that the diffeomorphism has to be an isometry?

 

[haushofer]

Just my two cents (maybe I already gave them, but never mind) In my thesis I consider at page 26-29 the hole-argument explicitly for the Schwarzschild solution,

http://www.rug.nl/research/portal/e...ed(fb063f36-42dc-4529-a070-9c801238689a).html

General covariance means that one can change coordinates in a general way, after which the solution has the same physical meaning in the new coordinates. Isometries mean you fix your coordinate system and look for directions in which the metric does not change. The big confusion (for me) was that both the general coordinate transformation and the isometry are described by a Lie derivative: general covariance means you can add the Lie derivative of the metric to a metric solution wrt a general vector field without changing the physical meaning of that solution, while isometries are found by putting the Lie derivative of the metric to zero and solving for the vector field.

So, general covariance means that physics does not change under
$$g_{\mu\nu} \rightarrow g_{\mu\nu} + 2 \nabla_{(\mu}\xi_{\nu)}$$
while isometries of g are given by the solution of the vector field xi of
$$2 \nabla_{(\mu}\xi_{\nu)} = 0$$
That makes the distinction between isometries and general covariance, I think, a bit blurred. Hope this helps.

 

[vanhees71]

Yes, and that's gauge symmetry (in fact you gauge Lorentz invariance, but that you know much better than I do :-)).

 

[Frank Castle]

Just my two cents (maybe I already gave them, but never mind) In my thesis I consider at page 26-29 the hole-argument explicitly for the Schwarzschild solution,

http://www.rug.nl/research/portal/e...ed(fb063f36-42dc-4529-a070-9c801238689a).html

General covariance means that one can change coordinates in a general way, after which the solution has the same physical meaning in the new coordinates. Isometries mean you fix your coordinate system and look for directions in which the metric does not change. The big confusion (for me) was that both the general coordinate transformation and the isometry are described by a Lie derivative: general covariance means you can add the Lie derivative of the metric to a metric solution wrt a general vector field without changing the physical meaning of that solution, while isometries are found by putting the Lie derivative of the metric to zero and solving for the vector field.

So, general covariance means that physics does not change under
$$g_{\mu\nu} \rightarrow g_{\mu\nu} + 2 \nabla_{(\mu}\xi_{\nu)}$$
while isometries of g are given by the solution of the vector field xi of
$$2 \nabla_{(\mu}\xi_{\nu)} = 0$$
That makes the distinction between isometries and general covariance, I think, a bit blurred. Hope this helps.

Thanks for the link to your thesis, I have had a read and it has cleared up a few things. However, I'm still a bit unsure on a few points. For example:

general covariance means you can add the Lie derivative of the metric to a metric solution wrt a general vector field without changing the physical meaning of that solution

Why is this so? How does one show mathematically that the Einstein-Hilbert action and hence the EFE are invariant under this transformation, $g_{\mu\nu}(x)\rightarrow g'_{\mu\nu}(x)=g_{\mu\nu}(x)+\mathcal{L}_{\xi}g_{\mu\nu}(x)=g_{\mu\nu}(x)+2\nabla_{(\mu}\xi_{\nu)}$? Is the point that for an active diffeomorphism, the points of the underlying manifold are "moved" around, and all of the tensorial objects are "dragged" along with them. When we do this we "move" the points, but remain in the same coordinate system, such that we evaluate the "old" metric $g_{\mu\nu}(x(p))$ at point $p$ with coordinate value $x^{\mu}(p)$, and the "new" metric $g'_{\mu\nu}(x(p'))$ at the point $p'$ which has the same coordinate value $x^{\mu}(p')$ as $p$. Since the mapping is a diffeomorphism $g_{\mu\nu}(x(p))$ and $g'_{\mu\nu}(x(p'))$ has the same value, and furthermore satisfy the same EFE?

 

[PeterDonis]

I think that you are talking around each other by everyone else assuming this metric and you assuming that there already exists a metric on the target manifold.

