BdS:

It depends on what model of "space" you want to use, I guess.

For example, in the theory of relativity there are spatial and time coordinates (which transform from one reference frame to the next). Then there are various tensors (e.g. the curvature tensor) and other quantities (e.g. the spacetime metric) that describe how the coordinates interact and transform, as well as how distances and time intervals are to be measured using the coordinates. I guess that, if you like, you could call all of these things "properties of space". I'm not sure if I would.

The spacetime manifold is curving. Curvature describes a set of mathematical relationships in the model, essentially.

Are you being sarcastic? Where have I claimed that I already know everything? There are lots of things I don't know.

It is not true that nobody understands the general theory of relativity. However, a more than superficial understanding of it requires quite a high level of mathematics as a prerequisite. Most people do not have the mathematical background required to understand the mathematics of the theory. But some do.

I'll give you a simpler example, as an analogy. Consider a curve in two dimensions. We can represent that curve on an x-y plot as a function $y=f(x)$.

Without showing you how it is derived, I can tell you that the curvature at any point on such a curve is given by the following mathematical formula:

$$C=\frac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$$

where $f'(x)$ is the first derivative of the function and $f''(x)$ is the second derivative.

This looks complicated, but I'll give you a specific example or two.

Consider the line $y=f(x)=2$. Clearly, on an x-y graph, that is just a straight line. Applying the formula above, we have $f'(x)=f''(x)=0$, and so we easily see that $C=0$. Our line has zero curvature.

Next, consider a slightly more complicated straight line: $y=f(x)=7x$. For that line we have $f'(x)=7, f''(x)=0$. Plugging into the formula for the curvature, we again see that $C=0$, as expected.

How about a parabola, then? Consider $y=f(x)=x^2$. The derivatives in that case are $f'(x)=2x, f''(x)=2$. So, for the parabola, the curvature is:

$$C=\frac{2}{[1+4x^2]^{3/2}}$$

Notice that the curvature is no longer a constant; rather, it varies depending on where you are on the curve. For instance, at $x=0$ (the apex of the parabola), the curvature is $C=2$. The magnitude of C, 2 in this case, is a measure of how "curvy" it is at that point. If we pick a point a little to the right, at $x=1$ and calculate we find:

$$C=\frac{2}{5^{3/2}}=0.18$$

At $x=1$ the parabola is not as tightly curved, compared to $x=0$. If we look at the general curvature formula, we can see that our parabola becomes less and less "curvy" as $x\rightarrow \infty$.

So, you get the idea that this measure, C, quantifies our common-sense notion of how curved a line is.

In general relativity, things are more complicated. We're dealing with curvature in four dimensions (curvature of a spacetime manifold), and the spacetime at a single point can potentially curve in different directions by different amounts.

Of course, all of this is just maths. The power of general relativity is that it tells us

*why* spacetime curves. Matter and energy cause the curvature, and we can quantify exactly how much and what kind of curvature a given amount of mass, say, will cause. GR also allows us to calculate how objects will move in the curved spacetime (under the influence of gravity).

??

What am I doing with you now?