The Newtonian vacuum field equation $\nabla^2 \phi = \rho$ where $\phi$ is the gravitational potential and $\rho$ is proportional to mass density also has non-trivial vacuum solutions, for example $\phi = -1/r$ for $r$ outside some spherical surface. The Maxwell equations also have non-trivial solutions. In electrostatics, precisely the same as classical gravitation, in elctrodynamics, also radiative solutions of various kinds.

It is not strange that a field theory has non-trivial vacuum solutions. From a mathematical point of view, if it did not, it would not be possible to solve boundary value problems otherwise. Physically, a (local) field theory is supposed to provide a way for spatially separated matter to interact without spooky action at a distance. If interactions were unable to propagate through a region of vacuum we would have a very boring field theory!

If we want to be a little more specific to general relativity, let us note that this theory actually consists of **two** field equations. The most famous one is Einstein's, $$
R_{\mu\nu} = 8\pi T_{\mu\nu}
$$
which says that matter is the source for the field $\Gamma^{\mu}{}_{\nu\sigma}$ -- the Christoffel symbols. **This equation alone does not contain the fundamental characterization of general relativity**. It is just an equation for some field. For this field to actually correspond to the curvature it must also satisfy **the Bianchi identity** $$
R_{\mu\nu[\sigma\tau;\rho]} = 0.
$$

The Bianchi identity is redundant if the Christoffel symbols are defined the way they are in terms of the metric. This is actually analogous with electrodynamics (and for a very good reason, because ED is also a theory of curvature). The Maxwell equations are $$F^{\mu\nu}{}_{,\nu} = j^\mu $$ $$F_{[\mu\nu,\sigma]} = 0$$
and the first equation is the one that couples the electromagnetic field to matter. The second equation is redundant if $F_{\mu\nu}$ is defined in terms of the vector potential.

Now, the electromagnetic field has 6 components but as you can see only 4 of them really couple to matter directly. The second equation represents the freedom for the electromagnetic field to propagate in vacuum. (In fact if you do Fourier analysis to find radiation solutions to the Maxwell equations the first only tells you that radiation is transverse, and the second is the one that actually determines the radiation.) The components are naturally not independent since matter and radiation interact, but I think that this is a nice way to think about why of the classical Maxwell equations $$\begin{matrix}
\nabla \cdot \mathbf{E} = & \sigma\\
\nabla \cdot \mathbf{B} = & 0 \\
\nabla \times \mathbf{E} = & -\frac{\partial \mathbf B}{\partial t} \\
\nabla \times \mathbf{B} = & \mathbf{j} + \frac{\partial \mathbf E}{\partial t}
\end{matrix}$$
two involve only the fields and two involve matter.

Similarly for Einstein's general relativity, in the Einstein field equation $$R_{\mu\nu} = 8\pi T_{\mu\nu}$$ matter only couples to 10 components out of the 20 components in the Riemann curvature tensor. (The Riemann tensor is the physically observable quantity in general relativity.) The other 10 components are in the Weyl tensor. They are the part of the gravitational field that is present in vacuum, so they must include at least the Newtonian potential. By analogy with electrodynamics they also include gravitational radiation.

In the specific case of the Schwarschild and Kerr metrics, not only are all the components of the Ricci tensor 0, one can in fact arrange for all the components of the Weyl tensor except one to be 0 also. This is sort of analogous to how in electrostatics you can always choose the gauge so that the vector potential $\mathbf A = 0$. Perhaps you can think of this as saying that these metrics do not radiate, so only the part of the gravitational field whose limit is the Newtonian potential exists. (But there are radiating metrics with the same property, so maybe this isn't a good way to think.).

There are other vacuum metrics where fewer of the Weyl tensor's components can be made 0, or some gauge freedom remains. It is common to classify metrics along this scheme, which is called Petrov type. In a really famous paper Newman and Penrose show that the Petrov type of gravitational radiation has a near field - transition zone - radiation zone behavior, where more components of the Weyl tensor become irrelevant the further away from the source you go. (This is analogous with electrodynamics again, since in the radiation zone the EM field is transverse, but in the near field it is not.)

I think the essential problem lies in the difference between the mathematical meaning of *curvature*, and the way in which we actually describe a manifold, or a curved space (or spacetime).

Although we describe the universe as having spacetime curvature (which is mathematically true), *curvature* refers to the Riemann curvature tensor, which is a rank-4 tensor, meaning that it has $4^4 =256$ components, of which (due to various symmetries) $20$ are independent. This is far too cumbersome for even mathematicians to think about, but what is certainly true is that you cannot separate it nicely into space curvature and time curvature. As @G.Smith says in comments, "temporal curvature" does not make any sense. Time is a single dimension, and a one-dimensional subspace does not have any Riemannian curvature.

In other words, we use the mathematics of curved spacetime, but we don't actually describe anything directly in terms of Riemannian curvature. We do write Einstein's equation for gravity using the Einstein curvature tensor (or Ricci) but since this is zero except in the presence of mass-energy (the source of gravity), it does not directly tell us about the geometry of spacetime; to know that we have to solve Einstein's equation.

When we do solve Einstein's equation, we do not find curvature as such. Instead we find the **metric**. The metric is much easier to think about than curvature (we can write down a formula from which we could calculate curvature given the metric, but actually we never bother with that horrible calculation).

Rather than think about curvature, we think about scaling distortions in maps. In other words, we choose a coordinate system, and think about how actual or **proper** quantities appear in those coordinates. Proper quantities are the physical properties which would be measured by an observer moving with the object being measured.

We can compare this to scaling distortions in maps of the surface of the Earth. Any number of different maps are possible. The metric for the map tells us how to compare apparent distances on the map to actual distances as measured by someone on the ground.

So, rather than talk of curvature, talk of scaling distortions in maps. Then your question makes sense. For example, we cannot directly measure scaling distortions in Euclidean geometry in the region of the Earth, because they are too small. But we can, and do, measure scaling distortions in time. Clocks on GPS satellites measure the same unit of time as identical clocks on Earth. They measure exactly one second per second (as required by the general principle of relativity). But they *appear* on Earth to run at a different rate, because of the scaling distortion in the map used to describe them. Indeed, we can explain Newtonian gravity completely in terms of the scaling distortion of the time component, the scaling distortions of the space components being too small to have any impact.

## Best Answer

The Einstein equation allows spacetime to have an inherent curvature, but this is an adjustable parameter. That is, general relativity does not predict what the inherent curvature would be, only that it could exist and could take any value. The only way we can tell whether spacetime has an inherent curvature or not is by observation.

One of the terms allowed in the Einstein equation is a cosmological constant, normally written as $\Lambda$. If $\Lambda$ is non-zero then spacetime has a scalar curvature (the Ricci scalar) given by:

$$ R = 4\Lambda $$

And this is exactly the sort of curvature that you are asking about because it is "built in" to spacetime and exists even in a universe completely empty of matter or energy.

In fact does appear that the universe could have a cosmological constant. When we observe the motion of supernovae in the universe it looks as if the universe is expanding faster than it should, and this could be due to a cosmological constant. The trouble is that it could also be due to a form of energy called dark energy, and at the moment we cannot tell which (if either) of these is the case. One way to tell would be to see if the effect changes with the age of the universe. A cosmological constant would be ... well ...

constant, while dark energy could change with time. However at the moment our measurements are not precise enough to tell if the effect has changed over the age of the universe.