F b e a lpha. α n z n. n. ±n>0

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1 APPENDIX A: THE VERTEX ALGEBRA OF AN INTEGRAL LATTICE. In this section we generalise the vertex operator construction found for fermions and ŝl 2 at level 1, to any integral lattice. Let X = R N be Euclidean space with the usual inner product and let Λ be an integral lattice in X, i.e. a subgroup isomorphic to Z m with m N and such that (α, β) Z for α, β Λ. As before, we say Λ is even if (α, α) is even for each α Λ. Setting B(α, β) = ( 1) (α,β) we can find a normalised cocycle ε(α, β) such that B(α, β) = ε(α, β)ε(β, α). The corresponding cocycle representation on l 2 (Λ), with orthonormal basis (e α ), is defined by U α e β = e α+β ε(α, β). We have U α+β = ε(α, β)u α U β and U β U α = B(α, β)u α U β. Let C[Λ] be the subspace of l 2 (Λ) spanned by the e α s. There is a bosonic system associated to X, namely operators v n depending linearly on v X such that [v m, u n ] = m(v, u)δ m+n,0 I. As usual we have the derivation d with [d, v n ] = nv n. Taking an orthonormal basis of V, this gives n commuting families of bosonic operators (a n ) each satisfying [a m, a n ] = mδ m+n,0 I. Since each has a unique positive energy vacuum representation with a 0 = 0, so too does their direct sum. It is just the tensor product of the N individual vacuum representations. It is uniquely characterised by the existence of a vacuum vector Ω with v n Ω = 0 for n 0. As usual v n = v n. We denote it by F b. Let V = F b C[Λ) = α Λ F b e a lpha. The operators v n for n 0 act on F b e α exactly as they act on F b, The operators v 0 acts scalars on F b e α, v n (ξ e α ) = (v n ξ) e α. v 0 (ξ e α ) = (v, α)ξ e α. For each v V we thus have a field v(z) = v n z n 1. Clearly v(z)u(w) = u(w)v(z) and the matrix coefficients lie in R from the computations in the previous section. For each α Λ we define where Φ α (z) = U α z α0 E + (α, z)e (α, z), E ± (α, z) = exp ±n>0 α n z n. n We wish to check the locality relations between different Φ α (z) s and between the Φ α (z) and v(z) s, We begin by noting that the vertex operators Φ α (z) satisfy the Fubini Veneziano relations for the operator L n given by the direst sum of the Virasoro operators constructed from the N commuting bosons. In addition [v n, Φ α (z)] = (v, α)z n Φ α (z). We shall also need the following result to compute two point functions. Lemma. If A and B are formal power series of operators in z and w 1 such that C = [A, B] is a multiple of the identity operator, then e A e B = e C e B e A. Proof. By a previous lemma, [e A, B] = Ce A so that e A Be A = B + C. Hence e A e B e A = expe A Be A = e B+C = e C e B. Corollary. E + (α, z)e (β, w) = (1 w/z) (α,β) E (β, w)e + (α, z). 1

2 Proof. Let A = n>0 α nz n /n and B = n<0 β nw n /n. Then C = [A, B] = (α, β) n>0 z n w n /n = (α, β)log(1 w z ), so that e C = exp (α, β)log(1 w z ) = (1 w z ) (α,β). Theorem. The operators Φ α (z) and v(z) generate a vertex algebra on V. Proof. Only the locality conditions need to checked. We start by showing that (X(z)E α (w)ω, Ω) and (E α (w)x(z)ω, Ω) define the same rational function in R. In fact note that Then we have on the one hand (v(z)φ α (w)ω e β, Ω e α+β ) = µ + n>0 (E α (z)ω e β, Ω e α+β ) = 1. (a n E α (w)ω, Ω)z n 1 = µ + n>0 αw n z n 1 = µ + α w z 2 (1 w z ) 1, while on the other hand (Φ α (w)v(z)ω e β, Ω e α+β ) = µ + n>0(e α (w)a n Ω, Ω)z n 1 = µ α w n z n 1 = µ αw 1 (1 z/w) 1. In view of the formulas for the commutators [u m, v(z)] and [u m, Φ α (z)] for u X, the two point functions define the same function in R and hence Φ α (z) and v(z) are mutually local. Next we check that, if α + β + γ = δ, then (Φ α (z)φ β (w)ω e γ, Ω e δ ) = ε(α, β)ε(α + β, γ)(z w) (α,β) z (α+β,γ). (1) For these lowest energy vectors, this gives the locality relation