\section{Introduction}
\label{sec:Pse_Introduction}

The question posed by Davis ``When is the underlying space of a smooth orbifold a topological manifold?" \cite[p.~9]{MR2883685} amounts to the classification of finite subgroups of the orthogonal group $\Or_n$ for which the quotient space $\R^n/G$ is a topological manifold and has been completely answered in \cite{Lange_Mik}, \cite{Lange2} and \cite{Lange1}. The quotient space $\R^n/G$ not only inherits a topology from $\R^n$ but also other structures, i.e. a metric and a piecewise linear structure. One may as well ask when $\R^n/G$ is a manifold with respect to such an additional structure. In case of the quotient metric this task translates to the question of when a Riemannian orbifold is a Lipschitz manifold. Moreover, it makes sense to admit manifolds with boundary in the formulation of Davis's question.

We call a finite group generated by \emph{reflections} and \emph{rotations} in a finite-dimensional Euclidean space, i.e. by orthogonal transformations with codimension \emph{one} and \emph{two} fixed point subspaces, a \emph{reflection-rotation group}. We call it a rotation group if it is generated by rotations. A classification of reflection-rotation groups is contained in \cite{Lange_Mik}. Based on earlier results (cf. \cite{Mikhailova}, \cite{Lange2,Lange1}) we answer the question posed in the title in the following ways. Theorem A has already been obtained in \cite{Lange2}, we state it for completeness.

\begin{thmAB} \label{thm:mikhailovas_theorem} For a finite subgroup $G<\Or_n$ the quotient space $\R^n/G$ is a PL (Lipschitz) manifold with boundary if and only if $G$ is a reflection-rotation group. In this case $\R^n/G$ is either PL (bi-Lipschitz) homeomorphic to the half space $\R^{n-1}\times\R_{\geq 0}$ and $G$ contains a reflection or $\R^n/G$ is PL (bi-Lipschitz) homeomorphic to $\R^n$ and $G$ does not contain a reflection.
\end{thmAB}

In particular, the underlying space of a Riemannian orbifold is a Lipschitz manifold if and only if all its local groups are reflection-rotation groups. The quotient space $S^3/P$ of a $3$-sphere by the binary icosahedral group $P<\SOr_4$ is Poincar\'e's homology sphere and it follows from Cannon's double suspension theorem that its double suspension $\Sigma^2(S^3/P)$ is a topological $5$-sphere \cite{MR541330}. We refer to the realization of the binary icosahedral group in $\SOr_4$ as \emph{Poincar\'e group}. The answer in the topological category reads as follows. 

\begin{thmC}\label{thm:theorem_C} For a finite subgroup $G < \Or_n$ the quotient space $\R^n/G$ is a topological manifold with boundary if and only if $G$ has the form
\begin{eqnarray*}
		G = G_{rr} \times P_1 \times \ldots \times P_k
\end{eqnarray*}
for a reflection-rotation group $G_{rr}$ and Poincar\'e groups $P_i<\SOr_4$, $i=1,\ldots,k$, such that the factors act in pairwise orthogonal spaces and such that $n>4$ if $k=1$ and $n>5$ if $G$ contains in addition a reflection. In this case $\R^n/G$ is either homeomorphic to the half space $\R^{n-1}\times\R_{\geq 0}$ and $G$ contains a reflection or $\R^n/G$ is homeomorphic to $\R^n$ and $G$ does not contain a reflection.
\end{thmC}

The if directions of Theorem B and C follow from the if direction of Theorem A which has been proven in \cite{Lange2} (cf. \cite{Mikhailova}) and the double suspension theorem. To show the only-if direction we first prove

\begin{thmD}\label{thm:theorem_CD} For a finite subgroup $G < \Or_n$ the quotient space $\R^n/G$ is a homology manifold with boundary if and only if $G$ has the form
\begin{eqnarray*}
		G = G_{rr} \times P_1 \times \ldots \times P_k
\end{eqnarray*}
for a reflection-rotation group $G_{rr}$ and Poincar\'e groups $P_i<\SOr_4$, $i=1,\ldots,k$, such that the factors act in pairwise orthogonal spaces. In this case the boundary of $\R^n/G$ is nonempty if and only if $G_{rr}$ contains a reflection.
\end{thmD}

