# Isometrically embed $K$-gon into $\mathbb{R}^N$

In this post, the regular $K$-gon is the metric space $M = \{1,\dots,K\}$ such that, for $k,l \in M$, $$d(k,l) := \begin{cases} |k-l|, & |k-l| \leq K/2, \\ K-|k-l|, & |k-l| > K/2. \end{cases}$$ This means that adjacent vertices have distance one and all other vertices have distance according to the length of the shortest path along the edges of the $K$-gon.

Let $N(K)$ be the minimal dimension that admits a norm $\|\cdot\|$ on $\mathbb{R}^{N(K)}$ such that the regular $K$-gon is isometrically embedded into $(\mathbb{R}^{N(K)},\|\cdot\|)$.

Since any metric space with $K$ elements can be isometrically embedded into $\mathbb{R}^K$ with $l^\infty$-norm, there is at least the upper bound $N(K) \leq K$ (e.g. by the Kuratowski embedding).

Questions:

1. Is $N(K)$ non-decreasing?

2. Is $N(K)$ unbounded for $K \to \infty$?

3. If $N(K)$ is unbounded, do we have $N(K)/K \to 0$?

4. What is $N(K)$?

What we know already:

• The square in the above sense (regular $4$-gon) can be isometrically embedded into $\mathbb{R}^2$ (but not into $\mathbb{R}^1$) with $\|\cdot\|_1$-norm (Manhattan distance) as $(0,0),(0,1),(1,1),(1,0)$.

• The pentagon (regular $5$-gon) can be embedded into $\mathbb{R}^3$, but not into $\mathbb{R}^2$, see Isometrically embed a pentagon into $\mathbb{R}^2$

• The hexagon (regular $6$-gon) can be embedded into $\mathbb{R}^3$ as $(1,0,0)$, $(1,1,0)$, $(0,1,0)$, $(0,1,1)$, $(0,0,1)$, $(1,0,1)$ with $l^1$-norm. Can it be embedded into $\mathbb{R}^2$?

• Together: $N(1) = N(2) = 1$, $N(3) = N(4) = 2$, $N(5) = 3$, $N(6) \leq 3$, $\dots$?

• When looking at an embedding $f\colon M \to \mathbb{R}^N$, we can assume $f(1) = (0,\dots,0)$ by translation invariance of norms. For the critical $N$ such that $M$ can be embedded into $\mathbb{R}^N$ but not into $\mathbb{R}^{N-1}$, we can assume in addition, that $N$ of the vectors $f(k)$ are linearly independent. By a linear transformation of the points and modifying the norm accordingly, we can assume that $N$ of the vectors $f(k)$ are mapped to the standard normal vectors $e_n = (0,\dots,0,1,0,\dots,0)$. This fixes already $N+1$ of the $K$ vertices.

This is a generalization of the question Isometrically embed a pentagon into $\mathbb{R}^2$

• I added some seemingly simpler questions like: Is $N(K)$ unbounded? If it's bounded, that would mean that there is a critical $\mathbb{R}^N$ that allows embeddings of all regular $K$-gons. That would be mind-blowing... – thomas Jun 19 '18 at 15:30