Let $E\to\Sigma$ be a real rank-two vector bundle with a Lorentzian bundle metric. I've heard that such a bundle must necessarily be trivial. This seems surprising since the same is not true with a positive-definite bundle metric. (Consider the tangent bundle of any two-dimensional surface for which there doesn't exist a global nonvanishing vector field.) Why is it the case? Is it possible to directly construct a nonvanishing section?
(In particular, this implies that there exists a nonvanishing null normal vector field along any spacelike embedded codimension-2 submanifold of a Lorentzian manifold.)
This is not true. For instance, given any pair of line bundles $L_+$ and $L_-$, the bundle $L_+\oplus L_-$ has a Lorentzian metric by just taking the direct sum of a positive definite metric on $L_+$ and a negative definite metric on $L_-$. It's certainly not true that a sum of line bundles is always trivial; in particular, $w_2(L_+\oplus L_-)=w_1(L_+)w_1(L_-)$ and $w_1(L_+\oplus L_-)=w_1(L_+)+w_1(L_-)$ and if either of these are nonzero then $L_+\oplus L_-$ is nontrivial (and $w_1(L_+)$ and $w_1(L_-)$ can be any classes at all in $H^1(\Sigma;\mathbb{Z}/(2))$). In particular, $L_+\oplus L_-$ can be nontrivial even if it is orientable: orientability means that $w_1=0$, so $w_1(L_+)$ and $w_1(L_-)$ must be the same, but then $w_2$ could still be nontrivial if the class $w_1(L_+)$ has nonzero square.
It is true if $H^1(\Sigma;\mathbb{Z}/(2))$ is trivial, and so in particular if $\Sigma$ is simply connected. Indeed, Lorentzian plane bundles have structure group $O(1,1)$ which deformation-retracts onto $O(1)\times O(1)\cong \mathbb{Z}/(2)\times\mathbb{Z}/2$. So, a Lorentzian plane bundle on $\Sigma$ is classified by an element of $H^1(\Sigma;\mathbb{Z}/(2)\times\mathbb{Z}/(2))\cong H^1(\Sigma;\mathbb{Z}/(2))\oplus H^1(\Sigma;\mathbb{Z}/(2))$, and thus is always trivial if $H^1(\Sigma;\mathbb{Z}/(2))$ is trivial.
(In more concrete terms, this is saying every example is of the form in the first paragraph: a Lorentzian plane bundle can always be split up as $L_+\oplus L_-$ where the metric is positive definite on $L_+$ and negative definite on $L_-$. If $H^1(\Sigma;\mathbb{Z}/(2))$ is trivial, then every line bundle on $\Sigma$ is trivial, and thus $L_+\oplus L_-$ is trivial.)