The following related question was asked by Vinberg.

\begin{qst} Let $G<\Or_n$ be a rotation group. Is it always possible to find $n$ polynomials $f_1,\ldots,f_n \in \R[x_1,\ldots,x_n]^{G}$ in the invariant ring of $G$ for which the induced map
\[
		f= [f_1,\ldots,f_n] : \R^n/G \To \R^n
\]
defines a homeomorphism?
\end{qst} 

\emph{Acknowledgements.} I thank Alexander Lytchak for introducing me to the problem and for helpful discussions.

\section{Preliminaries}

\subsection{Homology manifolds}
Homology manifolds are generalizations of topological manifolds (cf. \cite{MR0494113}). In order to simplify our proofs we make the following modified definition.

\begin{dfn}\label{dfn:hom_mfd} We say that a Hausdorff space $X$ is a \emph{homology $n$-manifold}, if all its local homology groups coincide with the local homology groups of $\R^n$, i.e. if for all $x \in X$
\[
	H_i(X, X-\{x\})=\begin{cases}
  0,  & \text{for }i\neq n\\
  \Z, & \text{for } i=n \\
\end{cases} 
\]
holds.
\end{dfn}

\begin{rem}
All spaces occurring in this paper are finite-dimensional simplicial complexes. Every such space is a finite-dimensional absolute neighborhood retract \cite[Application 18.4, p.~61]{MR0007094}. Therefore, our modification of other stricter definitions of homology manifolds (cf. \cite{MR0494113}, \cite{MR1886687}) does not make any difference for the formulation of Theorem D.
\end{rem}

For a topological space $X$ and a subspace $Y \subset X$ we define the \emph{double of $X$ along $Y$} to be
\[
	 2_Y X = X \times \{0,1\} / \sim \text{ where } (y,0)\sim (y,1) \text{ for all }y\in Y	
\]
endowed with the quotient topology and we simply denote it by $2X$ if the meaning of the subspace is clear. In order to deal with Davis's question for manifolds with boundary, we define homology manifolds with boundary in the following way.

\begin{dfn}\label{def:hom_man_boun} We say that a Hausdorff space $X$ is a \emph{homology $(n+1)$-manifold with boundary}, if it can be decomposed into a nonempty set of interior points $\mathring{X}$ and a set of boundary points $\partial X$ such that its double $2X$ along its boundary is a homology $(n+1)$-manifold, its boundary $\partial X$ is either empty or a homology $n$-manifold and the local homology groups at boundary points coincide with those of $0\in \partial (\R^n\times \R_{\leq 0})$, i.e. for all $x \in \partial X$ and all $i\geq0$ we have $H_i(X, X-\{x\})= 0$.
\end{dfn}

\begin{rem}	If the space $X$ in the definition of a homology manifold with boundary is sufficiently nice, then its boundary and its double are automatically homology manifolds. This is for example the case if $X$ is a PL space (cf. \cite[p.~510]{MR1402473}, \cite[Prop.~5.4.11, p.~188]{MR1019276}) and so it holds for all spaces we are working with in this paper (cf. \cite{Lange2}).
\end{rem}

The open cone of a topological space $X$ is defined to be $CX=(X \times [0,1)) / (X \times \{0\})$. Homology manifolds share the following properties. Proofs are standard computations (cf. \cite[Cor. VI.12.10, p.~181]{MR606196}, \cite[Thm.~16.11, p.~378]{MR1481706}, \cite[p.~117]{MR1867354}).
\begin{lem} \label{lem:characterization_of_homology_manifolds}
For Hausdorff spaces $X$ and $Y$ the following statements hold for integers $n\geq0$.
\begin{compactenum}
	\item $X$ and $Y$ are homology manifolds, if and only if $X\times Y$ is a homology manifold.
	\item If $X$ is a homology manifold, then $X\times Y$ is a homology manifold with boundary, if and only if $Y$ is a homology manifold with boundary. Moreover, we have $\partial (X\times Y) = X \times \partial Y$.
 	\item $CX$ is a homology $(n+1)$-manifold if and only if $X$ is a homology $n$-manifold and $H_*(X)=H_*(S^n)$.
	\item $CX$ is a homology $(n+2)$-manifold with nonempty boundary $C(\partial X)$ if and only if $X$ is a homology $(n+1)$-manifold with nonempty boundary and $H_*(X)=H_*(\{*\})$, $H_*(\partial X)=H_*(S^n)$.
\end{compactenum}
\end{lem}

