18.276 Lecture Note 14
Lecture by Victor Kac, Scribed by Dongkwan Kim
April 2, 2015
14.1
Some Properties of Vertex Algebras
In this lecture, we construct various kinds of vertex algebras from existing ones. To this
end, we first need to check some properties as follows.
Proposition 14.1. Let V be a vertex algebra.
(a) Any T -invariant left ideal of V is a two-sided ideal.
(b) Suppose V is generated by a subset {aj }j∈J , i.e.
D
E
V = aj(n1 1 ) · · · aj(ns s ) |0i | j1 , · · · , js ∈ J, n1 , · · · , ns ∈ Z
as a vector space. Then I ⊂ V is a left ideal if and only if it is a subspace invariant
under all aj(n) for j ∈ J and n ∈ Z.
L
(c) If L ∈ V is a conformal vector such that V = n≥0 Vn is the eigenspace decomposition
of V with respect to L0 and V0 = F|0i, then V has a unique maximal ideal.
Proof. For (a), we use the skewsymmetry of V (in Lecture 7,) i.e.
a(z)b = (−1)p(a)p(b) ezT b(−z)a.
More precisely, if I ⊂ V is a T -invariant left ideal, then for any a ∈ I and b ∈ V we have
a(z)b = (−1)p(a)p(b) ezT b(−z)a, thus a(n) b ∈ I for all n ∈ Z by the condition on I. For (b),
note that the only if part is trivial. But the converse is also straightforward since {aj }j∈J
generates V .
For (c), first note that since L−1 = T by the definition of a conformal vector, every
left ideal of V is two-sided by (a), which we will simply
call ideals. Now any ideal I ⊂ V
L
is graded with respect to the decomposition V = n≥0 Vn since it is L0 -invariant. Also,
I ∩ V0 = 0 since V0 = F|0i. Now we let M ⊂ V to be the subspace of V spanned by all
proper ideals of V . Then it is graded and still a proper ideal since M ∩ V0 = 0, and it is
clearly a unique maximal ideal of V .
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14.2
Tensor Product of Vertex Algebras
Now we will define a tensor product of two vertex algebras as follows.
Proposition 14.2. Let (Vi , |0ii , Ti )i=1,2 be vertex algebras. Then we have a vertex algebra
(V, |0i, T ) defined as follows;
• V = V1 ⊗ V2 .
• |0i = |0i1 ⊗ |0i2 .
• T = T1 ⊗ IV2 + IV1 ⊗ T2 .
• For a ∈ V1 and b ∈ V2 , (a ⊗ b)(z) := a(z) ⊗ b(z) =
P
n,m∈Z a(n)
⊗ b(m) z −n−m−2 .
Note that in the last part, for operators A and B, A ⊗ B is defined by
(A ⊗ B)(a ⊗ b) = (−1)p(B)p(a) (A(a) ⊗ B(b)).
P
−n−m−2
Proof. (Exercise 14.1) First of all, note that the expression
P n,m∈Z a(n) ⊗ b(m) z
is well-defined even if for each fixed N ∈ Z, the sum
−n−m−2=N a(n) ⊗ b(m) is infinite.
Namely,
since
a,
b
are
both
quantum
fields,
for
any
v ⊗ w ∈ VPwe see that
P
n,m∈Z a(n) ⊗
−n−m−2=N (a(n) ⊗ b(m) )(v ⊗ w) is a finite sum. Also by the same reason
b(m) z −n−m−2 is a quantum field.
Now we check that V satisfies the axioms of a vertex algebra. Indeed,
[T, (a ⊗ b)(z)] = [T1 , a(z)] ⊗ b(z) + a(z) ⊗ [T2 , b(z)]
= ∂z a(z) ⊗ b(z) + a(z) ⊗ ∂z b(z)
= ∂z (a(z) ⊗ b(z)),
which proves translation covariance. Also,
T |0i = (T1 ⊗ I + I ⊗ T2 )(|0i1 ⊗ |0i2 ) = 0 + 0 = 0,
|0i(z) = |0i1 (z) ⊗ |0i2 (z) = IV1 ⊗ IV2 = I,
(a ⊗ b)(z)|0i|z=0 = a ⊗ b,
which is the vacuum axiom. To check locality, it suffices only to consider single terms
a ⊗ b, c ⊗ b ∈ V . Then there exists N, M ∈ N such that (z − w)N [a(z), c(w)] = 0, (z −
2
w)M [b(z), d(w)] = 0. Meanwhile, we have
[a(z) ⊗ b(z), c(w) ⊗ d(w)]
= (a(z) ⊗ b(z))(c(w) ⊗ d(w)) − (−1)p(a⊗b)p(c⊗d) (c(w) ⊗ d(w))(a(z) ⊗ b(z))
= (−1)p(b)p(c) (a(z)c(w)) ⊗ (b(z)d(w)) − (−1)p(a⊗b)p(c⊗d)+p(a)p(d) (c(w)a(z)) ⊗ (d(w)b(z))
= (−1)p(b)p(c) a(z)c(w) − (−1)p(a)p(c) c(w)a(z) ⊗ (b(z)d(w))
+ (−1)p(a)p(c)+p(b)p(c) (c(w)a(z)) ⊗ b(z)d(w) − (−1)p(b)p(d) d(w)b(z)
= (−1)p(b)p(c) [a(z), c(w)] ⊗ (b(z)d(w)) + (−1)p(a)p(c)+p(b)p(c) (c(w)a(z)) ⊗ [b(z), d(w)].
