0

According to the answer of this related question, isometry actions on any manifold form a locally compact group and are closely related to Hilbert-Smith conjecture. As the $3$-dimensional case is solved for the conjecture, I would like to know how tight Thurston's geometrization restricts on the isometry actions on a $3$-manifold $M$.

A (possible) geometric structure on $M$ consists of a triple $(G,X; \Gamma)$, such that $G$ is a maximal Lie group acting faithfully and transitively on the simply connected space $X$ with compact stabilizers, $\Gamma < G$ is discrete and acts freely on $X$. Furthermore, there is a homeomorphism $X/ \Gamma \cong M$. The geometric structure on $M$ needs not exist or be unique.

Suppose $H$ acts faithfully and isometrically on $(M,d)$, where $d$ is a complete metric on $M$. Is it necessary that $H$ should come from certain geometric structure on $M$, i.e. $H$ is a subgroup of $G$ acting on $X$ that is also $\Gamma$-equivariant ($\forall g \in H, \exists h \in H, g \Gamma = \Gamma h$)? A positive answer to this question would imply that manifolds with no geometric structure can only have a trivial faithful group action. The answer is shown to be negative cause for every regular covering map $p:M \rightarrow N$ the deck transformation group $\mathrm{Aut}(p)$ acts freely on $M$ and has an invariant complete metric, regardless of the geometric structure on $M$

If the answer is negative, then can we explicitly calculate all possible maximal isometry actions through the topological information of $M$? A counterexample candidate could be $\Bbb{T}^3 \# \Bbb{T}^3$, which I know no geometric structure on but at least admits a $\Bbb{Z}/2\Bbb{Z} \times \Bbb{Z}/2\Bbb{Z}$ isometry action when properly embedded in $\Bbb{R}^4$ and with inherited Riemann structure.

Zerox
  • 2,010
  • 1
    No, it's not true that such $M$ admits a group action: You have $M=K\backslash G/\Gamma$, where $K$ is a certain compact subgroup of $G$ (the $G$-stabilizer of a point in $X$). In some cases, indeed, you get $M=G/\Gamma$, but this is rather exceptional. What kind of faithful actions are you interested in? Proper actions? General actions? Note that every smooth manifold of positive dimension admits a smooth faithful action of ${\mathbb R}$. – Moishe Kohan Nov 30 '21 at 00:27
  • 1
    No, $G$ only acts on $G/\Gamma$, not on the biquotient. Just try to write down in detail the actual action on the biquotient (whose elements are double cosets $Kg\Gamma$) to see this. – Moishe Kohan Nov 30 '21 at 15:11
  • For the new version, the answer, I think, is positive in the case of compact manifolds $M$. I am less sure about noncompact manifolds. – Moishe Kohan Nov 30 '21 at 15:48
  • No, it does not admit a geometric structure. – Moishe Kohan Nov 30 '21 at 15:53
  • 1
    The connected sum $T^3# T^3$ does not admit any effective action of a compact connected nontrivial Lie group. – Moishe Kohan Nov 30 '21 at 16:17
  • Well, I don't require $H$ to be connected (because I want to include cases like closed hyperbolic surface with $g \ge 2$). I just want to know if $H$ is embedded into some $G$, and if not, where it comes from. – Zerox Nov 30 '21 at 16:19
  • 1
    If you do not require connectivity, there are many nongeometric 3-manifolds which admit effective actions of finite groups. Such actions are absolutely impossible to classify. For instance, take a nontrivial regular finiste-sheeted covering map between two 3-manifolds (any compact manifold apart from the sphere, admits such coverings). – Moishe Kohan Nov 30 '21 at 16:21
  • 1
    Every action of finite group on a compact manifold has a complete invariant metric. Just think for 5 minutes instead of 5 seconds. – Moishe Kohan Nov 30 '21 at 16:26
  • @MoisheKohan Sorry for the interference. You mean a deck transformation group on $M$? That requires $M$ to cover some other manifold $N$ and can not be chosen arbitrarily. – Zerox Nov 30 '21 at 16:33
  • @MoisheKohan: Is there an easy way to see your statement about actions on $T^3\sharp T^3$? – Jason DeVito - on hiatus Nov 30 '21 at 17:01
  • @JasonDeVito No easy way, but see my answer below. – Moishe Kohan Nov 30 '21 at 21:56

1 Answers1

2

Here is an answer in the setting of compact (boundaryless) 3-manifolds. Note that if $H$ is a compact topological group acting continuously on a compact metric space $M$, then there is a $H$-invariant, necessarily complete, metric on $M$. Thus, the existence of an invariant metric is not a restriction.

First, consider actions of finite groups $H$. Even if you assume that the action is free, there is virtually nothing of interest that can be said about manifolds admitting such actions:

Suppose that $M$ is a compact connected 3-manifold which is not $S^3$. Then $M$ has a nontrivial regular covering $p: M'\to M$ of finite degree. The deck-transformation group of $p$ will act freely and effectively on $M'$. If $M$ has infinite fundamental group then there is an infinite tower of such coverings. This existence result is nontrivial, we know it only due to the proof of Thurston's Geometrization Conjecture. For instance, given any finite group $H$, there is a compact 3-manifold on which $H$ acts freely (and there is enormous supply of such manifolds, both geometric and nongeometric). There is no hope classifying such actions. The best one can do is:

Suppose that $H$ is a finite group acting smoothly and effectively on a compact geometrizable manifold $M$. Then there is a geometric structure on $M$ (in Thurston's sense) invariant under the action of $H$.

