The frame bundle of a smooth manifold

Assume ${\mathrm{A}\mathrm{C}}_{\omega }$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)), inherited from the smooth tangent-bundle theorem. Let $M$ be a smooth $m$-manifold with $m\ge 1$. Its tangent bundle $TM={⨆}_{x\in M}{T}_{x}M$ carries the canonical smooth $2m$-manifold structure of Assuming countable choice, the tangent bundle has a canonical smooth 2n-manifold structure (The tangent bundle as a disjoint union, Smooth manifolds and their smooth charts). The frame bundle of $M$ is the frame bundle of the tangent bundle in the sense of Frame bundles and associated vector bundles, $B(M):=\mathrm{Fr}(TM)=\{(x,b):x\in M,b:{\mathbb{R}}^m\to T_{x}M\text{ a linear isomorphism}\},$ the total space of the principal ${\mathrm{G}\mathrm{L}}_m(\mathbb{R})$-bundle of Invertible matrices and the general linear group ${\mathrm{GL}}_n(F)$ associated with $TM$. It carries the smooth structure induced by the linear bundle charts of $TM$ (locally $U\times {\mathrm{G}\mathrm{L}}_m(\mathbb{R})$, the second factor an open subset of the matrix space), the smooth projection $\pi :B(M)\to M$, $\pi (x,b)=x$, and the free smooth right action $(x,b)\cdot A=(x,b\circ A)$ of ${\mathrm{G}\mathrm{L}}_m(\mathbb{R})$, whose orbits are exactly the fibres $B(M{)}_x$; each fibre is therefore a ${\mathrm{G}\mathrm{L}}_m(\mathbb{R})$-torsor. A framing of the point $x\in M$ is an element $(x,b)$ of the fibre $B(M{)}_x$, equivalently a linear isomorphism $b:{\mathbb{R}}^m\to T_{x}M$.

The fibre $B(M{)}_x$ has exactly two path components, the two orientation classes of bases of $T_{x}M$ (Orientation of a finite-dimensional real vector space). Indeed the determinant $\det :{\mathrm{G}\mathrm{L}}_m(\mathbb{R})\to {\mathbb{R}}^{\times }$ is a surjective continuous group homomorphism (Determinant is a group homomorphism $\mathrm{GL}(V)\to F^{\times }$, and $\det (T^{-1})=\det (T{)}^{-1}$), so its sign separates ${\mathrm{G}\mathrm{L}}_m(\mathbb{R})$ into the nonempty open sets of positive and negative determinant, and the torsor action identifies these with $B(M{)}_x$; left multiplication by $\mathrm{diag}(-1,1,\dots ,1)$ identifies the negative-determinant matrices with the positive-determinant matrices, which are path-connected by Positively oriented bases of an oriented vector space are path-connected, so these are exactly the two path components. When $M$ is oriented, a chart of $M$ whose coordinate frame is positive at a point gives the identification of $B(M{)}_x$ with the positive and negative bases used here (Oriented smooth manifolds and oriented charts).

A framing of a closed $0$-dimensional submanifold $N\subseteq M$ is a framing of $N$ in $M$ in the sense of the normal-quotient convention: since $\dim N=0$, the normal bundle $\nu (N\subseteq M)={\coprod }_{x\in N}{T}_{x}M\mathrm{/}{T}_{x}N$ is ${\coprod }_{x\in N}{T}_{x}M$ over the discrete set $N$ (Normal and conormal bundles of an embedded submanifold), so a framing of $N$ is exactly a family $(b_x{)}_{x\in N}$ of linear isomorphisms $b_x:{\mathbb{R}}^m\to T_{x}M$, that is, a family of framings of the individual points $x\in N$. If $N$ is compact it is finite: its singleton subsets form an open cover and admit a finite subcover.

Finally, if $M$ is oriented, the sign of a framing $(x,b)$ is $\epsilon (b)=+1$ when the isomorphism $b$ carries the standard orientation of ${\mathbb{R}}^m$ (the one for which the standard basis is positive) to the given orientation of $T_{x}M$, and $\epsilon (b)=-1$ otherwise. Two framings of the same point have the same sign exactly when they lie in the same component of $B(M{)}_x$: the sign is constant on a component because it is a continuous function with values in $\{\pm 1\}$, and the two components are the positive and the negative bases for the given orientation. No orientation of $M$ is needed for the definition of $B(M)$ or of a framing, and this definition selects nothing beyond the supplied chart data of $M$.