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Let $S$ be a (unit) sphere in a Hilbert Space $H$ with $\dim H \ge 3$. Let $A \subset S$ have the following properties:

  1. $A$ is connected;

  2. The affine hull of $A$ is the whole space;

  3. For every $x,y\in A$ there is a unitary operator $T:H\to H$ (or, alternatively, an isometry $T:S\to S$), such that $TA=A$, and such that $Tx=y$.

Is it true that $A=S$?

You can assume that $A$ with the induced strong or weak topology is locally compact and locally linearly connected if needed.

I think that the problem may be meaningful even in dimension $3$, where it would be nice to have an elementary solution.

The problem I have started with is slightly more general, in case of the affirmative answer for this one, there may be a follow-up.

Thank you.

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  • $\begingroup$ In dimension 3 (2-sphere), it easy to see that $A$ should be dense (discussing on the closure of the isometry group). Under the local compactness assumption, it should easily follow that $A$ is the whole sphere. In general, I'm less sure. $\endgroup$
    – YCor
    Commented Jan 3, 2017 at 16:19
  • $\begingroup$ @YCor: could you please explain the argument about the closure of the isometry group? Also, I wonder if density, homogeneity and connectedness may imply closeness without local compactness. $\endgroup$
    – erz
    Commented Jan 3, 2017 at 18:56
  • $\begingroup$ Oops I was meaning the 3-dim real Hilbert space (and orthogonal maps). If you mean complex Hilbert space I have no claim. $\endgroup$
    – YCor
    Commented Jan 3, 2017 at 20:37
  • $\begingroup$ In real dim 3, let $G$ the group of orthogonals preserving $A$. Then $G$ is transitive on $A$. By compactness, it follows that the closure $G'$ of $G$ is transitive on the closure $A'$ of $A$. If $G'$ contains $SO(3)$ it follows that $A'$ is the whole sphere. Otherwise $G'$ is either finite (and hence $A'$ is finite, hence a singleton by connectedness), or $G'$ is up to index 2 a group of rotations. Its orbits are circles, pairs of circles or of points. By connectedness it's a circle or a point, but then its affine hull is not the whole space, contradiction. $\endgroup$
    – YCor
    Commented Jan 3, 2017 at 20:38
  • $\begingroup$ How did you deduce these two alternatives for $G'$(either finite, or up to index 2 a group of rotations)? Is this where the dimension $3$ come into the picture? $\endgroup$
    – erz
    Commented Jan 3, 2017 at 23:53

1 Answer 1

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The answer is 'no'.

For example, let $H$ be the (5-dimensional) space of symmetric, traceless $3$-by-$3$ matrices, where the Hilbert inner product is $\langle x,y\rangle = \tfrac12\mathrm{tr}(xy)$. Let $A\subset S$ be the subset consisting of those matrices of unit norm for which $0$ is an eigenvalue. Then $A$ is connected and $\mathrm{SO}(3)$ acts transitively on $A$ by the action $g\cdot x = gxg^T$. Moreover, the affine hull of $A$ must be a linear subspace of $H$ containing $A$ that is invariant under the action of $\mathrm{SO}(3)$, and this implies that this affine hull is all of $H$.

More generally, if $G$ is a connected compact Lie group acting irreducibly by isometries on a (finite-dimensional) Hilbert space $H$, then each of the orbits of $G$ in the unit sphere $S\subset H$ will be a subset $A\subset S$ with the OP's desired properties, and such orbits will not be all of $S$ whenever $G$ does not act transitively on $S$.

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  • $\begingroup$ Thank you for the answer. Is it possible to extend this construction (described in the last paragraph) to the infinite-dimensional case? I understand that compactness should be dropped, but maybe the group action can still be a homeomorphism in both variables? $\endgroup$
    – erz
    Commented Jan 5, 2017 at 3:05

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