Yes, I think this is at least part of it. The point I was trying to make is that just saying "GR is invariant under diffeomorphisms" can have multiple meanings, and that terminology might not be the best way to think about things anyway. We haven't even dug into the hole argument.

 

[Frank Castle]

I think that you are talking around each other by everyone else assuming this metric and you assuming that there already exists a metric on the target manifold.

So is the argument essentially that one can have two manifolds $M$ with metric $g_{\mu\nu}$, then another manifold $N$ is physically equivalent if it satisfies $N=\phi M$ and has a metric endowed on it given by $\tilde{g}=\phi^{\ast}g$?

 

[haushofer]

Thanks for the link to your thesis, I have had a read and it has cleared up a few things. However, I'm still a bit unsure on a few points. For example:
Why is this so? How does one show mathematically that the Einstein-Hilbert action and hence the EFE are invariant under this transformation, $g_{\mu\nu}(x)\rightarrow g'_{\mu\nu}(x)=g_{\mu\nu}(x)+\mathcal{L}_{\xi}g_{\mu\nu}(x)=g_{\mu\nu}(x)+2\nabla_{(\mu}\xi_{\nu)}$? Is the point that for an active diffeomorphism, the points of the underlying manifold are "moved" around, and all of the tensorial objects are "dragged" along with them. When we do this we "move" the points, but remain in the same coordinate system, such that we evaluate the "old" metric $g_{\mu\nu}(x(p))$ at point $p$ with coordinate value $x^{\mu}(p)$, and the "new" metric $g'_{\mu\nu}(x(p'))$ at the point $p'$ which has the same coordinate value $x^{\mu}(p')$ as $p$. Since the mapping is a diffeomorphism $g_{\mu\nu}(x(p))$ and $g'_{\mu\nu}(x(p'))$ has the same value, and furthermore satisfy the same EFE?

I have to think more about this whole dragging along thing, but if you take e.g. the vacuum EFE G=0, then the Lie derivative of G=0 is necessarily also 0.

 

[Orodruin]

If the metrics on the two manifolds are "incompatible" this just means that ther is no isomorphism between the two manifolds. Diffeomorphisms are the isomorphisms beetween differentiable manifolds but not necessarily between Riemannian ones, where in addition to being a diffeomorphism they must also keep the metric the same to be an isomorphism.

Last time I checked, the defining property of a ($C_\infty$) diffeomorphism is that it is an invertible smooth map between two manifolds with a smooth inverse? That makes no reference whatsoever to the metric tensor, only to the topological and smooth structure of the manifolds.

 

[vanhees71]

Exactly, and the diffeomorphisms are precisely the isomorphisms of differentiable manifolds without additional structure.

If you have a (pseudo-)Riemannian manifold only those diffeomorphisms that keep the fundamental form, or (pseudo-)metric, invariant are isomorphisms. That's because a (pseudo-)Riemannian manifold has the fundamental form as additional defining structure.

A simpler analogon are, e.g., real finite-dimensional vector spaces. The plane vanilla $d$-dimensional vector spaces are all equivalent with all invertible linear maps as isomorphisms (they form the general linear group $\mathrm{GL}(d,\mathbb{R})$. If you define a fundamental form (i.e., a non-degenerate bilinear form) on it you have a (pseudo-)Euclidean space, and only those invertible linear maps are isomorphisms that leave the fundamental form unchanged, which reduces the group to $\mathrm{O}(n_1,n_2)$, where $n_1+n_2=d$ denotes the signature of the fundamental form. If on top you define an orientation on the vector space the isomorphisms are the $\mathrm{SO}(n_1,n_2)$ transformations, i.e., those with positive determinant and so on. In other words, the more "extra structure" you impose on the vector space the smaller becomes the set (in this case forming groups with composition as group product) of isomorphisms.

 

[Frank Castle]

I have to think more about this whole dragging along thing

Have you had any further thoughts on this? I'm still a bit confused about the whole thing if I'm honest.

