I use here the term Euclidean space in the rigorous sense of an affine space over $\mathbb{R}^n$, equipped with the Euclidean inner product (see here). Let $\mathbb{E}^n$ be the Euclidean $n-$space.

Consider the set $S$ of all isometries on $\mathbb{E}^n$. Given any element $f\in S$, intuitively I would expect $f$ to be equivalent to a composition $g$ of the three "elementary" isometries, i.e., rotations, translation and reflections. With equivalent, I mean that $f(P)=g(P) \ \forall P \in \mathbb{E}^n$.

  • Is this true? Since "elementary" isometries are affine functions (they can all be represented by a multiplication by a matrix or by addition of a constant, i.e., a translation), and since the composition of two affine functions is affine (I think), this would imply that all isometries on $\mathbb{E}^n$ are affine functions, right?
  • If it is, how many "elementary" isometries are needed to generate any isometry? For example, can we say that for $n\ge2$, any isometry is a composition of a rotation $or$ a reflection plus a translation?

In a Euclidean vector space of dimension $n$, any isometry is the product of at most $n$ reflections.

The proof consists in defining by induction $r_1, \dots, r_n$ reflections such that $r_1 \circ f$ stabilizes a line, $r_2 \circ r_1 \circ f$ stabilizes à plane and so on where $f$ is the isometry that you are looking to decompose as a product of reflections.

Knowing that a translation is a product of $2$ reflections, you can also conclude that any affine isometry in an affine space of dimension $n$ is the product of affine reflections,at most $n+2$.

  • $\begingroup$ I was going to edit my question a bit, but you answered before I could do that :-) anyway, I think your answer will answer also my edited question, with just a minor edit. $\endgroup$ – DeltaIV Jun 16 '18 at 8:43
  • $\begingroup$ Hmmm, sorry but you say that "any affine isometry in an affine space of dimension n is the product of affine reflections,at most n+2". I was talking about (bijective) isometries in general, not just affine ones. Of course, if all bijective isometrics are affine, then your answer is fine, but could you either prove your assertion or provide a reference to the proof? $\endgroup$ – DeltaIV Jun 16 '18 at 11:40
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    $\begingroup$ Two comments. First any isometry in a finite dimensional Euclidean space is linear. See math.stackexchange.com/questions/570139/…. Second, the notion of affine reflection exists. Finally it is easy to write a translation as a produçt of two affine reflections with parallel hyperplanes. $\endgroup$ – mathcounterexamples.net Jun 16 '18 at 12:42
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    $\begingroup$ Please try to be a bit flexible... An isometry of an "Euclidean affine space" induces an isometry of the underlying Euclidean vector space. This one is linear as proved in the link I provided. Hence the first place isometry of the affine space is affine. And so on... $\endgroup$ – mathcounterexamples.net Jun 16 '18 at 13:20
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    $\begingroup$ math.stackexchange.com/questions/138020/… $\endgroup$ – mathcounterexamples.net Jun 16 '18 at 14:47

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