# I Diffeomorphism invariance of GR

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1. Jul 29, 2017

### Frank Castle

it is often stated in texts on general relativity that the theory is diffeomorphism invariant, i.e. if the universe is represented by a manifold $\mathcal{M}$ with metric $g_{\mu\nu}$ and matter fields $\psi$ and $\phi:\mathcal{M}\rightarrow\mathcal{M}$ is a diffeomorphism, then the sets $(\mathcal{M},\,g_{\mu\nu},\,\psi)$ and $(\mathcal{M},\,\phi^{\ast}g_{\mu\nu},\,\phi^{\ast}\psi)$ represent the same physical situation.

Given this, how does one show explicitly that the Einstein-Hilbert action $$S_{EH}[g]=M^{2}_{Pl}\int d^{4}x\sqrt{-g}R$$ is diffeomorphism invariant?

I know that under an infinitesimal diffeomorphism $x^{\mu}\rightarrow y^{\mu}=x^{\mu}+tX^{\mu}$, generated by some vector field $X=X^{\mu}\partial_{\mu}$, the metric transforms such that $$\delta_{X}g_{\mu\nu}=\mathcal{L}_{X}g_{\mu\nu}=2\nabla_{(\mu}X_{\nu)}$$ and so $$\delta_{X}\sqrt{-g}=\sqrt{-g}g^{\mu\nu}\nabla_{(\mu}X_{\nu)}$$ Furthermore, the Ricci scalar transforms such that $$\delta_{X}R=\mathcal{L}_{X}R=X^{\mu}\nabla_{\mu}R$$ This is all well and good, but I'm unsure how the volume element $d^{4}x$ transforms under diffeomorphisms?! I'm inclined to think that it doesn't transform, since if I've understood things correctly, under a diffeomorphism, the points on the manifold are mapped to new positions, but simultaneously, the coordinate maps are "pulled back", such that the coordinates of the point at its new position in the new coordinate chart are the same as the coordinates of the point at its old position in the old coordinate chart. If this is correct, then I think I may be able to show that $S_{EH}$ is diffeomorphism invariant as follows: $$\delta_{X}S_{EH}=\int d^{4}x\left[(\delta_{X}\sqrt{-g})R+\sqrt{-g}(\delta_{X}R)\right]=\int\,d^{4}x\sqrt{-g}\left[\nabla_{\mu}X^{\mu}R+X^{\mu}\nabla_{\mu}R\right]\\ =\int d^{4}x\sqrt{-g}\nabla_{\mu}\left(X^{\mu}R\right)=\int d^{3}\Sigma\,n_{\mu}X^{\mu}R$$ where I have used Stokes' theorem in the penultimate equality, in which $d^{3}\Sigma$ is the surface element of the 3-dimensional boundary hypersurface to the manifold and $n^{\mu}$ a unit vector normal to this hypersurface. Now, assuming that $X$ has compact support, such that $X^{\mu}\rightarrow 0$ on the hypersurface $\Sigma$, the we find that $\delta_{X}S_{EH}=0$, i.e. the Einstein-Hilbert action is diffeomorphism invariant.

I'm not sure if this is correct at all, particularly my argument about where or not the volume element $d^{4}x$ transforms or not? Any help would be much appreciated.

2. Jul 30, 2017

### Geometry_dude

As you are looking at a local expression, you may view the diffeomorphism as a change of coordinates from $x$ to $x'$.

When doing this change, you will see that the volume form
$$\mu = \sqrt{-g} \, d^4 x$$
and the Ricci Scalar $R$ stay invariant under this coordinate transformation. From a more abstract perspektive, this is trivial as they are invariant objects (tensor fields).

3. Jul 30, 2017

### Frank Castle

I've read though, that under an active transformation (i.e. actually moving the points around on the manifold), that $d^{4}x$ is invariant, whereas $\sqrt{-g}$ is not? (I understand that under a passive coordinate transformation $d^{4}x$ is not invariant, but $d^{4}x\sqrt{-g}$ is). It is not necessarily true that tensor fields are invariant under (active) diffeomorphisms.

