I am self learning GR. This is a rather long post but I needed to clarify few things about the effect of general coordinate transformations on the global symmetries of metric. Any comments, insights are much appreciated.

To be concrete let's consider that $g_{\mu\nu}$ represents an axially symmetric spacetime i.e a Kerr black hole. In Boyer-Lindquist coordinates ($t,\, r,\, \theta,\, \phi $) metric has no dependence on $t$ and $\phi$ (they are cyclic coordinates):

$$ g_{\mu\nu} dx^{\mu}dx^{\nu}=-(1-\frac{2Mr}{r^2+a^2\cos^2{\theta}})\,dt^2-\frac{r^2+a^2\cos^2{\theta}}{r^2+2Mr +a^2} dr^2 +(r^2+a^2\cos^2{\theta})\, d\theta^2 \\ \,\,\,+ \left( r^2+a^2+ \frac{2Ma^2 r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}}\right) \sin^2{\theta}\, d\phi^2 - \frac{2Ma r \sin^2{\theta}}{r^2+a^2\cos^2{\theta}} dt\, d\phi $$

Consequently particle's energy $E_0$ and angular momentum $L_0$ are conserved. Suppose I make a coordinate transformation: $d \bar{x}^{\mu} = \Lambda^{\mu}_{\,\,\nu}\,dx^{\nu}$ where, if I am not mistaken, $\Lambda$ is a matrix, which is an element of the general linear group GL$\,(4,\,R)$ with a non-zero determinant. In the new coordinates metric may not have cyclic coordinates, a good example would be representation of the above metric in Kerr-Schild coordinates:

$$\begin{align} g_{\mu\nu} dx^{\mu}dx^{\nu}=&-d\bar{t}^2+dx^2+dy^2+dz^2 \\ &+ \frac{2Mr^3}{r^4+a^2z^2} \left[d\bar{t} + \frac{r(x \,dx+y \,dy)+a(x\,dy-y\,dx)}{r^2+a^2} + \frac{z\,dz}{r} \right]^2\\ &\text{where}\qquad r^4+ (x^2+y^2+z^2)\, r^2 -a^2 z^2=0 \end{align} $$

Now there is only one cyclic coordinate, $\bar{t}$. I presume we may in principle introduce new coordinates where none of the coordinates appear to be cyclic in functional form of $g_{\mu\nu}\, dx^{\mu}dx^{\nu}$. Suppose I make such a coordinate transformation. My questions are:

  1. Does the new metric still have global symmetries i.e conserved quantities that correspond to $E_0$ and $L_0$?

  2. If the answer to the first question is yes, then suppose I gave this new metric to someone without telling about the coordinate transformation and ask if there are any symmetries. Will she/he able to find symmetries by finding closed orbits in the geodesic flow?

To me the answer to both questions above seemed to be yes. I think that the integrability (in the Liouville sense) of the metric should not depend on one's definition of the coordinates. In other words, because of the global symmetries, we expect bounded geodesics around the black hole. In the old coordinates we may easily calculate such closed trajectories and according to my thinking closed geodesics exist (objects revolve around the black hole) regardless of how we label the coordinates.

But I could not be sure of this. To explain my confusion let me write:

$$ \frac{d x^{\mu}}{ds^2} + \Gamma^{\mu}_{\nu \lambda} \frac{d x^{\nu}}{ds} \frac{d x^{\lambda}}{ds}=0 $$

which gives the geodesic flow. The existence of the bounded trajectories here depends on the number of zeros and poles of the Christoffel symbol, and also on their relative locations. The thing is, one can mess with these parameters by making an arbitrary but an invertible coordinate transformation to the metric and therby change the flow. Also, perhaps somewhat related to this, it was pointed out here that a coordinate transformation, when viewed as a diffeomorphism, does not always map geodesics to geodesics unless it is an isometry.

  1. So what is the correct way to think about this? Do all coordinate transformations/diffeomorphisms conserve global symmetries? or a subgroup of them?

