What topology has that static spherically spacetime? For a static spherically symmetric metric below:
\begin{equation}
ds^2=(r/r_{0})^4~c^2dt^2-dr^2-r^2(d\theta^2+sin^2\theta~d\phi^2),\tag{1}
\end{equation}
where $r$ is sectional curvature radius and $r_{0}$ some constant,
the obtained scalar curvature $S$ yields:
\begin{equation}
S= 12/r^2.\tag{2}
\end{equation}
What topology has the manifold characterized by that expression, is it $\mathbb{R}\times\mathbb{S}^3$ or  $\mathbb{S}^4$ ?
The idea to understand it as a $\mathbb{S}^4$ originates from the paper on "Riemann manifolds dual to static spacetime" https://arxiv.org/abs/2004.10505. The metric (1) can be written as
\begin{equation}
ds^2+dr^2+r^2(d\theta^2+sin^2\theta~d\phi^2)=(r/r_{0})^4~c^2dt^2,\tag{3}
\end{equation}
where $s, r, \theta, \phi$ are coordinates of a four-dimensional Euclidean (Riemann) space.
 A: Not a full answer, but some information about the connection between local and global information on manifolds:
In general you cannot infer the topology (global structure) of a manifold from a given metric tensor (local structure). However, often there are obstructions to the existence of certain metrics on certain manifolds (particularly, for Lorentzian metrics) or additional assumptions, which give you information about the topology using some deeper theorems:

*

*The topology is certainly not $\mathbb{S}^4$ because there are no Lorentzian metrics on even-dimensional spheres. This follows from the Euler numbers or the hairy ball theorem. See e.g. O'Neill, Semi-Riemannian Geometry, Prop. 5.37

*As the metric you gave is stationary (even static), you can employ some mild causality assumptions (being distinguishing suffices) to infer that the manifold splits as $\mathbb{R}\times M^3$ with $M^3$ being a three-dimensional manifold. See, e.g.: https://arxiv.org/pdf/0806.0812.pdf

*Note that you need some additional assumptions, as otherwise "your $t$-direction could wrap-around", i.e., the topology could be $\mathbb{S}^1\times M^3$.

*As soon as you know the manifold splits as $\mathbb{R}\times M^3$ you can still think about the topology of $M^3$ using the Riemannian part of your metric. I am not sure why it should be $\mathbb{S}^3$, but maybe this is part of your assumptions.

A: The simplest topology is clearly $\mathbb{R} \times (\mathbb{R}^3 \setminus \{ 0 \})$. The singularity corresponds to $r=0$, and is a (timelike) naked singularity. The paper you mention has no bearing on this question, it just establishes a correspondence between static spacetimes and Riemannian 4-manifolds which maps geodesics to geodesics.
