Frame bundle of a manifold - Imported Knowlpedia Content

Let $M$ be an $n$ -dimensional smooth manifold with tangent bundle $\pi:TM\to M$ .

The frame bundle of $M$ , denoted $\mathrm{Fr}(TM)$ , is the set of all ordered bases (frames) of tangent spaces:

\mathrm{Fr}(TM)=\{(x,(e_1,\dots,e_n)):\ x\in M,\ (e_1,\dots,e_n)\text{ is a basis of }T_xM\}.

The projection $\mathrm{Fr}(TM)\to M$ sends a frame to its basepoint. There is a right action of $\mathrm{GL}(n,\mathbb R)$ by change of basis:

(x,(e_1,\dots,e_n))\cdot A := (x,(\sum_j e_j A_{j1},\dots,\sum_j e_j A_{jn})).

With this structure, $\mathrm{Fr}(TM)\to M$ is a principal G-bundle with structure group $\mathrm{GL}(n,\mathbb R)$ .

Connections on $TM$ can be equivalently encoded as principal connections on $\mathrm{Fr}(TM)$ (see connections via frame bundles).

Equivalent characterizations

Equivalently, a point of $\mathrm{Fr}(TM)$ can be viewed as a linear isomorphism $u:\mathbb R^n\to T_xM$ .

Examples

Euclidean space. For $M=\mathbb R^n$ , choosing the standard coordinate frame identifies $\mathrm{Fr}(T\mathbb R^n)\cong \mathbb R^n\times \mathrm{GL}(n,\mathbb R)$ .

Parallelizable manifolds. If $M$ admits a global frame of vector fields (a global trivialization of $TM$ ), then $\mathrm{Fr}(TM)$ is globally a product $M\times \mathrm{GL}(n,\mathbb R)$ .

The 2-sphere. For $M=S^2$ , the tangent bundle is nontrivial, so $\mathrm{Fr}(TS^2)$ is not isomorphic to $S^2\times \mathrm{GL}(2,\mathbb R)$ . This is a bundle-level reflection of the fact that $TS^2$ has no global frame.