Can this isometric embedding $\mathbb{R}^2/S_2 \hookrightarrow \mathbb{R}^2$ be generalised to higher dimensions?

Question

Endow $\mathbb{R}^2$ with the Euclidean metric. The symmetric group with two elements, $S_2$, acts on $\mathbb{R}^2$ by swapping the coordinates. Endow $\mathbb{R}^2/S_2$ with the quotient metric, which is again a metric space. Then the map $$ \begin{matrix} \mathbb{R}^2/S_2 &\rightarrow& \mathbb{R}^2 \\ [(x,y)] &\mapsto& (\min(x,y), \max(x,y)) \end{matrix} $$ is an isometry. Is there an extension of this to one of the following settings?

1. The symmetric group $S_n$ acting on $\mathbb{R}^n$ by permuting coordinates
2. The cyclic group with $n$ elements $\mathbb{Z}_n$ acting on $\mathbb{R}^n$ by cyclic permutation of coordinates
3. The group actions of $S_n$ and $\mathbb{Z}_n$ on $\mathbb{R}^n_+$, i.e. vectors in $\mathbb{R}^n$ with all entries being positive real numbers

Answer 1

You can get an isometry $\Bbb R^n/S_n\to \Bbb R^n$ by sorting the elements of the input tuple. More percisely, you map

$$[(x_1,...,x_n)]\;\mapsto\;(x_{i_1},...,x_{i_n})$$

so that $x_{i_1}\le x_{i_2} \le\cdots \le x_{i_n}$.

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The quotient $\Bbb R^n/\Bbb Z_n$ is essentially isometric to a product of 2-dimensional cones, which is not isometric to any subset of $\Bbb R^n$ (except for $n=2$), so there is no such isometry.

Here is what I mean precisely: let $C_k$ be the infinite 2-dimensional cone surface with circular link and apex angle $\alpha=2\arcsin(1/k)$, a finite part of which is depicted below.

The apex angle is chosen in such a way, so that when you unroll the cone, the mantle will go around $2\pi/k$ of the full circle.

$$\Bbb R^n/\Bbb Z_n\cong \begin{cases} [0,\infty]\times\Bbb R \times\prod_{k=1}^{\lfloor n/2\rfloor-1} C_{n/\!\gcd(n,k)} & \text{if $n$ is even}\\ \phantom{[0,\infty]\times}\,\,\,\!\Bbb R \times \prod_{k=1}^{\lfloor n/2\rfloor} C_{n/\!\gcd(n,k)} & \text{if $n$ is odd}\\ \end{cases},$$

So for example, you will find

\begin{align} \Bbb R^2/\Bbb Z_2&\cong \Bbb R\times\Bbb [0,\infty], \\ \Bbb R^3/\Bbb Z_3&\cong C_3\times\Bbb R, \\ \Bbb R^4/\Bbb Z_4&\cong C_4\times \Bbb R\times \Bbb [0,\infty], \\ \Bbb R^5/\Bbb Z_5&\cong C_5\times C_5 \times \Bbb R, \\ \Bbb R^6/\Bbb Z_6&\cong C_6\times C_3\times \Bbb R\times\Bbb [0,\infty], \\ &\;\vdots \end{align}

This decomposition results from the decomposition of $\Bbb R^n$ into (real) irreducible invariant subspaces of $\Bbb R^n$ w.r.t. the action of $\Bbb Z_n$.