Timeline for Almost All Hyperplanes are Not Tangent [closed]
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 25, 2017 at 17:30 | vote | accept | J126 | ||
| Jun 23, 2017 at 21:18 | comment | added | Deane Yang | I don't see any explicit mention of the fact that there seems to be two different definitions of "hyperplane" being used here. The hint above seems to indicate that in the statement of the problem, it is assumed that "hyperplane" means a linear subspace of dimension $n$. In that case, the assertion is obviously incorrect. However, in all of the answers below, "hyperplane" is assumed to be "affine hyperplane", in which case the assertion is correct for the reasons given below. | |
| Jun 23, 2017 at 17:58 | comment | added | Wlod AA | The problem was: Let $M$ be an $n$-dimensional manifold embedded in $R^{n+1}$. Then almost every hyperplane in $R^{n+1}$ is not tangent to MM at any point. ***** A stronger result, which admits a simple, ELEMENTARY and elegant proof is: there are at the most countably many $a\in\mathbb R$ such that the affine $n$-plane $\{x\in R^{n+1}:\pi_0(x)=a\}$ is tangent to at least one point of $M$ (where $\pi_1(x_0\ldots x_n):=x_0$ is the projection onto the $0$th coordinate). | |
| Jun 23, 2017 at 14:52 | comment | added | Wlod AA | In view of certain confusion caused by the Question, I'd keep this post, let it serve people who sometimes hurry. | |
| Jun 23, 2017 at 11:23 | review | Reopen votes | |||
| Jun 24, 2017 at 13:37 | |||||
| Jun 23, 2017 at 7:48 | history | closed |
Chris Gerig Anton Petrunin abx Loïc Teyssier Mikhail Katz |
Not suitable for this site | |
| Jun 23, 2017 at 7:32 | answer | added | alvarezpaiva | timeline score: 6 | |
| Jun 23, 2017 at 6:30 | comment | added | Francesco Polizzi | Added my previous comment as a answer. | |
| Jun 23, 2017 at 6:27 | answer | added | Francesco Polizzi | timeline score: 5 | |
| Jun 23, 2017 at 1:03 | comment | added | macbeth | Following up on previous comments, the space of hyperplanes of $\mathbb{R}^{n+1}$ (not exactly a Grassmannian) is $\mathbb{RP}^{n+1}\setminus \{pt\}$. For any given unit vector, there is a 1-parameter family of hyperplanes with that unit normal. | |
| Jun 23, 2017 at 0:44 | comment | added | Francesco Polizzi | Any normal hyperplane is uniquely determined by its normal versor, that can be identified with a point is $S^n$. So the tangent hyperplanes form a submanifold of dimension at most $n$ in the space of hyperplanes of $R^{n+1}$, which has dimension $n+1$. This implies the result. | |
| Jun 23, 2017 at 0:41 | review | Close votes | |||
| Jun 23, 2017 at 7:51 | |||||
| Jun 23, 2017 at 0:24 | comment | added | Steven Gubkin | Hmm. What about looking at the map from $M$ to the Grassmanian of hyperplanes? (Diff Geo is not really my area, so forgive me if I have my names wrong). Then I think the dimension of the graddmanian will just be too large. | |
| Jun 23, 2017 at 0:12 | history | asked | J126 | CC BY-SA 3.0 |