$2+1$-dimensional Einstein gravity has no local degrees of freedom. This can be proved in two different ways:

  1. In $D$-dimensional spacetime, a symmetric metric tensor appears to have $\frac{D(D+1)}{2}$ degrees of freedom satisfying $\frac{D(D+1)}{2}$ apparently independent Einstein field equations. However, there is a set of $D$ constraints on the equations due to the invariance of the equations under differmorphisms, and a second set of $D$ constraints due to the conservation of the stress-energy tensor. Therefore, there are really only

$$\frac{D(D+1)}{2} - D - D = \frac{D(D-3)}{2}$$

degrees of freedom of the metric tensor satisfying $\frac{D(D-3)}{2}$ independent Einstein field equations.

  1. In the ADM formulation in $D$-dimensional spacetime, the metric induced on the spacelike hypersurfaces appears to have $\frac{D(D-1)}{2}$ degrees of freedom. However, there is a set of $D$ constraints due to the $D$ Lagrangian multipliers in the Hamiltonian. Therefore, there are really only

$$\frac{D(D-1)}{2} - D = \frac{D(D-3)}{2}$$

degrees of freedom of the metric tensor.

The metric tensor of a manifold encodes information about the infinitesimal distance between nearby points on the manifold, so the are $\frac{D(D-3)}{2}$ degrees of freedom are all local degrees of freedom.

Therefore, it is said that $2+1$-dimensional Einstein gravity is trivial locally.

But what does it mean to say that $2+1$-dimensional Einstein gravity is non-trivial globally?

Why is the word topological used to describe $2+1$-dimensional Einstein gravity?


2 Answers 2


Let's start with a simpler example: $1+1$ dimensional gravity. This is actually a pretty important example to understand because all of string theory takes place in this framework!Within $1+1$ dimensional gravity, just as in $2+1$ dimensions, there are no local interactions. Thus, it seems like a pretty good starting place to talk about this kind of stuff.

If we denote by $h$ our metric tensor and by $R$ the scalar curvature of our manifold $\mathcal{M}$, then we can write


where here $h$ denotes the scalar $\det{h^{\alpha\beta}}$ and $\sigma$ is our choice of coordinates on open patches of $\mathcal{M}$. There is a beautiful theorem, known as the Gauss-Bonnet theorem, which states that this integral can be solved without the knowledge of the coordinates or the metric. It states that


where $\chi(\mathcal{M})$ is the so-called Euler characteristic of the manifold. This is a purely topological expression. Intuitively, it counts the number of "holes" in a continuous manifold (namely, $\chi(\mathcal{M})=2-2g$, where $g$ is the number of holes in the manifold -- for instance, a torus has one hole, and so has $\chi=0$, while a sphere has no holes and has $\chi=2$). Thus, since the action itself is independent of local features of the manifold but is dependent of the global (read: topological) features, we say that $1+1$ dimensional gravity is locally trivial but globally nontrivial.

The confusion may be caused by your (understandable) thinking that gravity is all about forces between objects. But in a modern sense, gravity is identical to geometry. This includes topological statements as well as local ones! We are used to talking about gravity as a local thing because the standard treatment of GR is a local one. But gravity determines the topology of our manifold as well as the local geometry!

This is all good and well, but why do we care? Well, in $1+1$ dimensional quantum gravity (read: string theory), the topology plays an important role! This is because the path integral treatment, we have (in a Euclidean signature),


where $\phi$ is a placeholder for all of the fields propagating on the manifold, $\mathcal{O}$ is some operator function of our fields, and


where $\lambda$ is an effective coupling constant and $\mathcal{L}(\phi,h)$ is the Lagrangian that defines the gravitational interaction of the fields $\phi$. The sum over $g$ indicates that we are summing over all possible topologies of the theory by counting over the genus. Using the Gauss-Bonnet theorem, we can immediately write down


Assuming $\lambda$ is positive, this tells us that we can essentially treat expectation values as being perturbative expansions in possible topologies of our manifold!

