I learned about metachirality through an amazing Vihart series ([1] [2]), and the only examples I have encountered are infinitely long screws, and this infinite graph Wikipedia mentions.

Are there finite examples of metachirality? My intuition says no, because I can't even think of finite examples with glide symmetry or screw symmetry, both of which feel like a weaker thing to ask for. I also suspect that any example of metachirality must in some way rely on translation, and if I can somehow formalise and prove that notion, I think I'll be done. (Because I know that translational invariance implies that, at least in one direction, the object is infinite [3])

In the positive case, I am looking for examples of finite symmetry groups (so maybe the object to consider is infinite, but the symmetry group is somehow still finite), but examples with finite objects would be cool too.

In the negative case, are there any proofs that all examples of metachirality need to be finite? And do such proofs have anything to say about glide/screw symmetries?

  • $\begingroup$ I'm confused by your use of "finite" here. Do you mean bounded? $\endgroup$ Dec 5, 2020 at 6:37
  • $\begingroup$ I care more about the finiteness of the symmetry group (which seems hard/impossible to construct in the first place). I would be happy with bounded or unbounded objects, but if there are examples with bounded objects that would truly surprise me. $\endgroup$ Dec 5, 2020 at 17:09

2 Answers 2


In Conway & Smith's book "On Quaternions and Octonions", Section 4.6, it is stated that all the finite subgroups of O(3) coincide with their mirror images, i.e., are not metachiral. There is no proof given, but earlier in the chapter they enumerate all the finite subgroups of O(3), so you can just verify each one.

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    $\begingroup$ Aha this is an excellent reference, thank you! I have a follow-on clarification - why is considering O(3) enough? A priori, it may be possible that there exists a (finite) symmetry group that is metachiral, but fixes no point. Is the argument that if no point is fixed, there must be some translational element (and that can be repeated forever to embed $\mathbb{Z}$ in the group, thus breaking finiteness)? $\endgroup$ Dec 5, 2020 at 17:06
  • $\begingroup$ *I should clarify that I don't actually know how to prove that fixing no point implies there exists some transnational element in the group. I'm wondering if there's a way of formalising that? $\endgroup$ Dec 5, 2020 at 17:11
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    $\begingroup$ If $G$ is a finite symmetry group, then for any point $x$, the orbit centroid $\frac{1}{|G|}\sum_{g \in G} gx$ is a fixed point. $\endgroup$
    – Ted
    Dec 5, 2020 at 18:43
  • $\begingroup$ Very neat, thank you - will accept your answer :) $\endgroup$ Dec 5, 2020 at 18:48

I can't even think of finite examples with glide symmetry or screw symmetry, both of which feel like a weaker thing to ask for.

Your intuition here is correct. In the case of either glide or screw symmetry, you can consider the infimum and supremum of any finite object along its projection onto an axis parallel to the glide or screw motion. Since the motion would translate these bounds, it can't leave the object unchanged. So any bounded set in $\mathbb{R}^3$ cannot be invariant under such an isometry of the plane. (The same is true of translation, by the way, which perhaps is a more intuitively obvious case.)

This sort of argument would be more complicated in higher dimensions, and might fail altogether, with the greater variety of isometries and rotations available; it's possible that you could find examples in spaces beyond $\mathbb{R}^3$.

  • $\begingroup$ Thank you! That make sense to me, and I think it's progress towards a proof. The point you raise about higher dimensions is very curious too! $\endgroup$ Dec 5, 2020 at 17:18

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