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I am reading a proof of reflection representations of isometries that fix the origin in $\mathbb{R}^n$. It is a simple induction on the dimension $n$. For $f(0)=0$ we have some $v\neq w=f(v)$. The reflection in the hyperplane orthogonal to $u=v-w$, $r_u$, suffices $r_uf(u)=-u$ and therefore $r_uf(v)=v$. How do we know $\mathbb{R}v$ is left pointwise invariant under $r_uf$? Do we need to know that $r_uf$ is affine? *Edit: we could know that any isometry is affine without knowing whether it is a product of reflections.

Now $r_uf$ is the identity on $\mathbb{R}v$. Then it seems that I need to prove $r_uf$ is also an isometry on the hyperplane that goes through $O$ and orthogonal to $\mathbb{R}v$. I find this very difficult to understand. It seems to me that this is a result of the fact that $r_uf$ is the identity on $\mathbb{R}v$. But I could approach this conclusion. Why is $r_uf$ an isometry on the hyperplane?

These questions perplexed me when I am reading Stillwell's Naive Lie Theory. I tried to do some online studying but I couldn't understand the most essential parts of the proofs:

My linear algebra background is poor, so can somebody explain this to me with some easy concepts and notations? Thanks in advance. Any help will be appreciated.

  • You can just add the answer's link by clicking on share. So people won't have to scroll – UmbQbify Jul 21 '20 at 08:10
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    @UmbQbify-Key20- thank you for your kind note. Will keep that in mind. –  Jul 21 '20 at 08:11
  • This argument would be enough to answer the first question – Ben Grossmann Jul 21 '20 at 09:19
  • Thanks, @BenGrossmann. I am aware of that. –  Jul 21 '20 at 09:21
  • For anyone who finds this question because of Stillwell, note that the proof should in fact read "the map $r_uf$ ($f$ followed by $r_u$) is the identity on the subspace $\mathbb{R}v$ of $\mathbb{R}^k$. My copy says $\mathbb{R}u$, which is an unfortunate typo. I'd advise anyone to consider the specific case of 90 degree rotation of $\mathbb{R}^2$ with a suitably chosen $v$ to get a feel for the proof (and see why $\mathbb{R}u$ is clearly a typo, while $\mathbb{R}v$ is the only real choice). – Thierry Mar 26 '23 at 00:42

1 Answers1

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I assume that you are aware of the fact that the "isometries of $\Bbb R^n$ that fix the origin" are exactly the "linear orthogonal transformation in $\mathrm O(\Bbb R^n)$".

So, besides $u=v-w$, let us define $u'=v+w$, which satisfies

$$\langle u,u'\rangle = \langle v-w,v+w\rangle = \|v\|^2-\|w\|^2 = 0.$$

We used that $f$ is an isometry via $\|w\|=\|w-0\|=\|f(v)-f(0)\| = \|v-0\|=\|v\|$. So we see that $u'$ must be contained in the reflection hyperplane of $r_u$, and so we have $r_uu'=u'$. Thus

$$r_u f(v) = r_u w = r_u(-u/2+u'/2) = u/2 + u'/2 = v.$$

So indeed, $r_u f$ fixes $v$, and since $r_u$ and $f$ are linear, it also fixes $\Bbb R v$ pointwise.

The next is a general result from representation theory (of finite groups), but it applies to single transformations as well:

If an orthogonal map, let's call it $r\in\mathrm O(\Bbb R^n)$, fixes a subspace $U\subseteq\Bbb R^n$ setwise, then it also fixes its orthogonal complement $U^\bot$ setwise: for every $u\in U$ and $u'\in U^\bot$, we have $r^{-1}u\in U$ and $$\langle u, ru'\rangle = \langle r^{-1}u,u'\rangle = 0.$$ Thus $r u'\in U^\bot$.

So since $r_u f$ fixes $\Bbb Rv$ pointwise (but also setwise), it also fixes the orthogonal complement of that (setwise), in particular, it restricts to an orthogonal map on this complement, and is thus an isometry on it.

M. Winter
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  • What a great answer. –  Jul 21 '20 at 10:57
  • Can you elaborate on the claim that "it restricts to an orthogonal map on this complement"? Am I correct to interpret this as: "since $r_uf$ fixes $\mathbb{R}v$ setwise then $r_uf$ also fixes $\mathbb{R}v^\perp$ setwise, which means $r_uf(w)\in\mathbb{R}v^\perp$ if $w\in\mathbb{R}v^\perp$ and thus $r_uf$ is an isometry on $\mathbb{R}v^\perp$"? –  Jul 21 '20 at 11:04
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    @GoogleME You said it perfectly. $r_u f$ can be considered as a map $\Bbb Rv^\bot\to\Bbb R v^\bot$. And obviously it stays an isometry when restricting it in this way. – M. Winter Jul 21 '20 at 11:29