I'm trying to show the curvature of a one dimensional vector bundle with a Riemannian metric vanishes, no matter what the connection is. I found this can be done for orientable bundles, because an orientable Riemannian line bundle is trivial.


My questions are,


1. Are Riemannian line bundles trivial?
2. If not, how to prove their curvature vanish?



Riemannian vector bundle
Given a connection $A\_{i\alpha}^\beta$ the curvature is
$$F\_{ij\alpha}^\beta=\frac{\partial A\_{j\alpha}^\beta}{\partial x^i}-\frac{\partial A\_{i\alpha}^\beta}{\partial x^j}+A\_{i\gamma}^\beta A\_{j\alpha}^\gamma-A\_{j\gamma}^\beta A\_{i\alpha}^\gamma.$$
In the case of a line bundle one can only have $\alpha=\beta=1$ so denoting $A\_i=A\_{i1}^1$ you get
$$F\_{ij}=\frac{\partial A\_j}{\partial x^i}-\frac{\partial A\_i}{\partial x^j}.$$
In general this is nonzero.


Now, given a bundle metric $g\_{\alpha\beta}$, metric-compatibility of the connection means that $A\_{i\alpha}^\gamma g\_{\beta\gamma}+A\_{i\beta}^\gamma g\_{\alpha\gamma}=\partial\_i g\_{\alpha\beta}.$ Differentiating relative to $x^j$, and subtracting off the same equation with $i$ and $j$ swapped, you get the skew-symmetry
$$F\_{ij\alpha}^\gamma g\_{\beta\gamma}+F\_{ij\beta}^\gamma g\_{\alpha\gamma}=0.$$
Returning again to the case of a line bundle, this says $2F\_{ij}=0$ i.e. $F=0.$