4. Jul 30, 2017

### vanhees71

$\mathrm{d}^4 x$ is not invariant, because you need the Jacobian when transforming it
$$\mathrm{d}^4 x' = \mathrm{d}^4 x \left |\mathrm{det} \left (\frac{\partial x^{\prime \mu}}{\partial x^{\nu}} \right ) \right|=\mathrm{d}^4 x |J|.$$
Now
$$g'=\frac{1}{J^2} g,$$
and thus
$$\sqrt{-g'}=\frac{1}{|J|} \sqrt{-g}.$$
So
$$\mathrm{d}^4 x \sqrt{-g}$$
is a scalar under general diffeomorphisms and thus invariant volume integrals should have integrands of the form $\sqrt{-g} \times \text{scalar field}$, and that's the case for the Einstein-Hilbert action.

5. Jul 30, 2017

### Orodruin

Staff Emeritus
A coordinate free way of seeing this is to write the integral as an integral of a volume 4-form $\eta$. Going from an integral of a 4-form to the coordinate expression, one can rather easily convince oneself that
$$\int \ldots \eta = \int \ldots \sqrt{-g} d^4 x,$$
where $\ldots$ represents a scalar expression.

6. Jul 30, 2017

### Frank Castle

What confuses me is that I've been reading this set of notes http://web.mit.edu/edbert/GR/gr5.pdf and from page 18 onwards they discuss the diffeomorphism invariance of the Einstein-Hilbert action and they note that under passive coordinate transformations $\sqrt{-g}d^{4}x$ is invariant whereas under active coordinate transformations $d^{4}x$ is invariant whereas $\sqrt{-g}$ is not.

7. Jul 30, 2017

### haushofer

I've read those notes too. Without jumping into detail, I just want to warn you that this discussion is overloaded with bad notation and different usages of terminology. I guess your paper's important statement is the following:

"In other words, we transform the coordinates so that the new coordinates of the new
trajectory are the same as the old coordinates of the old trajectory. The pushforward
changes the trajectory; the coordinate transformation covers our tracks."

So what he does as I understand it, is the interpretation of the Lie derivative as "active coordinate transformation plus a passive one".

8. Jul 30, 2017

### vanhees71

I have no clue, what this "active" vs. "passive" debate is about. A transformation is a transformation, and the mathematical "objects" transform as they transform. A diffeomorphism from one set of coordinates to leads to the transformation properties for $\mathrm{d}^4 x$ as given in my previous posting. For the metric components you get the transformation properties by the demand that the line element is invariant (by definition), i.e.,
$$g_{\mu \nu}'\mathrm{d} x^{\prime \mu} \mathrm{d} x^{\prime \nu}=g_{\rho \sigma} \mathrm{d} x^{\rho} \mathrm{d} x^{\sigma} = g_{\rho \sigma} \frac{\partial x^{\rho}}{\partial x^{\prime \mu}} \frac{\partial x^{\sigma}}{\partial x^{\prime \nu}}\mathrm{d} x^{\prime \mu} \mathrm{d} x^{\prime \nu},$$
i.e.,
$$g_{\mu \nu}'=g_{\rho \sigma} \frac{\partial x^{\rho}}{\partial x^{\prime \mu}} \frac{\partial x^{\sigma}}{\partial x^{\prime \nu}},$$
from which you get by taking the determinant of this equation (with the notation as in my previous posting)
$$\sqrt{-g'}=\frac{1}{J^2} \sqrt{-g}.$$

9. Jul 30, 2017

### Frank Castle

By active I was meaning that there is some mapping $\phi:\mathcal{M}\rightarrow\mathcal{M}$ such that $p\mapsto p'=\phi(p)$, i.e. the points on the manifold are "moved" around. This seems different from a passive transformation in which we simply change from one coordinate chart $\psi:\mathcal{M}\rightarrow\mathbb{R}^{n}$ to $\psi':\mathcal{M}\rightarrow\mathbb{R}^{n}$ by $\psi\circ\psi':\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ such that $x^{\mu}(p)\rightarrow y^{\mu}(x(p))$?!

10. Jul 30, 2017

### stevendaryl

Staff Emeritus
If you're talking about a single object, then it's pretty clear the difference between a passive and an active transformation is. I have some object--say, a pencil. We can simplify it's description, and say that it's described by a pair $(\mathcal{P}, \vec{V})$, where $\mathcal{P}$ is the location of one end, and $\vec{V}$ is the vector pointing from that end to the other. We can pick a set of basis vectors for the tangent space at $\mathcal{P}$ and using that basis, the vector $\vec{V}$ can be described by three numbers: $(V^x, V^y, V^z)$

There is certainly a distinction between
• Rotating the pencil so that it points in a different direction, thus changing its description from $V^x, V^y, V^z$ to $V'^{x}, V'^{y}, V'^{z}$
• Changing to a different basis, so that $V^x, V^y, V^z$ change to $V'^{x}, V'^{y}, V'^{z}$.
In both cases, you end up with a new description, $V'^x, V'^y, V'^z$. But the first change is changing the physical state of the pencil, while the second one is leaving the pencil unchanged, and is only changing your description of the pencil. That's the distinction that I understand between active and passive transformations. An active transformation changes the physical situation, while a passive transformation only changes your description of the situation.

