Timeline for Almost All Hyperplanes are Not Tangent
Current License: CC BY-SA 3.0
8 events
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| Jun 23, 2017 at 15:47 | comment | added | Francesco Polizzi | I think that in the smooth case (when M is not a plane) there is at least one point where the differential of the Gauss map has maximal rank. This implies that it is generically finite onto its image. I will look for references. | |
| Jun 23, 2017 at 14:18 | comment | added | Wlod AA | I don't feel that the above argument is correct. At least, it is not complete. Tangent planes at different points can be different but parallel. Then these parallel tangents are represented by the same point of the Gauss sphere. Along the line of the above argument, if Gauss sphere had dimension $0$ (instead of $n$) then this would make things still easier, and this is false--it would show that the presented logical argument were a complete miss. | |
| Jun 23, 2017 at 10:09 | comment | added | Francesco Polizzi | Well, my argument shows that in any case the dimension of the space of tangent hyperplanes is the same as the dimension of the image of the Gauss map, and this is at most n (and exactly n when the Gauss map is generically finite into its image). | |
| Jun 23, 2017 at 10:06 | comment | added | Francesco Polizzi | Yes, "tangent". Corrected, thanks. | |
| Jun 23, 2017 at 10:05 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Jun 23, 2017 at 7:37 | comment | added | Duchamp Gérard H. E. | did you mean "Any <tangent> hyperplane is uniquely determined by its normal versor" ? It seems also that you must keep track of the base point, no ? | |
| Jun 23, 2017 at 6:33 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Jun 23, 2017 at 6:27 | history | answered | Francesco Polizzi | CC BY-SA 3.0 |