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(1) The flat plane always has isometry group $ E_2 = O_2 \ltimes \mathbb{R}^2 $.

(2) A full classification of the four possible isometry groups of a flat torus are given in Isometry of Torus

The isometry group of a product of (irreducible, non isometric) Riemannian manifolds is the product of the isometry groups. This point is basically addressed in How does one prove that $\text{Isom}(\mathbb{S}^2 \times \mathbb{R}) = \text{Isom}(\mathbb{S}^2) \times \text{Isom}(\mathbb{R})$? and https://mathoverflow.net/questions/351646/the-isometry-group-of-a-product-of-two-riemannian-manifolds so

(3) So the isometry group of a flat cylinder $ \mathbb{R} \times S^1 $ is always $ Iso(\mathbb{R}) \times Iso(S^1)= E_1 \times O_2=(O_1 \ltimes \mathbb{R} ) \times O_2 $

Now my question: is a full classification of isometry groups, as done for the torus, possible for the other flat surfaces:

(4) The flat Moebius band (I think the isometry group is always $ Iso(\mathbb{R})=E_1=O_1 \ltimes \mathbb{R} $, just isometries of the fiber)

(5) The flat Klein bottle (I think the isometry group is always $ Iso(S^1)=O_2 $, just isometries of the fiber)

EDIT: Using the hint from Kajelad and applying it to coverings of both of these manifolds by the flat cylinder I compute the following.

The Moebius strip is the quotient of the cylinder by an isometry acting on points in the cylinder $ (t,z) \in \mathbb{R}\times U_1 $ by $$ (t,z) \to (t,\overline{z}) $$ this generates a two element cyclic group $ \Gamma $ of isometries which intersects two of the connected components of the isometry group $ (O_1 \ltimes \mathbb{R} ) \times O_2 $ of the flat cylinder. Since $ \Gamma $ has only two elements the normalizer is the same as the centralizer. And the centralizer is the full isometry group of the line together with the isometries of the circle $ z \to -z $, $ z \to \overline{z} $ and product. So modding out $ N(\Gamma)/\Gamma $ is the isometries of the fiber (line) direct product a cyclic two group acting on the base (circle) so the full isometry group of the flat Moebius strip is $ O_1 \ltimes \mathbb{R} ) \times O_1 $.

The Klein bottle is the quotient of the cylinder by an isometry acting on points in the cylinder $ (t,z) \in \mathbb{R}\times U_1 $ by $$ (t,z) \to (t+1,\overline{z}) $$ this generates an infinite cyclic group $ \Gamma $ of isometries which intersects two of the connected components of the isometry group $ (O_1 \ltimes \mathbb{R} ) \times O_2 $ of the flat cylinder. The normalizer of $ \Gamma $ is a group of isometries consisting of all the translations in the $ \mathbb{R} $ factor together with the isometries $ z \mapsto \overline{z} $ and $ z \mapsto -z $ in the circle factor. The quotient $ N(\Gamma)/ \Gamma $ is isomorphic to $ O_2 $ (since $ \Gamma $ intersects two different connected components it cuts down the four connected components of the normalizer to just two in the quotient and since $ \Gamma $ is cyclic it cuts down the $ \mathbb{R} $ translation factor in the normalizer to just a compact piece $ \mathbb{R}/\mathbb{Z} $ in the quotient. So in conclusion the isometry group of the Klein bottle is just $ O_2 $ the isometries of the circle fiber.

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    This question needs more context: What have you tried? How do the listed groups act on the corresponding manifolds? Why do you think they are the full isometry groups? – Kajelad Dec 09 '21 at 02:25

1 Answers1

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If you're familiar with covering spaces, then there's an approach which should be useful for each of these:

Given a Riemannian manifold $(M,g)$, the universal cover $\widetilde{M}$ can be equipped with a metric $\widetilde{g}$ such that the covering map $\widetilde{M}\mapsto M$ coincides with the quotient map $\widetilde{M}\mapsto\widetilde{M}/\Gamma$ where $\Gamma$ is a discrete subgroup of isometries of $\widetilde{M}$. Using among other things the lifting property of universal coverings, you can show that the isometry groups are related by the following expression: $$ \operatorname{Isom}(M)\cong N_{\operatorname{Isom}(\widetilde{M})}(\Gamma)/\Gamma $$ (here $N$ denotes the normalizer)

Each of the listed manifolds are universally covered by $\mathbb{R}^2$, so they can all be described as quotients of the Euclidean plane by discrete subgroups of $E(2)$. Actually computing these will depend on which flat metrics you're interested in.

Kajelad
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  • Great formula! Never seen it before and I love it! I tried this hint. See the update to my question. I'm still a bit stuck on the isometry group of the Klein bottle. Is it discrete? Are there any Klein bottles with isometry group with connected component $ SO_2(\mathbb{R}) $? – Ian Gershon Teixeira Dec 10 '21 at 18:35
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    @IanGershonTeixeira A few points: in each case there are multiple choices of subgroup that give the same manifold, but different subgroups will lead to flat metrics with potentially different isometry groups, so you will want to be deliberate in your choice of subgroup/metric. Also, the subgroup $\langle \tau h,\tau v\rangle$ doesn't lead to a Klein bottle (notice it doesn't even act freely); the "standard" choice is $\langle\tau h,v\rangle$. You may also want to be a bit more systematic about computing normalizers: $O(2,\mathbb{Z})$ does not normalize the cylinder subgroup $\langle h\rangle$. – Kajelad Dec 10 '21 at 23:25