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Let $\Delta$ be a simplicial sphere, that is, a finite (abstract) simplicial complex whose canonical geometric realization $|\Delta|$ is homeomorphic to a sphere $\mathbf S^d\subset\Bbb R^{d+1}$.

Question: Can the homeomorphism $\phi :|\Delta|\to\mathbf S^d$ be chosen in a way, so that all combinatorial symmetries of $\Delta$ are realized geometrically?

That is, if $\sigma :\Delta\to\Delta$ is a combinatorial automorphism of $\Delta$ (a bijective simplicial map) I want there to be an isometry $\smash{f_\sigma:\mathbf S^d\to\mathbf S^d}$ so that $$\phi\circ \sigma = f_\sigma\circ \phi.$$

You can think of this as a subdivision of $\mathbf S^d$ that has the same symmetries as the abstract simplicial complex $\Delta$. If we consider the sphere embedded in $\smash{\Bbb R^{d+1}}$, the isometries are exactly the orthogonal transformations restricted to $\smash{\mathbf S^d}$.

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    $\begingroup$ No: There are smooth fake real-projective spaces. $\endgroup$ – Moishe Kohan Apr 19 at 15:05
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    $\begingroup$ @MoisheKohan Can you elaborate? What are those and how do they answer the question? $\endgroup$ – M. Winter Apr 19 at 15:06
  • $\begingroup$ Manifolds homeomorphic to $\mathbb RP^n$ but not diffeomorphic to it. They have double-cover the standard $S^n$ and the covering transformation is a smooth involution that is not isotopic (as an involution) to the antipodal map. $\endgroup$ – Ryan Budney Apr 19 at 15:07
  • $\begingroup$ Even if we assume that $\Delta$ is polytopal (i.e. is the simplicial complex corresponding to the boundary of a simplicial convex polytope), it's not clear to me that your question has an affirmative answer. Do you know the answer in this case? $\endgroup$ – Sam Hopkins Apr 19 at 23:22
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    $\begingroup$ @Sam No, and in fact, this motivated my question. There are boundary complexes of simplicial polytopes that cannot be realized as simplicial polytopes with all symmetries. I wondered whether this can be "fixed" by allowing more general realizations. I now wonder whether any complex described in the answer by Moishe can be obtained as a polytopal boundary complex. $\endgroup$ – M. Winter Apr 20 at 11:10
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The answer is negative. Already in dimension 4 there are fake real-projective spaces, which are smooth 4-manifolds homotopy-equivalent but not homeomorphic to $RP^4$. These correspond to smooth free involutions $\sigma: S^4\to S^4$ which are not topologically conjugate to orthogonal transformations. Similar examples exist in higher dimensions. See manifold atlas for references.

However, in dimensions $n\le 3$ indeed, every finite group of PL homeomorphisms of $S^n$, is PL conjugate to a finite subgroup of the orthogonal group. This is easy in dimension 2 and hard in dimension 3 (a consequence of the "orbifold geometrization theorem").

Edit. A nice, although dated, survey is

M.Davis, A survey of results in higher dimensions, In "The Smith Conjecture", (editors: J. W. Morgan and H. Bass), Academic Press, New York, 1984.

dealing with examples of "exotic" actions of compact (in particular, finite) groups on spheres. Some of the examples he discusses are PL. Note that smooth actions of finite groups can be always made PL.

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  • $\begingroup$ Thank you. And is it that such a "non-orthogonal" free involution $\sigma:\mathbf S^4\to\mathbf S^4$ can be obtained as an automorphism of a simplicial sphere? (sorry if this is implicitly clear, this is not my expertise). $\endgroup$ – M. Winter Apr 19 at 15:19
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    $\begingroup$ Yes: You lift a triangulation from the quotient $RP^4$ to $S^4$. You may also wonder if this simplicial $S^4$ is PL homeomorphic to the standard one, and it is. (It's the last remaining open version of the Poincare Conjecture/Problem: Does $S^4$ admit unique PL structure? Their examples do not disprove the conjecture.) $\endgroup$ – Moishe Kohan Apr 19 at 15:46

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