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?


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 $w_1(\ell_1) + w_1(\ell_2) = 0$, 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 $\ell$. Then $w_1(\xi) = 0$, so $\xi$ is orientable, and $w_2(\xi) = w_1(\ell)^2$. If we can choose $\ell$ such that $w_1(\ell)^2 \neq 0$, then $\xi$ will be non-trivial. Such an $\ell$ exists on any closed orientable surface of positive genus.

For example, on $S^1\times S^1$, let $\ell = \pi_1^*\gamma\otimes\pi_2^*\gamma$ where $\pi_i : S^1\times S^1 \to S^1$ is projection onto the $i^{\text{th}}$ factor, and $\gamma$ is the unique non-trivial real line bundle on $S^1$ which is nothing but the tautological line bundle on $\mathbb{RP}^1 = S^1$. We have $H^*(S^1\times S^1; \mathbb{Z}_2) \cong \mathbb{Z}_2[\alpha, \beta]/(\alpha^2, \beta^2)$ and $w_1(\ell) = \alpha + \beta \neq 0$, and $w_1(\ell)^2 = \alpha\beta \neq 0$. Therefore $\xi = \ell\oplus\ell$ is an orientable non-trivial bundle.

For a higher genus surface $\Sigma_g$, let $f : \Sigma_g \to \Sigma_1$ be a degree one map. Then $f^*\ell$ is a line bundle of the desired form on $\Sigma_g$.


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