\subsection{Piecewise linear manifolds} For a discussion of piecewise linear manifolds, we refer to the preliminary section of \cite{Lange2}.

\subsection{Lipschitz manifolds}\label{sec:Lipschitz-man}
We say that a metric space $X$ is a \emph{Lipschitz manifold} if it is a topological manifold and the coordinate maps can be chosen to be bi-Lipschitz. For a finite subgroup $G < \Or_n$ the quotient $\R^n/G$ inherits a metric from $\R^n$, the so-called \emph{quotient metric}, where the distance between two points in $\R^n/G$ is defined to be the distance of the corresponding orbits in $\R^n$. With respect to this metric $\R^n/G$ is a \emph{length space}, i.e. the distance between two points is the infimum of the lengths of all rectifiable paths connecting these points (cf. \cite{MR1835418}). We would like to know when it is a Lipschitz manifold. Since $\R^n/G$ can be triangulated by a simplicial complex $K/G$ (cf. \cite{Lange2}), the metric on $\R^n/G$ can be recovered from the flat metrics on the simplices of $K/G$. Therefore, $\R^n/G$ is a Lipschitz manifold with boundary if it is a PL manifold with boundary. In particular, $\R^n/G$ is a Lipschitz manifold with boundary if $G$ is a reflection-rotation group by Theorem A which has been proven in \cite{Lange2}.

Siebenmann and Sullivan established the following necessary and sufficient condition for a simplicial complex to be a Lipschitz manifold \cite[Thm.~1,  p.~504; Thm.~2, p.~506 in combination with Remark $(i)$, p.~507)]{MR537747}.
\begin{thm}\label{thm:characterization_lip_manifolds}
A locally finite simplicial complex $K$ with a length metric induced by flat metrics on its simplices is a Lipschitz manifold, if and only if the link of every simplex of $K$ is a homotopy sphere and a Lipschitz manifold with respect to its induced length metric.
\end{thm}

According to this result, the same argument as in the PL category shows that $G$ is a rotation group, if $\R^n/G$ is a Lipschitz manifold (cf. \cite{Lange2}). First, the fact that the link of the origin in the triangulation $K$ of $\R^n$ is again a Lipschitz manifold, implies by induction that all proper isotropy groups in $G$ are rotation groups. Then the simply connectedness of this link implies that $G$ is generated by its isotropy groups and thus is a rotation group itself. Necessary conditions for the case that $\R^n/G$ is a Lipschitz manifold with nonempty boundary will be deduced in the next section.

\section{Proof of the main results}

For manifolds without boundary Theorem D has been proven in \cite{Lange1}. To prove it also in the general case for manifolds with boundary we first show a lemma based on the following concepts.

\begin{dfn}\label{dfn:orbifold} A metric space $\Orb$ is called a \emph{Riemannian orbifold} of dimension $n$, if for each point $x \in \Orb$ there exists an open neighborhood $U$ of $x$ in $\Orb$ and an $n$-dimensional Riemannian manifold $M$ together with a finite group $G$ acting isometrically on $M$ such that $U$ and $M/G$ are isometric.
\end{dfn}

For a set of data $(x,U,M,G)$ as in the definition denote by $\pi_x : M \To U$ the composition of the natural projection from $M$ to $M/G$ and an isometry from $M/G$ to $U$ and let $p\in \pi_x^{-1}(x)$. Every Riemannian orbifold in the above sense is a smooth orbifold in the sense of \cite{MR2883685, Thurston} with the $(B_r(0)\subset T_p M,G_p,B_r(x),\pi_x\circ \exp_p)$ for sufficiently small $r$ as orbifold charts.

