# Duality between passive VS active transfo's

1. Dec 17, 2008

### haushofer

Hi, I have a rather conceptual question about the exact duality between passive and active transformations in general relativity in order to fully understand the concept of diffeomorphism invariance ( I've read most of the topics about this subject here on PF ).
I found an article about this by John Norton, called "Coordinates and covariance: Einstein's view of space-time and the modern view" ( foundations of physics, Vol.19, No 10, 1989 ). I will briefly outline the relevant parts of this article concerning my question.

He contends that general relativity has a geometric structure:

$$M2 \equiv \ <M,\ g_{ab}, \ T_{ab}>$$

where I use the abstract index notation of Wald; a and b are NOT indices. This geometric structure has a coordinate representation as follows:

$$M3 \equiv \ <A, \ g_{\mu\nu}, \ T_{\mu\nu}>$$

A is here an open subset of $R^{4}$, and $g_{\mu\nu}$ and $T_{\mu\nu}$ are the 4x4 matrices representing the metric and energy momentum tensor. Now to the duality between transformation on M2 and M3.

We have:
- a coordinate chart $x^{\mu}: M \rightarrow A$
-an invertible function $f: A \rightarrow A$
-a diffeomorphism $h: M \rightarrow M$

The dual transformations are:

Active: h maps $p \in M$ with coordinate $x^{\mu}(p)$ to $hp \in M$ with coordinate $x^{\mu}(hp)$ ( because we are working in 1 chart ). This last coordinate can also be written as $x^{\mu}(hp) = fx^{\mu}(p)$.

Passive: the point p stays untouched; we simply go from one chart $x^{\mu}(p)$ to another $x^{'\mu}(p) = (fx^{\mu})(p)$. If we apply this to our metric tensor, we have:

Active: h induces the carry-along map h* which maps geometric fields on the manifold to geometric fields on the manifold. For scalars we have for instance h*f(hp)=f(p), for vectors we have $h*V^{a}(h*f)=V^{a}(f)$ etc. So the metric is carried along, and we get $h*g_{ab}$.

Passive: f defines the transformation between the objects of geometric fields in the coordinate charts $x^{\mu}$ and $x^{'\mu}$. For our tensor $g_{ab}$ this means that

$$g^{'}_{\mu\nu} = \frac{\partial x^{\alpha}}{\partial x^{'\mu}}\frac{\partial x^{\beta}}{\partial x^{'\nu}}g_{\alpha\beta}$$

, the tensor coordinate transformation rule. Now he states that "the duality becomes most apparent with the aid of the notion of a carried-along coordinate chart". He defines this carried-along coordinate chart via

$$hx^{\mu}(hp) \equiv x^{\mu}(p)$$

So this is what I don't understand. I understand that in the beginning we could state for an active transformation that on the manifold $h \rightarrow hp$ we induce a coordinate transformation $x^{\mu}(p) \rightarrow x^{\mu}(hp) = fx^{\mu}(p)$, but now he uses the diffeomorphism on the manifold to act on the coordinates! Why is this allowed? I don't see how one could possibly define this.

He then continues: the components of the carried-along object h*O at hp in the carried along coordinate chart hx equal numerically the components of the original object O at p in the coordinate chart x. For the metric this means that the carry-along is defined by

$$(h*g)_{'\mu '\nu}(hp) = g_{\mu\nu}(p)$$

The statement then is that the matrix components of the LHS and RHS of this equation are defined by the tensor coordinate transformation rule, and so that this rule covers both passive and active transformations.

In my head moving the point on the manifold is something very different than switching your coordinatelabels. For instance, in the Schwarzschild case, going from Cartesian to polar coordinates is something very different than transforming actively from 100 kilometers from the singularity to 10 kilometers from the singularity.

Can someone clarify this? I'm sure more questions will follow (like: are the so-called "gauge transformations" related to relabeling your coordinates and using the tensor transformation law, or something else? ), but I would appreciate some help, because this is frustrating me already quite some time :D

2. Dec 17, 2008

### Fredrik

Staff Emeritus
The only way I can make sense of this equation is if the first h is the pullback corresponding to h-1. If it is, I'd write the equation as

$$h^{-1}^*x(h(p))=x\circ h^{-1}(h(p))=x(p)$$

The first step is just the definition of a pullback. Note that h-1* acts on the function x, not on the n-tuple of real numbers x(h(p)). I'm writing h(p) instead of hp because it's only when a function is linear that it's conventional to drop the parentheses.

