# Is the set of collections of directions satisfying the following convexity conditions “geometric”?

Let $$X_n$$ denote the set of all collections $$(v_{ij})$$ of points on the sphere $$S^2$$, for $$1 \leq i,j \leq n$$, $$i \neq j$$, such that:

1. $$v_{ji} = -v_{ij}$$,

2. the origin $$O$$ in $$\mathbb{R}^3$$ is in the convex hull of $$v_{ab}$$, $$v_{bc}$$ and $$v_{ca}$$ for all $$1 \leq a < b < c \leq n$$.

3. the origin $$O$$ in $$\mathbb{R}^3$$ is in the convex hull of $$v_{ab}$$, $$v_{bc}$$, $$v_{cd}$$ and $$v_{da}$$, for all $$1 \leq a < b < c < d \leq n$$. and so on.

Let us call conditions 1, 2, etc. the convexity conditions.

Let $$C_n$$ be the subset of $$X_n$$ consisting of all collections $$(v_{ij})$$ arising from configurations $$(\mathbf{x}_i)$$, $$1 \leq i \leq n$$, of $$n$$ distinct points in $$\mathbb{R}^3$$, using the formulas:

$$v_{ij} = \frac{\mathbf{x}_j-\mathbf{x}_i}{|\mathbf{x}_j-\mathbf{x}_i|}$$

for $$1 \leq i,j \leq n$$, $$i \neq j$$. One can indeed check that elements of $$C_n$$ satisfy the convexity conditions above.

Let us call the closure $$\bar{C}_n$$ of $$C_n$$ the set of "geometric" collections of pairwise directions. My question can now be stated. Is every element of $$X_n$$ "geometric"? It is easy to show that $$\bar{C}_n \subseteq X_n$$. My question is whether or not $$X_n$$ is equal to $$\bar{C}_n$$.

Note: I have simplified my post, compared to previous versions, as this is the crucial point remaining. I apologize for that. I hope this shorter version would be easier to read.

• Can you explain the need to take the closure $\bar C_n$? It's not obvious to me that $C_n$ itself is not closed. – user856 Jul 13 '19 at 6:37
• @Rahul, suppose you have a configuration $\mathbf{x} = (\mathbf{x}_i)$ of $n$ distinct points, and let two of the points collide, keeping track of the direction of their collision, so to speak. This can be done in a compactification of the configuration space. The collection of pairwise directions of such a "degenerate" configuration is in the closure of $C_n$, without being in $C_n$ itself. – Malkoun Jul 13 '19 at 6:41

## 1 Answer

First, let's show that conditions 3, 4, etc follow from 1 and 2.

For simplicity, I'll only consider "generic solutions"/"general positions": ie no two points are the same, and no four points lie in a plane. Resolving these cases should be a mere technical matter.

For a triple of points in condition 2, there must be a relation $$rv_{ij}+sv_{jk}+tv_{ki}=0$$ for some $$r,s,t>0$$. This leads to $$v_{ik} = -v_{ki} = r'v_{ij}+s'v_{jk}\quad\text{where}\quad r'=r/t>0 \text{ and } s'=s/t>0.$$ Also, note that condition 2 does not actually depend on the ordering $$i: it applies to distinct $$i,j,k$$ of any order.

For a quadruplet of points in condition 3, we may use the above result on the triplets $$a,b,c$$ and $$c,d,a$$ to obtain $$v_{ac}=pv_{ab}+qv_{bc}\quad\text{and}\quad v_{ca}=rv_{cd}+sv_{da} \quad\text{for some}\quad p,q,r,s>0.$$ Since $$v_{ac}+v_{ca}=0$$, this makes $$pv_{ab}+qv_{bc}+rv_{cd}+sv_{da}=0$$ placing the origin in the convex hull. Again, the order of $$a,b,c,d$$ does not matter.

The same approach could be used to reduce the condition on $$n+1$$ points to that on $$n$$ points.

Next, we try to express the $$v_{ij}$$ in terms of vectors $$x_i$$. As observed in the problem statement, if we start with vectors $$x_i$$ and define the derived points $$x_{ij}=\frac{x_j-x_i}{|x_j-x_i|}$$ on the plane, the set of points $$v_{ij}=x_{ij}$$ satisfy the convexity conditions.

Given a set of points, $$v_{ij}$$, that satisfy the convexity conditions, I'll construct vectors $$x_i$$ which result in $$v_{ij}=x_{ij}$$.

Note that the vectors $$x'_i=sx_i+u$$, where $$s>0$$ and $$u$$ is a fixed vector, result in $$x'_{ij}=x_{ij}$$. So given $$v_{12}$$, we are free to select $$x_1=0$$ and $$x_2=v_{12}$$ without loss of generality, ensuring $$x_{12}=v_{12}$$.

For $$k>2$$, $$x_{1k}=v_{1k}$$ and $$x_{2k}=v_{2k}$$ uniquely determines $$x_k$$. This is basically just that given $$x_1$$ and $$x_2$$ and the directions $$x_{1k}$$ and $$x_{2k}$$ towards $$x_k$$, the point $$x_k$$ is determined.

So, we now have $$x_1, \ldots, x_n$$ (uniquely determined up to translation and scaling) resulting in $$x_{12}=v_{12}$$, $$x_{1k}=v_{1k}$$, and $$x_{2k}=v_{2k}$$, and need to show the remaining $$x_{ij}=v_{ij}$$.

The point $$v_{ij}$$ lies on the plane spanned by the vectors $$v_{1i}$$, $$v_{1j}$$ since we can write $$rv_{ij}+sv_{j1}+tv_{1i}=rv_{ij}-sv_{1j}+tv_{1i}=0$$. Similarly, $$v_{ij}$$ lies on the plane spanned by $$v_{2i}$$, $$v_{2j}$$. These two planes intersect in a line, which intersects the sphere in two points, determining $$v_{ij}$$ uniquely up to change of sign. The requirement that $$rv_{ij}-sv_{1j}+tv_{1i}=0$$ for $$r,s,t>0$$, is enough to determine which of the two points.

However, since the $$x_{ij}$$ also satisfy the convexity conditions, the same result applies to $$x_{ij}$$: it lies on the same two planes as $$x_{1i}=v_{1i}$$, etc. So, as the set of points is uniquely determined from $$v_{12}$$, $$v_{1k}$$, and $$v_{2k}$$, and are those are identical for $$v$$ and $$x$$, the remaining must also be the same: ie, $$v_{ij}=x_{ij}$$ for all $$i,j$$.

• Yes true. Thank you for this nice argument. – Malkoun Jul 30 '19 at 3:23
• @Malkoun: I think I have the full proof of the conjecture, and in a fairly "constructive" manner. Upon review, it still looks correct, which is a good sign. – Einar Rødland Aug 16 '19 at 22:16
• I checked it, and it seems fine. I mean, the only outstanding thing is what to do with degenerate cases, but those should follow by some kind of continuity/density argument, being limiting cases of sequences of configurations in general position. Ok, nice! – Malkoun Aug 17 '19 at 10:23