between the operators Φ α (z) and Φ β (z), with a sign ( 1) (α,β), since ε(α, β)ε(β, α) = ( 1) (α,β). As before we can derive the general locality relation by inserting operators u m using their commutation relations with the vertex operators. To prove (1), we write (Φ α (z)φ β (w)ω e γ, Ω e δ ) = (U α z α0 E (α, z)e + (α, z)u β w β0 E (β, w)e + (β, w)ω e γ, Ω e δ ) = ε(α, β)(1 w/z) (α,β) z (α,β) (U α+β z α0+β0 E (α, z)e (β, w)e + (α, z)e + (β, w)ω e γ, Ω e δ ) = ε(α, β)ε(α + β, γ)(1 w/z) (α,β) z (α,β) z (α+β,γ). In the section on vertex operators, we already proved the Funibi Veneziano relations: [L 1, Φ α ] = dφ α dz, [L 0, Φ α ] = z dφ α dz + α 2 2 Φ alpha. We derive here the operator product expansion for Φ α (z)φ β (w) in the case root vectors using the vertex algebra formalism. Proposition. (a) Φ α (z)φ β (w) 0 if α, β are roots and α + β is not a root. (b) Φ α (z)φ β (w) ε(α, β)(z w) 1 Φ α+β (w) if α + β is a root. (c) Φ α (z)φ α (w) (z w) 2 + (z w) 1 α(w). Proof. We have On the other hand Φ α (z)φ β (w) = V (Φ α (z w)ω e β, w). Φ α (z)ω e β = ε(α, β)z (α,β) (1 + α( 1)z + )Ω e α+β. 2

3 This has no singular terms if (α, β) 0. If (α, β) = 1, it has a singular term εω e α+β corresponding to the operator ε(α, β)φ α+β (w) if α + β is a root. When β = α, we get the identity operator and V (α( 1)Ω e 0, z) = α(z). Remark. Note that in lattice vertex algebra, the vectors e α Ω are singular vectors of energy α 2 /2. So the corresponding fields Φ α (z) = V (e α Ω, z) are primary fields of conformal dimension h = α 2 /2. Examples. (a) Complex fermions correspond to the lattice Z R. (b) ŝl 2 at level one corresponds to the lattice 2Z R. This is because E(z) = ψ(z) φ(z) and F(z) = φ(z) ψ(z) are clearly vertex operators constructed using the bosonic fields H(z) = X(z) I I X(z). The vertex algebra is the corresponding to the sublattice {(n, n) n Z} of Z 2. (c) Any time we get a vertex algebra containing a Virasoro field T(z), we can take the vertex subalgebra generated by the single field T(z). There is also a coset or centraliser construction (see below). (d) The A,D,E lattices give vertex algebras corresponding to affine Kac Moody algebras. Taking the operators X(0) for X V and Φ α (0) for α 2 = 2, the commutation relations can be worked out directly from the either the Fubini Veneziano relation or the Borcherds commutation relations coming from the OPE. This gives a different way of exhibiting these simple Lie algebras directly as operators and the commutators as genuine commutators. APPENDIX B: THE KAC DETERMINANT FORMULA. Using the Poincaré Birkhoff Witt theorem we can construct a Verma module V (c, h) which is a largest possible representation of the Virasoro algebra with central charge c, generated by a cyclic vector ξ 0 such that L n ξ = 0 for n > 0 and L 0 ξ = hξ. It has the universal property that, for any representation generated by a cyclic vector satisfying similar relations, there is unique equivariant map sending ξ 0 to the cyclic vector. A basis of the Verma module is given by monomials in the raising operators L n k k Ln2 2 Ln1 1 ξ 0, where n i 0. Clearly the Verma module is a positive energy representation. We now assume that c and h are real. Let f : V (c, h) C be the linear map picking out the coefficient of ξ 0 and extend the involution L n = L n to a complex involution on the universal enveloping algebra, so that (AB) = B A. We can then define a hermitian form on V (c, h) by (Aξ 0, Bξ 0 ) = f(b Aξ 0 ). By definition it satisfies the invariance condition (L n ξ, η) = (ξ, L n η). Since L 0 = L 0, the eigenspaces of L 0 are orthogonal. Moreover (, ) is the unique invariant hermitian form on V (c, h) with (ξ 0, ξ 0 ) = 1: for the orthogonality conditions force (Aξ 0, Bξ 0 ) = (B Aξ 0, ξ 0 ) = f(b A). In particular if L(c, h) is a unitary