 

[Geometry_dude]

If you were a bit more specific, we might be able to help you.

If you want to stick to a coordinate point of view, isometries are those non-trivial 'coordinate transformations' that don't change the appearance of the $g_{\mu \nu}$, while more general 'diffeomorphisms' might look like they change these functions, but actually it's just a reformulation of the theory. I used the parantheses to indicate that the statement is not really mathematically correct, but I though it might help you to get on the right track.

 

[haushofer]

Have you had any further thoughts on this? I'm still a bit confused about the whole thing if I'm honest.

Everything follows from the interpretation of a diffeomorphism on the metric; this is just a Lie-derivative, and the action of a diff. on the Einstein tensor is just a Lie derivative of the Einstein tensor. So you should stick to the metric, and the dragging along interpretation of the Lie derivative on a tensor can be found in any decent textbook, e.g. Zee or Carroll. The appendix of Wald covers the notion between passive and active coordinate transformations, if you like to open up that gate to hell also. The confusion arises when you see the metric as just another field, which can be dragged along with respect to the spacetime points. Well, that's wrong: these points don't have any meaning without the metric. So 'decoupling' the metric from spacetime and dragging the metric along with respect to spacetime points does not have any meaning at all if you haven't fixed your coordinate system. Only after the metric has been introduced you can interpret spacetime points and their corresponding coordinates.

Once you have done this and fixed your coordinates, the dragging along of the metric does tell you something: it gives you the isometries of the spacetime in the particular coordinate system.

Hope this helps.

 

[vanhees71]

Well, as Wald emphasizes, it's "philosophically differerent" whether you interpret a diffeomorphism as active or passive, but mathematically and physically it's equivalent. That answers the question, why there is so much confusion about "active" vs. "passive". It's only a philosophical confusion!

 

[Frank Castle]

If you want to stick to a coordinate point of view, isometries are those non-trivial 'coordinate transformations' that don't change the appearance of the gμνg_{\mu \nu}, while more general 'diffeomorphisms' might look like they change these functions, but actually it's just a reformulation of the theory. I used the parantheses to indicate that the statement is not really mathematically correct, but I though it might help you to get on the right track.

In what sense is the statement mathematically incorrect?
To be honest, I think the set of notes that I read (http://web.mit.edu/edbert/GR/gr5.pdf) really threw a spanner in the works for me. It has confused me about the difference between a diffeomorphism and a general coordinate transformation. Are they mathematically and physically equivalent (just philosophically different)? Is the Einstein-Hilbert automatically diffeomorphism invariant by virtue of the fact that it is invariant under general coordinate transformations?

 

[haushofer]

Yes. Numerically, a general coordinate transfo brings you from a tensor T(x) to T'(x'). Given T'(x') you can reset the coordinates' value giving you T'(x). Comparing this with T(x) is exactly what you do if you interpret the gct actively, giving you Lie derivatives.

Note that infinitesimal coordinate transfo's are often used synonymous with the action of Lie derivatives.

 

[Frank Castle]

Yes. Numerically, a general coordinate transfo brings you from a tensor T(x) to T'(x'). Given T'(x') you can reset the coordinates' value giving you T'(x). Comparing this with T(x) is exactly what you do if you interpret the gct actively, giving you Lie derivatives.

Note that infinitesimal coordinate transfo's are often used synonymous with the action of Lie derivatives.

Ah ok, so in the notes that I linked to, is the author simply interpreting a diffeomorphism as a combination of a mapping $p\mapsto\phi(p)=q$ (such that in coordinates $x^{\mu}(p)\mapsto y^{\mu}(q)=(\phi^{\ast}x)^{\mu}(p)$), and a coordinate transformation $y^{\mu}\mapsto x'^{\mu}$, such that the points $p\in M$ on the manifold are moved around, but the coordinates are not, i.e. the coordinates of $q=\phi(p)$ are the same $q$, $x'^{\mu}(q)=x^{\mu}(p)$. Is this what the author is getting at when they claim that under an active coordinate transformation $d^{4}x$ is invariant, but $\sqrt{-g}$ is not?

