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](https://arxiv.org/abs/1602.03933) 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.



Diameter of a quotient of the infinite dimensional sphere
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.




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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 $\|f - e^{\pi i/n}\cdot 1\_{[0,1]}\|\_2 \geq \beta$.