irreducible representation and T : V (c, h) L(c, h) is the canonical map, then the invariant hermitian form on V (c, h) is just the pull back of the inner product on L(c, h). Let K = {ξ V (c, h) (ξ, V (c, h)) = 0}. Then K is invariant under the Virasoro algebra and the hermitian form passes to a non degenerate invariant hermitian form on L = V (h, c)/k. Since K is invariant under L 0, L is itself a positive energy representation. We claim that L is irreducible. In fact let L be a submodule of L and let v 0 be the image of ξ 0 in L. L can be written as the direct sum of eigenspaces L 0. Choose v 0 in L, the image of Aξ 0. Thus (Aξ 0, Bξ 0 ) 0 for some monomial B, by nondegeneracy. Hence (B Av, v 0 ) 0. But then L (0) 0, so that v 0 lies in L and thus L = L. It follows that the Verma module V (c, h) is irreducible iff (, ) is non degenerate. In particular this happens iff (, ) is non degenerate on every energy subspace V (N). Let M N (c, h) be the P(N) P(N) matrix (L ip L i1 ξ 0, L jq L j1 ξ 0 ) where 1 i 1 i p and 1 j 1 j q with N = i s = j t. The Kac determinant det N (c, h) is the determinant of this matrix. Note that L(c, h) is unitary iff M N (c, h) is positive semi definite for all N. In this case, it is is necessary that det N (c, h) 0 for all N. Using raising and lowering operators to compute the entries of M N (c, h), we see that they are all polynomials in c and h if c, h R. Examples. (0) det 0 (x, h) = ξ 0 2 = 1. (1) det 1 (c, h) = (L 1 ξ 0, L 1 ξ 0 ) = 2h. 3

4 ( ) 4h + c/2 6h (2) det 2 (c, h) = 6h 8h 2 = 2h(16h + 4h 2 + 2hc 10h + c). Lemma. If L(c, h) is unitary, then h 0 and c 0. Proof. The computation of det 1 = 2h shows that h 0. Now for n > 0 we compute L n ξ 0 2 = ([L n, L n ]ξ 0, ξ 0 ) = 2nh + c(n 3 n)/12. For this to be positive for all values of n, we must have c 0. Proposition. For fixed c, det N is a polynomial in h of degree 1 rs N p(n rs). (The coefficient of the highest power of h is independent of c.) Proof. We prove the result by degenerating to bosons. For c fixed, let h = t 2 and a 0 = h 1 L 0, a n = (2h) 1/2 L n. Thus a 0 ξ 0 = ξ 0, [a m, a m ] = ma 0 +t 2 c(m 3 m)/12 and [a m, a n ] = (m n)ta m+n if m n, 0, [a 0, a m ] = mt 2 a m. Moreover a n = a n. If we look at monomials (a ip a i1 ξ 0, a jq a j1 ξ 0 ), these are polynomials in t. We extend these to t = 0; this obviously gives the leading order terms in h in the original problem. In the limit t = 0, we get the system of oscillators a 0 = I, [a m, a m ] = mi. For this bosonic system it is immediate that x = (a mp p am1 1 ξ 0, a nq q an1 1 ξ 0) is zero unless m s = n s for all s, in which case x = m s!s ms. This is independent of c. If we substitute these terms into the determinant for det N, we see that the off diagonal terms vanish when t = 0, so the determinant is given by the product of the diagonal entries, all non zero. Thus lim h h M det N 0 is indepedent of c, where M is the sum of all j k s with j k k = N. Let m(r, s) be the number of partitions of N in which r appears exactly s times. Cearly M = 1 rs N s m(r, s). Now the number of partitions of N in which r appears s times is P(N rs). Thus m(r, s) = P(N rs) P(N r(s + 1)) (where P(0) = 1 and P( k) = 0 for k > 0). Thus M = s m(r, s) = [ n r ] s (P(n rs) P(n r(s + 1)) = s=1 1 rs N P(N rs). Since h M is evidently the highest power of h with a non zero coefficient in det N, the result follows. Definitions. Let h p,q (c) = 1 48 [(13 c)(p2 + q 2 ) + (c 1)(c 25)(p 2 q 2 ) 24pq 2 + 2c]. Set ϕ p,p (c, h) = h h p,p (c) = h + (p 2 1)(c 1)/24 and ϕ p,q (c, h) = (h h p,q (c))(h h q,p (c)) = (h (p q) 2 /4) 2 + h 24 (p2 + q 2 2)(c 1) (p2 1)(q 2 1)(c 1) (c 1)(p q)2 (pq + 1). ( ) If we parametrise c as c = 1 6/m(m + 1), then h p,q (c) = ((m + 1)p mq)2 1. 4m(m + 1) Kac determinant formula. det N (c, h) = C N 1 rs N (h h rs(c)) P(N rs), where C N > 0 is independent of c and h. Lemma 1. If t A(t) is a polynomial mapping into N N matrices and dimkera(t 0 ) = k, then (t t 0 ) k divides det A(t). Proof. Take a basis v i such that A(t 0 )v i = 0 for i = 1,...,k. Thus the first k columns of A(t) are divisible by t t 0 and hence (t t 0 ) k divides deta(t). 4

