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Hello I wanted to prove the following statement.

Let $M$ be a compact connected surface in $\mathbb{R}^3$ such that for all $d\in S^2$ there exists a plane called $\pi_d$, such that is orthogonal to $d$ and it's a plane of symmetry of $M$, then $M=S^2$.

So far, I could prove that if the origin is contained in every normal line to $M$, then $M$ is a sphere centered at the origin. So now I tried to see that if I have the symmetry plane given by the statement, every such plane contains the origin, and this way I think I could conclude, but I am not able to prove this last assertion.

M. Winter
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J. Salieri
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    What is a plane of symmetry of $M$ ? Can you give a definition of it ? – HK Lee Aug 11 '17 at 17:50
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    Yes, it means that if you take the reflection associated to a plane of symmetry of the surface, the surface keeps invariant. For example if you take $S^2$ and you apply to it the reflection associated to the plane $z=0$, you get again $S^2$. – J. Salieri Aug 11 '17 at 17:55
  • Not necesarilly, I consider all the planes which are orthogonal to a given direction. – J. Salieri Aug 11 '17 at 17:59
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    Hint: Two consecutive reflections are a rotation, i.e. your object as also lots of rotational symmetries. – M. Winter Aug 11 '17 at 18:21
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    Do you mean to say $d \in S^2$ or $d \in M$? – Robert Lewis Aug 11 '17 at 19:05
  • You cannot prove your last assertion because it is not true. E.g. take a sphere which is not centered at the origin. It still has a symmetry plane for each normal $d\in S^2$. But most of the planes will not contains the origin. – M. Winter Aug 12 '17 at 15:22

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Let $n\in S^2$ and $\alpha$ and angle. Choose two vectors $n_1,n_2\in S^2$ with $n_1\times n_2=n$ and $2\arccos\langle n_1,n_2\rangle=\alpha$. Then the consecutive reflection at the planes $\pi_{n_1}$ and $\pi_{n_2}$ will give a rotation around $n$ with angle $\alpha$. As the used reflections are symmetries, so is the rotation. A surface for which any rotation is a symmetry must be a sphere.

M. Winter
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  • But these symmetries that you are taking are centered at the origin? How can I know that the origin is contained in every plane of symmetry? – J. Salieri Aug 12 '17 at 14:30
  • @J.Salieri The planes $\pi_{n_i}$ do not have to contain the origin. Why do you think so? They are the planes that you describe in your question, which should exist for your surface. – M. Winter Aug 12 '17 at 15:18
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Like you mentioned, it's all about showing that the different planes of symmetry all pass through the same point. Take three symmetries $\sigma_1$, $\sigma_2$, $\sigma_3$in pairwise perpedicular planes, interesecting at $0$. Let another plane of symmetry be $\pi$. Now check that for a symmetry $\sigma$ ( in fact, any isometric $\pi$) and another symmetry $S_{\pi}$ with respect to the plane $\pi$ we have $$\sigma \circ S_{\pi} \circ \sigma^{-1} = S_{\sigma(\pi)}$$ the reflection w r to the plane $\sigma(\pi)$.

Since $\sigma_1$, $\sigma_2$, $\sigma_3$ and $S_{\pi}$ invariante $M$ we conclude that $S_{\sigma_1 \sigma_2 \sigma_3(\pi)}$ also invariante it. But $\pi' \colon = \sigma_1 \sigma_2 \sigma_3 (\pi)$ is the symmetric of $\pi$ with respect to the point $0$. It is a plane parallel to $\pi$, and, if $\pi$ does not contain $0$, distinct from $\pi$. But then also $S_{\pi'} \circ \S_{\pi}$ invariantes $M$, and this is a translation by a $\ne 0$ vector, not possible.

Obs: $M$ matters just a bit, rather the whole thing is about the group generated by these reflection. Since it cannot contain not trivial translation, it will be the group of isometries fixing a certain point $0$ ( the fact that the reflections generate that is rather standard.

orangeskid
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Assume that $M$ is 2-dimensional orientable closed surface in $\mathbb{R}^3$.

(1) For $x\in (M,\rho)$, where $\rho$ is a metric in $M$, we have a plane $P$ and a reflection $r_P$ wrt the plane $P$.

So we have a claim that $M\cap P$ is an image of geodesic path.

Proof : Assume that injectivity radius of $x$ is $\epsilon$ and $\rho(x,y)<\varepsilon $.

If a shortest path $c$ between $x$ and $y$ is not in $M\cap P$, then $r_P(c)\neq c$ is a shortest path between $x$ and $y$. It is a contradiction.

(2) Similarly, for $x'\in M$ we have a plane $P'$.

If $M\cap P,\ M\cap P'$ do not intersect, then by using $r_P,\ r_{P'}$, there is infinite isometric copies in $M$ of $M\cap P,\ M\cap P'$ s.t. they do not intersect pairwisely. Hence $M$ has an infinite diameter. It is a contradiction.

Hence $M\cap P,\ M\cap P'$ intersect at $x$ with an angle $t$. Here a composition of $r_P,\ r_{P'}$ is a $2t$-rotation (M. Winter commented).

(3) Consider another point $x''$ so that we have a plane $P''$. That is from $M\cap P,\ M\cap P',\ M\cap P''$ we have a geodesic triangle $\Delta$. In further, at each vertex we have rotation isometry and at each side we have reflection isometry. From these isometries, we consider all copies of $\Delta$ so that they cover $M$. They may overlap.

If some two distinct copies overlap in nonzero measure, then there is a geodesic triangle $\Delta' $, which is the overlapping. Hence we consider a cover of copies of $\Delta'$.

If any two distinct copies do not overlap in nonzero measure, then consider another point $x'''$ so that we obtain $\Delta ''$ which has smaller area than that of $\Delta$.

That is, $M$ has a cover of copies of a geodesic triangle $\Delta$ where area of $\Delta$ can be arbitrary small. If $x\in \Delta$, then any point $y$ in $M$ has a point $y'$ with small $ \rho(y,y')$ s.t. $x,\ y'$ has same curvature. Hence $M$ is a sphere of constant curvature.

HK Lee
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