Direct sum of non-orientable bundles is orientable?

Question

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.

Answer 1

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.