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Let $K\subset\Bbb R^n$ be a non-empty compact subset of $\Bbb R^n$. A symmetry of $K$ is an isometry of $\Bbb R^n$ that fixes $K$ set-wise. Since $K$ is compact, there is always a point $x\in\Bbb R^n$ fixed by all symmetries.

Question: Can the following three things be true at the same time:

  1. $K$ is contractible.
  2. $x$ is the only point fixed by all symmetries of $K$.
  3. $x\not\in K$.

The following image shows that such a $K$ exists if we ignore one of the properties:

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  • $\begingroup$ I'm sure it's obvious to anyone who might be in a position to answer this question, but why does the compactness of $K$ imply that $G = \operatorname{Isom}(\mathbb R^n, K)$ has a fixed point on $\mathbb R^n$? I could imagine buying it if we had some sort of appropriate compactness for $G$; do we? $\endgroup$
    – LSpice
    Commented May 12, 2022 at 15:52
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    $\begingroup$ @LSpice I considered this folklore, but I have no quick argument at hand. My vague intuition is that if $g\in G$ has not fixed point then it sends $x\mapsto Ax + b$ with $A$ fixing the orthogonal complement of $b$. So no orbit of $G$ can be compact. $\endgroup$
    – M. Winter
    Commented May 12, 2022 at 16:07
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    $\begingroup$ @LSpice a result of Jung (1905) is that every nonempty bounded subset of a Hilbert space is contained in a unique ball of minimal radius. In particular the center of this ball is fixed by the isometry group. $\endgroup$
    – YCor
    Commented May 12, 2022 at 16:08
  • $\begingroup$ That seems only to argue that every $g$ has a fixed point, not that all of $G$ does; but @YCor's citation does the job. I guess the latter is a special case of the Bruhat–Tits fixed-point theorem, which I should have thought of since I recently asked a question about it. Thanks! $\endgroup$
    – LSpice
    Commented May 12, 2022 at 16:20
  • $\begingroup$ I'm probably making a stupid mistake, but isn't half sphere in $\mathbb R^3$ such an example? $\endgroup$
    – Nick S
    Commented May 14, 2022 at 19:09

2 Answers 2

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I posted a refined/generalized question Does a compact contractible metric space have a point that is fixed by all isometries? and received an answer that contained all essential ingredients to provide an affirmative answer to this question. The related question Homeomorphic to the disk implies existence of fixed point common to all isometries? has an answer with further very interesting details.

There are two ingredients:

  1. there exists a group $G$ that acts without fixed points on the $n$-dimensional ball (for an appropriate $n>5$).
  2. the ball can be smoothly embedded into some Euclidean space $\Bbb R^m$ so that the action of $G$ is now by linear transformations and only fixes the origin (Mostow's embedding theorem).

If we choose $K$ to be this embedded ball and $x$ to be the origin, then this configuration satisfies all the conditions from my question.

References to the claims 1 and 2 can be found in the linked answers, except for the assertion that the linear transformations only fix the origin, which I found in the original paper "Equivariant Embeddings in Euclidean Space" by Mostow.

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  • $\begingroup$ So under this embedding, the disc does not contain the origin? $\endgroup$
    – Bma
    Commented May 14, 2022 at 18:05
  • $\begingroup$ @Bma Yes, because otherwise the action of $G$ on the disc would have a fixed point. $\endgroup$
    – M. Winter
    Commented May 14, 2022 at 18:06
  • $\begingroup$ You refer to "the linked answers", but you link only one. Did you mean to link others? $\endgroup$
    – LSpice
    Commented May 14, 2022 at 21:21
  • $\begingroup$ @LSpice I mean the answers below the linked questions, but mostly the first linked question. $\endgroup$
    – M. Winter
    Commented May 14, 2022 at 21:23
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    $\begingroup$ @Yaakov I don't know how $m$ depends on $n$. I haven't checked how explicit Mostow's theorem is about this. A comment below the linked answer states that you cannot have examples for $n\le 4$ as every group acting on a ball in dimension up to four has a fixed point. The case $n=5$ is open. And of course $m\ge n$. $\endgroup$
    – M. Winter
    Commented May 15, 2022 at 9:03
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As I see, the answer is “no”. If you have the single fixed point $X$ not from your set, then the symmetry of your compact is a reflection to the line $l$ which contains $X$.

It is easy to see that you cannot have another reflection line. Because if you have it, then their composition — rotation, is also a symmetry of your compact. But if you have a rotation, then $X$ is a centre of the corresponding circle and it's easy to the сontradiction to contractability. And because of contractibility defined line l must intersect K. And that would be another fixed point which is in K.

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    $\begingroup$ I don't understand your claim "then the symmetry...". For instance, if the group of symmetries consists of $\{\mathrm{id},-\mathrm{id}\}$ (so $x=0$), the assumption is $K=-K$, $0\notin K$. In particular a subquestion is whether there exists a nonempty contractible compact subset $K$ of $\mathbf{R}^n$ such that $0\notin K$ and $K=-K$. $\endgroup$
    – YCor
    Commented May 12, 2022 at 15:48

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