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Suppose a group $\Gamma$ acts by isometries on the Hilbert space $\mathbb{H}^\infty$ and it fixes the origin. So $\Gamma$ acts on the unit sphere $\mathbb{S}^\infty$ as well.

Assume that the action $\Gamma$ on $\mathbb{S}^\infty$ has no dense orbits. Is there a universal constant $\varepsilon >0$ such that there are two orbits of $\Gamma$ on distance at least $\varepsilon$ from each other?

In other words, is there a constant $\varepsilon>0$ such that $$\mathrm{diam}\, (\mathbb{S}^\infty/\Gamma) >0 \quad\Longrightarrow\quad \mathrm{diam}\, (\mathbb{S}^\infty/\Gamma) > \varepsilon\ ?$$

Comments. The statement is related to the following results in finite dimension. It was proved in [Gre] that, on the $n$-sphere for $n\ge 2$, there is such a lower bound $\varepsilon_n>0$ (choose it as optimal). It is expected (see the introduction of this arXiv paper by Gorodski and Lytchak) that $\inf_{n\ge 2}\varepsilon_n>0$. This is announced to hold by Claudio Gorodski, Christian Lange, Alexander Lytchak and Ricardo Mendes (it is not published yet). Namely, for some universal constant $\varepsilon>0$, for any $n\ge 2$ and for any isometric group action of any compact group on the unit $n$-sphere, the orbit space of the action is either a point or has diameter $\ge\varepsilon$.

[Gre] S. J. Greenwald, Diameters of spherical Alexandrov spaces and curvature one orbifolds, Indiana Univ. Math. J. 49 (2000), no. 4, 1449–1479.

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  • $\begingroup$ To be clear, I think you mean: is there a constant $\varepsilon>0$ such that for every $\Gamma$ and every action, we have the given inequality. Also note that the statement "the orbit $X$ is at distance at least $\varepsilon$ from some other orbit" is equivalent to the statement "there exists a point $v$ (in the 1-sphere) at distance $\ge\varepsilon$ of $X$". This is true for arbitrary isometric actions on arbitrary metric spaces. $\endgroup$
    – YCor
    Commented Sep 22, 2018 at 17:31
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    $\begingroup$ If the representation is reducible then there exist 2 orbits at distance $\sqrt{2}$ (or geodesic distance $\pi/2$), so one has to consider irreducibles. Already, this gives a definite answer for virtually abelian groups, which have no infinite-dimensional orthogonal irreducibles. $\endgroup$
    – YCor
    Commented Sep 22, 2018 at 17:38
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    $\begingroup$ I am trying to understand what is going on for the following classes of group actions indexed over $n \in \mathbb{N}$. Let $\Gamma_n$ be generated by rotations in the first two coordinates by $2*\pi/n$ and all coordinate permutations. If I would allow all rotations, the group action is has dense orbits. My hope is that by making $n$ large, we could beat every $\varepsilon$. Has $\Gamma_n$ dense orbits ? $\endgroup$ Commented Sep 22, 2018 at 19:10
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    $\begingroup$ @HenrikRüping Yes, it does have dense orbits. This follows from the case of the 2-sphere: there exists $t_0>0$ such that having $t_0$-dense orbits on the 2-sphere implies having dense orbits. Apply this to the subgroup of your group $\Gamma_n$ generated by the rotations and the permutations of the first 3 variables. For $n\ge n_0$, it has $t_0$-dense orbits on the 2-sphere and hence had dense orbits; more precisely the closure is all of $O(3)$ (non-dense subgroups of $SO(3)$ don't have dense orbits). So the closure contains all rotations, and hence has dense orbits. $\endgroup$
    – YCor
    Commented Sep 22, 2018 at 21:38
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    $\begingroup$ I'm confused about the cited result. Consider the group of rotations by integer multiples of $2\pi/n$ acting on the unit circle = 1-sphere. Doesn't the diameter of the orbit space go to zero as $n \to \infty$? $\endgroup$
    – Nik Weaver
    Commented Sep 29, 2018 at 3:52

1 Answer 1

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There is no such universal constant $\epsilon > 0$. Work with the complex Hilbert space $L^2[0,1]$ (which of course is also a real Hilbert space). Fix $n \in \mathbb{N}$.

