Given a vector bundle, I am a bit hazy about the difference between the notions of its orientability as a bundle and as a manifold. I think I know that the following are true,

  • A tangent bundle of a manifold is orientable if and only if the manifold is orientable.

  • The tangent bundle of any manifold thought of as a new manifold is in some sense always "trivially" orientable.

And also I vaguely remember there being something special about the orientability of the tangent bundle of the tangent bundle of the manifold.

I am not aware if there is any generic relationship between the orientability of a vector bundle and its base manifold.

It would be helpful if someone can help me tie up these loose ends and help me see the full picture.


One way of thinking about orientability of an arbitrary rank $k$ vector bundle $E$ is to think about it analogously to the orientability of the tangent bundle.

Recall that the definition of an orientable tangent bundle is that the manifold is orientable, which is equivalent to the existence of a nowhere vanishing top form, aka a volume form. We can do something similar to the vector bundle $E$. If $E$ has rank $k$, we can form the top exterior power $\wedge^kE$. Then we say, analogous to the case of the tangent bundle, that the vector bundle $E$ is orientable iff there exists a nowhere vanishing section of $\wedge^kE$.

There is no direct relation between the orientability of a vector bundle and the base manifold. The reason that there is one between the tangent bundle is that the two notions (orientable manifold and orientable tangent bundle) mean exactly the same thing! So to ask for the relation of orientability of an arbitrary vector bundle and of the base manifold is to ask for the relation between an arbitrary vector bundle and the tangent bundle; and in general there isn't one.

(Just to illustrate: consider the Moebius strip formed by taking $[0,1]\times[0,1]$ and giving a half-twist. If you embed the Moebius strip in $\mathbb{R}^3$, and pull-back the tangent bundle of $\mathbb{R}^3$ to the strip, it is an orientable vector bundle when the Moebius strip itself is not orientable. On the other hand, consider the image of $S^1$ in the midline of the Moebius strip. Its normal bundle relative to the Moebius strip is not orientable, while $S^1$ is.)

About that double tangent bundle: if you treat $TTM$ as a bundle over $TM$, then for arbitrary $M$, since $TM$ is orientable as a manifold, clearly $TTM$, as its tangent bundle, is orientable as a bundle.

Let me add one more thing: if you are familiar with the proof that $TM$ is always orientable as a manifold, you can use the same method to prove the following theorem:

Theorem 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.

  • $\begingroup$ Sorry for the late reply. I am a bit confused about these two statements of yours. Firstly you say that a tangent bundle is defined to be orientable iff the manifold is orientable. But towards the end you say that for arbitrary manifolds its tangent bundle is orientable as a manifold. Is the second statement a provable statement and not just a definition? And is the first statement a definition of orientability as a bundle? $\endgroup$ – Anirbit Jul 21 '11 at 13:16
  • $\begingroup$ It would be great if you can give some reference along these directions. Like I don't have the familiarity that you assumed in your addition below the horizontal bar. Its very rarely that I have had to encounter this issue of conflict in orientability as a bundle and as a manifold. $\endgroup$ – Anirbit Jul 21 '11 at 13:19
  • $\begingroup$ That $TM$ is always orientable as a manifold is usually an exercise in most differential geometry textbooks. I suggest you try to work it out yourself. If you get stuck, try this and other links you get if you search for tangent bundle orientable manifold on Google. $\endgroup$ – Willie Wong Jul 21 '11 at 14:36
  • $\begingroup$ I'm sorry to bother both of you due to my lack of imagination but I was trying to prove the theorem proposed by Willie Wong, in particular the implication $(i) + (ii) \Rightarrow (iii)$. I can't see how to build triviliazation maps such that their transiction maps on the fibers have positive determinant. I want to show this withous using exterior powers\forms. Do you have any hints? Looking for good trivialization maps is the right path(like in the proof for $TM$)? $\endgroup$ – Ker Apr 30 at 20:42
  • $\begingroup$ @ker: consider that the definition I stated uses nowhere vanishing top degree forms, the easiest proofs are via non-vanishing of forms of appropriate degree. But if you want to use transition maps: Let $(\hat{U},\Phi)$ be a chart in the oriented atlas of $M$, and $(\hat{V},\Psi)$ be a chart in the oriented atlas of $E$, such that $\pi\circ\Psi(\hat{V}) \cap \Phi(\hat{U})$ is non-empty. Let $\tau: \Phi(\hat{U})\times\mathbb{R}^k \to \pi^{-1}(\Phi(\hat{U}))$ be any local-trivialization, the map (where it is defined) $\Psi^{-1}\circ\tau\circ\Phi$ has either positive or negative determinant... $\endgroup$ – Willie Wong May 3 at 14:06

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