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Let the cyclic group $\mathbb{Z}_n$ act on $\mathbb{C}^n$ (or on $\mathbb{R}^n$, I'm interested in both) by permuting coordinates. What does the topological quotient $Q$ by this group action look like? More explicitly, I'd like to identify points under the equivalence relation $x\sim y\Leftrightarrow x=g\cdot y$ for some $g\in\mathbb{Z}_n$. Is there some nice way to embed it as a subset of $\mathbb{C}^m$ for some $m$?

So far all I've thought of is that if $f: \mathbb{C}\to\mathbb{C}$ is any function, $e_f: \mathbb{C}^n\to\mathbb{C}^n$ is the application of $f$ elementwise, and $\mathcal{F}$ is the discrete Fourier transform, then the map $\mathcal{F}_f^n: \mathbb{C}^n\to\mathbb{C}^n$ defined by $x\mapsto \left(\mathcal{F}\left(e_f(x)\right)\right)^n$ is fixed by the group action and so defines a map out of $Q$. It's easy to see that this map is not injective, but perhaps by concatenating $\mathcal{F}_f^n$ for several different $f$ one can get an injective map and by choosing nicely-behaved $f$ one can get a nice embedding.

But I'd bet there are more illuminating embeddings. I'm happy to consider related questions where some bad points are removed from $\mathbb{C}^n$ (such as multiples of the all-ones vector), $n$ is assumed to be prime, etc. I'm also interested in other (abelian, so far) group actions, like $\mathbb{Z}_n\times\mathbb{Z}_n$ acting on $\mathbb{C}^{n\times n}$, so if there is a general theory of such quotient spaces, I'd be interested to learn about it. It seems like this sort of question must be well-studied, but I am not sure where exactly it fits, so feel free to re-tag.

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    $\begingroup$ Do you want the quotient as a topological space? A scheme? A stack...? $\endgroup$ Dec 1, 2015 at 23:57
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    $\begingroup$ I would like to understand it as a topological space or even just as a set. Or to understand why that is a bad idea. $\endgroup$
    – Noah Stein
    Dec 2, 2015 at 0:02
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    $\begingroup$ Maybe too complicated if you are interested in topological embeddings, but $\mathbb{C}^n$ is an affine variety, the quotient is also an affine variety, so there is an algebraic (hence also topological) embedding of $Q$ into some $\mathbb{C}^m$. Finding an algebraic embedding is equivalent to finding a set of algebra generators for the ring of polynomials in $n$ variables invariant under cyclic permutation of the variables, this question has some good info $\endgroup$ Dec 2, 2015 at 2:10
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    $\begingroup$ @MikeShulman: That won't work because (in the $n=2$ case, say) $(0,2)$ should be identified with $(2,0)$ but not with $(1,1)$. $\endgroup$
    – Noah Stein
    Dec 2, 2015 at 19:01
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    $\begingroup$ @Mike: Noah wants to take the quotient as a topological space, not as a vector space. $\endgroup$ Dec 2, 2015 at 22:02

1 Answer 1

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The action of $\mathbb{Z}_n$ on $\mathbb{C}^n$ can be diagonalized: it's conjugate to the action sending a vector $(z_0, z_1, \dots z_{n-1}) \in \mathbb{C}^n$ to

$$(z_0, \zeta_n z_1, \zeta_n^2 z_2, \dots \zeta_n^{n-1} z_{n-1}).$$

The quotient can be stratified according to which of the $z_i$ are nonzero (starting with $z_1$). On the open stratum where $z_1 \neq 0$, there's a unique representative of each orbit where $0 \le \text{arg}(z_1) < \frac{2\pi}{n}$. In general, on the stratum where $z_1 = \dots = z_{k-1} = 0$ and $z_k \neq 0$, there's a unique representative of each orbit where $0 \le \text{arg}(z_k) < \frac{2 \pi \gcd(k, n)}{n}$. Each stratum lies in the closure of the previous stratum.

There should be a similar and more complicated story for the action of $\mathbb{Z}_n$ on $\mathbb{R}^n$, coming from the decomposition of the latter into real irreducible representations.

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    $\begingroup$ Your orbit representatives aren't quite unique because of the possibility $z_1=0$. $\endgroup$ Dec 2, 2015 at 1:24
  • $\begingroup$ I would like to easily define continuous maps out of $Q$, so having a continuous embedding of $Q$ into some vector space would make this trivial. This conjugacy was what brought the Fourier transform into the picture, but it wasn't immediately obvious to me how to choose representatives in a way that everything would be continuous. $\endgroup$
    – Noah Stein
    Dec 2, 2015 at 1:25
  • $\begingroup$ @Julian: thanks for the correction. I'll edit. $\endgroup$ Dec 2, 2015 at 1:30
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    $\begingroup$ 5 years on it’s surely unimportant, but $\mathbb C^{n - 1}$ should be $\mathbb C^n$. $\endgroup$
    – LSpice
    Oct 8, 2020 at 0:34
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    $\begingroup$ Yep, thank you. $\endgroup$ Oct 8, 2020 at 0:39

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