Contemporary Mathematics Volume 00, 0000

Spinor Construction of the c = 1/2 Minimal Model

ALEX J. FEINGOLD, JOHN F. X. RIES, MICHAEL D. WEINER

1. Introduction The representation theories of affine Kac-Moody Lie algebras and the Virasoro

algebra have been found to have interesting connections with many other parts of mathematics and have played a vital role in the development of conformal field theory in theoretical physics.(See [BPZ], [GO], [BMP], [MS], [TK], the Introduction and references in [FLM].) An important contribution to Kac-Moody representation theory was the rigorous mathematical construction of representations using vertex operators, discovered independently by mathematicians and physicists, but known earlier by physicists. Once the connection was discovered, an exciting dialog began which has enriched both sides. The precise axiomatic definition of vertex operator algebras (VOA’s) by mathematicians [FLM] gave a rigorous foundation to the algebraic aspects of conformal field theory called chiral algebras by physicists. This theory includes, unifies and vastly extends the representation theories of affine Kac-Moody algebras and the Virasoro algebra.

In the theory of VOA’s one has the notion of a module and of intertwining operators going between modules [FHL], [F]. The main axiom for a VOA is an identity called the Jacobi-Cauchy identity because it combines the usual Jacobi identity for Lie algebras with the Cauchy residue formula for rational functions whose possible poles are limited to three points, 0, 1 and ∞. A slight modification is needed for the appropriate definition of a module, and intertwining operators are defined by a similar axiom relating them to vertex operators. We believe that

\begin{matrix}c \\ ◯ \end{matrix}0000 American Mathematical Society 0000-0000/00 $1.00 + $.25 per page

1991 Mathematics Subject Classification. Primary 17B65, 17B67, 81R10; Secondary 17B68, 17A70, 81T40.

A.J.F. was partially supported by the National Security Agency under Grant number MDA904-94-H-2019. The United States Government is authorized to reproduce and distribute reprints notwithstanding any copyright notation heron.

J.F.X.R. died July 6, 1993. This paper is dedicated to his memory.

This paper is in final form and no version of it will be submitted for publication elsewhere.

an important next step is the understanding of a new kind of “matrix” Jacobi-Cauchy identity relating any two intertwining operators. This would lead to a larger unifying structure, incorporating a VOA, its modules and its intertwining operators.

There are several new features which appear when trying to understand intertwining operators and the new kind of Jacobi-Cauchy identity they obey. First one has to deal with fusion rules,

N (M_{1}, M_{2}, M_{3}) = dim(I(M_{1}, M_{2}, M_{3})),

which give the dimension of the space of intertwining operators determined by a triple of modules. It is a basic principle of VOA’s that there is a one-to-one correspondence between vectors v in a simple VOA V and vertex operators Y_{M}(v, z) acting on an irreducible V -module M. One can think of this as a map

Y_{M}(·, z) : V → End(M)[[z, z^{−1}]]

which obeys the various axioms defining a V -module. The fusion rule in this case is always N (V, M, M) = 1 and one axiom normalizes Y_{M} so it is uniquely determined. Given three V -modules, M_{1}, M_{2}, M_{3}, one can think of an intertwining operator as a map

Y (·, z) : M_{1} → Hom(M_{2}, M_{3}){{z, z^{−1}}}

which obeys the axioms for intertwining operators. (The notation {{z, z^{−1}}} indicates rational powers of z.) It is quite possible that the fusion rule is N (M_{1}, M_{2}, M_{3}) = n > 1, and in that case one does not have a one-to-one correspondence between vectors w in module M_{1} and operators Y (w, z) whose components send M_{2} to M_{3}. It would seem that this is a kind of labeling problem, there not being enough “copies” of the vectors in M_{1} to distinguish the n linearly independent intertwiners which could be taken as a basis for the space of all intertwiners. It is also possible to have four modules M_{1}, ⋯, M_{4} with M_{3} ≠ M_{4}, with fusion rules N (M_{1}, M_{2}, M_{3}) ≥ 1 and N (M_{1}, M_{2}, M_{4}) ≥ 1. This also indicates a labeling problem, showing the inadequacy of the notation Y (w_{1}, z)w_{2}, where knowing that w_{i} ∈ M_{i} still does not determine which module contains the outcome.

Another new feature which appears is the nature of the correlation functions,

(Y (w_{1}, z_{1})Y (w_{2}, z_{2})w_{3}, w_{4}) and (Y (Y (w_{1}, z)w_{2}, z_{2})w_{3}, w_{4}),

made from two intertwiners. These are series which converge in certain domains to functions which, after factoring out some rational powers of z_{1} and z_{2}, can be expressed as power series in z_{2}/z_{1} and z/z_{2} satisfying certain differential equations. One way of thinking of the usual Jacobi-Cauchy identity for a VOA V is as follows. The three series

(Y (v_{1}, z_{1})Y (v_{2}, z_{2})v_{3}, v_{4}), (Y (v_{2}, z_{2})Y (v_{1}, z_{1})v_{3}, v_{4})

SPINOR CONSTRUCTION OF THE c = 1/2 MINIMAL MODEL 3

and

(Y (Y (v_{1}, z_{1} − z_{2})v_{2}, z_{2})v_{3}, v_{4})

converge in their respective domains, |z_{1}| > |z_{2}|, |z_{2}| > |z_{1}|, and |z_{2}| > |z_{1} − z_{2}|, to the same rational function f(z_{1}, z_{2}) in the ring

R = ℂ[z_{1}, z_{1}^{−1}, z_{2}, z_{2}^{−1}, (z_{1} − z_{2})^{−1}].

For fixed z_{2} ≠ 0, these are functions of z_{1} with possible poles only at z_{1} = 0, z_{1} = ∞ and z_{1} = z_{2}. The Jacobi-Cauchy identity is just the statement of the Cauchy residue theorem applied to f(z_{1}, z_{2})g(z_{1}, z_{2}), where g(z_{1}, z_{2}) ∈ R is arbitrary, and the residues at each of the three points are computed from the three series giving f and appropriate series expansions of g. But the kinds of correlation functions obtained from intertwiners generally involve hypergeometric functions to which the treatment just described does not apply.

The purpose of this paper is to show how both of these new features can be handled in a simple but nontrivial example which may have important applications. In the monograph [FFR] we constructed certain vertex operator superalgebras (VOSA’s) and their twisted modules from Clifford algebras and their spinor representations. These “fermionic” constructions extend the corresponding constructions of the orthogonal affine Kac-Moody algebras of type D_{n}^{(1)} in [FF]. (See [Fr1], [Fr2].) In [W] vertex operator para-algebras were constructed from the bosonic constructions of level −\frac{1}{2} representations of symplectic affine Kac-Moody algebras of type C_{n}^{(1)}. It is well known to physicists, and follows immediately from the work in [FFR], that a spinor construction from one fermion gives a VOSA and a twisted module for it. These are known in the physics literature as the Neveu-Schwarz and Ramond sectors, respectively, and they each decompose into two irreducible modules for the Virasoro algebra with central charge c = \frac{1}{2}. Using the usual notation for labeling Virasoro modules, the two components into which the Neveu-Schwarz sector decomposes are labeled by h = 0 and h = \frac{1}{2}, and the two coming from the Ramond sector are both h = \frac{1}{16}. It is very interesting and important that in this construction one naturally has two copies of the h = \frac{1}{16} Virasoro module. That enables us to define unique intertwining operators for each vector in the Ramond sector, such that the usual Ising fusion rules for just three modules are replaced by fusion rules given by the group ℤ_{4}. This behavior is just like the behavior of vertex operator paraalgebras defined in [FFR]. (See also [DL].) It means that the VOA V and its modules are indexed by a finite abelian group Γ such that V = V_{0} (0 ∈ Γ is the identity element) and the fusion rules are N (V_{a}, V_{b}, V_{c}) = 1 if a+b = c in Γ, zero otherwise. The important question is whether this is a rare special situation, or if there are other natural constructions of VOA’s where multiple copies of the modules allow unique labeling of intertwining operators and where fusion rules are replaced by a group law.

In the example studied here we find that the hypergeometric functions to which the correlation function converge, come in pairs, as bases for the two dimensional spaces of solutions of certain differential equations. In order to relate them to each other, we must use Kummer’s quadratic transformation formula. This involves certain substitutions which lift the correlation functions of t = z_{2}/z_{1} to functions of x on a four-sheeted covering of the t-sphere, branched at t = 0, 1, ∞ with possible poles only at x = α ∈ {0, ∞, 1, −1, i, −i}. We may then find the 2 × 2 matrices B_{α} which relate the 2 × 4 matrices of transformed correlation functions at x = 0 to those at x = α. The matrix-valued functions have possible poles only at those six points, so after multiplying the matrixvalued functions by any function g(x) in the ring

R_{x} = ℂ[x, x^{−1}(x^{4} − 1)^{−1}]

the Cauchy residue theorem gives that the sum of their residues at the six points adds up to zero. Expressing the residues in terms of the series and expanding the function g(x) as an appropriate series, we get a “matrix” Jacobi-Cauchy identity. It is, of course, actually a generating function of identities, equivalent to infinitely many identities for the components of the intertwining operators. It may take some time to sort out which ones are the most important, but we can already point out some interesting ones.

A very interesting aspect of this example is the structure of the matrices B_{α}. In order to find those matrices one must choose some linear fractional transformations relating the variable x = x_{0} to the variables x_{α} which are local variables at the poles. There are several sign choices which can be made, and these correspond to choices of braiding. It is not so surprising that these matrices provide a representation of a braid group. The details of this aspect of the example have not been completely worked out yet, and will be studied later. There should be an interesting connection with the work in [MaS].

As a final motivation for the detailed study of this special example, we would like to mention the possible application to a spinor construction of the moonshine module for the monster group. It has been noted by Dong, Mason and Zhu [DMZ] that there are 48 commuting c = \frac{1}{2} Virasoro algebras in the moonshine module V^{♮} considered as a VOA. It means that V^{♮} decomposes into a sum of tensor products of 48 Virasoro modules, each of which is one of the h = 0, h = \frac{1}{2} or h = \frac{1}{16} modules. Although some information is known about this decomposition, the complete picture is not clear. But since there is a spinor construction of each of these modules, there is some hope that a spinor construction of V^{♮} is possible. In fact, we hope that the new light we have shed on the c = \frac{1}{2} minimal model will be of help in achieving that goal.

2. Construction of Vertex Operator Superalgebra and Module

Let E = ℂe with ⟨e, e⟩ = 2 and let Z = ℤ or Z = ℤ + \frac{1}{2}. Let E(Z) be the vector space with basis {e(m) | m ∈ Z} and the symmetric form

⟨e(m), e(n)⟩ = ⟨e, e⟩δ_{m,−n} = 2δ_{m,−n}.

Let Cliff(Z) be the Clifford algebra generated by E(Z) and that form. Let E(Z) = E(Z)^{+} ⊕E(Z)^{−} be the polarization where E(Z)^{+} is spanned by {e(m) | 0 < m ∈ Z} and E(Z)^{−} is spanned by {e(m) | 0 ≥ m ∈ Z}. Define ℑ(Z) to be the left ideal in Cliff(Z) generated by E(Z)^{+}, so that

CM(Z) = Cliff(Z)/ℑ(Z)

is a left Cliff(Z)-module. One has the parity decomposition

CM(Z) = CM(Z)^{0} ⊕ CM(Z)^{1}

where CM(Z)^{i}, i = 0, 1, is the subspace with basis

{e(−m_{1}) … e(−m_{r})vac(Z) | m_{1} > … > m_{r} ≥ 0, m_{1}, …, m_{r} ∈ Z, r ≡ i mod 2}

and vac(Z) = 1 + ℑ(Z). Define the vacuum space

VAC(Z) = {v ∈ CM(Z) | E(Z)^{+} · v = 0}.

Then VAC(ℤ + \frac{1}{2}) is one-dimensional, spanned by

vac = vac(ℤ + \frac{1}{2}) = 1 + ℑ(ℤ + \frac{1}{2}),

and VAC(ℤ) is two-dimensional, spanned by

vac′ = vac(ℤ) = 1 + ℑ(ℤ)  and  e(0)vac′.

It is easy to see that CM(ℤ + \frac{1}{2}) is an irreducible Cliff(ℤ + \frac{1}{2})-module, but CM(ℤ) decomposes into two irreducible Cliff(ℤ)-modules. Note that

vac′_{+} = vac′ + e(0)vac′  and  vac′_{−} = vac′ − e(0)vac′

are eigenvectors for e(0) with eigenvalues +1 and −1, respectively, because the relations in Cliff(ℤ) give e(0)^{2} = 1. We then have the alternative decomposition into two irreducible Cliff(ℤ)-modules,

CM(ℤ) = CM(ℤ)^{+} ⊕ CM(ℤ)^{−}

where CM(ℤ)^{±} is the subspace with basis

{e(−m_{1}) … e(−m_{r})vac′_{±} | m_{1} > … > m_{r} > 0, m_{1}, …, m_{r} ∈ ℤ}.

Later we will need the vector space isomorphism

θ : CM(ℤ) → CM(ℤ)

defined by

θ(vac′) = e(0)vac′  and  θ(e(m)v) = e(m)θ(v)

for any m ∈ ℤ and v ∈ CM(ℤ). It is clear that θ is an involution switching CM(ℤ)^{0} with CM(ℤ)^{1}, and that CM(ℤ)^{+} and CM(ℤ)^{−} are the +1 and −1 eigenspaces, respectively, for θ.

Define the Vir operators

L(k) = −\frac{1}{4} ∑_{n∈Z}(n + \frac{1}{2}) ∘ ∘e(n)e(k − n)∘ ∘ for k ≠ 0,

L(0) = \frac{1 + ι}{32} − \frac{1}{4} ∑_{n∈Z}(n + \frac{1}{2}) ∘ ∘e(n)e(−n)∘ ∘

where ι = 1 if Z = ℤ and ι = −1 if Z = ℤ + \frac{1}{2}.

Theorem 1. The operators L(k), k ∈ ℤ, and the identity operator, represent a c = \frac{1}{2} Virasoro algebra Vir on CM(Z). In particular, for k, n ∈ ℤ, m ∈ Z, we have

[L(k), e(m)] = −(m + \frac{1}{2}k) e(m + k),
[L(k), L(n)] = (k − n)L(k + n) + \frac{1}{24}(k^{3} − k)δ_{k,−n}1.

The parity decomposition of CM(Z) is a decomposition into two irreducible Virmodules. The highest weight vectors in these modules are vac, e(−\frac{1}{2})vac, vac′ and e(0)vac′, whose weights are 0, \frac{1}{2}, \frac{1}{16} and \frac{1}{16}, respectively. The decomposition of CM(ℤ) into two irreducible Cliff(ℤ)-modules is also a decomposition into two irreducible Vir-modules. The highest weight vectors in these modules are vac′_{+} and vac′_{−}, whose weights are both \frac{1}{16}. The operators L(k) commute with θ on CM(ℤ).

As usual we can define a positive Hermitian form ( , ) on CM(Z) such that e(m)^{∗} = e(−m) and L(k)^{∗} = L(−k), where ∗ denotes adjoint. The eigenspaces of L(0) provide CM(Z) with a grading. If u is an eigenvector for L(0) write wt(u) for the eigenvalue and write (CM(Z))_{n} for the n-eigenspace. For u = e(−m_{1}) … e(−m_{r})vac(Z) ∈ CM(Z) we have

L(0)u = (m_{1} + … + m_{r} + \frac{1 + ι}{32}) u.

Let W be any subspace of CM(Z) which is a direct sum of L(0) eigenspaces, (W)_{n}. The homogeneous character of W is defined to be

ch(W) = ∑_{n} dim(W)_{n}q^{n},

a formal series in q. The homogeneous characters of the Clifford modules are then

ch(CM(ℤ + \frac{1}{2})) = ∏_{0≤n∈ℤ}(1 + q^{n+ \frac{1}{2}} )

and

ch(CM(ℤ)) = 2q^{1/16} ∏_{1≤n∈ℤ}(1 + q^{n}).

As in the spinor construction of D_{n}^{(1)} we construct a vertex operator superalgebra on CM(ℤ + \frac{1}{2}) and a (twisted) representation on CM(ℤ). This is done by defining a vertex operator Y (v, ζ) on CM(ℤ + \frac{1}{2}) and on CM(ℤ) for any v ∈ CM(ℤ + \frac{1}{2}). Actually Y (v, ζ) is a generating function of operators

Y (v, ζ) = ∑_{n∈ \frac{1}{2} ℤ} Y_{n+1−wt(v)}(v)ζ^{−n−1} = ∑_{n∈ \frac{1}{2} ℤ}\{v\}_{n}ζ^{−n−1}

where

wt(Y_{m}(v)w) = wt(w) − m.

The definition of Y (v, ζ) on CM(ℤ) is more complicated than it is on CM(ℤ+ \frac{1}{2}), but there is a common part which we denote by Ȳ (v, ζ).

Recall that vac = vac(ℤ + \frac{1}{2}) and vac′ = vac(ℤ). On CM(Z) define Ȳ (vac, ζ) = 1 (the identity operator), and for 0 ≤ n ∈ ℤ let

Ȳ (e(−n − \frac{1}{2})vac, ζ) = n!^{−1}(d/dζ)^{n} ∑_{m∈Z}e(m)ζ^{−m− \frac{1}{2}} .

Using the fermionic normal ordering ∘ ∘e(n_{1}) ⋯ e(n_{r})∘ ∘ of a product of Clifford generators, for vectors of the form

v = e(−n_{1} − \frac{1}{2}) ⋯ e(−n_{r} − \frac{1}{2})vac ∈ CM(ℤ + \frac{1}{2})

we define

Ȳ (v, ζ) = ∘ ∘Ȳ (e(−n_{1} − \frac{1}{2})vac, ζ) ⋯ Ȳ (e(−n_{r} − \frac{1}{2})vac, ζ)∘ ∘

and extend the definition to all v ∈ CM(ℤ + \frac{1}{2}) by linearity. Then

Ȳ_{m}(e(−\frac{1}{2})vac) = e(m)

is a Clifford generator. Furthermore, with

ω = L(−2)vac = \frac{1}{4}e(−\frac{3}{2})e(−\frac{1}{2})vac,

we have

Ȳ (ω, ζ) = \{ \begin{matrix}L(ζ) & on CM(ℤ + \frac{1}{2}) \\ L(ζ) − \frac{1}{16}ζ^{−2} & on CM(ℤ) \end{matrix}

where L(ζ) = ∑_{k∈ℤ}L(k)ζ^{−k−2} is the generating function of the Virasoro operators.

In order to define the operators Y (v, ζ) we need the additional quadratic operator

Δ(ζ) = \frac{1}{4} ∑_{0≤m,n∈ℤ} C_{mn}e(m + \frac{1}{2})e(n + \frac{1}{2})ζ^{−m−n−1}

whose definition involves the combinatorial coefficients

C_{mn} = \frac{1}{2} \frac{m − n}{m + n + 1}\binom{−\frac{1}{2}}{m}\binom{−\frac{1}{2}}{n}.

For v ∈ CM(ℤ + \frac{1}{2}) we define

Y (v, ζ) = \{ \begin{matrix}Ȳ (v, ζ) & on CM(ℤ + \frac{1}{2}) \\ Ȳ (exp(Δ(ζ))v, ζ) & on CM(ℤ). \end{matrix}

Note that exp(Δ(ζ))v is a finite sum of vectors whose weights are between 0 and wt(v). In particular,

exp(Δ(ζ))ω = ω + \frac{1}{16}ζ^{−2}vac,

so Y (ω, ζ) = L(ζ) on both CM(ℤ + \frac{1}{2}) and CM(ℤ).

Theorem 2. We have (CM(ℤ+ \frac{1}{2}), Y ( , z), vac, ω) is a vertex operator superalgebra and (CM(ℤ), Y ( , z)) is a (twisted) vertex operator superalgebra module.

Corollary 3. Let v_{i} ∈ CM(ℤ + \frac{1}{2})^{α_{i}} for i = 1, 2. Then for any r ∈ ℤ, for m, n ∈ ℤ on CM(ℤ + \frac{1}{2}), and for m ∈ ℤ + \frac{1}{2}α_{1}, n ∈ ℤ + \frac{1}{2}α_{2} on CM(ℤ), we have

\begin{matrix}∑_{0≤i∈ℤ}(−1)^{i}\binom{r}{i}(\{v_{1}\}_{m+r−i}\{v_{2}\}_{n+i} − (−1)^{α_{1}α_{2}+r}\{v_{2}\}_{n+r−i}\{v_{1}\}_{m+i}) \\ = ∑_{0≤k∈ℤ} \binom{m}{k}\{\{v_{1}\}_{r+k}v_{2}\}_{m+n−k} . \end{matrix}

The sum over k is finite, and we may take 0 ≤ k ≤ wt(v_{1}) + wt(v_{2}) − r − 1.

