Algebraic Topology – Fiber Bundle Orientability vs Manifold Orientability

algebraic-topologyfiber-bundlesmanifoldsprincipal-bundlessmooth-manifolds

I read this question about vector bundles

Bundle orientability vs manifold orientability

In the answer to this question the last sentence states the following (I think fairly well known) result about vector bundles

"Let E be a vector bundle over M. Consider the statements (i) M is orientable as a manifold (ii) E is orientable as a manifold (iii) E is orientable as a vector bundle. Any two of the statements being true will imply the third."

I am curious if this statement generalizes to all fiber bundles with orientable fiber.

In other words, is it true that: Let $ F \to E \to M $ be a fiber bundle over M, with $ F $ an orientable manifold. Consider the statements (i) M is orientable as a manifold (ii) E is orientable as a manifold (iii) E is orientable as a fiber bundle. Any two of the statements being true will imply the third.

If not, is a similar 2-out-of-3 theorem true for sphere bundles or some other more restricted class of fiber bundles?

Some stuff that I tried (focused on circle bundles in which case the bundle being orientable is equivalent to it being $ U_1 $ principal see https://mathoverflow.net/questions/144092/is-every-orientable-circle-bundle-principal)(Note that this is a complete list of all circle bundles over $ S^1,S^2,T^2,K^2,\mathbb{R}P^2 $):

Base $ S^1 $:

  • All three: $ S^1 \to T^2 \to S^1 $ (in general any trivial bundle $ S^1 \times M $ for any orientable manifold $ M $ has base, bundle and total space all orientable)

  • Only base orientable: $ S^1 \to K^2 \to S^1 $ where $ K^2 $ is the Klein bottle.

Base $ S^2 $:

  • All three: The lens spaces $ S^1 \to L_{n,1} \to S^2 $, where $ n $ is the Euler class of the bundle (for small values of $ n $ we have $ L_{0,1}= S^1 \times S^2, L_{1,1}\cong S^3, L_{2,1}\cong \mathbb{R}P^3 $). $ E^1 \times S^2 $ geometry for $ n=0 $, $ S^3 $ geometry otherwise.

Base $ T^2 $:

  • All three: The circle bundles $ S^1 \to MT(\begin{bmatrix} 1 & r \\ 0 & 1 \end{bmatrix}) \to T^2 $ where $ MT(\begin{bmatrix} 1 & r \\ 0 & 1 \end{bmatrix}) $ denotes the mapping torus of $ T^2 $ corresponding to the mapping class $ \begin{bmatrix} 1 & r \\ 0 & 1 \end{bmatrix} $, which is the $ r $th power of the Dehn twist $ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} $. For $ r=0 $ this is $ T^3 $ and admits $ E^3 $ (flat) geometry while for $ r \neq 0
    $     these are the nilmanifolds $ N_r $ described in Is every Nil manifold a nilmanifold? and they admit Nil geometry. $ E^3 $ geometry for $ r=0 $ otherwise Nil geometry.

  • Only base orientable: $ S^1 \rtimes_b T^2 $ two of the four flat compact non orientable three manifolds. For $ b=0 $ this is $ S^1 \times K^2 $ with first homology $ \mathbb{Z}^2 \times C_2 $, for $ b=1 $ this is the mapping torus of the Dehn twist diffeomorphism of $ K^2 $ with first homology $ \mathbb{Z}^2 $. The total space is not orientable. These coincide with the two $ U_1 $ principal bundles over $ K^2 $. $ E^3 $ geometry.

Base $ \mathbb{R}P^2 $:

  • Only bundle orientable: $ S^1 \to S^1 \times \mathbb{R}P^2 \to \mathbb{R}P^2 $. (in general any trivial bundle $ S^1 \times M $ for any non orientable manifold $ M $ has only the bundle orientable). $ E^1 \times S^2 $ geometry.

  • Only bundle orientable: $ S^1 \to (S^2 \times S^1)/(-1,-1) \to \mathbb{R}P^2 $. This is the mapping torus of the antipodal map of $ S^2 $. It is the unique nontrivial $ U_1 $ principal bundle over $ \mathbb{R}P^2 $. $ E^1 \times S^2 $ geometry.

  • Only total space orientable: $ S^1 \to P_{4n,1} \to \mathbb{R}P^2 $ where $ P_{4n,1} $ is the standard prism manifold with $ 4n $ element dicyclic fundamental group. $ S^3 $ geometry.

  • Only total space orientable: $ S^1 \to UT(\mathbb{R}P^2) \cong L_{4,1} \to \mathbb{R}P^2 $, the unit tangent bundle of $ \mathbb{R}P^2 $. $ S^3 $ geometry.

