Table Of ContentUse of Nilpotent weights in Logarithmic Conformal Field
Theories
∗
S. Moghimi-Araghi †, S. Rouhani ‡and M. Saadat §
2 Department of Physics, Sharif University of Technology,
0 Tehran, P.O.Box: 11365-9161, Iran
0
Institute for studies in Theoretical physics and Mathematics,
2
Tehran, P.O.Box: 19395-5531, Iran
n
a
J
5
1
Abstract
1 We show that logarithmic conformal field theories may be derived using nilpotent
v
scaletransformation. UsingsuchnilpotentweightswederivepropertiesofLCFT’s,such
9
as twoandthreepoint correlationfunctions solelyfromsymmetryarguments. Singular
9
0 vectors and the Kac determinant may also be obtained using these nilpotent variables,
1 hence the structure of the four point functions can also derived. This leads to non ho-
0 mogeneous hypergeometricfunctions. We can construct”superfields” using a nilpotent
2 variable. Using this construct we show that the superfield of conformal weight zero,
0
composedof the identity and the pseudo identity is relatedto a superfieldof conformal
/
h dimension two, which comprises of energy momentum tensor and its logarithmic part-
t
- ner. Thisdevicealsoallowsustoderivetheoperatorproductexpansionforlogarithmic
p
operators. Finally we consider LCFT’s near a boundary.
e
h
: Keywords: Field theory, Conformal, Logarithmic.
v
i
X
1 Introduction
r
a
Logarithmic conformal field theories (LCFT) were first introduced by Gurarie [1] in the
context of c = 2 conformal field theory (CFT). The difference between an LCFT and a
−
CFT [2], lies in the appearance of logarithmic as well powers in the singular behavior of
the correlation functions. In an LCFT, degenerate groups of operators may exist which all
have the same conformal weight. They form a Jordan cell under the action of L . In the
0
simplest case a pair of operators exist which transform according to
φ(λz) = λ−∆φ(z),
ψ(λz) = λ−∆[ψ(z) φ(z)lnλ]. (1)
−
The correlation function of this pair were derived in [3, 4, 5]. Now using nilpotent variables
[6]
θ2 = 0,
i
θ θ = θ θ , (2)
i j j i
∗Talk delivered in school and workshop on Logarithmic Conformal Field Theory, Tehran, Iran,2001
†e-mail: [email protected]
‡e-mail: [email protected]
§e-mail: [email protected]
1
and postulating that the conformal weights may have a nilpotent part
Φ(λz,θ) = λ−(∆+θ)Φ(z,θ), (3)
and the construct Φ(z,θ) = φ(z)+θψ(z), we arrive at equation (1). In order to construct
bigger Jordan cells it is sufficient to expand our construct, using θn = 0 and
Φ(z,θ) = φ (z)+φ (z)θ+φ (z)θ2+...+φ (z)θn−1. (4)
0 1 2 n−1
We will however restrict ourselves to rank two cell, and generalization to higher order cells
is straight forward.
2 Two and three point correlation functions
Now consider the following two point function
G(z ,z ,θ ,θ ) = Φ (z ,θ )Φ (z ,θ ) , (5)
1 2 1 2 1 1 1 2 2 2
h i
invariance under translation and rotation force dependence of G(z ,z ,θ ,θ ) to be on
1 2 1 2
z z . Under scale transformation we have
1 2
−
G(λ(z z ),θ ,θ ) = λ−(∆1+θ1)λ−(∆2+θ2)G((z z ),θ ,θ ), (6)
1 2 1 2 1 2 1 2
− −
and under special conformal transformation z z we find that ∆ = ∆ =: ∆ and
−→ 1+bz 1 2
constant term in the expansion of G(z ,z ,θ ,θ ) vanishes. Then
1 2 1 2
1
Φ(z ,θ )Φ(z ,θ ) = (a (θ +θ )+a θ θ ). (7)
h 1 1 2 2 i (z z )2∆+(θ1+θ2) 1 1 2 12 1 2
1 2
−
Expanding both sides in terms of θ and θ leads to all possible correlation functions in-
1 2
cluding φ and ψ
φ(z′)φ(z) = 0,
1
(cid:10)ψ(z′)ψ(z)(cid:11) = ( 2a ln(z′ z)+a ),
(z′ z)2∆ − 1 − 12
−
(cid:10) (cid:11) a
ψ(z′)φ(z) = 1 . (8)
(z′ z)2∆
−
(cid:10) (cid:11)
which is consistent with previous works [3].
