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Mathematics
Differential Geometry
Fixing an orientation for a connected smooth surface
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[QUOTE="Unconscious, post: 6382170, member: 656732"] The day after posting the question I figured out how to show what I needed (that it is sufficient to consider the first chart of an oriented atlas to determine the orientation), and that's just the way you told me. I also understood that, equivalent to displaying a single chart to fix the orientation of a connected smooth manifold, it is also possible to fix a frame at any point of that manifold. But now, I'd like to go one step further and understand why, in addition to everything I've said above, my book also adds: [ATTACH type="full" alt="zmmz0.png"]268089[/ATTACH]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 ##. [/QUOTE]
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Fixing an orientation for a connected smooth surface
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