3. Dec 17, 2008

### atyy

I have never understood most of the literature on this. I like MTW's discussion: 1) general covariance is meaningless, because even Newton's laws can be made generally covariant 2) the distinction is between applying the diffeomorphism to "everything" or only to "dynamical objects", with GR being distinguished by a theory in which "everything is dynamical", ie. no prior geometry. There are still difficulties with these, as this very clear and hilarious article shows:

Some remarks on the notions of general covariance and background independence
Domenico Giulini
http://arxiv.org/abs/gr-qc/0603087

4. Dec 18, 2008

### haushofer

Yes, this makes sense, also if you draw a picture of the manifold with the points p and h(p) and the corresponding coordinate functions. But then it's not a definition; it's just a quite trivial property of the pullback you're using here. What does this mean "physically"? And how does this give the tensor transformation law both a passive and an active meaning?

A related question: if the "gauge freedom" of GR is all about the equivalence of 2 metrics connected by the tensor transformation law, is it then justified to think about this "gauge equivalence" in both a passive and an active way? This doesn't seem to make a lot of sense to me, except if you apply the active transformation to ALL the dynamical fields ( vector potentials, scalar fields etc which are present in our space-time ). Then I can understand the invariance: you shift actively all the dynamical fields, including the metric, which on its turn "brings along the geometry". ( so there is no one-to-one correspondence between the points on the manifold representing space-time and space-time itself )

5. Dec 18, 2008

### haushofer

It's good to hear I'm not the only one having problems with this :) It frustrates me, because even though I work with tensors a lot, this gives me the feeling I never quite understood the exact meaning of their transformation properties :S I'll read your article, thanks for the link!

6. Dec 20, 2008

### haushofer

Let me ask another question: are the Noether currents resulting from diffeomorphism invariance results of reparametrization invariance, or of actual space-time symmetries? According to this duality, I expect the answer "both", but the first is just the way we write things down ( tensorially ), while the second interpretation says something about the present symmetries of space-time. Those two things look very different from my point of view.

7. Dec 22, 2008

### atyy

What is the Noether current associated with diffeomorphism invariance?

8. Dec 22, 2008

### haushofer

A Lagrangian which is diffeomorphism invariant can be varied as

$$\delta L = E \delta\phi + d\Theta$$

and with this one can construct a current which is conserved if the equations of motion hold:

$$j = \Theta - \xi\cdot L$$

, where $\xi$ is the vector field defining this variation. See for instance "black hole entropy is Noether charge", by Wald.

But also in standard GR textbooks we can read the statement that "due to diffeomorphism invariance of the action, the energy-momentum tensor is conserved". I know that in special relativistic field theories, we can use Noether's theorem to derive the energy momentum tensor. I allways regarded this procedure as active; it says something about the present space-time symmetries, not about how we write our equations down ( tensorially ).

How do other people here look at these things?

9. Dec 22, 2008

### haushofer

The way Rovelli states it in "Loop quantum gravity and the meaning of diffeomorphism invariance" is that GR is special from other theories in that it is invariant under ACTIVE diffeomorphisms. Being invariant under passive diffeomorphisms is not special, because every theorie can be written down in such a way.

How can this be distinguished if the tensor transformation law exhibits both passive and active interpretations?

I think it's clear this is confusing me big time :')

10. Dec 22, 2008

### atyy

Bertschinger's notes 5 and 7 seem to have helpful comments on the distinction between coordinate transformtions and diffeomorphism invariance in Lagrangians. Also, diffeomorphism invariance in the Lagrangian apparently leads to Killing's equations, but as these don't have solutions in all spacetimes, there isn't always a globally conserved energy-momentum tensor. Somehow there is always a globally conserved energy-momentum pseudotensor.
http://web.mit.edu/edbert/GR/

Last edited by a moderator: Apr 24, 2017
11. Dec 22, 2008

### atyy

Perhaps useful.

