18.276 Lecture Note 6
Lecture by Victor Kac, Scribed by Dongkwan Kim
March 5, 2015
Note. In this lecture note we deal with super-cases. But if we decree that all the
parities are even, then we get the corresponding formulae for ”ordinary” cases.
6.1
Expression of Operator Product Extension
In conformal field theory, physicists write the operator product extension (OPE) of two
local fields as follows:
X a(w)(j) b(w)
a(z)b(w) ” = ”
(1)
(z − w)j+1
j∈Z
To justify this statement, first we show the following lemma [1, Theorem 2.3].
Lemma 6.1. The following are equivalent.
(i) the commutator formula:
[a(z), b(w)] =
j
X a(w)(j) b(w)∂w
δ(z − w)
j≥0
(2)
j!
(ii)
a(z)b(w) =
X
(a(w)(j) b(w))iz
j≥0
(−1)p(a)p(b) b(w)a(z) =
X
1
+ : a(z)b(w) :
(z − w)j+1
(a(w)(j) b(w))iw
j≥0
1
+ : a(z)b(w) :
(z − w)j+1
(3)
(4)
j
1
1
Proof. (ii) ⇒ (i) is obvious, since we have ∂w δ(z−w)
= iz (z−w)
j+1 −iw (z−w)j+1 . To show the
j!
converse (Exercise 6.1) first we decompose the commutator
formula
(2) into the parts of
nonnegative(resp. negative) powers of z as follows, using
1
j
∂w
δ(z−w)
j!
+
1
= −iw (z−w)
j+1 and
j
∂w
δ(z−w)
j!
−
1
= iz (z−w)
j+1 where the subscript +(resp. −) means the nonnegative (resp.
negative) part of the expression with respect to the power of z.
X
(a(w)(j) b(w))iw
(a(z)b(w) − (−1)p(a)p(b) b(w)a(z))+ = [a(z), b(w)]+ = −
j≥0
X
(a(w)(j) b(w))iz
(a(z)b(w) − (−1)p(a)p(b) b(w)a(z))− = [a(z), b(w)]− =
j≥0
1
(z − w)j+1
1
(z − w)j+1
Thus we have
a(z)b(w)− : a(z)b(w) : = a(z)− b(w) − (−1)p(a)p(b) b(w)a(z)−
X
1
(a(w)(j) b(w))iz
=
(z − w)j+1
j≥0
(−1)p(a)p(b) b(w)a(z)− : a(z)b(w) : = −a(z)+ b(w) + (−1)p(a)p(b) b(w)a(z)+
X
1
=
(a(w)(j) b(w))iw
(z − w)j+1
j≥0
from which (3) and (4) follow.
Now recall that we have Taylor’s formula
iz a(z + w) =
X ∂zj a(z)
j!
j≥0
wj .
If we replace z by w and w by z − w, and regard z as z = w + (z − w), then this formula
becomes
j
X ∂w
a(w)
iw a(w + (z − w)) =
(z − w)j .
j!
j≥0
Therefore, if we apply this formula to a(z)+ and a(z)− , respectively, in the region |w| >
|z − w| we have
: a(z)b(w) : =
j
j
X ∂w
(a(w)+ )b(w) + (−1)p(a)p(b) b(w)∂w
(a(w)− )
j!
j≥0
=
j
j
X (∂w
a(w))+ b(w) + (−1)p(a)p(b) b(w)(∂w
a(w))−
j!
j≥0
=
X
(a(w)(−j−1) b(w))(z − w)j .
j≥0
Now we see that this result and (3) gives a justification of (1).
2
(z − w)j .
(z − w)j .
6.2
Detailed Formulation and Proof of the Extension Theorem
In this section we state and prove
theorem. Suppose we have a pre-vertex
P the jextension
j
−n−1
algebra (V, |0i, T, F = {a (z) = n∈Z a(n) z
}j∈J ). We denote by Ftr.cov the space of all
T -covariant EndV -valued quantum fields. Now we will prove a series of lemmae as follows.
Lemma 6.2. If a(z) ∈ Ftr.cov , then a(z)|0i ∈ V [[z]] and a(z)|0i = ezT (a(z)|0i|z=0 ).
This is already proved in Lecture 1. Thus, the map
f s : Ftr.cov → V : a(z) 7→ a(z)|0i|z=0
is well-defined, and the last statement can be reformulated as a(z)|0i = ezT f s(a(z)).
Lemma 6.3. Ftr.cov contains the identity map IV : V → V and is closed under n-th
products for all n ∈ Z. Also, it is ∂z -invariant.