Corollary 4. For v ∈ CM(ℤ + \frac{1}{2}), m ∈ ℤ, on CM(Z) we have

[L(m), Y (v, z)] = ∑_{0≤k∈ℤ}\binom{m + 1}{k}z^{m+1−k}Y (L(k − 1)v, z)

and the sum is finite. If L(n)v = 0 for all n > 0, then we have

[L(m), Y (v, z)] = z^{m+1}(d/dz)Y (v, z) + (m + 1)z^{m}Y (L(0)v, z)

which means that

[L(m), Y_{n−m}(v)] = (−n + m wt(v))Y_{n}(v).

Corollary 5. For v ∈ CM(ℤ + \frac{1}{2}), m ∈ ℤ, on CM(Z) we have

\begin{matrix}Y (L(−m − 1)v, z) = \\ ∑_{0≤i∈ℤ} \binom{m + i − 1}{i}[z^{i}L(−m − i − 1)Y (v, z) − (−1)^{m}z^{−m−i}Y (v, z)L(i − 1)]. \end{matrix}

Let

V_{0} = CM(ℤ + \frac{1}{2})^{0}, V_{2} = CM(ℤ + \frac{1}{2})^{1},
V_{1} = CM(ℤ)^{0}, V_{3} = CM(ℤ)^{1},
V = V_{0} ⊕ V_{1} ⊕ V_{2} ⊕ V_{3},
V = V_{0} ⊕ V_{2},

and extend the Hermitian form ( , ) to V so that CM(ℤ + \frac{1}{2}) and CM(ℤ) are orthogonal. We give the set of subscripts {0, 1, 2, 3} used to index these Virmodules (sectors) the group structure of ℤ_{4}. In the following table we give an orthonormal basis for the space (V_{i})_{Δ_{i}} of vectors of minimal weight Δ_{i} in sector V_{i}. These are vacuum vectors for Vir.

V_{0}  u_{0} = vac  Δ_{0} = 0
V_{1}  u_{1} = vac′  Δ_{1} = \frac{1}{16}
V_{2}  u_{2} = \frac{1}{\sqrt{2}}e(−\frac{1}{2})vac  Δ_{2} = \frac{1}{2}
V_{3}  u_{3} = e(0)vac′  Δ_{3} = \frac{1}{16}

For m, n ∈ ℤ_{4}, define

Δ(m, n) = Δ_{m} + Δ_{n} − Δ_{m+n}.

Then we have Δ(0, n) = 0 and

Δ(1, 1) = −\frac{3}{8},  Δ(1, 2) = \frac{1}{2},  Δ(1, 3) = \frac{1}{8},
Δ(2, 2) = 1,  Δ(2, 3) = \frac{1}{2},  Δ(3, 3) = −\frac{3}{8}.

For m, n, p ∈ ℤ_{4}, define the totally symmetric function

Δ(m, n, p) = Δ(m, n) + Δ(m, p) − Δ(m, n + p)
= Δ_{m} + Δ_{n} + Δ_{p} − Δ_{m+n} − Δ_{m+p} − Δ_{n+p} + Δ_{m+n+p}.

Then we have Δ(0, n, p) = 0 and

Δ(1, 1, 1) = −\frac{5}{4}, Δ(1, 1, 2) = 0, Δ(1, 1, 3) = −\frac{1}{4},
Δ(1, 2, 2) = 1, Δ(1, 2, 3) = 1, Δ(1, 3, 3) = −\frac{1}{4},
Δ(2, 2, 2) = 2, Δ(2, 2, 3) = 1, Δ(2, 3, 3) = 0, Δ(3, 3, 3) = −\frac{5}{4}.

3. Intertwining Operators

We wish to extend the definition of vertex operators so that Y (v_{1}, z)v_{2} is defined for all v_{1}, v_{2} ∈ V. The new operators we need to define, when v_{1} ∈ V_{1} ⊕ V_{3} are called “intertwining operators”. If v_{i} ∈ V_{n_{i}} then our basic assumptions are that the components of Y (v_{1}, z)v_{2} are in V_{n_{1}+n_{2}}, that

[L(−1), Y (v_{1}, z)] = \frac{d}{dz} Y (v_{1}, z)

and that Corollaries 4 and 5 are valid for all v ∈ V. Let us see to what extent these assumptions determine the intertwiners. Since the form (·, ·) on V is nondegenerate, Y (v_{1}, z)v_{2} is determined if the one-point function (Y (v_{1}, z)v_{2}, v_{3}) is known for all v_{3} ∈ V. Suppose v_{i} ∈ V_{n_{i}} for 1 ≤ i ≤ 3, L(0)v_{i} = λ_{i}v_{i} and n_{1} + n_{2} = n_{3} (otherwise the one-point function is zero). For any 0 < m ∈ ℤ we have

(Y (v_{1}, z)v_{2}, L(−m)v_{3}) = (L(m)Y (v_{1}, z)v_{2}, v_{3})

= (Y (v_{1}, z)L(m)v_{2}, v_{3}) + ([L(m), Y (v_{1}, z)]v_{2}, v_{3})

= (Y (v_{1}, z)L(m)v_{2}, v_{3}) + ∑_{0≤k∈ℤ} \binom{m + 1}{k}z^{m+1−k}(Y (L(k − 1)v_{1}, z)v_{2}, v_{3})

= (Y (v_{1}, z)L(m)v_{2}, v_{3}) + z^{m+1} \frac{d}{dz} (Y (v_{1}, z)v_{2}, v_{3})

+ (m + 1)z^{m}λ_{1}(Y (v_{1}, z)v_{2}, v_{3})

+ ∑_{1<k∈ℤ} \binom{m + 1}{k}z^{m+1−k}(Y (L(k − 1)v_{1}, z)v_{2}, v_{3})

where we have used

Y (L(−1)v_{1}, z) = [L(−1), Y (v_{1}, z)] = \frac{d}{dz} Y (v_{1}, z)

which comes from Corollary 5 with m = 0, and our assumptions. This shows that (Y (v_{1}, z)v_{2}, L(−m)v_{3}) is determined by the one-point functions (Y (v′_{1}, z)v′_{2}, v_{3}) where wt(v′_{1}) ≤ wt(v_{1}) and wt(v′_{2}) ≤ wt(v_{2}). This reduces the problem to the case when v_{3} has minimal weight in its sector, that is, when v_{3} is a vacuum vector for Vir.

Assuming that L(k)v_{3} = 0 for all 0 < k ∈ ℤ, for 0 ≤ m ∈ ℤ we have

(Y (L(−m − 1)v_{1}, z)v_{2}, v_{3}) =

∑_{0≤i∈ℤ} \binom{m + i − 1}{i}[z^{i}(L(−m − i − 1)Y (v_{1}, z)v_{2}, v_{3})

− (−1)^{m}z^{−m−i}(Y (v_{1}, z)L(i − 1)v_{2}, v_{3})]

= ∑_{0≤i∈ℤ} \binom{m + i − 1}{i}[z^{i}(Y (v_{1}, z)v_{2}, L(m + i + 1)v_{3})

− (−1)^{m}z^{−m−i}(Y (v_{1}, z)L(i − 1)v_{2}, v_{3})].

But since m ≥ 0 and i ≥ 0, L(m + i + 1)v_{3} = 0 by assumption, so the above equals

\begin{matrix}− (−1)^{m} ∑_{1<i∈ℤ} \binom{m + i − 1}{i}z^{−m−i}(Y (v_{1}, z)L(i − 1)v_{2}, v_{3}) \\ − (−1)^{m}z^{−m}(Y (v_{1}, z)L(−1)v_{2}, v_{3}) − (−1)^{m}z^{−m−1}λ_{2}(Y (v_{1}, z)v_{2}, v_{3}). \end{matrix}

Now we use the fact that

(Y (v_{1}, z)L(−1)v_{2}, v_{3}) = (L(−1)Y (v_{1}, z)v_{2}, v_{3}) − ([L(−1), Y (v_{1}, z)]v_{2}, v_{3})

= (Y (v_{1}, z)v_{2}, L(1)v_{3}) − \frac{d}{dz} (Y (v_{1}, z)v_{2}, v_{3})

= −\frac{d}{dz} (Y (v_{1}, z)v_{2}, v_{3}).

So (Y (L(−m − 1)v_{1}, z)v_{2}, v_{3}) for m ≥ 0 and v_{3} a vacuum vector is determined by the one-point functions (Y (v_{1}, z)v′_{2}, v_{3}) for wt(v′_{2}) ≤ wt(v_{2}). Thus we are reduced to the case when v_{1} and v_{3} are vacuum vectors.

With that assumption, for 0 < m ∈ ℤ, we have

(Y (v_{1}, z)L(−m)v_{2}, v_{3}) = (L(−m)Y (v_{1}, z)v_{2}, v_{3}) − ([L(−m), Y (v_{1}, z)]v_{2}, v_{3})

(Y (v_{1}, z)v_{2}, L(m)v_{3}) − ∑_{0≤k∈ℤ} \binom{−m + 1}{k}z^{−m+1−k}(Y (L(k − 1)v_{1}, z)v_{2}, v_{3})

= −z^{−m+1}(Y (L(−1)v_{1}, z)v_{2}, v_{3}) − (−m + 1)z^{−m}(Y (L(0)v_{1}, z)v_{2}, v_{3})

= −z^{−m+1} \frac{d}{dz} (Y (v_{1}, z)v_{2}, v_{3}) + (m − 1)z^{−m}λ_{1}(Y (v_{1}, z)v_{2}, v_{3})

showing that (Y (v_{1}, z)L(−m)v_{2}, v_{3}) is determined by (Y (v_{1}, z)v_{2}, v_{3}). Thus, we have reduced the general one-point function to the special case when v_{1}, v_{2} and v_{3} are vacuum vectors. These complex numbers are called the structure constants, and we will see if they can be consistently determined by conditions we want for the two-point functions.

4. Two-Point Functions and Hypergeometric Differential Equations

For 1 ≤ i ≤ 4 let v_{i} ∈ V_{n_{i}} with wt(v_{i}) = |v_{i}| = N_{i} + Δ_{n_{i}} for 0 ≤ N_{i} ∈ ℤ and suppose n_{1} + n_{2} + n_{3} = n_{4}. Let

G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2}) = (Y (v_{1}, z_{1})Y (v_{2}, z_{2})v_{3}, v_{4}),
H(v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2}) = (Y (Y (v_{1}, z)v_{2}, z_{2})v_{3}, v_{4}).

Then we have

\begin{matrix}G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2}) = \\ z_{1}^{N_{4}−N_{1}−Δ(n_{1},n_{2}+n_{3})}z_{2}^{−N_{2}−N_{3}−Δ(n_{2},n_{3})} ∑_{0≤k∈ℤ} (\frac{z_{2}}{z_{1}} )^{k} Φ_{k}(v_{1}, v_{2}, v_{3}, v_{4}) \end{matrix}

where

Φ_{k} = Φ_{k}(v_{1}, v_{2}, v_{3}, v_{4}) = (Y_{k+Δ_{n_{2}+n_{3}} −|v_{4}|}(v_{1}) Y_{−k−Δ_{n_{2}+n_{3}} +|v_{3}|}(v_{2})v_{3}, v_{4})

and

\begin{matrix}H(v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2}) = \\ z^{−N_{1}−N_{2}−Δ(n_{1},n_{2})}z_{2}^{N_{4}−N_{3}−Δ(n_{1}+n_{2},n_{3})} ∑_{0≤k∈ℤ} (\frac{z}{z_{2}} )^{k} Ψ_{k}(v_{1}, v_{2}, v_{3}, v_{4}) \end{matrix}

where

Ψ_{k} = Ψ_{k}(v_{1}, v_{2}, v_{3}, v_{4}) = (Y_{|v_{3}|−|v_{4}|}(Y_{−k−Δ_{n_{1}+n_{2}} +|v_{2}|}(v_{1})v_{2})v_{3}, v_{4}).

Definition 6. For m, n, p ∈ ℤ_{4} let

A_{mn} = (Y_{Δ_{n}−Δ_{m+n}}(u_{m})u_{n}, u_{m+n}),
K_{mnp} = Φ_{0}(u_{m}, u_{n}, u_{p}, u_{m+n+p}) = A_{np}A_{m,n+p},
M_{mnp} = Ψ_{0}(u_{m}, u_{n}, u_{p}, u_{m+n+p}) = A_{mn}A_{m+n,p}.

Lemma 7. Let v ∈ V_{i}, 0 ≤ i ≤ 3, be of weight Δ_{i}. Then we have L(−1)^{2}v = γL(−2)v where γ is given by the following table:

i  0  1  2  3
γ  0  \frac{3}{4}  \frac{4}{3}  \frac{3}{4}

Theorem 8. For 1 ≤ i ≤ 4 let v_{i} ∈ V_{n_{i}} be vacuum vectors for Vir. For i = 1, 2 let Y_{i} = Y (v_{i}, z_{i}), suppose that L(−1)^{2}v_{3} = γL(−2)v_{3} and let

G = G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2}) = (Y_{1}Y_{2}v_{3}, v_{4}).

Then G satisfies the partial differential equation

(∂_{1} + ∂_{2})^{2}G + γ(z_{1}^{−1}∂_{1} + z_{2}^{−1}∂_{2})G − γ(z_{1}^{−2}Δ_{n_{1}} + z_{2}^{−2}Δ_{n_{2}})G = 0.

Proof. We have

(Y_{1}Y_{2}L(−1)^{2}v_{3}, v_{4}) = (Y_{1}L(−1)Y_{2}L(−1)v_{3}, v_{4}) − (Y_{1}[L(−1), Y_{2}]L(−1)v_{3}, v_{4})
= (L(−1)Y_{1}Y_{2}L(−1)v_{3}, v_{4}) − ([L(−1), Y_{1}]Y_{2}L(−1)v_{3}, v_{4})
− ∂_{2}(Y_{1}Y_{2}L(−1)v_{3}, v_{4})
= −(∂_{1} + ∂_{2})(Y_{1}Y_{2}L(−1)v_{3}, v_{4})
= (∂_{1} + ∂_{2})^{2}(Y_{1}Y_{2}v_{3}, v_{4})

and

(Y_{1}Y_{2}L(−2)v_{3}, v_{4}) = (Y_{1}L(−2)Y_{2}v_{3}, v_{4}) − (Y_{1}[L(−2), Y_{2}]v_{3}, v_{4})
= −([L(−2), Y_{1}]Y_{2}v_{3}, v_{4}) − (Y_{1}[L(−2), Y_{2}]v_{3}, v_{4})
= −(z_{1}^{−1}∂_{1} − z_{1}^{−2}Δ_{n_{1}})(Y_{1}Y_{2}v_{3}, v_{4})
− (z_{2}^{−1}∂_{2} − z_{2}^{−2}Δ_{n_{2}})(Y_{1}Y_{2}v_{3}, v_{4})

so the relation

(Y_{1}Y_{2}L(−1)^{2}v_{3}, v_{4}) = γ(Y_{1}Y_{2}L(−2)v_{3}, v_{4})

gives the result. ■ Let v_{i} = u_{n_{i}} for 1 ≤ i ≤ 4, and suppose n_{4} = n_{1} + n_{2} + n_{3}. If we write

G_{n_{1}n_{2}n_{3}}(z_{1}, z_{2}) = K_{n_{1}n_{2}n_{3}}z_{1}^{−A}z_{2}^{−B} (1 − \frac{z_{2}}{z_{1}} )^{−C} F (z_{2}/z_{1})

for

A = Δ(n_{1}, n_{2} + n_{3}), B = Δ(n_{2}, n_{3}), C = Δ(n_{1}, n_{2})

then the differential equation for G becomes a differential equation for F. Letting x = z_{2}/z_{1} and using the fact that

Δ(n, n_{3})(Δ(n, n_{3}) + 1) = γ(Δ(n, n_{3}) + Δ_{n})

for any n = 0, 1, 2, 3, we get the ordinary differential equation

x(1 − x)F′′ + [(γ − 2B) − (2A + 2 − 2C − γ)x]F′ + [2AB − 2BC + γC]F = 0

for F. The hypergeometric function

2F_{1}(a, b, c; z) = ∑_{n=0}^{∞} \frac{(a)_{n} (b)_{n}}{n! (c)_{n}} z^{n}

where (a)_{n} = a(a+1) … (a+n−1), is a solution to the hypergeometric differential equation

x(1 − x)w′′ + (c − (a + b + 1)x)w′ − abw = 0.

Therefore, we find that

G_{n_{1}n_{2}n_{3}}(z_{1}, z_{2}) = K_{n_{1}n_{2}n_{3}}z_{1}^{−A}z_{2}^{−B} (1 − \frac{z_{2}}{z_{1}} )^{−C} _{2}F_{1}(a, b, c; z_{2}/z_{1})

where

c = γ − 2B,  ab = (2B − γ)C − 2AB,  a + b = 2A − 2C − γ + 1.

In fact, we find that a = −Δ(n_{1}, n_{2}, n_{3}).

Theorem 9. For 1 ≤ i ≤ 4 let v_{i} ∈ V_{n_{i}} be vacuum vectors for Vir. Suppose that L(−1)^{2}v_{3} = γL(−2)v_{3} and let

H = H(v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2}) = (Y (Y (v_{1}, z)v_{2}, z_{2})v_{3}, v_{4}).

Then H satisfies the partial differential equation

∂_{2}^{2}H + γ(z_{2}^{−1}∂_{2} − (z_{2} + z)^{−1}z_{2}^{−1}z∂)H − γ((z_{2} + z)^{−2}Δ_{n_{1}} + z_{2}^{−2}Δ_{n_{2}})H = 0.

Let v_{i} = u_{n_{i}} for 1 ≤ i ≤ 4, and suppose n_{4} = n_{1} + n_{2} + n_{3}. If we write

H_{n_{1}n_{2}n_{3}}(z, z_{2}) = M_{n_{1}n_{2}n_{3}}z_{2}^{−A′}z^{−B′} (1 + \frac{z}{z_{2}} )^{−C′} F (z/z_{2})

for

A′ = Δ(n_{1} + n_{2}, n_{3}),  B′ = Δ(n_{1}, n_{2}),  C′ = Δ(n_{1}, n_{3})

then the differential equation for H becomes a differential equation for F. Letting x = z/z_{2} and using the facts that

Δ(n, n_{3})(Δ(n, n_{3}) + 1) = γ(Δ(n, n_{3}) + Δ_{n})

for any n = 0, 1, 2, 3, and

(A′ − C′)(A′ + C′ + 1 − γ) = γ(Δ_{n_{2}} − B′)

we get the ordinary differential equation

x(1+x)F′′ +[(2A′ +2−2γ)+(2A′ +2−2C′ −γ)x]F′ −[2(A′ −C′)C′ +γB′]F = 0

for F. Therefore, we find that

H_{n_{1}n_{2}n_{3}}(z, z_{2}) = M_{n_{1}n_{2}n_{3}}z_{2}^{−A′}z^{−B′} (1 + \frac{z}{z_{2}} )^{−C′} _{2}F_{1}(a′, b′, c′; −z/z_{2})

where

c′ = 2(A′ +1−γ),  a′b′ = 2(C′ −A′)C′ −γB′,  a′ +b′ = 2A′ −2C′ −γ +1.

In fact, we find that a′ = a = −Δ(n_{1}, n_{2}, n_{3}).

For n_{1}, n_{2} ∈ {1, 3} we give the values for a, b, c, a′, b′, c′, A, B, C, A′, B′, C′ as n_{3} = 0, 1, 2, 3 in tables at the end of the paper.

5. Transformation Formulas and Rationalization

The following Lemmas are well-known results in the theory of hypergeometric functions. The formula in Lemma 10 can easily be obtained from formula (2) on page 92 of [L], and the formulas in Lemma 11 are the formulas numbered (1.4.5) and (1.4.13) on page 14 of [S].