Energy-Momentum Conservation in General Relativity
Dongsu Bak, D. Cangemi, R. Jackiw
http://arxiv.org/abs/hep-th/9310025

"Moreover, as we have seen, a variety of conserved currents may be derived, depending on whether one uses Einstein-Hilbert or Palatini formulations, whether the coordinate invariance is viewed as diffeomorphisms of geometrical variables or a gauge transformations on gauge connections. Finally observe that our expressions are neither diffeomorphism nor gauge invariant. At the same time, in all instances one Noether tensor gives the “correct” integrated expressions."

12. Dec 23, 2008

### haushofer

Last edited by a moderator: Apr 24, 2017
13. Dec 31, 2008

### atyy

"There are three general principles in general relativity relevant to Lorentz violation: general covariance (which implies both passive and active diffeomorphism invariance, the equivalence principle, and lack of prior geometry." Mattingly, http://relativity.livingreviews.org/Articles/lrr-2005-5/ [Broken]

Mattingly references Rovelli, but I wonder if you find Mattingly's terminology more satisfactory? If I remember right, I think Wald also says there's no big difference between active and passive diffeomorphisms in GR.

Last edited by a moderator: May 3, 2017
14. Jan 30, 2010

### jdstokes

I am in the middle of trying to understand the same problems of the people in this thread so I thought that I would resurrect it and throw in my 2 cents. I have also been trying to reconcile Bertschinger's notes with the rest of the literature on the subject such as Giulini's 06 article.

I gather that coordinate invariance is a property of any physical theory formulated on a manifold, whereas diffeomorphism invariance is a type of symmetry possessed by certain theories which are devoid on any non-dynamical geometrical structure.

I found that I was able to gain a lot of insight about coordinate and diffeomorphism invariance by thinking in terms of an action principle. Any equation of motion which is derivable from a finite-dimensional Lagrangian system is automatically generally covariant by covariance of the Euler-Lagrange equations under point-transformations. For GR this doesn't seem to carry much information, however, because the theory is infinite-dimensional and the coordinate invariance of the Einstein-Hilbert action is more like a type of parameterization invariance where the non-dynamical coordinate variables $d^4 x$ are treated like the parameter $d\tau$ appearing in the action for a point-particle in curved spacetime,

$S = \int d\tau \sqrt{g_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}}$.

Since it is clear how both $d^4 x$ and $\sqrt{-g}$ transform under general coordinate transformations, it is interesting to inquire what effect a diffeomorphism has from the action point of view. According to Bertschinger's notes, a diffeomorphism involves shifting all trajectories in the action $x^\mu \to y^\mu$, followed by a coordinate transformation such that the coordinates of the shifted points relative to the new coordinate system are identical to the coordinates of the original points relative to $x^\mu$.

I don't find this definition satisfying. The first part of the operation seems to convey the information of the diffeomorphism $\phi : M \to M$ which shifts points about in the manifold. Bertschinger argues that the second operation follows from a freedom to perform coordinate transformations arbitrarily due to the coordinate invariance of the theory. But where in this process do the operations of push-forwards and pull-backs of tensor fields enter?

In fact, for infinitesimal diffeomorphisms, I checked that Bertschinger's definition of diffeomorphism corresponds (for 1-forms) to applying the pull-back $\phi_\ast$, leaving the spacetime manifold fixed (i.e., not applying $\phi$ to $M$). This is consistent with the fact that the coordinate values do not change in Bertschinger's definition of diffeomorphism.

So if by diffeomorphism people actually mean pull-back (for 1-form), how does this fit in with Giulini?

According to him, a theory is diffeomorphism-invariant if the the dynamical (1-form) fields obey the same equations of motion after applying the pull-back operation. He also claims that non-dynamical geometrical data should remain intact.

From an action point of view, I would describe this process as pulling-back all 1-forms, leaving the non-dynamical geometrical data (i.e., the $d^4 x$) intact. Moreover, for infinitesimal diffeos, I claim this is precisely the process that is described by Bertschinger.

So there you have it, diffeomorphism invariance of GR is nothing other than invariance of the theory under the pull-back operation. Perhaps a better name would be pull-back invariance.