Let $(M,g)$ be a Riemannian connected manifold of dimension $m$ and let $O(M)$ be its orthonormal frame bundle. I would like to show that $O(M)$ is connected when $M$ is non-orientable.
My idea was to use parallel transport. I know that (Sharpe, Differential Geometry, Chap. 1, Corollary 1.18) $M$ is orientable if and only if, at a given point $p\in M$, all the loops $\lambda:[0,1]\longrightarrow M$ such that $\lambda(0)=\lambda(1)=p$ are orientation-preserving. Since $M$ is non-orientable, there exists a loop $\bar{\lambda}$ with base point $p$ which is orientation-reversing: this means that, given $X_1,...,X_m$ vector fields along $\bar{\lambda}$ such that $B_0=\{X_1(0),...,X_m(0)\}$ is an orthonormal basis of $T_pM$, the determinant of the transformation between $B_0$ and the orthonormal basis $B_1=\{X_1(1),...,X_m(1)\}$ is negative. This implies that there exists a path between the points $(p,X_1(0),...,X_m(0))$ and $(p,X_1(1),...,X_m(1))$ of $O(M$). I have some questions, though:
- Did I do some mistakes in this part of the proof?
- How could I extend this argument (if it is right) to all the paths $\gamma:[0,1]\longrightarrow M$ such that $\gamma(0)=p$ and $\gamma(1)=q$, in order to show that all points in $O(M)$ can be connected?
- What happens when $M$ is oriented?