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Quotienting $\mathbb R^2$ by different lattices isomorphic to $\mathbb Z^2$, we get different tori.

Somehow I think of the tori as having different "structures", but thinking more about it, I am not quite sure what different structures I am really thinking of. Two structures I am guessing at are complex structures and metrics.

Could someone explain how these differ? Also, am I thinking of the right kind of structure? Are there other structures which vary with the lattice chosen?

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1 may be relevant. – Willie Wong Jul 26 '12 at 10:46
Thanks. If I understand it correctly, points in the Teichmüller space for the torus correspond to complex structures, but the various metrics on the Teichmüller space don't correspond to metrics on the torus... How about the Riemannian metric inherited from $\mathbb R^2$. This one should vary with the lattice, I would assume. I just don't know how... – Earthliŋ Jul 26 '12 at 11:41
In so far as the metric is concerned, what you are looking for is actually the existence of isometries between the various tori. One quick way to see that you have different metrics is by consider the geodesic loops in various homotopy classes. In particular, if they have different lengths the two tori cannot be isometric. – Willie Wong Jul 26 '12 at 12:12
A torus gives a complex elliptic curve, which is a complex variety with a (compatible) structure of an abelian group. Changing the lattice changes the group structure, and hence the elliptic curve under consideration. – M Turgeon Jul 26 '12 at 12:54
@MTurgeon: yes, but the surprising thing is that the complex variety also changes. (Actually, the group structure doesn't add anything to the picture, since an elliptic curve is just a complex torus with a marked point). – PseudoNeo Jul 26 '12 at 13:11
up vote 5 down vote accepted

You're mostly right.

The relation between complex structures and metrics comes from their common passion about angles.

Basically, a complex structure on a Riemann surface is just a procedure for turning tangent vectors 90° counterclockwise. (Well, actually, that is an almost complex structure but they are the same thing in (complex) dimension 1). But that is one of the things a metric (with an orientation) allows you to do! So a metric on a surface defines canonically a complex structure. Of course, many metrics give the same structure (example 1: you can rescale the metric; example 2: the sphere $S^2$ has a lot of metrics but only one complex structure.)

The good notion is the notion of conformally equivalent metrics. Roughly speaking, two metrics are conformally equivalent if they define the same notions of angles between two tangent vectors. That means that they are proportional to each other (the ratio being a positive smooth function on the manifold.) So you get an important fact about surfaces: complex structures and conformal classes of metrics are basically the same thing. So, most of the Riemann surface theory can be stated equivalently in the holomorphic world or in the conformal world. (Example: the uniformisation theorem says either “Any Riemann surface is a quotient of $S^2$, $\mathbb C$ or $\mathbb H$” or “Any Riemannian metric on a surface is conformally equivalent to a metric of constant curvature.”) This polyvalence clearly is one of the riches of Riemann surface theory and explains partially the huge number of dedicated textbooks, as there is room for lots of different approaches.

So, here are two possible answers to your question: the relevant structures are “complex structures” or “conformal Riemannian structures”. This gives the same notion of “equivalent” lattices: two of them are equivalent if there is an affine map sending one to the other.

(One can imagine variants: for example, one could choose to call two lattices equivalent if there is a volume-preserving map sending one to the other. In that case, one has to enrich the relevant structures on the quotient. Of course, the finer the equivalence relation on the lattices, the richer the structure, but affine equivalence is a popular choice, because the two structures I mentioned are very important and very natural.)

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Maybe you should mention that every Riemannian metric on a surface is locally conformal to a flat metric, so we can say that these two structures on surfaces are equivalent. – Yuchen Liu Jul 26 '12 at 12:35
That is much more than I expected from any answer. Thank you so much! – Earthliŋ Jul 26 '12 at 12:42
I should said something about flat structures, I agree. (That's one of the reasons I mentioned the uniformisation theorem, but I should have mentioned that in the torus case, the curvature is 0). But it is not true that these two structures are equivalent: there are pairs of nonisometric flat tori that are conformally equivalent (you can rescale them, but I don't think that this the only possible construction). – PseudoNeo Jul 26 '12 at 12:42
@PseudoNeo But if you consider the conformal class of Riemannian metrics, then two metrics are in the same conformal class up to a orient-preserving (or orient-reversing) diffeomorphism if and only if they define the same (or conjugate) complex structure up to the same diffeomorphism. – Yuchen Liu Jul 26 '12 at 12:48
Ah, OK, you meant that “conformal classes of flat metrics” was still another equivalent answer! You're right, of course. Sorry for the confusion. – PseudoNeo Jul 26 '12 at 12:50

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