# Two dimensional spacetime and the Gauss Bonnet theorem

Generally two dimensional spacetimes are deemed to be static, as the Gauss Bonnet theorem implies that the Einstein Hilbert action would be a constant independent of $g$.

But as far as I can tell, the Gauss Bonnet theorem only applies to compact manifolds, even in the version for Lorentzian manifolds. The extension to non compact manifold is only an inequality, which does not specify that the result has to be independent of the metric.

Are two dimensional manifolds actually static as far as the metric goes, and if so, how to prove it?

As far as I can tell, I have to use the bounded version of the gravitational action, where, for a compact subset $U$ of the manifold, the action is

$$S = \int_U d^2x \sqrt{-g} R(x) + 2\int_{\partial U} d\lambda \sqrt{h}{h^a}_b(\lambda ) \nabla_a n^b(\lambda )$$

the second term being the Gibbons-Hawking-York term, $h$ the induced metric on the boundary and $n$ its normal unit vector.

For this to coincide with the Gauss Bonnet theorem, I need to show that

$$2\int_{\partial U} d\lambda \sqrt{h} {h^a}_b(x) \nabla_a n^b(x) = \int_{\partial U} k_g dA$$

with $k_g$ the geodesic curvature of the boundary, the norm of the tangent vector of the boundary.

• Should it not suffice, that the action is constant on all compact subsets of the manifold? Then you get the result for each compact subset, which say that subset is static. As you can cover your manifold with compact sets this implies the result. – Sebastian Riese Nov 21 '15 at 20:29
• Wouldn't a 2D FRWL space-time be an example of a non static space-time? – MBN Nov 22 '15 at 9:14
• "Static" as in "It does not depend on the matter content", not as in "It has a timelike Killing vector field" – Slereah Nov 22 '15 at 23:36