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Does every triangulable manifold have a vertex-transitive triangulation?

When I talk about a vertex-transitive triangulation of a manifold, I mean in the sense of realizing a manifold homeomorphically as a simplicial complex such that the automorphism group of the simplicial complex is transitive on the vertices or 0-simplexes of the complex. This could also be formulated for other categories, like the category of simplicial sets, or the category of maps on surfaces (once you define G-actions suitably) but we'll stick with graphs and simplicial complexes here.

Two of the reasons I was interested in this problem is the work that's been done on local computability of characteristic classes of manifolds by Gelfand and others, as well as the conjecture of Kahn, Saks and Sturtevant on vertex transitive nonevasive complexes.

I'm surprised no one has really raised the question in print before, to my knowledge. Once, I thought I had an idea for a proof that some 2-manifolds can't be suitably triangulated, by focusing on surfaces which can't be realized suitably by regular maps (or flag transitive triangulations) and showing some of these surfaces can't be vertex transitively triangulated in any other way. I hope to return to this and complete the details sometime soon.

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    $\begingroup$ Not every manifold is triangulable, so maybe you need to reformulate your question. $\endgroup$ Feb 25, 2023 at 22:38
  • $\begingroup$ Presumably this question is for PL-manifolds. $\endgroup$ Feb 25, 2023 at 22:53
  • $\begingroup$ Yes, thanks. I'm not going to go into the complexities for topological manifolds. $\endgroup$
    – Mike
    Feb 25, 2023 at 22:54
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    $\begingroup$ I would imagine the answer is no. For example, you can find semi-simplicial triangulated (delta complex) manifolds so that the automorphism group is trivial. You could then barycentrically subdivide that triangulation twice to turn it into a proper simplicial complex. I don't imagine the automorphism group becomes any larger, but I suppose there's something to check there. $\endgroup$ Feb 25, 2023 at 22:57
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    $\begingroup$ Most compact surfaces have flag transitive triangulations, This is connected with the theory of regular maps and Hurwitz groups (Conder, Macbeath, Singerman, Siran, Tucker and others). Here's an example of what I'm talking about: sciencedirect.com/science/article/pii/0095895685900620 $\endgroup$
    – Mike
    Feb 26, 2023 at 14:52

2 Answers 2

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There exists many closed connected hyperbolic 3-manifolds $M$ with trivial symmetry group, and hence trivial mapping class group. $M$ cannot be homeomorphic to a simplicial complex $\tau$ which admits a vertex-transitive automorphism group $G$ (which must be non-trivial). Then the quotient $M/G = \tau/G$ is a 3-orbifold which by the orbifold theorem must be hyperbolic. But this implies that the symmetry group was conjugate in the mapping class group to a group of isometries, a contradiction.

I imagine that this ought to be true for symmetry-free hyperbolic manifolds in any dimension >2, but I’m not quite sure how to prove that the automorphism group of the triangulation induces non-trivial outer automorphisms of the fundamental group (equivalently non-trivial isometries up to homotopy by Mostow rigidity).

For the 2-dimensional case, it does appear to be open in general which closed surfaces admit a vertex-transitive triangulation. This paper states that there are at most four exceptions with $\chi \geq -127$.

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This is to supplement Ian's answer and get examples in all dimensions $\ge 3$.

Let $M={\mathbb H}^n/\Gamma$ be a compact hyperbolic $n$-manifold; suppose that $f\in Homeo(M)$ is a homeomorphism of prime order $p$. Assume that $f$ is homotopic to the identity. Our goal is to reach a contradiction. Let $\tilde f: {\mathbb H}^n\to {\mathbb H}^n$ denote a lift of $f$ to the universal covering of $M$. Then, there exists $g\in \Gamma$, a deck-transformation of ${\mathbb H}^n\to M$, such that $g$ and $\tilde f$ induce (by conjugation) the same automorphism of $\Gamma$. Extending $\tilde f$ to the boundary sphere of ${\mathbb H}^n$, we obtain that $\tilde f= g$ on $S^{n-1}$. Consider the composition $h=g^{-1}\circ \tilde f$. I claim that $h=id$, i.e. $\tilde f=g$. That would imply that $f=id_M$, contradicting our assumptions that $f$ has period $p\ge 2$.

The map $h$ restricts to the identity on $S^{n-1}$. Since $f^p=id_M$, there exists $\gamma\in \Gamma$ such that $h^p=\gamma$. Since $h^p$ restricts to the identity on $S^{n-1}$, it follows that $\gamma=1\in \Gamma$. Thus, $h$ is periodic. I will be using the conformal model of ${\mathbb H}^n$, identified with with the open unit ball $B^n$ in ${\mathbb R}^{n}$. Thus, extend $h$ by the identity to the exterior of $B^n$ in $S^n={\mathbb R}^{n}\cup \{\infty\}$. The extension, again denoted $h$, is $p$-periodic and its fixed-point set has nonempty interior in $S^n$. Unless $h=id$, this contradicts Newman's theorem [1] or, if you prefer, the contradicts the Smith Theory [2], according to which the fixed-point set of a $p$-periodic homeomorphism of $S^n$ is either empty or is a $p$-homology $k$-manifold and $p$-homology $k$-sphere for some $k$.

The conclusion, therefore, is that if $f: M\to M$ is $p$-periodic, $p$ is prime, then $f$ cannot be homotopic to the identity, hence, by Mostow Rigidity, is homotopic to a $p$-periodic isometry of $M$. Now, one uses the existence of compact hyperbolic $n$-manifolds for all $n\ge 3$, with trivial isometry group, just as in Ian's argument. To conclude: For all $n\ge 3$ there exist smooth compact $n$-manifolds without nontrivial periodic self-homeomorphisms.

[1] Newman, M. H. A., A theorem on periodic transformations of spaces, Q. J. Math., Oxf. Ser. 2, 1-8 (1931). ZBL0001.22703.

[2] Bredon, Glen E., Introduction to compact transformation groups, Pure and Applied Mathematics, 46. New York-London: Academic Press. XIII,459 p. (1972). ZBL0246.57017.

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  • $\begingroup$ Okay, I was thinking that something along these lines ought to work. To apply Newman’s theorem, I guess you can just shrink hyperbolic space down by a conformal transformation so that the period translates as little as you like, a contradiction? I was trying to apply a theorem of Smith which says that the fixed point set of a periodic homemorphism on euclidean space has the mod p homology of a point, but I missed that there is a local version. $\endgroup$
    – Ian Agol
    May 18, 2023 at 20:16
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    $\begingroup$ @IanAgol: It is theorem 9.5 in Bredon's book (that he attributes to Newman) that the interior has to be empty. Bredon also explains a reduction to Newman’s theorem on the size of the displacement. Yes, in the nutshell, the idea is to shrink the support set of a period $p$-map using conformal rescaling in the spherical case. But the theorem also works for all manifolds, not just spheres. $\endgroup$
    – Misha
    May 18, 2023 at 21:40

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