Let us next consider the three point function
G(z ,z ,z ,θ ,θ ,θ ) = Φ (z ,θ )Φ (z ,θ )Φ (z ,θ ) . (9)
1 2 3 1 2 3 1 1 1 2 2 2 3 3 3
h i
Again it is clear that if one obtains this correlation function, all the correlators such as
φ φ φ , φ φ ψ , can be calculated readily by expanding this correlation function in
1 2 3 1 2 3
h i h i ···
terms of θ , θ and θ . Also note that these fields may belong to different Jordan cells. The
1 2 3
procedure of finding this correlator is just the same as the one we did for the two point
function. Like ordinary CFTs, the three point function is obtained up to some constants.
Of course, in our case it is found up to a function of θ ’s, i.e.
i
G(z ,z ,z ,θ ,θ ,θ ) = f(θ ,θ ,θ )z−a12z−a23z−a31, (10)
1 2 3 1 2 3 1 2 3 12 23 31
2
where z = (z z ) and a = ∆ +∆ ∆ +(θ +θ θ ). There are some constraints
ij i j ij i j k i j k
− − −
on f(θ ,θ ,θ ), but further reduction requires specification the rank of Jordan cell. As we
1 2 3
have taken it to be 2, we have
3
f(θ ,θ ,θ ) = C θ + C θ θ +C θ θ θ . (11)
1 2 3 i i ij i j 123 1 2 3
i=1 1≤i<j≤3
X X
While symmetry considerations do not rule out a constant term on the right hand side
of equation (11), but a consistent OPE forces this constant to vanish [7, 8]. This form
together with the equations (10) and (11) leads to correlation functions already obtained
in the literature. This is also consistent with the observation that in all the known LCFT’s
so far the three point function of the first field in the Jordan cell vanishes.
In a similar fashion one can derive the form of the four point functions. But before this
is done, we need to address the question of singular vectors in an LCFT.
3 Hilbert Space
Considering the infinitesimal transformation consistent with equation (3) we have
δΦ = ǫ∂Φ (∆+θ)Φ∂ǫ. (12)
− −
This defines the action of the generators of the Virasoro algebra on the primary fields and
points to the existence of a highest weight vector with nilpotent eigenvalue
L ∆+θ = (∆+θ)∆+θ ,
0
| i | i
L ∆+θ = 0, n 1 . (13)
n
| i ≥
Nilpotent state ∆+θ can be considered as
| i
∆+θ = Φ(0,θ)0 = [φ(0)+θψ(0)]0 ,
| i | i | i
= ∆,0 +θ ∆,1 . (14)
| i | i
Itcanbeeasily seenthatthelawwritteninequation (14), leadstothewellknownequations
L ∆,0 = ∆ ∆,0 ,
0
| i | i
L ∆,1 = ∆ ∆,1 + ∆,0 . (15)
0
| i | i | i
Now we define an out state in an LCFT
∆+θ = 0Φ†(0,θ),
h | h |
= 0[φ(0)+θψ(0)]†,
h |
(16)
where dagger means adjoint of fields, just the same as CFT. So
∆+θ = lim 0 φ(z´)z´2∆ +θ¯[ψ(z´)+lnz´2φ(z´)]z´2∆ , (17)
h | z´→∞h |
