Undergrad Assumptions about the convex hull of a closed path in a 4D space

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The discussion centers on understanding the convex hull of a closed path in 4D space with a constant Euclidean norm. The author posits that three points from the path may suffice to span the entire 4D volume, while also suggesting that the 3D surface of this volume can be reached using combinations of two points. There is uncertainty about whether three distinct terms can represent points on the surface, as it appears that at least one term must vanish or two angles must coincide. Suggestions include exploring simpler cases in lower dimensions to build intuition for the 4D scenario. The conversation emphasizes the need for further exploration and validation of these assumptions.
Arne
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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
 
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I suggest first trying to prove these hypotheses for the simpler case of lower dimension. For hypothesis 1, try to prove that all points on the convex hull of a simple, closed, non-planar curve on the surface of an ordinary sphere can be expressed as a linear combination of two points on the curve. Following your intuition above, two points have three independent parameters, which can parametrise a 3D space. But can it parametrise that particular 3D space, ie the convex hull?
If you're able to prove that, it may give some intuitions on how to approach the 4D case.
 

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