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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 $3$?

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    $\begingroup$ 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? $\endgroup$ – Olivier Bégassat Mar 20 '20 at 23:42
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    $\begingroup$ $A$ 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 $\Bbb{R}^2$. $\endgroup$ – Olivier Bégassat Mar 21 '20 at 0:00
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    $\begingroup$ @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. $\endgroup$ – YCor Mar 21 '20 at 0:12
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    $\begingroup$ 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.) $\endgroup$ – Anton Petrunin Mar 21 '20 at 5:25
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    $\begingroup$ @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. $\endgroup$ – YCor Mar 21 '20 at 7:17
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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.

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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 $\sup_{x\in T}\inf_{t\in [0,20]}d(x,x_t)\le 1$.

Long torus picture

(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 $x_0$.

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    $\begingroup$ 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. $\endgroup$ – Pierre PC Mar 22 '20 at 15:57

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