Lemma 10. (Kummer’s Quadratic Transformation Formula) For |4z| < |1 − z|^{2} we have

2F_{1}(a, b, 1 + a − b; z) = (1 − z)^{−a} _{2}F_{1}(\frac{1}{2}a, \frac{1}{2} + \frac{1}{2}a − b, 1 + a − b; −4z/(1 − z)^{2}).

Lemma 11. (Gauss Recurrence Relations for Contiguous Functions) For a and b nonzero, we have

(c − a − 1) _{2}F_{1}(a, b, c; z) + a _{2}F_{1}(a + 1, b, c; z) = (c − 1) _{2}F_{1}(a, b, c − 1; z),

c(1 − z) _{2}F_{1}(a, b, c; z) + (c − a)z _{2}F_{1}(a, b, c + 1; z) = c _{2}F_{1}(a, b − 1, c; z).

Corollary 12. For |4z| < |1 − z|^{2} we have

2F_{1}(\frac{1}{4}, \frac{3}{4}, \frac{3}{2}; −4z/(1 − z)^{2}) = (1 − z)^{\frac{1}{2}} ,

2F_{1}(−\frac{1}{4}, \frac{1}{4}, \frac{1}{2}; −4z/(1 − z)^{2}) = (1 − z)^{− \frac{1}{2}} ,

2F_{1}(\frac{5}{4}, \frac{3}{4}, \frac{3}{2}; −4z/(1 − z)^{2}) = \frac{(1 − z)^{\frac{3}{2}}}{1 + z} ,

2F_{1}(\frac{1}{4}, \frac{3}{4}, \frac{1}{2}; −4z/(1 − z)^{2}) = \frac{(1 − z)^{\frac{1}{2}}}{1 + z} .

Proof. The first two equations follow from Kummer’s quadratic transformation formula by taking a = \frac{1}{2} or a = −\frac{1}{2}, b = 0. The last two equations follow from the Gauss recurrence relations by taking a = \frac{1}{4}, b = \frac{3}{4}, c = \frac{3}{2} or c = \frac{1}{2}.

We wish to rationalize the correlation functions so that we can apply the usual contour integration techniques to obtain algebraic relations for the vertex operators. To do this we will use substitutions which give four sheeted coverings of the t-plane, where t = z_{2}/z_{1}. The functions G and H have possible poles at t = 0, t = 1 and t = ∞, as well as various cuts, but the four sheeted coverings are branched at these points, and each of these three points has only two points lying above it. First, in G_{n_{1}n_{2}n_{3}}(z_{1}, z_{2}), we use the substitution

t^{1/2} = (\frac{z_{2}}{z_{1}} )^{\frac{1}{2}} = \frac{2x_{0}}{1 + x_{0}^{2}} = \frac{2x_{∞}}{1 + x_{∞}^{2}} .

In G_{n_{2}n_{1}n_{3}}(z_{2}, z_{1}) we use the substitution

t^{−1/2} = (\frac{z_{1}}{z_{2}} )^{\frac{1}{2}} = \frac{2x_{i}}{1 + x_{i}^{2}} = \frac{2x_{−i}}{1 + x_{−i}^{2}} .

In H_{n_{1}n_{2}n_{3}}(z, z_{2}) we use the substitution

(t^{−1} − 1)^{1/2} = (\frac{z_{1} − z_{2}}{z_{2}} )^{\frac{1}{2}} = \frac{2x_{1}}{1 − x_{1}^{2}} = \frac{2x_{−1}}{1 − x_{−1}^{2}} .

In this notation, for α ∈ {0, ∞, i, −i, 1, −1}, the variable x_{α} is local at the point α on the four sheeted cover. These points are all related to each other by linear fractional transformations, and we let x be a global variable on the genus zero covering. Below we will discuss the regions of absolute convergence after the substitutions have been made. Let

R_{x} = ℂ[x, x^{−1}(x^{4} − 1)^{−1}]

be the ring of rational functions in x with possible poles at those six points. Note that if f(x) ∈ R_{x} and μ(x) is any of the linear fractional transformations I(x) = x,

A(x) = \frac{1}{x}, B(x) = \frac{x + i}{ix + 1}, C(x) = \frac{ix + 1}{x + i}, D(x) = \frac{x − 1}{x + 1} or E(x) = \frac{1 + x}{1 − x}

then f(μ(x)) ∈ R_{x}. Also note that R_{x} is preserved by the operator d/dx. Note that

1 − \frac{z_{2}}{z_{1}} = 1 − t = \frac{(1 − x_{0}^{2})^{2}}{(1 + x_{0}^{2})^{2}} = \frac{(1 − x_{0}^{2})^{4}}{(1 − x_{0}^{4})^{2}}

and

1 + \frac{z}{z_{2}} = t^{−1} = \frac{(1 + x_{1}^{2})^{2}}{(1 − x_{1}^{2})^{2}} = \frac{(1 + x_{1}^{2})^{4}}{(1 − x_{1}^{4})^{2}} .

The S_{3} permutation group acting on the three points t = 0, 1, ∞, lifts to an S_{4} permutation group acting on the six points 0, ∞, i, −i, 1, −1. For a ∈ {1, −1, i, −i} let F_{a}(x) = ax. The group of 24 linear fractional transformations giving these permutations is generated by B(x) and D(x), each of which has order 4. It consists of the compositions of the four functions F_{a}, with the six functions I, A, B, C, D and E. One has the relations BDB^{−1} = F_{i}, DBD^{−1} = F_{−i}, F_{−1}DF_{−1} = D^{−1} and F_{−i}DF_{i} = B. It is easy to check that the correspondence B ↔ (1, 2, 3, 4), D ↔ (1, 3, 2, 4) and F_{i} ↔ (1, 2, 4, 3) determines an isomorphism between this group of 24 transformations and the permutation group S_{4}. In this paper we will not use all 24 transformations, but we will just choose enough to relate local variables at each of the six points. However, we believe that it will be necessary to use all 24 transformations in order to fully understand the algebraic structure of the 2 × 2 B matrices defined just before Theorem 15. That will be a subject for future investigation.

We can choose local variables x_{α} such that

x_{∞} = \frac{1}{x_{0}} ,  x_{i} = \frac{1}{x_{−i}} ,  x_{1} = \frac{−1}{x_{−1}} .

The relations between x_{0} and x_{α} are determined, up to some sign choices, by their relationship to t. We make the choices

x_{0} = \frac{x_{i} + i}{ix_{i} + 1} so x_{0} = \frac{−x_{−i} + i}{ix_{−i} − 1}, x_{0} = \frac{x_{1} + 1}{−x_{1} + 1} so x_{0} = \frac{x_{−1} − 1}{x_{−1} + 1},

so that

x_{i} = \frac{−x_{0} + i}{ix_{0} − 1} and x_{1} = \frac{x_{0} − 1}{x_{0} + 1}.

Until further notice we assume that n_{1}, n_{2}, n_{3} ∈ {1, 3}.

Theorem 13. For α = 0 or α = ∞, after the substitution (z_{2}/z_{1})^{1/2} =

2x_{α}/(1 + x_{α}^{2}), the series z \frac{1}{8} 2 G_{n_{1}n_{2}n_{3}}(z_{1}, z_{2}) converges absolutely to a function of x_{α} in the domain |x_{α}| < \sqrt{3 − 2\sqrt{2}} and in that domain we have

(1 − x_{α}^{4})^{\frac{1}{4}} z \frac{1}{8} 2 G_{n_{1}n_{2}n_{3}}(z_{1}, z_{2}) = \{ \begin{matrix}2x_{α}K_{n_{1}n_{2}n_{3}} & if n_{2} = n_{3} \\ K_{n_{1}n_{2}n_{3}} & if n_{2} ≠ n_{3} \end{matrix}

For α = i or α = −i the above assertions with z_{1} and z_{2} switched and with n_{1} and n_{2} switched, are true.

Proof. In the expression for G_{n_{1}n_{2}n_{3}}(z_{1}, z_{2}) obtained after Theorem 8 we see it as a series in t = z_{2}/z_{1} which converges absolutely for |t| < 1. But in order to use Corollary 12 with z = −x_{α}^{2} we need \frac{|2x_{α}|}{|1+x_{α}^{2}|} < 1. Using polar coordinates x_{α} = re^{iθ} this condition is equivalent to 4r^{2} < 1 + 2r^{2} cos(2θ) + r^{4}. This is certainly true when 4r^{2} < 1 − 2r^{2} + r^{4}, that is, when 0 < r^{4} − 6r^{2} + 1. The parabola y = x^{2} − 6x + 1 is positive for x < 3 − 2\sqrt{2} so with 0 < r < \sqrt{3 − 2\sqrt{2}} we are guaranteed to have the desired condition.

Using the values of A, B, C, a, b, c in the tables, and Corollary 12, after some algebra one gets the explicit formula stated in the theorem. ■

Theorem 14. For α = 1 or α = −1, after the substitution (z/z_{1})^{1/2} = 2x_{α}/(1 − x_{α}^{2}), the series z^{\frac{1}{8}} H_{n_{1}n_{2}n_{3}}(z, z_{2}) converges absolutely to a function of x_{α} in the domain |x_{α}| < \sqrt{3 − 2\sqrt{2}} and in that domain we have

(1 − x_{α}^{4})^{\frac{1}{4}} z^{\frac{1}{8}} H_{n_{1}n_{2}n_{3}}(z, z_{2}) = \{ \begin{matrix}2x_{α}M_{n_{1}n_{2}n_{3}} & if n_{1} = n_{2} \\ M_{n_{1}n_{2}n_{3}} & if n_{1} ≠ n_{2} \end{matrix}

Proof. In the expression for H_{n_{1}n_{2}n_{3}}(z, z_{2}) obtained after Theorem 9 we see it as a series in t^{−1} − 1 = z/z_{2} which converges absolutely for |z/z_{2}| < 1. But in order to use Corollary 12 with z = x_{α}^{2} we need \frac{|2x_{α}|}{|1−x_{α}^{2}|} < 1. Using polar coordinates x_{α} = re^{iθ} this condition is equivalent to 4r^{2} < 1 − 2r^{2} cos(2θ) + r^{4}. This is certainly true when 4r^{2} < 1 − 2r^{2} + r^{4}. So the analysis proceeds just as in Theorem 13.

Using the values of A′, B′, C′, a′, b′, c′ in the tables, and Corollary 12, after some algebra one gets the explicit formula stated in the theorem. ■

In order to get the most general form of the Jacobi identity, we will eventually apply the Cauchy residue theorem to matrix valued differential forms on the four sheeted covering. Let

[G(z_{1}, z_{2})] = [ \begin{matrix}G_{111}(z_{1}, z_{2}) & G_{311}(z_{1}, z_{2}) & G_{333}(z_{1}, z_{2}) & G_{133}(z_{1}, z_{2}) \\ G_{331}(z_{1}, z_{2}) & G_{131}(z_{1}, z_{2}) & G_{113}(z_{1}, z_{2}) & G_{313}(z_{1}, z_{2}) \end{matrix} ] ,

[G(z_{2}, z_{1})] = [ \begin{matrix}G_{111}(z_{2}, z_{1}) & G_{311}(z_{2}, z_{1}) & G_{333}(z_{2}, z_{1}) & G_{133}(z_{2}, z_{1}) \\ G_{331}(z_{2}, z_{1}) & G_{131}(z_{2}, z_{1}) & G_{113}(z_{2}, z_{1}) & G_{313}(z_{2}, z_{1}) \end{matrix} ]

and

[H(z, z_{2})] = [ \begin{matrix}H_{111}(z, z_{2}) & H_{331}(z, z_{2}) & H_{113}(z, z_{2}) & H_{333}(z, z_{2}) \\ H_{311}(z, z_{2}) & H_{131}(z, z_{2}) & H_{313}(z, z_{2}) & H_{133}(z, z_{2}) \end{matrix} ] .

Note that in the matrix [G(z_{1}, z_{2})] = [G_{n_{1}n_{2}n_{3}}(z_{1}, z_{2})] the first row has n_{2} = n_{3}, the second row has n_{2} ≠ n_{3}, while each column has the same n_{3} but different values of n_{1}. The [G(z_{2}, z_{1})] matrix has the same pattern of subscripts as the [G(z_{1}, z_{2})] matrix, but the variables z_{1} and z_{2} are switched. In the matrix [H(z, z_{2})] the first row has n_{1} = n_{2}, the second row has n_{1} ≠ n_{2}, while each column has the same n_{2} and the same n_{3}. Then, for α = 0 or α = ∞, we have

(1 − x_{α}^{4})^{\frac{1}{4}} z \frac{1}{8} 2 [G(z_{1}, z_{2})]_{α} = [ \begin{matrix}2K_{111}x_{α} & 2K_{311}x_{α} & 2K_{333}x_{α} & 2K_{133}x_{α} \\ K_{331} & K_{131} & K_{113} & K_{313} \end{matrix} ] ,

for α = i or α = −i, we have

(1 − x_{α}^{4})^{\frac{1}{4}} z \frac{1}{8} 1 [G(z_{2}, z_{1})]_{α} = [ \begin{matrix}2K_{111}x_{α} & 2K_{311}x_{α} & 2K_{333}x_{α} & 2K_{133}x_{α} \\ K_{331} & K_{131} & K_{113} & K_{313} \end{matrix} ] ,

and for α = 1 or α = −1, we have

(1 − x_{α}^{4})^{\frac{1}{4}} z^{\frac{1}{8}} [H(z, z_{2})]_{α} = [ \begin{matrix}2M_{111}x_{α} & 2M_{331}x_{α} & 2M_{113}x_{α} & 2M_{333}x_{α} \\ M_{311} & M_{131} & M_{313} & M_{133} \end{matrix} ] .

These equalities mean that the left sides are series which, after an appropriate substitution, converge absolutely in a small disk around the appropriate x_{α} to the globally defined matrix valued functions given on the right sides.

For any f(x) ∈ R_{x} we can use linear fractional transformations to express the globally defined matrix valued differential form

[ \begin{matrix}2K_{111}x & 2K_{311}x & 2K_{333}x & 2K_{133}x \\ K_{331} & K_{131} & K_{113} & K_{313} \end{matrix} ] f(x)dx

in terms of the appropriate local coordinate variable x_{α} at each of the six possible poles. The residue at each pole is then easily found. By the residue theorem, the sum of all the residues is zero, giving us a relation among the correlation functions. That relation is the generalization of the Jacobi Identity which we seek. The residue at x = 0 can be found immediately from the global expression. To find the residue at x = ∞, use x_{∞} = x_{0}^{−1} and find the residue at x_{∞} = 0. Leaving off the function f(x) and the differential dx for the moment, we have

[ \begin{matrix}2K_{111}x_{0} & 2K_{311}x_{0} & 2K_{333}x_{0} & 2K_{133}x_{0} \\ K_{331} & K_{131} & K_{113} & K_{313} \end{matrix} ]
= [ \begin{matrix}2K_{111}x_{∞}^{−1} & 2K_{311}x_{∞}^{−1} & 2K_{333}x_{∞}^{−1} & 2K_{133}x_{∞}^{−1} \\ K_{331} & K_{131} & K_{113} & K_{313} \end{matrix} ]
=\frac{1}{x_{∞}} [ \begin{matrix}2K_{111} & 2K_{311} & 2K_{333} & 2K_{133} \\ K_{331}x_{∞} & K_{131}x_{∞} & K_{113}x_{∞} & K_{313}x_{∞} \end{matrix} ]
=\frac{1}{x_{∞}} [ \begin{matrix}0 & \frac{2K_{111}}{K_{331}} \\ \frac{K_{331}}{2K_{111}} & 0 \end{matrix} ] [ \begin{matrix}2K_{111}x_{∞} & 2K_{311}x_{∞} & 2K_{333}x_{∞} & 2K_{133}x_{∞} \\ K_{331} & K_{131} & K_{113} & K_{313} \end{matrix} ]

= (1 − x_{∞}^{4})^{\frac{1}{4}} z \frac{1}{8} 2 \frac{1}{x_{∞}} [ \begin{matrix}0 & \frac{2K_{111}}{K_{331}} \\ \frac{K_{331}}{2K_{111}} & 0 \end{matrix} ] [G(z_{1}, z_{2})]_{∞}

if and only if the following conditions are consistent:

K_{111}K_{131} = K_{331}K_{311},  K_{111}K_{113} = K_{331}K_{333},  K_{111}K_{313} = K_{331}K_{133}.

To find the residue at x = i use x_{0} = (x_{i} + i)/(ix_{i} + 1) and find the residue at x_{i} = 0. Without imposing any further conditions, we have

[ \begin{matrix}2K_{111}x_{0} & 2K_{311}x_{0} & 2K_{333}x_{0} & 2K_{133}x_{0} \\ K_{331} & K_{131} & K_{113} & K_{313} \end{matrix} ]
= [ \begin{matrix}2K_{111} \frac{x_{i}+i}{ix_{i}+1} & 2K_{311} \frac{x_{i}+i}{ix_{i}+1} & 2K_{333} \frac{x_{i}+i}{ix_{i}+1} & 2K_{133} \frac{x_{i}+i}{ix_{i}+1} \\ K_{331} & K_{131} & K_{113} & K_{313} \end{matrix} ]
=\frac{1}{ix_{i} + 1} [ \begin{matrix}2K_{111}(x_{i} + i) & 2K_{311}(x_{i} + i) & 2K_{333}(x_{i} + i) & 2K_{133}(x_{i} + i) \\ K_{331}(ix_{i} + 1) & K_{131}(ix_{i} + 1) & K_{113}(ix_{i} + 1) & K_{313}(ix_{i} + 1) \end{matrix} ]
=\frac{1}{ix_{i} + 1} [ \begin{matrix}1 & \frac{2iK_{111}}{K_{331}} \\ \frac{iK_{331}}{2K_{111}} & 1 \end{matrix} ] [ \begin{matrix}2K_{111}x_{i} & 2K_{311}x_{i} & 2K_{333}x_{i} & 2K_{133}x_{i} \\ K_{331} & K_{131} & K_{113} & K_{313} \end{matrix} ]

= (1 − x_{i}^{4})^{\frac{1}{4}} z \frac{1}{8} 1 \frac{1}{ix_{i} + 1} [ \begin{matrix}1 & \frac{2iK_{111}}{K_{331}} \\ \frac{iK_{331}}{2K_{111}} & 1 \end{matrix} ] [G(z_{2}, z_{1})]_{i}.

To find the residue at x = −i use x_{0} = (−x_{−i} + i)/(ix_{−i} − 1) and find the residue at x_{−i} = 0. Without imposing any new conditions, we have

[ \begin{matrix}2K_{111}x_{0} & 2K_{311}x_{0} & 2K_{333}x_{0} & 2K_{133}x_{0} \\ K_{331} & K_{131} & K_{113} & K_{313} \end{matrix} ]
=\frac{−1}{ix_{−i} −1}[ \begin{matrix}2K_{111}(x_{−i} − i) & 2K_{311}(x_{−i} − i) & 2K_{333}(x_{−i} − i) & 2K_{133}(x_{−i} − i) \\ K_{331}(−ix_{−i}+1) & K_{131}(−ix_{−i}+1) & K_{113}(−ix_{−i} +1) & K_{313}(−ix_{−i} +1) \end{matrix} ]
=\frac{1}{ix_{−i} − 1} [ \begin{matrix}−1 & \frac{2iK_{111}}{K_{331}} \\ \frac{iK_{331}}{2K_{111}} & −1 \end{matrix} ] [ \begin{matrix}2K_{111}x_{−i} & 2K_{311}x_{−i} & 2K_{333}x_{−i} & 2K_{133}x_{−i} \\ K_{331} & K_{131} & K_{113} & K_{313} \end{matrix} ]

= (1 − x_{−i}^{4})^{\frac{1}{4}} z \frac{1}{8} 1 \frac{1}{ix_{−i} − 1} [ \begin{matrix}−1 & \frac{2iK_{111}}{K_{331}} \\ \frac{iK_{331}}{2K_{111}} & −1 \end{matrix} ] [G(z_{2}, z_{1})]_{−i}.

