# Riemannian manifolds which admit a smooth free $\mathbb{Z}/3\mathbb{Z}$ action but do not admit an equilateral triangle action

A free action of $$\mathbb{Z}/3\mathbb{Z}$$ on a Riemannian manifold $$(M, g)$$ is called an equilateral action if for every $$x\in M$$ all three points of orbit of $$x$$ have the same distance from each other.

What is an example of a Riemannian manifold which does not admit such an action but it already admit a smooth free action by cyclic group of order $$3$$?

• Are you assuming compactness or completeness? If you don't I think you can find uninteresting counter examples of the form $M=\Bbb{R}^2\setminus\{0,p, Ap, A^2p\}$ where $A$ is a matrix that is conjugate to a rotation by $2\pi/3$, $p$ is any nonzero point and the Riemannian structure is induced from that of $\Bbb{R}^2$. I would expect you can make this into a complete example by tampering with the metric. So maybe you should assume compactness? – Olivier Bégassat Mar 20 '20 at 23:42
• $A$ is only conjugate to a rotation (i.e. satisfies $A^3=I_2$) but it's not an isometry. The Z/3Z action is that of $A$ on $M$. I'm removing two orbits from the action of $A$ on $\Bbb{R}^2$. – Olivier Bégassat Mar 21 '20 at 0:00
• @OlivierBégassat for sure it works for the plane minus 4 points, none 3 of which form an equilateral triangle.If $f$ is a homeomorphism, it acts on the end compactification, which is a 2-sphere. If it fixes all end points, then this is a homeo of order 3 with exactly 5 fixed points and this doesn't exist. So it permutes 3 and fixes the last two. A limit argument shows that the point at infinity is fixed, and that the three moved ones form an equilateral triangle. – YCor Mar 21 '20 at 0:12
• I guess you want a manifold with a free action of $\mathbb{Z}_3$ such that there is no equilateral action for ANY Riemanninan metric. (Otherwise the question has no sense.) – Anton Petrunin Mar 21 '20 at 5:25
• @AntonPetrunin for any metric space an equilateral action makes sense (a free continuous action of $C_3$ in which every orbit is equilateral). So asking for a Riemannian manifold with no such action is perfectly meaningful. While your modified question is quite trivial: if there's a free $C_3$-action, you can average to get an invariant Riemannian metric. – YCor Mar 21 '20 at 7:17

Consider a torus with three handles, where one handle is much larger than the others, and with a smooth and free $$\mathbb{Z}_3$$ action which permutes the handles.

Let $$\gamma$$ be a small geodesic loop going through one of the small handles. Let $$z\gamma$$ be an image of $$\gamma$$ under the group action which goes through the large handle. Let $$p$$ be a point on $$\gamma$$ such that $$zp$$ is far out on the large handle.

Now the distance between $$p$$ and $$zp$$ is almost the diameter of the manifold, and there is no third point $$z^2p$$ which makes an equilateral triangle with them. So there is no smooth and equilateral $$\mathbb{Z}_3$$ action on the manifold.

Here's a compact example without boundary (a 2-torus).

Choose a topological disc $$D$$ on the two-torus $$T$$, and choose a Riemannian metric so that $$D$$ has very close Gromov-Hausdorff distance (say $$\le 1$$) to a segment of length $$20$$, while $$T\smallsetminus D$$ has diameter $$\le 1$$ (so metrically $$D$$ is predominant, while all the homotopic part lies in $$T\smallsetminus D$$. Also assume (just to fix ideas) that there exists $$x_0\in D$$ such that $$d(x_0,D)=20$$ (so $$x_0$$ is the "opposite tip" of the segment). Again to fix ideas, suppose that $$x_0$$ is part a geodesic segment $$(x_t)_{0\le t\le 10}$$ such that $$\sup_{x\in T}\inf_{t\in [0,20]}d(x,x_t)\le 1$$.

(On the picture, the complement of $$D$$ is the little left tip, including the handle, and $$x_0$$ is at the right tip.)

Then $$T$$ (with this metric) has no large equilateral triangle: indeed if $$r$$ is the size of an equilateral triangle, such a triangle would be 1-close to a "triangle" $$\{x_{t_1},x_{t_2},x_{t_3}\}$$, with $$t_1\le t_2\le t_3$$ and $$||t_i-t_j|-a\le 2$$ for all $$i\neq j$$. So $$t_3-t_1\le a+2$$, $$t_3-t_2,t_2-t_1\ge a-2$$, hence $$t_3-t_1\ge 2a-4$$, so $$2a-4\le a+2$$, i.e., $$a\le 6$$.

Hence, if $$f$$ is an "equilateral" self-homeomorphism, we have $$d(f(x),x)\le 6$$ for all $$x$$. Let $$B$$ be the open $$7$$-ball around $$x_0$$. Then the open subset $$U=B\cup f(B)\cup f^{-1}(B)$$ is $$f$$-invariant, and contained in the $$13$$-ball around $$x_0$$. The connected component $$U'\subset U$$ of $$x_0$$ contains the $$f$$-orbit of $$x_0$$.

Let $$T'$$ be the quotient $$T/\langle f\rangle$$, so $$T\to T'$$ is a connected covering of degree 3. Since $$\pi_1(U',x_0)\to \pi_1(T,x_0)$$ is trivial, the covering is trivial in restriction to $$U'$$. We get a contradiction, since $$U'$$ is connected and contains as fiber the orbit of $$x_0$$.

• I think you mean $a$ for the size of the equilateral triangle, and $x_0$ satisfies $d(x_0,T\setminus D)=20$. Very nice answer by the way, this is also how I read the OP's question. – Pierre PC Mar 22 '20 at 15:57