2-plane bundle over a surface splitting as line bundles

Let $$F$$ be a closed orientable surface and let $$\xi$$ be an orientable 2-plane bundle over $$F$$. Is it possible to have $$\xi = l_1 \oplus l_2$$ for line bundles $$l_1$$ and $$l_2$$ and not have $$\xi$$ be trivial?

The line bundles $$l_1$$ and $$l_2$$ would need to either both be orientable or both be nonorientable, since $$w_1(l_1) + w_1(l_2) = w_1(\xi) = 0$$. In the case where they are both orientable, they must both be trivial (since there is no 2-torsion in $$H^1(F;\mathbb{Z})$$ every oriented line bundle is trivial) so $$\xi$$ is trivial.

Therefore my question is equivalent to: Can I add two nonorientable line bundles on $$F$$ together to get a nontrivial orientable 2-plane bundle?

• The previous version of my answer was incorrect. The answer to the question at the end of your post is no. (The mistake I made in my initial answer was the computation of $w_1(\ell)^2$.) Jan 3 at 22:50

First of all, an orientable line bundle is always trivial, and $$H^1(F; \mathbb{Z})$$ never has two-torsion (it's a free abelian group).
As $$w_1(\xi) = w_1(\ell_1) + w_1(\ell_2) = 0$$, we see that $$w_1(\ell_1) = w_1(\ell_2)$$ and hence $$\ell_1 \cong \ell_2$$. Suppose then that $$\xi \cong \ell\oplus\ell$$ for some real line bundle $$\ell$$. Now, as alluded to here, the Euler class of $$\xi$$ is $$e(\xi) = e(\ell\oplus\ell) = \beta(w_1(\ell)) \in H^2(F; \mathbb{Z}) \cong \mathbb{Z}$$. But $$\beta(w_1(\ell))$$ is $$2$$-torsion, so it must be zero and therefore $$\xi$$ is trivial.