To find the residue at x = −1 use x_{0} = (x_{−1} − 1)/(x_{−1} + 1) and find the residue at x_{−1} = 0. Assuming the previous consistency conditions are true, we have

[ \begin{matrix}2K_{111}x_{0} & 2K_{311}x_{0} & 2K_{333}x_{0} & 2K_{133}x_{0} \\ K_{331} & K_{131} & K_{113} & K_{313} \end{matrix} ]
=\frac{1}{x_{−1} + 1} [ \begin{matrix}2K_{111}(x_{−1} − 1) & 2K_{311}(x_{−1} − 1) & 2K_{333}(x_{−1} − 1) & 2K_{133}(x_{−1} − 1) \\ K_{331}(x_{−1} + 1) & K_{131}(x_{−1} + 1) & K_{113}(x_{−1} + 1) & K_{313}(x_{−1} + 1) \end{matrix} ]
=\frac{1}{x_{−1} + 1} [ \begin{matrix}\frac{K_{111}}{M_{111}} & \frac{−2K_{111}}{M_{311}} \\ \frac{K_{331}}{2M_{111}} & \frac{K_{331}}{M_{311}} \end{matrix} ] [ \begin{matrix}2M_{111}x_{−1} & 2M_{331}x_{−1} & 2M_{113}x_{−1} & 2M_{333}x_{−1} \\ M_{311} & M_{131} & M_{313} & M_{133} \end{matrix} ]
=(1 − x_{−1}^{4})^{\frac{1}{4}} z^{\frac{1}{8}} \frac{1}{x_{−1} + 1} [ \begin{matrix}\frac{K_{111}}{M_{111}} & \frac{−2K_{111}}{M_{311}} \\ \frac{K_{331}}{2M_{111}} & \frac{K_{331}}{M_{311}} \end{matrix} ] [H(z, z_{2})]_{−1}

if and only if the following conditions hold:

\frac{K_{111}}{K_{133}} = \frac{M_{111}}{M_{333}} = \frac{M_{311}}{M_{133}} , \frac{K_{111}}{K_{311}} = \frac{M_{111}}{M_{331}} = \frac{M_{311}}{M_{131}} , \frac{K_{111}}{K_{333}} = \frac{M_{111}}{M_{113}} = \frac{M_{311}}{M_{313}} .

To find the residue at x = 1 use x_{0} = (x_{1} + 1)/(−x_{1} + 1) and find the residue at x_{1} = 0. Without imposing any new conditions, we have

[ \begin{matrix}2K_{111}x_{0} & 2K_{311}x_{0} & 2K_{333}x_{0} & 2K_{133}x_{0} \\ K_{331} & K_{131} & K_{113} & K_{313} \end{matrix} ]
=\frac{1}{−x_{1} + 1} [ \begin{matrix}2K_{111}(x_{1} + 1) & 2K_{311}(x_{1} + 1) & 2K_{333}(x_{1} + 1) & 2K_{133}(x_{1} + 1) \\ K_{331}(−x_{1} + 1) & K_{131}(−x_{1} + 1) & K_{113}(−x_{1} + 1) & K_{313}(−x_{1} + 1) \end{matrix} ]
=\frac{1}{−x_{1} + 1} [ \begin{matrix}\frac{K_{111}}{M_{111}} & \frac{2K_{111}}{M_{311}} \\ \frac{−K_{331}}{2M_{111}} & \frac{K_{331}}{M_{311}} \end{matrix} ] [ \begin{matrix}2M_{111}x_{1} & 2M_{331}x_{1} & 2M_{113}x_{1} & 2M_{333}x_{1} \\ M_{311} & M_{131} & M_{313} & M_{133} \end{matrix} ]

= (1 − x_{1}^{4})^{\frac{1}{4}} z^{\frac{1}{8}} \frac{1}{−x_{1} + 1} [ \begin{matrix}\frac{K_{111}}{M_{111}} & \frac{2K_{111}}{M_{311}} \\ \frac{−K_{331}}{2M_{111}} & \frac{K_{331}}{M_{311}} \end{matrix} ] [H(z, z_{2})]_{1}.

Using Definition 6, the consistency conditions we have found translate into the following conditions on the structure constants A_{mn}:

\frac{A_{30}}{A_{10}} = \frac{A_{12}}{A_{32}} = \frac{A_{11}}{A_{33}} = \frac{A_{31}}{A_{13}} , \frac{A_{30}A_{31}}{A_{10}A_{13}} = 1, A_{21} = A_{23}, A_{01} = A_{03}.

In fact, we know quite a bit more about the structure constants A_{mn}. Since Y (u_{0}, z) = Y (vac, z) = 1, we have A_{0n} = 1 for any n. We also want

Y (v, z)vac = e^{zL(−1)}v ∈ V[z]

which implies the “creation property”

lim_{z→0}Y (v, z)vac = Y_{−wt(v)}(v)vac = v.

These give A_{m0} = 1 for any m. We also have

A_{21} = (Y_{0}(u_{2})u_{1}, u_{3}) = \frac{1}{\sqrt{2}}(e(0)vac′, e(0)vac′) = \frac{1}{\sqrt{2}},

A_{22} = (Y_{1/2}(u_{2})u_{2}, u_{0}) = \frac{1}{2}(e(\frac{1}{2})e(−\frac{1}{2})vac, vac) = \frac{1}{2}⟨e, e⟩(vac, vac) = 1,

A_{23} = (Y_{0}(u_{2})u_{3}, u_{1}) = \frac{1}{\sqrt{2}}(e(0)e(0)vac′, vac′) = \frac{1}{\sqrt{2}}.

We also want the “symmetry” condition, generalizing what is called “skewsymmetry” in [FHL],

z^{Δ(n_{1},n_{2})}Y (v_{1}, z)v_{2} = (−z)^{Δ(n_{1},n_{2})}e^{zL(−1)}Y (v_{2}, −z)v_{1}

for v_{i} ∈ V_{n_{i}}. With v_{1} = u_{m} and v_{2} = u_{n}, after pairing with u_{m+n}, this gives A_{mn} = A_{nm}. Along with the above constraints, this determines all the structure constants A_{mn} except A_{11}, which we normalize to 1/\sqrt{2}, and A_{13}, which we normalize to 1. Here is a table summarizing the results.

A_{mn}  n = 0  n = 1  n = 2  n = 3
m = 0  1  1  1  1
m = 1  1  \frac{1}{\sqrt{2}}  \frac{1}{\sqrt{2}}  1
m = 2  1  \frac{1}{\sqrt{2}}  1  \frac{1}{\sqrt{2}}
m = 3  1  1  \frac{1}{\sqrt{2}}  \frac{1}{\sqrt{2}}

Definition. Define the matrices

\begin{aligned}B_{0} & = [ \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ] ,  B_{∞} = [ \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ] ,  B_{i} = [ \begin{matrix}1 & i \\ i & 1 \end{matrix} ] , \\ B_{−i} & = [ \begin{matrix}−1 & i \\ i & −1 \end{matrix} ] ,  B_{−1} = [ \begin{matrix}1 & −1 \\ 1 & 1 \end{matrix} ] ,  B_{1} = [ \begin{matrix}1 & 1 \\ −1 & 1 \end{matrix} ] . \end{aligned}

Theorem 15. With A_{mn} as given in the above table, and with notations as defined above, we have

(1 − x_{0}^{4})^{\frac{1}{4}} z \frac{1}{8} 2 [G(z_{1}, z_{2})]_{0}
∼(1 − x_{∞}^{4})^{\frac{1}{4}} z \frac{1}{8} 2 x_{∞}^{−1}B_{∞} [G(z_{1}, z_{2})]_{∞}
∼(1 − x_{i}^{4})^{\frac{1}{4}} z \frac{1}{8} 1 (ix_{i} + 1)^{−1}B_{i} [G(z_{2}, z_{1})]_{i}
∼(1 − x_{−i}^{4})^{\frac{1}{4}} z \frac{1}{8} 1 (ix_{−i} − 1)^{−1}B_{−i} [G(z_{2}, z_{1})]_{−i}
∼(1 − x_{−1}^{4})^{\frac{1}{4}} z^{\frac{1}{8}} (x_{−1} + 1)^{−1}B_{−1} [H(z, z_{2})]_{−1}
∼(1 − x_{1}^{4})^{\frac{1}{4}} z^{\frac{1}{8}} (−x_{1} + 1)^{−1}B_{1} [H(z, z_{2})]_{1}

where ∼ means that these series converge absolutely in their appropriate domains to the same globally defined matrix valued function

[ \begin{matrix}x_{0} & x_{0} & x_{0} & x_{0} \\ 1 & 1 & 1 & 1 \end{matrix} ]

on the four sheeted covering.

Note that if we associate to a matrix M = [ \begin{matrix}a & b \\ c & d \end{matrix} ] a linear fractional trans-

formation f(x) = \frac{ax+b}{cx+d} then we associate to the matrix B_{α} the linear fractional transformation

f_{α}(x_{α}) = x_{0} = \frac{ax_{α} + b}{cx_{α} + d}.

6. Inductive formulas

Now let us see how the general two-point functions are determined inductively.

Theorem 16. Let v_{i} ∈ V_{n_{i}}, 1 ≤ i ≤ 4, be eigenvectors for L(0) with L(0)v_{i} = wt(v_{i})v_{i} = |v_{i}|v_{i}. For i = 1, 2 let Y_{i} = Y (v_{i}, z_{i}), ∂_{i} = ∂/∂z_{i}, and recall the notation

G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2}) = (Y_{1}Y_{2}v_{3}, v_{4}).

Then for any k ∈ ℤ we have

G(v_{1}, v_{2}, v_{3}, L(−k)v_{4}; z_{1}, z_{2}) = G(v_{1}, v_{2}, L(k)v_{3}, v_{4}; z_{1}, z_{2})

+ (z_{1}^{k+1}∂_{1} + z_{2}^{k+1}∂_{2})G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2})

+ (k + 1)(z_{1}^{k}|v_{1}| + z_{2}^{k}|v_{2}|)G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2})

+ ∑_{i≥1} \binom{k + 1}{i + 1}z_{1}^{k−i}G(L(i)v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2})

+ ∑_{i≥1} \binom{k + 1}{i + 1}z_{2}^{k−i}G(v_{1}, L(i)v_{2}, v_{3}, v_{4}; z_{1}, z_{2}).

Proof. We have

G(v_{1},v_{2}, v_{3}, L(−k)v_{4}; z_{1}, z_{2}) = (L(k)Y_{1}Y_{2}v_{3}, v_{4})

= ([L(k), Y_{1}]Y_{2}v_{3}, v_{4}) + (Y_{1}[L(k), Y_{2}]v_{3}, v_{4}) + (Y_{1}Y_{2}L(k)v_{3}, v_{4})

= ∑_{i≥0} \binom{k + 1}{i}z_{1}^{k+1−i}(Y (L(i − 1)v_{1}, z_{1})Y_{2}v_{3}, v_{4})

+ ∑_{i≥0} \binom{k + 1}{i}z_{2}^{k+1−i}(Y_{1}Y (L(i − 1)v_{2}, z_{2})v_{3}, v_{4})

+ (Y_{1}Y_{2}L(k)v_{3}, v_{4})

giving the result after separating the i = 0 and the i = 1 terms and reindexing. ■

Theorem 17. Let v_{i} ∈ V_{n_{i}}, 1 ≤ i ≤ 4, be eigenvectors for L(0) with L(0)v_{i} = wt(v_{i})v_{i} = |v_{i}|v_{i}. Let ∂_{1} = ∂/∂z_{1}, ∂ = ∂/∂z, and recall the notation

H(v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2}) = (Y (Y (v_{1}, z)v_{2}, z_{2})v_{3}, v_{4}).

SPINOR CONSTRUCTION OF THE c = 1/2 MINIMAL MODEL 23

Then for any k ∈ ℤ we have H(v_{1}, v_{2}, v_{3}, L(−k)v_{4}; z, z_{2}) = H(v_{1}, v_{2}, L(k)v_{3}, v_{4}; z, z_{2}) + z_{2}^{k+1}∂_{2}H(v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2}) + (k + 1)(z_{2}^{k}|v_{2}| + (z_{2} + z)^{k}|v_{1}|)H(v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2}) + [(z_{2} + z)^{k+1} − z_{2}^{k+1}]∂H(v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2})

+ ∑_{i≥1} \binom{k + 1}{i + 1}z_{2}^{k−i}H(v_{1}, L(i)v_{2}, v_{3}, v_{4}; z, z_{2})

+ ∑_{j≥1} \binom{k + 1}{j + 1}(z_{2} + z)^{k−j}H(L(j)v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2}).

Proof. For brevity, let us write

H = H(v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2}), H(L(k)v_{3}) = H(v_{1}, v_{2}, L(k)v_{3}, v_{4}; z, z_{2}),

H(L(j)v_{1}) = H(L(j)v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2})

and

H(L(i)v_{2}) = H(v_{1}, L(i)v_{2}, v_{3}, v_{4}; z, z_{2}).

We have H(v_{1}, v_{2}, v_{3}, L(−k)v_{4}; z, z_{2}) = ([L(k), Y (Y (v_{1}, z)v_{2}, z_{2})]v_{3}, v_{4}) + (Y (Y (v_{1}, z)v_{2}, z_{2})L(k)v_{3}, v_{4})

= ∑_{i≥0} \binom{k + 1}{i}z_{2}^{k+1−i}(Y (L(i − 1)Y (v_{1}, z)v_{2}, z_{2})v_{3}, v_{4}) + H(L(k)v_{3})

= z_{2}^{k+1}∂_{2}H + H(L(k)v_{3}) + ∑_{i≥0} \binom{k + 1}{i + 1}z_{2}^{k−i}(Y ([L(i), Y (v_{1}, z)]v_{2}, z_{2})v_{3}, v_{4})

+ (k + 1)z_{2}^{k}|v_{2}|H + ∑_{i≥1} \binom{k + 1}{i + 1}z_{2}^{k−i}(Y (Y (v_{1}, z)L(i)v_{2}, z_{2})v_{3}, v_{4})

= z_{2}^{k+1}∂_{2}H + H(L(k)v_{3})

+ ∑_{i≥0} \binom{k + 1}{i + 1}z_{2}^{k−i} (z^{i+1}∂H + (i + 1)z^{i}|v_{1}|H + ∑_{j≥1} \binom{i + 1}{j + 1}z^{i−j}H(L(j)v_{1}))

+ (k + 1)z_{2}^{k}|v_{2}|H + ∑_{i≥1} \binom{k + 1}{i + 1}z_{2}^{k−i}H(L(i)v_{2})

= z_{2}^{k+1}∂_{2}H + H(L(k)v_{3}) + z_{2}^{k+1} ∑_{i≥0} \binom{k + 1}{i + 1} (\frac{z}{z_{2}} )^{i+1} ∂H

+ |v_{1}|z_{2}^{k}H ∑_{i≥0} \binom{k + 1}{i + 1}(i + 1) (\frac{z}{z_{2}} )^{i}

+ z_{2}^{k−j} ∑_{j≥1} H(L(j)v_{1}) ∑_{i≥0} \binom{k + 1}{i + 1}\binom{i + 1}{j + 1} (\frac{z}{z_{2}} )^{i−j}

+ (k + 1)z_{2}^{k}|v_{2}|H + ∑_{i≥1} \binom{k + 1}{i + 1}z_{2}^{k−i}H(L(i)v_{2})

which gives the result after using

\binom{k + 1}{i + 1}(i + 1) = (k + 1)\binom{k}{i}

and

(z_{2} + z)^{m} = z_{2}^{m} (1 + \frac{z}{z_{2}} )^{m} = z_{2}^{m} ∑_{i≥0} \binom{m}{i} (\frac{z}{z_{2}} )^{i} .

■

For 1 ≤ i ≤ 4 let v_{i} ∈ V_{n_{i}} with wt(v_{i}) = |v_{i}| = N_{i} + Δ_{n_{i}} for 0 ≤ N_{i} ∈ ℤ. For n_{1}, n_{2}, n_{3} ∈ {1, 3} we have

Δ(n_{1}, n_{2}+n_{3}) = \{ \begin{matrix}0 & if n_{2} + n_{3} = 0 \\ \frac{1}{2} & if n_{2} + n_{3} = 2 \end{matrix} and Δ(n_{2}, n_{3}) = \{ \begin{matrix}\frac{1}{8} & if n_{2} + n_{3} = 0 \\ −\frac{3}{8} & if n_{2} + n_{3} = 2 \end{matrix}

so that

Δ(n_{1}, n_{2} + n_{3}) + Δ(n_{2}, n_{3}) = \frac{1}{8}

for n_{1}, n_{2}, n_{3} ∈ {1, 3}. Let N = N_{1} + N_{2} + N_{3} − N_{4}. With t = z_{2}/z_{1} and Δ = Δ(n_{1}, n_{2} + n_{3}) we have

G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2}) = t^{N_{1}−N_{4}+Δ}z_{2}^{−N− \frac{1}{8}} Φ(v_{1}, v_{2}, v_{3}, v_{4}; t)

where

Φ(v_{1}, v_{2}, v_{3}, v_{4}; t) = ∑_{0≤k∈ℤ} t^{k}Φ_{k}(v_{1}, v_{2}, v_{3}, v_{4})

is a power series in t. Similarly, with s = z/z_{2} = t^{−1} − 1 and Δ̄ = Δ(n_{1} + n_{2}, n_{3}) we have

H(v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2}) = s^{N_{3}−N_{4}+Δ̄}z^{−N− \frac{1}{8}} Ψ(v_{1}, v_{2}, v_{3}, v_{4}; s)

where

Ψ(v_{1}, v_{2}, v_{3}, v_{4}; s) = ∑_{0≤k∈ℤ} s^{k}Ψ_{k}(v_{1}, v_{2}, v_{3}, v_{4})

is a power series in s.

We choose the relationship s^{1/2} = (t^{−1}−1)^{1/2} such that we have (1−x_{0}^{2})/2x_{0} = −2x_{1}/(1 − x_{1}^{2}). We wish to rewrite the recursions for G and H given in the last two theorems as recursions for the functions

\begin{aligned}Φ̃(v_{1}, v_{2}, v_{3}, v_{4}; x) & = (1 − x^{4})^{\frac{1}{4}} z_{2}^{N+ \frac{1}{8}} G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2}) \\ & = (1 − x^{4})^{\frac{1}{4}} t^{N_{1}−N_{4}+Δ}Φ(v_{1}, v_{2}, v_{3}, v_{4}; t) \end{aligned}

where x = x_{0} or x = x_{∞} and t^{1/2} = 2x/(1 + x^{2}), and

\begin{aligned}Ψ̃(v_{1}, v_{2}, v_{3}, v_{4}; w) & = (1 − w^{4})^{\frac{1}{4}} z^{N+ \frac{1}{8}} H(v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2}) \\ & = (1 − w^{4})^{\frac{1}{4}} s^{N_{3}−N_{4}+Δ̄}Ψ(v_{1}, v_{2}, v_{3}, v_{4}; s) \end{aligned}

where w = x_{1} or w = x_{−1} and s^{1/2} = 2w/(1 − w^{2}).

