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Consider the configuration space $\text{Conf}_n(\mathbb R^k)$, and consider the subgroup $G=\mathbb R^k\rtimes \mathbb R^{\times}\leq \mathbb R^k\rtimes \text{GL}(k,\mathbb R)=\text{Aff}(\mathbb R^k)$ of dilations and translations in affine group, then there is a natural smooth action $G\times \text{Conf}_n(\mathbb R^k)\to \text{Conf}_n(\mathbb R^k)$. Is the orbit space $\text{Conf}_n(\mathbb R^k)/G$ a smooth manifold of dimension $k(n-1)-1$ when $n\geq 2$?

This action is free but not proper, so I cannot use quotient manifold theorem. I also tried to immbed it in $(\mathbb R^k)^n/G$. Though $\text{Conf}_n(\mathbb R^k)$ is an open submanifold of $(\mathbb R^k)^n$, unfortunately $(\mathbb R^k)^n/G$ is not a smooth manifold, because the zero point is dense.

Rmk: I should have considered the action of $\mathbb R^k \rtimes\mathbb R^{+}$.

Eric
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    I think you want $\mathbb{R}^{\times}$ instead of $\mathbb{R}$? Dilations are an action of the multiplicative group. – Thorgott Feb 14 '24 at 13:29
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    You have a global cross-section of the $G$-action given by $x_1={\mathbb 0}$, $||x_2||=1$ (where $(x_1,...,x_n)$ are elements of $Conf_n(R^k)$). This cross-section is a manifold naturally homeomorphic to the quotient space. – Moishe Kohan Feb 14 '24 at 13:30

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Let $X$ be a topological manifold, $G$ a Lie group, $\mu: G\times X\to X$ a continuous action.

Definition. A submanifold $C\subset X$ is called a cross-section for the $G$-action on $X$ if each $G$-orbit in $X$ intersects $C$ in exactly one point (more precisely, for each $x\in X$ there exists $y\in C$ and $g\in G$ such that $\mu(g,x)=y$, and such pair $(g,y)$ is unique).

Note that, in general, cross-sections do not exist. The simplest example is the action of ${\mathbb Z}_2$ on $S^n$ generated by the antipodal map.

Lemma. If $C$ is a cross-section, then the map $G\times C\to X$, $f: (g,c)\mapsto \mu(g,c)$ is a homeomorphism.

Proof. Since $C$ is a cross-section and $\mu$ is continuous, the map $f$ is a continuous bijection. Since both $G, C, X$ are manifolds, the invariance of domain theorem (combined with the Baire category theorem) implies that $f$ is a homeomorphism. qed

In particular, the $G$-action on $X$ is proper.

Corollary. The quotient $X/G$ is homeomorphic to $C$.

Now, to your question. Consider the subset $$ C=\{(x_1={\mathbf 0}, x_2,...,x_n): ||x_2||=1, (x_1, x_2,..., x_n)\in X=Conf_n({\mathbb R}^k)\}\subset X=Conf_n({\mathbb R}^k). $$ This subset is clearly a submanifold in $X$ (of dimension $(n-1)k-1$). It is also clear that $C$ is a cross-section of the $G$-action on $X$, where $G_0$ is the semidirect product ${\mathbb R}^k \rtimes {\mathbb R}$, where $t\in {\mathbb R}$ acts on ${\mathbb R}^k$ via multiplication by the scalar $e^t$. (This is the identity component of your group $G$.) Thus the quotient $X/G_0$ is homeomorphic to $C$, hence, is a manifold of dimension $(n-1)k-1$. To get the quotient by the full group $G$ one further divides $C$ by the order 2 group acting freely and generated by the antipodal map. Thus, $X/G$ is still a manifold.

Remark. I have no idea how did you conclude that the $G$-action is not proper.

Moishe Kohan
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  • I was under the impression they considered the action of $\mathbb{R}^{\times}$ by scalar multiplication. In that case, one would have to additionally quotient by the induced $\mathbb{Z}/2\mathbb{Z}$-action on $C$ (simply given as multiplication by $-1$), which is free. – Thorgott Feb 16 '24 at 01:50
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    @Thorgott: I see. The question was edited to change the group $G$. – Moishe Kohan Feb 16 '24 at 02:10
  • @Moishe Kohan: So according to your lemma, the configuration space $\text{Conf}_n(\mathbb R^k)=G\times C$? I wonder if this can be generalized, for example on arbitary manifold $M$, $\text{Homeo}(M)$ acts on $\text{Conf}_n(M)$, but it seems that there is in general no cross-section. – Eric Feb 16 '24 at 05:55
  • @EricLey: Yes. In general one should not expect existence of a cross-section. – Moishe Kohan Feb 16 '24 at 06:10