Is it possible to have an S³ smooth manifold of constant curvature? Note: I am a mathematics enthusiast and may not be able to respond to any queries at the level they may be written at.
I believe it is possible to create an (S¹)³ space that has constant curvature at all points, as we could make the dimensions form Clifford Toruses pair-wise.
However, S³ is quite different, especially if I further require it to have differentiation defined within it (I believe that is what "smooth" means?), and my knowledge doesn't extend that far.
Is it possible that an S³ space be a smooth manifold with constant curvature (i.e. identical to an Euclidean space).
Could anyone be kind enough to satisfy my curiosity?
Thank you.
 A: Not every smooth manifold with constant curvature is even homeomorphic to Euclidean space. Example: the "flat torus", consisting of all points of the form
$$
(\sin t, \cos t, \sin s, \cos s),
$$
where $0 \le s,t \le 2\pi$ has constant curvature zero, but is a torus (so in particular, it's compact, which Euclidean space is not).
The standard sphere in 4-space has constant curvature (although not zero-curvature; thanks, @Didier, for pointing out the impossibility of this!).
A: The sphere $\mathbb{S}^3:=\{x \in \mathbb{R}^4:|x|=1\}$ with the induced metric has constant positive curvature. Note however that $\mathbb{S}^3$ is not identical to $\mathbb{R}^n$ for any $n$. $\mathbb{R}^n$ has constant zero curvature, rather than positive curvature. Also, $\mathbb{S}^3$ is bounded, in the sense that it has a finite maximum distance between two points, but it is boundaryless in the sense that it is a manifold without boundary.
$\mathbb{S}^3$ is a smooth manifold. Remember that a set $M$ is a 3-dimensional manifold (without boundary) if for every $p \in M$, there is an open set $U_p \subset M$ and a homeomorphism (continuous invertible function) $\phi_p:U_p \to \phi_p(U_p)$, where $\phi_p(U_p)$ is a subset of $\mathbb{R}^3$. A (3-dimensional) manifold $M$ is a smooth manifold if there is a "smooth atlas" of homeomorphisms $\phi_p : p \in M$, i.e. for each $p,q \in M$ such that $U_p \cap U_q \ne \emptyset$, $\phi_q \cap \phi_p^{-1} : \phi_p(U_p \cap U_q) \to \phi_q (U_p \cap U_q)$ is a smooth function. Everyone has their own favourite textbook for this - I would recommend John M Lee's Introduction to Smooth Manifolds as a good starting point.