We will later apply the first of these recursions to the functions

\begin{aligned}Φ̃(v_{2}, v_{1}, v_{3}, v_{4}; y) & = (1 − y^{4})^{\frac{1}{4}} z_{1}^{N+ \frac{1}{8}} G(v_{2}, v_{1}, v_{3}, v_{4}; z_{2}, z_{1}) \\ & = (1 − y^{4})^{\frac{1}{4}} t^{−N_{2}+N_{4}−Δ′}Φ(v_{2}, v_{1}, v_{3}, v_{4}; t^{−1}) \end{aligned}

where y = x_{i} or y = x_{−i} and Δ′ = Δ(n_{2}, n_{1} + n_{3}). These will be obtained from the first kind of recursions by switching v_{1} with v_{2}, z_{1} with z_{2}, and n_{1} with n_{2}, and by replacing t by t^{−1}. Define the matrices

[Φ̃_{top}]_{α} = (1 − x_{α}^{4})^{\frac{1}{4}} z \frac{1}{8} 2 [G(z_{1}, z_{2})]_{α}

for α = 0 or α = ∞,

[Φ̃_{top}]_{α} = (1 − x_{α}^{4})^{\frac{1}{4}} z \frac{1}{8} 1 [G(z_{2}, z_{1})]_{α}

for α = i or α = −i, and

[Ψ̃_{top}]_{α} = (1 − x_{α}^{4})^{\frac{1}{4}} z^{\frac{1}{8}} [H(z, z_{2})]_{α}

for α = 1 or α = −1. Then our earlier work, which will give the base cases of our inductions, can be written as

[Φ̃_{top}]_{0} ∼ x_{∞}^{−1}B_{∞} [Φ̃_{top}]_{∞}
∼ (ix_{i} + 1)^{−1}B_{i} [Φ̃_{top}]_{i} ∼ (ix_{−i} − 1)^{−1}B_{−i} [Φ̃_{top}]_{−i}
∼ (x_{−1} + 1)^{−1}B_{−1} [Ψ̃_{top}]_{−1} ∼ (−x_{1} + 1)^{−1}B_{1} [Ψ̃_{top}]_{1}

The general case requires us to define 2 × 4 matrices which are analogs of [Φ̃_{top}]_{α} and [Ψ̃_{top}]_{α} with the “top” vectors u_{n_{i}} replaced by general vectors. There is no loss of generality in assuming that these vectors v_{i} are L(0)-eignevectors with wt(v_{i}) = N_{i} + Δ_{n_{i}}, 1 ≤ i ≤ 4. We must do this in such a way that our induction formulas imply that the above top ∼ relations hold for the more general matrices. This means that we need a more general description than the pattern of subscripts defining the matrices [G(z_{1}, z_{2})], [G(z_{2}, z_{1})] and [H(z, z_{2})]. We will use the isomorphism θ on CM(ℤ) defined at the beginning of this paper to give such a description. Assume that v_{1}, v_{2}, v_{3} ∈ V_{1} and v_{4} ∈ V_{3} are L(0)-eigenvectors with weights as given above. In the entries of the following matrices we will write the functions Φ̃(v_{1}, v_{2}, v_{3}, v_{4}; x_{α}) and Ψ̃(v_{1}, v_{2}, v_{3}, v_{4}; x_{α}) more briefly as just Φ̃(v_{1}, v_{2}, v_{3}, v_{4}) and Ψ̃(v_{1}, v_{2}, v_{3}, v_{4}). For α = 0 or α = ∞, define the matrix

[Φ̃]_{α} = [Φ̃(v_{1}, v_{2}, v_{3}, v_{4}; x_{α})] =

[ \begin{matrix}Φ̃(v_{1}, v_{2}, v_{3}, v_{4}) & Φ̃(θv_{1}, v_{2}, v_{3}, θv_{4}) & Φ̃(θv_{1}, θv_{2}, θv_{3}, θv_{4}) & Φ̃(v_{1}, θv_{2}, θv_{3}, v_{4}) \\ Φ̃(θv_{1}, θv_{2}, v_{3}, v_{4}) & Φ̃(v_{1}, θv_{2}, v_{3}, θv_{4}) & Φ̃(v_{1}, v_{2}, θv_{3}, θv_{4}) & Φ̃(θv_{1}, v_{2}, θv_{3}, v_{4}) \end{matrix} ] .

For α = i or α = −i define the matrix

[Φ̃]_{α} = [Φ̃(v_{1}, v_{2}, v_{3}, v_{4}; x_{α})] =

[ \begin{matrix}Φ̃(v_{2}, v_{1}, v_{3}, v_{4}) & Φ̃(θv_{2}, v_{1}, v_{3}, θv_{4}) & Φ̃(θv_{2}, θv_{1}, θv_{3}, θv_{4}) & Φ̃(v_{2}, θv_{1}, θv_{3}, v_{4}) \\ Φ̃(θv_{2}, θv_{1}, v_{3}, v_{4}) & Φ̃(v_{2}, θv_{1}, v_{3}, θv_{4}) & Φ̃(v_{2}, v_{1}, θv_{3}, θv_{4}) & Φ̃(θv_{2}, v_{1}, θv_{3}, v_{4}) \end{matrix} ] .

For α = 1 or α = −1 define the matrix

[Ψ̃]_{α} = [Ψ̃(v_{1}, v_{2}, v_{3}, v_{4}; x_{α})] =

[ \begin{matrix}Ψ̃(v_{1}, v_{2}, v_{3}, v_{4}) & Ψ̃(θv_{1}, θv_{2}, v_{3}, v_{4}) & Ψ̃(v_{1}, v_{2}, θv_{3}, θv_{4}) & Ψ̃(θv_{1}, θv_{2}, θv_{3}, θv_{4}) \\ Ψ̃(θv_{1}, v_{2}, v_{3}, θv_{4}) & Ψ̃(v_{1}, θv_{2}, v_{3}, θv_{4}) & Ψ̃(θv_{1}, v_{2}, θv_{3}, v_{4}) & Ψ̃(v_{1}, θv_{2}, θv_{3}, v_{4}) \end{matrix} ] .

In our inductions we will see such matrices with one of the vectors, say v_{i}, replaced by L(−k)v_{i}. We will denote the above matrices with such a replacement by [Φ̃(L(−k)v_{i})]_{α} and [Ψ̃(L(−k)v_{i})]_{α}.

Let Φ′ = \frac{∂}{∂t}Φ, Ψ′ = \frac{∂}{∂s}Ψ and let us use notations such as Φ̃(L(i)v_{1}) = Φ̃(L(i)v_{1}, v_{2}, v_{3}, v_{4}; x) as we did in the proof of Theorem 17. We have

∂_{1}G = \frac{N_{4} − N_{1} − Δ}{z_{1}}G − \frac{z_{2}}{z_{1}^{2}} t^{N_{1}−N_{4}+Δ}z_{2}^{−N− \frac{1}{8}} Φ′

∂_{2}G = \frac{Δ − N_{2} − N_{3} − \frac{1}{8}}{z_{2}} G + \frac{1}{z_{1}} t^{N_{1}−N_{4}+Δ}z_{2}^{−N− \frac{1}{8}} Φ′

∂H = \frac{−N_{1} − N_{2} − \frac{1}{8} + Δ̄}{z}H + \frac{1}{z_{2}} s^{N_{3}−N_{4}+Δ̄}z^{−N− \frac{1}{8}} Ψ′

∂_{2}H = \frac{N_{4} − N_{3} − Δ̄}{z_{2}}H − \frac{z}{z_{2}^{2}} s^{N_{3}−N_{4}+Δ̄}z^{−N− \frac{1}{8}} Ψ′.

Theorem 18. Define the operator

L_{x} = \frac{x(1 + x^{2})}{2(1 − x^{2})} \frac{d}{dx} + \frac{x^{4}}{2(1 − x^{2})^{2}} = \frac{x(1 + x^{2})}{2(1 − x^{2})} [\frac{d}{dx} + \frac{x^{3}}{1 − x^{4}} ] .

Then for any k ∈ ℤ we have

\begin{aligned}Φ̃(L(−k)v_{4}) = & Φ̃(L(k)v_{3}) − (N + \frac{1}{8})Φ̃ + (1 − t^{−k})L_{x}Φ̃ \\ & + ∑_{i≥0} \binom{k + 1}{i + 1} (t^{i−k}Φ̃(L(i)v_{1}) + Φ̃(L(i)v_{2})) . \end{aligned}

SPINOR CONSTRUCTION OF THE c = 1/2 MINIMAL MODEL 27

Proof. For any k ∈ ℤ we have

Φ̃(L(−k)v_{4}) = (1 − x^{4})^{\frac{1}{4}} z_{2}^{N−k+ \frac{1}{8}} G(v_{1}, v_{2}, v_{3}, L(−k)v_{4}; z_{1}, z_{2})

= (1 − x^{4})^{\frac{1}{4}} z_{2}^{N−k+ \frac{1}{8}} G(v_{1}, v_{2}, L(k)v_{3}, v_{4}; z_{1}, z_{2})

+ (1 − x^{4})^{\frac{1}{4}} z_{2}^{N−k+ \frac{1}{8}} (z_{1}^{k+1}∂_{1} + z_{2}^{k+1}∂_{2})G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2}) + (1 − x^{4})^{\frac{1}{4}} z_{2}^{N−k+ \frac{1}{8}} (k + 1)(z_{1}^{k}|v_{1}| + z_{2}^{k}|v_{2}|)G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2})

+ (1 − x^{4})^{\frac{1}{4}} z_{2}^{N−k+ \frac{1}{8}} ∑_{i≥1} \binom{k + 1}{i + 1}z_{1}^{k−i}G(L(i)v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2})

+ (1 − x^{4})^{\frac{1}{4}} z_{2}^{N−k+ \frac{1}{8}} ∑_{i≥1} \binom{k + 1}{i + 1}z_{2}^{k−i}G(v_{1}, L(i)v_{2}, v_{3}, v_{4}; z_{1}, z_{2})

= Φ̃(L(k)v_{3}) + ((N_{4} − N_{1} − Δ)t^{−k} + (Δ − N_{2} − N_{3} − \frac{1}{8}))Φ̃ + (t − t^{1−k})t^{N_{1}−N_{4}+Δ}(1 − x^{4})^{\frac{1}{4}} Φ′ + (k + 1)(t^{−k}|v_{1}| + |v_{2}|)Φ̃

+ ∑_{i≥1} \binom{k + 1}{i + 1}t^{i−k}Φ̃(L(i)v_{1}) + ∑_{i≥1} \binom{k + 1}{i + 1}Φ̃(L(i)v_{2}).

Note that

\frac{dΦ}{dx} = \frac{dΦ}{dt} \frac{dt}{dx} and \frac{dt}{dx} = \frac{8x(1 − x^{2})}{(1 + x^{2})^{3}} = \frac{8x(1 − x^{4})}{(1 + x^{2})^{4}}

so that

Φ′ = \frac{dΦ}{dt} = \frac{dΦ}{dx} \frac{(1 + x^{2})^{4}}{8x(1 − x^{4})}.

From

\frac{d}{dx}Φ̃(x) = \frac{d}{dx} ((1 − x^{4})^{\frac{1}{4}} t^{N_{1}−N_{4}+Δ}Φ(t))

= \frac{−x^{3}}{1 − x^{4}} Φ̃(x) + \frac{N_{1} − N_{4} + Δ}{t} \frac{dt}{dx}Φ̃(x) + (1 − x^{4})^{\frac{1}{4}} t^{N_{1}−N_{4}+Δ} \frac{dΦ}{dx}

we get

(1 − x^{4})^{\frac{1}{4}} t^{N_{1}−N_{4}+Δ} \frac{dΦ}{dt} = (\frac{dt}{dx})^{−1} \frac{dΦ̃}{dx} + \frac{x^{3}}{1 − x^{4}} (\frac{dt}{dx})^{−1} Φ̃ − \frac{N_{1} − N_{4} + Δ}{t}Φ̃.

This gives us (t − t^{1−k})t^{N_{1}−N_{4}+Δ}(1 − x^{4})^{\frac{1}{4}} Φ′

= (1 − t^{−k}) (\frac{x(1 + x^{2})}{2(1 − x^{2})} \frac{d}{dx} + \frac{x^{4}}{2(1 − x^{2})^{2}} + (N_{4} − N_{1} − Δ)) Φ̃

= (1 − t^{−k})(L_{x} + (N_{4} − N_{1} − Δ))Φ̃

which gives the result. ■

28 FEINGOLD, RIES, WEINER

Note that the operator L_{x} and multiplication by any integral power of t^{\frac{1}{2}} preserve the ring R_{x}.

Lemma 19. For any g ∈ ℂ(x), if x = u^{−1}, we have

L_{x}(xg(x^{−1})) = u^{−1}L_{u}(g(u)).

Proof. We have

L_{x}(xg(x^{−1})) = \frac{x(1 + x^{2})}{2(1 − x^{2})} (x\frac{dg}{du} \frac{du}{dx} + g(x^{−1}) + \frac{x^{4}}{1 − x^{4}} g(x^{−1}))

= \frac{u^{2} + 1}{2u(u^{2} − 1)} (−u\frac{dg}{du} + g(u) + \frac{1}{u^{4} − 1}g(u))

= \frac{1 + u^{2}}{2(1 − u^{2})} \frac{dg}{du} + \frac{u^{3}}{2(1 − u^{2})^{2}} g(u)

= u^{−1}L_{u}(g(u)).

■

Lemma 20. For any g ∈ ℂ(x), if x = \frac{±y+i}{iy±1}, t^{1/2} = \frac{2x}{1+x^{2}} and N ∈ ℤ, we have

L_{x}(t^{N}(iy ± 1)^{−1}g(x)) = \frac{−t^{N}}{iy ± 1}(L_{y} − N − \frac{1}{8})g (\frac{±y + i}{iy ± 1}) .

Proof. We have L_{x}(t^{N}(iy ± 1)^{−1}g(x))

= \frac{x(1 + x^{2})}{2(1 − x^{2})} [\frac{t^{N}}{iy ± 1} \frac{dg}{dy} \frac{dy}{dx} + t^{N}g(x)\frac{d}{dx}(iy ± 1)^{−1}

+ Nt^{N−1} \frac{dt}{dx}(iy ± 1)^{−1}g(x) + t^{N}(iy ± 1)^{−1} \frac{x^{3}}{1 − x^{4}} g(x)]

= \frac{x(1 + x^{2})}{2(1 − x^{2})} [\frac{t^{N}}{iy ± 1} \frac{(iy ± 1)^{2}}{2} \frac{d}{dy} − t^{N} \frac{i}{2}

+ Nt^{N−1}2\frac{2x}{1 + x^{2}} \frac{2(1 − x^{2})}{(1 + x^{2})^{2}}(iy ± 1)^{−1} + t^{N}(iy ± 1)^{−1} \frac{x^{3}}{1 − x^{4}} ] g(x)

= \frac{−t^{N}}{iy ± 1} \frac{y(1 + y^{2})}{2(1 − y^{2})} [\frac{d}{dy} − \frac{i}{iy ± 1} − \frac{2N(1 − y^{2})}{y(1 + y^{2})}

+ \frac{(±y + i)^{3}}{±4yi(1 − y^{2})(iy ± 1)}] g (\frac{±y + i}{iy ± 1})

= \frac{−t^{N}}{iy ± 1} \frac{y(1 + y^{2})}{2(1 − y^{2})} [\frac{d}{dy} + \frac{3y^{2} − 1}{4y(1 − y^{2})} − \frac{2N(1 − y^{2})}{y(1 + y^{2})}] g (\frac{±y + i}{iy ± 1})

SPINOR CONSTRUCTION OF THE c = 1/2 MINIMAL MODEL 29

= \frac{−t^{N}}{iy ± 1} \frac{y(1 + y^{2})}{2(1 − y^{2})} [\frac{d}{dy} + \frac{y^{3}}{1 − y^{4}} − \frac{1 − y^{2}}{4y(1 + y^{2})} − \frac{2N(1 − y^{2})}{y(1 + y^{2})}] g (\frac{±y + i}{iy ± 1})

= \frac{−t^{N}}{iy ± 1} \frac{y(1 + y^{2})}{2(1 − y^{2})} [\frac{d}{dy} + \frac{y^{3}}{1 − y^{4}} − \frac{(1 − y^{2})(1 + 8N)}{4y(1 + y^{2})}] g (\frac{±y + i}{iy ± 1})

= \frac{−t^{N}}{iy ± 1} [L_{y} − \frac{y(1 + y^{2})}{2(1 − y^{2})} \frac{(1 − y^{2})(1 + 8N)}{4y(1 + y^{2})}] g (\frac{±y + i}{iy ± 1})

= \frac{−t^{N}}{iy ± 1}(L_{y} − N − \frac{1}{8})g (\frac{±y + i}{iy ± 1})

■

Lemma 21. For any g ∈ ℂ(x), if x = \frac{w±1}{∓w+1}, s^{1/2} = \frac{2w}{1−w^{2}} and N ∈ ℤ, we have

L_{x}(s^{−N}(∓w + 1)^{−1}g(x)) = \frac{−s^{−N}}{∓w + 1}(s^{−1}L_{w} − (N + \frac{1}{8})(1 + s^{−1}))g (\frac{w ± 1}{∓w + 1}) .

Proof. We have L_{x}(s^{−N}(∓w + 1)^{−1}g(x))

= \frac{x(1 + x^{2})}{2(1 − x^{2})}[\frac{s^{−N}}{∓w + 1} \frac{dg}{dw} \frac{dw}{dx} + s^{−N}g(x)\frac{dw}{dx} \frac{d}{dw} \frac{1}{∓w + 1}

+ \frac{1}{∓w + 1}g(x)\frac{dw}{dx} \frac{d}{dw} s^{−N} + \frac{s^{−N}}{(∓w + 1)} \frac{x^{3}}{(1 − x^{4})}g(x)]

= \frac{x(1 + x^{2})}{2(1 − x^{2})}[\frac{s^{−N}}{∓w + 1} \frac{(∓w + 1)^{2}}{2} \frac{d}{dw} ± \frac{s^{−N}}{2}

− \frac{1}{∓w + 1} \frac{(∓w + 1)^{2}}{2} Ns^{−N} \frac{2(1 + w^{2})}{w(1 − w^{2})} + \frac{s^{−N}}{(∓w + 1)} \frac{x^{3}}{(1 − x^{4})}]g(x)

= \frac{x(1 + x^{2})}{2(1 − x^{2})} \frac{s^{−N}}{∓w + 1}[\frac{(∓w + 1)^{2}}{2} \frac{d}{dw} ± \frac{∓w + 1}{2}

− \frac{N(∓w + 1)^{2}(1 + w^{2})}{w(1 − w^{2})} + \frac{x^{3}}{1 − x^{4}} ]g(x)

= \frac{x(1 + x^{2})(∓w + 1)^{2}}{4(1 − x^{2})} \frac{s^{−N}}{∓w + 1}[\frac{d}{dw} ± \frac{1}{∓w + 1}

− \frac{2N(1 + w^{2})}{w(1 − w^{2})} + \frac{2x^{3}}{(1 − x^{4})(∓w + 1)^{2}} ]g(x)

= \frac{−s^{−N}}{∓w + 1} \frac{(1 − w^{2})(1 + w^{2})}{8w}[\frac{d}{dw} ± \frac{1}{∓w + 1}

− \frac{2N(1 + w^{2})}{w(1 − w^{2})} ∓ \frac{(w ± 1)^{3}}{4(∓w + 1)w(1 + w^{2})}]g(x)

30 FEINGOLD, RIES, WEINER

= \frac{−s^{−N}}{∓w + 1} \frac{(1 − w^{2})(1 + w^{2})}{8w}[\frac{d}{dw} + \frac{w^{3}}{1 − w^{4}} − \frac{(8N + 1)(1 + w^{2})}{4w(1 − w^{2})}]g(x)

= \frac{−s^{−N}}{∓w + 1}[\frac{(1 − w^{2})(1 + w^{2})}{8w}(\frac{d}{dw} + \frac{w^{3}}{1 − w^{4}} ) − (N + \frac{1}{8})\frac{(1 + w^{2})^{2}}{4w^{2}} ]g(x)

= \frac{−s^{−N}}{∓w + 1}[s^{−1}L_{w} − (N + \frac{1}{8})(1 + s^{−1})]g (\frac{w ± 1}{∓w + 1})

■

Theorem 22. For any k ∈ ℤ we have Ψ̃(L(−k)v_{4}) = Ψ̃(L(k)v_{3}) − (N + \frac{1}{8})s^{−k−1}[(1 + s)^{k+1} − 1]Ψ̃ + s^{−k−1}[(1 + s)^{k} − 1]L_{w}Ψ̃

+ ∑_{i≥0} \binom{k + 1}{i + 1} [s^{i−k}Ψ̃(L(i)v_{2}) + (\frac{1 + s}{s})^{k−i} Ψ̃(L(i)v_{1})] .

