- #1
Afonso Campos
- 29
- 0
##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?
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?