While I didn't really talk about $2+1$ dimensional gravity, the basic takeaway should be this:

  • While $2+1$ gravity doesn't have local degrees of freedom, the topology is still a geometric property of the spacetime manifold $\mathcal{M}$, and is thus still important and nontrivial.

  • Global nontrivialities can play an important role in the quantum version of the gravitational theory you're working in.

Finally, to answer your question in one sentence (as if I haven't babbled on enough): $2+1$ dimensional Einstein gravity is called "topological" because its only possible (vacuum) nontrivialities are all topological in nature.

[Note: throughout I assumed a vacuum or "pure" gravitational theory. Einstien gravity in $2+1$ dimensions with point masses is actually very interesting and has the same geometry as a cone. This can lead to some very interesting and unintuitive results, the most famous of which is the Gott time machine (worth a read).]

I hope this helped!

  • $\begingroup$ I think you could have elaborated on the details of the existence of non-trivial vacuum solutions being topological, rather than addressing $1+1$ dimensional gravity, which while interesting (it is part of my field), doesn't address the question, nor does it motivate your final answer to the question. $\endgroup$
    – JamalS
    Jun 18, 2017 at 19:35
  • $\begingroup$ Notational comment: in the beginning, you use $h$ for the metric tensor and two lines later as a determinant of the metric. Question: If in the 2D case we can use Gauss-Bonnet what is being done in the 3D case to show the topological nontriviality of the gravity theory? $\endgroup$ Jun 3, 2019 at 10:36
  • 1
    $\begingroup$ @RadekSuchánek The use of $h$ as both the metric tensor and its determinant is pretty unambiguous in context, since the latter almost always appears in a square root next to the volume element. Since the Gauss-Bonnet theorem doesn't apply in three dimensions, it is harder to see how topologically nontrivial spacetimes contribute to 3-dimensional Einstein gravity. However, the fact that there exist topologically nontrivial 3-dimensional Lorenzian manifolds with vanishing Ricci scalar tells us we do get these contributions. Furthermore, BTZ black holes exist, which are topological in nature. $\endgroup$ Jun 3, 2019 at 10:45

Bob Knighton's answer is very detailed. But I want to add a few remarks. You can try to prove this identity

$$ R_{\mu\nu\rho\sigma} = \frac{R}{D(D-1)} (g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho}). $$

Then, it's straightforward to show that for 1+1 spacetime, the Einstein tensor $R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu}$ is identically zero. That is, the vacuum Einstein field equation is trivially satisfied, no matter how much matter content you put. This means, there is no coupling between gravity and matter.

Similar consequences also happen on the 2+1 case. In fact, people had believed that 2+1 gravity was just as trivial as 1+1 gravity. However, in around 1990, it was discovered that in presence of a negative cosmological constant, 2+1 gravity is a rich subject (there are BTZ black hole solutions, etc).

  • 1
    $\begingroup$ That is not an identity. It only holds if the spacetime is maximally symmetric. $\endgroup$
    – Winther
    Jun 18, 2017 at 12:25
  • $\begingroup$ @Winther That's right. But what I wanted to mean is that in low dimensional spacetime, the Riemann tensor can be neatly expressed by the metric and Ricci scalar (because of few degrees of freedom), whence one can find trivial identities about the Einstein tensor. $\endgroup$
    – JamieBondi
    Jun 18, 2017 at 18:39
  • $\begingroup$ A short comment only. The answer by JamieBondi above is incorrect. The formula he writes down is valid only in D=2. In D=3, the Weyl tensor vanishes, and then the Riemann tensor can be written as a sum of the Ricci tensor times the metric, plus the Ricci scalar times the metric squared (the expression posted above). Standing like this, the answer given above is hence incorrect. $\endgroup$
    – Jens
    Oct 20, 2017 at 21:58

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