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let $C^\Bbb N$ be the $\Bbb C$-vector space of complexe sequences $(u_i )_{i∈\Bbb N}$ with finite support, together with a norme $||(u_i )_{i∈\Bbb N}|| = ( \sum_{i∈\Bbb N} |u_i|^2 )^{1/2}$.

Let's denote $S^∞$ the $C^\Bbb N$ topological subspace of sequences with norm $1$.

Let $n ∈ \Bbb N, n ≥ 2$.

build a free action of $\Bbb Z/n\Bbb Z$ on $S^∞$ such that $p$ the canonical projection on the quotient space is a covering with $Aut(p) \cong \Bbb Z/n\Bbb Z$.

What I did:

I took the continuous function: $d: S^∞ \to S^∞$ defined by:

$d(u)_0 = 0$ et $d(u)_i = u_{i−1}$ for $i > 0$.

Then the following action is well defined:

$\Bbb Z/n\Bbb Z \times S^∞ \to S^∞: k.(u) = d^k(u)$.

This action is free. I need to prove that it is proper and that $S^∞$ is $T_2$ separated and locally compact.

My intuition is that $S^∞$ is compact, which implies that it is $T_2$ separated and locally compact and that the action is proper(since $\Bbb Z/n\Bbb Z$ is also compact).

Thanks for your help and comments.

Conjecture
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  • I just figured out that $S^∞$ is a metric space with a distance induced but the norm defined on $S^∞$, therefore $S^∞$ has a normal Hausdorff, so it is hausdorff – Conjecture Feb 20 '19 at 00:44
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    If you apply the function $d$ a total of $n$ times, why is it the identity function? – Jason DeVito - on hiatus Feb 20 '19 at 02:45
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    Can you find an action of $S^1$ on this space? If so, you can restrict that action to the subgroup of $n$-th roots of unity. – William Feb 20 '19 at 04:41
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    Also $S^\infty$ is not compact: the unit sphere in a (real or complex) vector space is compact iff the space is finite dimensional. (see for example this question: https://math.stackexchange.com/questions/287360/is-it-true-that-the-unit-ball-is-compact-in-a-normed-linear-space-iff-the-space ) – William Feb 20 '19 at 04:47
  • Right Jason! my action is not well defined. Right William, completely forgot Riesz theorem. Thank you! – Conjecture Feb 20 '19 at 13:48

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As Jason DeVito points out your function $d$ will not induce an action of $\mathbb{Z}/n\mathbb{Z}$ since $d^n$ is not the identity map. Moreover $d$ is not a homeomorphism of $S^\infty$ since it is not surjective (it misses the first basis vector).

Since $S^\infty \subset C^\mathbb{N}$ and $\|r\cdot v\|=|r|\cdot \|v\|$ for all $r\in\mathbb{C}$, there is an $S^1$ action on $S^\infty$ built into the complex vector space structure. In analogy to lower dimensions

$$ S^\infty/S^1 = \mathbb{CP}^\infty$$

which is a $K(\mathbb{Z}, 2)$.

When the action of $S^1$ is restricted to $\mathbb{Z}/n\mathbb{Z}$ (the discrete subgroup of $n$-th roots of unity) the quotient is $K(\mathbb{Z}/n\mathbb{Z}, 1)$.

William
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  • Thanks @William, now that the action is free, is it enough to say that the quotient map is a Galois covering? – Conjecture Feb 20 '19 at 15:18
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    The covering will be Galois by virtue of the fact that it is a quotient by a group action. If $G$ acts on $X$ in such a way that $q\colon X\to X/G$ is a covering map, then the covering will be Galois because the $G$ action induces a homomorphism $G\to Aut(q)$ and of course $G$ acts transitively in each orbit, so $Aut(q)$ is transitive in each fibre. Moreover since $S^\infty$ is simply connected (and actually contractible) the quotient $p \colon S^\infty \to K(\mathbb{Z}/n\mathbb{Z}, 1)$ is a universal covering space, and in particular $\mathbb{Z}/n\mathbb{Z}\cong Aut(p)$. – William Feb 20 '19 at 18:17