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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?
DeltaIV
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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 $n+2$.

mathcounterexamples.net
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  • 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. – DeltaIV Jun 16 '18 at 08:43
  • 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? – DeltaIV Jun 16 '18 at 11:40
  • Finally, "a translation is a product of 2 reflections". This seems weird - reflections are linear transformations,since they can be represented by an orthogonal matrix whose determinant is -1. If a translation is the product of two reflections (product $\equiv$ composition here, right??) then this would mean that all affine transformations are linear...it sounds wrong – DeltaIV Jun 16 '18 at 11:46
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    Two comments. First any isometry in a finite dimensional Euclidean space is linear. See https://math.stackexchange.com/questions/570139/should-isometries-be-linear. 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. – mathcounterexamples.net Jun 16 '18 at 12:42
  • Well, no. The question you link to seems to explicitly say that any isometry onto $\mathbb{E}^n$ is affine, not linear. As a matter of fact, both the OP corrected the text of her/his question ("Is it true that $T$ is linear affine?" a dashed line covers "linear" in the original), and also the accepted answer starts by saying that "I suppose you meant that $T$ is affine rather than merely linear." – DeltaIV Jun 16 '18 at 13:16
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    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... – mathcounterexamples.net Jun 16 '18 at 13:20
  • Concerning the fact that a translation can be produced by composing two reflections, I thought about it and it's obviously true in the plane (just let the reflections be around two parallel axes). I'll trust you on the fact that it's true $\forall n$. – DeltaIV Jun 16 '18 at 13:22
  • Ok, I did clarify in my original question that with Euclidean space I was referring to an affine space, but I didn't know that by extension you could call also the underlying vector space Euclidean. My bad. Also, in meantime I found this and, from there, this, which I didn't know about. This again confirms what you said, i.e., any isometry is affine (even for infinite-dimensional spaces). – DeltaIV Jun 16 '18 at 13:35
  • Ps just for clarity, I didn't downvote your answer: I upvoted it. But I'd really appreciate if you could expand the proof a little bit, or provide a reference: I'm not sure how "stabilization" (meaning that all points on the line, say, are transformed into themselves, right)? is sufficient to say that any isometry is a product of reflections. I like your answer, but I would just be sure I understood it properly. Apologies for being a bother. – DeltaIV Jun 16 '18 at 13:41
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    https://math.stackexchange.com/questions/138020/proving-that-every-isometry-of-mathbbrn-is-of-the-form-of-a-composition-of – mathcounterexamples.net Jun 16 '18 at 14:47
  • I'll read that in more detail, but if I understand correctly, your link proves that any isometry on a affine space of dimension $n$ is the product of $n+1$ reflections. Correct? – DeltaIV Jun 16 '18 at 14:58
  • Read it please! – mathcounterexamples.net Jun 16 '18 at 14:59
  • Ok! I wasn't going to skip it, it's just that it will take me time (the question and the answers are quite long), so don't expect me to reply soon. In meantime, if you want, you may have a look at my other question. – DeltaIV Jun 16 '18 at 15:16