(cid:16) (cid:17)
whichtogether ’in’statedefinedinequation(14)andusingformofthetwopointcorrelation
functions in equation (8) leads to
∆+θ ∆+θ = θ+θ¯+d θ¯θ, (18)
h | i
3
where d is a if we normalize a to one. By expanding left hand side of the last equation,
12 1
we find
∆,0∆,0 = 0, ∆,0∆,1 = ∆,1∆,0 = 1, ∆,1∆,1 = d. (19)
h | i h | i h | i h | i
In addition to these highest weight states, there are descendants which can be obtained by
applying L ’s on the highest weight vectors
−n
∆+n +n + +n +θ = L L L ∆+θ . (20)
| 1 2 ··· k i −n1 −n2··· −nk| i
Most general state of Hilbert space at level n can be constructed as follows
χ(n)(θ) = b~nL ∆+θ ,
| ∆,c i −~n| i
|~nX|=n
= b(n1,n2,···,nk)L L ...,L ∆+θ . (21)
−n1 −n2 −nk| i
{n1+n2+X...+nk=n}
As an application of the above definitions we compute the character formula
χ∆(θ,θ¯) = N +∆+θ ηL0−2c4 N +∆+θ , (22)
h | | i
N
X
which by equation (20) simplifies to
χ∆(θ,θ¯) = η∆+θ−2c4 ηNp(N,θ) ∆+θ ∆+θ . (23)
h | i
N
X
Writing p(N,θ)= p (N)+θp (N) we obtain four characters
0 1
(φ,φ)
χ = 0,
∆
χ(∆φ,ψ) = χ(∆ψ,φ) = η∆−2c4 ηNp0(N),
N
X
χ∆(ψ,ψ) = η∆−2c4 ηN [p1(N)+(d+lnη)p0(N)]. (24)
N
X
Appearance of logarithms in character formula have been discussed in [9, 10].
4 Singular Vectors in LCFT
(n)
We definea singular vector at level n by χ as a vector that is orthogonal to all vectors
| ∆,ciS
in its level
(n) (n)
χ χ = 0. (25)
h ∆′,c| ∆,ciS
As a result it has a zero norm and is orthogonal to any vector at higher levels. According
to equation (20), last condition is equivalent to
′ (n) ′ ′
∆ +θ L χ = 0, ~n : ~n = n. (26)
h | ~n′| ∆,ciS ∀ | |
The number of L ’s is p(n), the number of partitions of the integer n. So equation (26)
~n′
is equivalent to p(n) equation with p(n) unknown coefficients. By putting equation (21) in
the last equation we have
∆′ +θ L χ(n) = b~n ∆′ +θ L L ∆+θ = 0, ~n′ : ~n′ = n, (27)
h | ~n′| ∆,ciS h | ~n′ −~n| i ∀ | |
|~nX|=n
4
Since L ∆+θ = 0 (for k 1) and using Virasoro algebra we find
k
| i ≥
′
L L ∆+θ = α~n ,~n(c)Lm ∆+θ , (28)
~n′ −~n| i m 0 | i
m=0
X
′
where coefficients α~n ,~n(c) are numbers or constants dependent on central charge c. By
m
putting this expression in equation (27)
b~n α~n′,~n(c)(∆+θ)m ∆′ +θ ∆+θ = 0, ~n′ : ~n′ = n, (29)
m !h | i ∀ | |
|~nX|=n mX=0
where (∆ + θ)m is eigenvalue of Lm ∆ + θ . Non zero solutions for b~n’s leads to Kac
0 | i
determinant in LCFT
det α~n′,~n(c)(∆+θ)m = 0 ~n′ = n, ~n = n. (30)
m ! ∀ | | | |
m=0
X
Singular vectors exist for those values of ∆ and c that Kac determinant is zero just the
same as CFT.
