Let $M$ be a smooth manifold, and let $E_1,E_2$ be two non-orientable vector bundles over $M$.

Is $E_1 \oplus E_2$ orientable?

I am sure there is an easy answer, but somehow my search didn't result with anything useful.

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    $\begingroup$ No; consider line bundles, which are orientable iff they’re trivializable. $\endgroup$ – Qiaochu Yuan Nov 19 '17 at 9:04
  • $\begingroup$ OK, how do you use this observation to construct a counterexample? $\endgroup$ – Asaf Shachar Nov 19 '17 at 9:45
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    $\begingroup$ Take any manifold $M$ such that $H^1(M, \mathbb{F}_2)$ has dimension at least two ($\mathbb{RP}^2 \times \mathbb{RP}^2$ will suffice); then you can find two line bundles with different nonzero first Stiefel-Whitney classes, and their sum will also have a nonzero first Stiefel-Whitney class. In general I've found it very strange that you've asked all these questions about bundles and orientability and nobody involved has mentioned Stiefel-Whitney classes yet. $\endgroup$ – Qiaochu Yuan Nov 19 '17 at 18:08
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    $\begingroup$ Here are some useful facts that I can't tell if you know: 1) a real vector bundle $E$ has a characteristic class $w_1(E) \in H^1(-, \mathbb{F}_2)$ called its first Stiefel-Whitney class, which is zero iff $E$ is orientable, 2) if $E$ is a line bundle, it is classified by this class, and taking tensor products of line bundles corresponds to adding classes, 3) $w_1(E)$ is the characteristic class classifying the top exterior power $\Lambda^{\dim E}(E)$ of $E$, which is a line bundle, and 4) it follows from 2) and 3) that $w_1(E_1 \oplus E_2) = w_1(E_1) + w_1(E_2)$. $\endgroup$ – Qiaochu Yuan Nov 19 '17 at 18:12
  • $\begingroup$ Thanks, I admit my ignorance, though lately I decided I need to learn some things about Stiefel–Whitney classes and characteristic classes in general. What is your favourite reference? $\endgroup$ – Asaf Shachar Nov 20 '17 at 8:17

As Qiaochu says, the answer is no, in general. However, there is more to be said here.

An orientation on a vector bundle is a global section of the associated orientation bundle. So, a vector bundle is orientable if and only if its associated orientation bundle (which is a double cover) is trivial.

Now, an observation: Let $V$ and $V'$ be two real vector spaces. Then an orientation on $V\oplus V'$ is equivalent to a bijection $\mathrm{or}(V)\to\mathrm{or}(V')$, where $\mathrm{or}(\cdot)$ denotes the set of orientations.

As follows from the above observation, an orientation on the vector bundle $E_1\oplus E_2$ is equivalent to a bundle-isomorphism $\mathrm{or}(E_1)\to\mathrm{or}(E_2)$. In other words, $E_1\oplus E_2$ is orientable if and only if the orientation bundles of $E_1$ and $E_2$ are isomorphic. This is usually not the case.

Of course, the answer may change if the topology of $M$ is simple in some way. For example, if $M$ is the circle, then it has exactly two different double covers. Hence, every two non-orientable vector bundles have the same orientation bundle, and consequently, their direct sum is orientable.

Edit: As an example, let $M$ be the twice punctured plane, $$M=\mathbb{C}\setminus\{0,1\}.$$ Set $$P_0:=\left.\left\{(x,y)\in\mathbb{C}\times M\right|x^2=y\right\},\quad P_1:=\{(x,y)\in\mathbb{C}\times M|x^2=y-1\}.$$ So $P_0$ and $P_1$ are two different non-trivial double covers of $M$. Each one of them has an associated line bundle. Each of these line bundles is non-orientable, and so is their direct sum.

  • $\begingroup$ Thanks, this is great. Do you have a concrete example for non-orientable bundles with non-isomorphic orientation bundles? $\endgroup$ – Asaf Shachar Nov 19 '17 at 10:07
  • $\begingroup$ @AsafShachar Check the edit. $\endgroup$ – Amitai Yuval Nov 19 '17 at 12:20

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