Note that we used p(a⊗b)p(c⊗d) = (p(a)+p(b))(p(c)+p(d)). Thus (z −w)max(N,M ) [a(z)⊗
b(z), c(w) ⊗ d(w)] = 0, which we want to get.
14.3
Derivation and Homology Vertex Algebra
In this section, we will define a derivation of a vertex algebra and check some properties.
Also, if its square is zero, then we can consider its homology which is again naturally a
vertex algebra.
Definition 14.3. For a vertex algebra V , D ∈ EndV is called a derivation of parity p(D)
if for any a, b ∈ V and n ∈ N, it satisfies D(a(n) b) = (Da)(n) b + (−1)p(D)p(a) a(n) (Db).
Proposition 14.4. Let V be a vertex algebra.
(a) If D is a derivation of parity p(D), then ker D is a vertex subalgebra of V .
(b) For any a ∈ V , a(0) is a derivation of V of parity p(a), which is called an inner
derivation.
Proof. (Exercise 14.3) For (a), first D(|0i) = 0, thus |0i ∈ ker D. Also if a, b ∈ ker D,
then clearly a(z)b ∈ (ker D)((z)). On the other hand, for a ∈ ker D and b ∈ V we have
0 = (Da)(n) b = D(a(n) b) − (−1)p(D)p(a) a(n) (Db), whence [D, a(n) ] = 0. Thus,
(D(T a))(n) b = D((T a)(n) b) − (−1)p(D)p(T a) (T a)(n) (Db)
= D[T, a(n) ]b − (−1)p(D)p(a) [T, a(n) ]Db
= −n(Da(n−1) b − (−1)p(D)p(a) a(n−1) Db)
= −n[D, a(n−1) ]b = 0,
from which it follows that T (ker D) ⊂ ker D. The rest is obvious. For (b), we use the
following identity (from Lecture 8.)
[aλ b(z)c] = eλz [aλ b](z)c + (−1)p(a)p(b) b(z)[aλ c]
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(∗)
Thus when λ = 0, we have a(0) (b(z)c) = (a(0) b)(z)c + (−1)p(a)p(b) b(z)(a(0) c), which implies
that a(0) is a derivation of parity p(a).
d2(0)
Suppose V = V0̄ ⊕ V1̄ is a vertex superalgebra and d ∈ V1̄ is such that d(0) d = 0. Then
= 21 [d(0) , d(0) ] since d is odd. Now we recall the commutator formula;
[a(m) , b(n) ] =
X m
j
j≥0
(a(j) b)(m+n−j)
Thus we have d2(0) = 21 (d(0) d)(0) = 0, which means im d(0) ⊂ ker d(0) . Meanwhile, ker d(0)
is a subalgebra of V by Proposition 14.4(a), and we claim im d(0) is a two-sided ideal of
ker d(0) . Indeed, for any a ∈ ker d(0) and d(0) b ∈ im d(0) we have
d(0) (a(n) b) = (d(0) a)(n) b + (−1)p(d)p(a) a(n) (d(0) b) = (−1)p(d)p(a) a(n) (d(0) b),
from which a(n) (d(0) b) ∈ im d(0) . Also, since [T, d(0) ] = 0, we have T (d(0) b) = d(0) (T b),
which means that im d(0) is T -invariant. Thus by Proposition 14.1(a), we conclude that
im d(0) is a two-sided ideal of ker d(0) . In sum, it is possible to define a vertex superalgebra
H(V, d(0) ) := ker d(0) / im d(0) .
