How to derive the Euclidean metric from translation and rotation invariance?

Question

I want to know which assumptions are sufficient to derive the Euclidean metric.

Suppose $d(\mathbf{x}, \mathbf{y})$ is a metric on $\mathbb{R}^n$. Also suppose that:

1. $d$ is translation invariant, i.e. $d(\mathbf{x}, \mathbf{y}) = d(\mathbf{x} + \mathbf{c}, \mathbf{y} + \mathbf{c})$ for all constant $\mathbf{c}$;
2. $d$ is rotation invariant, i.e. $d(\mathbf{x}, \mathbf{y}) = d(R\mathbf{x}, R\mathbf{y})$ for all $SO(n)$ matrices.

Does it follow from these assumptions that $d$ is the Euclidean metric?

If the answer is 'no', would it follow if we added a further assumption:

3. $d$ is absolutely homogeneous, i.e. $d(\alpha\mathbf{x}, \alpha\mathbf{y}) = \alpha^2 d(\mathbf{x}, \mathbf{y})$

Please let me know if you know the answer!

Answer 1

The discrete metric ($d(x,y) = 1$ if $x \neq y$ and $d(x,y) = 0$ if $x = y$) would satisfy the first two assumption (since translation and rotation are injective functions). So the answer to your first question is no.

If you add in the third assumption then it is still technically no since scaling the Euclidean norm by a positive number would still satisfy the 3 assumptions (for example you could scale it by 2 which would mean the points on the unit sphere are actually a distance 2 away from the origin).

If you add a 4th assumption saying that the distance from the origin to a unit coordinate vector is 1 then the metric would be unique and it would be the Euclidean metric. You can see this by noticing that assumption #1 implies that defining the metric at the origin defines it everywhere (since you can just shift to the origin: $d(x,y) = d(0, y - x)$). Assumption #2 tells you that the distance from the origin to any point on an n-dimensional sphere is the same. Adding in assumption #4 means that the distance from the origin to any point in the unit sphere is $1$. Any point, $\textbf{x}$, can be written as $\alpha \textbf{u}$ where $\alpha$ is a positive real and $\textbf{u}$ is a unit vector (take $\alpha = ||\textbf{x}||$ and $\textbf{u} = \frac{\textbf{x}}{||\textbf{x}||}$). Using assumption #3 we get that $d(\textbf{0}, \textbf{x}) = d(\textbf{0}, \alpha \textbf{u}) = \alpha d(\textbf{0}, \textbf{u}) = \alpha = ||\textbf{x}||$. So the metric has to be Euclidean at the origin and hence is Euclidean everywhere.