What curves spacetime in Schwarzschild metric? I understand that the Schwarzschild solution is valid in the outside region of a massive object, with no other masses involved. Therefore the energy-momentum tensor is 0. But then: what curves space? In other words, in a vacuum without the presence of a massive object, the energy-momentum tensor is also 0, but that space is not curved.
Sorry if the question seems trivial, but I just don't understand.
 A: Stationary space-times like the Schwarzschild metric have a time-like Killing vector field $\xi^a$. A Killing vector field is a generator of isometries of the metric, so you can think of a time-like Killing vector field as generating the symmetry that corresponds to energy conservation. Every space-time where the coefficients of the metric are not explicitly time depend has a time-like Killing vector field.
The so-called Komar integral gives rise to the total energy, i.e. mass within a stationary space-time and is given by
$$M=-\frac{1}{8\pi G}\int_S\epsilon_{abcd}\nabla^c\xi^d,$$
where you integrate the so-called Hodge dual of the derivative of $\xi^a$ over the 2-sphere $S$ at space-like infinity ($r\to\infty$). (Note that there is no mess up with the indices, but the integrand is a two-form integrated over a 2-sphere.) The derivation of this can for example be found in GR by Wald from page 285 on.
Although in the Schwarzschild metric the energy momentum tensor is zero everywhere, being a vacuum solution, the Komar integral is not zero, showing that with this interpretation mass is a property of the geometry itself. As result we get the parameter from the metric which is usually interpreted as the mass of the black hole for which the metric is modelling it's surrounding (or also the mass of a star, if the metric is applied to one, but then it is only considered valid for $r$ larger than the star's radius)
A: As you correctly mentioned, the Schwarzschild (exterior) solution is valid only in the outside region of a massive object. However, the solution of Einstein field equations, the metric, constitues the whole space-time. Thus, in situation you describe, there is always an interior region of the space-time. The space-time is curved due to energy density and pressure contained there. In the simplest case the interior metric is given by so-called interior Schwarzschild solution. Because, I assume, you speak about a black hole, we must see what happens when the central pressure of the interior solution growths unbounded. The established conclusion is that the static interior and exterior space-time regions become transient, i.e. the object collapses, and thereafter two physically decoupled space-time regions remain: the static exterior, your Schwarzschild vacuum solution, and the transient vacuum interior due to space to time conversion. Because in the whole process the total mass (energy density) has not been changed, the space-time remains curved. Where is this energy located is an open question. I think it is the static external space-time. These two papers [1,2] seem to show that.
[1] D. Lynden-Bell and J. Katz, “Gravitational field energy density for spheres and black holes”, https://articles.adsabs.harvard.edu/pdf/1985MNRAS.213P..21L
[2] Aharon Davidson and Ilya Gurwich, “Hollography Driven Holography: Black hole with vanishing volume interior”
https://arxiv.org/abs/1007.1170
