# Local trivializations for orthonormal frame bundle

Let $(E,\pi, M)$ be a real vector bundle of Rank $N$. Then one can define its frame bundle $GL(E)$ as follows:

$GL(E)_x:=\{\text{ordered bases of }E_x\}$ (for $x\in M$).

$GL(E):=\bigcup_{x\in M} GL(E)_x$ (disjoint union)

$\hat{\pi}\colon GL(E)\rightarrow M,$ $b\mapsto x$, where $b$ is an ordered basis of $E_x$.

Then one can define a manifold structure and a $Gl(N)$-action on on $GL(E)$ sucht that $(GL(E),\hat{\pi}, M, Gl(N))$ turns into a $Gl(N)$-principal fibre bundle.

The manifold structure is defined so that the following holds: If $\varphi\colon \pi^{-1}(U)\rightarrow U\times \mathbb{R}^N$ is a local trivialization of $E$, then

$$\psi\colon \hat{\pi}^{-1}(U)\rightarrow U \times Gl(N),\qquad b\mapsto (x,(\varphi(b_1),\ldots,\varphi(b_N))$$

(where $b=(b_1,\ldots, b_N)$ is a ordered basis of $E_x$) is a local trivialization of $GL(E)$.

If $E$ is oriented and has a metric, the bundle $SO(E)$ can be defined in a similar way with

$$SO(E)_x:=\{\text{positively oriented orthonormal bases of } E_x\}.$$

Question: How can I get local trivialistions for $SO(E)$?

The above construction doesn't work any longer, since $\varphi|_{E_x}$ is only a linear isomorphism, so if you take a positively oriented basis $(b_1,\ldots,b_N)$ the matrix $(\varphi(b_1),\ldots,\varphi(b_N))$ isn't necessarily orthogonal with determinant 1.

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I think the same thing will work if you just ensure that $\phi$ preserves the metric and orientation. You could do this by taking $N$ orthonormal local sections (using Grahm Schmidt). Does that make sense? –  Tim kinsella Nov 11 '13 at 9:11