Let $(X, d)$ be a metric topology and define $\nabla^x$ be the set $\{\{x,y,z\} \subset X^3\ \text{such that } d(x,y)=d(y,z)=d(z,x)\}$. My question is classifying $\nabla^{S^n}$ and $\nabla^{\mathbb R P^n}$. Hopefully you can see the curly braces in the definition - but in case you can't, it is the set of points in $X^3$ forming an equilateral triangle via $X$'s metric $d$.
So for $S^1$, we easily have $\nabla^{S^1}$. There is a sense in which the natural isomorphism $f : \mathbb Z \to \mathbb Z$ between $\tilde H_1(S^1) = \mathbb Z$ and $\tilde H_1(\nabla^{S^1}) = \mathbb Z$ where $\tilde H_1$ is reduced homology should satisfy $f(n)=3n$. That's because by rotating a third of a circle, the triangle you were spanning is now the same object in nabla. Keep in mind that Nabla doesn't care about the ordering of the vertices of the triangle - so that is the sense in which the covering map is 3-fold.
For $S^2$, If three vertices chosen on the sphere cannot form a great circle, then, we can contract them continuously down to a point. Since planes through the origin is described by a sphere with antipodal points identified, it looks like $\tilde H_1(\mathbb RP^2) \subset \tilde H_1(\nabla^{S^2})$ and is maybe equal but I might be missing something.
So how can we solve this for a non-orientable surface, where a bit more rigor is required and you aren't able to cleanly visualize the surface embedded in 3 spacial dimensions like our own universe.. Not to mention, I'm not sure exactly how to solve the higher dimensional variants. Not sure which methods will be needed to solve this, so I added some tags..