In the following we determine the null vectors at level 2 for a Jordan cell of rank 2. We
thus have
χ(2)(θ) = b(1,1)L2 +b(2)L ∆+θ . (31)
| ∆,c iS −1 −2 | i
(cid:16) (cid:17)
Equation (29) is equivalent to
4(2∆2 +∆)+4(4∆+1)θ 6(∆+θ) b(1,1)
= 0, (32)
6(∆+θ) 4(∆+θ)+ c b(2)
2 ! !
which leads to
1 1
∆ 8∆2 5∆+c(∆+ ) + 24∆2 10∆+c(2∆+ ) θ = 0. (33)
− 2 − 2
(cid:20) (cid:21) (cid:20) (cid:21)
So Kac determinant vanishes for (∆,c) = (0,0),( 5,25),(1,1). Because of homogeneity of
−4 4
the equation (32) at least one of the coefficients is arbitrary. Therefore we write
b(1,1) = 3, b(2) = [4(∆+θ)+2], (34)
−
(2) (2) (2)
and since for a Jordan cell of rank 2, χ (θ) = χ (0) +θ χ (1) we find
| ∆,c iS | ∆,c iS | ∆,c iS
χ(2)(1) = 3L2 2(2∆+1)L ∆,1 4L ∆,0 , (35)
| ∆,c iS −1− −2 | i− −2| i
h i
for∆ = 1, 5 asalogarithmicsingularvector. ∆ = 0doesnotleadtoalogarithmicsingular
4 −4
(2)
vector because ∆,1 does not appear in χ (1) . By the same technique logarithmic
| i | ∆,c iS
singular vectors can be obtained at higher levels which is consistent with [11].
5 Roots of Kac determinant in LCFT
In the previous section we saw that in an LCFT singular vectors exist for those values of
(∆,c) which Kac determinant vanishes just the same as CFT. If we compare equations (28)
5
and (30) with their counterparts in CFT,we see that ∆ in CFT has beenreplaced by ∆+θ.
One thus concludes that in an LCFT and at level n Kac determinant has the form
n
det (c,∆+θ)= (∆+θ ∆ (c))p(n−rs), (36)
n r,s
−
r,s=1;1≤rs≤n
Y
where ∆ (c) is
r,s
1 2 1 c
∆ (c) = (r+s)√1 c+(r s)√25 c − . (37)
r,s
96 − − − − 24
h i
andp(n rs)isthenumberof partitions oftheinteger n rs. InanordinaryCFTwhenthe
− −
Kac determinant vanishes we have a singular vector. In our case and for a rank 2 Jordan
cell the condition of vanishing det (c,∆+θ) are:
n
(i) If p(n rs) 2 for some r and s, Kac determinant vanishes for all values of ∆ that
− ≥
satisfy in ∆ = ∆ (c).
r,s
(ii) If p(n rs) = 1 for some pairs of (r,s) = (r ,s ),(r ,s ), we can have vanishing
1 1 2 2
− ···
determinant if at least ∆ = ∆ (c) = ∆ (c). In this case unlike (i) we are limited to
ri,si rj,sj
special values for ∆ and c which last condition is held. As an example we consider Kac
determinant at level 2. Since
1 r=1 , s=1
p(2 rs)= , (38)
− 1 r=1 , s=2 or r=2 , s=1
(cid:26)
all of them are cases of (ii). So
c = 1, ∆ = 1
∆ = ∆ 4
1,2 2,1 ⇒ ( c = 25, ∆ = −45,
∆ = ∆ c= 0,∆ = 0. (39)
1,2 1,1
⇒
Thisapproachcan beextendedeasily tohigherlevels andJordancells of biggerrank. These
results are consistent with those of [11].
6 Four point functions
To obtain further information about the theory with which we are concerned, such as
surface critical exponents, OPE structure, monodromy group and etc. one should compute
four point correlation functions. In the language we have developed so far, the four point
correlation functions depend on four θ’s in addition to the coordinates of points
G(z ,z ,z ,z ,θ ,θ ,θ ,θ ) = Φ (z ,θ )...Φ (z ,θ )
1 2 3 4 1 2 3 4 1 1 1 4 4 4
h i
= f(η,θ ,θ ,θ ,θ ) zµij. (40)
1 2 3 4 ij
1≤i<j≤4
Y
where
4
1 z z
41 23
µ = (∆ +θ ) (∆ +θ ) (∆ +θ ), η = . (41)
ij k k i i j j
3 − − z z
43 21
k=1
X
This form is invariant under all conformal transformations. Although there is no other
restrictions on G due to symmetry considerations, but because of OPE structure, the four-
point function φφφφ should vanish [7, 8], that is, the term independent of θ ’s in G is
i
h i
6
zero. Thus in addition to the differential equations which should be satisfied by G, one
must impose the condition φφφφ = 0 on the solution derived.
h i
If there is a singular vector in the theory, a differential equation can be derived for
f(η,θ ,θ ,θ ,θ ). Let us consider a theory which contains a singular vector at level two.
