Definition (Central extension)

Let be a group and be an abelian group. A central extension of by is a group which contains as a central normal subgroup (i.e., ) such that . That is, there is a short exact sequence of groups

Theorem (Classification by group cohomology)

Central extensions of by are classified by , where is a -module by letting act trivially on , i.e., for all and .

Proof

Choose a section (just a map of sets) such that where is the surjection.

We compare and . Note that they have the same image under since is a group homomorphism and

So and must differ by an element of : there exists ( is the fibre) such that

This defines a function (which depends on ). By associativity of group multiplication in , we now derive a cocycle condition: for any , we expand in two different ways:

Since , we can commute past in , and therefore we get the cocycle condition: for all ,

Conversely, we can start with ^0d656f () and define a central extension: defining a group structure on the underlying set by

gives a central extension if satisfies the cocycle condition . #TODO This requires some checking.

When are two central extensions isomorphic?

For two choices of sections above coming from the same central extension , we get two different cocycles . But, again, since and live in the same fibre, i.e.,

and must differ by an element of : there exists such that

We now look at how , , and relate to figure out the equivalence relation on cocycles: we have

Again, since , , and all map into , we can commute the , , terms past the other terms. Hence, we will say that if there exists such that (note here all the terms are in abelian)

called the coboundary condition.

There are still details to be filled, but we conclude that central extensions of by are classified by satisfying ^0d656f () modulo the above equivalence relation equivalence relation.

Q: How does this relate to ?
A: The cocycles satisfying ^0d656f () will be elements of the kernel and the coboundary condition coboundary condition will come from elements of the image in the bar resolution.

Additive notation

In additive notation:

the cocycle condition cocycle condition is
the coboundary condition coboundary condition for is

Bar resolutions (continued)

Recall that we had a bar resolution, which was a projective resolution of the trivial -module :

where all tensors are over . We want to compute , and hence we drop the , and apply :

We need to work out the maps. We note the following:

We have via $⟻$
is a free abelian group with basis for all . So we see that it is a free -module with basis for all . So we have an isomorphism of abelian groups $⟻$

Recall that the boundary map is (up to sign)

So we have the boundary map after taking is

Acting on an element , we get

In other words, maps to the function under the identifications and .

This matches our expectations because we want to be -invariants, and if and only if for all .

For the boundary map , we have

Hence

We conclude that:

the -cocycles are (from ^a81998 ())
the -coboundaries are
.

For , we compute :

Let , and analogous to the argument above above, we define and element of by

Now,

We need this to be to be an element of , i.e.,

We see that this is precisely the cocycle condition cocycle condition, where the only difference is that in the discussion above above, trivially.

Further, consists of of the form in ^a81998 (): there exists such that

which corresponds to the condition above above, but trivially.

Indeed, classifies the central extensions.

Line bundles on an elliptic curve

How do we construct a line bundle over an elliptic curve over ?

An elliptic curve is a torus where is a lattice such that . #TODO diagram

We construct a line bundle via

where the equivalence relation is defined by choosing a function (note that ) and setting

Here, is a scaling on each fibre (which is ). But now, for , the following elements are all related:

so we need to impose the condition

This is a $1$-cocycle condition -cocycle condition where and (non-vanishing holomorphic functions on ). This is because even though , we can take the adjunction :

where acts on by translation: .