2 Answers 2


Continuous symmetries are the one-parameter groups of isometries generated by Killing vector fields. A Killing vector field $X$ is defined by the requirement (${\cal L}_X$ is the standard Lie derivative along $X$) $${\cal L}_X g =0\tag{1}$$ which is equivalent to the famous Killing equation $$\nabla_a X_b + \nabla_b X_a =0\:.\tag{1'}$$

These equations are intrinsic, so they are valid in every coordinate system.

It is possible to prove that the space of Killing vectors is a finite dimensional vector space and thus it admits a basis. Determining a basis of Killing fields fixes all continuous symmetries of your spacetime.

If you fix a Killing field $X$ and integrate it, you have a congruence of curves through the spacetime. In a neighborhood of every point you can complete these curves with other $n-1$ curves to construct a coordinate system $x^1, \ldots, x^n$ where $X = \frac{\partial}{\partial x^1}$. In that coordinate system, (1) takes a simple form $$\frac{\partial g_{ab}}{\partial x^1}=0\:.$$ That is the reason why symmetries can be seen looking at the components of the metric in suitable coordinate systems. It should be evident that changing coordinates does not preserve this property. If actively interpreting changes of coordinates, this is the same as saying that diffeomorphisms generally do not preserve the symmetries.

However there is a fundamental reason why this procedure (trying to represent symmetries by means of a suitable choice of coordinates) cannot exhibit all continuous symmetries of the metric simultaneously.

If composing all continuous symmetries in all possible ways, you obtain a Lie group $S$. It turns out that the Lie algebra $s$ of $S$ is represented by a basis of Killing fields: if $t_1,\ldots, t_k$ is a basis of $s$ and the commutation relations hold $$[t_i,t_j] = \sum_{k=1}^k c_{ij}^k t_k \tag{2}$$ the one parameter group generated by the $t_j$ define continuous symmetries generated by corresponding Killing fields $X_j$. The map $t_j \to X_j$ linearly extends to a Lie-algebra isomorphism because it turns out that (the constants $c_{ij}^k $ are the same as before) $$\{X_i,X_j\} = \sum_{k=1}^k c_{ij}^k X_k \tag{3}\:,$$ where $\{\cdot, \cdot \}$ is the standard Lie commutator of vector fields.

Here comes the problem with coordinates. Barring essentially trivial cases, $S$ is not Abelian and thus some of the constants $c_{ij}^k $ do not vanish. If $X_i$ and $X_j$ were tangent vectors to corresponding coordinates of a coordinate system, we would have $$\{X_i,X_j\} = \left\{\frac{\partial }{\partial x^i},\frac{\partial }{\partial x^j}\right\} =0 \tag{3'}\:$$ instead. So, the use of coordinate is not so useful and direct procedures to determine a basis of solutions of (1) or (1') are surely more effective.

As a trivial example of the problem, think of the standard Euclidean flat metric of $\mathbb R^3$. It admits the full group of rotations around the origin as symmetry (sub)group. It is however impossible to construct a coordinate system corresponding to the action of rotations around the three axes simultaneosly. At most one coordinate may be the integral line of the action of rotations around an axis (it happens in spherical coordinates and $\phi$ is such coordinate describing rotations around the $z$ axis and the component of the metric in spherical coordinates do not depend on $\phi$). The reason is that the group of rotations $SO(3)$ is not Abelian. Conversely, standard orthonormal Cartesian coordinates simultaneously represent the action of the three subgroup of translations: This is not a problem as the translations of $\mathbb R^3$ form an Abelian group.