1 2 3 4
As seen in previous section the singular vector in such a theory is
χ(2)(z ,θ )= 3L2 (2(2∆ +1)+4θ )L Φ (z ,θ ). (42)
4 4 −1− 4 4 −2 4 4 4
h i
As this vector is orthogonal to all the other operators in the Verma module
Φ Φ Φ χ(2) = 0, (43)
1 2 3
h i
one immediately is led to the differential equation
3
∆ +θ ∂
3∂2 (2(2∆ +1)+4θ ) i i zi Φ Φ Φ Φ = 0. (44)
" z4 − 4 4 (zi z4)2 − zi z4#h 1 2 3 4i
Xi=1 − −
By sending points to z = 0, z = 1, z and z = η, we find
1 2 3 4
−→ ∞
2µ 2µ 2η 1 µ (µ 1) µ (µ 1)
∂2f +[ 14 24 α − ]∂ f +[ 14 14− + 24 24−
η η − 1 η − η(1 η) η η2 (1 η)2
− − −
2µ µ α(∆ +θ µ ) α(∆ +θ µ ) αµ
14 24 1 1 14 2 2 24 12
− − + ]f = 0, (45)
−η(1 η) − η2 − (1 η)2 η(1 η)
− − −
where α = 1[2(2∆ +1)+4θ ]. Renormalizing using
3 4 4
H(η,θ ,θ ,θ ,θ ) = η−β1+µ14(1 η)−β2+µ24f(η,θ ,θ ,θ ,θ ), (46)
1 2 3 4 1 2 3 4
−
we find that β satisfy
i
β (β 1)+α(β ∆ θ ) = 0, i = 1,2 (47)
i i i i i
− − −
and H satisfies the hypergeometric equation
d2H dH
η(1 η) +[c (a+b+1)η] abH = 0, (48)
− dη2 − dη −
where
ab = (β +β )(β +β +2α 1)+α(∆ ∆ +θ θ ),
1 2 1 2 4 3 4 3
− − −
a+b+1 = 2(β +β +α),
1 2
c = 2β +α. (49)
1
We can now write down the solution of equation (48) in terms of the hypergeometric series
H(a,b,c;η) = K(θ ,θ ,θ ,θ )h(a,b,c;η), (50)
1 2 3 4
where
∞
(a) (b)
h(a,b,c;η) = n nηn, (51)
n!(c)
n=0 n
X
with (x) = x(x+1)...(x+n 1) ,(x) =1 . Also
n 0
−
4
K(θ ,θ ,θ ,θ ) = k θ + k θ θ + k θ θ θ +k θ θ θ θ . (52)
1 2 3 4 i i ij i j ijk i j k 1234 1 2 3 4
i=1 1≤i<j≤4 1≤i<j<k≤4
X X X
7
Note that the coefficients in equation (51) contain nilpotent terms. Therefore equation
(51) actually describes morethan onesolution. Thisexpansion results in16 functions. This
is natural because as seen in the case of two and three point functions, inside the correlator
(40)thereexistsixteendistinctcorrelationfunctions. Ofcourse,ifallthefieldsbelongtothe
sameJordancell, only fourof them may beindependentand therestarerelated by crossing
symmetry. Also, one of them which only contains φ fields, vanishes and hence only three
independent functions remain. The form of these functions may be obtained by expanding
equation (50) and collecting powers of θ ’s. Note that expanding equation (48) leads to
i
sixteen differential equations. The general form of these equations is given in appendix. As
an example we solve equation (48) for the special case of ∆ = ∆ = ∆ = ∆ = 1. In this
1 2 3 4 4
case one of the solutions of equation (47) is
1 1 2
β = +θ θ + θ θ ,
1 1 4 1 4
2 − 3 3
1 1 2
β = +θ θ + θ θ . (53)
2 2 4 2 4
2 − 3 3
We then find from equation (49)
2
a = 2+θ +θ +θ +θ + (θ +θ +θ )θ ,
1 2 3 4 1 2 3 4
3
1 2
b = 1+θ +θ θ + θ + (θ +θ θ )θ ,
1 2 3 4 1 2 3 4
− 3 3 −
2 4
c = 2+2θ + θ + θ θ . (54)
1 4 1 4
3 3
Finally from equations (40), (46) and (50) we get expressions for the various four point
functions. For example
2 2 −1
hψ(z1)φ(z2)φ(z3)φ(z4)i = k1η3(1−η)3h0(η) zij6. (55)
1≤i<j≤4
Y
Where h (η) is given in the appendix. The other four point functions having 2, 3 or 4 ψ’s
0
can be calculated in the same way. In these functions, depending on how many ψ’s are
present in the correlators, different functions appear on the right hand side. If there is no ψ
in the correlator, the correlator is zero, if there is one ψ(z ), only H (= k h (η)) appears, if