I can find the precise reference if you are interested, but it is a very hard result, a culmination of work of many people. Even the case of finite cyclic group actions on $S^3$ was unknown for about 40 years and called the Smith Conjecture.

Things get only worse if we allow non-free topological actions. For instance, $S^3$ admits an involution whose fixed-point set is a wild sphere (this is due to Bing). This action of $Z_2$ on $S^3$ cannot possibly come from a subgroup of a Lie group acting smoothly on $S^3$.

I will, therefore, assume that the compact Lie group $H$ has positive dimension. Let $H^0$ denote the identity component of $H$. Thus, we have a nontrivial connected compact Lie group $H^0$ acting effectively on our manifold $M$. In what follows, I will restrict my attention to such actions, so $H=H^0$. These were classified in 1960s in the work of Neumann, Orlik and Raymond. Their papers appeared in

Mostert, Paul S. (ed.), Proceedings of the conference on transformation groups, New Orleans, 1967, Berlin-Heidelberg-New York: Springer-Verlag. xii, 456 p. (1968). ZBL0167.00104.

Note that a connected compact Lie group always contains $S^1=SO(2)$. The hardest part of classification is to classify such actions. This was done in

Orlik, Peter; Raymond, Frank, Actions of SO(2) on 3-manifolds, Proc. Conf. Transform. Groups, New Orleans 1967, 297-318 (1968). ZBL0172.25402.

For simplicity, I will assume that our manifold is orientable (see their paper for details on the nonorientable case). Then, if $M$ is a compact 3-manifold which admits an effective action of $SO(2)$, then:

a. The case of an $S^1$-action fixing at least one point. Then $M$ is a connected sum where each summand is either $S^2\times S^1$ or a lens space (one can have a mixture of both). Moreover, the converse is also true: Every such connected sum admits an effective $S^1$-action with nonempty fixed-point set.

With few exceptions, these manifolds are non-geometrizable: The only (geometrizable) exceptions are $S^2\times S^1$, lens spaces and $P^3 \# P^3$.

b. Suppose that the action does not have a fixed point. Then $M$ is a Seifert manifold, hence, is geometrizable, and the action comes from the Seifert fibration (which is unique apart from a few exceptions). Seifert manifolds are geometric and one can show by case-by-case analysis that the $S^1$-action comes from a geometric structure in Thurston's case, but you may have to switch from compact group to a noncompact group. (The quoted papers were written before Thurston's work, so they did not use the geometric language.) For instance, if $M$ is an $H^2\times R$-manifold, then the circle action comes from the action of the $R$-factor acting by translations. (Lifting the circle action to the universal cover, one gets an $R$-action.)

From the list given by Orlik and Raymond, you can see that $T^3\# T^3$ does not admit an effective circle group action, hence, no action of a compact Lie group of positive dimension.

The actions of compact groups on 3-manifolds whose principal orbits are 2-dimensional were analyzed by Neumann (completing and correcting the earlier work by Mostert) in

Neumann, W. D., 3-dimensional G-manifolds with 2-dimensional orbits, Proc. Conf. Transform. Groups, New Orleans 1967, 220-222 (1968). ZBL0177.52101.

The class of manifolds is even more restrictive: These are lens spaces, $T^3$, $S^2\times S^1$, $P^3\# P^3$. (There are few more non-orientable examples.) Neumann also classifies the actions themselves. In particular, all these manifolds are geometric. From what I saw, my impression is that, for some choice of a locally homogeneous metric from Thurston's list all these actions come from actions of subgroups of the full isometry group of that metric. But it's worth verifying this.

Lastly, one should consider actions with 3-dimensional principal orbits. I did not chase down the references; since you are interested in the question, you should read the references given in the papers above to find out. I think, the only manifolds admitting such actions are $T^3$ and some spherical space-forms.

Moishe Kohan
  • 97,719
  • Thanks for your patient answers! I only have one last question that you did not seem to include: is it always guaranteed that the invarient metric induced by a finite group action $H$ can be chosen to have an isometry group precisely $H$ for every $3$-manifold? I originally wanted to classify maximal isometries, not merely group actions. – Zerox Dec 01 '21 at 14:26
  • 1
    @Zerox: No, you cannot guarantee this (even in the case of surfaces!). What's true is that for every finite group $F$ there is a compact hyperbolic manifold $M$ of the given dimension such that $Isom(M)\cong F$. – Moishe Kohan Dec 01 '21 at 14:28
  • So is it possible to classify all the maximal isometries for a given closed $3$-manifold? That's my second bold question. – Zerox Dec 01 '21 at 14:31
  • I'd also like to know the reference of the mentioned "best result" on smooth and effective finite actions. – Zerox Dec 01 '21 at 14:37