5 Lemma 2. Fix c and regard det N (c, h) as a polynomial in h. If det k vanishes at h = h 0, then (h h 0 ) P(N k) divides det N. Proof. We may take k minimal subject to det k (c, h 0 ) = 0. Thus V (c, h 0 ) has a singular vector v at energy level k. By the Poincaré Birkhoff Witt theorem, the vectors L it L i1 v are all linearly independent for i t i 1 1. So at level N, this submodule has dimension P(N k). On the other hand this submodule is contained in ther kernel of (, ). Thus M N has a kernel of dimension at least P(N k) at h 0. The assertion therefore follows from Lemma 1. Lemma 3. det N vanishes at h r,s (c) for 1 rs N. Proof. By the GKO construction, ch L(c (m), h (m) r,s ) qh ϕ(q) (1 qrs q r s + ), where r = m r, s = m + 1 s and the inequality is to be understood in terms of coefficients of q i. It follows that the kernel of (, ) in V (c, h) has a non zero component at each energy level N min(rs, r s ). Thus det N vanishes at h (m) rs for m sufficiently large. But then det N vanishes at infinitely many points of the curve ϕ rs (c, h) = 0 (namely (c (m), h (m) r,s )). Since ϕ r,s (c, h) is irreducible in C[c, h], we see that ϕ r,s divides det N for N rs. Thus det N vanishes at h r,s (c) for 1 rs N. Proof of determinant formula (Kac Wakimoto). By Lemmas 2 and 3, det N is divisible by (h h p,q (c)) P(N pq), 1 pq N since the h p,q (c) s are distinct for generic c. Since both sides have the same degree in h and the highest order term in h is indepedent of c (by the Proposition), the result follows. Corollary. (a) For c > 1 and h 0 all L(c, h) are unitary and the Verma modules M(c, h) are irreducible. (b) For c = 1 all L(c, h) with h 0 are unitary. In this case det M pq N (h (p q)2 /4) P (N pq). If h k 2 with k a non negative half integer, then M(c, h) is irreducible. Otherwise if h = k 2, then L(1, h) occurs in the sl 2 multiplicity space of a level one representation of ŝl 2 and M(c, h) is not irreducible. APPENDIX C. THE FRIEDAN QIU SHENKER UNITARITY CRITERION. We prove the easy part of the FQS criterion for unitarity, putting restrictions only on h for 0 < c < 1. FQS Theorem. Let L(c, h) be a unitary representation of the Virasoro algebra with c = 1 6/m(m + 1) for m 3. Then h = h p,q with 1 q p m 1 and h p,q = [(p(m + 1) qm) 2 1]/4m(m + 1). Proof. The factors that appear in the Kac determinant formula the form ϕ p,q (c, h) = (h h p,q (c))(h h q,p (c)) and ϕ p,p (c, h) = h h p,p (c). These define curves C p,q in the region 0 c 1 and h 0. For φ p,p, the curve is a straight line through (1, 0) that intersects the h axis at (p 2 1)/24. The curve defined by ϕ p,q (c, h) = 0 touches the line c = 1 at h = (p q) 2 /4 and is parametrised by a real variable m in c = 1 6/m(m + 1), h p,q (c) = ((m + 1)p mq)2 1. 4m(m + 1) These parametrisations show that the φ p,q is strictly positive when c > 1 and h 0, since m(m + 1) < 0 and the values of h p,q and h q,p are then complex conjugates with non zero imaginary part. If we use the variable x = m + 1/2 to parametrise the curve, then c = 1 6/(x ), h p,q(c) = [(x(p q) (p + q))2 1]/(4x 2 1). This parametrisation applies equally well when p = q. The figure below shows the possible curves schematically (with a rescaling in the h direction). 5

