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I feel slightly embarrassed to ask this, but I've managed to thoroughly confuse myself about the following.

Consider $\mathbb{R}^2$ together with the lattice $\Lambda=\{(n,m): n,m\in \mathbb{Z}\}$.

Clearly, the torus $\mathbb{T}^2=\mathbb{R}^2/\Lambda$ admits the obvious flat metric that makes it a square torus. Geometrically, this corresponds to taking the fundamental domain $[0,1]\times [0,1]$ with the metric $dx^2+dy^2$ and identifying the appropriate sides.

My confusion is that one could also think of the domain given by the parallelogram $\{(x,y): 0\leq y \leq 1, y\leq x \leq y+1\}$ (i.e. determined by the elements $(1,0)$ and $(1,1)$ that also generate the lattice). If I understand things correctly, on this domain the metric $dx^2+dy^2$ would induce a different metric on the quotient given by identifying the appropriate sides than the metric you would get by taking the appropriate quotient metric (which here would correspond to $dx^2-dxdy-dydx+2dy^2$).

This is causing me some confusion and doubt about how well I actually understand the situation.

I guess for a concrete question: Is there something special about the square as a fundamental domain that makes the quotient metric given by identifying opposite sides agree with the quotient metric of the lattice? Or am I misunderstanding things?

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  • $\begingroup$ Of course they induce different metrics. Why do you expect anything else? $\endgroup$
    – Trebor
    Sep 10, 2021 at 0:38
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    $\begingroup$ The two tori are not even equal as sets, why should they have the same metric? You're being a victim of the abuse of notation of denoting the quotient coordinates on both tori by $(x,y)$ again. I address the (possibly correct) question regarding isometries in my answer below. $\endgroup$
    – Ivo Terek
    Sep 10, 2021 at 0:40
  • $\begingroup$ @IvoTerek So is the issue is that the one forms $dx$ and $dy$ don't descend to the quotient by the lattice without some additional choice of data? $\endgroup$
    – RBega2
    Sep 10, 2021 at 13:38

2 Answers 2

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So thinking about this some more I realized that the correct answer is that both the square quotient and the parallelogram quotient are isometric. The part that confused me was that none of the possible isometries lift to an isometry of $\mathbb{R}^2$ and instead look something like

$$ (x,y)\mapsto \left\{ \begin{array}{cc} (x,y) & 0\leq x\leq 1 \\ (x-1, y) & 1\leq x \leq 2. \end{array} \right. $$ which maps the parallelogram to the square and is a smooth isometry on the quotient.

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  • $\begingroup$ (+1), but as you note, the lattices are the same, so the identity mapping on the torus is an isometry that lifts to an isometry (in fact, to infinitely many isometries) of the plane. That said, no isometry of the torus maps the open square to the open parallelogram. $\endgroup$ Sep 11, 2021 at 2:30
  • $\begingroup$ @AndrewD.Hwang I think part of my confusion was that while there is a "global" map between the subsets in the plane (i.e. the affine map) it just isn't an isometry. I'm still a little unclear if there is some deeper meaning to why some geometric structures (e.g. volume or symplectic forms) admit what seems to me to be a nicer map while others (e.g. metric or almost complex structure) don't. $\endgroup$
    – RBega2
    Sep 11, 2021 at 12:44
  • $\begingroup$ Not exactly an explanation, but volume/symplectic structures have infinite-dimensional symmetry groups, while metric and holomorphic structures on compact manifolds don't. <> In this situation, the fundamental parallelogram is something of a red herring, an extra datum not determined by any of the geometric structures, but (as you've noted) a choice of basis for first homology. The way I think of it is, since the choice of basis is not determined "geometrically", there's no reason to expect symmetries of a geometric structure to act transitively on the set of bases. $\endgroup$ Sep 11, 2021 at 15:08
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If you choose two different lattices $\Lambda_1$ and $\Lambda_2$, the resulting tori $\Bbb R^2/\Lambda_1$ and $\Bbb R^2/\Lambda_2$ will be diffeomorphic, sure. Both of them inherit flat metrics from $\Bbb R^2$ (because lattices act on $\Bbb R^2$ by translations, and translations are isometries for the standard flat metric in $\Bbb R^2$). Then, with such quotient flat metrics, the two tori are locally isometric for very general reasons, but this is not natural (in a very precise sense). If you can map one fundamental domain for $\Lambda_1$ to a fundamental domain for $\Lambda_2$ via a rigid motion of $\Bbb R^2$, then such rigid motion survives in the quotient as a global isometry between the two tori, and I suppose this is as general as it gets. I.e., taking generators of the lattices, the matrix of base change should be in ${\rm SO}(2,\mathbb{Z})$.

In your example, the two tori are not isometric simply because the two parallelograms playing the role of the fundamental domains do not have the same internal angles, say, so you cannot move one to the other via a rigid motion of the plane.

Compare this with the situation where you think of $\Bbb R^2$ as $\Bbb C$. The tori will be diffeomorphic, but they'll be biholomorphic to each other (as Riemann surfaces) precisely when the matrix of base change is in ${\rm SL}(2,\mathbb{Z})$.

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  • $\begingroup$ In my question the two lattices are the same (I've just chosen different generators) so the quotient construction of a metric gives the same metric. Both the square and the parallelogram are fundamental domains for this lattice, but they are not related by an isometry. In particular, the quotient metric agrees with the ``gluing" metric on the square but does not for the parallelogram. I am trying to clarify what is going on in an elementary way (I suspect it has to do with generators of the first homology group of the torus). $\endgroup$
    – RBega2
    Sep 10, 2021 at 13:16

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