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.

  • 1
    $\begingroup$ 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

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 $\mathbb R_{\ge 0}$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.