6 The first curve in the picture corresponds to the degenerate case p = q and cuts the h axis at (p 2 1)/24. The factor ϕ p,q appears in det N for the first time when N = pq, i.e. at level N. Each curve has an interior where φ p,q (c, h) < 0 which we denote U p,q, an open subset of the strip [0, 1] [0, ). Its closure is compact and connected. Let X N = pq N U p,q. We will show that thia is the region bounded a explicit sequence of pieces of the curves C p,q as depicted schematically in the diagram below, with the h direction rescaled to make the special points equally spaced on the line c = 1. Fixing a level N, at each of the points M 2 /4 M = 0, 1, 2,..., N 1 we can pick out curves C pq that touch the line c = 1 at that point of highest possible level. In this case pq M < (p+1)(q +1). The highest level curve through the special point with (1, (M +1) 2 /4 is C p+1,q if (p+1)q N and C p,q 1 otherwise. The highest level curve (1, (M 1) 2 /4 is C p,q+1 if p(q + 1) N and C p 1,q otherwise. Note that if p(q + 1) > N then necessarily (p + 1)q > N since p q, so that for are are at most three possibilities for the highest level curves through the two adjacent special points. In particular if a highest curve through a special point has level N, then the highest level curves through the adjacent points have level < N. We consider the piecewise smooth curve formed by the axes and the segments between the highest level curves through the special points M 2 /4 at level N. These bound a region X N. Proposition. X N = X N. Proof. We assume the result by induction on N. 6

7 At level N when a new curve is added at a point M 2 /4 either it is at the top end through (1, (N 1) 2 /4) and forms a curved strip as shown in the diagram; or it interects the two adjacent curved parts of the boundary, forming a shape like a bow tie. In fact it is easy check that the upper part of C N,1 does not intersect any of the other curves of level N in the strip. Moreover for the other upper and lower parts of the curves at level N, including the lower of C N,1, it is easy to check that they first meet the curved part of the boundary of X N 1 at a point on the curve through the adjacent point (M ± 1) 2 /4 and have no further intersections with the curved part of the boundary. Proof of FQS Theorem. We first prove that for fixed c (0, 1) with c = 1 6/m(m + 1), the only values of h for which L(c, h) can be unitary are the h (m) p,q with p, q 1 and p + q m. At level N define the Nth excluded region R N as being the open set where det n is negative for some n N. We claim that RN = {(c (m), h (m) p,q ) : m > p + q 1}. Note first that every point (c, h) with 0 < c < 1 and h > 0 lies in the interior of one of the curves C p,q for pq sufficiently large. Indeed the regions U p,p sweep out the region since the point of intersection of C p,p with the h axis is at (p 2 1)/4, and thus tends to infinity. We shall prove that the only parts of X N that might not lie in R N are the new parts of the boundary added at level N. It is easy to check that if the piece of the boundary formed by C p,q with pq = N corresponds to values of the parameter x with x > p + q 1/2. In fact the functions det N do not vanish for c > 1 and therefore have the same sign. Since the Segal Sugawara construction gives positive values for all N, it follows that det N is positive for all c > 1, h > 0 and N 0. Hence if p q, near (1, (p q) 2 /4), det N is positive for c > 1. Thus on one side of the curve C p,q near (1, (p q) 2 /4), det N is positive. Note that near (1, (p q) 2 /4), the function det N changes sign as C p,q is crossed, because at the Nth stage h p,q is a simple zero of det N. We can therefore exclude the interior of the new part of the region at the level N, so only the boundary curves remain, where det N vanishes. Thus we have shown that the only possible points of unitarity in h > 0 and c (0, 1) have h = h (m) p,q with p + q < m + 1, except possibly for points added on the boundary of the strip formed by C N,1. A separate argument is required in that