 

[vanhees71]

Yes. Numerically, a general coordinate transfo brings you from a tensor T(x) to T'(x'). Given T'(x') you can reset the coordinates' value giving you T'(x). Comparing this with T(x) is exactly what you do if you interpret the gct actively, giving you Lie derivatives.

Note that infinitesimal coordinate transfo's are often used synonymous with the action of Lie derivatives.

No, a coordinate transformation doesn't change your tensors at all. That's the key to tensor analysis! It's basis independent!

 

[Frank Castle]

No, a coordinate transformation doesn't change your tensors at all. That's the key to tensor analysis! It's basis independent!

But the components of a tensor will change, right (i.e. $T^{\mu\nu}(x)\rightarrow T'^{\mu\nu}(x')$).

 

[haushofer]

No, a coordinate transformation doesn't change your tensors at all. That's the key to tensor analysis! It's basis independent!

The components do change. That's what I was implying in my shorthand notation and I'm assuming that in this discussion this basic fact about tensors is clear to anyone.

By the way, in the common language of "infinitesimal coordinate transformation" a tensor does change; it is synonymous to the Lie derivative of the tensor. That's why I was also warning the TS for the different usages of the terminology.

 

[haushofer]

But the components of a tensor will change, right (i.e. $T^{\mu\nu}(x)\rightarrow T'^{\mu\nu}(x')$).

Yes.

 

[Frank Castle]

Yes.

Cool, that's what I thought. Is any of what I wrote in my previous post #53 correct at all?

 

[haushofer]

Ah ok, so in the notes that I linked to, is the author simply interpreting a diffeomorphism as a combination of a mapping $p\mapsto\phi(p)=q$ (such that in coordinates $x^{\mu}(p)\mapsto y^{\mu}(q)=(\phi^{\ast}x)^{\mu}(p)$), and a coordinate transformation $y^{\mu}\mapsto x'^{\mu}$, such that the points $p\in M$ on the manifold are moved around, but the coordinates are not, i.e. the coordinates of $q=\phi(p)$ are the same $q$, $x'^{\mu}(q)=x^{\mu}(p)$. Is this what the author is getting at when they claim that under an active coordinate transformation $d^{4}x$ is invariant, but $\sqrt{-g}$ is not?

As far as I can tell, that author uses the word coordinate transformation for T(x) --> T'(x'), as is common, and "diffeomorphism" for what I would call T(x) --> T'(x), so a coordinate transformation with the coordinate x' set back to x . See the last few lines on page 9: "the volume element is invariant under a coordinate transformation but not under a diffeomorphism". I read "invariant" as "scalar".

 

[haushofer]

Cool, that's what I thought. Is any of what I wrote in my previous post #53 correct at all?

I think it does; see also Wald's discussion in his appendix on active v.s. passive.

 

[vanhees71]

But the components of a tensor will change, right (i.e. $T^{\mu\nu}(x)\rightarrow T'^{\mu\nu}(x')$).

Of course. It's easy to remember: $\mathrm{d} x^{\mu}$ are vector components. So you get
$$\mathrm{d} x^{\prime \mu}=\frac{\partial x^{\prime \mu}}{\partial x^{\nu}} \mathrm{d} x^{\nu},$$
and each contravariant component of a multilinear indicates transformation like this, i.e.,
$$T^{\prime \mu \nu}= \frac{\partial x^{\prime \mu}}{\partial x^{\rho}}\frac{\partial x^{\prime \nu}}{\partial x^{\sigma}} T^{\rho \sigma},$$
where I use the holonomous basis
$$\boldsymbol{b}_{\mu} = \partial_{\mu},$$
which thus transforms as
$$\boldsymbol{b}_{\mu}'=\boldsymbol{b}_{\nu} \frac{\partial x^{\prime \nu}}{\partial x^{\mu}},$$
i.e., contragrediently to contravariant (upper index) components.