Let $\Gamma_0$ be the set of continuous piecewise linear increasing bijections from $[0,1]$ to itself. [1] It is a group with composition as product. [2] It acts by isometries of $L^2[0,1]$ by the map $f \mapsto \sqrt{\phi'}\cdot (f\circ\phi)$ for $f \in L^2[0,1]$ and $\phi \in \Gamma_0$. Also let $\Gamma_1 \subset L^\infty[0,1]$ consist of the measurable functions from $[0,1]$ to $\mathbb{T}_n = \{e^{2\pi i k/n}: 0 \leq k < n\}$, identifying functions which differ on a null set. This is a group under pointwise product and [3] it acts isometrically by multiplication on $L^2[0,1]$. Let $\Gamma$ be the group of isometries of $L^2[0,1]$ generated by $\Gamma_0$ and $\Gamma_1$ under these actions. ([4] This is a semidirect product of $\Gamma_0$ and $\Gamma_1$.)

[5] The $\Gamma_0$ action takes the unit vector $1_{[0,1]}$ to any piecewise constant strictly positive unit vector $f \in L^2[0,1]$. (If $f$ takes the value $c$ on an interval $I$, let $\phi$ have slope $c^2$ on this interval.) [6] These functions are dense in the positive part of the unit sphere. [7] Applying the action of $\Gamma_1$ then gets us to arbitrarily close to any unit vector in $L^2[0,1]$ whose argument lies almost everywhere in $\mathbb{T}_n$. [8] It follows that the distance from $1_{[0,1]}$ to any other orbit is at most $\alpha = |1 - e^{\pi i/n}|$ ($\approx \frac{\pi}{n}$ for large $n$). [9] It follows straightforwardly that the same is true for any positive unit vector in place of $1_{[0,1]}$, and then [10] that the distance between any two orbits is at most $2\alpha$.

[11] On the other hand, the distance from the orbit of $1_{[0,1]}$ to the vector $e^{\pi i/n}\cdot 1_{[0,1]}$ is at least the distance from $(1,0) \in \mathbb{R}^2$ to the line through the origin of slope $\frac{\pi}{n}$ (again approximately $\frac{\pi}{n}$ for large $n$), so this orbit is not dense and since the action is isometric no orbit is dense.


Edit: maybe people want more details.

[1] The composition of two continuous functions is continuous, of two PL functions is PL, of two increasing functions is increasing, of two bijections is a bijection. The inverse of a continuous PL increasing bijection is a continuous PL increasing bijection.

[2] $\|\sqrt{\phi'}\cdot (f\circ \phi)\|_2^2 = \int_0^1 \phi'|f\circ\phi|^2\, dt = \int_0^1 |f|^2\, dt = \|f\|_2^2$.

[3] If $h \in \Gamma_1$ then $\|hf\|_2^2 = \int_0^1 |hf|^2\, dt = \int_0^1 |f|^2\, dt = \|f\|_2^2$ since $|h| = 1$ a.e.

[4] This isn't needed, but anyway if $\phi \in \Gamma_0$ and $h \in \Gamma_1$ then $\sqrt{\phi'}\cdot (hf\circ \phi) = (h\circ \phi)\cdot \sqrt{\phi'}\cdot(f\circ\phi)$.

[5] Let $f$ be a piecewise constant strictly positive unit vector in $L^2[0,1]$. Then $f = a_0\cdot 1_{[0,t_1)} + \cdots + a_k\cdot 1_{[t_k,1)}$ a.e. for some $0 < t_1 < \cdots < t_k < 1$ and some $a_0, \ldots, a_k > 0$. The unit norm condition means that $\sum_{i=1}^k a_i^2(t_{i+1} - t_i) = 1$. Now define $\phi: [0,1] \to \mathbb{R}$ so that $\phi(0) = 0$, $\phi$ is continuous, and $\phi$ is linear with slope $a_i^2$ on $[t_{i-1},t_i]$. The unit norm condition just detailed shows that $\phi(1) = 1$, i.e., $\phi$ is a continuous PL increasing bijection. We have $\sqrt{\phi'}\cdot (1_{[0,1]}\circ \phi) = \sqrt{\phi'}$, which takes the value $a_i$ constantly on $(t_{i-1},t_i)$. So $1_{[0,1]}$ is taken to $f$.