case, which is simple version of the arguments used to prove restrictions on c. Lemma. For h 0 and c > 0, the number of negative eigenvalues of the matrix of inner products M N (c, h) is non decreasing as N increases. Proof (Kac). Let e = L 1 and f = L 1 and h = [e, f] = 2L 0. Thus e = f, h = h, [h, e] = 2e and [h, f] = 2f. The representation on V (h, c) is positive energy, so breaks up as a direct sum of orthogonal irreducible modules for this Lie algebra, with respect to the invariant hermitian form. each generated by a lowest energy vector v i with L 0 v i = h i v i. On any module the form is either positive definite, negative definite or zero. Unless v i is a vacuum vector (h = 0), each module is infinite dimensional with a basis given by the vectors L m 1 v i (m 0). A basis of vectors which contribute negative eigenvalues at level N is therefore formed of vectors L m 1v i with (v i, v i ) < 0 and m + h i = N. The result follows. We now use this lemma to rule out the part of C N,1 that lies in the interior of C N+1,1, i.e. the upper part of C N,1 with m > N. In the interior of strip formed between C N+1,1 and C N,1, the number of negative 7

8 eigenvalues of M N+1 (c, h) is at least one and det N+1 is negative. Crossing the boundary formed by C N,1 changes the sign of the determinant, so it is positive there. By the lemma, M N+1 (c, h) has at least one negative eigenvalue there. So positivity of the determinant implies it has at least two negative eigenvalues there. Taking a sequence of point tending to h N,1 (c) and unit orthogonal eigenvectors there for the two negative eigenvalues, using compactness it follows that M N+1 (c, h) has two orthogonal eigenvectors with non positive eigenvalues. On the other hand, for c fixed, h N,1 (c) is a simple root of det N (c, h), so that 0 is an eigenvalue of M N+1,1 (c, h) of multiplicity one at h = h N,1 (c). Hence M N+1 (h, c) has a negative eigenvalue there, as required. Finally we specialise to the case when m 3 is an integer. We have shown that if L(c, h) is unitary then h = h (m) p,q = [(p(m + 1) qm) 2 1]/4m(m + 1) with p, q 1 and p + q m. We want to show that h = h (m) rs with 1 s r m 1. Note that since p, q 1 and p + q m, we have p, q m 1. If p q, we take r = p, s = q. Otherwise q p Let p = m p and q = m + 1 q. Then h (m) pq = h (m) p,q and 1 q p m 1. so in this case we may take r = p, s = q. This completes the proof of the FQS theorem. Corollary. For c = 1 6/m(m + 1) with m 3, the values of h from which L(c, h) is unitary are given by h = h r,s = [(r(m + 1) sm) 2 1]/4m(m + 1) with 1 s r m 1. We next prove the complete version of the Friedan Qiu Shenker unitarity theorem. Theorem (Friedan Qiu Shenker). If 0 < c < 1, then a representation with central charge c is unitary iff c = 1 6/m(m + 1) with m 3. Remark. From Section 10, the permitted values of h for a particular m 3 are h p,q (m) with 1 q p m 1. We already know that any point not on a curve C p,q cannot be unitary. The points h p,q (m) are precisely the points of the boundaries of the regions X N where the adjacent curves C p,q intersect. Each intersection P may be described by first choosing a curve C p,q with P C p,q and pq minimal and then choosinf C p,q with P C p,q and p q minimal. It is easy to check that these intersections are obtained by taking p = p + k, q = q +k for k 1 and m = p+q +k 1. We now rule out all the points between these intersection points. Note that every curve C p,q touches c = 1 at h = (p q) 2 /4. Thus if C p,q intersects C p,q, then the part of C p,q on the c < 1 side of C p,q is the part with h decreasing if h < h and h increasing if h > h. Take a point P 0 on C p,q corresponding to p, q as above. Let N = pq < N = p q. Starting from the asymptote at c = 1 through (p q ) 2 /4, we may follows the C p,q curve as it