i i i 0
there are two ψ’s, H and H appear, and so on. The situation is just the same for the two
i ij
or three point functions. For example the two point function φ(z)ψ(0) is written in terms
h i
of the function z−2∆ and in the correlator ψ(z)ψ(0) there exist both functions z−2∆ and
h i
z−2∆lnz.
7 Energy momentum tensor
Two central operators in a CFT are theenergy momentum tensor, T with conformal weight
∆ = 2andtheidentity operator, I withconformalweight∆ = 0. However, T isasecondary
field of the identity, because L I =T.
−2
In an LCFT degenerate operators exist which form a Jordan cell under conformal trans-
formation. This holds true for the identity as well. The existence of a logarithmic identity
operator has been discussed by a number of authors [1, 12, 13].
Now consider the identity operator Ω and its logarithmic partner ω. According to equation
(1) this pair transforms as
Ω(λz) = Ω(z),
ω(λz) = ω(z) Ω(z)lnλ. (56)
−
8
So according to our convention, we define a primary field Φ (z,θ)
0
Φ (z,θ) = Ω(z)+θω(z), (57)
0
with conformal weight θ. Under scaling, Φ (z,θ) transforms according to equation (3).
0
Thus we have
L Ω(z) = 0,
0
L ω(z) = Ω(z), (58)
0
where was first observed in c= 2 theory by Gurarie [1].
−
Here we wish to find the field T(z,θ) with conformal weight 2+θ which is a secondary
of Φ (z,θ) in the sense
0
L Φ (z,θ) = T(z,θ). (59)
−2 0
By writing T(z,θ) = T (z)+θt(z) and since L T(z,θ) = (2+θ)T(z,θ) we have
0 0
L T (z) = 2T (z),
0 0 0
L t(z) = 2t(z)+T (z). (60)
0 0
This points to the existence of an extra energy momentum tensor [12, 13, 14]. By applying
L on both sides of equation (59) we have
2
c
L T (z) = Ω(z),
2 0
2
c
L t(z) = ω(z)+4Ω(z). (61)
2
2
The first of this pair exists in an ordinary CFT, so T (z) leads to the Virasoro algebra,
0
while t(z) must leads to a new algebra [14]. We now attempt at finding the OPE of the
T (z′) with ψ(z) and extra energy momentum tensor, t(z). Because of OPE’s invariance
0
under scaling and according to our convention it is sufficient to change conformal weight of
each field to ∆+θ. Consider the following OPE
∆+θ ∂ Φ(z,θ)
T (z′)Φ(z,θ) = Φ(z,θ)+ z + . (62)
0 (z′ z)2 z′ z ···
− −
This relation leads to the familiar OPE for T (z′)φ(z) and a new OPE
0
φ(z)+∆ψ(z) ∂ ψ(z)
T (z′)ψ(z) = + z + . (63)
0 (z′ z)2 z′ z ···
− −
Also
c(θ)
Φ (z,θ) 2+θ ∂ T(z,θ)
T (z′)T(z,θ) = 2 0 + T(z,θ)+ z + , (64)
0 (z′ z)4 (z′ z)2 z′ z ···
− − −
wherec(θ) = c +θc . Again weobtain two OPE,oneof themisT (z′)T (z)which isknown
1 2 0 0
from CFT and the other is
c1ω(z)+ c2Ω(z) T (z)+2t(z) ∂ t(z)
T (z′)t(z) = 2 2 + 0 + z + . (65)
0 (z′ z)4 (z′ z)2 z′ z ···
− − −
The emergence of an extra energy momentum tensor and central charge have been
noticedbyGurarieandLudwig[14],althoughourapproachisverydifferent. Recently ithas
been argued [15] that in theories with zero central charge (c = 0) t, the logarithmic partner
of T is not a descendant of any other field. So these theories can have non degenerate
0
9
vacua. Such theories will not fit into the framework presented here. Theories with non zero
central charge (c = 0) behave very differently. In these theories there exist a logarithmic
6
partner for T only if the vacua is degenerate and so t is a descendant field [15].