[6] First, positive piecewise constant functions can uniformly approximate any continuous function on $[0,1]$, and since the positive continuous functions are dense for the $L^2$ norm in the positive part of $L^2[0,1]$, this shows that positive piecewise constant functions are dense in the positive part of $L^2[0,1]$. Given a positive $f \in L^2[0,1]$ with unit norm, find a sequence $(f_k)$ of positive piecewise constant functions with $f_k \to f$ in $L^2[0,1]$. Then $\|f_k\|_2 \to 1$ so $\frac{1}{\|f_k\|_2}f_k \to f$. Thus, any positive unit vector is approximated by positive piecewise constant unit vectors.

[7] Since multiplying by $h \in \Gamma_1$ is an isometry, if $f = h|f| \in L^2[0,1]$ is a unit vector whose argument $h$ lies in $\mathbb{T}_n$ a.e. and $g$ is a positive piecewise constant unit vector which is close to $|f|$, then $hg$ will be equally close to $f$.

[8] Given any unit vector $f = h|f| \in L^2[0,1]$, we can find $\tilde{h} \in \Gamma_1$ such that $|h(t) - \tilde{h}(t)| \leq \alpha$ a.e. As we just saw that the orbit of $1_{[0,1]}$ comes arbitrarily close to $\tilde{h}|f|$, it follows that the distance from $f$ to this orbit is at most $\|f - \tilde{h}|f|\|_2 = \|(h - \tilde{h})|f|\|_2 \leq \alpha \|f\|_2 = \alpha$.

[9] We saw already that any positive unit vector $f$ is approximated by elements in the orbit of $1_{[0,1]}$. Since the action is isometric, this means that $1_{[0,1]}$ is approximated by elements in the orbit of $f$. Again by isometric action, since we can take $1_{[0,1]}$ to within $\alpha'$ of any unit vector, for any $\alpha' > \alpha$, the same is then true of $f$.

[10] Any two unit vectors lie within $\alpha'$ of the orbit of $1_{[0,1]}$, for any $\alpha' > \alpha$. So (by isometric action again) one lies within $2\alpha'$ of the orbit of the other.

[11] The argument of any vector $f$ in the orbit of $1_{[0,1]}$ lies pointwise a.e. in $\mathbb{T}_n$. So $|f(t) - e^{\pi i/n}| \geq \beta$ pointwise, where $\beta$ is the distance from $(1,0) \in \mathbb{R}^2$ to the line through the origin of slope $\frac{\pi}{n}$ (= the distance from $e^{\pi i/n} \in \mathbb{C}$ to the union of the lines through the origin of slopes $\frac{2k\pi}{n}$). Thus $\|f - e^{\pi i/n}\cdot 1_{[0,1]}\|_2 \geq \beta$.

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    $\begingroup$ Note that in [2] you don't say a word about being an action. Actually, defining $T_\phi(f)=\sqrt{\phi}(f\circ\phi)$ indeed satisfies $T_{\phi\circ\psi}=T_\psi\circ T_\phi$. Thus you get a (left) action by $\phi\cdot f=T_{\phi^{-1}}(f)$. $\endgroup$
    – YCor
    Commented Sep 29, 2018 at 21:52
  • $\begingroup$ Good point (but it is $\sqrt{\phi'}$, not $\sqrt{\phi}$). $\endgroup$
    – Nik Weaver
    Commented Sep 29, 2018 at 21:54
  • $\begingroup$ Yes sure, it's a typo on my behalf (but I checked the computation with the correct formula). $\endgroup$
    – YCor
    Commented Sep 29, 2018 at 22:10
  • $\begingroup$ Thank you very much. I wonder, is this example has some relatives? I mean are there other problems with similar counterexamples? $\endgroup$ Commented Oct 2, 2018 at 1:51
  • $\begingroup$ I don't know. The construction seems specific to the unit sphere of a Hilbert space, but I will think about it. (And thank you for your generous donation of points. Not necessary, but certainly made me feel appreciated.) $\endgroup$
    – Nik Weaver
    Commented Oct 2, 2018 at 2:24

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