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I'm looking for a compatibility result which links two types of structures that could be imposed on a topological space $X$:

  1. Call $X$ triangulable if there exists a finite simplicial complex $K$ whose geometric realization $|K|$ admits a homeomorphism to $X$.
  2. Call $X$ involutive if it admits a nontrivial $\mathbb{Z}/2$-action, in the sense that there is a continuous $\iota:X \to X$ satisfying $\iota \circ \iota = \text{id} \neq \iota$.

Here is the question: let's say a given space $X$ admits a triangulation $K$ and an involution $\iota$ in the senses described above. Is it true that we can find another triangulation $K'$, ideally a subdivision of $K$, with respect to which $\iota$ is a simplicial map?

There are more general versions of this question about which I'm curious, obtained by replacing $\mathbb{Z}/2$ by another finite group $G$ acting on a triangulable $X$. If the desired statement is true, I would love to be pointed to a reference; and if it is false, I wonder if there is a nice obstruction to the existence of $K'$.

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    $\begingroup$ I imagined Bredon's book on transformation groups discusses this, but I just skimmed it without luck. What if we pass to the quotient $X/G$, try to put a simplicial complex on that, and then lift it? (A related result is that a CW-complex of the base of a regular covering map lifts to a G-CW-complex on the cover, where G is the deck group.) $\endgroup$
    – Chris Gerig
    Commented Sep 11, 2020 at 19:33
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    $\begingroup$ Try shareok.org/handle/11244/319144. The page has a pdf link to J.Dover's thesis "Equivariant Piecewise-Linear Topology and Combinatorial Applications". $\endgroup$
    – Igor Belegradek
    Commented Sep 11, 2020 at 19:38
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    $\begingroup$ In the very particular case when $X$ is a smooth compact manifold, see the "equivariant triangulation theorem" at ncatlab.org/nlab/show/equivariant+triangulation+theorem and link.springer.com/article/10.1007/BF01405351. There appears to also be a version where $X$ is allowed to have boundary. $\endgroup$
    – Henry Adams
    Commented Sep 11, 2020 at 21:19

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There are involutions $\sigma$ of the 3-sphere, whose fixed-point sets are wild 2-spheres: The fixed-point set cannot be a subcomplex of any triangulation, hence, $\sigma$ cannot be PL in any triangulation.

Bing, R. H., A homeomorphism between the 3-sphere and the sum of two solid horned spheres, Ann. Math. (2) 56, 354-362 (1952). ZBL0049.40401.

See also here for Calegari's take on Bing's proof.

Edit. There is even (unique in some sense) free involution of the 4-sphere which cannot preserve a triangulation (this is due to Ruberman). Thus, a bad fixed point set is not the only obstruction.

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