My question is the natural continuation of this.
In that topic I understood why, for a connected $ S \subset \mathbb {R} ^ n $ smooth manifold, fixing its orientation is equivalent to:
- show any chart;
- fix a frame applied to any point of $ S $.
Having established this, I would like to understand why my textbook (Zorich, Mathematical Analysis II, 1st ed., Page 175) adds:
My doubt is: if every time I fix a normal $ \mathbf {n} $, I force the other $ n-1 $ vectors $ (\xi_1, ..., \xi_{n-1}) $ (generators of the tangent space) to be such that the frame $ F (\mathbf {n}) = (\mathbf {n}, \xi_1, ..., \xi_ {n-1}) $ is in the same equivalence class as the frame $ F_0 = (\mathbf {e} _1, ..., \mathbf {e} _ {n}) $ (i.e., $ \det M_ {F_0 \to F (\mathbf {n})}> 0 $, with $ M_ {F_0 \to F (\mathbf {n})} $ frame transition matrix), then if I take any two normals $ \mathbf {n} _1 $ and $ \mathbf {n} _2 $, I will always have that:
$$\det M_{F(\mathbf{n}_2)\to F(\mathbf{n}_2)}=\det M_{F(\mathbf{n}_2)\to F_0}\cdot \det M_{F_0\to F(\mathbf{n}_1)}=\frac{\det M_{F_0\to F(\mathbf{n}_1)}}{\det M_{F_0\to F(\mathbf{n}_2)}}>0$$
by construction, that is, I will always fix only one of the possible orientations of $ S $. I would have expected instead that, following the procedure described by Zorich, reversing the normal I would have found a frame in the opposite equivalence class to that of the first frame.