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Suppose you have a finite group $G<SO(5)$ such that $S^4/G$ is homeomorphic to $S^4$ and such that $S^4/G$ is a PL manifold with respect to a PL structure induced by a standard structure on $S^4$. Is there a general argument by which $S^4/G$ is a standard PL $4$-sphere?

More precisely: $\mathbb{R}^5$ has a canonical PL structure which induces a PL structure on $\mathbb{R}^5/G$ (cf. https://eudml.org/doc/210089). $S^4/G$ is then meant to be the link of $\overline{0}$ in $\mathbb{R}^5/G$.

Example: Take $G$ to be the exotic image of the alternating group $A_5$ in the symmetric group $S_6<SO(5)$. In this case one can check by hand that $S^4/G$ is a standard PL $4$-sphere, but in the way I do it it is very cumbersome.

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    $\begingroup$ Can you explain this example? $\endgroup$
    – Daniele Zuddas
    Commented Sep 27, 2014 at 15:00
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    $\begingroup$ In this special example one can construct a fundamental domain for the action of $G$ on $S^4$. Examining the identifications on the boundary that have to be carried out to obtain $S^4/G$ carefully shows that $S^4/G$ is a standard PL sphere (cf. mathnet.ru/php/…, Theorem 1.3) $\endgroup$
    – Christian Lange
    Commented Sep 27, 2014 at 16:39

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