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## Main Question or Discussion Point

##2+1##-dimensional Einstein gravity has no

1. In ##D##-dimensional spacetime, a symmetric metric tensor

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

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

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

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

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

But what does it

Why is the word

*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?