Infinite Dimensional Lie Algebras February 9, 2005

Gabriel C. Drummond-Cole

March 18, 2005

Last time: we say that a lattice is integral if (L, L) ∈ ℤ. What is the order of the set {λ_{0}+L, (λ_{0}, L) ∈ ℤ}? That was the question. So L = ⊕ℤe_{i}, A = (e_{i}, e_{j}), and |L^{∗}/L| = det A.

For example, let L = ⟨e_{1}, e_{2}⟩ be the root lattice for A_{2}. Then L^{∗} is the weight lattice,

A = ( \begin{matrix}2 & 1 \\ 1 & 2 \end{matrix} ) and |L^{∗}/L| = 3.

Okay, so 𝔥̂ = 𝔥 ⊕ ℂ[t, t^{−1}] ⊕ ℂ subject to the relations [xt^{n}, yt^{m}] = n(x, y)δ_{n,−m}. Here xt^{n} = x[n]. So when 𝔥 = ℂ then a[n] = a_{n}.

Now W = V ect(S^{1})⊗ℂ = ⟨d_{i}⟩. Here d_{n} = −t^{n+1} \frac{d}{dt}. The relations are [d_{n}, d_{m}] = (n−m)d_{n+m}.

So W acts on 𝔥̂ by d_{n}x[k] = −kx[k + n].

A question: can we define the action of W in representations of 𝔥̂? So d_{n} → L_{n} ∈ End(ℱ_{λ}). Here [L_{n}, x[k]] = d_{n}x[k] = −kx[k + n]. For h = ℂ we have L_{n} = \frac{1}{2} ∑_{k+l=n} : a_{k}a_{l} : . Oh, L_{0} = H.

So why this formula? Well, acting by a decreases the number of a by one so it should be quadratic. By degree considerations, L_{n} must be of degree n. So k + l = n. So it must be ∑ c_{kl} : a_{k}a_{l} : and then solving for constants gives this formula.

Now

[L_{n}, L_{m}] = \frac{1}{2}[L_{n}, ∑ : a_{k}a_{l} :] = \frac{1}{2} ∑([L_{n}, a_{k}]a_{l} + a_{k}[L_{n}, a_{l}])

= \frac{1}{2} ∑(−ka_{n+k}a_{l} − la_{k}a_{l+n}).

Call this *. Now, if n − m ≠ 0 this becomes

\frac{1}{2} ∑ −ka_{n+k}a_{l} + \frac{1}{2} ∑ −la_{k}a_{l+n}

= \frac{1}{2} ∑_{i+j=m+n}(n − i)a_{i}a_{j} + \frac{1}{2} ∑(i − m)a_{i}a_{k}

= \frac{1}{2} ∑_{i+j=m+n}(n − m)a_{i}a_{j} = (n − m)L_{m+n}.

When m + n = 0 * makes sense since for all but finitely many terms we get cancellation. There we have

\frac{1}{2} ∑((n − i)a_{i}a_{−i} + ia_{i−n}a_{n−i})

which we turn into

= \frac{1}{2} ∑ : ((n − i)a_{i}a_{−i} + ia_{i−n}a_{n−i}) : +\frac{1}{2} ∑(n − i)ix(i) + i(i − n)x(i − n)

= 2nL_{0} + \frac{1}{2} ∑_{i=0}^{n}(n − i)i = 2nL_{0} + \frac{1}{12}(n^{3} − n).

Here x(i) is zero for negative i and 1 otherwise so that : a_{i}a_{−i} : +ix(i) = a_{i}a_{−i}. So [L_{n}, L_{m}] = (n − m)L_{n+m} + δ_{n,−m} \frac{n^{3}−n}{12}.

Definition 1 V ir = W ⊕ℂc = ⟨L_{n}, c⟩ with commutation relations [L_{n}, L_{m}] = (n0m)L_{n+m}+ δ_{n,−m} \frac{n^{3}−n}{12}c and [c, L_{n}] = 0.

Theorem 1 The action of W on 𝔥̂ extends to an action of V ir on ℱ_{λ} such that c acts by dim 𝔥. [i.e., a representation with (Virasoro) central charge dim 𝔥]

Instead of writing commutation relations between a_{n}, L_{n}, consider a(z) = ∑_{n∈ℤ} a_{n}z^{−n−1}, that is, what physicists call functions in z with operator values. Then

[a(z), a(w)] = ∑[a_{n}, a_{m}]z^{−n−1}w^{−m−1} = ∑_{n∈ℤ} nz^{−n−1}w^{n−1} = z^{−1}δ_{w}(∑_{n∈ℤ}(w/z)^{n})

where this last sum is the Fourier series for the δ function at w/z. Then this is z^{−1}δ_{w}δ(w/z), where δ(z) = ∑ z^{n}.

Exercise 1 For P a Laurent polynomial in z and w, P (z, w)δ(z/w) = P (z, z)δ(z/w) = P (w, w)δ(z/w). This is similar to f(x)δ(y) = f(y)δ(x).

Now L(z) = ∑ L_{n}z^{−n−2} = \frac{1}{2} ∑_{k,l} : a_{k}a_{l} : z^{−(k+l)−2} = \frac{1}{2} : a^{2}(z) : . The commutation relations are

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

= ∑ −ka_{n+k}z^{−n−2}w^{−k−1} = ∑ a_{i}( ∑_{n+k=i} −kz^{−n−2}w^{−k−1})

Try to figure out where to go from here.

Okay, now

\frac{1}{2}[: a^{2}(z) :, a(w)] = \frac{1}{2}[a^{2}(z), a_{w})] = \frac{1}{2}([a(z), a(w)]a(z) + a(z)[a(z), a(w)])

= \frac{1}{2}(z^{−1}δ_{w}δ(w/z)a(z) + a(z)z^{−1}δ_{w}δ(w/z))

= z^{−1}a(z)δ_{w}δ(w/z).

As homework, try to write [L(z), L(w)]. Here is a reference: E. Frenkel and B. Zvi, “Vertex Algebras and Algebraic Curves,” and V. Kac, “Vertex Algebras for Beginners.”

The operators “almost” commute except when w = z.