(Exercise 14.4) Now suppose L ∈ V is a conformal vector and d ∈ V1̄ as above is such
that d(z) is primary of conformal weight ∆d = 1 (defined in Lecture 10.) Then for any
m ∈ Z, by commutator formula we have
X m + 1 [Lm , d(0) ] = [L(m+1) , d(0) ] =
(L(j) d)(m+1−j)
j
j≥0
X m + 1
=
(Lj−1 d)(m+1−j)
j
j≥0
m+1
m+1
=
(L−1 d)(m+1) +
(L0 d)(m)
0
1
= (T d)(m+1) + (m + 1)d(m) = 0
since d is primary and (T d)(m+1) = [T, d(m+1) ] = −(m + 1)d(m) . Thus also [d(0) , Lm ] = 0.
But it means that (d(0) L)(m+1) = [d(0) , L(m+1) ] = [d(0) , Lm ] = 0. Thus d(0) L = 0 and we
have L ∈ ker d(0) .
Thus we may consider its image L̄ in H(V, d(0) ). We claim that it is also a conformal
vector. Indeed, clearly L̄ is even and L̄(z) is a Virasoro field with central charge c. Also,
clearly L̄−1 = T and L̄0 is diagonalizable on H(V, d(0) ).
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14.4
Orbifold Model
Here is another construction of a vertex algebra. Suppose we have a vertex algebra V and
a group G acting on V . Then V G is also a vertex algebra by the following proposition. If
G is finite, then it is called an orbifold model by physicists.
Proposition 14.5. Let V be a vertex algebra and G ⊂ Aut(V ). Then V G is again a vertex
algebra with the structure inherited from V .
Proof. (Exercise 14.2) It is almost straightforward. First of all, G fixes |0i, thus |0i ∈ V G .
Also, if a, b ∈ V G , then clearly a(z)b ∈ V G ((z)) and T a ∈ V G . The rest is obvious.
14.5
Coset Model
In this section, we define a coset model which is a centralizer of a subspace of a vertex
algebra. Namely, we have the following
Definition 14.6. let V be a vertex algebra and U ⊂ V be a subspace. We define
CV (U ) := {b ∈ V | [aλ b] = 0 for all a ∈ U }
= {b ∈ V | a(n) b = 0 for all a ∈ U, n ∈ Z≥0 },
which is called a coset model by physicists, and the centralizer of U in V by mathematicians.
(Note that the formula (∗) justifies that it is a subalgebra of V .)
14.6
Application to Universal Affine Vertex Algebra
We will apply this construction to the universal affine vertex algebra associated to some
Lie algebra. To this end, let g be a simple or abelian Lie algebra with a nondegenerate
symmetric invariant bilinear form ( | ) and V k (g) be the universal affine vertex algebra
of g such that k 6= −h∨ . Then we have Lg ∈ V k (g) given by the Sugarawa construction
(Lecture 13.)
Also let p ⊂ g be a Lie subalgebra which is reductive and the restriction ( | )|p is
nondegenerate.1 Then we may decompose p = p0 ⊕ p1 ⊕ · · · ⊕ pl where p0 is abelian and
pi are simple for 1 ≤ i ≤ l. Now we define Lp := Lp0 + Lp1 + · · · + Lpl where each Lpi is
given by the Sugawara construction in V k (pi ), which is considered naturally as a subspace
of V k (p). Then since likewise V k (p) ⊂ V k (g) naturally, we may regard Lp as in V k (g).
Now we will consider the coset model of this subspace as follows.
Proposition 14.7. Let Lg , Lp be as above, and define C := CV k (g) (V k (p)) (the centralizer
g
p
of V k (p) in V k (g).)
PlThenpi L − L ∈ C, and it is a conformal vector of C with central
g
charge c = c (k) − i=0 c (k).
1
Note that the latter condition does not imply that p is reductive. A counterexample can be found in
http://en.wikipedia.org/wiki/Cartan’s_criterion.
