It is widely accepted that the singularity of the Schwarzschild metric at the event horizon is purely an artifact of the coordinates and no physical singularity exists at the horizon. However, as Karlhede had shown in 1982, the Karlhede's scalar $R^{ijkl;m}R_{ijkl;m}$ (the square of the covariant derivative of the Riemann tensor) changes sign at the Schwarzschild horizon and therefore, in principle, a freely falling observer can detect the moment of crossing the horizon by local measurements (see http://arxiv.org/abs/1404.1845 , Karlhede's invariant and the black hole firewall proposal, by J. W. Moffat and V. T. Toth).

What are geometric meanings of the Karlhede's scalar and black hole horizon (if the latter indeed can be defined in an invariant way)?

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    Any arguments on the geometry of black hole spacetimes based entirely on the Schwarzschild example should be taken with a heavy grain of salt: it simply has too many symmetries. That said: for asymptotically flat space-times the even horizon certainly can be defined invariantly, as the boundary of $\mathcal{J}^-(\mathscr{I}^+)$. The apparent horizon, on the other hand, is much more difficult to define. The people studying dynamical horizons often use MOTS, but there are some problems with that even in Schwarzschild. – Willie Wong Sep 1 '14 at 8:43
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    If you look at the arXiv pre-print you cited (page 2, near eq. 18), for something as simple as Kerr, Karlhede's invariant changes sign on the ergosphere, and not the event/apparent horizon. To me this is evidence enough that while the invariant may have some physical meaning, that its vanishing coincides with the event horizon in Schwarzschild is exactly that: a coincidence. – Willie Wong Sep 1 '14 at 8:52
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    As far as I can tell their notion of Schwarzschild-like means the union of Schwarzschild, Reissner-Nordstrom, and Sch-dS; all are spherically symmetric and static. In other words, where I said Schwarzschild in my first comment you can equivalently read "Schwarzschild-like", using this notion. – Willie Wong Sep 1 '14 at 9:48
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    The fundamental reason that this has to be a coincidence is that event horizons are observer-dependent. A good example is an observer with constant proper acceleration in Minkowski space. Such an observer sees an event horizon, which clearly can't be connected to any intrinsic property of the spacetime, since the Riemann tensor vanishes identically. The horizon that we customarily talk about in the case of the Schwarzschild spacetime just happens to be the one seen by an observer at null infinity. – Ben Crowell Oct 9 '15 at 3:52
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    For those interested, there is an answer to this question on PhysicsOverflow. – Dilaton Nov 16 '15 at 14:21

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