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Let $ (F,g) $ be a Riemannian manifold. Let $ G:=Iso(F,g) $ be the isometry group. Let $ M $ be the mapping torus of some isometry of $ F $. So we have a bundle $$ F \to M \to S^1 $$ $ M $ has Riemannian cover $ F \times \mathbb{R} $ and there is natural action of $ G \times \mathbb{R} $ on $ F \times \mathbb{R} $. If the mapping torus is trivial $ M\cong F \times S^1 $ then this action on the cover descends to a natural action on the mapping torus. What if the mapping torus is nontrivial? When is there a natural action of $ G \times \mathbb{R} $ on $ M $? I am especially interested in the case where $ F $ is Riemannian homogeneous and this action on the mapping torus is transitive. I was inspired to ask this by a claim in this question

https://mathoverflow.net/questions/410547/exact-condition-for-smooth-homogeneous-to-imply-riemannian-homogeneous

and a similar claim in this question

https://mathoverflow.net/questions/413409/mapping-torus-of-orientation-reversing-isometry-of-the-sphere

that there is a natural action of the group $ O_{n+1}(\mathbb{R}) \times \mathbb{R} $ on the mapping torus of the antipodal map on $ S^n $.

Ian Gershon Teixeira
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A sufficient condition is that the isometry $\varphi$ is central in $G$, since in that case the action of $G\times\mathbb{R}$ preserves fibers of the covering. This is what happens for the mapping tori of the antipodal maps on round $S^n$.

Kajelad
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  • What do you mean by $ F $ is central in $ G $? $ F $ is a manifold and $ G $ is the isometry group of $ F $. – Ian Gershon Teixeira Jan 19 '22 at 21:32
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    @IanGershonTeixeira: In this context, centrality means that the (cyclic) group of covering transformations of $F\times {\mathbb R}\to M$ commutes with $G$. Equivalently, if $\phi\in G$ is the isometry of $M$ defining the mapping torus, you want $\phi$ to be contained in the center of $G$. This condition is sufficient and almost necessary. The necessary and sufficient condition is that the subgroup generated by $\phi$ is normal in $G$. – Moishe Kohan Jan 19 '22 at 22:25
  • @MoisheKohan oh wow necessary and sufficient is always better than sufficient. Would you mind making this comment into an answer? It seems even better than Kajelad's already great contribution. Does the action of $ G×\mathbb{R} $ on M factors through an action of $ G×SO_2(\mathbb{R}) $? Also does $ F $ homogeneous imply the action is transitive? I'm just double checking transitivity because what you guys say suggests that mapping torus of antipodal map of $ S^2 $ was erroneously omitted from the classification in https://www.academia.edu/9710510/Three-dimensional_homogeneous_spaces – Ian Gershon Teixeira Jan 20 '22 at 00:10
  • @IanGershonTeixeira I misread "Some isometry of $F$" as "Some isometry $F$"; edited. – Kajelad Jan 20 '22 at 01:16
  • @Kajelad For the case of the mapping tori of antipodal maps of round $ S^n $ is this action faithful? Or does it factor through a compact group like $ O_{n+1} \times SO_2 $? – Ian Gershon Teixeira Jan 21 '22 at 14:48
  • @IanGershonTeixeira If it descends to the quotient, the action of $G\times\mathbb{R}$ is never faithful, since $(e,1)$ acts equivalently to $(\varphi,0)$. – Kajelad Jan 22 '22 at 17:32
  • @Kajelad Just confirming that you are claiming the kernel is the infinite discrete group generated by $ (\phi, -1) $. Am I correct that discrete group is cocompact? And so (for even spheres where the antipodal map is orientation reversing) the action descends to a transitive action of the compact group $ SO_{n+1} \times SO_2 $ and therefore the mapping torus of the orientation reversing map on an even sphere admits a Riemannian homogeneous metric? – Ian Gershon Teixeira Jan 22 '22 at 17:37
  • @IanGershonTeixeira Strictly speaking only claimed that the kernel contains $\langle (\varphi,-1)\rangle$, though you are correct that this is the entire kernel in the case that $G$ acts faithfully on $F$ as it does here. You're also correct that this mapping torus admits a homogeneous metric, though in my opinion it's easier to show this by direct construction. – Kajelad Jan 22 '22 at 17:51
  • @Kajelad Haha I'm sure you are right it is easier to do it by direct construction. I would be very interested to see that. perhaps you could post that as an answer here https://math.stackexchange.com/questions/4362827/mapping-torus-of-the-antipodal-map-of-s2 Also just wondering about the mapping torus of the orientation reversing map of $ S^1 $ (its the Klein bottle). That manifold is not Riemannian homogeneous but oddly it is smooth homogeneous for Euclidean group $ E_2 $. Discussion about shows that natural action does not descend for odd spheres. But other transitive actions exist? When? – Ian Gershon Teixeira Jan 22 '22 at 18:08
  • @IanGershonTeixeira All this approach gives you is a subgroup of the isometries of the canonical metric on the mapping torus of a central isometry of a Riemannian manifold. It says nothing about case of non-central isometries (such as the Klein bottle) nor does it say anything about the full isometry group, much less the other possible (RIemannian) homogeneous structures on the mapping torus. – Kajelad Jan 23 '22 at 03:27