travels to P 0 C p,q. We get a stright line parametrisation of C p,q by taking y = m as coordinate. Along the way to P 0, the curve will cross other curves C p,q transversely and simply at points P 1,...,P k. At level N, the dimension of the null space is 1 on C p,q away from the intersection points. Near the asymptote on the c > 1 side of C p,q, the matrix of inner products A(c, h) is positive definite. We shall find a open neighbourhood U of the part of the curve C p,q above P 0 (containing P 1,...,P k and part of the asymptote to c = 1) and a rank one spectral projection P(c, h) of A(c, h) in this strip depending continuously on (c, h) such that P(c, h)a(c, h) = λ(c, h)p(c, h) with λ(c, h) = 0 only on C p,q. It will follow that λ(c, h) < 0 on the c < 1 side of C p,q. In particular A(c, h) will have a negative eigenvalue on C p,q in the segment between P 0 and the next intersection. This clearly will prove the unitarity theorem. Using m as parameter, we can replace the curve C p,q by the y axis. The following result (with M = 1) shows the existence and uniqueness of the spectral projection P(z) for z in an open neighbourhood of the y axis with z P i. (This open neighbourhood should of course contain P 0,...,P k.) Lemma 1. Let A(z) be a continuous self adjoint matrix valued function on a topological space Z such that ker(a(z)) has constant rank 1 (or rank M more generally) for z Z 0, a closed subset of Z, and is invertible otherwise. For each z Z 0, there is an open neigbourhood U of z such that if P(z) is the orthogonal projection onto (Mth) the lowest eigenspace(s) of A(z) 2, then z P(z) is continuous on U. If Z is an open subset of R n and A(z) is a smooth (or analytic) function of z, then z P(z) is also smooth (or analytic) on U. 8

9 Proof. Take z Z 0. By minimax there is a neighbourhood U of z such that the lowest eigenvalue of A(t) 2 is less than r/2 and the next eigenvalue is greater than are greater than 2r > 0. Let χ be a continuous bump function supported in ( r, r) with χ(0) = 1. Then P(z) = χ(a(z) 2 ) for t U, since the only the lowest eigenvalue of A(z) occurs in ( r, r). Since χ can uniformly approximated by polynomials on any compact interval, it follows that z A(z) is continuous on U. The second assertion follows immediately from the contour integral expression for the spectral projection P(z): P(z) = 1 (wi A(z)) 1 dw. 2πi w =ε Corollary. There is an open subset U of Z containing Z 0 on which P(z) can be defined (uniquely). Proof. Take an open neighbourhood U z for each point z Z 0 and set U = z Z 0 U z. By uniqueness the different P(z) s must agree on intersections of these opens. We next need to use information from the Kac character formula to continue P(z) across the points P i. We simply have to define P(z) in an open neighbourhood of each point P i. Since the intersection at P i is transverse, we may assume that the transverse curve is the x axis and P = P i corresponds to the point (0, 0). Proposition. In an open neighbourhood of P = P i there are (a) a unique continuous determination of a rank one projection P(z) such that P(z) is a spectral projection of A(z) coinciding with the projection onto the kernel of A(z) for (0, y) with y 0. (b) a unique continuous determination of a rank m projection Q(x) on y = 0 such that Q(x) is the spectral projection onto the kernel of A(x, 0). Moreover P and Q are orthogonal on y = 0 and P(0) + Q(0) is the projection onto the kernel of A(0, 0). Proof. Note that Q(z) could be constructed on y = 0, x 0 using the method of the previous lemma. We need a variant of this construction. Let C = C p,q and let the transverse curve be C = C p,q with N = p q < N. The kernel of M N is rank one on C. We choose parameters such that P = (0, 0), C is the y axis and C the x axis. As in the previous lemma, let