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$(\mathbb{E}^n,+)$ is embedded in the Group $(M(n),\circ)$ as its Normal subgroup.

Define $T:\mathbb{E}^n \to M(n)$ be the map $T(v):=T_v$, the Translation defined by the point $v$. i.e. $T_v(x)=x+v$ for all $x \in \mathbb{E}^n.$

Claim 1: $T$ is a group homomorphism between $\mathbb{E}^n$ and the group $M(n)$

Let $u,v \in \mathbb{E}^n.$ then $T(u+v)=T_{u+v}, T(u)=T_u$ and $T(v)=T_v$. Now for any $x \in \mathbb{E}^n$ we have the following: $$T_{u+v}(x) = x+(u+v) = x+(v+u) = (x+v)+u = T_{v}(x)+u = T_u(T_v(x)) = T_u \circ T_v(x)$$ $$\implies T(u+v) = T(u) \circ T(v) \ \text{for all} \ u,v \in \mathbb{E}^n $$ Therefore $T$ is a Group homomorphism.

Claim 2: $T$ is One-One and $T(\mathbb{E}^n)$ is a Normal subgroup of $M(n)$.

For $u,v \in \mathbb{E}^n$ we have $$T_u = T_v \Rightarrow T_u(x) = T_v(x) \ \text{for all} \ x \in \mathbb{E}^n \Rightarrow T_u(0) = T_v(0) \Rightarrow u=v.$$ Hence $T$ is One-One.

Note that $T(\mathbb{E}^n)=\{T_u:u \in \mathbb{E}^n\}$ i.e all Translations of $\mathbb{E}^n$. Let $S \in M(n)$ and choose any $T_u$ from the image set, then I have to show that $S\circ T_u \circ S^{-1}$ is a Translation. i.e.

$$S\circ T_u \circ S^{-1} (x) = x+v = T_v(x) \ \text{for some} \ v \in \mathbb{E}^n.$$

How can I show that?

Although there are a few answers here, but if someone helps me solve the way I proceeded above. It will be very helpful. Thank you.

The link to those answers are: here, here and here.

Saikat
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  • Should be $T(u)\circ T(v)$, not $T(u)+T(v)$. –  Apr 23 '21 at 09:15
  • Proving normality is straight-forward if you can assume that $M(n)$ is a set of affine maps. If you can't take that for granted, then it's still true that the the maps $\Bbb R^n\to\Bbb R^n$ that preserve the Euclidean distance are affine, and it's essentially the easy case of Mazur-Ulam. –  Apr 23 '21 at 09:26
  • @Gae.S. : Thank you for pointing it out. I have corrected it. – Saikat Apr 24 '21 at 01:44

1 Answers1

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An isometry of euclidean space is an affine transformation $S$ of the form $S(x) = Ox + p$ where $O$ is an orthogonal matrix. The inverse can be computed explicitly (in terms of $O$ and $p$) as

$$ S^{-1}(x) = O^{-1}x - O^{-1}p. $$

With this in mind, you just have to compute the conjugation of $T_u$ by $S$,

$$ ST_u S^{-1}(x) = ST_u(O^{-1}x - O^{-1}p) = S(O^{-1}x - O^{-1}p + u) = O(O^{-1}x - O^{-1}p + u) + p = T_{Ou}(x). $$

Edit: an isometry $f$ is an affine map because of the following: take $x,y$ in Euclidean space and let $d = \|x-y\|$. By preservation of distances, for each $t \in [0,1]$ we have

$$ f(tx+(1-t)y) = f(\overline{B}(x,td) \cap \overline{B}(y,(1-t)d)) = \overline{B}(f(x),td) \cap \overline{B}(f(y),(1-t)d) = tf(x) + (1-t)f(y). $$

Now, the map $fT_{-f(0)}$ is still an isometry and it is linear. A standar argument in linear algebra shows that it is represented by an orthogonal matrix, thus $f(x) = Ox+p$ where $O \in O(n)$ and $p=f(0)$.

qualcuno
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  • It is unclear to me if the OP is taking for granted that $M(n)$ is a set of affine maps, or if he's essentially hitting his head over an attempt to prove Mazur-Ulam. –  Apr 23 '21 at 09:34
  • Given that OP didn't even define $M(n)$ in their post... I will wait for them to ask for clarification, if needed. – qualcuno Apr 23 '21 at 09:36
  • (by the way: I don't think one needs to appeal to that much technology for the case of $\mathbb R^n$. Iirc in a strictly convex space the proof is not hard) – qualcuno Apr 23 '21 at 09:40
  • I agree, but it's still something you would hit your head on. –  Apr 23 '21 at 09:43
  • @Gae.S. there, added a (sketch of) proof. I agree, it's far from obvious. – qualcuno Apr 23 '21 at 10:22