\begin{dfn}\label{dfn:orbifold_covering} A map $\varphi : \Orb \To \Orb'$ between Riemannian orbifolds is called an orbifold covering if any point $x\in \Orb'$ has a neighborhood $U$ isometric to some $M/G$ (cf. Definition \ref{dfn:orbifold}) such that each component $U_i$ of $\varphi^{-1}(U)$ contains precisely one preimage of $x$ and is isometric to $M/G_i$ for some subgroup $G_i<G$.
\end{dfn}

Now we can show

\begin{lem}\label{lem:boundary-double-reflection}
Let $G<\Or_n$ be a finite subgroup with orientation preserving subgroup $\Gp$ and assume that $\R^n/G$ is a homology manifold with nonempty boundary. Then $G$ contains a reflection and there exists an isometry $\varphi$ from the double $2(\R^n/G)$ with its induced length metric to $\R^n/\Gp$ such that $p_0=p_1\circ \varphi$ where $p_0$ and $p_1$ are the natural projections from $2(\R^n/G)$ and $\R^n/\Gp$ to $\R^n/G$.
\end{lem}
\begin{proof} The proof is by induction on $n$. For $n=1,2$ the claim is clear. Assume it holds for some fixed $n>1$ and let $G<\Or_{n+1}$ be a finite subgroup such that $\R^{n+1}/G$ is a homology manifold with nonempty boundary. Then $S^n/G$ is also a homology manifold with nonempty boundary by Lemma \ref{lem:characterization_of_homology_manifolds}. For a point $x \in S^n$ whose coset lies in the boundary of $S^n/G$ the quotient space $T_x S^n /G_x$ is a homology manifold with nonempty boundary. Therefore, it follows by induction that $G_x\subset G$ contains a reflection and that there exists an isometry $\tilde{\theta}: 2(T_xS^n /G_x) \To T_xS^n /\Gp_x$ with the property stated in the lemma. Using the exponential map, we obtain an equivariant bijection $\theta: B_r(x_0) \To B_r(x_1)$ between small balls $B_r(x_0)$ and $B_r(x_1)$ about the cosets $x_0$ and $x_1$ of $x$ in $2(S^n /G_x)$ and $S^n /\Gp_x$, respectively. By construction the map $\theta$ descends to an isometry, in fact to the identity, between the quotients of $B_r(x_0)$ and $B_r(x_1)$ by the respective reflection. Since the metrics on $S^n /G_x$, $2(S^n /G_x)$ and $S^n /\Gp_x$ are length metrics this implies that the restriction $\theta: B_{r/4}(x_0) \To B_{r/4}(x_1)$ is an isometry. In particular, we see that $2(S^n /G)$ is a Riemannian orbifold and that the natural projection $p_0 : 2(S^n /G) \To S^n /G$ is a covering of Riemannian orbifolds. By the assumption $n>1$ the sphere $S^n$ is simply connected. Therefore there exists an index $2$ subgroup $\tilde{G}<G$ and a homeomorphism $\varphi: 2(S^n/G) \To S^n/\tilde{G}$ with $p_0=p_1\circ \varphi$ where $p_1:S^n/\tilde{G}\To S^n/G$ is the natural projection \cite[Chapt.~13, p.~305]{Thurston}. Similar as above we see that $\varphi$ is an isometry. Moreover, since $2(S^n/G) = S^n/\tilde{G}$ has the integral homology of a sphere (cf. Definition \ref{def:hom_man_boun} and Lemma \ref{lem:characterization_of_homology_manifolds}), the subgroup $\tilde{G}$ preserves the orientation \cite[Lem.~2.14]{Lange1}. The fact that both $\Gp$ and $\tilde{G}$ are orientation preserving subgroups of index 2 in $G$ implies $\Gp=\tilde{G}$. The linear extension $\varphi: 2(\R^{n+1}/G) \To \R^{n+1}/\Gp$ of $\varphi: 2(S^n/G) \To S^n/\Gp$ is an isometry that satisfies the desired property (note that $C(2(S^n/G))=2(C(S^n/G))=2((CS^n)/G)$ as metric spaces, cf. \cite[Sect. 3.6.2.]{MR1835418}).
\end{proof}

