Timeline for Which vector configurations generate as zonotope the regular $2n$-gon?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 6, 2023 at 21:42 | comment | added | BerndM | For $n=3$, take $X=( (1,0), (\cos\frac{2\pi}3,\sin\frac{2\pi}3), (\cos\frac{4\pi}3,\sin\frac{4\pi}3) )$. | |
| Jan 6, 2023 at 21:29 | comment | added | Sam Hopkins | By the way, your definition of zonotope implicitly assumes that the generating vectors are based at the origin, but with this restriction you will never be able to get something centered at the origin. Better to define a zonotope to be a Minkowski sum of line segments. | |
| Jan 6, 2023 at 21:28 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
added 7 characters in body
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| Jan 6, 2023 at 21:24 | history | edited | BerndM | CC BY-SA 4.0 |
deleted 2 characters in body
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| Jan 6, 2023 at 20:09 | comment | added | Sam Hopkins | The vertices of this $2n$-gon are $(\cos(\theta),\sin(\theta))$ for $\theta=\frac{0}{2n}\pi, \frac{1}{2n}\pi,\ldots, \frac{2n-1}{2n}\pi$. Choose $n+1$ consecutive of these points and make your vectors be the consecutive differences. Anyways, I don't think this is a research-level math question, so probably belongs on math.stackexchange.com instead... | |
| Jan 6, 2023 at 20:05 | comment | added | BerndM | Thanks. See edit. | |
| Jan 6, 2023 at 20:04 | history | edited | BerndM | CC BY-SA 4.0 |
Clarified the condition.
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| Jan 6, 2023 at 19:00 | comment | added | Sam Hopkins | You take vectors at angles $\frac{0}{n} \pi, \frac{1}{n} \pi, \ldots, \frac{n-1}{n} \pi$ of appropriate equal length (which should be easy to work out based on the unit disk requirement, but I am too lazy to do it...) | |
| Jan 6, 2023 at 18:53 | history | asked | BerndM | CC BY-SA 4.0 |