A ##2+1##-D Einstein gravity is topological and only non-trivial locally

[Afonso Campos]

$2+1$-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 diffeomorphisms, 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.

2. 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 $\frac{D(D-3)}{2}$ 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?

 

[atyy]

I am not sure, but one possible reason is that 3D gravity is associated with things such as BF and Turaev-Viro theories. Derek Wise's PhD thesis says, "In fact, 3d general relativity is a special case of 'BF theory', which we now describe more generally."
http://math.ucr.edu/home/baez/theses.html#derek
https://arxiv.org/abs/hep-th/9505027
https://arxiv.org/abs/hep-th/9304164

BF theory and 4D gravity are related if one adds a constraint.
https://arxiv.org/abs/1004.5371
https://arxiv.org/abs/1201.4247

One topological aspect of BF theory is that its ground state degeneracy depends on the topology of the manifold on which it is defined.
https://arxiv.org/abs/cond-mat/0404327v1
https://arxiv.org/abs/1006.0412v1

 

[haushofer]

$2+1$-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 diffeomorphisms, 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.

2. 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 $\frac{D(D-3)}{2}$ 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?

Because the Einstein equations of GR tell you something about the metric of spacetime, not about the topology of it. In 2+1 dimensions the vacuum Einstein equations tell you that the Riemann tensor vanishes (or is constant, with a cosmological constant), hence you don't have gravitational waves (local degrees of freedom). This is confirmed by the fact that the Newtonian limit of 2+1 GR tells you there is no gravitational interaction between masses. So the only thing to play around with is the topology of spacetime (the global structure of spacetime).

Topology and the geometry of angles and distances (the 'metric') are two separate, independent things!