## Friday, May 25, 2012

### Vertex algebras

Think about your favorite Lie algebra Lie(G). We have a mapping on it, namely, the adjoint representation:

ad:Lie(G) → End[Lie(G)]

where "End[Lie(G)]" are the endomorphisms of the Lie algebra Lie(G).

Normally this is of the form "ad(u)v∈Lie(G)" and is shorthand for "ad(u)=[u,-]".

The Jacobi identity looks like:

ad(u)ad(v)-ad(v)ad(u)=ad(ad(u)v).

This is the most important identity. Vertex operator algebras are an algebra with a similar property.

A vertex operator algebra consists of a vector space V equipped with a mapping usually denoted

Y:V→(End V)[[x,x-1]].

In this form, it looks like left-multiplication operator...or that's the intuition anyways. So if "v∈V", we should think Y(v,x) belongs to "(End V)[[x,x-1]]" and acts on the left.

Really through currying this should be thought of as "V⊗V→V[[x,x-1]]", i.e., a sort of multiplication operator with a parameter "x". (This is related to the "state-operator correspondence" physicists speak of with conformal field theories.)

Just like a Lie algebra, the Vertex Operator algebra satisfies a Jacobi identity and it is the most important defining property for the VOA.

Lets stop and look at this structure again:

Y:V→(End V)[[x,x-1]].

What's the codomain exactly? Well, it's a formal distribution (not a mere formal power series!).

So what does one look like? Consider δ(z-1) = Σ zn where the summation ranges over n∈ℤ. This series representation is a formal distribution, and behaves in the obvious way. Lets prove this!

Desired Property: δ(z-1) vanishes almost everywhere.

Consider the geometric series f(z) = Σzn where n is any non-negative integer (n=0,1,...).

Observe that δ(z-1) = f(z) + z-1f(z-1). Lets now substitute in the resulting geometric series:

δ(z-1) = [1/(1-z)] + z-1[1/(1-z-1)]

and after some simple arithmetic we see for z≠1 we have δ(z-1)=0.

Desired Property: for any Laurent polynomial f(z) we have δ(z-1)f(z)=δ(z-1)f(1).

This turns out to be true, thanks to the magic of infinite series; but due to html formatting, I omit the proof. The proof is left as an exercise to the reader (the basic sketch is consider δ(z-1)zn, then prove linearity, and you're done).

## Friday, May 18, 2012

### Finite Field with Four Elements

Small note to myself on notational problems when facing finite groups.

Recall the finite field with four elements is ℤ2[x]/(1+x+x2).

People often write ω = 1+x and ω=x. Observe then that ω2 = ω, and ω2 = ω. Moreover ωω=1 and 1+ω+ω=0.

I have only seen this ω notation specified in Pless' Error Correcting Codes, Third ed., page 102 et seq.