  • Only total space orientable: $ S^1 \to \mathbb{R}P^3 \# \mathbb{R}P^3 \to \mathbb{R}P^2 $. $ E^1 \times S^2 $ geometry.

Base $ K^2 $:

  • Only the bundle is orientable: The two principal $ U_1 $ bundles over $ K^2 $ coincide with the two non principal $ S^1 $ bundles over $ T^2 $. These are two of the four non orientable compact flat three manifolds they can also be viewed as two of the four mapping tori of $ K^2 $. $ E^3 $ geometry.

  • none of 3: The other two of the four non orientable compact flat three manifolds, they can also be viewed as the other two of the four mapping tori of $ K^2 $. They are non principal bundles $ S^1 \to S^1 \rtimes_b K^2 \to K^2 $ both with non orientable total space. For $ b=0 $ this is the mapping torus of the Y homoeomorphism of $ K^2 $, it has first homology $ \mathbb{Z}^2 \times C_2 \times C_2 $. For $ b=1 $ this is the mapping torus of $ K^2 $ for the mapping class corresponding to the combination of a Dehn twist and a Y homoemorphism. It has first homology $ \mathbb{Z}^2 \times C_4 $. $ E^3 $ geometry.

  • Only total space orientable: The circle bundles $ S^1 \to MT(\begin{bmatrix} -1 & -r \\ 0 & -1 \end{bmatrix}) \to K^2 $ where $ MT(\begin{bmatrix} -1 & -r \\ 0 & -1 \end{bmatrix}) $ denotes the mapping torus of $ T^2 $ corresponding to the mapping class $ \begin{bmatrix} -1 & -r \\ 0 & -1 \end{bmatrix} $. These manifolds are double covered by $ MT(\begin{bmatrix} 1 & 2r \\ 0 & 1 \end{bmatrix}) $. For $ r=0 $ this is the unit tangent bundle of the Klein $ UT(K^2) $, which admits $ E^3 $(flat) geometry, while for $ r \neq 0
    $     these admit Nil geometry. $ E^3 $ geometry for $ r=0 $, Nil geometry otherwise.

Best Answer

I'm late to the party, though I wanted to share a quick proof using differential geometry.

Let $F$ be an orientable, connected manifold.

Any fiber bundle $F\hookrightarrow E \overset{\pi}{\to} B$ admits a connection (see for example here) this is just a splitting of the tangent bundle of the total space $TE\simeq H\oplus V$ where $V = \ker d\pi$ is the vertical subbundle (given by $\pi$) and $H$ is the horizontal subbundle (what gives the connection). Any fiber bundle admits a connection, so we pick one and compute the first Steifel-Whitney class*: $$ w_1(TE)= w_1(H) + w_1(V) = \pi^*(w_1(TB))+ \pi^*(w_1(E))= \pi^*( w_1(TB)+ w_1(E))\in H^1(E;\mathbb Z/2) $$

where we used the isomorphism $H \overset{d\pi}{\to} TB$ implying that $w_1(H)=\pi^*(w_1(TB))$ and that the obstruction to orient the bundle $\pi^*(E)\to E$ is $w_1(V)$ (see **)

Corollary: i) if $B$ is orientable then the total space $E$ is orientable (as manifold) iff the bundle $E\to B$ is orientable.

ii) if $B$ is not orientable, i.d. $w_1(TB)\neq 0$, then $E$ is orientable (as manifold) iff $w_1(E) = w_1(TB)$ (note that $\pi^*: H^1(B; \mathbb Z/2) \to H^1(E; \mathbb Z/2)$ is injective because $\pi_1(E)\to \pi_1(B)$ is surjective, abelianization is right exact and $Hom(\cdot, \mathbb Z/2)$ is left exact).

*To avoid confusion, I will AVOID the shorthand notation $w_1(X) \sim w_1(TX)$, so when I write $w_1(\xi)$, it is clear that $\xi$ is a bundle.

** If $E$ is vector bundle we have that $\pi^* E \simeq V$. For a general fiber bundle, notice that if $\gamma\subset E$ is a loop, and I want to check if the bundle given by $\bigcup_{t} E|_{\pi(\gamma(t))}\to \mathbb S^1$ is orientable, then (since $F$ is orientable and connected) this is the same as checking what happens at a tangent space to the fiber at one point (does the monodromy reverse the orientation?) but this is exactly the same as looking at the monodromy of $\gamma^*(V)\to \mathbb S^1$.

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