Update on Overleaf.

main.tex

@@ -136,7 +136,7 @@ hence $Ax \in \mathbb{S}^3$ implying $A|_{\mathbb{S}^3}$ is a well-defined map a
 
 $$\|Ax - Ay\| = \|A(x-y)\| = \|x-y\|,$$ hence $A$ is an isometry.
 
-Nex, for any subgroup $H \leq SO(4)$ we can consider the orbits of the action of $H$ on $\mathbb{S}^3$. We recall from group theory that this produces an equivalence relation on $\mathbb{S}^3$ where $x \sim y$ whenever $x$ and $y$ are in the same orbit of $H$. That is, the set of equivalency classes of $\mathbb S^3$ with
+Next, for any subgroup $H \leq SO(4)$ we can consider the orbits of the action of $H$ on $\mathbb{S}^3$. We recall from group theory that this produces an equivalence relation on $\mathbb{S}^3$ where $x \sim y$ whenever $x$ and $y$ are in the same orbit of $H$. That is, the set of equivalency classes of $\mathbb S^3$ with
 $$x \sim y \,\iff\, \exists\, g\in H: x = g\cdot y.$$
 
 The set $\mathbb{S}^3/_\sim$ can be given a manifold structure if certain conditions on the projection $\pi: \mathbb{S}^3 \to\mathbb{S}^3 /_\sim $ are satisfied, the details can be found in \cite{serri}. For instance, if $H$ is a finite subgroup of $SO(4)$ then the manifold $\mathbb{S}^3 /_\sim$ is a 3-manifold which has positive curvature, similar to $\mathbb{S}^3$.