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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.

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What action of $\mathbb Z_3$? –  Mariano Suárez-Alvarez Jun 30 '11 at 14:24
    
@ Mariano Suárez-Alvarez: $\mathbb Z_3$ acts by permuting coordinates. –  palio Jun 30 '11 at 14:35
    
Do you mean $S_3$? Else you'd have to specify how $\mathbb Z_3$ permutes the coordinates. –  joriki Jun 30 '11 at 15:12
    
@ joriki : $\mathbb Z_3$ is a subgroup of $S_3$ generated by $(123)$ –  palio Jun 30 '11 at 15:17
    
@palio: That still doesn't specify the action. It was already clear that only this subgroup of $S_3$ is isomorphic $\mathbb Z_3$, but you need to specify one of the two possible isomorphisms. By the way, people don't get pinged if you put a space between the '@' and the name. –  joriki Jun 30 '11 at 15:20
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1 Answer

up vote 3 down vote accepted

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 $C(S^3/G)$.

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DeVito: Thank you!! we can see that $H\cong \mathbb C^2=\mathbb R^4$ so can we say some thing for more general $H$ and $G$ : if $H$ is subspace homeomorphic to $\mathbb C^n$ on which a group $G$ acts equivariantly then $H/G$ is homeo to $C(S/G)$ where $S$ is the sphere in $H$? –  palio Jul 1 '11 at 9:39
    
@palio: You'd need to know 2 additional facts: 1. that $G$ acts linearly and 2. that $G$ preserves the spheres in some metric. When $G$ is a finite (or more generally, compact), then 2. is automatic. It's hard to find something not satisfying 1, but they do exist (even for finite groups I believe - I think there's a weird action of $\mathbb{Z}_2 on \mathbb{R}^6$ which isn't conjugate to a linear action though it preserves all spheres.) –  Jason DeVito Jul 1 '11 at 13:57
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