Proof. For any k ∈ ℤ we have Ψ̃(L(−k)v_{4}) = (1 − w^{4})^{\frac{1}{4}} z^{N−k+ \frac{1}{8}} H(v_{1}, v_{2}, v_{3}, L(−k)v_{4}; z, z_{2}) = (1 − w^{4})^{\frac{1}{4}} z^{N−k+ \frac{1}{8}} H(v_{1}, v_{2}, L(k)v_{3}, v_{4}; z, z_{2}) + (1 − w^{4})^{\frac{1}{4}} z^{N−k+ \frac{1}{8}} z_{2}^{k+1}∂_{2}H(v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2}) + (1 − w^{4})^{\frac{1}{4}} z^{N−k+ \frac{1}{8}} [(z_{2} + z)^{k+1} − z_{2}^{k+1}]∂H(v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2})

+ (1 − w^{4})^{\frac{1}{4}} z^{N−k+ \frac{1}{8}} ∑_{i≥0} \binom{k + 1}{i + 1}z_{2}^{k−i}H(v_{1}, L(i)v_{2}, v_{3}, v_{4}; z, z_{2})

+ (1 − w^{4})^{\frac{1}{4}} z^{N−k+ \frac{1}{8}} ∑_{i≥0} \binom{k + 1}{i + 1}(z_{2} + z)^{k−i}H(L(i)v_{1}, v_{2}, v_{3}, v_{4}; z, z_{2})

= Ψ̃(L(k)v_{3}) + (N_{4} − N_{3} − Δ̄)s^{−k}Ψ̃ − s^{1−k}(1 − w^{4})^{\frac{1}{4}} s^{N_{3}−N_{4}+Δ̄}Ψ′ − (N_{1} + N_{2} + \frac{1}{8} − Δ̄)s^{−k−1}[(1 + s)^{k+1} − 1]Ψ̃ + s^{−k}[(1 + s)^{k+1} − 1]s^{N_{3}−N_{4}+Δ̄}(1 − w^{4})^{\frac{1}{4}} Ψ′

+ ∑_{i≥0} \binom{k + 1}{i + 1}s^{i−k}Ψ̃(L(i)v_{2}) + ∑_{i≥0} \binom{k + 1}{i + 1}s^{i−k}(1 + s)^{k−i}Ψ̃(L(i)v_{1})

Note that

\frac{dΨ}{dw} = \frac{dΨ}{ds} \frac{ds}{dw} and \frac{ds}{dw} = \frac{8w(1 + w^{2})}{(1 − w^{2})^{3}} = \frac{8w(1 − w^{4})}{(1 − w^{2})^{4}}

so that

Ψ′ = \frac{dΨ}{ds} = \frac{dΨ}{dw} \frac{(1 − w^{2})^{4}}{8w(1 − w^{4})}.

From

\frac{d}{dw} Ψ̃(w) = \frac{d}{dw} ((1 − w^{4})^{\frac{1}{4}} s^{N_{3}−N_{4}+Δ̄}Ψ(s))

= \frac{−w^{3}}{1 − w^{4}} Ψ̃(w) + \frac{N_{3} − N_{4} + Δ̄}{s} \frac{ds}{dw} Ψ̃(w) + (1 − w^{4})^{\frac{1}{4}} s^{N_{3}−N_{4}+Δ̄} \frac{dΨ}{dw}

we get

(1 − w^{4})^{\frac{1}{4}} s^{N_{3}−N_{4}+Δ̄} \frac{dΨ}{ds} = (\frac{ds}{dw} )^{−1} \frac{dΨ̃}{dw} + \frac{w^{3}}{1 − w^{4}} (\frac{ds}{dw} )^{−1} Ψ̃ − \frac{N_{3} − N_{4} + Δ̄}{s}Ψ̃.

The two terms involving Ψ′ in the above expression for Ψ̃(L(−k)v_{4}) combine to give

s^{−k}(1 + s)[(1 + s)^{k} − 1](1 − w^{4})^{\frac{1}{4}} s^{N_{3}−N_{4}+Δ̄} \frac{dΨ}{ds}

and since 1 + s = \frac{(1+w^{2})^{2}}{(1−w^{2})^{2}} this gives us

s^{−k}(1 + s)[(1 + s)^{k} − 1](1 − w^{4})^{\frac{1}{4}} s^{N_{3}−N_{4}+Δ̄}Ψ′

= s^{−k}[(1 + s)^{k} − 1] (\frac{1 − w^{4}}{8w} \frac{d}{dw} + \frac{w^{2}}{8} − \frac{(N_{3} − N_{4} + Δ̄)(1 + s)}{s}) Ψ̃

= s^{−k}[(1 + s)^{k} − 1] (\frac{L_{w}}{s} − \frac{(N_{3} − N_{4} + Δ̄)(1 + s)}{s}) Ψ̃.

Substituting this in the earlier expression, after some simplification, we get the result. ■

Note that the operator L_{w} and multiplication by any rational function in s^{\frac{1}{2}} preserve the ring R_{w}.

Now let us return to the problem of determining the general two-point functions inductively.

Theorem 23. For 1 ≤ i ≤ 4 let v_{i} ∈ V_{n_{i}} be eigenvectors for L(0) with L(0)v_{i} = wt(v_{i})v_{i} = |v_{i}|v_{i}. For i = 1, 2 let Y_{i} = Y (v_{i}, z_{i}), ∂_{i} = ∂/∂z_{i}. Then for any k ∈ ℤ we have

G(L(−k)v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2})

= ∑_{i≥0} \binom{k + i − 2}{i}z_{1}^{i}G(v_{1}, v_{2}, v_{3}, L(k + i)v_{4}; z_{1}, z_{2})

+ (−1)^{k}(z_{1} − z_{2})^{1−k}∂_{2}G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2})

+ (−1)^{k} ∑_{j≥0}(−1)^{j+1}\binom{1 − k}{j + 1}(z_{1} − z_{2})^{−k−j}G(v_{1}, L(j)v_{2}, v_{3}, v_{4}; z_{1}, z_{2})

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}z_{1}^{−k−i}G(v_{1}, v_{2}, L(i)v_{3}, v_{4}; z_{1}, z_{2})

+ (−1)^{k}z_{1}^{1−k}(G(v_{1}, v_{2}, v_{3}, L(1)v_{4}; z_{1}, z_{2}) − (∂_{1} + ∂_{2})G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2})).

32 FEINGOLD, RIES, WEINER

Proof. We have G(L(−k)v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2}) = (Y (L(−k)v_{1}, z_{1})Y_{2}v_{3}, v_{4})

= ∑_{i≥0} \binom{k + i − 2}{i}z_{1}^{i}(L(−k − i)Y_{1}Y_{2}v_{3}, v_{4})

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 2}{i}z_{1}^{−k+1−i}(Y_{1}L(i − 1)Y_{2}v_{3}, v_{4})

= ∑_{i≥0} \binom{k + i − 2}{i}z_{1}^{i}G(v_{1}, v_{2}, v_{3}, L(k + i)v_{4}; z_{1}, z_{2})

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 2}{i}z_{1}^{−k+1−i}[(Y_{1}[L(i − 1), Y_{2}]v_{3}, v_{4})

+ (Y_{1}Y_{2}L(i − 1)v_{3}, v_{4})]

= ∑_{i≥0} \binom{k + i − 2}{i}z_{1}^{i}G(v_{1}, v_{2}, v_{3}, L(k + i)v_{4}; z_{1}, z_{2})

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 2}{i}z_{1}^{−k+1−i} ∑_{j≥0} \binom{i}{j}z_{2}^{i−j}(Y_{1}Y (L(j − 1)v_{2}, z_{2})v_{3}, v_{4})

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 2}{i}z_{1}^{−k+1−i}(Y_{1}Y_{2}L(i − 1)v_{3}, v_{4})

= ∑_{i≥0} \binom{k + i − 2}{i}z_{1}^{i}G(L(k + i)v_{4}) + (−1)^{k}z_{1}^{1−k} ∑_{i≥0} \binom{k + i − 2}{i} (\frac{z_{2}}{z_{1}} )^{i} ∂_{2}G

+ (−1)^{k} ∑_{j≥0} ∑_{i≥0} \binom{k + i − 2}{i}\binom{i}{j + 1}z_{1}^{1−k}z_{2}^{−j−1} (\frac{z_{2}}{z_{1}} )^{i} G(L(j)v_{2})

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}z_{1}^{−k−i}G(L(i)v_{3}) + (−1)^{k}z_{1}^{1−k}(Y_{1}Y_{2}L(−1)v_{3}, v_{4})

= ∑_{i≥0} \binom{k + i − 2}{i}z_{1}^{i}G(L(k + i)v_{4}) + (−1)^{k}z_{1}^{1−k}∂_{2}G ∑_{i≥0} \binom{k + i − 2}{i} (\frac{z_{2}}{z_{1}} )^{i}

+ (−1)^{k} ∑_{j≥0} z_{1}^{−j−k}G(L(j)v_{2}) ∑_{i≥0} \binom{k + i − 2}{i}\binom{i}{j + 1} (\frac{z_{2}}{z_{1}} )^{i−j−1}

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}z_{1}^{−k−i}G(L(i)v_{3})

+ (−1)^{k}z_{1}^{1−k}(G(L(1)v_{4}) − ([L(−1), Y_{1}]Y_{2}v_{3}, v_{4}) − (Y_{1}[L(−1), Y_{2}]v_{3}, v_{4})). We may write the last line as

(−1)^{k}z_{1}^{1−k}(G(L(1)v_{4}) − (∂_{1} + ∂_{2})G).

SPINOR CONSTRUCTION OF THE c = 1/2 MINIMAL MODEL 33

Also, note that

\binom{k + i − 2}{i} = (−1)^{i} \binom{1 − k}{i}

so

∑_{i≥0} \binom{k + i − 2}{i} (\frac{z_{2}}{z_{1}} )^{i} = ∑_{i≥0}(−1)^{i}\binom{1 − k}{i} (\frac{z_{2}}{z_{1}} )^{i} = (1 − \frac{z_{2}}{z_{1}} )^{1−k}

and therefore,

(−1)^{k}z_{1}^{1−k}∂_{2}G ∑_{i≥0} \binom{k + i − 2}{i} (\frac{z_{2}}{z_{1}} )^{i} = (−1)^{k}(z_{1} − z_{2})^{1−k}∂_{2}G.

In addition, we have

\begin{matrix}∑_{i≥0}(−1)^{i}\binom{1 − k}{i}\binom{i}{j + 1}t^{i−j−1} = \frac{1}{(j + 1)!}∂_{t}^{j+1} ∑_{i≥0}(−1)^{i}\binom{1 − k}{i}t^{i} \\ = \frac{1}{(j + 1)!}∂_{t}^{j+1}(1 − t)^{1−k} = (−1)^{j+1}\binom{1 − k}{j + 1}(1 − t)^{−j−k} \end{matrix}

so that

(−1)^{k} ∑_{j≥0} z_{1}^{−j−k}G(L(j)v_{2}) ∑_{i≥0} \binom{k + i − 2}{i}\binom{i}{j + 1} (\frac{z_{2}}{z_{1}} )^{i−j−1}

= (−1)^{k} ∑_{j≥0} z_{1}^{−j−k}G(L(j)v_{2}) ∑_{i≥0}(−1)^{i}\binom{1 − k}{i}\binom{i}{j + 1} (\frac{z_{2}}{z_{1}} )^{i−j−1}

= (−1)^{k} ∑_{j≥0} z_{1}^{−j−k}G(L(j)v_{2})(−1)^{j+1}\binom{1 − k}{j + 1} (1 − \frac{z_{2}}{z_{1}} )^{−j−k}

= (−1)^{k} ∑_{j≥0} G(L(j)v_{2})(−1)^{j+1}\binom{1 − k}{j + 1}(z_{1} − z_{2})^{−j−k}

which gives the result. ■

Theorem 24. For 1 ≤ i ≤ 4 let v_{i} ∈ V_{n_{i}} be eigenvectors for L(0) with L(0)v_{i} = wt(v_{i})v_{i} = |v_{i}|v_{i}. For i = 1, 2 let Y_{i} = Y (v_{i}, z_{i}), ∂_{i} = ∂/∂z_{i}. Then for

34 FEINGOLD, RIES, WEINER

any k ∈ ℤ we have

G(v_{1}, L(−k)v_{2}, v_{3}, v_{4}; z_{1}, z_{2}) = ∑_{i≥0} \binom{k + i − 2}{i}z_{2}^{i}G(v_{1}, v_{2}, v_{3}, L(k + i)v_{4}; z_{1}, z_{2})

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}z_{2}^{−k−i}G(v_{1}, v_{2}, L(i)v_{3}, v_{4}; z_{1}, z_{2}) − (∂_{1}G)(z_{1} − z_{2})^{1−k}

+ ∑_{j≥0} G(L(j)v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2})(−1)^{j}z_{2}^{−j−k}·

· ∑_{i≥0}(−1)^{i}\binom{1 − k}{i}\binom{i + j + k − 1}{j + 1} (\frac{z_{2}}{z_{1}} )^{i+j+k}

+ (−1)^{k}z_{2}^{1−k}(G(v_{1}, v_{2}, v_{3}, L(1)v_{4}; z_{1}, z_{2}) − (∂_{1} + ∂_{2})G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2})).

Proof. We have

G(v_{1}, L(−k)v_{2}, v_{3}, v_{4}; z_{1}, z_{2}) = (Y_{1}Y (L(−k)v_{2}, z_{2})v_{3}, v_{4})

= ∑_{i≥0} \binom{k + i − 2}{i}z_{2}^{i}(Y_{1}L(−k − i)Y_{2}v_{3}, v_{4})

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 2}{i}z_{2}^{−k+1−i}(Y_{1}Y_{2}L(i − 1)v_{3}, v_{4})

= ∑_{i≥0} \binom{k + i − 2}{i}z_{2}^{i}(L(−k − i)Y_{1}Y_{2}v_{3}, v_{4})

− ∑_{i≥0} \binom{k + i − 2}{i}z_{2}^{i}([L(−k − i), Y_{1}]Y_{2}v_{3}, v_{4})

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}z_{2}^{−k−i}G(L(i)v_{3})

+ (−1)^{k}z_{2}^{1−k}(G(L(1)v_{4}) − (∂_{1} + ∂_{2})G)

= ∑_{i≥0} \binom{k + i − 2}{i}z_{2}^{i}G(L(k + i)v_{4})

− ∑_{i≥0} \binom{k + i − 2}{i}z_{2}^{i} ∑_{j≥0} \binom{−k − i + 1}{j}z_{1}^{1−i−j−k}(Y (L(j − 1)v_{1}, z_{1})Y_{2}v_{3}, v_{4})

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}z_{2}^{−k−i}G(L(i)v_{3})

+ (−1)^{k}z_{2}^{1−k}(G(L(1)v_{4}) − (∂_{1} + ∂_{2})G).

SPINOR CONSTRUCTION OF THE c = 1/2 MINIMAL MODEL 35

We may rewrite the second line in the last expression as

− ∑_{j≥0} G(L(j − 1)v_{1})(−1)^{j}z_{2}^{1−j−k}·

· ∑_{i≥0}(−1)^{i}\binom{1 − k}{i}\binom{i + j + k − 2}{j} (\frac{z_{2}}{z_{1}} )^{i+j+k−1}

= −(∂_{1}G)(z_{1} − z_{2})^{1−k}

+ ∑_{j≥0} G(L(j)v_{1})(−1)^{j}z_{2}^{−j−k} ∑_{i≥0}(−1)^{i}\binom{1 − k}{i}\binom{i + j + k − 1}{j + 1} (\frac{z_{2}}{z_{1}} )^{i+j+k}

giving the result. Note that the j = 0 term in the last line simplifies to (k − 1)|v_{1}|G(z_{1} − z_{2})^{−k}. ■

Theorem 25. For any k ∈ ℤ we have

H(L(−k)v_{1}) =

∑_{i≥0}(−1)^{i}\binom{1 − k}{i}(z_{2} + z)^{i}H(L(k + i)v_{4})

+ (−1)^{k}(z_{2} + z)^{1−k}(H(L(1)v_{4}) − ∂_{2}H)

+ (−1)^{k} ∑_{j≥0} H(L(j)v_{3})z^{−j−k} ∑_{i≥0} \binom{1 − k}{i}\binom{i + j + k − 1}{j + 1} (\frac{z}{z_{2}} )^{i+j+k}

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}z^{−k−i}H(L(i)v_{2})

+ (−1)^{k}z^{1−k}(∂_{2}H − ∂H)

and

H(L(−k)v_{2}) = ∑_{i≥0} \binom{k + i − 2}{i}z_{2}^{i}H(L(k + i)v_{4})

+ (−1)^{k}z_{2}^{1−k}(H(L(1)v_{4}) − ∂_{2}H)

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}z_{2}^{−k−i}H(L(i)v_{3})

− z^{1−k}∂H − ∑_{i≥0} \binom{1 − k}{i + 1}z^{−k−i}H(L(i)v_{1})

36 FEINGOLD, RIES, WEINER

Theorem 26. For any k ∈ ℤ we have

Φ̃(L(−k)v_{1}) = ∑_{i≥0} \binom{k + i − 2}{i}t^{−i}Φ̃(L(k + i)v_{4})

+ (−1)^{k}(N + \frac{1}{8})t^{k−1}[1 − (1 − t)^{1−k}]Φ̃ + (−1)^{k}t^{k−1}(1 − t)[(1 − t)^{−k} − 1]L_{x}Φ̃

+ (−1)^{k} ∑_{j≥0}(−1)^{j+1}\binom{1 − k}{j + 1} (\frac{t}{1 − t})^{j+k} Φ̃(L(j)v_{2})

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}t^{k+i}Φ̃(L(i)v_{3})

+ (−1)^{k}t^{k−1}Φ̃(L(1)v_{4})

and

Φ̃(L(−k)v_{2}) = ∑_{i≥0} \binom{k + i − 2}{i}Φ̃(L(k + i)v_{4})

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}Φ̃(L(i)v_{3})

+ [t^{k}(1 − t)^{−k} − (−1)^{k}](1 − t)L_{x}Φ̃

+ ∑_{j≥0}(−1)^{j}Φ̃(L(j)v_{1}) ∑_{i≥0}(−1)^{i}\binom{1 − k}{i}\binom{i + j + k − 1}{j + 1}t^{i+j+k}

+ (−1)^{k}Φ̃(L(1)v_{4}) + (−1)^{k}(N + \frac{1}{8})Φ̃.

Theorem 27. For any k ∈ ℤ we have

Ψ̃(L(−k)v_{1}) = ∑_{i≥0}(−1)^{i}\binom{1 − k}{i} (\frac{1 + s}{s})^{i} Ψ̃(L(k + i)v_{4})

+ (−1)^{k} (\frac{1 + s}{s})^{1−k} Ψ̃(L(1)v_{4})

+ (−1)^{k} ∑_{j≥0} Ψ̃(L(j)v_{3}) ∑_{i≥0} \binom{1 − k}{i}\binom{i + j + k − 1}{j + 1}s^{i+j+k}

+ (−1)^{k} [(\frac{s}{1 + s})^{k} − 1] L_{w}Ψ̃

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}Ψ̃(L(i)v_{2}) + (−1)^{k}(N + \frac{1}{8})Ψ̃

and

Ψ̃(L(−k)v_{2}) = ∑_{i≥0} \binom{k + i − 2}{i}s^{−i}Ψ̃(L(k + i)v_{4})

+ (−1)^{k}s^{k−1}Ψ̃(L(1)v_{4}) + (N + \frac{1}{8})Ψ̃

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}s^{k+i}Ψ̃(L(i)v_{3})

− ∑_{i≥0} \binom{1 − k}{i + 1}Ψ̃(L(i)v_{1}) + \frac{1}{1 + s}[(−1)^{k}s^{k} − 1]L_{w}Ψ̃.

Theorem 28. (Rationality) For 1 ≤ i ≤ 4 let v_{i} ∈ V_{n_{i}} be eigenvectors for L(0) with L(0)v_{i} = wt(v_{i})v_{i} = |v_{i}|v_{i}. Suppose that n_{i} ∈ {1, 3}, n_{4} = n_{1} +n_{2} +n_{3} (mod 4), wt(v_{i}) = N_{i} + Δ_{n_{i}} and N = N_{1} + N_{2} + N_{3} − N_{4}. Then, after the substitution (z_{2}/z_{1})^{1/2} = 2x/(1 + x^{2}), the series

Φ̃(v_{1}, v_{2}, v_{3}, v_{4}; x) = (1 − x^{4})^{\frac{1}{4}} z_{2}^{N+ \frac{1}{8}} G(v_{1}, v_{2}, v_{3}, v_{4}; z_{1}, z_{2})

converges absolutely in the domain |x| < \sqrt{3 − 2\sqrt{2}} to a rational function in the ring R_{x} = ℂ[x, x^{−1}(x^{4} − 1)^{−1}].

Proof. We proof the statement by induction on M = N_{1} + N_{2} + N_{3} + N_{4}. The base case, when N_{i} = 0 for 1 ≤ i ≤ 4, is given by Theorem 13. The result of Theorem 18 with k > 0 inductively reduces the general case to the case where N_{4} = 0, and with k < 0 it reduces further to the case where N_{3} = 0 also. The two results of Theorem 26 with k > 0 further reduce one to the base case. ■

Theorem 29. For any v_{1}, v_{2}, v_{3} ∈ V_{1}, v_{4} ∈ V_{3}, we have

[Φ̃]_{0} ∼ \frac{1}{x_{∞}} B_{∞}[Φ̃]_{∞}.

