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The mapping torus of a Klein bottle $ K $ is a compact flat 3 manifold.

The mapping class group of the Klein bottle $ K $ is the Klein four group $ C_2 \times C_2 $. See proposition 20 of

https://arxiv.org/abs/1410.1123

There are exactly four compact flat non-orientable 3 manifolds, one of which is $ K \times S^1 $, the mapping torus of the trivial mapping class of $ K $.

Now for the four compact flat non orientable 3 manifolds. These are distinguished by their first homology $ H_1(M;\mathbb{Z}) $ (see Wolf) $$ \mathbb{Z}^2 \times C_2, \mathbb{Z}^2, \mathbb{Z}\times C_2 \times C_2, \mathbb{Z} \times C_4 $$ They correspond to the trivial mapping torus $ S^1 \times K $, the mapping torus of the Dehn twist, the mapping torus of the Y homeomorphism and the mapping torus of the Dehn twist plus Y homeomorphism, respectively.

See the wikipedia page for Seifert fibration with positive orbifold euler characteristic.

I'm curious how these homology groups are computed from the geometry of the mapping torus perspective.

Ian Gershon Teixeira
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  • cool question +1 . My guess is no. One flat three manifold is $(S^1)^3$, and by guess is that the long exact sequence in homology will present an obstruction to this working out just for rank of $H_1$ reasons – Andres Mejia Feb 03 '22 at 22:04
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    See here for the long exact sequence. For example, we should have a long exact sequence $ \tilde{H}_1(K) \to \tilde{H}_1(M_f) \to 0$ . Plugging in the first homology of $(S^1)^3$, we can see that this would give $\mathbb Z \oplus \mathbb Z/2 \to \mathbb Z^3 \to 0$, but there is no such surjection. – Andres Mejia Feb 03 '22 at 22:09
  • @AndresMejia I'm glad you like my question! I'm asking about non-orientable manifolds so not $ (S^1)^3=T^3 $ . There are actually 10 flat compact 3 manifolds, 6 orientable and 4 non orientable. The 6 orientable ones include the torus $ T^3 $ which is the trivial mapping torus of $ T^2 $ as well as the mapping tori of the four other finite order elements of the mapping class group of $ T^2 $ (order 2,3,4,6) and a sixth exceptional compact flat orientable manifold with monodromy $ C_2 \times C_2 $. I am asking if there is a similar story about the four nonorientable ones and the Klein bottle. – Ian Gershon Teixeira Feb 04 '22 at 02:18
  • Oops no doubt. Glad I left it as a comment lol. Maybe there still is some homological obstruction. Thanks for the comment! – Andres Mejia Feb 04 '22 at 02:19
  • Now posted to MO, https://mathoverflow.net/questions/420017/mapping-torus-of-klein-bottle – Gerry Myerson Apr 09 '22 at 23:51

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