Orthonormal frame bundle - Imported Knowlpedia Content

Let $\pi:E\to M$ be a real vector bundle of rank $n$ over a smooth manifold and let $\langle\cdot,\cdot\rangle$ be a bundle metric on $E$ . The orthonormal frame bundle of $(E,\langle\cdot,\cdot\rangle)$ , denoted $\mathrm{O}(E)$ , is the submanifold of the frame bundle consisting of frames that are orthonormal in each fiber:

\mathrm{O}(E):=\{(e_1,\dots,e_n)\in \mathrm{Fr}(E)\ :\ \langle e_i,e_j\rangle = \delta_{ij}\ \text{fiberwise}\}.

The right action of $\mathrm{GL}(n,\mathbb R)$ on $\mathrm{Fr}(E)$ restricts to a right action of the orthogonal group $\mathrm{O}(n)$ on $\mathrm{O}(E)$ , and $(\mathrm{O}(E),p)$ is a principal G-bundle with structure group $\mathrm{O}(n)$ .

This construction realizes a reduction of structure group from $\mathrm{GL}(n,\mathbb R)$ to $\mathrm{O}(n)$ determined by the metric.

Examples

Riemannian orthonormal frames. If $E=TM$ and the metric comes from a Riemannian metric on $M$ , then $\mathrm{O}(TM)$ is the usual bundle of orthonormal tangent frames.

Trivial bundle with Euclidean metric. On $E=M\times\mathbb R^n$ with the standard inner product, $\mathrm{O}(E)\cong M\times \mathrm{O}(n)$ .

Rank-one case. If $n=1$ , then $\mathrm{O}(1)=\{\pm 1\}$ and $\mathrm{O}(E)$ is a double cover of $M$ (the bundle of unit vectors in each 1-dimensional fiber).