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Let $\xi \colon E \to B$ be a (finite dim.) vector bundle and let $\pi \colon S(E)\to B$ the restriction to its sphere bundle. In particular, if $i\colon S(E)\to E$ is the embedding, we have $\xi \circ i = \pi$.

It is known (Bott- Tu Prop 11.2) that orientability of $\xi$ is equivalent to orientability of $\pi$ (i.e. orientability in the sense of bundles).

I would like to go on and say something more about the orientability of $E$ and $S(E)$ (as manifolds).

My intuition tells me that if the base space is oriented, then from the above proposition, we can conclude that $E$ orientable if and only if $S(E)$ orientable.

Can we go on and claim something like $$w_1(TS(E))=w_1(TE_{|S(E)})$$ always under the assumption that the base space is orientable?

Edit I think this could be a counterexample: the tautological bundle $\gamma\colon E\to S^1$ is clearly not orientable as a bundle and since $S^1$ is orientable, $E$ cannot be orientable as a mfld (in fact it is the Moebius strip). Now the sphere bundle $S(E)$ is the boundary of the Moebius strip, therefore it is homeo to $S^1$. But this would disprove the claim (but not the proposition above)

Luigi M
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  • If the base space $B$ is a manifold, and if $B$ is orientable, then yes, orientability of $\xi : E \to B$ and $\pi : S(E) \to B$ are equivalent. But keep in mind, the base space of a vector bundle is not required to be a manifold. – Lee Mosher Mar 04 '17 at 14:25
  • Dear @LeeMosher, just to be sure, do you intend orientability as manifolds? And what about the relation between sw-classes? – Luigi M Mar 05 '17 at 06:59
  • The source of my doubts is the counterexample I just added to my question (and that's why I would like to be sure when speaking about bundle orientation/mfld orientation) – Luigi M Mar 05 '17 at 07:26
  • In your counterexample, $S(E)$ is a "$0$-sphere bundle", and one has to decide what is meant by orientation of $0$-spheres. What works is to define an orientation of a $0$-sphere --- a 2-point set --- to be a $+$ sign on one point and a $-$ sign on the other point. So in your example $S(E)$ is not an orientable $0$-sphere bundle. – Lee Mosher Mar 05 '17 at 14:35
  • @LeeMosher ok I agree about the $S^0$-orientation but then the "equality" regarding the mfld orientation of $E$ and $S(E)$ (over an oriented mfld base space) isn't true anymore. Since $S(E)$ is mfld-orientable (it's the circle) but $E$ no (Moebius strip). – Luigi M Mar 05 '17 at 17:54
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    You already have an example. Are you asking if the claim holds whenever rank(E)>1? Then the answer is positive. This is even true for bundles whose fibers are connected manifolds. – Moishe Kohan Mar 20 '17 at 13:46
  • Dear @Moishe Cohen, I thought about this example, but I was not enough confident with that. So you are claiming that under the assumption that rank(E)>1 then my claim is true? Is that correct? – Luigi M Mar 20 '17 at 14:03
  • @LuigiM: That's right. The key point is that if you have an oriented manifold with boundary $M$ and a diffeomorphism (or even homeomorphism) $f: M\to M$ then $f$ preserves orientation if and only if its restriction to $\partial M$ does (for the induced orientation). – Moishe Kohan Mar 20 '17 at 14:32
  • Dear @MoisheCohen, I tried using your reasoning but to not avail. Looking forward to your answer. (not clear how do you use your hint) – Luigi M Mar 20 '17 at 20:20
  • OK, I will write one a bit later. – Moishe Kohan Mar 20 '17 at 21:16
  • dear @MoisheCohen, I don't want to sound unpolite, but I'm looking forward to see your answer :) – Luigi M Mar 23 '17 at 10:18

1 Answers1

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Let me start with the case of smooth bundles $E\to S^1$ over the circle (with connected oriented manifold fiber $F$). Every such bundle is diffeomorphic to the mapping torus of a diffeomorphism $h: F \to F$ (defined up to smooth isotopy). Then you show that if $F$ is connected then the total space $E$ is orientable if and only $h$ is orientation-preserving (this sentence makes sense only if $F$ is connected), equivalently $h_*:H_n(F)\to H_n(F)$ is the identity map, where $n=dim(F)$. You can see this, for instance by computing $n+1$-homology of $E$ using the M-V sequence, or by observing that since $F$ is connected, you can isotope $h$ to fix a point $x$ in $F$. (Then $x\times [0,1]$ projects to a loop in $E$ which preserves orientation iff $h$ does.) Next, if $M$ if a connected smooth oriented manifold with connected nonempty boundary, you check that $h: M\to M$ is orientation-preserving if and only if $h$ restricted to $\partial M$ is. Suppose now that $\xi: E\to B$ is a bundle with fiber $F$ (an oriented connected manifold with connected boundary) and manifold base $B$. Since orientability is a determined by the restriction of the tangent bundle to the 1-sketelon of a suitable CW complex structure on your manifold, $\xi$ (as a bundle) is orientable if and only if its restriction $\xi_1$ of $\xi$ to 1-skeleton of $B$ is; similarly, the total space $E$ is orientable if and only if this is true for $E(\xi_1)$ (the total space of $\xi_1$). Now, you apply the above discussion over bundles with circle base to $\xi_1$ and conclude that $E(\xi_1)$ is orientable if and only if $\xi_1$ is, hence, $E$ is orientable if and only if $\partial E$ is, if and only if the bundle $\xi$ is orientable. Now to your question in the case of vector bundles $\xi: E\to B$: You compactify the fibers as the closed balls (the boundary sphere is identified with the space of rays from the origin). The boundary sphere bundle $S(\xi)$ is the unit sphere bundle you are interested in. It follows from the general manifold bundle discussion that the total space of $E$ is orientable if and only if the total space $S(E)$ is, provided $rank(E)\ge 2$ (since $S^n$ is connected iff $n\ge 1$).

Moishe Kohan
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