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I am trying to better understand real line bundles. The real line bundles on a compact manifold are classified by $ H^1(M,C_2) $. See

https://mathoverflow.net/a/113944/387190

So, for example, every real line bundle on a simply connected manifold is trivial.

Here is a list of some low dimensional base spaces together with all possible real line bundles up to homeomorphism of the total space

  • $ S^1 $: trivial $ \mathbb{R} \times S^1 $, and tautological bundle on $ \mathbb{R}P^1 $ (coincides with the Moebius strip $ \text{Mo} $ )

  • $ \mathbb{R} $: trivial $ \mathbb{R}^2 $

  • $ S^2 $: trivial $ \mathbb{R} \times S^2 $

  • $ \mathbb{R}P^2 $: trivial $ \mathbb{R} \times \mathbb{R}P^2 $, and nontrivial coincides with the tautological bundle on $ \mathbb{R}P^2 $

  • $ \mathbb{R}^2 $: trivial $ \mathbb{R}^3 $

  • $ T^2 $: trivial $ T^2 \times \mathbb{R} $ and nontrivial which coincides with $ \text{Mo} \times S^1 $

  • $ \text{Mo} $: trivial $ \text{Mo} \times \mathbb{R} $ and nontrivial which coincides with $ \mathbb{R}^2 \times S^1 $

  • $ \mathbb{R} \times S^1 $: trivial $ \mathbb{R}^2 \times S^1 $ and nontrivial which coincides with $ \text{Mo} \times \mathbb{R} $

Now I'm trying to understand the real line bundles on the Klein bottle $ K $. $$ H^1(K,C_2) \cong C_2 \times C_2 $$ So there are at most four real line bundles over $ K $ up to homeomorphism of the total space. One is the trivial bundle $ K \times \mathbb{R} $. One nontrivial line bundle over $ K $ has total space $ M_0^3 $ as described in table 1 of

https://www.researchgate.net/publication/242949481_Three-dimensional_homogeneous_spaces

What about the other two real line bundles over the Klein bottle?

Ian Gershon Teixeira
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  • If the real line bundles over $M$ are classified by $H^1 (M, \mathbb{Z}_2 )$, and the first such cohomology group of the torus $T^2$ is $\mathbb{Z}_2 \times \mathbb{Z}_2$, shouldn't there be $4$ distinct isomorphism classes of real line bundles over $T^2$, instead of just two as you list? If not, what am I missing here? – Paul Cusson Feb 07 '24 at 07:44
  • @PaulCusson the three nontrivial line bundle over the torus happen to have homeomorphic total space. Usually topologists are interested in the finer notion of equivalence of bundles, in which case the three are distinct, but I am only interested in the homeomorphism type of the total space of the bundle. – Ian Gershon Teixeira Feb 07 '24 at 13:33

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I haven't checked all the details but arguing by analogy with the complex case one should have a picture like this.

Consider $K$ as an $S^1$-bundle over $S^1$, pulling back the trivial bundle and tautological bundle of the base appears to give two bundles. I presume that this $S^1$ -bundle as a projectivisation of a rank 2 real bundle, Then I guess tensoring with the tautologival bundle of the projective bundle should give all 4.

Nick L
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  • "Consider K as an $S^1 $-bundle over $S^1$, pulling back the trivial bundle and tautological bundle of the base appears to give two bundles." I understand this, The pullback with respect to the trivial (cylinder) bundle is ( I assume ) the trivial line bundle over $ K $. The pullback with respect to the tautological (Moebius) bundle gives exactly the nontrivial line bundle over $ K $ described in the reference (the reference calls it $ M_0^3 $) . – Ian Gershon Teixeira Apr 05 '22 at 21:45
  • "I presume that this $S^1 $-bundle as a projectivisation of a rank 2 real bundle." I'm not sure what you mean here. Could you clarify or expand on this statement? – Ian Gershon Teixeira Apr 05 '22 at 21:51
  • "Then I guess tensoring with the tautologival bundle of the projective bundle should give all 4." I understand this, the Picard group here is the Klein four group $ C_2 \times C_2 $ so tensoring any two nontrivial line bundles will give the third nontrivial line bundle. But what are the two nontrivial bundles you are using here? One nontrivial bundle is certainly the pullback of $ K \to S^1 $ using the Moebius bundle. Are you claiming that if we pullback $ K \to S^1 $ using the trivial line bundle over the circle then we will get a nontrivial line bundle over the Klein bottle? – Ian Gershon Teixeira Apr 05 '22 at 21:52
  • For any real $k$-bundle $E$ there is a $\mathbb{RP}^{k-1}$ bundle over the base, the fibre over $p$ is the space of lines in the fibre $E_{p}$. The projective bundle comes with a natural line bundle on it. I.e. the fibre over a point in the projctive space is the line which that point represents. Similar to the tautolgical bundle of a projective space but performed in a fibre-wise way. This is a central idea in complex algebraic geometry, – Nick L Apr 06 '22 at 08:35