# Diameter of a quotient of the infinite dimensional sphere

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.

• 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. – YCor Sep 22 '18 at 17:31
• 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. – YCor Sep 22 '18 at 17:38
• 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 ? – HenrikRüping Sep 22 '18 at 19:10
• @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. – YCor Sep 22 '18 at 21:38
• 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 $n \to \infty$? – Nik Weaver Sep 29 '18 at 3:52

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.  It is a group with composition as product.  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  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. ( This is a semidirect product of $$\Gamma_0$$ and $$\Gamma_1$$.)

 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.)  These functions are dense in the positive part of the unit sphere.  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 $$\mathbb{T}_n$$.  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 $$n$$).  It follows straightforwardly that the same is true for any positive unit vector in place of $$1_{[0,1]}$$, and then  that the distance between any two orbits is at most $$2\alpha$$.

 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.

 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.

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

 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.

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

 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 $$f$$.

 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.

 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 $$f$$.

 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 $$\|f - \tilde{h}|f|\|_2 = \|(h - \tilde{h})|f|\|_2 \leq \alpha \|f\|_2 = \alpha$$.

 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 $$f$$.

 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.

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

• Note that in  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 $\phi\cdot f=T_{\phi^{-1}}(f)$. – YCor Sep 29 '18 at 21:52
• Good point (but it is $\sqrt{\phi'}$, not $\sqrt{\phi}$). – Nik Weaver Sep 29 '18 at 21:54
• Yes sure, it's a typo on my behalf (but I checked the computation with the correct formula). – YCor Sep 29 '18 at 22:10
• Thank you very much. I wonder, is this example has some relatives? I mean are there other problems with similar counterexamples? – Anton Petrunin Oct 2 '18 at 1:51
• 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.) – Nik Weaver Oct 2 '18 at 2:24