Timeline for The projection of density $1$ point on a rectifiable set
Current License: CC BY-SA 3.0
19 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| S Jan 22, 2016 at 5:04 | history | bounty ended | CommunityBot | ||
| S Jan 22, 2016 at 5:04 | history | notice removed | CommunityBot | ||
| Jan 19, 2016 at 23:51 | comment | added | rozu | @tankonetoone: Yes I'm quite certain about that, following the sketch of a proof above (and you assume that $\Gamma$ has locally finite $\mathcal H^{N-1}$-measure). | |
| Jan 19, 2016 at 19:30 | comment | added | JumpJump | @rozu: Thank you! So what I can have is that for this $\Gamma$, a.e. $x\in\Gamma$ has density 1 in normal sense and also in the sense as I defined in my post right? | |
| Jan 19, 2016 at 11:02 | comment | added | rozu | Although this argument shows that your statement is true for almost every point $x$, you cannot fix a point x as you say and only assume that the density and a weak tangent plane at $x$ exists. To see this consider the $1$-rectifiable set $\Gamma$ in $\mathbb{R}^2$ obtained as the union over all integers $n \in \mathbb{Z}$ of the vertical segments $\{\frac{1}{n}\} \times [0,\frac{1}{|n|(|n|+1)}]$. Then the horizontal axis is a weak tangent plane at $(0,0)$ (with the right density) but the projection of $\Gamma$ onto the horizontal axis has measure zero. | |
| Jan 18, 2016 at 22:35 | comment | added | rozu | As a rectifiable set, $\Gamma$ can be covered up to a negligible set by countably many $C^1$ submanifolds $M_k$ of the same dimension. The statement should be correct for each $M_k$ separately. I think it then follows rather quickly for a density point of $\Gamma \cap M_k \setminus \bigcup_{i \neq k} M_i$ for any $k$. This should imply your statement. | |
| Jan 14, 2016 at 14:43 | history | edited | JumpJump | CC BY-SA 3.0 |
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| Jan 14, 2016 at 4:06 | history | edited | JumpJump | CC BY-SA 3.0 |
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| S Jan 14, 2016 at 3:22 | history | bounty started | JumpJump | ||
| S Jan 14, 2016 at 3:22 | history | notice added | JumpJump | Draw attention | |
| Jan 11, 2016 at 16:15 | history | edited | JumpJump | CC BY-SA 3.0 |
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| Jan 11, 2016 at 16:15 | comment | added | JumpJump | Sure sure. You are right. $T_x$ should be the tangent hyperplane of $\Gamma$, and $\nu(x)$ is a vector normal to $\Gamma$ | |
| Jan 11, 2016 at 16:10 | comment | added | Silvia Ghinassi | So it's the normal vector to the tangent hyperplane of $x$, not the tangent vector. Which makes $T_x$ the tangent hyperplane. | |
| Jan 11, 2016 at 16:09 | comment | added | JumpJump | @SilviaGhinassi The tangent vector $\nu(x)$ is the vector normal to $\Gamma$ at $x$. Sorry for the confusion. | |
| Jan 11, 2016 at 16:07 | comment | added | Silvia Ghinassi | Question: what is the tangent vector $\nu(x)$? There should be an hyperplane tangent to $x \in \Gamma$, or am I misunderstanding? | |
| S Jan 10, 2016 at 17:56 | history | suggested | Silvia Ghinassi | CC BY-SA 3.0 |
fixed grammar, fixed typo (switched parenthesis in the last formula)
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| Jan 10, 2016 at 17:48 | review | Suggested edits | |||
| S Jan 10, 2016 at 17:56 | |||||
| Jan 10, 2016 at 3:19 | history | asked | JumpJump | CC BY-SA 3.0 |