What is the isometry group of a torus given a flat metric? I know $ O(1) \times O(1) $ should be a subgroup of it. Is there any other possible isometries? What if the metric is not flat?


2 Answers 2


I'm going to consider the flat torus $\mathbb T = \mathbb R^2 / \Gamma$ for some plane lattice $$\Gamma_{e_1, e_2} = \{ z_1 e_1 + z_2 e_2 : z_1, z_2 \in \mathbb Z \}.$$ (Here $e_1, e_2$ is any basis of $\mathbb R^2$.) Since any isometry of the torus lifts to an isometry of $\mathbb R^2,$ we just need to determine which isometries of the plane descend to diffeomorphisms of $\mathbb T.$

A map $\phi: \mathbb R^2 \to \mathbb R^2$ descends iff it satisfies $$\phi(x+z) - \phi(x) \in \mathbb \Gamma \textrm{ for all } x \in \mathbb R^2, z \in \mathbb \Gamma.\tag 1 $$ Since the isometries of the plane can be written as $\phi(x) = Ax + c$ for $A \in O(2), c \in \mathbb R^2,$ this condition reduces to the fact that $Az = z$ for all $z,$ i.e. $A(\Gamma) \subset \Gamma.$ Finally, to have an inverse, $\phi^{-1}$ must also descend, and thus the condition is $A(\Gamma) = \Gamma;$ so $\phi(x) = Ax + c$ is an isometry of $\mathbb T$ iff the linear isometry $A$ is also an automorphism of the lattice $\Gamma.$

Note that two isometries induce identical maps on $\mathbb T$ if their constant terms differ by an element of $\Gamma,$ and composition of translations corresponds to addition of constant terms; so we can conclude that $$\mathrm{Isom}(\mathbb T) = (O(2)\cap \operatorname{Aut}(\Gamma)) \ltimes (\mathbb T, +). $$ (Compare to $\operatorname{Isom}(\mathbb R^2) =O(2) \ltimes (\mathbb R^2, +).$)

The nature of the normal subgroup $G = O(2)\cap \operatorname{Aut}(\Gamma)$ depends on the isometry class of the lattice (or equivalently on the shape of the fundamental domain):

  • For a square, $G$ is the isometry group of the square, $D_4.$
  • For a rhombus with angle $\pi/3$ (which fits on a hexagonal grid), $G$ is the isometry group of the hexagon, $D_6.$
  • For any other rectangle, rhombus, or parallelogram with $\pi/3$ angle, $G$ is the product $C_2 \times C_2$ of cyclic groups (acting as orthogonal reflections).
  • For a generic parallelogram, $G = C_2$ (acting as the antipodal map).
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    $\begingroup$ For completeness, you may want to add the complete classification of possible groups $O(2)\cap Aut(\Gamma)$ (as abstract groups): There are only four different groups that occur with $Z_2$ (generated by the antipodal map) as the generic case. $\endgroup$ Commented Mar 20, 2020 at 14:55
  • $\begingroup$ @MoisheKohan: thanks for the correction, not sure why my brain thought the $C_2\times C_2$ reflections were always isometries. Is there a fourth group I've missed? $\endgroup$ Commented Mar 21, 2020 at 0:30
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    $\begingroup$ Yes, there is one more group, the dihedral group of order 12. A fundamental domain is a rhombus with the vertex angle $\pi/3$. For $Z_2^2$ there is one more possible type of fundamental domain: A generic parallologram with the vertex angle $\pi/3$. $\endgroup$ Commented Mar 21, 2020 at 10:13
  • $\begingroup$ Ah, my mental model of lattices was clearly too simple - I missed that e.g. $e_1, e_1 + e_2$ also generates $\Gamma,$ so the automorphisms are more plentiful than I thought. I don't have time to amend this with a good argument for the classification now, but I'll correct my list. $\endgroup$ Commented Mar 23, 2020 at 5:03

Consider $$ (\theta, \phi) \mapsto (\pm\phi, \pm\theta), $$

In short: consider all affine maps on the plane that map the integer lattice (including the lattice "edges") in the plane to itself in a 1-1 way, and project them to the torus, and that gets you some more isometries.

(I'm assuming your flat torus is the quotient of $\mathbb R^2$ by the integer grid $\mathbb Z^2$, with the metric inherited from the quotient map.)

I have this feeling that rotations of the plane should work in some form as well, but haven't worked it out (and after screwing up once today, I'm gun-shy.)

  • $\begingroup$ So let's take $a = c = d = 1$ and $b = 0$, How does $(\theta, \phi) \mapsto (\theta, \theta + \phi)$ induce an isometry on the torus? $\endgroup$
    – Rob Arthan
    Commented Nov 9, 2016 at 21:59
  • $\begingroup$ It doesn't. It induces an area-preserving map. And I'm an idiot. :( I'll edit. $\endgroup$ Commented Nov 9, 2016 at 22:18
  • $\begingroup$ (+1) Actually, you're correct if "affine" is strengthened to be an affine isometry that preserves the torus's lattice: If $f$ is an isometry of $T = \mathbf{C}/\Lambda$ and $\pi:\mathbf{C} \to T$ is the projection (a local isometry), then $f \circ \pi:\mathbf{C} \to T$ lifts to an isometry $\tilde{f}:\mathbf{C} \to \mathbf{C}$ that maps $\Lambda$ to $\Lambda$. (Whether or not there are rotations other than a half-turn depends on whether the lattice is square/hexagonal or not.) $\endgroup$ Commented Nov 9, 2016 at 22:45
  • $\begingroup$ You are implicitly assuming that your metric is the product metric. There are other flat metric on the tori, whose isometry groups are a bit different. Even for the product tori the answer depends on the lengths of the circles you are multipying. If they are different, you found all the isometries. If they are the same, you only found an index 2 subgroup: There is an extension coming from an isometry which swaps the factors. $\endgroup$ Commented Nov 9, 2016 at 23:12
  • $\begingroup$ Absolutely right: I was assuming the quotient of the plane by the standard integer lattice, and the pushforward metric that produces. And when I said "the integer lattice" I was thinking of lines on graph paper, i.e., not just the nodes, but the edges too. (Which knocks out a lot of what Andrew suggests, but does ensure isometry.) I've edited to reflect this. $\endgroup$ Commented Nov 9, 2016 at 23:40

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