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What can we say about the orientability of the unit tangent bundle $ UTM $ of $ M $?

The unit tangent bundle of the sphere $ S^2 $ is $ \mathbb{R}P^3 $ see A question on the unit tangent bundle of the sphere and $SO(3)$

The unit tangent bundle of $ T^2 $ is $ T^3 $ (torus is Lie group so parallelizable so unit tangent bundle is trivial)

The unit tangent bundle of an orientable surface $ \Sigma_g $ of genus $ g \geq 2 $ is $$ UT(\Sigma_g) \cong SL_2(\mathbb{R})/\pi_1(\Sigma_g) $$

For the non orientable surfaces the story is similar. Every unit tangent bundle is double covered by the unit tangent bundle of its orientable double cover.

For example the unit tangent bundle of the projective plane is the homogeneous (q=1) lens space with fundamental group $ C_4 $ cyclic of order 4. $$ UT(\mathbb{R}P^2) \cong L_{4,1} \cong SU_2/C_4 $$ So in particular $ UT(\mathbb{R}P^2) $ is orientable. See https://www.projecteuclid.org/journals/nihonkai-mathematical-journal/volume-13/issue-1/Unit-Tangent-Bundle-over-Two-Dimensional-Real-Projective-Space/nihmj/1273779621.full

On the other hand, the unit tangent bundle of the Klein bottle is this 3 manifold double covered by the torus

Unit (co)tangent bundle of Klein bottle

Is the unit tangent bundle always orientable?

edit: Orientability of the total space of a vector bundle and total space of its sphere bundle

Ian Gershon Teixeira
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1 Answers1

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  1. The total space $TM$ of tangent bundle to any manifold $M$ (orientable or not) is orientable, see here.

  2. The total space $UTM$ of the unit tangent bundle is a hypersurface in $TM$ which admits a nowehere vanishing normal vector field (at each point $(x,v)\in UTM$ take the normal vector $((x,v),v)$).

  3. Thus, $UTM$ is a cooriented hypersurface in an orientable manifold, hence, is itself orientable.

Moishe Kohan
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