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Say that $A\subset \mathbb{R}^n$ is measurable and of positive, finite measure.

I'm trying to see if the following is true.

If $A$ is invariant under all orthogonal reflections across $(n-1)$ dimensional subspaces in $\mathbb{R}^n$, then $A$ is a ball (centered at the origin).

I can see that for $n=1$ this is false. But I've been told its true for $n>1$. Can somebody please shed a bit of light on this one for me? There must be some kind of a elegant symmetry argument that I don't see.

If needed we can also assume that $A$ is open.

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    $\begingroup$ For $n=2$ an annulus around the origin also has this property. Analogous examples (ball minus a smaller ball) work in higher dimensions. $\endgroup$ – moonlight Sep 1 '15 at 6:50
  • $\begingroup$ It's easy to see that if $A$ and $B$ are such sets, $A\cup B$, $A\cap B$ and $A-B$ are such sets too. $\endgroup$ – Mohsen Shahriari Sep 1 '15 at 7:02
  • $\begingroup$ Reflections generate O(n), so an invariant set would be rotationally symmetric. Conversely, rotational symmetry is sufficient. $\endgroup$ – whacka Sep 1 '15 at 7:11
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By the Cartan–Dieudonné theorem, reflections around $(n-1)$-dimensional hyperplanes generate the entire orthogonal group $O(n)$ of $\mathbb R^n$. So, asking that a subset $A \subset \mathbb R^n$ is closed under reflections is the same as asking that it is closed under the natural operation by $O(n)$. Thus, $A$ is a union of $O(n)$-orbits. Now $O(n)$ contains the special orthogonal group $SO(n)$ as subgroup. The orbit of a point under $SO(n)$ is simple an $(n-1)$-sphere, and an $(n-1)$-sphere is also closed under reflections. Hence the $O(n)$-orbit of a point is an $(n-1)$-sphere. It follows that $A$ is a union of $(n-1)$-spheres.

Taking polar coordinates, we have more or less $\mathbb R^n \cong S^{(n-1)} \times \mathbb R_{\ge 0}$. (Technically, we have to remove some measure zero-sets to get a diffeomorphism.) Under this map $A$ takes the form $S^{(n-1)} \times A_0$ with $A_0 \subset \mathbb R_{\ge 0}$. From this it should follow that $A$ has positive, finite measure if and only if $A_0$ has.

Altogether, $A$ is a union of $(n-1)$-spheres with the radii forming a measureable subset of finite, positive measure of $\mathbb R_{\ge 0}$.

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  • $\begingroup$ Thank you, can you say a few words on how to prove that theorem? Is it easy to see in $n=2$? $\endgroup$ – user265974 Sep 1 '15 at 11:48
  • $\begingroup$ It can be shown by induction on $n$. If $\varphi \in O(n)$ and $\varphi\ne \operatorname{id}$, there exists $v \in \mathbb R^n$ with $\varphi(v) \ne v$. Let $\sigma$ denote the reflection about the hyperplane orthogonal to $u:=\varphi(v)-v$. Then $\sigma(v)=\varphi(v)$, and so $\sigma^{-1}\varphi$ fixes the subspace $\mathbb R u$. Now consider $\sigma^{-1}\varphi$ restricted to the orthogonal complement of $\mathbb Ru$. Write it as product of reflections using the induction hypothesis; extend them to $\mathbb R^n$. See John Stillwell, Naive Lie Theory, p.36-37; or books on quadratic forms. $\endgroup$ – moonlight Sep 1 '15 at 12:25
  • $\begingroup$ Jean Gallier, Geometric Methods and Applications: For Computer Science and Engineering, p.202-205 gives a more detailed proof, while most books on quadratic forms (e.g. by T.Y. Lam) contain the abstract version for regular quadratic spaces. $\endgroup$ – moonlight Sep 1 '15 at 12:29
  • $\begingroup$ Thanks! You were a great help! $\endgroup$ – user265974 Sep 1 '15 at 22:34

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