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

• I
• Arne
In summary: For hypothesis 2, try to prove that the extremal points of the convex hull are the points that can be reached with at most two terms. You could try to do this by showing that the extremal points are the points that are the closest to two given points on the curve, or that the extremal points are the points that are furthest from two given points on the curve. For hypothesis 3, try to show that the points on the surface can't be parametrized by three terms.
Arne
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

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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.

## 1. What is the convex hull of a closed path in a 4D space?

The convex hull of a closed path in a 4D space is the smallest convex set that contains all points on the path. In other words, it is the smallest shape that encloses the entire path without any indentations or concavities.

## 2. Why is the convex hull important in 4D space?

The convex hull is important in 4D space because it helps us understand the shape and structure of a closed path. It also has practical applications in fields such as computer graphics, where it is used to create 3D models from 2D images.

## 3. How is the convex hull of a closed path in 4D space calculated?

The convex hull of a closed path in 4D space can be calculated using various algorithms, such as the Graham scan or the Quickhull algorithm. These algorithms use the coordinates of the points on the path to determine the smallest convex shape that encloses them.

## 4. Can the convex hull of a closed path in 4D space be a complex shape?

Yes, the convex hull of a closed path in 4D space can be a complex shape. It can take on various forms, such as a pyramid, a cone, or even a curved shape. The shape of the convex hull depends on the arrangement of points on the closed path.

## 5. What assumptions are made when calculating the convex hull of a closed path in 4D space?

One of the main assumptions made when calculating the convex hull of a closed path in 4D space is that the path is a set of points with four coordinates. Additionally, it is assumed that the path is closed, meaning the starting and ending points are the same. Lastly, the algorithm assumes that there are no duplicate points on the path.

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