
Last week I promised to explain some mysterious connections between elliptic curves, string theory, and the number 24. I claimed that it all boils down to the fact that there are two especially symmetric lattices in the plane, namely the square lattice:
* * * * * * * * * * * *with 4fold symmetry, and the hexagonal lattice:
* * * * * * * * * * *with 6fold symmetry. Now it's time for me to start backing up those claims.
First I need to talk a bit about lattices and SL(2,Z). As I explained in "week66", a lattice in the complex plane consists of all points that are integer linear combinations of two complex numbers, say ω_{1} and ω_{2}. However, we can change these numbers without changing the lattice by letting
ω'_{1} = a ω_{1} + b ω_{2} ω'_{2} = c ω_{1} + d ω_{2}where
a b c dis a 2×2 invertible matrix of integers whose inverse again consists of integers. Usually it's good to require that our transformation preserve the handedness of the basis (ω_{1}, ω_{2}), which means that this matrix should have determinant 1. Such matrices form a group called SL(2,Z). In the context of elliptic curves it's also called the "modular group".
Now associated to the square lattice is a special element of SL(2,Z) that corresponds to a 90 degree rotation. Everyone calls it S:
0 1 S = 1 0Associated to the hexagonal lattice is a special element of SL(2,Z) that corresponds to a 60 degree rotation. Everyone calls it ST:
0 1 ST = 1 1(See, there's a matrix they already call T, and ST is the product of S and that one.) Now, you may complain that the matrix ST doesn't look like a rotation, but you have to be careful! What I mean is, if you take the hexagonal lattice and pick a basis for it like this:
ω_{2} * * * * ω_{1} * 0* * * * * *then in this basis the matrix ST represents a 60 degree rotation.
So far this is pretty straightforward, but now come some surprises. First, it turns out that SL(2,Z) is generated by S and ST. In other words, every 2×2 integer matrix with determinant 1 can be written as a product of a bunch of copies of S, ST, and their inverses. Second, all the relations satisfied by S and ST follow from these obvious ones:
S^{4} = 1
(ST)^{6} = 1
together with
S^{2} = (ST)^{3}
which holds because both sides describe a 180 degree rotation.
Right away this implies that SL(2,Z) has a certain inherent "12ness" to it. Let me explain. SL(2,Z) is a nonabelian group  this is how someone with a Ph.D. says that matrix multiplication doesn't commute  but suppose we abelianize it by imposing extra relations forcing commutativity. Then we get a group generated by S and ST, satisfying the above relations together with an extra one saying that S and ST commute. This is the group Z/12, which has 12 elements!
This "12ness" has a lot to do with the magic properties of the number 24 in string theory. But to see how this "12ness" affects string theory, we need to talk about elliptic curves a bit more. It will take forever unless I raise the mathematical sophistication level a little. So....
We can define an elliptic curve to be a torus C/L formed by taking the complex plane C and modding out by a lattice L. Since C is an abelian group and L is a subgroup, this torus is an abelian group, but in the theory of elliptic curves we consider it not just as a group but also as a complex manifold. Thus two elliptic curves C/L and C/L' are considered isomorphic if there is a complexanalytic function from one to the other that's also an isomorphism of groups. This happens precisely when there is a nonzero number z such that zL = L', or in other words, whenever L' is a rotated and/or dilated version of L.
There's a wonderful space called the "moduli space" of elliptic curves: each point on it corresponds to an isomorphism class of elliptic curves. In physics, we think of each point in it as describing the geometry of a torusshaped string worldsheet. Thus in the pathintegral approach to string theory we need to integrate over this space, together with a bunch of other moduli spaces corresponding to string worldsheets with different topologies. All these moduli spaces are important and interesting, but the moduli space of elliptic curves is a nice simple example when you're first trying to learn this stuff. What does this space look like?
Well, suppose we have an elliptic curve C/L. We can take our lattice L and describe it in terms of a righthanded basis (ω_{1}, ω_{2}). For the purposes of classifying the describing the elliptic curve up to isomorphism, it doesn't matter if we multiply these basis elements by some number z, so all that really matters is the ratio
τ = ω_{1}/ω_{2}.Since our basis was righthanded, τ lives in the upper halfplane, which people like to call H.
Okay, so now we have described our elliptic curve in terms of a complex number τ lying in H. But the problem is, we could have chosen a different righthanded basis for our lattice L and gotten a different number τ. We've got to think about that. Luckily, we've already seen how we can change bases without changing the lattice: we just apply a matrix in SL(2,Z), getting a new basis
ω'_{1} = a ω_{1} + b ω_{2} ω'_{2} = c ω_{1} + d ω_{2}This has the effect of changing τ to
τ' = (a τ + b)/(c τ + d).If you don't see why, figure it out  you've gotta understand this to understand elliptic curves!
Anyway, two numbers τ and τ' describe isomorphic elliptic curves if and only if they differ by the above sort of transformation. So we've figured out the moduli space of elliptic curves: it's the quotient space H/SL(2,Z), where SL(2,Z) acts on H as above!
