Can a $1d$ space never be curved? I was wondering about this: Wikipedia article I refer to (here I refer to the first part: metric)
This wikipedia article claims that this hyperbolic space model has constant curvature $-1.$ I believe they are talking about sectional curvature here, because they write down a metric in $n-$ coordinates. Now, I have two questions:
What is about $n=1$: In that case, I think a Riemann curvature tensor will always be zero due to its symmetry properties, so there cannot be any curvature for the $1d$ case, right?
Is there a way to argue that since the sectional curvature is $-1$ in the case of $n=2$ (Gauss curvature of Poincaré disk is -1) it has to be $-1$ for $n \ge 3$ and any plane, too? By a symmetry argument, the sectional curvature should be independent of the plane we are considering at a point for $n \ge 3$(because the metric is the same in all coordinates). Actually, I would assume that a metric that is conformal to the identity always has this property. Despite, I don't see why it has to be $-1$ too for $n \ge 3$ without carrying out a cumbersome calculation.  
 A: For the first question: Correct -- a one-dimensional space cannot have any intrinsic curvature; the Riemann tensor always vanishes.
For the second: In 3-dimensional hyperbolic space, the Gaussian curvature of a geodesic plane will be $-1$, because it it itself a 2-dimensional hyperbolic space. This is most easily seen by considering the hyperplane that is represented in the ball model by a plane through the center of the ball. Since points in hyperbolic space are all equally good (something that is worth proving independently, by showing an isometry that moves any given point into the center spot), this holds for every geodesic plane through hyperbolic 3-space.
However, it is slightly problematic to extend this to say that the Gaussian curvature of hyperbolic space is $-1$. Namely, the very idea of "Gaussian curvature" is inherently about 2-dimensional manifolds -- in a manifold of higher dimension a single number is not in general enough to describe the curvature at a particular point. (Now there are too many degrees of freedom in the Riemann tensor. You can contract them away to get a scalar curvature number, but it will no longer tell everything about the intrinsic geometry of the space).
