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For a Reissner–Nordström or Kerr black hole there is an analytic continuation through the event horizon and back out. Assuming this is physically meaningful (various site members hereabouts think not!) does this change the genus of spacetime?

In the 2D rubber sheet analogy a loop drawn around the event horizon wouldn't be contractible because it would go through the event horizon and back out and still have the black hole in it's centre. I don't know if the same argument applies to 4D spacetime.

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    $\begingroup$ @ChrisWhite - no, for a charged or rotating black hole it's possible to find time-like trajectories that pass through the event horizon, miss the singularity then pass back out through the event horizon. $\endgroup$
    – John Rennie
    Commented Dec 19, 2012 at 9:45
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    $\begingroup$ But in thinking about the topology of the manifold, shouldn't we ignore the event horizon because it's nothing special. For topology genus questions wouldn't we be looking at shrinking any closed curves, regardless of their being spacelike or timelike? $\endgroup$
    – twistor59
    Commented Dec 19, 2012 at 10:27
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    $\begingroup$ Ok, ignore the event horizon. $\endgroup$
    – John Rennie
    Commented Dec 19, 2012 at 11:37
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    $\begingroup$ I agree with @twistor59 that the causal classification of the curves shouldn't enter if we're just doing pure topology. $\endgroup$
    – user10851
    Commented Dec 19, 2012 at 20:32
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    $\begingroup$ The correct spelling of centre has now been restored :-) $\endgroup$
    – John Rennie
    Commented May 17, 2013 at 15:23

2 Answers 2

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Much more simply than Chay Peterson's answer, a look at the conformal diagrams for the two spacetimes in question:

Schwarzschild spacetime

Kerr-Newman spacetime (non-extremal)

It should be readily apparent that the global topological structure of these two spacetimes is radically different. In addition to this, the Schwarzschild spacetime has a pointlike singularity, while the Kerr spacetime has a ring singularity. Since the singularities cannot be included in the spacetime itself, these spacetimes will have different fundamental groups (a closed curve going through the ring singularity will not be smoothly deformable to a point), and thus be different topologically.

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To rephrase the question slightly, you are asking for one of the Betti numbers of the (3+1)-dimensional manifold corresponding to one of the solutions of the Einstein field equations that corresponds to charged or rotating black hole.

The Betti numbers of a manifold are topological invariants that intuitively represent the number of non-contractible d-dimensional "handles" on that manifold (or formally, the number of generators of the d-th homology group), and for a D dimensional manifold there are D+1 nonzero Betti numbers, $B_0,B_1, ... ,B_D$. The zero-th, $B_0$, corresponds to the number of connected components, $B_1$ the number of one-dimensional handles, and so on.

For example, the Betti numbers of a single torus $\mathbb{T}^2$ are: $B_0=1$, $B_1=2$, and $B_2=1$; and for a sphere $\mathbb{S}^2$ they are $B_0=1$, $B_1=0$, and $B_2=1$. These can be related directly back to simpler invariants you may already have heard of, like the Euler characteristic $\chi$, and the genus $g$, and we can see that for closed two-dimensional manifolds with no boundaries like the above, $g$ is related to the first Betti number:

$B_1 = 2g$

For surfaces where $B_1$ is not an even number, that is, for surfaces which are open or aren't orientable, this definition breaks down, and we talk instead of a non-orientable genus $k=B_1$. For higher dimensional manifolds we can take $k=B_{D-1}$.

So to return to your question, we want to know what the Betti numbers of the spacetime you're describing are. Assuming the topology of timelike curves is simple enough (i.e. no closed timelike curves) or that the solution is independent of time, it's easiest to look at the topology of the purely spacelike part of the solution; $D=3$, so $k=B_2$.

@Siva pointed out that the only relevant non-contractible surface that is introduced by the event horizon is one sphere, $\mathbb{S}^2$, so I would guess* that $B_2 = 1$. This would mean that $k=1$, which is a genus, but because $B_2$ is not an even number, essentially because asymptotically flat spacetimes like these are open, not closed, we can't interpret it in the same sense as the genus of a torus or another 2-d handlebody. But it is a genus in some modified sense, it's a non-orientable genus.

So, I would say that $k=1$ in this case, and clearly $k=0$ for an asymptotically flat spacetime with no black holes in it.

EDIT: To directly answer the question "Does a charged or rotating black hole change the genus of spacetime?"; yes, it increases the non-orientable genus ($k = B_2$) from $0$ to $1$.

*If there is a string theorist or someone else who can compute homology groups better than me, this might need checking. In 2D there's an isomorphism between the first homotopy group (which is really what Siva's sphere argument is pointing to) and the first homology group but this doesn't necessarily hold in higher dimensions IIRC. I did a quick calculation with a cellular homology that looked reasonable but I might be oversimplifying.

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  • $\begingroup$ I do not understand your point about $B_2$ being not an integer. It is the rank of a certain homology group (or equivalently, the vectorspace dimension of the same homology taken with rational coefficients). If it is not infinite, how can it be non-integer? $\endgroup$
    – Peter Kravchuk
    Commented May 17, 2013 at 19:35
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    $\begingroup$ If you consider the space time to have the only peculiarity -- the point of the singularity being removed, that is a whole time-like path cut out from the 3+1D space, then the homology groups are easily calculated. Homology groups are homopoty-invariant, so you only need to contract the full space-time to a constant-t slice, then it is an $\mathbb{R}^3$ without a point. The latter is contractible onto an $S^2$. So the homology groups are that of a two-sphere, i.e. $H_0(X,\mathbb{Z})=\mathbb{Z},\,H_1(X,\mathbb{Z})=0,\,H_2(X,\mathbb{Z})=\mathbb{Z}$, which says $b_0=1,\,b_1=0,\,b_2=1,\,\chi=2$. $\endgroup$
    – Peter Kravchuk
    Commented May 17, 2013 at 20:03
  • $\begingroup$ @PeterKravchuk Ah, that's a typo. I mean to write "even number", so it can't be identified with $g$. $b_3 = 0$ as well unless spacetime is asymptotically spherical, which would also change $b_1$ by Poincaré duality IIRC. $\endgroup$
    – Chay Paterson
    Commented May 18, 2013 at 11:42
  • $\begingroup$ My point is that we can't identify genus as the $g$ in $\chi=2-2g$, since e.g. $S^3$ has $\chi=0$, but as a sphere we still want to say its genus is zero (it's genus zero under a standard splitting). So I'm picking $b_2$, but since it's not an even number, it can't be genus in the same sense as a handlebody, but still counts as a nonorientable genus. $\endgroup$
    – Chay Paterson
    Commented May 18, 2013 at 12:22
  • $\begingroup$ About $b_3$, I wanted to write "and all other numbers are zero", but there was no space left.) $\endgroup$
    – Peter Kravchuk
    Commented May 18, 2013 at 19:28

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