Proof. It sufficies to prove this for homogeneous vectors v_{i} with wt(v_{i}) = N_{i} + Δ_{n_{i}}, where v_{i} ∈ V_{n_{i}}. We will prove this using Theorem 18, Theorem 26, and Lemma 19, by induction on N_{1} + N_{2} + N_{3} + N_{4}. The base case, when all N_{i} = 0, has been established already. First note that since wt(v_{i}) = wt(θ(v_{i})), the N in Theorems 18 and 26 is the same for each of the entries in the matrices [Φ̃]_{0} and [Φ̃]_{∞}. Also, t = 4x_{0}^{2}/(1 + x_{0}^{2})^{2} = 4x_{∞}^{2}/(1 + x_{∞}^{2})^{2}, so the t in Theorems 18 and 26 is the same whether x is x_{0} or x_{∞}. First we will do the inductive step which reduces the weight of v_{4}. Assume the statement is true as stated for all choices of four vectors whose weights add up to be less than or equal to N_{1} + N_{2} + N_{3} + N_{4}. We will show that it is then true with v_{4} replaced by L(−k)v_{4} for any 0 < k ∈ ℤ. We have

[Φ̃(L(−k)v_{4})]_{0} =

[Φ̃(L(k)v_{3})]_{0} − (N + \frac{1}{8})[Φ̃]_{0} + (1 − t^{−k})L_{x_{0}}[Φ̃]_{0}

+ ∑_{i≥0} \binom{k + 1}{i + 1} (t^{i−k}[Φ̃(L(i)v_{1})]_{0} + [Φ̃(L(i)v_{2})]_{0})

∼ \frac{1}{x_{∞}} B_{∞}[Φ̃(L(k)v_{3})]_{∞} − (N + \frac{1}{8})\frac{1}{x_{∞}} B_{∞}[Φ̃]_{∞} + (1 − t^{−k})L_{x_{0}} \frac{1}{x_{∞}} B_{∞}[Φ̃]_{∞}

+ ∑_{i≥0} \binom{k + 1}{i + 1}\frac{1}{x_{∞}} B_{∞} (t^{i−k}[Φ̃(L(i)v_{1})]_{∞} + [Φ̃(L(i)v_{2})]_{∞})

= \frac{1}{x_{∞}} B_{∞} [[Φ̃(L(k)v_{3})]_{∞} − (N + \frac{1}{8})[Φ̃]_{∞} + (1 − t^{−k})L_{x_{∞}}[Φ̃]_{∞}]

+ \frac{1}{x_{∞}} B_{∞} ∑_{i≥0} \binom{k + 1}{i + 1} (t^{i−k}[Φ̃(L(i)v_{1})]_{∞} + [Φ̃(L(i)v_{2})]_{∞})

= \frac{1}{x_{∞}} B_{∞}[Φ̃(L(−k)v_{4})]_{∞}.

But the same calculation with 0 > k ∈ ℤ shows that the statement is also true with v_{3} replaced by L(k)v_{3}.

Applying the two parts of Theorem 26 for 0 < k ∈ ℤ, one similarly gets that the statement is true with v_{i} replaced by L(−k)v_{i}, for i = 1, 2. Lemma 19 plays a crucial role, giving

L_{x_{0}} \frac{1}{x_{∞}} [Φ̃]_{∞} = \frac{1}{x_{∞}} L_{x_{∞}}[Φ̃]_{∞}.

■

Theorem 30. Let v_{1}, v_{2}, v_{3} ∈ V_{1} and v_{4} ∈ V_{3} with wt(v_{i}) = |v_{i}| = N_{i} + Δ_{n_{i}} and N = N_{1} + N_{2} + N_{3} − N_{4}. Then we have

[Φ̃]_{0} ∼ \frac{t^{N}}{ix_{i} + 1}B_{i}[Φ̃]_{i} ∼ \frac{t^{N}}{ix_{−i} − 1}B_{−i}[Φ̃]_{−i}.

Proof. We will prove this using Theorem 18, Theorem 26, and Lemma 20, by induction on N_{1} + N_{2} + N_{3} + N_{4}. The base case, when all N_{i} = 0, has been established already. Since wt(v_{i}) = wt(θ(v_{i})), the N in Theorems 18 and 26 is the same for each of the entries in the matrices [Φ̃]_{0} and [Φ̃]_{±i}. First we will do the inductive step which reduces the weight of v_{4}. Assume the statement is true as stated for all choices of four vectors whose weights add up to be less than or equal to N_{1} + N_{2} + N_{3} + N_{4}. We will show that it is then true with v_{4} replaced

SPINOR CONSTRUCTION OF THE c = 1/2 MINIMAL MODEL 39

by L(−k)v_{4} for any 0 < k ∈ ℤ. We have [Φ̃(L(−k)v_{4})]_{0} = [Φ̃(L(k)v_{3})]_{0} − (N + \frac{1}{8})[Φ̃]_{0} + (1 − t^{−k})L_{x_{0}}[Φ̃]_{0}

+ ∑_{i≥0} \binom{k + 1}{i + 1} (t^{i−k}[Φ̃(L(i)v_{1})]_{0} + [Φ̃(L(i)v_{2})]_{0})

∼ \frac{t^{N−k}}{ix_{i} + 1}B_{i}[Φ̃(L(k)v_{3})]_{i} − (N + \frac{1}{8})\frac{t^{N}}{ix_{i} + 1}B_{i}[Φ̃]_{i} + (1 − t^{−k})L_{x_{0}} \frac{t^{N}}{ix_{i} + 1}B_{i}[Φ̃]_{i}

+ ∑_{i≥0} \binom{k + 1}{i + 1}\frac{t^{N−i}}{ix_{i} + 1}B_{i} (t^{i−k}[Φ̃(L(i)v_{1})]_{i} + [Φ̃(L(i)v_{2})]_{i})

= \frac{t^{N−k}}{ix_{i} + 1}B_{i}[[Φ̃(L(k)v_{3})]_{i} − (N + \frac{1}{8})t^{k}[Φ̃]_{i} + (1 − t^{−k}) (−t^{k}(L_{x_{i}} − N − \frac{1}{8})[Φ̃]_{i})

+ ∑_{i≥0} \binom{k + 1}{i + 1} ([Φ̃(L(i)v_{1})]_{i} + t^{k−i}[Φ̃(L(i)v_{2})]_{i}) ]

= \frac{t^{N−k}}{ix_{i} + 1}B_{i}[[Φ̃(L(k)v_{3})]_{i} + (1 − t^{k})L_{x_{i}}[Φ̃]_{i} − (N + \frac{1}{8})[Φ̃]_{i}

+ ∑_{i≥0} \binom{k + 1}{i + 1} ([Φ̃(L(i)v_{1})]_{i} + t^{k−i}[Φ̃(L(i)v_{2})]_{i}) ]

= \frac{t^{N−k}}{ix_{i} + 1}B_{i}[Φ̃(L(−k)v_{4})]_{i}.

In the last step we used Theorem 18 with v_{1} and v_{2} switched, z_{1} and z_{2} switched, and therefore, t replaced by t^{−1}. The same calculation with 0 > k ∈ ℤ shows that the statement is also true with v_{3} replaced by L(k)v_{3}.

We now apply the two parts of Theorem 26 for 0 < k ∈ ℤ to get that the statement is true with v_{i} replaced by L(−k)v_{i}, for i = 1, 2. Lemma 20 plays a crucial role. We have

[Φ̃(L(−k)v_{1})]_{0}

= ∑_{i≥0} \binom{k + i − 2}{i}t^{−i}[Φ̃(L(k + i)v_{4})]_{0}

+ (−1)^{k}(N + \frac{1}{8})t^{k−1}[1 − (1 − t)^{1−k}][Φ̃]_{0} + (−1)^{k}t^{k−1}(1 − t)[(1 − t)^{−k} − 1]L_{x_{0}}[Φ̃]_{0}

+ (−1)^{k} ∑_{j≥0}(−1)^{j+1}\binom{1 − k}{j + 1} (\frac{t}{1 − t})^{j+k} [Φ̃(L(j)v_{2})]_{0}

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}t^{k+i}[Φ̃(L(i)v_{3})]_{0} + (−1)^{k}t^{k−1}[Φ̃(L(1)v_{4})]_{0}

40 FEINGOLD, RIES, WEINER

∼ ∑_{i≥0} \binom{k + i − 2}{i}t^{−i} \frac{t^{N+k+i}}{ix_{i} + 1} B_{i}[Φ̃(L(k + i)v_{4})]_{i}

+ (−1)^{k}(N + \frac{1}{8})t^{k−1}[1 − (1 − t)^{1−k}]\frac{t^{N}}{ix_{i} + 1}B_{i}[Φ̃]_{i}

+ (−1)^{k}t^{k−1}(1 − t)[(1 − t)^{−k} − 1]L_{x_{0}} \frac{t^{N}}{ix_{i} + 1}B_{i}[Φ̃]_{i}

+ (−1)^{k} ∑_{j≥0}(−1)^{j+1}\binom{1 − k}{j + 1} (\frac{t}{1 − t})^{j+k} \frac{t^{N−j}}{ix_{i} + 1}B_{i}[Φ̃(L(j)v_{2})]_{i}

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}t^{k+i} \frac{t^{N−i}}{ix_{i} + 1}B_{i}[Φ̃(L(i)v_{3})]_{i}

+ (−1)^{k}t^{k−1} \frac{t^{N+1}}{ix_{i} + 1}B_{i}[Φ̃(L(1)v_{4})]_{i}

= ∑_{i≥0} \binom{k + i − 2}{i}\frac{t^{N+k}}{ix_{i} + 1}B_{i}[Φ̃(L(k + i)v_{4})]_{i}

+ (−1)^{k}(N + \frac{1}{8})t^{k−1}[1 − (1 − t)^{1−k}]\frac{t^{N}}{ix_{i} + 1}B_{i}[Φ̃]_{i}

+ (−1)^{k}t^{k−1}(1 − t)[(1 − t)^{−k} − 1] [\frac{−t^{N}}{ix_{i} + 1}B_{i}(L_{x_{i}} − N − \frac{1}{8})] [Φ̃]_{i}

+ (−1)^{k} ∑_{j≥0}(−1)^{j+1}\binom{1 − k}{j + 1}(1 − t)^{−j−k} \frac{t^{N+k}}{ix_{i} + 1}B_{i}[Φ̃(L(j)v_{2})]_{i}

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}\frac{t^{N+k}}{ix_{i} + 1}B_{i}[Φ̃(L(i)v_{3})]_{i} + (−1)^{k} \frac{t^{N+k}}{ix_{i} + 1}B_{i}[Φ̃(L(1)v_{4})]_{i}

= \frac{t^{N+k}}{ix_{i} + 1}B_{i}[ ∑_{i≥0} \binom{k + i − 2}{i}[Φ̃(L(k + i)v_{4})]_{i}

+ (−1)^{k}(N + \frac{1}{8})t^{−1}[1 − (1 − t)^{1−k}][Φ̃]_{i} − (−1)^{k}t^{−1}(1 − t)[(1 − t)^{−k} − 1](L_{x_{i}} − N − \frac{1}{8})[Φ̃]_{i}

+ (−1)^{k} ∑_{j≥0}(−1)^{j+1}\binom{1 − k}{j + 1}(1 − t)^{−j−k}[Φ̃(L(j)v_{2})]_{i}

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}[Φ̃(L(i)v_{3})]_{i} + (−1)^{k}[Φ̃(L(1)v_{4})]_{i}]

= \frac{t^{N+k}}{ix_{i} + 1}B_{i}[ ∑_{i≥0} \binom{k + i − 2}{i}[Φ̃(L(k + i)v_{4})]_{i}

+ (−1)^{k}(N + \frac{1}{8})[Φ̃]_{i} + [t^{−k}(1 − t^{−1})^{−k} − (−1)^{k}](1 − t^{−1})L_{x_{i}}[Φ̃]_{i}

+ ∑_{j≥0}(−1)^{j}[Φ̃(L(j)v_{2})]_{i} ∑_{i≥0} \binom{1 − k}{i}\binom{i + j + k − 1}{j + 1}t^{−i−j−k}

+ (−1)^{k} ∑_{i≥0} \binom{k + i − 1}{i + 1}[Φ̃(L(i)v_{3})]_{i} + (−1)^{k}[Φ̃(L(1)v_{4})]_{i}]

= \frac{t^{N+k}}{ix_{i} + 1}B_{i}[Φ̃(L(−k)v_{1})]_{i}.

In the last step we used the second part of Theorem 26 with v_{1} and v_{2} switched, z_{1} and z_{2} switched, and therefore, t replaced by t^{−1}.

The calculation for [Φ̃(L(−k)v_{2})]_{0} is similar, completing the inductive proof of the first part of the theorem. The second part, with x_{−i} in place of x_{i}, B_{−i} in place of B_{i}, and certain sign changes, is proved in exactly the same way. ■

Theorem 31. Let v_{1}, v_{2}, v_{3} ∈ V_{1} and v_{4} ∈ V_{3} with wt(v_{i}) = |v_{i}| = N_{i} + Δ_{n_{i}} and N = N_{1} + N_{2} + N_{3} − N_{4}. Then we have

[Φ̃]_{0} ∼ \frac{s^{−N}}{−x_{1} + 1}B_{1}[Ψ̃]_{1} ∼ \frac{s^{−N}}{x_{−1} + 1}B_{−1}[Ψ̃]_{−1}.

Proof. We will prove this using Theorems 18, 22, 27, and Lemma 21, by induction on N_{1} + N_{2} + N_{3} + N_{4}. The base case, when all N_{i} = 0, has been established already. First note that since wt(v_{i}) = wt(θ(v_{i})), the N in Theorems 18 and 27 is the same for each of the entries in the matrices [Φ̃]_{0} and [Ψ̃]_{±1}. First we will do the inductive step which reduces the weight of v_{4}. Assume the statement is true as stated for all choices of four vectors whose weights add up to be less than or equal to N_{1} + N_{2} + N_{3} + N_{4}. We will show that it is then true with v_{4} replaced by L(−k)v_{4} for any 0 < k ∈ ℤ. We have

[Φ̃(L(−k)v_{4})]_{0}

= [Φ̃(L(k)v_{3})]_{0} − (N + \frac{1}{8})[Φ̃]_{0} + (1 − t^{−k})L_{x_{0}}[Φ̃]_{0}

+ ∑_{i≥0} \binom{k + 1}{i + 1} (t^{i−k}[Φ̃(L(i)v_{1})]_{0} + [Φ̃(L(i)v_{2})]_{0})

∼ \frac{s^{−N+k}}{−x_{1} + 1}B_{1}[Ψ̃(L(k)v_{3})]_{1} − (N + \frac{1}{8})\frac{s^{−N}}{−x_{1} + 1}B_{1}[Ψ̃]_{1}

+ (1 − t^{−k})L_{x_{0}} \frac{s^{−N}}{−x_{1} + 1}B_{1}[Ψ̃]_{1}

+ ∑_{i≥0} \binom{k + 1}{i + 1}\frac{s^{−N+i}}{−x_{1} + 1}B_{1} (t^{i−k}[Ψ̃(L(i)v_{1})]_{1} + [Ψ̃(L(i)v_{2})]_{1})

= \frac{s^{−N+k}}{−x_{1} + 1}B_{1}[[Ψ̃(L(k)v_{3})]_{1} − (N + \frac{1}{8})s^{−k}[Ψ̃]_{1}

− (1 − t^{−k})s^{−k} (s^{−1}L_{x_{1}} − (N + \frac{1}{8})(1 + s^{−1})) [Ψ̃]_{1}

+ ∑_{i≥0} \binom{k + 1}{i + 1} [(\frac{s + 1}{s})^{k−i} [Ψ̃(L(i)v_{1})]_{1} + s^{i−k}[Ψ̃(L(i)v_{2})]_{1}] ]

= \frac{s^{−N+k}}{−x_{1} + 1}B_{1}[[Ψ̃(L(k)v_{3})]_{1} − (N + \frac{1}{8})s^{−k}[Ψ̃]_{1} + ((1 + s)^{k} − 1)s^{−k−1}L_{x_{1}}[Ψ̃]_{1}

+ (1 − (1 + s)^{k})s^{−k}(1 + s^{−1})(N + \frac{1}{8})[Ψ̃]_{1}

+ ∑_{i≥0} \binom{k + 1}{i + 1} [(\frac{s + 1}{s})^{k−i} [Ψ̃(L(i)v_{1})]_{1} + s^{i−k}[Ψ̃(L(i)v_{2})]_{1}] ]

= \frac{s^{−N+k}}{−x_{1} + 1}B_{1}[[Ψ̃(L(k)v_{3})]_{1} − (N + \frac{1}{8})s^{−k−1}((1 + s)^{k+1} − 1)[Ψ̃]_{1}

+ s^{−k−1}((1 + s)^{k} − 1)L_{x_{1}}[Ψ̃]_{1}

+ ∑_{i≥0} \binom{k + 1}{i + 1} [(\frac{s + 1}{s})^{k−i} [Ψ̃(L(i)v_{1})]_{1} + s^{i−k}[Ψ̃(L(i)v_{2})]_{1}] ]

= \frac{s^{−N+k}}{−x_{1} + 1}B_{1}[Ψ̃(L(−k)v_{4})]_{1}

We used Theorem 22 in the last step.

The same calculation with 0 > k ∈ ℤ shows that the statement is also true with v_{3} replaced by L(k)v_{3}.

Applying the two parts of Theorem 27 for 0 < k ∈ ℤ, one similarly gets that the statement is true with v_{i} replaced by L(−k)v_{i}, for i = 1, 2. Lemma 21 plays a crucial role. ■

We are, at last, ready to state the new “matrix” Jacobi-Cauchy Identity which is valid for the c = \frac{1}{2} minimal model. This is the main objective of this paper, but considerable further work remains to be done in order to understand other minimal models and the WZW models. That will be the subject of future investigations.

Theorem 32. (Matrix Jacobi-Cauchy Identity) Let v_{1}, v_{2}, v_{3} ∈ V_{1} and v_{4} ∈ V_{3} with wt(v_{i}) = |v_{i}| = N_{i} + Δ_{n_{i}} and N = N_{1} + N_{2} + N_{3} − N_{4}. Let f(x) be any function in R_{x}. Let C_{0} be a small positively oriented circle with center x_{0} = 0 and for α ∈ {∞, 1, −1, i, −i} let C_{α} be the circle obtained from C_{0} by the appropriate Möbius transformation sending 0 to α. Then we have

0 = ∮_{C_{0}}[Φ]_{0}f(x_{0})dx_{0}

+ ∮_{C_{∞}} \frac{−1}{x_{∞}^{3}} B_{∞}[Φ̃]_{∞}f(1/x_{∞}) dx_{∞}

+ ∮_{C_{i}} \frac{2t^{N}}{(ix_{i} + 1)^{3}} B_{i}[Φ̃]_{i}f (\frac{x_{i} + i}{ix_{i} + 1}) dx_{i}

+ ∮_{C_{−i}} \frac{2t^{N}}{(ix_{−i} − 1)^{3}} B_{−i}[Φ̃]_{−i}f (\frac{−x_{−i} + i}{ix_{−i} − 1}) dx_{−i}

+ ∮_{C_{1}} \frac{2s^{−N}}{(−x_{1} + 1)^{3}} B_{1}[Ψ̃]_{1}f (\frac{x_{1} + 1}{−x_{1} + 1}) dx_{1}

+ ∮_{C_{−1}} \frac{2s^{−N}}{(x_{−1} + 1)^{3}} B_{−1}[Ψ̃]_{−1}f (\frac{x_{−1} − 1}{x_{−1} + 1}) dx_{−1}.

Proof. Apply the Cauchy residue theorem to f(x) times the globally defined matrix valued function G(x) to which each of the expressions in Theorems 29 - 31 converge. It says that the sum of the residues at the six possible poles is zero, that is,

0 = ∑_{α} ∮_{C_{α}} G(x_{0})f(x_{0})dx_{0}.

