Infinite Dimensional Lie Algebras February 14, 2005

Gabriel C. Drummond-Cole

February 21, 2005

Okay, let’s see what’s going on. We have ℋ = ⟨a_{n}, n ∈ ℤ⟩, subject to the commutation relations [a_{n}, a_{m}] = nδ_{n,−m} and then a(z) = ∑ a_{n}z^{−n−1} with relations [a(z), a(w)] = z^{−1}δ_{w}δ(w/z).

[something about vector fields on S^{1}]

Then there are L_{n} with [L_{n}, a_{k}] = −ka_{n+k} and L(z) = ∑ L_{n}z^{−n−2} = \frac{1}{2} : a^{2}(z) : and then we have [L(z), a(w) = z^{−1}a(w)δ_{w}δ(w/z).

We want to be able to think of a(z), L(z) as operator valued functions, i.e., a(z) as a function on ℂ^{×} with values in End(ℱ).

So a(z)|0⟩ = ∑ a_{n}|0⟩z^{−n−1} = ∑_{n<0} a_{n}|0⟩z^{−n−1}. This is bad because it’s not in ℱ . We have ℱ = ℂ[a_{−1}, a_{−2}, ⋯]|0⟩. We want to work with ℱ̄ = ℂ[[a_{−1}, a_{−2}, ⋯]]|0⟩, that is, infinite linear combinations. So this is ∑ v_{k} where the degree of v_{k} goes to negative infinity. So a(z) is a function on ℂ^{×} with values in Hom(ℱ , ℱ̄).

There is a problem, as usual. Such operators cannot be multiplied. E.g., can a(z)a(w)|0⟩ ∈ ℱ̄ be defined? So a(z)a(w)|0⟩ = a(z) ∑ a_{−n}w^{n−1} = ∑_{n>0} nz^{−n−1}w^{n−1}|0⟩ plus other terms. This last part can be written z^{−1}w^{−1} ∑ n(\frac{w}{z})^{n}. Recall that a_{n}a_{−n}|0⟩ = n|0⟩ for n > 0.

Theorem 1 1. For |w| < |z| a(z)a(w) =: a(z)a(w) : +\frac{1}{(z−w)^{2}} where the fraction is expanded in a power series in w/z.

2. : a(z)a(w) : is a well-defined operator ℱ → m̄athscrF for any z, w ∈ ℂ^{∗}.

Corollary 1 a(z)a(w) can be defined as a function of z, w ∈ ℂ^{∗} for |w| < |z| with values in Hom(ℱ , ℱ̄).

So for the proof of the theorem, for the first part we have

a(z)a(w) = ∑ a_{n}a_{m}z^{−n−1}w^{−m−1};

: a(z)a(w) := ∑ : a_{n}a_{m} : z^{−n−1}w^{−m−1}.

The only places these differ are when the first has a_{n}a_{−n}z^{−n−1}w^{n−1} and the second a_{−n}a_{n}z^{−n−1}w^{n−1}. So

a(z)a(w)− : a(z)a(w) := ∑_{n>0} nz^{−n−1}w^{n−1}

= z^{−1} ∑_{n≥0} \frac{nw^{n−1}}{z^{n}} = z^{−1}δ_{w}(∑_{n≥0}(\frac{w}{z})^{n})

= z^{−1}δ_{w}(\frac{1}{1 − \frac{w}{z}} ) = δ_{w} \frac{1}{z − w} = \frac{1}{(z − w)^{2}} .

Corollary 2 Matrix coefficients of a(z)a(w) can be analytically continued to (ℂ^{∗})^{2}\diagonal and they will have a second order pole at z = w. Physicists write this as a(z)a(w) ∼ \frac{1}{(z−w)^{2}} .

So this is a(w)a(z) =: a(w)a(z) : +\frac{1}{(z−w)^{2}} for |z| < |w| and a(z)a(w) =: a(z)a(w) : +\frac{1}{(z−w)^{2}} for |z| > |w|. Now, this means that a(z)a(w) − a(w)a(z) is nowhere defined. As formal power series this would be the expansion of \frac{1}{(z−w)^{2}} for |z| < |w| less the expansion of the same function for |w| > |z|.

So without analytic continuation [a(z), a(w)] is nowhere convergent; with analytic continuation [a(z), a(w)] = 0. The moral is that all information about [a_{n}, a_{k}] can be obtained from the polar part of a(z)a(w) ∼ \frac{1}{(z−w)^{2}} .

Example 1 a(z)L(w) =: a(z)L(w) : +\frac{a(w)}{(z−w)^{2}} for |w| < |z|.

Definition 1 : a_{n}L_{k} := a_{n}L_{k} for n < 0, L_{k}a_{n} for n ≥ 0.

So

a(z)L(w)− : a(z)L(w) := ∑_{n≥0,k∈ℤ}[a_{n}, L_{k}]z^{−n−1}w^{−k−2}

= ∑ na_{n+k}z^{−n−1}w^{−k−2} = ∑ nz^{−n−1}w^{n−1}a_{n+k}w^{−n−k−1}

= (z^{−1} ∑_{n≥0} \frac{nw^{n−1}}{z^{n}})(∑ a_{l}w^{−l−1}) = \frac{1}{(z − w)^{2}} a(w).

So a(z)L(w) =: a(z)L(w) : +\frac{a(w)}{(z−w)^{2}} for |w| < |z| and L(w)a(z) =: a(z)L(w) : \frac{a(w)}{(z−w)^{2}} for

|w| > |z|. Then a(z)L(w) ∼ \frac{a(w)}{(z−w)^{2}} .