How do I show that some metric $d$ on the set $\text{I}(\mathbb{R}^n)$ gives the usual metric on $\mathbb{R}^n$?

Question

The problem from "Notes on Geometry" by Rees:

On the subset of $\text{I}(\mathbb{R}^n)$ (The set of all isometries of $\mathbb{R}^n$) consisting of all the translations, show that $d$ gives the usual metric on $\mathbb{R}^n$.

It didn't mention which $d$ we are considering, but a metric $d$ on $\text{I}(\mathbb{R}^n)$ is defined in the book:

Choose a set of $(n+1)$ affinely independent points $a_0, a_1, ..., a_n$ in $\mathbb{R}^n$ and let $d(f,g)= \max_{0 \leq i \leq n}d(f(a_i),g(a_i))$ for $f,g \in \text{I}(\mathbb{R}^n)$.

I'm actually not sure if this is what this problem wants me to show, but I interpreted it as "Show that $d$ (which is defined as above) is a metric on $\text{I}(\mathbb{R}^n)$."

So, I tried to solve it as follows:

Let $f,g,h \in \text{I}(\mathbb{R}^n)$ be randomly chosen isometries such that for any $x \in \mathbb{R}^n$ $f(x)=Ax +a, g(x)=B(x)+b, h(x)=C(x)+c$ where $A,B,C \in \text{O}(n)$ and $a,b,c \in \mathbb{R}^n$.

Let $a_0, a_1, ..., a_n$ be affinely independent points in $\mathbb{R}^n$.

• $d(f,g)=d(f(a_i),g(a_i))$ for some $i \in \{0,1,2,...,n\}$.

$\Leftrightarrow d(f,g)= d(x,y)=0 \Leftrightarrow x=y \Leftrightarrow f=g $ (where $f(a_i)=x, g(a_i)=y$.)

Thus, $d(f,g)=0 \Leftrightarrow f=g$ as we wanted.

• $d(f,g)=d_{\mathbb{R}^n}(x,y)=d_{\mathbb{R}^n}(y,x)=d(g,f)$ $\Rightarrow d(f,g)=d(g,f)$ for any $f,g \in \text{I}(\mathbb{R}^n)$.

• $d(f,g)=d_{\mathbb{R}^n}(x,y), d(f,h)=d_{\mathbb{R}^n}(x,z), d(h,g)= d_{\mathbb{R}^n}(z,y)$ where $f(a_i)=x, g(a_i)=y, h(a_i)=z$ for some $i$.

But, we know that $d_{\mathbb{R}^n}(x,y) \leq d_{\mathbb{R}^n}(x,z) + d_{\mathbb{R}^n}(z,y)$ by the triangle inequality.

Thus, $d(f,g) \leq d(f,h) + d(h,g)$ for any $f,g,h$.

Hence, $d$ satisfies all properties of metric. Therefore, it gives a usual metric on $\mathbb{R}^n$.

Question:

First of all, I wonder if my interpretation of this problem is correct. If so, is my answer correct? Can I really conclude like this? I feel like I'm missing something before the conclusion, but I don't know what that is.

Answer 1

First, your interpretation of the problem is not correct: The problems wants you to prove the following: Let $T(\mathbf R^n) \subseteq I(\mathbf R^n)$ the set of all translations, i.$\,$e. the isometries $\tau_a: x \mapsto x + a$, $a \in \mathbf R^n$. Note that the map $\tau\colon\mathbf R^n \to T(\mathbf R^n)$, $a \mapsto \tau_a$ is a group isomorphism. The problem wants you to prove that the metric $d$ induced on $T(\mathbf R^n)$ induces the standard metric on $\mathbf R^n$ via $\tau$, or to put it in other words: $\tau$ is an isometry, i.$\,$e. $$ d(\tau_a, \tau_b) = d_{\mathbf R^n}(a,b), \qquad a,b \in \mathbf R^n. $$ Second, your argument in showing $d$ being a metric is not fully correct. The idea, to use the metric properties of $d_{\mathbf R^n}$ is correct, but your are missing some points: The equivalence of $x=y$ and $f=g$ does not follow directly. In the symmetry part, you must at least say why $d(f,g) = d_{\mathbf R^n}(x,y)$ implies $d(g, f) = d_{\mathbf R^n}(y, x)$ (note that you choose $x$ and $y$). For the last part, in general it is not true, that you can choose the same point $a_i$ for both the distance $d(f, g)$ and $d(g, h)$, there are $i, j$ such that $d(f, g) = d_{\mathbf R^n}((f(a_i), g(a_i))$ and $d(g, h) = d_{\mathbf R^n}((g(a_j), h(a_j))$, so you cannot just write $y$ for both $g(a_i)$ and $g(a_j)$.