Proof. (Exercise 6.2) First note that we may regard IV as in (EndV )[[z]]. Then it is
clearly a quantum field and satisfies [T, IV ] = 0 = ∂z IV . Thus IV ∈ Ftr.cov . Also if
a(z), b(z) ∈ Ftr.cov , then (note that p(T ) = 0)
[T, a(z)(n) b(z)] = [T, Res[a(w), b(z)](w − z)n dw]
= Res[T, [a(w), b(z)]](w − z)n dw
= Res([[T, a(w)], b(z)] + [a(w), [T, b(z)]])(w − z)n dw
= Res([∂w a(w), b(z)] + [a(w), ∂z b(z)])(w − z)n dw
= −Res[a(w), b(z)]∂w (w − z)n dw + Res[a(w), ∂z b(z)](w − z)n dw
= −nRes[a(w), b(z)](w − z)n−1 dw + Res[a(w), ∂z b(z)](w − z)n dw
= ∂z (Res[a(w), b(z)](w − z)n dw)
= ∂z (a(z)(n) b(z))
Here we use the fact that Res[∂w a(w), b(z)](w − z)n dw + Res[a(w), b(z)]∂w (w − z)n dw =
Res∂w ([a(w), b(z)](w−z)n )dw = 0. Thus Ftr.cov is closed under every n-th product. Finally,
we clearly have [T, ∂z a(z)] = ∂z [T, a(z)] = ∂z2 a(z), thus Ftr.cov is ∂z -invariant.
Lemma 6.4 (Dong’s Lemma). If a(z), b(z), c(z) are pairwise local quantum fields, then
(c(z), a(z)(n) b(z)) is again a local pair.
Proof. (Exercise 6.3) This is given in [1, pp.84-85]. We will give a proof later.
Lemma 6.5. Let a(z), b(z) ∈ Ftr.cov , a = f s(a(z)), b = f s(b(z)). Then we have
(i) fs (IV ) = |0i.
(ii) fs (∂z a(z)) = T a.
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(iii) f s(a(z)(n) b(z)) = a(n) b.
Proof. (Exercise 6.4) (i) is obvious. Also note that (∂z a(z))|0i|z=0 = [T, a(z)]|0i|z=0 = T a,
from which (ii) follows (since T |0i = 0.) For (iii), first recall that we have
(a(w)(n) b(w)) = Res(a(z)b(w)iz (z − w)n − (−1)p(a)p(b) b(w)a(z)iw (z − w)n )dz.
Now by Lemma 6.3, a(w)(n) b(w) is T -covariant, thus we have
f s(a(w)(n) b(w)) = Res(a(z)b(w)iz (z − w)n − (−1)p(a)p(b) b(w)a(z)iw (z − w)n )dz|0i|w=0
= Res(a(z)bz n )dz − (−1)p(a)p(b) Res(b(w)a(z)iw (z − w)n )dz|0i|w=0
= a(n) b − (−1)p(a)p(b) Res(b(w)a(z)iw (z − w)n )dz|0i|w=0 .
Thus it suffices to show Res(b(w)a(z)iw (z − w)n )dz|0i|w=0 = 0. Indeed, a(z)|0i ∈ V [[z]] by
Lemma 6.2, and iw (z − w)n ∈ F[[w, w−1 ]] ⊗ V [[z]]. Thus b(w)a(z)iw (z − w)n |0i does not
contain any z −1 term, whence the residue is zero by definition.
Lemma 6.6 (Uniqueness Lemma). Let F 0 ⊂ Ftr.cov be a subset such that f s(F 0 ) = V ,
and suppose a(z) ∈ Ftr.cov such that f s(a(z)) = 0 and (a(z), b(z)) is a local pair for all
b(z) ∈ F 0 . Then a(z) = 0.
Proof. For any b(z) ∈ F 0 , there exists n ∈ N such that (−1)p(a)p(b) (z − w)N b(w)a(z)|0i =
(z − w)N a(z)b(w)|0i by locality. By assumption and Lemma 6.2, it equals
0 = (−1)p(a)p(b) (z − w)N b(w)ezT f s(a(z)) = (z − w)N a(z)ewT f s(b(z)).
Now we put w = 0, and we have z N a(z)f s(b(z)) = 0, whence a(z)f s(b(z)) = 0. Since
f s(F 0 ) = V , a(z)V = 0, hence a(z) = 0 as desired.
Now we define some subspaces of Ftr.cov as follows.
Fmax = {T -covariant quantum fields a(z) | (a(z), aj (z)) is local for all aj (z) ∈ F}
j1
j2
js
ji
Fmin = F-span{a (z)(n1 ) (a (z)(n2 ) (· · · (a (z)(ns ) IV ) · · · )) | a (z) ∈ F, ni ∈ Z}
(5)
(6)
Then we have the following inclusion relations.
