Supersymmetric isolated horizons

Details Supersymmetric isolated horizons Supersymmetric isolated horizons

2007-12-19

arxiv.org/abs/0712.3308v2

Class.Quant.Grav.25:105020,2008

Astronomy and Astrophysics

Astrophysics

General Relativity and Quantum Cosmology

26 pages, 1 figure; v2: typos corrected, topology arguments corrected, discussion of black rings and dipole charge added, refe

Scientific paper

10.1088/0264-9381/25/10/105020

We construct a covariant phase space for rotating weakly isolated horizons in Einstein-Maxwell-Chern-Simons theory in all (odd) $D\geq5$ dimensions. In particular, we show that horizons on the corresponding phase space satisfy the zeroth and first laws of black-hole mechanics. We show that the existence of a Killing spinor on an isolated horizon in four dimensions (when the Chern-Simons term is dropped) and in five dimensions requires that the induced (normal) connection on the horizon has to vanish, and this in turn implies that the surface gravity and rotation one-form are zero. This means that the gravitational component of the horizon angular momentum is zero, while the electromagnetic component (which is attributed to the bulk radiation field) is unconstrained. It follows that an isolated horizon is supersymmetric only if it is extremal and nonrotating. A remarkable property of these horizons is that the Killing spinor only has to exist on the horizon itself. It does not have to exist off the horizon. In addition, we find that the limit when the surface gravity of the horizon goes to zero provides a topological constraint. Specifically, the integral of the scalar curvature of the cross sections of the horizon has to be positive when the dominant energy condition is satisfied and the cosmological constant $\Lambda$ is zero or positive, and in particular rules out the torus topology for supersymmetric isolated horizons (unless $\Lambda<0$) if and only if the stress-energy tensor $T_{ab}$ is of the form such that $T_{ab}\ell^{a}n^{b}=0$ for any two null vectors $\ell$ and $n$ with normalization $\ell_{a}n^{a}=-1$ on the horizon.

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