Curvature of space near a black hole

I read on Wikipedia that “The topology of the event horizon of a black hole at equilibrium is always spherical.”

This answer clarifies what that statement means. It means that if we start with any black hole in 4d spacetime, then consider the horizon as a 3d manifold by itself, this manifold has the topology $S^2\times \mathbb{R}$, where $S^2$ is a two-sphere (the surface of a ball) and $\mathbb{R}$ is a line. It's a statement about topology, not about geometry. In particular, the statement says (almost) nothing about geodesics (or parallel lines).

By the way, the statement is specific to black holes in 4d spacetime. In 5d spacetime, a black hole can have an event horizon with non-spherical topology.


Consider the Schwarzschild metric in 4d spacetime. The line element for spacelike worldlines is $$ ds^2 = -A(r) dt^2 + \frac{dr^2}{A(r)}+r^2d\Omega^2 \tag{1} $$ where $A(r)$ goes to zero on the horizon. The notation $d\Omega^2$ is an abbreviation for the spherical-coordinate part: without the factor of $A$, the combination ${dr^2}+r^2d\Omega^2$ would be the line element of flat 3d euclidean space in spherical coordinates. Any fixed value of $r$ defines a 3d submanifold of the 4d spacetime. If $A(r)\neq 0$, the induced metric on this manifold is $$ ds^2 = -A(r) dt^2 +r^2d\Omega^2 \tag{2} $$ where now $r$ and $A(r)$ are constants. This is the standard metric on $S^2\times\mathbb{R}$, where the factor $\mathbb{R}$ accounts for the extra coordinate $t$. On the horizon, we have $A(r)=0$, and equation (1) doesn't make sense there. The smooth manifold still makes sense there, but the components of the metric don't. We can address this in either of two ways:

  • Take $r$ to be arbitrarily close to this value. That's good enough to see what the topology of the $A(r)=0$ manifold will be. Equation (1) says that the $dt^2$ disappears on the horizon, which corresponds to the fact that the horizon is a null hypersurface: displacements in the $t$-direction are lightlike (have zero length).

  • Even better, we can use a different coordinate system so that the 4d metric is well-defined on the horizon. In Kerr-Schild coordinates, the Schwarzschild metric has the form $$ ds^2 = -dt^2+dr^2+r^2d\Omega^2 + V(r)(dt+dr)^2 \tag{3} $$ where $V(r)$ is well-defined everywhere except at $r=0$. The horizon corresponds to $V(r)=1$, where the $dt^2$ term disappears. Setting $r$ equal to this special value gives the induced metric $$ ds^2 = r^2d\Omega^2. \tag{4} $$ This is the standard metric on $S^2$, but the topology is actually $S^2\times\mathbb{R}$, where the $\mathbb{R}$ factor accounts for the $t$-coordinate. There is no $dt^2$ term in (4) because the horizon is a null hypersurface: displacements in the $t$-direction have zero length. This is the same conclusion we reached before, but now we have reached it more directly because the metric (3) is well-defined on the horizon.

I don't think it would be right to describe the spacetime near a black hole as "spherical". For one thing, the curvature of space changes depending on how close you are to the black hole. For a sphere, the curvature is a constant and does not vary with location. Also, you need more than a single real number to specify the curvature of space-times with dimensions higher than 2. (This is because you might have a space where the angles of a triangle oriented in one direction add up to less than 180 degrees, but the angles of a triangle oriented in a different direction add up to more than 180 degrees.) Also, the gravitational field of the black hole depends in large part on the fact that spacetime is curved, not just on spatial curvature.

You could probably still classify the curvature of spacetime based on the signs of various components of the curvature tensor, but the classification would be more complicated than spherical vs. flat vs. hyperbolic.