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Dec 11, 2023 at 15:53 comment added Sidharth Ghoshal This question is closely related: math.stackexchange.com/questions/4758159/…
Jan 22, 2022 at 20:13 comment added Lee Mosher @IanAgol: W. Thurston used to call this the "Möbius band beyond infinity" (thinking of unoriented lines in one-one correspondence with points of $\mathbb RP^2 - \overline K$ where $K$ is the Klein disc model).
Dec 8, 2021 at 18:37 comment added Ian Agol There is a homogeneous Lorentzian metric, the space of unoriented lines in the hyperbolic plane.
Dec 8, 2021 at 17:39 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 8, 2021 at 16:34 vote accept Ian Gershon Teixeira
Dec 8, 2021 at 15:00 answer added YCor timeline score: 2
Dec 8, 2021 at 14:34 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 8, 2021 at 14:27 comment added Ian Gershon Teixeira And thanks for your argument that the Moebius strip is not Riemannian homogeneous! The reference: "Isometries of 2-Dimensional Riemannian Manifolds into Themselves" by Sumner Byron Myers agrees with you but does not offer an explanation. I'll update my question to reflect this.
Dec 8, 2021 at 14:23 comment added Ian Gershon Teixeira @Ben McKay Thanks for the recommendation of such a great paper! It's actually already been pointed out to me by some other folks in other places on MSE/MO. I looked through it and love the classification result! However it does not address the case of isometries- just smooth homogeneous .
Dec 8, 2021 at 7:51 comment added Ben McKay The length of the image of the zero section would have to remain the same, but the identity component would act locally transitively, so move some points of the zero section far away, so I think there is no Riemannian metric on the Moebius strip invariant under a transitive group action.
Dec 8, 2021 at 7:49 comment added Ben McKay The classification of homogeneous connected surfaces with connected Lie groups acting on them is in George Daniel Mostow, The Extensibility of Local Lie Groups of Transformations and Groups on Surfaces, Annals of Math., Vol. 52, No. 3 (Nov., 1950), pp. 606-636.
Dec 8, 2021 at 7:44 comment added Ben McKay For the Moebius strip, the zero section has nontrivial self intersection, so the isometry group has to move it to another section which intersects it. Hence every isometry has a fixed point. So the isometry group cannot have 2-dimensional identity component, so has to have dimension 3, so the metric is a constant curvature Riemannian metric.
Dec 8, 2021 at 5:56 comment added Kapil One can produce examples of "inequivariant" bundles on $\mathbb{P}_{\mathbb{C}}^n$ for $n\geq 2$ where the automorphisms of the base do not lift to the bundle. These would give "counter-examples" if you leave out the equivariance hypothesis.
Dec 8, 2021 at 4:34 history edited LSpice CC BY-SA 4.0
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Dec 8, 2021 at 4:30 history edited Ian Gershon Teixeira CC BY-SA 4.0
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Dec 8, 2021 at 3:54 history asked Ian Gershon Teixeira CC BY-SA 4.0