4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Rahul Jul 13 at 6:37
  • 1
    $\begingroup$ @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. $\endgroup$ – Malkoun Jul 13 at 6:41
0
$\begingroup$

I have a somewhat complicated argument by contradiction that $X_n = \bar{C}_n$. Roughly speaking, here is an outline. Assume there is a point $v = (v_{ij}) \in X_n$ that is not in $\bar{C}_n$. Let $R$ be the set of all assignments $r = (r_{ij})$, with $r_{ij} \geq 0$ of "pairwise distances" such that

$$\max_{i\neq j} r_{ij} = 1.$$

Note that $R$ is compact. Given $r \in R$, and $1 \leq i<j<k \leq n$, define

$$\operatorname{hol}_{ijk}(v,r) = |r_{ij} v_{ij} + r_{jk} v_{jk} + r_{ki} v_{ki}|,$$

where $|\_|$ denotes the Euclidean norm in $\mathbb{R}^3$. Also define

$$\operatorname{hol}(v,r) = \operatorname{max}_{1 \leq i<j<k \leq n} \operatorname{hol}_{ijk}(v,r)$$

and call it the $2$-holonomy of $(v,r)$. Let

$$m = \operatorname{min}\{ \operatorname{hol}(v,r); r \in R \}$$

and let $r_{\min} \in R$ be an assignment of pairwise distances at which this minimum $m$ is attained. Note that $m>0$, otherwise $v$ would be geometric. We will construct, for the same $v$, an $r' \in R$, for which

$$\operatorname{hol}(v,r')<\operatorname{hol}(v,r_{\min})$$

thus reaching a contradiction.

Consider the set $S$ of all $(i,j,k)$, with $1 \leq i<j<k \leq n$, for which

$$\operatorname{hol}_{ijk}(v,r_{\min}) = \operatorname{hol}(v,r_{\min}).$$

We take each $(i,j,k) \in S$, and perturb the corresponding "distances" $(r_{\min})_{ij}$, $(r_{\min})_{jk}$, $(r_{\min})_{ki}$, so that $\operatorname{hol}_{ijk}(v,r')$ is smaller than the original value of $\operatorname{hol}_{ijk}(v,r_{\min})$, and we do that carefully while respecting that the $\max_{i\neq j} r'_{ij}$ is $1$. This requires considering several cases, depending on whether the number of $1$s among the values of the $3$ pairwise distances is $0$, $1$, $2$ or $3$.

After at most $|S|$ steps, we are guaranteed to have lowered the $2$-holonomy, while respecting that $\max_{i \neq j} r'_{ij} = 1$. We have thus reached a contradiction, and proved that $X_n = \bar{C}_n$.

I apologize for having skipped the case-by-case analysis step, but it should be easy to fill out the blanks by an interested reader.

Observation: interestingly, we only used conditions $1$ and $2$ ($2$-convexity and $3$-convexity respectively) in our proof. Thus the "higher" convexity conditions turn out to be redundant, in that $X_n$ could have been defined using only conditions $1$ and $2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.