Now we are in the position to finish the proof of Theorem D. \renewcommand*{\proofname}{Proof of Theorem D}
\begin{proof} The if direction follows from Lemma \ref{lem:characterization_of_homology_manifolds}, $(ii)$, and the result in \cite{Lange2} (cf. \cite{Lange1}). Conversely, assume that $G<\Or_n$ is a finite subgroup such that $\R^n/G$ is a homology manifold with boundary. According to Lemma \ref{lem:boundary-double-reflection}, our Definition \ref{def:hom_man_boun} (cf. the subsequent remark) and the result in \cite{Lange1}, the orientation preserving subgroup $\Gp$ of $G$ is a product of a rotation group and a certain number of Poincar\'e groups. So we are done, if $G$ itself preserves the orientation. Otherwise $G$ contains a reflection $s$ by Lemma \ref{lem:boundary-double-reflection} which normalizes $\Gp$. This reflection can only act in one of the factors of $\Gp$. Therefore the claim follows, if we can show that $\R^4/\tilde{P}$ is not a homology manifold for $\tilde{P}=\left\langle P,s \right\rangle$ where $s$ is one of the existing reflections in the normalizer of $P$ in $\Or_4$. The coset of $s$ in $\tilde{P}/P$ acts as an orientation reversing isometry on $S^3/P$. Hence, its fixed point subspace is a disjoint union of points and totally geodesic embedded surfaces. If $\R^4/\tilde{P}$ were a homology manifold, then only a single embedded sphere could occur and $S^3/P$ would be the double of $S^3/\tilde{P}$ along this sphere by Lemma \ref{lem:boundary-double-reflection}. In this case $P$ would be a free product of isomorphic groups due to the theorem of Seifert and van Kampen on fundamental groups \cite[Thm.~1.20., p.~43]{MR1867354} (the boundary of $S^3/\tilde{P}$ admits a collar). This is a contradiction, since $P$ is neither trivial nor infinite and thus the claim follows. 
\end{proof}
\renewcommand*{\proofname}{Proof}

Now Theorem C can be proven as well.
\renewcommand*{\proofname}{Proof of Theorem C}
\begin{proof} According to the results in \cite{Lange1} and \cite{Lange2} the quotient space $\R^n/G$ is a topological manifold with boundary for all groups described in Theorem C. Conversely, suppose $G<\Or_n$ is a finite subgroup for which $\R^n/G$ is a topological manifold with boundary. Then $G$ has the form $G = G_{rr} \times P_1 \times \ldots \times P_k$ as in Theorem $D$ by that theorem with $n>4$ for $k=1$ (cf. \cite{Lange1}). Moreover, the additional condition $n>5$ for $k=1$ in the case that $G$ contains a reflection also holds, since $\R^4/P$, which would have to be the boundary of $\R_{\geq 0} \times \R^4/P$ by homological reasons (cf. Lemma \ref{lem:characterization_of_homology_manifolds}), is not a topological manifold.
\end{proof}

We conclude with a proof of Theorem B.
\renewcommand*{\proofname}{Proof of Theorem B}
\begin{proof} The if direction is a direct consequence of Theorem A (cf. Section \ref{sec:Lipschitz-man}). The only-if direction in the case in which the boundary is empty has been proven in Section \ref{sec:Lipschitz-man}. So assume that $G<\Or_n$ is a finite subgroup for which $\R^n/G$ is a Lipschitz manifold with nonempty boundary. In view of Lemma \ref{lem:boundary-double-reflection} it suffices to observe that the double of a length space that is a Lipschitz manifold with boundary with its induced length metric is a Lipschitz manifold without boundary.
\end{proof}
\renewcommand*{\proofname}{Proof}



