In Special Relativity, the Lorentz Group is the set of matrices that preserve the metric, i.e. $\Lambda \eta \Lambda^T=\eta$.

Is there any equivalent in General Relativity, like: $\Lambda g \Lambda^T=g$?

(We could at least take locally $g\approx\eta$, so we recover the Lorentz group, but I don't know whether we could extend this property globally.)

Why does Group Theory have much less importance in General Relativity than in QFT and particle physics?

  • 1
    $\begingroup$ Reference: Moshe Carmeli's Group Theory and General Relativity: Representations of the Lorentz Group and Their Applications to the Gravitational Field studies this question in quite some depth $\endgroup$ – Alex Nelson May 25 '13 at 16:19

As you point out, the Minkowski metric $\eta = \mathrm{diag}(-1,+1, \dots, +1)$ in $d+1$ dimensions possesses a global Lorentz symmetry. A highbrow way of saying this is that the (global) isometry group of the metric is the Lorentz group. Well, translations are also isometries of Minkowski, so the full isometry group is the Poincare group.

The general notion of isometry that applies to arbitrary spacetimes is defined as follows. Let $(M,g)$ be a semi-Riemannian manifold, then any diffeomorphism $f:M\to M$ (coordinate transformation essentially) that leaves the metric invariant is called an isometry of this manifold.

A closely related notion that is often useful in relation to isometries is that of Killing vectors. Intuitively a killing vector of a metric generates an "infinitesimal" isometry of a given metric. Intuitively this means that they change very little under the action of the transformations generated by the Killing vectors.

Isometries and Killing vectors are a big reason for which group theory is relevant in GR. Killing vectors often satisfy vector field commutator relations that form a Lie algebra of some Lie Group.

Addendum (May 28, 2013). Remarks on symmetric spaces and physics.

One can show that in $D$ dimensions, a metric can posses at most $D(D+1)/2$ independent killing vectors. Any metric that has this maximum number of killing vectors is said to be maximally symmetric.

Example. Consider $4$-dimensional Minkowski space $\mathbb R^{3,1}$. The isometry group of this space, the Poincare group, has dimension $10$ since there are $4$ translations, $3$ rotations, and $3$ boosts. On the other hand, in this case we have $D=4$ so that the maximum number of independent killing vectors is $4(4+1)/2 = 10$. It turns out, in fact, that each rotation, translation, and boost gives rise to an independent Killing vector field, so that Minkowski is maximally symmetric.

One can in fact show that there are (up to isometry) precisely three distinct maximally symmetric spacetimes: $\mathrm{AdS}_{d+1}, \mathbb R^{d,1}, \mathrm{dS}_{d+1}$ called anti de-Sitter space, Minkowski space, and de-Sitter space respectively, and that these spacetimes all have constant negative, zero and positive curvature respectively. The isometry groups of these spacetimes are well-studied, and these spacetimes form the backbone of a lot of physics. In particular, the whole edifice of $\mathrm{AdS}/\mathrm{CFT}$ relies on the fact that $\mathrm{AdS}$ has a special isometry group that is related to the conformal group of Minkowski space.

| cite | improve this answer | |
  • 3
    $\begingroup$ It's a quibble, but weren't Killing vectors named after a chap called Killing, so they should have a capital K? $\endgroup$ – John Rennie May 24 '13 at 6:39
  • 2
    $\begingroup$ @JohnRennie I hear Lie was way cooler than Killing. $\endgroup$ – joshphysics May 24 '13 at 15:18
  • $\begingroup$ @joshphysics Nitpick When you say "Minkowski metric" in the first line of your answer, you mean "Minkowski metric tensor" and when you say "(global) isometry group of the metric" you mean the metric itself and not the metric tensor. Is that right? I feel that one should distinguish between a metric and metric tensor. $\endgroup$ – SRS May 27 '18 at 13:00
  • $\begingroup$ @SRS In this context, I don't personally think it's necessary to be that pedantic. $\endgroup$ – joshphysics May 27 '18 at 19:38

Symmetry is just as important in General Relativity as it is in Special Relativity.

