- #1

Arne

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- TL;DR Summary
- I have a closed path in 4d space, and I would like to prove some assumption about it's convex hull.

Hello everyone,

I am struggling to get insight into a certain set in 4D space. Given is a closed path in 4D-space with constant Euclidean norm

$$\vec{\gamma} (\theta):[0,2\pi]\to\mathbb{R}^4, \ \ \vec{\gamma}(0)=\vec{\gamma}(2\pi), \ \ ||\vec{\gamma}(\theta)||_2 = \mathrm{const.}$$

I am looking at a set which is defined as the convex hull of these points

$$S = \left\{\sum_k \alpha_k \vec{\gamma}(\theta_k): \ \ 0\leq\alpha_k\leq 1, \ \ \sum_k \alpha_k = 1, \ \ \theta_k \in [0,2\pi]\right\}$$

I'm also assuming, that the path is injective (##\vec{\gamma} (\theta_1)\neq\vec{\gamma} (\theta_2)## for ##\theta_1 \neq \theta_2##), and it does NOT lie in a 2- or 3-dimensional hyperplane; the set thus forms a 4-dimensional volume. Due to the constant norm, the path lies on the 3-dimensional boundary/surface of the volume, but since there are no straight sections in the path, it should also form the extremal points of the convex hull.

Now I have some assumptions about this set, that I can support with heuristic test. However, I would like to make sure, that these things are actually correct (I don't necessarily need to do a formal prove, but at least some sufficient arguments, why I can regard these assumptions to be true with some certainty).

1. It seems, that three terms ##\sum_{k=1}^3 \alpha_k \vec{\gamma}(\theta_k)## are enough to span the whole volume. This would give 5 independent parameters, which can be enough to parameterize a 4d volume. But it could still be, that 4 terms might be able to reach points, that are not possible with 3 terms.

2. It seems, that the 3-dimensional surface of the 4-dimensional volume are exactly the points that can be reached with at most 2 terms ##\alpha_1 \vec{\gamma}(\theta_1)+\alpha_2 \vec{\gamma}(\theta_2)##. Two terms would give 3 independent parameters, that can span a 3-dimensional surface, but I don't know, why this would be the surface of the 4d-volume.

3. It seems, that the points on the surface cannot be written with 3 distinct terms (which means for 3 terms, either a ##\alpha## vanishes, or two of the angles ##\theta## are the same).

Of course, I am not expecting any complete proves. But after a frustrating lengthy google search, I would be happy over any fresh ideas or pointers for where to look further. Thanks!Arne

I am struggling to get insight into a certain set in 4D space. Given is a closed path in 4D-space with constant Euclidean norm

$$\vec{\gamma} (\theta):[0,2\pi]\to\mathbb{R}^4, \ \ \vec{\gamma}(0)=\vec{\gamma}(2\pi), \ \ ||\vec{\gamma}(\theta)||_2 = \mathrm{const.}$$

I am looking at a set which is defined as the convex hull of these points

$$S = \left\{\sum_k \alpha_k \vec{\gamma}(\theta_k): \ \ 0\leq\alpha_k\leq 1, \ \ \sum_k \alpha_k = 1, \ \ \theta_k \in [0,2\pi]\right\}$$

I'm also assuming, that the path is injective (##\vec{\gamma} (\theta_1)\neq\vec{\gamma} (\theta_2)## for ##\theta_1 \neq \theta_2##), and it does NOT lie in a 2- or 3-dimensional hyperplane; the set thus forms a 4-dimensional volume. Due to the constant norm, the path lies on the 3-dimensional boundary/surface of the volume, but since there are no straight sections in the path, it should also form the extremal points of the convex hull.

Now I have some assumptions about this set, that I can support with heuristic test. However, I would like to make sure, that these things are actually correct (I don't necessarily need to do a formal prove, but at least some sufficient arguments, why I can regard these assumptions to be true with some certainty).

1. It seems, that three terms ##\sum_{k=1}^3 \alpha_k \vec{\gamma}(\theta_k)## are enough to span the whole volume. This would give 5 independent parameters, which can be enough to parameterize a 4d volume. But it could still be, that 4 terms might be able to reach points, that are not possible with 3 terms.

2. It seems, that the 3-dimensional surface of the 4-dimensional volume are exactly the points that can be reached with at most 2 terms ##\alpha_1 \vec{\gamma}(\theta_1)+\alpha_2 \vec{\gamma}(\theta_2)##. Two terms would give 3 independent parameters, that can span a 3-dimensional surface, but I don't know, why this would be the surface of the 4d-volume.

3. It seems, that the points on the surface cannot be written with 3 distinct terms (which means for 3 terms, either a ##\alpha## vanishes, or two of the angles ##\theta## are the same).

Of course, I am not expecting any complete proves. But after a frustrating lengthy google search, I would be happy over any fresh ideas or pointers for where to look further. Thanks!Arne

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