The distinction becomes fuzzier, or disappears altogether, if you are transforming the entire universe. If you rotate the pencil, and also move every other object in the universe so that it keeps the same orientation relative to the pencil, that total transformation is, I suppose, no different than a change of coordinates.

In the case of GR, if you mess with all the matter and fields in a continuous way, and simultaneously adjust the metric, the result is physically the same as before.

Last edited: Jul 30, 2017
11. Jul 30, 2017

### Geometry_dude

Well, that's the idea, yes. Physically, however, it doesn't matter how you call the points, either. It's the geometric structures (metrics, volume forms, curves, etc.) mutually relating those different spacetime points that matter. Hence "Diffeomorphism invariance", which is really better termed "taking a categorical perspective":

Using category theory, which is a subfield of logic, one can make rigorous the notion for what it means for two spacetimes to be 'isomorphic'. Being isomorphic in some category is a fancy way of saying that two things constitute one and the same mathematical object in this context. That's all behind " diffeomorphism invariance":

The physics shouldn't depend on how you write it down and which conventions you use.

12. Jul 30, 2017

### vanhees71

Ok, then what I was discussing is a passive transformation, describing the same physical situation with different (local) coordinates, which are connected by a (local) diffeomorphism with each other. By construction GR is diffeomorphism invariant, which is a way to realize Einstein's (strong) principle of equivalence.

It's another thing if you describe symmetries, where you consider mappings of the points of the manifold. This you can describe (locally) in a fixed set of coordinates.

Considering infinitesimal versions of these different kind of transformations leads, e.g. to different ideas of derivatives. The former leads to covariant derivatives the latter to Lie derivatives.

13. Jul 30, 2017

### Geometry_dude

Well, this stuff is quite subtle so maybe the following remarks help:
The points of a manifold can be and are usually marked by more numbers than the dimension of the manifold. Take as an example the $2$-sphere
$$\mathbb{S}^2 = \left\lbrace \vec x \in \mathbb{R}^3 \middle\vert {\vec x}^2 =1 \right\rbrace \, \, .$$
Each point is marked by three numbers, yet the dimension of the manifold is two. If you pick a chart on it, e.g. spherical coordinates or stereographic projection, then you get two numbers, of course. People in GR usually do not care about this, because usually the manifolds of interest admit (almost) global coordinates - making it possible to identify points on the manifold with their coordinate values.

It is (tautologically) true that chart mappings and transition mappings between different charts are diffeomorphisms, but the word diffeomorphism is usually understood to be an active transformation', not a passive' coordinate transformation. An example of a diffeomorphism would be a rotation of the sphere
$$\varphi_A \colon \mathbb{S}^2 \to \mathbb{S}^2 \colon \vec x \to \varphi_A (\vec x) = A \cdot \vec x$$
with $A \in \text{SO}_3$.

Also note that a local diffeomorphism is, by definition, a (smooth) map between (smooth) manifolds of the same dimension, whose differential has full rank at each point. So, as opposed to a diffeomorphism, it need neither be injective nor surjective. Chart transition mappings are always both.

Diffeomorphism invariance' really refers to what I said above and doesn't (directly) have anything to do with the equivalence principle. The mathematical implementation of the equivalence principle in GR is done via the metric: Because gravity cannot be detected by local experiments', it cannot be a force (something that makes accelerometers spot acceleration) and thus must be described differently -> in GR via spacetime geometry.

Symmetries mathematically refers to Lie group actions, so, yes, active transformations'. Take, for instance, the mapping
$$\varphi \colon \text{SO}_3 \times \mathbb{S}^2 \to \mathbb{S}^2 \colon \left( A, \vec x \right) \to \varphi_A (\vec x) = A \cdot \vec x \, \, .$$
This is a (left) Lie group action and it can be shown that it preserves the metric $\mathbb{S}^2$ inherits from $\mathbb{R}^3$ via pullback, i.e. it is a symmetry of the Riemannian manifold $\mathbb{S}^2$ equipped with this metric.