Now, the quotient space H/SL(2,Z) is not a smooth manifold, because while the upper halfplane H is a manifold and the group SL(2,Z) is discrete, the action of SL(2,Z) on H is not free: i.e., certain points in H don't move when you hit them with certain elements of SL(2,Z).
If you don't see why this causes trouble, think about a simpler example, like the group G = Z/n acting as rotations of the complex plane, C. Most points in the plane move when you rotate them, but the origin doesn't. The quotient space C/G is a cone with its tip corresponding to the origin. It's smooth everywhere except the tip, where it has a "conical singularity". The moral of the story is that when we mod out a manifold by a group of symmetries, we get a space with singularities corresponding to especially symmetrical points in the original manifold.
So we expect that H/SL(2,Z) has singularities corresponding to points in H corresponding to especially symmetrical lattices. These, of course, are our friends the square and hexagonal lattices!
But let's be a bit more careful. First of all, nothing in H moves when you hit it with the matrix 1. But that's no big deal: we can just replace the group SL(2,Z) by
PSL(2,Z) = SL(2,Z)/{±1}Since 1 doesn't move any points of H, the action of SL(2,Z) on H gives an action of PSL(2,Z), and the moduli space of elliptic curves is H/PSL(2,Z).
Now most points in H aren't preserved by any element of PSL(2,Z). However, certain points are! The point
τ = icorresponding to the square lattice, is preserved by S and all its powers. And the point
τ = exp(2πi/3)corresponding to the hexagonal lattice, is preserved by ST and all its powers. These give rise to two conical singularities in the moduli space of elliptic curves. Away from these points, the moduli space is smooth.
Lest you get the wrong impression, I should hasten to reassure you that the moduli space is not all that complicated: it looks almost like the complex plane! There's a famous onetoone and onto function from the moduli space to the complex plane: it's called the "modular function" and denoted by j. So the moduli space is topologically just like the complex plane; the only difference is that it fails to be smooth at two points, where there are conical singularities.
This may seem a bit hard to visualize, but it's actually not too hard. Here's one way. Start with the region in the upper halfplane outside the unit circle and between the vertical lines x = 1/2 and x = 1/2. It looks sort of like this:
..................... ..................... ..................... ..................... ..................... ..................... ..................... ..................... ..........A.......... ..... ..... ... ... . . B B'Then glue the vertical line starting at B to the one starting at B', and glue the arc AB to the arc AB'. We get a space that's smooth everywhere except at the points A and B = B', where there are conical singularities. The total angle around the point A is just 180 degrees  half what it would be if the moduli space were smooth there. The total angle around B is just 120 degrees  one third what it would be if the moduli space were smooth there.
The reason this works is that the region shown above is a "fundamental domain" for the action of PSL(2,Z) on H. In other words, every elliptic curve is isomorphic to one where the parameter τ lies in this region. The point A is where τ = i, and the point B is where τ = exp(2πi/3).
Now let's see where the "12ness" comes into this picture. Minhyong Kim explained this to me in a very nice way, but to tell you what he said, I'll have to turn up the level of mathematical sophistication another notch. (Needless to say, all the errors will be mine.)
So, I'll assume you know what a "complex line bundle" is  this is just another name for a 1dimensional complex vector bundle. Locally a section of a complex line bundle looks a lot like a complexvalued function, but this isn't true globally unless your line bundle is trivial. If you aren't careful, sometimes you may think you have a function defined on a space, only to discover later that it's actually a section of a line bundle. This sort of thing happens all the time in physics. In string theory, when you're doing path integrals on moduli space, you have to make sure that what you're integrating is really a function! So it's important to understand all the line bundles on moduli space.
Now, given any sort of space, we can form the set of all isomorphism classes of line bundles over this space. This is actually an abelian group, since when we tensor two line bundles we get another line bundle, and when you tensor any line bundle with its dual, you get the trivial line bundle, which plays the role of the multiplicative identity for tensor products. This group is called the "Picard group" of your space.
What's the Picard group of the moduli space of elliptic curves? Well, when I said "any sort of space" I was hinting that there are all sorts of spaces  topological spaces, smooth manifolds, algebraic varieties, and so on  each one of which comes with its own particular notion of line bundle. Thus, before studying the Picard group of moduli space we need to decide what context we're going to work in! As a mere topological space, we've seen that the moduli space of elliptic curves is indistinguishable from the plane, and every topological line bundle over the plane is trivial, so in this context the Picard group is the trivial group  boring!
But the moduli space is actually much more than a mere topological space. It's not a smooth manifold, but it's awfully close: it's the quotient of the smooth manifold H by the discrete group SL(2,Z), and its singularities are pretty mild in nature.
Somehow we should take advantage of this when defining the Picard group of the moduli space. One way to do so involves the theory of "stacks". Without getting into the details of this theory, let me just vaguely sketch what it does for us here. For a much more careful treatment, with more of an algebraic geometry flavor, try:
1) David Mumford, Picard groups of moduli problems, in Arithmetical Algebraic Geometry, ed. O. F. G. Schilling, Harper and Row, New York, 1965.
