# Are there simplicial spheres with "non-geometric symmetries"?

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 $$\smash{\mathbf S^d}$$.

• No: There are smooth fake real-projective spaces. Apr 19 '21 at 15:05
• @MoisheKohan Can you elaborate? What are those and how do they answer the question? Apr 19 '21 at 15:06
• 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. Apr 19 '21 at 15:07
• 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? Apr 19 '21 at 23:22
• @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. Apr 20 '21 at 11:10

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").
• 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). Apr 19 '21 at 15:19
• 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.) Apr 19 '21 at 15:46