  • $\begingroup$ Valter, thanks for pointing out the connection between the non-cyclic coordinates and underlying nonabelian group structure. It is true that Killing equation is the way to go but to find solutions for $X_a$, I assume you have to fix the coordinates anyway..So let me ask this in a more refined way. Do you think that invertible otherwise arbitrary coordinate transformations always preserve the number of independent solutions to the Killing equation ? $\endgroup$
    – user91411
    Nov 2, 2016 at 0:08
  • $\begingroup$ Yes, the number of linearly independent Killing vector fields does not depend on the used coordinates, it is an intrinsic notion. $\endgroup$ Nov 2, 2016 at 6:27
  • $\begingroup$ Thanks for clarification. A follow up question(s): In the case of Kerr metric we have only two conserved quantities. Do they still form a lie algebra ? Such conserved quantities can not be valid observables because their numerical value depends on the choice of coordinates, right ? $\endgroup$
    – user91411
    Nov 5, 2016 at 12:09
  • $\begingroup$ I do not understand what you mean by "conserved quantities" so I cannot answer to your second question. If you mean Killing vectors and the group of isometries they generate, yes the group of continuous isometries is always a Lie group. Maybe you have more than two Killing vectors: If you have two Killing vectors their commutator is still a Killing vector (possibly the trivial one). $\endgroup$ Nov 5, 2016 at 12:14
  • $\begingroup$ I meant the values of $E_0$ and $L_0$ as observables. $\endgroup$
    – user91411
    Nov 5, 2016 at 12:21

Well, coordinates do not exist in the physics: they are just labels we need so we can talk about the physics. That being the case, a coordinate transformation can't make any difference to the physical system it is describing: it can only describe the physical system more-or-less well. Good examples of 'less well' are where the coordinates become degenerate and you get an artificial singularity: the famous example of this is the Schwarzschild solution.

So any physical symmetries of the system must be entirely unaffected by choice of coordinates. So the answer to your first question is that yes, the metric will still have whatever symmetries it had, because these symmetries are properties of the physics, not the coordinate system.

However these symmetries may be deeply opaque, since there's really no limit to how bad a choice of coordinates I can make: so long as $\Lambda^\mu{}_\nu$ is nonsingular I can really pick any suitably smooth $\Lambda^\mu{}_\nu(x^\xi)$ that I like (here I'm using $x^\xi$ to mean 'all the coordinates', so really $\left\{x^\xi\right\}$: it's not an index you can sum over).

So, in general, if I give you a metric expressed in some coordinate system, you can't even know if it's the same physical system as some other metric. This is quite literally the case: there is no algorithm which will tell you if two metrics describe the same physical system.

So the answer to your second question, I think, is no: I can express a metric in such an awful way that it is impossible to know what is going on.

However things are not as bad as this in practice. In particular you're never actually going to choose some horrible transformation to make things obscure. And even if they are obscure there are a bunch of heuristic tricks you can use to try and work the metric into a form where the symmetries are apparent. It is a long time since I knew enough about this, but I think the sort of things you want to look for are Petrov classification, and also anything you can find on the classification of exact solutions (I can't find anything that isn't either on paper or behind a paywall, but there are a bunch of papers and books on this subject).

  • 1
    $\begingroup$ @CountTo10 I think so, yes. However I am working from fairly distant memory as well. $\endgroup$
    – user107153
    Nov 1, 2016 at 14:16
  • $\begingroup$ This is plain wrong. There is no way you can differentiate between active or passive diffeomorphism (relabelling). Changing coordinates do change the metric and hence the physics as well. Take for example case of a 2 sphere. Stretching it is active diffeormorphism while changing the metric (by coordinate transformation) in such a way that the sphere appears stretched is passive diffeomorphism. Since metric is part of the Riemannian manifolds there is no way you can differentiate between the two transformations. $\endgroup$ Nov 29, 2019 at 9:25
  • $\begingroup$ There is indeed an algorithm which does tell if two metrics describe the same physical system or not. It's called Cartan-Karlhede algorithm. $\endgroup$ Nov 29, 2019 at 10:26
  • $\begingroup$ @amateurRebel: not being able to distinguish between changes of coordinates (which clearly don't matter) and (active) diffeomorphisms (which do) is indeed part of the problem. Another part is that things like the Cartan-Karlhede 'algorithm' aren't: all of these things involve being able to answer questions like whether $E = 0$ for some expression $E$, and that's not computable in general. $\endgroup$
    – user107153
    Nov 30, 2019 at 14:30
  • $\begingroup$ @amateurRebel: in particular, for any particular algorithm, it is possible to pick a coordinate system for flat spacetime that algorithm can't know that the curvature is identically zero. $\endgroup$
    – user107153
    Nov 30, 2019 at 15:17

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