# Isometrically embed a pentagon into $\mathbb{R}^2$

In this post, the regular pentagon is the metric space $M=\{1,2,3,4,5\}$ with $$d(1,2) = d(2,3) = d(3,4) = d(4,5) = d(5,1) = 1,$$ $$d(1,3) = d(2,4) = d(3,5) = d(4,1) = d(5,2) = 2.$$

Questions:

1. Is there a norm $\|\cdot\|$ on $\newcommand{\R}{\mathbb{R}}\R^2$ such that the regular pentagon can be isometrically embedded into $(\R^2,\|\cdot\|)$?

2. If not, what is the minimal dimension $n$ that admits a norm $\|\cdot\|$ on $\R^n$ such that the regular pentagon is isometrically embedded into $(\R^n,\|\cdot\|)$?

What I know already:

• Since any metric space with $k$ elements can be isometrically embedded into $\R^k$ with $l^\infty$-norm, there is at least an embedding into $\R^5$ (e.g. by the Kuratowski embedding).

• For illustration: The square in the above sense (regular 4-gon) can be isometrically embedded into $\R^2$ with $\|\cdot\|_1$-norm (Manhattan distance) as $(0,0),(0,1),(1,1),(1,0)$.

For a generalization of this question, see Isometrically embed $K$-gon into $\mathbb{R}^N$

## 1 Answer

Notation : $f: M\rightarrow (\mathbb{R}^n,\|\ \|),\ f(i)=x_i$

$[xy]$ line segment

$[xy)$ ray starting at $x$ and passing through $y$

$(xy)$ line containing $x,\ y$

$|xy|$ distance between $x$ and $y$

EXE : $n\neq 2$

Proof : Step 1 : Here $\Delta =[ x_1x_3x_4]$ is a triangle. We will prove that a point $x_2$ is not in the triangle $\Delta$.

If $x_2\in \Delta$, then let $$[x_4x_2)\bigcap [x_1x_3] =\{z\}$$

By convexity of distance function $d(x)= |x_4x|$, since $|x_4x_1|,\ |x_4x_3|\leq 2$, then $|x_4z|\leq 2$. By assumption, $z= x_2$ and $|x_4z|=2$.

By assumption of $x_i$, $z$ is a midpoint of $[x_1x_3]$. If $z'$ is a midpoint of $[x_1x_4]$, then $$2=|zx_4| \leq |z'x_4|+|zz'|=\frac{3}{2}$$ so that it is a contraction.

Step 2 : $(x_ix_j)$ where $x_i\in \{x_1,x_3,x_4\}$ divides $\mathbb{R}^2$ into 7 regions

$R_i$ is truncated cone whose vertex is $x_i$ and $S_i$ is a cone whose vertex is $x_i$.

1) $x_2$ is not in $S_1,\ S_3,\ S_4$ and their boundaries :

If $x_2$ is in $S_1$, then we will use the convexity of the function $d(x)=|xx_3|$ Recall that $d(x_1)=2,\ d(x_4)=1$ If we define $$[x_2x_3]\bigcap (x_1x_4) =\{z\}$$

then $$d(x_2)\geq d(z)\geq d(x_1)\geq 2$$

For $S_3,\ S_4$, we do the same argument.

2) $x_2$ is not in $R_3$ : If $x_2$ is in $R_3$, then define $[x_2x_3]\bigcap [x_1x_4]=\{z\}$. Then $$2=|x_4x_2| \leq |zx_4| + |zx_2|$$

so that $|zx_4| =1+\epsilon$. Hence $$2=|x_1x_3|\leq |zx_1| +|zx_3| <2$$ since $|zx_1|=1-\epsilon$. Contradiction.

3) $x_2$ is not in $R_1$ : Now, assume that $x_2\in R_1$.

If $[x_1 x_2]\bigcap [x_3x_4]=\{z\}$, then $|x_1x_3|\leq |x_1z| +|zx_3| <2$. Contradiction.

4) $x_2$ is not in $R_4$ :

If $z,\ z'$ are mid points in $[x_1x_3],\ [x_1x_4]$, then $$2=|x_2x_4|\leq |x_2z| +|zz'| +|z'x_4|=|x_2z| +\frac{3}{2}$$

so that $|x_2z|\geq \frac{1}{2}$.

Since $|x_1x_2|,\ |x_1z|,\ |x_1z'|$ are 1, so $x_2$ is in a closed half plane containing $(zz'),\ x_1$.

If $[zx_4]\bigcap [z'x_3]=\{w\}$, then $[wzz'],\ [wx_4x_3]$ are similar with ratio 2.

Since $|zx_4|\leq \frac{3}{2}$, so $$|zw|\leq \frac{1}{2},\ |wx_4|\leq 1$$

If $x_2x_1x_2'x_3$ is parallelogram, then $x_2'$ is not in $[zz'w]$.

In further, $|x_3x_2'|=1$, then $x_2'$ is not in $[zx_3x_4]$. For $x_5$ we have $x_5'$ so that $[zx_2']\bigcap [z'x_5']$ is in $[zz'w]$.

This contradicts to the fact that $[x_2x_2']$ is in the boundary of unit ball whose center is $x_1$.

EXE : In $\mathbb{R}^3$, there exists an example : $$(0,1,1),\ (-1,1,1),\ (-1,0,1),\ (-\frac{1}{2},0,\frac{3}{2} ),\ (0,\frac{1}{2},\frac{3}{2} )$$ in $\|\ \|_1$

EXE : In $\mathbb{R}^4$, there exists an example : $$(0,0,0,0),\ (1,0,1,1),\ (2,-1,0,0),\ (2,0,0,-1),\ (1,1,-1,0)$$ in $\|\ \|_\infty$

• Thanks! However, some things remain unclear to me: In step 2, how do you get that $x_2$ is not in $S_1$, $S_3$, $S_4$? You prove in step 1, that $x_2$ is not in $\Delta$. And you want to prove in step 2 that $x_2$ is neither in $R_1$, $R_3$, $R_4$ nor in $S_1$, $S_3$, $S_4$. That means there is no place left where $x_2$ could be. Why do you need a third step then? What do you try to prove in step 3? – thomas Jun 18 '18 at 8:13
• I add more explanation. And order of proof is a thing you said (There is no meaning in Step 3 and I will rewrite). – HK Lee Jun 18 '18 at 8:25
• Thanks for the improvements. There are still open questions: i) In Step 2, part (1): In general, $|x_3z| \geq 2$ is wrong under the given assumptions (namely, if the angle of $\Delta$ at $x_1$ is larger than $90^\circ$). ii) I don't understand what happens in step 2, part (4) and I don't understand why the assertion there would end the proof. – thomas Jun 18 '18 at 11:55
• Now, I have questions regarding Step 2, part(4): You define $z''$, but never use it. Is there a reason for this? I don't understand the definition of $x_5'$. Does it depend on the position of $x_5$? Where does $x_5$ come into play in the proof? – thomas Jun 19 '18 at 12:17
• I edited. And $z,\ z'$ is a midpoints of $[x_2x_2'],\ [x_5x_5']$. (Here like a point $x_2$, we have $x_5\in R_3$ so that we have $x_5'$) For the proof, the key is that for any $t\in [x_2x_2']$ we have $|x_1t|=1$. Similarly, for any $t\in [x_5x_5']$ we have $|x_1t|=1$. And then we derive a contradiction. – HK Lee Jun 19 '18 at 12:25