Prove that $(\mathbb{R}^2,d_2)$ and $(\mathbb{R}^2,d_1)$ are not isometric, where $d_2$ is euclidean metric and $d_1$ is absolute value metric.

Question

Prove that $(\mathbb{R}^2,d_2)$ and $(\mathbb{R}^2,d_1)$, are not isometric where $d_2$ is euclidean metric and $d_1$ is absolute value metric.

I have to show that there cannot exist a map from $(\mathbb{R}^2,d_2)$ to $(\mathbb{R}^2,d_1)$ that preserves length. But I didn't get any idea.

Answer 1

With $d_1\bigl((x,y),(u,v)\bigr)=|x-u|+|y-v|$, we have $$ (0,0),\quad (1,0), \quad (\tfrac12,\tfrac12)$$ and $$ (0,0),\quad (0,1), \quad (\tfrac12,\tfrac12),$$ that is, two equilateral triangles with side length $1$ and common edge and the other vertices have distance $2$.

Under the Euclidean metric, we would always have distance $\sqrt 3$ instead of $2$.

Answer 2

Consider the points $p=(1,0)$ and $q=(0,1)$. Then $d_1(p,q)=2$ and both points $(0,0)$ and $(1,1)$ are such that their distance both to $p$ and to $q$ is equal to $1$.

However, in $(\mathbb R^2,d_2)$, whenever you have two points $p$ and $q$ such that $d_2(p,q)=2$, the only point whose distance to both of them is $1$ is $\frac{p+q}2$; there is no other point for which this is true.