5
Proof. For a ∈ p ⊂ V k (p) ⊂ V k (g), [aλ Lg ] = λa = [aλ Lp ] (proved in Lecture 13,) thus
[aλ (Lg − Lp )] = 0. Since V k (p) is generated by p, it means that Lg − Lp ∈ C. In order to
check that it is a Virasoro field, first note that [Lp λ (Lg − Lp )] = 0 since Lp ∈ V k (p). Also,
by skew-commutativity of λ-bracket2 (in Lecture 9) we have [(Lg − Lp )λ Lp ] = 0. Thus,
[(Lg − Lp )λ (Lg − Lp )] = [Lg λ Lg ] − [Lp λ (Lg − Lp )] − [(Lg − Lp )λ Lp ] − [Lp λ Lp ]
= [Lg λ Lg ] − [Lp λ Lp ]
= (T + 2λ)(Lg − Lp ) +
λ3 g
(c (k) − cp (k))
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since Lg and Lp arePVirasoro fields. But it means Lg − Lp is a Virasoro field with central
charge c = cg (k) − li=0 cpi (k). It remains to show that (Lg − Lp )−1 = T and (Lg − Lp )0
is diagonal on C. But by the definition of C we have Lp−1 b = Lp0 b = 0 for b ∈ C, thus the
result follows form the fact that Lg−1 = T and Lg0 is diagonalizable on V k (g).
On the other hand, we can also consider the orbifold model corresponding to the universal affine vertex algebra as follows. Recall that [aλ Lg ] = λa for a ∈ g, thus a(0) Lg = 0.
By commutator formula, this implies that
[a(0) , Lg0 ] = [a(0) , Lg(1) ] = (a(0) Lg )(1) = 0,
which means that a(0) and Lg0 commute. Now recall that V = ⊕n≥0 Vn with respect to
Lg0 , and dim Vn < ∞. By the previous argument, each Vn is a(0) -invariant. Since a(0) is a
derivation, it follows that ea(0) converges on each Vn to an automorphism of V . Thus the
action of g on V by derivation a(0) exponentiates to the action of the group G on V by
automorphism.
Now here is an open problem: for a finite subset Γ ⊂ G, is V k (g)Γ finitely generated?
Also, for such an Lie subalgebra p ⊂ g we discussed before, is C = CV k (g) (V k (p)) finitely
generated?
Remark. We still assume the setting above, i.e. g is simple and p is reductive, such that
( | )|p is nondegenerate. Recall that Lp ∈ V k (p) by Sugawara construction is defined by
Pdim p
1
i
i
Lp = 2(k+h
∨)
i=1 : ai a :, where {ai } is the basis of p and {a } is its dual basis, i.e.
(ai |aj ) = δij . Now we let {ai } be the orthonormal basis of the compact form of p, so that
it is self-dual. (Note that Lp is independent of the choice of a basis of p.) Then since
: a(z)b(z) :(n) =
−∞
X
a(j) b(n−j−1) + (−1)p(a)p(b)
j=−1
2
∞
X
b(n−j−1) a(j) ,
j=0
For a local pair (a, b), we have [bλ a] = −(−1)p(a)p(b) ← [a−λ−∂ b] =
6
P
n≥0
(−λ−∂)n
(a(n) b).
n!
we have
Lp0 =
=
1
2(k + h∨ )
1
(k + h∨ )
dim
Xp


i=1
dim
Xp
i=1
−∞
X
(ai tj )(ai t−j ) +
j=−1
∞
X

(ai t−j )(ai tj )
j=0

 1 (ai )2 +
2
∞
X

(ai t−j )(ai tj ) .
j=1
Meanwhile, note that there exists the contravariant Hermitian form on V k (g) and Vk (g)
(the simple quotient of V k (g)) unique up to real scalar, which is called the Shapovalov
form. Now we claim that if the Shapovalov form on Vk (g) is positive definite, then Lp0 is
diagonalizable. Indeed, first note that (Lp0 )∗ = Lp0 with respect to Chevalley involution,
since (ai t−n )∗ = ai tn . Thus in particular Lp0 is normal, i.e. it commutes with (Lp0 )∗ .
Now recall that [(Lg − Lp )λ Lp ] = 0, thus especially we have (Lg−1 − Lp−1 )Lp = (Lg0 −
p
L0 )Lp = 0. But by commutator formula,
[Lg0 , Lp0 ] = (Lg−1 Lp )(2) + (Lg0 Lp )(1)
= (Lp−1 Lp )(2) + (Lp0 Lp )(1)
= [Lp0 , Lp0 ] = 0,
thus every eigenspace Vk (g)n of Lg0 is also Lp0 -invariant. Since it is finite dimensional, we can
diagonalize Lp0 on each Vk (g)n using the fact that the Shapovalov form is positive definite,
and it gives the diagonalization of Lp0 on the whole space Vk (g).
References
[1] Kac, Victor G. Vertex algebras for beginners. Vol. 10. Providence: American Mathematical Society, 1998.
[2] Kac, Victor G. Infinite-dimensional Lie algebras. Vol. 44. Cambridge university press,
1994.
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