Let

x_{0} = μ_{α}(x_{α}) = \frac{a_{α}x_{α} + b_{α}}{c_{α}x_{α} + d_{α}}

denote the Möbius transformation chosen before Theorem 13 to relate x_{0} to x_{α}. Then the chain rule gives dx_{0} = \frac{a_{α}d_{α}−b_{α}c_{α}}{(c_{α}x_{α}+d_{α})^{2}} dx_{α}. For each α we write the corresponding integral with G(x_{0})f(x_{0}) represented by the series in x_{α} which converges to it in the neighborhood of x_{0} = α, and we write dx_{0} as the appropriate chain rule factor times dx_{α}. ■

Just as Corollary 3 was obtained from Theorem 2, we can obtain infinitely many explicit identities for products of components of intertwining operators from Theorem 32 by making explicit choices for the test function f(x). One computes the integral in each term of the Matrix Jacobi-Cauchy Identity (MJCI) by expanding the integrand as a series in the local variable x_{α} and then finding the residue as the coefficient of x_{α}^{−1}. The algebra involved is tedious but straightforward, so we will not present it here. The most general function f(x) ∈ R_{x} is a linear combination of functions of the form

f(x) = x^{m_{0}}(1 + ix)^{m_{i}}(1 − ix)^{m_{−i}}(1 − x)^{m_{1}}(1 + x)^{m_{−1}}

where m_{0}, m_{i}, m_{−i}, m_{1}, m_{−1} ∈ ℤ are five independent parameters. Since the MJCI is linear in f(x), it suffices to compute the identity just for f(x) of that form. In fact, using a few simple combinatorial identities, there is a considerable simplification of the result if we restrict f(x) to be a rational function of the variable t, say t^{r}(1−t)^{s}. This seems a natural restriction because the correlation functions made from intertwining operators were defined from series in z_{1} and z_{2} which could be written in terms of t. Only later did we have to express them in terms of the local variables x_{α} in order to use properties of the hypergeometric functions to relate the functions at the six poles. It makes the local variables seem like a necessary but artificial technical construction, and one might think that the final answer should reflect the fact that there were only three possible poles in the t-plane. This may not be the last word on that subject, but our results seem to show that with that restriction on f there are six terms in the MJCI, but the second row of the matrix yields a trivial identity. We find a slightly simpler result if we restrict the test function to be of the form f(x) = x^{2}t^{r}(1 − t)^{s}. We give the results in the next three corollaries.

If we were to write out each entry of each 2 × 4 matrix in the identity, then the result would not fit across the page, and the pattern would be very similar in each column. So we will just show the first column of the answer in Corollary 33. The other columns are easily obtained from the first one by modifying the pattern of vectors v_{1}, v_{2}, v_{3}, v_{4} in the first row and θv_{1}, θv_{2}, v_{3}, v_{4} in the second row, as shown in the matrices [Φ̃]_{α} and [Ψ̃]_{α} defined before Theorem 18. In Corollaries 34 and 35, since the second row of the matrix yields a trivial identity, we only give the identity coming from the first row.

Corollary 33. Let v_{1}, v_{2}, v_{3} ∈ V_{1} and v_{4} ∈ V_{3} with wt(v_{i}) = |v_{i}| = N_{i} + Δ_{n_{i}} and N = N_{1} + N_{2} + N_{3} − N_{4}. Let

f(x) = x^{m_{0}}(1 + ix)^{m_{i}}(1 − ix)^{m_{−i}}(1 − x)^{m_{1}}(1 + x)^{m_{−1}}

for m_{0}, m_{i}, m_{−i}, m_{1}, m_{−1} ∈ ℤ, and let m = m_{0} + m_{i} + m_{−i} + m_{1} + m_{−1}. Define the following constants in terms of those parameters:

A_{∞} = (−1)^{m_{1}+m_{−i}+1}(i)^{m_{i}+m_{−i}},
A_{i} = 2^{m_{1}+m_{i}+m_{−i}+1}(i)^{m_{0}+m_{i}}(1 + i)^{m_{−1}−m_{1}},
A_{−i} = 2^{m_{−1}+m_{i}+m_{−i}+1}(−1)^{m_{0}+m_{−i}+1}(i)^{m_{0}+m_{−i}}(1 + i)^{m_{1}−m_{−1}},
A_{1} = 2^{m_{1}+m_{−1}+m_{−i}+1}(−1)^{m_{1}}(1 + i)^{m_{i}−m_{−i}},
A_{−1} = 2^{m_{1}+m_{−1}+m_{i}+1}(−1)^{m_{0}}(1 + i)^{m_{−i}−m_{i}},

and for each k ≥ 0 define

Q_{k} = 2(k + N_{1} − N_{4}),  R_{k} = 2(k − N_{1} − N_{3}),  S_{k} = 2(k − N_{1} − N_{2}).

In the following formula, for each k ≥ 0, and for each α ∈ {0, ∞, i, −i, 1, −1}, ∑_{α} and ∑′_{α} denote summations over all integers p_{1}, p_{−1}, p_{i}, p_{−i} ≥ 0 such that the sum p_{1}+p_{−1}+p_{i}+p_{−i} equals a fixed value, p̂, depending on k and some of the given parameters. The fixed values are as follows. For ∑_{0}, p̂ = −m_{0} − 2 − Q_{k}, for ∑′_{0}, p̂ = −m_{0}−1−Q_{k}, for ∑_{∞}, p̂ = m+1−Q_{k}, for ∑′_{∞}, p̂ = m+2−Q_{k}, for ∑_{i}, p̂ = −m_{i}−2−R_{k}, for ∑′_{i}, p̂ = −m_{i}−1−R_{k}, for ∑_{−i}, p̂ = −m_{−i}−2−R_{k}, for ∑′_{−i}, p̂ = −m_{−i} −1−R_{k}, for ∑_{1}, p̂ = −m_{1} −2−S_{k}, for ∑′_{1}, p̂ = −m_{1} −1−S_{k},

SPINOR CONSTRUCTION OF THE c = 1/2 MINIMAL MODEL 45

for ∑_{−1}, p̂ = −m_{−1} − 2 − S_{k}, and for ∑′_{−1}, p̂ = −m_{−1} − 1 − S_{k}. Within the summations we use the notations

C_{1} = \binom{\frac{1}{4} + m_{1}}{p_{1}}, C_{−1} = \binom{\frac{1}{4} + m_{−1}}{p_{−1}}, C_{i} = \binom{\frac{1}{4} + m_{i}}{p_{i}}, C_{−i} = \binom{\frac{1}{4} + m_{−i}}{p_{−i}},

and

D = (−1)^{p_{1}+p_{−i}}(i)^{p_{i}+p_{−i}}.

To shorten our notation for the correlation functions, we write Φ_{k} = Φ_{k}(v_{1}, v_{2}, v_{3}, v_{4}) = (Y_{k−N_{4}+ \frac{7}{16}} (v_{1})Y_{−k+N_{3}− \frac{7}{16}} (v_{2})v_{3}, v_{4}) ,

Φ_{k}^{θ} = Φ_{k}(θv_{1}, θv_{2}, v_{3}, v_{4}) = (Y_{k−N_{4}− \frac{1}{16}} (θv_{1})Y_{−k+N_{3}+ \frac{1}{16}} (θv_{2})v_{3}, v_{4}) ,

Φ_{k}^{∗} = Φ_{k}(v_{2}, v_{1}, v_{3}, v_{4}) = (Y_{k−N_{4}+ \frac{7}{16}} (v_{2})Y_{−k+N_{3}− \frac{7}{16}} (v_{1})v_{3}, v_{4}) ,

Φ_{k}^{∗θ} = Φ_{k}(θv_{2}, θv_{1}, v_{3}, v_{4}) = (Y_{k−N_{4}− \frac{1}{16}} (θv_{2})Y_{−k+N_{3}+ \frac{1}{16}} (θv_{1})v_{3}, v_{4}) ,

Ψ_{k} = Ψ_{k}(v_{1}, v_{2}, v_{3}, v_{4}) = (Y_{N_{3}−N_{4}}(Y_{−k+N_{2}− \frac{7}{16}} (v_{1})v_{2})v_{3}, v_{4}) ,
Ψ_{k}^{θ} = Ψ_{k}(θv_{1}, v_{2}, v_{3}, θv_{4}) = (Y_{N_{3}−N_{4}}(Y_{−k+N_{2}+ \frac{1}{16}} (θv_{1})v_{2})v_{3}, θv_{4}) .

Then we have the identity

0 = ∑_{0≤k∈ℤ} (2^{Q_{k}} [ \begin{matrix}∑_{0} C_{1}C_{−1}\binom{\frac{1}{4} +m_{i}−Q_{k}−1}{p_{i}}\binom{\frac{1}{4} +m_{−i}−Q_{k}−1}{p_{−i}}D2Φ_{k} \\ ∑′_{0} C_{1}C_{−1}\binom{\frac{1}{4} +m_{i}−Q_{k}}{p_{i}}\binom{\frac{1}{4} +m_{−i}−Q_{k}}{p_{−i}}DΦ_{k}^{θ} \end{matrix} ]

+2^{Q_{k}}A_{∞}B_{∞} [ \begin{matrix}∑_{∞} C_{1}C_{−1}\binom{\frac{1}{4} +m_{−i}−Q_{k}−1}{p_{i}}\binom{\frac{1}{4} +m_{i}−Q_{k}−1}{p_{−i}}D2Φ_{k} \\ ∑′_{∞} C_{1}C_{−1}\binom{\frac{1}{4} +m_{−i}−Q_{k}}{p_{i}}\binom{\frac{1}{4} +m_{i}−Q_{k}}{p_{−i}}DΦ_{k}^{θ} \end{matrix} ]

+2^{R_{k}}A_{i}B_{i} [ \begin{matrix}∑_{i} C_{1}C_{−1}\binom{\frac{1}{4} −m−R_{k}−4}{p_{i}}\binom{\frac{1}{4} +m_{0}−R_{k}−1}{p_{−i}}D2Φ_{k}^{∗} \\ ∑′_{i} C_{1}C_{−1}\binom{\frac{1}{4} −m−R_{k}−3}{p_{i}}\binom{\frac{1}{4} +m_{0}−R_{k}}{p_{−i}}DΦ_{k}^{∗θ} \end{matrix} ]

+2^{R_{k}}A_{−i}B_{−i} [ \begin{matrix}∑_{−i} C_{1}C_{−1}\binom{\frac{1}{4} +m_{0}−R_{k}−1}{p_{i}}\binom{\frac{1}{4} −m−R_{k}−4}{p_{−i}}D2Φ_{k}^{∗} \\ ∑′_{−i} C_{1}C_{−1}\binom{\frac{1}{4} +m_{0}−R_{k}}{p_{i}}\binom{\frac{1}{4} −m−R_{k}−3}{p_{−i}}DΦ_{k}^{∗θ} \end{matrix} ]

+2^{S_{k}}A_{1}B_{1} [ \begin{matrix}∑_{1} C_{i}C_{−i}\binom{\frac{1}{4} −m−S_{k}−4}{p_{1}}\binom{\frac{1}{4} +m_{0}−S_{k}−1}{p_{−1}}DΨ_{k} \\ ∑′_{1} C_{i}C_{−i}\binom{\frac{1}{4} −m−S_{k}−3}{p_{1}}\binom{\frac{1}{4} +m_{0}−S_{k}}{p_{−1}}DΨ_{k}^{θ} \end{matrix} ]

+2^{S_{k}}A_{−1}B_{−1} [ \begin{matrix}∑_{−1} C_{i}C_{−i}\binom{\frac{1}{4} +m_{0}−S_{k}−1}{p_{1}}\binom{\frac{1}{4} −m−S_{k}−4}{p_{−1}}DΨ_{k} \\ ∑′_{−1} C_{i}C_{−i}\binom{\frac{1}{4} +m_{0}−S_{k}}{p_{1}}\binom{\frac{1}{4} −m−S_{k}−3}{p_{−1}}DΨ_{k}^{θ} \end{matrix} ] ).

Corollary 34. Let v_{1}, v_{2}, v_{3} ∈ V_{1} and v_{4} ∈ V_{3} with wt(v_{i}) = |v_{i}| = N_{i} + Δ_{n_{i}} and N = N_{1} + N_{2} + N_{3} − N_{4}. For any r, s ∈ ℤ, define

a = \frac{1}{4} + 2r − 4,  b = r + s − 1,  c = −s − 1.

With notation as in Corollary 33, for any r, s ∈ ℤ, we have

0 = 2^{2r+2(N_{1}−N_{4})} ∑_{0≤k∈ℤ} 2^{−2k}Φ_{k} ∑_{0≤q∈ℤ} \binom{\frac{1}{4} + 2s}{q}\binom{\frac{1}{4} − 2(r + s) − Q_{k} − 1}{−Q_{k}/2 − r − 1 − q}(−1)^{q}

− 2^{2r+2(N_{1}−N_{4})} ∑_{0≤k∈ℤ} 2^{−2k}Φ_{k}^{θ} ∑_{0≤q∈ℤ} \binom{\frac{1}{4} + 2s}{q}\binom{\frac{1}{4} − 2(r + s) − Q_{k}}{−Q_{k}/2 − r + 1 − q}(−1)^{q}

+ (−1)^{s}2^{−2b−2(N_{1}+N_{3})} ∑_{0≤k∈ℤ} 2^{−2k}Φ_{k}^{∗}·

∑_{0≤q∈ℤ} \binom{\frac{1}{4} + 2s}{q}[\binom{a − R_{k}}{b − R_{k}/2 − q} − 3\binom{a − R_{k}}{b − R_{k}/2 − 1 − q}](−1)^{q}

− (−1)^{s}2^{−2b−2(N_{1}+N_{3})} ∑_{0≤k∈ℤ} 2^{−2k}Φ_{k}^{∗θ}·

∑_{0≤q∈ℤ} \binom{\frac{1}{4} + 2s}{q}[\binom{a − R_{k} + 1}{b − R_{k}/2 − 1 − q} − 3\binom{a − R_{k} + 1}{b − R_{k}/2 − q}](−1)^{q}

+ 2^{−2c−2(N_{1}+N_{2})} ∑_{0≤k∈ℤ} 2^{−2k}Ψ_{k}·

∑_{0≤q∈ℤ} \binom{\frac{1}{4} − 2(r + s)}{c − S_{k}/2 − q}[\binom{a − S_{k}}{q} − 3\binom{a − S_{k}}{q − 1}](−1)^{q}

− 2^{−2c−2(N_{1}+N_{2})} ∑_{0≤k∈ℤ} 2^{−2k}Ψ_{k}^{θ}·

∑_{0≤q∈ℤ} \binom{\frac{1}{4} − 2(r + s)}{c − S_{k}/2 − q}[\binom{a − S_{k} + 1}{q − 1} − 3\binom{a − S_{k} + 1}{q}](−1)^{q}.

Corollary 35. Let v_{1}, v_{2}, v_{3} ∈ V_{1} and v_{4} ∈ V_{3} with wt(v_{i}) = |v_{i}| = N_{i} + Δ_{n_{i}}, N = N_{1} + N_{2} + N_{3} − N_{4} and Γ = N_{1} + N_{2} + N_{3} + N_{4}. With notation as in Corollary 33, for any r, s ∈ ℤ, we have

∑_{0≤k∈ℤ} 2^{−2k} ∑_{0≤q≤k} \binom{\frac{1}{4} + 2r + 2(N_{1} + N_{2})}{q}(−1)^{q}·

[\binom{\frac{1}{4} − 2r + S_{k} − 1}{k − q}[Φ_{r+s+Γ−k} + (−1)^{r+N_{1}+N_{2}+1}Φ_{−s−1−k}^{∗}]

− \binom{\frac{1}{4} − 2r + S_{k}}{k − q}[Φ_{r+s+Γ−k}^{θ} + (−1)^{r+N_{1}+N_{2}+1}Φ_{−s−1−k}^{∗θ}]]

SPINOR CONSTRUCTION OF THE c = 1/2 MINIMAL MODEL 47

\begin{matrix}= ∑_{0≤k∈ℤ} 2^{−2k} ∑_{0≤q≤k} \binom{\frac{1}{4} + 2s + 2(N_{1} + N_{3})}{k − q}(−1)^{q}· \\ [\binom{\frac{1}{4} − 2s + R_{k}}{q}Ψ_{−r−1−k}^{θ} − \binom{\frac{1}{4} − 2s + R_{k} − 1}{q}Ψ_{−r−1−k}]. \end{matrix}

Tables of Constants from Section 4

n_{1} = 1  n_{2} = 1  n_{1} = 3  n_{2} = 3
n_{3}  0  1 2  3  n_{3}  0  1 2  3
a  0  \frac{5}{4}  0  \frac{1}{4}  a  0  \frac{1}{4}  0  \frac{5}{4}
b  1  \frac{3}{4}  \frac{2}{3}  \frac{3}{4}  b  1  \frac{3}{4}  \frac{2}{3}  \frac{3}{4}
c  0  \frac{3}{2}  \frac{1}{3}  \frac{1}{2}  c  0  \frac{1}{2}  \frac{1}{3}  \frac{3}{2}
a′  0  \frac{5}{4}  0  \frac{1}{4}  a′  0  \frac{1}{4}  0  \frac{5}{4}
b′  1  \frac{3}{4}  \frac{2}{3}  \frac{3}{4}  b′  1  \frac{3}{4}  \frac{2}{3}  \frac{3}{4}
c′  2  \frac{3}{2}  \frac{4}{3}  \frac{3}{2}  c′  2  \frac{3}{2}  \frac{4}{3}  \frac{3}{2}
A −\frac{3}{8}  \frac{1}{2}  \frac{1}{8}  0  A −\frac{3}{8}  0 \frac{1}{8}  \frac{1}{2}
B  0 −\frac{3}{8}  \frac{1}{2}  \frac{1}{8}  B  0  \frac{1}{8}  \frac{1}{2} −\frac{3}{8}
C −\frac{3}{8} −\frac{3}{8} −\frac{3}{8} −\frac{3}{8}  C −\frac{3}{8} −\frac{3}{8} −\frac{3}{8} −\frac{3}{8}
A′ 0 \frac{1}{2} 1 \frac{1}{2} A′ 0 \frac{1}{2} 1 \frac{1}{2}
B′ −\frac{3}{8} −\frac{3}{8} −\frac{3}{8} −\frac{3}{8}  B′ −\frac{3}{8} −\frac{3}{8} −\frac{3}{8} −\frac{3}{8}
C′ 0 −\frac{3}{8}  \frac{1}{2}  \frac{1}{8}  C′ 0  \frac{1}{8}  \frac{1}{2} −\frac{3}{8}
n_{1} = 1  n_{2} = 3  n_{1} = 3  n_{2} = 1
n_{3} 0  1 2  3  n_{3} 0  1 2  3
a  0  \frac{1}{4} −1  \frac{1}{4}  a  0  \frac{1}{4} −1  \frac{1}{4}
b  1 −\frac{1}{4} −\frac{1}{3}  \frac{3}{4}  b  1  \frac{3}{4} −\frac{1}{3} −\frac{1}{4}
c  0  \frac{1}{2}  \frac{1}{3}  \frac{3}{2}  c  0  \frac{3}{2}  \frac{1}{3}  \frac{1}{2}
a′ 0  \frac{1}{4} −1  \frac{1}{4}  a′ 0  \frac{1}{4} −1  \frac{1}{4}
b′ 1  \frac{3}{4} −\frac{1}{3} −\frac{1}{4}  b′ 1 −\frac{1}{4} −\frac{1}{3}  \frac{3}{4}
c′ 2  \frac{1}{2} −\frac{2}{3}  \frac{1}{2}  c′ 2  \frac{1}{2} −\frac{2}{3}  \frac{1}{2}
A \frac{1}{8}  0 −\frac{3}{8}  \frac{1}{2}  A \frac{1}{8}  \frac{1}{2} −\frac{3}{8}  0
B 0  \frac{1}{8}  \frac{1}{2} −\frac{3}{8}  B 0 −\frac{3}{8}  \frac{1}{2}  \frac{1}{8}
C \frac{1}{8}  \frac{1}{8}  \frac{1}{8}  \frac{1}{8}  C \frac{1}{8}  \frac{1}{8}  \frac{1}{8}  \frac{1}{8}
A′ 0  0 0  0  A′ 0  0 0  0
B′  \frac{1}{8}  \frac{1}{8}  \frac{1}{8}  \frac{1}{8}  B′  \frac{1}{8}  \frac{1}{8}  \frac{1}{8}  \frac{1}{8}
C′ 0 −\frac{3}{8}  \frac{1}{2}  \frac{1}{8}  C′ 0  \frac{1}{8}  \frac{1}{2} −\frac{3}{8}

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Dept. of Math. Sci., The State University of New York, Binghamton, New York 13902-6000

Math. Dept., East Carolina University, Greenville, North Carolina 27858-4353

Math. Dept., Broome Community College, Binghamton, New York 13902-1017