Suppose a discrete group G acts on a smooth manifold X. A "Gequivariant" line bundle on X is a line bundle equipped with an action of G that gets along with the action of G on X. If G acts freely on X, a line bundle on X/G is the same as a Gequivariant line bundle on X. This isn't true when the action of G on X isn't free. But we can still go ahead and define the Picard group of X/G to be the group of isomorphism classes of Gequivariant line bundles on X. Of course we should say something to let people know that we're using this funny definition. In our example, people call it the Picard group of the moduli stack of elliptic curves.
So what's this group, anyway?
Well, it turns out that you can get any SL(2,Z)equivariant line bundle on H, up to isomorphism, by taking the trivial line bundle on H and using a 1dimensional representation of SL(2,Z) to say how it acts on the fiber. So we just need to understand 1dimensional representations of SL(2,Z). The set of isomorphism classes of these forms a group under tensor product, and this is the group we're after.
Well, a 1dimensional representation of a group always factors through the abelianization of that group. We saw the abelianization of SL(2,Z) was Z/12. But everyone knows that the group of 1dimensional representations of Z/n is again Z/n  this is called Pontryagin duality. So: the Picard group of the moduli stack of elliptic curves is Z/12.
So we see again an inherent "12ness" built into the theory of elliptic curves! You may be wondering how this makes the number 24 so important in string theory. In particular, where does that extra factor of 2 come from? I'll say a little more about this next Week. I may or may not manage to tie together the loose ends!
You may also be wondering about "stacks". In this you're not alone. There's an amusing passage about stacks in the following book:
2) Joe Harris and Ian Morrison, Moduli of Curves, SpringerVerlag, New York, 1998.
They write:
"Of course, here I'm working with the moduli stack rather than with the moduli space. For those of you who aren't familiar with stacks, don't worry: basically, all it means is that I'm allowed to pretend that the moduli space is smooth and that there's a universal family over it."Actually Mumford's paper cited above gives a nice introduction to the theory of stacks without mentioning the dreaded word "stack". Alternatively, you can wait and read this book when it comes out:Who hasn't heard these words, or their equivalent, spoken in a talk? And who hasn't fantasized about grabbing the speaker by the lapels and shaking him until he says what  exactly  he means by them? But perhaps you're now thinking that all that is in the past, and that at long last you're going to learn what a stack is and what they do.
Fat chance.
3) K. Behrend, L. Fantechi, W. Fulton, L. Goettsche and A. Kresch, An Introduction to Stacks, in preparation.
But let me just briefly say a bit about stacks and the moduli stack of elliptic curves in particular. A stack is a weak sheaf of categories. For this to make sense you must already know what a sheaf is! In the simplest case, a sheaf over a topological space, the sheaf S gives you a set S(U) for each open set U, and gives you a function S(U,V): S(U) → S(V) whenever the open set U is contained in the open set V. These functions must satisfy some laws. The notion of "stack" is just a categorification of this idea. That is, a stack S over a topological space gives you a category S(U) for each open set U, and gives you a functor S(U,V): S(U) → S(V). These functors satisfy the same laws as before, but only up to specified natural isomorphism. And these natural isomorphisms must in turn satisfy some new laws of their own, socalled coherence laws.
In the case at hand there's a stack over the moduli space of elliptic curves. For any open set U in the moduli space, an object of S(U) is a family of elliptic curves over U, such that each elliptic curve in the family sits over the point in moduli space corresponding to its isomorphism class. Similarly, a morphism in S(U) is a family of isomorphisms of elliptic curves. This allows us to keep track of the fact that some elliptic curves have more automorphisms than others! And it takes care of the funny stuff that happens at the singular points in the moduli space.
By the way, this watereddown summary leaves out a lot of the algebraic geometry that you usually see when people talk about stacks.
Finally, one more thing  it looks like Kreimer and company are making great progress on understanding renormalization in a truly elegant way.
4) D. J. Broadhurst and D. Kreimer, Renormalization automated by Hopf algebra, preprint available as hepth/9810087.
Let me quote the abstract:
It was recently shown that the renormalization of quantum field theory is organized by the Hopf algebra of decorated rooted trees, whose coproduct identifies the divergences requiring subtraction and whose antipode achieves this. We automate this process in a few lines of recursive symbolic code, which deliver a finite renormalized expression for any Feynman diagram. We thus verify a representation of the operator product expansion, which generalizes Chen's lemma for iterated integrals. The subset of diagrams whose forest structure entails a unique primitive subdivergence provides a representation of the Hopf algebra H_{R} of undecorated rooted trees. Our undecorated Hopf algebra program is designed to process the 24,213,878 BPHZ contributions to the renormalization of 7,813 diagrams, with up to 12 loops. We consider 10 models, each in 9 renormalization schemes. The two simplest models reveal a notable feature of the subalgebra of Connes and Moscovici, corresponding to the commutative part of the Hopf algebra H_{T} of the diffeomorphism group: it assigns to Feynman diagrams those weights which remove ζ values from the counterterms of the minimal subtraction scheme. We devise a fast algorithm for these weights, whose squares are summed with a permutation factor, to give rational counterterms.