This implies that indeed the tensors themselves are invariant. You can easily check that indeed
$$\boldsymbol{T}=T^{\mu \nu} \boldsymbol{b}_{\mu} \otimes \boldsymbol{b}_{\nu} = T^{\mu \nu} \partial_{\mu} \partial_{\nu}$$
is invariant.

 

[haushofer]

Well, as Wald emphasizes, it's "philosophically differerent" whether you interpret a diffeomorphism as active or passive, but mathematically and physically it's equivalent. That answers the question, why there is so much confusion about "active" vs. "passive". It's only a philosophical confusion!

Well, maybe it's only philosphical, but I've confronted many of my former collegues and even some top-physicists having written textbooks on (quantum) gravity, and I can't say I've ever met a single person who had a clear, conceptual view on all of this and could tell me the relation between all the different interpretations. To me that's a sign that a lot of people can calculate with tensors, but interpreting is a whole different cookie.

To me this business is perhaps the most difficult problem I've ever encountered in my carreer as a PhD-student. In my experience, there are two kinds of reactions to this whole business: people who wave it away as being trivial, and people who admit right away they've never understood it properly in the first place. That explains also why somebody like Einstein was puzzled by the hole-problem for 2 years.

So my message to the Topic Starter would be: don't be discouraged by your confusion. It means you're thinking about. That's a start already.

 

[vanhees71]

Well, ok there's maybe indeed some subtle point in this if it comes to interpretation.

A drastic example is "time-reversal invariance" (let's stick to SR for this example). On the one hand you can easily define it in a passive way by saying, I let run "time in the other direction" by formally setting $t \rightarrow t'=-t, \vec{x} \rightarrow \vec{x}'=\vec{x}$. This implies the time-inversion properties of the usual quantities (if it can be defined at all) like $\vec{p} \rightarrow -\vec{p}$, $E \rightarrow E'=E$, etc. etc. So it's easily defined, and it's easily checked whether your Lagrangian in question is invariant under this transformation (e.g., electromagnetism is invariant, the weak interaction Lagrangian a la Glashow, Salam, Weinberg is not, as it should be).

Now, what's the "active" point of view of this? Of course, I cannot simply let run time backwards. A handwaving idea of making a movie of some process and let it run backwards, hints, however to the solution: What's meant is that if you look at a process under some dynamical law turning an initial state of affairs at time $t=0$ to a final state of affairs at time $t$, then you have time-reversal invariance, if you prepare the system in the exactly time-reversal of the final state at $t=0$ and let it go under the same dynamical laws you should end up at time $t$ with exactly the time-reversal of the initial state of the first setup.

I don't see any such problems between active and passive interpretations of symmetry principles for the usual continuous symmetries. The mathematics is finally all the same, no matter whether you prefer the active or passive point of view. In my opinion the passive point of view often is more easy to grasp: The physical laws shouldn't depend on our choice of coordinates and other men-made descriptions. On the other hand, the active point of view provides a more intuitive physical picture. E.g., you can state that SR is invariant under Poincare transformations of the coordinates because of the geometrical structure of Minkowski space or you can say (specializing to boost invariance): The physical laws look the same in all inertial reference frames or, even simpler, there is no absolute inertial reference frame since all physical laws are independent of the (constant) velocity of one inertial frame with respect to another.

 

[Frank Castle]

See the last few lines on page 9: "the volume element is invariant under a coordinate transformation but not under a diffeomorphism"

This is what I'm unsure about really. Is the point that the author is making here that under a diffeomorphism we are always free to choose a coordinate transformation, $x\rightarrow x'$, such that the coordinate values of the "new" point $q$ in the "new" coordinate system $x'$ are the same as the coordinate values of the "old" point $p$ in the "old" coordinate system $x$, i.e. $x'^{\mu}(q)=x^{\mu}(p)$. Hence, we have that $d^{4}x\rightarrow d^{4}x'=d^{4}x$ such that $d^{4}x$ is diffeomorphism invariant, however, $\sqrt{-g}$ transforms as $\sqrt{-g}\rightarrow\sqrt{-g}+\mathcal{L}_{X}\sqrt{-g}=\sqrt{-g}\left(1+g^{\mu\nu}\nabla_{\mu}X_{\nu}\right)$ (where $X$ is the vector field generating the diffeomorphism), and so is not invariant under diffeomorphisms?

