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:
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
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:
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).