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Suppose $d:\mathbb R^2\times \mathbb R^2\to \mathbb R$ is a metric on $\mathbb R^2$. Given arbitrary $\textbf k, \textbf x,\textbf {y}$, must it be true that $d(\textbf x+\textbf k,\textbf y+\textbf k)=d(\textbf x,\textbf y)$? The example metrics (Euclidean metric, $l_1$ metric, max metric, and discrete metric) agree with $d(\textbf x+\textbf k,\textbf y+\textbf k)=d(\textbf x,\textbf y)$, but are there other metrics in $\mathbb R^2$ such that $d(\textbf x+\textbf k,\textbf y+\textbf k)=d(\textbf x,\textbf y)$ does not hold?

Runyang Wang
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  • No, generic metrics on $\mathbb R^n$ are not translation invariant; the simplest example is probably $|x^2 - y^2|$ on $\mathbb R$ (and the same idea generalizes to higher dimensions). However, any metric space metric has an equivalent metric (i.e. inducing the same topology) that is translation invariant, and thus it is often natural to consider only translation invariant metrics. – Brevan Ellefsen Sep 10 '23 at 06:58
  • @BrevanEllefsen, $d(x,y)=\left|x^2-y^2\right|$ is not a metric on $\Bbb R$ indeed $d(1,-1)=0$ but $1\neq-1$. – Angelo Sep 10 '23 at 07:18
  • @BrevanEllefsen, if you write a simple example of metric on $\Bbb R$ which is not translation invariant, you could write $d(x,y)=\left|x^3-y^3\right|$. – Angelo Sep 10 '23 at 07:23
  • @BrevanEllefsen, if you want to write a simple example of metric on $\Bbb R$ which is not translation invariant, you could write $d(x,y)=\left|x^3-y^3\right|$, whereas a simple example on $\Bbb R^2$ is $d\big((x_1,y_1),(x_2,y_2)\big)=\sqrt{\left(x_2^3-x_1^3\right)^{!2}+\left(y_2^3-y_1^3\right)^{!2}}$. – Angelo Sep 10 '23 at 07:47
  • Sorry yeah, I meant cubing and not squaring. Misremembered and didn't check the calculation. Thanks for the correction @Angelo – Brevan Ellefsen Sep 10 '23 at 13:46

1 Answers1

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As mentioned in the comments, the answer is no. Here's an easy way to obtain a counterexample. Let $f: \mathbb{R}^2 \to \mathbb{R}^2$ be $f(x,y) = (|x|x,y)$, i.e. we stretch horizontally proportionally to the distance from the vertical axis. It is not hard to see that $f$ is a bijection. Let's define $d(x,y) = \|f(x) - f(y)\|$.

We now check the metric space axioms:

  1. $d(x,x) = \|f(x) - f(x)\| = 0$.
  2. $d(x,y) = 0$ implies $\|f(x) - f(y)\|$ implies $f(x) = f(y)$ implies $x=y$ because $f$ is a bijection.
  3. $d(x,y) = \|f(x) - f(y)\| = \|f(y) - f(x)\| = d(y,x)$.
  4. $d(x,z) = \|f(x) - f(z)\| \leq \|f(x) - f(y)\| + \|f(y) - f(z)\| = d(x,y) + d(x,z)$.

Finally, we show that $d$ is not translation-invariant.Let $x = (0,0)$ and $y = (1,0)$. Then $d(x,y) = \|(0,0) - (1,0)\| = 1$. But if $k=(1,0)$ then $d(x+k,y+k) = \|(1,0) - (4,0)\| = 3$.

This gives a general technique for generating many other metrics that fail to be translation-invariant, as we did not use any specific properties of $f$ to verify the metric space axioms, other than it being a bijection.

Elchanan Solomon
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    One caveat is that while this is a clever solution, the created metrics are still isometric to $\mathbb R^2$ with the standard metric (the map $f$ becomes an isometry between the spaces equipped with standard metric and the new metric), and while the original translations aren't isometries, the maps $T_v$ given by $x\mapsto f^{-1}(f(x)+v)$ are indeed isometries. Which is not to say there is anything wrong with this example at all, but its good to be aware that while the metric is very different in one sense, it is identical in another. – M W Sep 10 '23 at 08:04
  • @MW, why did you say that the map $f$ is an isometry? If $x=(2,1)$ and $y=(3,1)$, then $\lVert f(x)-f(y)\rVert=\lVert(4,1)-(9,1)\rVert=5\color{red}{\neq}1=\lVert x-y\rVert$. – Angelo Sep 10 '23 at 08:56
  • @Angelo Let $s$ be the standard metric $s(x,y)=||x-y||$. I'm not claiming $f$ is an isometry from $(\mathbb R^2,s)$ to itself. I'm claiming it is an isometry between $(\mathbb R^2,s)$ and $(\mathbb R^2,d)$, since $s(f(x),f(y))=||f(x)-f(y)||=d(x,y)$. – M W Sep 10 '23 at 09:01
  • @MW, I cannot understand why you say it is good to be aware that while the metric is very different in one sense, it is identical in another. In which sense is the metric identical ? Could you write an example of metric on $\Bbb R^2$ which is not identical to the standard one in the sense you mean ? – Angelo Sep 10 '23 at 09:15
  • @Angelo It perhaps would have been more proper to say in a certain sense the *metric space* is identical, meaning that the metric space with the new metric is isometric to the metric space with the old metric. This Is NOT the case with all metrics on $\mathbb R^2$. Take for example the discrete metric $d(x,y)=1$ whenever $x\neq y$. Or, if you'd rather have one that still preserves the topology, take a hyperbolic metric on $\mathbb R^2$. There is no isometry between hyperbolic plane and the Euclidean plane. – M W Sep 10 '23 at 09:22
  • @MW, could you write the hyperbolic metric on $\Bbb R^2$ ? Is it translation invariant ? – Angelo Sep 10 '23 at 09:46
  • Let us continue this discussion in chat. – M W Sep 10 '23 at 09:54