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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?

Michael Albanese
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user101010
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  • 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 $w_1(\ell)^2$.) – Michael Albanese Jan 03 '22 at 22:50

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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.

Michael Albanese
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