In SR the symmetry is the Poincare Group which is the group of mappings of space-time to itself that preserves the metric formula between events given by $d^2 = \Delta{x}^2+\Delta{y}^2 + \Delta{z}^2 - c^2\Delta{t}^2$ The Poincare group is an extension of the Lorentz group that also includes translations.

These group transformations are isometries of the fixed metric. The theory of isometries in GR on fixed backgrounds is described by Killing vector fields as in the answer by joshphysics. They form a Lie algebra using the Lie derivative and this generates the group of isometries, but this group is non-trivial only in special cases. Killing vectors are important for some applications but there is a larger symmetry in GR that is a symmetry of the full equations including a dynamic metric rather than a static background metric.

In GR the full symmetry is the diffeomorphism group of the manifold which is the group of all continuous and differentiable mappings (bijections) of the manifold to itself ("differentiable" means that functions of co-ordinates in patches are differentiable) The group is different for manifolds of different topologies but this is often overlooked. If the topology of the manifold is $ \mathbb{R}^4$ then you can embed the Poincare group into the diffeomorphism group in infinitely many ways.

The full equations describing physics in general relativity must be covariant under this diffeomorphism invariance. This is the case when you include the gravitational field equations as well as the equations of motion for matter. This is a much larger symmetry than the one in SR because the diffeomorphism group is infinite dimensional whereas the Poincare group is just ten dimensional.

| cite | improve this answer | |
  • $\begingroup$ Missing which point exactly? Yes, theories of gravity exhibit diffeomorphism invariance, and yes, the notions of isometry and Killing symmetry are relevant to fixed geometries, but these are the natural generalizations of Poincare invariance of $\mathbb R^{3,1}$; they are symmetries of a given metric. A huge amount of current research involves studying fixed backgrounds. What point is one missing in asserting that group theory is in large part important in GR because of the notions of isometry and Killing symmetry? $\endgroup$ – joshphysics May 24 '13 at 19:47
  • $\begingroup$ The key to his question is the last line which shows that he is really interested in how group theory applies to GR. He has been confused because he was looking only for a generalisation of isometries rather than the symmetry of the physical equations. If you answer just by telling him about the isometries that some background metrics have then you have just reinforced his misunderstanding that group theory is less important in GR, when in reality there is more symmetry and it is more important. $\endgroup$ – Philip Gibbs - inactive May 24 '13 at 20:32
  • $\begingroup$ You write "If you answer by telling him..." Huh? Essentially the entire relevance of Lie groups and Lie algebras to GR derives from these notions. Just open Weinberg's Gravitation and Cosmology and turn to the chapter on symmetric spaces or consider the entire field of AdS/CFT for evidence. I'm glad you discuss diffeomorphism invariance in your response because it's important, but I respectfully (and wholeheartedly) disagree that emphasizing isometries and Killing vectors reinforces any sort of misunderstanding about the importance (or lack thereof) of group theory in GR. $\endgroup$ – joshphysics May 24 '13 at 20:55
  • $\begingroup$ OK fair enough. I will edit my answer to make the point clearer and less contentious $\endgroup$ – Philip Gibbs - inactive May 25 '13 at 6:27
  • $\begingroup$ @PhilipGibbs Are these the type of diffeomorphisms studied in LQG? $\endgroup$ – jinawee May 26 '13 at 11:03

Group theory does play an important role in general relativity, and I'm aware of three different types of relevant symmetries:

First, there are the physical symmetries of specific solutions to the field equations, formalized by Killing fields, the generators of one-parameter groups of local isometries.

Second, there's general covariance. Mathematically, this means that general relativity needs to be formulated in terms of natural bundles where spacetime diffeomorphisms lift to bundle automorphisms called general covariant transformations. It's a type of gauge symmetry, but different from the one familiar from Yang-Mills-theory (the latter symmetry transformations are vertical and leave spacetime alone).

There's a third class of symmetries relevant in gauge gravity, the structural symmetries. If we start with an oriented spacetime, its tangent bundle is a vector bundle associated to a principal bundle with structure group $GL^+(4,\mathbb R)$. The pseudo-Riemannian metric is a classical Higgs field that breaks this symmetry by reduction to $SO(1,3)$. This symmetry can be further reduced to $SO(3)$, which yields space-time-decompositions as seen by physical observers.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.