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There are 10 compact flat 3 manifolds up to diffeomorphism, 6 orientable and 4 non orientable. I am looking to better understand how to construct the orientable ones.

The six orientable ones are determined by their holonomy groups $$ C_1,C_2,C_3,C_4,C_6 $$ and $$ C_2 \times C_2 $$ The five with cyclic holonomy all arise as the mapping torus of a mapping class of $ T^2 $ with the corresponding order: 1,2,3,4, or 6. These five Euclidean manifolds with cyclic holonomy can even be constructed as a quotient of the special Euclidean group $ SE_2 $ by a cocompact lattice constructed as the semidirect product of a lattice in $ \mathbb{R}^2 $ and a finite cyclic subgroup of $ SL_2(\mathbb{Z}) $ preserving that lattice. For example $ C_1 $ corresponds to the three torus $ T^3 $.

I am very curious about the compact flat orientable 3 manifold with holonomy $ C_2 \times C_2 $ (known as the Hantzsche-Wendt manifold). It is not a mapping torus of $ T^2 $ like the other five, but perhaps it is a mapping torus of the Klein bottle $ K $?

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The remaining orientable manifold is called the Hantzsche-Wendt manifold $M^{HW}$, and is not a mapping torus over the Klein bottle. It has first homology $H_1(M^{HW}, \mathbb{Z}) = \mathbb{Z}_4 \times \mathbb{Z}_4$. Any mapping torus $MT(M, f)$ of a manifold $M$ via the map $f: M \to M$ has $\pi_1(MT(M, f)) \cong \pi_1(M_0) \rtimes_{f_*} \mathbb{Z}$. Abelianizing and using Hurewicz's theorem yields that if $M$ is a mapping torus of the Klein bottle, $H_1(M, \mathbb{Z})$ must be of the form $\mathbb{Z} \times H$, where $H$ is a quotient of $H_1(K, \mathbb{Z}) = \mathbb{Z} \times \mathbb{Z}_2$, and hence cannot be $\mathbb{Z}_4 \times \mathbb{Z}_4$. It is worth mentioning that the $3$-torus is a normal covering space of all flat compact $3$-manifolds.

The Hantzsche-Wendt Manifold has a nice description as a branched cover of the complement of the Borromean rings, which is described in Zimmerman's paper On the Hantzsche-Wendt Manifold. The standard reference for a classification of compact flat 3-manifolds is J. Wolf's book Spaces of Constant Curvature, which gives a classification of them and explicit constructions as quotients of $\mathbb{R}^3$ by isometries.

A resource to develop visual intuition about these manifolds is Jeffrey Weeks' program Curved Spaces, which simulates how it would look to "fly around inside of" manifolds and has a number of flat 3 manifolds as pre-built examples, including the Hantzsche-Wendt manifold.

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  • $\begingroup$ I seem to remember that the Hantzsche--Wendt manifold also has a nice description as the result of gluing together two twisted interval bundles over the Klein bottle. $\endgroup$
    – HJRW
    Commented Feb 4, 2022 at 18:04
  • $\begingroup$ I know of a description of the Hantzsche-Wendt manifold as two cubes with appropriate face identifications, e.g. arxiv.org/pdf/2109.12172.pdf table 1. I'm not aware of a description of it as a gluing of twisted interval bundles over Klein bottles (but am not an expert on this sort of thing at all). $\endgroup$
    – Alex Nolte
    Commented Feb 4, 2022 at 18:25
  • $\begingroup$ The wikipedia page for Seifert fiber space says " {b; (n3, 2); } (b is 0 or 1) The other two non-orientable Euclidean Klein bottle bundles. The one with b = 1 is homeomorphic to {0; (n1, 1); (2, 1), (2, 1)}. The first homology is Z+Z/2Z+Z/2Z if b=0, and Z+Z/4Z if b=1. These two Klein bottle bundle are surface bundles associated to the y-homeomorphism and the product of this and the twist." Do you have any idea what this claim about Klein bottle bundles might refer to? Or is it just a case of wikipedia being wrong? $\endgroup$ Commented Feb 4, 2022 at 19:33
  • $\begingroup$ Also I understand that $ H_1 $ of the mapping torus must be a quotient of $ \mathbb{Z} \times H_1(K,\mathbb{Z}) \cong \mathbb{Z}^2 \times C_2 $ but could you explain why the abelianization must in particular be $ \mathbb{Z} $ times a quotient of $ H_1(K,\mathbb{Z}) $? $\endgroup$ Commented Feb 4, 2022 at 20:22
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    $\begingroup$ On the Abelianization: mathoverflow.net/questions/35713/…. $\endgroup$
    – Alex Nolte
    Commented Feb 5, 2022 at 7:15

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