 

[vanhees71]

If you have a diffeomorphic map of the manifold on itself, the map of a region doesn't need to have the same volume as this region, while a given region's volume of course shouldn't depend on the chosen coordinates.

The volume of a sphere in usual 3D Euclidean space doesn't change, only because you calculate it in terms of spherical coordinates (clever, because simple) or Cartesian ones (a bit lengthy but doable).

 

[Frank Castle]

If you have a diffeomorphic map of the manifold on itself, the map of a region doesn't need to have the same volume as this region, while a given region's volume of course shouldn't depend on the chosen coordinates.

Is this why $d^{4}x\sqrt{-g}$ isn't necessarily invariant under a diffeomorphism?

 

[haushofer]

This is what I'm unsure about really. Is the point that the author is making here that under a diffeomorphism we are always free to choose a coordinate transformation, $x\rightarrow x'$, such that the coordinate values of the "new" point $q$ in the "new" coordinate system $x'$ are the same as the coordinate values of the "old" point $p$ in the "old" coordinate system $x$, i.e. $x'^{\mu}(q)=x^{\mu}(p)$. Hence, we have that $d^{4}x\rightarrow d^{4}x'=d^{4}x$ such that $d^{4}x$ is diffeomorphism invariant, however, $\sqrt{-g}$ transforms as $\sqrt{-g}\rightarrow\sqrt{-g}+\mathcal{L}_{X}\sqrt{-g}=\sqrt{-g}\left(1+g^{\mu\nu}\nabla_{\mu}X_{\nu}\right)$ (where $X$ is the vector field generating the diffeomorphism), and so is not invariant under diffeomorphisms?

Yes. I think that's the point.

 

[Frank Castle]

Yes. I think that's the point.

Ok cool.

So, let me see if I have this right. If we have a diffeomorphism $\phi:M\rightarrow M$ such that $p\mapsto\phi(p)=q$, which in a local coordinate chart is given by, $x^{\mu}(p)\mapsto\left(\phi^{\ast}x\right)^{\mu}(p)=y^{\mu}(q)$. One can then introduce a coordinate transformation, $y^{\mu}(q)\mapsto x'^{\mu}(q)=x^{\mu}(p)$, such that the coordinates of $q=\phi(p)$ in the $x'$ coordinate chart are the same as the coordinates of $p$ in the original coordinate chart $x$.

If the above is correct, then there is still one thing I'm unsure about: say we have some tensor field $T$, then under the diffeomorphism $\phi$ it will transform as $\phi^{\ast}T$. When we make the coordinate transformation (described above) simultaneously with the diffeomorphism , do we not evaluate the transformed tensor $\phi^{\ast}T$ in a basis until we have made the coordinate transformation, such that $\left(\phi^{\ast}T\right)(x')=\left(\phi^{\ast}T\right)(x)$?

 

[haushofer]

I'm not sure what you mean by 'simultaneously', but I'll also be gone for some time and probably not able to respond. If I can with a proper answer, I'll do so.

 

[Frank Castle]

I'm not sure what you mean by 'simultaneously'

I was meaning that we carry out the diffeomorphism $x^{\mu}(p)\mapsto\left(\phi^{\ast}x\right)^{\mu}(p)=y^{\mu}(q)$ and the coordinate transformation $y^{\mu}(q)\mapsto x'^{\mu}(q)=x^{\mu}(p)$ at the same time.

If I can with a proper answer, I'll do so.

Ok. Thanks for all your help so far, it's much appreciated!