quotient of a hyperplane by the action of cyclic group let $H=\{(x,y,-x-y)\in \mathbb C^3\}$ and let $S^3$ the unit sphere in $H$. Why the following is true :

The linear action of $\mathbb Z_3$ on $S^3$ is free and  $H/\mathbb Z_3=C(M)$ the cone on $M$ where $M=S^3/\mathbb Z_3$ is a three-manifold
   (the apex at the origine and the cone extending to $\infty$)

Here $\mathbb Z_3$ is a subgroup of the symmetric group $S_3$ acting by permuting coordinates.
 A: First note that $S^3 = \{(x,y,-y-x)\mid |x|^2+|y|^2 + |x+y|^2 = 1\}$.  Since the action of $G=\mathbb{Z}_3$ permutes the coordinates, it's clear that $G$ preserves $S^3$.  Now, suppose $e\neq g\in G$ and that $g(x,y,-y-x) = (x,y,-y-x)$.  Then in particular, we also have $g^2(x,y,-y-x) = (x,y,-y-x)$.  It follows that we must have $x = y = -y-x$ and from this it follows that the fixed point was $(0,0,0)$, which is not an element of $S^3$.
Note that this proof doesn't use the fact that we're looking at the unit sphere, only that the sphere has nonzero radius.
Next, notice that $H \cong C(S^3) = S^3\times[0,\infty)/$~ where ~ collapses all of $S^3\times\{0\}$ to a point.  The map establishing this homemorphism can be defined as follows:  Every non $0$ point $q$ in $H$ determines a unique ray emanating from $0$.  This ray will pierce the sphere $S^3$ in precisely one point $f(q)$.  Now, map $H$ to $C(S^3)$ by sending $q$ to $(f(q), |q|^2)$ if $q\neq 0$ and sending $0$ to $S^3\times\{0\}$.  I leave it to you to prove this is a homeomorphism.
Further, this homemorphism is $G$ equivariant if $G$ acts on $C(S^3)$ by simply copying the $G$ action on each $S^3$.  That is, $g(p, t) = (gp, t)$.  (Again, I leave this to you to prove).  This should easily allow you to construct a homeomorphism from $H/G \cong C(S^3)/G$ to $C(S^3/G)$.