0
It is worth noting that equations (56) imply that Ω vanishes whereas ω = 1 . This
h i h i
immediately results in the vanishing of T T , even though the central charge may not
0 0
h i
vanish.
8 Boundary
Let usnow consider theproblem of LCFTnear a boundary. As shown in [16]in an ordinary
CFT,if thereal axis is taken to betheboundary,with certain boundarycondition thatT =
T on the real axis, the differential equation satisfied by n-point function near a boundary
are the same as the differential equations satisfied by 2n-point function in the bulk. This
trick may beusedin orderto derive correlations of an LCFT near aboundary[10,17]. Here
we rederive the same results using the nilpotent formalism. Again we consider an LCFT
with a rank 2 Jordan cell. First we find the one point functions of this theory. By applying
L ,L on the correlators, one obtains
0 ±1
(∂ +∂ ) Φ(z,z¯,θ) = 0,
z z¯
h i
(z∂ +z¯∂ +2(∆+θ) Φ(z,z¯,θ = 0,
z z¯
h i
(z2∂ +z¯2∂ +2z(∆+θ)+2z¯(∆+θ)) Φ(z,z,θ = 0. (66)
z z¯
h i
In these equations, we have assumed that Φ is a scalar field so that ∆ = ∆¯. The first
equation states Φ(z,z¯,θ) is a function of z z¯ and the solution to the second equation is
h i −
f(θ)
Φ(y,θ) = , (67)
h i y2(∆+θ)
where y = z z¯. The third line of equation (66) is automatically satisfied by this solution.
−
Expanding f(θ) as a+bθ one finds
a θ
Φ(y,θ) = + (b 2alny). (68)
h i y2∆ y2∆ −
As the field Φ(y,θ) is decomposed to φ(y) +θψ(y) one can read the one-point functions
φ(y) and ψ(y) from the equation (68)
h i h i
a
φ(y) = ,
h i y2∆
1
ψ(y) = (b 2alny). (69)
h i y2∆ −
To go further, one can investigate the two-point function
G(z ,z¯ ,z ,z¯ ,θ ,θ ) = Φ(z ,z¯ ,θ )Φ(z ,z¯ ,θ ) , (70)
1 1 2 2 1 2 1 1 1 2 2 2
h i
in the same theory. Invariance under the action of L implies
−1
(∂ +∂ +∂ +∂ )G = 0. (71)
z1 z¯1 z2 z¯2
The most general solution of this equation is G = G(y ,y ,x ,x ,θ ,θ ), where y = z
1 2 1 2 1 2 1 1
−
z¯ , y = z z¯ , x= x x and x = z +z¯. By invariance under the action of L we
1 2 2− 2 2− 1 i i i 0
should have
∂ ∂ ∂
y +y +x +2(∆+θ )+2(∆+θ ) G = 0, (72)
1 2 1 2
∂y ∂y ∂x
(cid:20) 1 2 (cid:21)
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