14. Jul 30, 2017

### vanhees71

There seems to be some clash of conventions between physicsts and mathematicians. I misunderstood (?) "diffeomorphism" to mean the transformation from one set of coordinates to another, while in #13 it's meant as a map of the manifold on itself. Of course, in the latter sense GR is not diffeomorphism invariant. Even flat Minkowski space or the highly symmetric FLRW universes are not invariant under all diffeomorphisms in this sense.

I guess, related with this is also the problem that physicists often say "$A^{\mu}$" is a vector field and transforms under that and that kind of transformations in that and that way. Of course, that's sloppy since $A^{\mu}$ are components of a vector field $\boldsymbol{A}=A^{\mu} \boldsymbol{b}_{\mu}$ which is independent of the choice of the basis $\boldsymbol{b}_{\mu}$.

15. Jul 30, 2017

### Geometry_dude

That's a common issue...

Well, as I said, it's subtle. By definition (by mathematicians), a (smooth) diffeomorphism is a smooth map $\varphi \colon M \to N$between two (smooth) manifolds $M, N$, that is bijective and whose inverse is also smooth. One can show (see for instance the book "Introduction to Smooth Manifolds" by Lee) that it is sufficient for the map to be bijective and for the differential to have full rank at each point. From this definition, it follows that $M$ and $N$ need formally not be the same, but as diffeomorphisms are isomorphisms in the category of smooth manifolds, $M$ and $N$ may be considered the same in this context. Of course, if you equip $M$ and $N$ with, for instance, different metrics $g$ and $h$ (different in the sense that $\varphi^*h \neq g$), then this is not a very useful point of view (different category/context). If $\varphi^*h= g$, then $(M,g)$ and $(N, h)$ are again isomorphic (in the category of pseudo-Riemannian manifolds).

So if you have (almost) global coordinates (in the sense that the set where the coordinates are not defined is negligible'), then applying a diffeomorphism is essentially the same as a coordinate transformation - as long as you transform the other structures (metric, ...) as well.

I didn't get this point.

Well, that is sloppiness, indeed. I think the main point behind everything is that the coordinate representation is not the actual mathematical object, but only a way to express it. Names are smoke and mirrors.

16. Jul 30, 2017

### Haelfix

The point is that a given metric only preserves a subset of all possible diffeomorphisms, namely those that are isometries of the particular manifold.. the only metric that is invariant under all diffeomorphisms in the sense above has zeros for all entries. Hence the difference between a discussion about a theory being invariant, vs a particular solution being invariant.
Edit: for clarity

Last edited: Jul 30, 2017
17. Jul 30, 2017

### Geometry_dude

I do not understand what you mean with "the metric preserving diffeomorphisms". Care to elaborate?

18. Jul 30, 2017

### Orodruin

Staff Emeritus
There are diffeomorphisms that preserve the form of the metric (eg, Lorentz transformations on Minkowski space) and there are those that do not - although they describe the same physical scenario.

19. Jul 30, 2017

### Geometry_dude

Ahh, I get it, you mean isometries. Yes, the group of isometries
$$\text{Aut} \left( \mathcal M, g \right) := \left\lbrace \varphi \colon \mathcal M \to \mathcal M \middle \vert \varphi \, \text{is a diffeomorphism and} \, \varphi^*g =g \right\rbrace$$
of a Lorentzian manifold $\left( \mathcal M, g \right)$ is a proper subgroup of the group of diffeomorphisms of $\mathcal M$.

Actually, there appears to be a proof that even in the general case $\text{Aut}\left( \mathcal M, g \right)$ is a Lie group (thus finite dimensional), while the group of diffeomorphisms is usually not a Lie group (it's too big).

That is really what physicists mean with "symmetries", not "diffeomorphism invariance". Symmetries are something special and nice to have - while it simply doesn't matter which coordinate system you choose to describe a situation.

EDIT: It might also help to understand that an isometry in one coordinate system stays an isometry in another coordinate system. So again, coordinates are just smoke and mirrors, it doesn't matter which ones you choose - though some are nicer for calculations than others.

Last edited: Jul 30, 2017
20. Jul 31, 2017

### vanhees71

Indeed, symmetries are something specfic to a given manifold, while general covariance refers to the covariance under changes of the coordinates.