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.