F ⊂ Fmin ⊂ Fmax ⊂ Ftr.cov
(7)
Indeed, the first inclusion comes from a(z)(−1) IV = a(z), the second follows form Lemma
6.3 and 6.4, and the last one is from the definition. Also note that the third inclusion
is strict; clearly T ∈ Ftr.cov , but (T, a(z)) need not be local for any a(z) ∈ F, thus
T ∈ Ftr.cov \ Fmax .
Now we are ready to state and prove our main theorem.
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Theorem 6.7 (Extension Theorem). We have
(i) Fmax = Fmin . We call it F.
(ii) f s : F → V : a(z) 7→ a(z)|0i|z=0 is well-defined and bijective. We denote its inverse
by sf : V → F.
(iii) We define a product on V with values in V ((z)) by (a, b) 7→ sf (a)b for a, b ∈ V . Then
this product endows V with a structure of a vertex algebra with the vacuum vector |0i
and the translation operator T .
(iv) Explicitly, the product on V is given by
sf (aj(n1 1 ) · · · aj(ns s ) |0i) = sf (aj1 )(n1 ) (sf (aj2 )(n2 ) (· · · (sf (ajs )(ns ) IV ) · · · )).
Remark. For the sake of (ii), we may abuse notations and write f s(a(z)) = a and sf (a) =
a(z). Also the product can be denoted by sf (a)b = a(z)b as usual. We will not use these
notations in this note.
Proof. First, note that f s : Fmin → V is surjective by completeness and Lemma 6.5. Also,
f s : Fmax → V is injective using Lemma 6.6 by setting F 0 = Fmin and by Lemma 6.4.
Thus (i) and (ii) are proved. For (iii), we need to check that (a, b) 7→ sf (a)(b) satisfies the
axioms for a vertex algebra, which are as follows.
• (vacuum axiom) sf (|0i)a = IV a = a, sf (a)|0i = ezT a = a + T (a)z + · · · by Lemma 6.2.
• (translation covariance) [T, a(z)] = ∂z a(z) for all a(z) ∈ F since F ⊂ Ftr.cov .
• (locality) for any a(z), b(z) ∈ F, (z − w)N [a(z), b(w)] = 0 for some N ∈ N, and it is
basically Lemma 6.4.
Thus we conclude that (V, |0i, T, F) gives a well-defined vertex algebra as desired. Finally,
(iv) is nothing but the reformulation of Lemma 6.5.
6.3
Proof of Dong’s Lemma
We will give a proof of Lemma 6.4 as promised.
Proof of Dong’s Lemma. We basically follow [1, pp. 84-85]. First we define
A = ix (x − z)n a(x)b(z)c(w) − (−1)p(a)p(b) iz (x − z)n b(z)a(x)c(w)
B = (−1)p(c)(p(a)+p(b)) (ix (x − z)n c(w)a(x)b(z) − (−1)p(a)p(b) iz (x − z)n c(w)b(z)a(x))
5
Then clearly ResAdx = (a(z)(n) b(z))c(w) and ResBdx = (−1)p(c)p(a(n) b) c(w)(a(z)(n) b(z)),
thus it suffices to show that there exists N ∈ N such that (z − w)N (A − B) = 0. Now by
assumption, there exists N0 ∈ N which satisfies N0 > −n and the following relations.
(x − z)N0 a(x)b(z) = (x − z)N0 (−1)p(a)p(b) b(z)a(x)
(8)
(z − w)N0 b(x)c(z) = (z − w)N0 (−1)p(b)p(c) c(w)b(z)
(9)
(x − w)
N0
N0
a(x)c(z) = (x − w)
p(a)p(c)
(−1)
c(w)a(x)
(10)
Now we take N = 4N0 . Then we have
(z − w)N A =
3N0 X
3N0
(z − x)3N0 −i (x − w)i
i
!
(z − w)N0 A
i=0
But (x − z)M ix (x − z)n = (x − z)M iz (x − z)n = (x − z)M +n whenever M + n ≥ 0, thus
for 0 ≤ i ≤ N0 we see that (z − x)3N0 −i A = 0 using (8) (since 3N0 − i + n ≥ N0 .) Thus,
indeed we have


3N0 X
3N
0
(z − w)N A = 
(z − x)3N0 −i (x − w)i  (z − w)N0 A
i
i=N0 +1
Similarly, we also have


3N0 X
3N0
(z − w)N B = 
(z − x)3N0 −i (x − w)i  (z − w)N0 B
i
i=N0 +1
Now the result follows from (9) and (10) by a similar argument.
References
[1] Kac, Victor G. Vertex algebras for beginners. Vol. 10. Providence: American Mathematical Society, 1998.
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