R(z) be a smooth or analytic determination of a spectral subspace of M N (z) giving the kernel on y = 0. Let u(z) = R(z)u 0 / R(z)u 0 be a smooth or analytic choice of eigenvector near z = 0. Define vectors u j (z) = L j1 L jr u(z) at level N for ji = N N. There are P(N N ) such vectors and they form a basis of A(z) for z = (x, 0) with x 0. The Gram schmidt orthonormalisation process shows that the orthogonal projection Q(z) onto the subspace spanned by the u j (z) s is smooth or analytic. Thus Q(z) is a projection of rank P(N N ) defined in a neighbourhood of 0. Note that Q(z) is a spectral projection of A(z) for y = 0, x 0, but not necessarily otherwise. Let B(z) = Q(z)A(z)Q(z) considered as a self adjoint operator on im(q(z)). Since im(q(x, 0)) ker(a(x, 0)), we must have B(x, 0) = 0. Hence B(x, y) = yb 0 (x, y) where B 0 is smooth or analytic. The next lemma shows that B 0 (0) is invertible. Lemma 2. detb 0 (0) 0. Proof. We claim that detm N (c, h + pq) 0 where N = N N. In fact if the determinant vanished, (c, N + pq) would have to lie on some C rs with rs N = pq p q. Thus By assumption p = q 1 + k, q = p + k for some k 1. Hence (m + 1)p + mq = ±[(m + 1)r ms]. (1) rs p q pq = m(m + 1) (m + 1)p mq. (2) Combining (1) and (2) yields rs ± (m + 1)r ms) m(m + 1), so that (r ± m)(s (m + 1)) 0. Since 1 r, s m, it follows that r = m or s = m + 1 and equality holds in (2). Reducing modulo m or m + 1, we deduce that p = m or q = m + 1, neither of which is compatible with m = p + q + k 1. Thus the claim holds. 9

10 Let u(x) be the null vector at level N and as above extend u to u(z). Since the submodule generated by u(z) is isomorphic to the Verma module M(c, h + pq), the corresponding matrix of inner products is ψ(z) M N N where ψ(z) = (u(z), u(z)). But ψ(x, 0) = 0, so that y ψ(x, y). From the Kac determinant formula, detm N (x, y) = yf(x, y) with f(0) 0. Hence ψ(z) = yg(z) with g(0) 0. But then detb(z) = (u(z), u(z)) P(N N ) h(z) where h(0) 0. Thus y P(N N ) is the highest power of y dividing det B(z). Since B(z) = yb 0 (z), we must have det B 0 (0) 0, as required. By continuity we deduce that B 0 (z) is invertible (possibly by shrinking the neighbourhood of 0). With ( respect to the orthogonal ) decomposition corresponding to I = Q(z) ( (I Q(z)), we) may write A(z) = B(z) C(z) yb0 (z) yc C(z). As above we have C(z) = yc D(z) 0 (z), so that A(z) = 0 (z) yc 0 (z). We have already ( ) D(z) a seen that B 0 (z) is invertible. If we try to solve A(z)v = 0 with v =, we find a = B 1 b 0 C 0b and (D yc B 1 C)b = 0. Looking at the kernel at (0, 0), we see that D(0, 0) has one dimensional kernel. Likewise D(0, y) has one dimensional kernel for y 0. On the other hand D(x, y) must have zero kernel for x 0. Thus we may define P(z) near z = 0 as the spectral projection of F(z) = D yc B 1 B corresponding to the lowest eigenvalue, just as in Lemma 1. By definition F(z) and hence P(z) is orthogonal to Q(z) on the x axis. By construction P(z) and Q(z) have all the required properties. Remark. Recall that if P and Q are orthogonal projections with P Q < 1/2, then T = PQ+(I P)(I Q) satisfies check that T I < 1 and PT = TQ. Thus T is an invertible operator conjugating P into Q. Since TT commutes with P, U = (TT ) 1/2 T gives a unitary such that UQU = P. This means that on a sufficiently small neighbourhood of 0 the projections (I Q(z)) can be identified using a unitary gauge change. This is not true for self adjoint maps A(z)! Corollary. The rank one projection valued function P(z) can be defined on a neighbourhood of the C p,q containing the points P 0,..., P k. It is continuous and satisfies P(z)A(z) = A(z)P(z) = λ(z)p(z) with λ(z) = 0 iff z C p,q. λ(z) < 0 on the c < 1 side of C p,q. Proof. These first part follows taking the open to be a (finite) union of neighbourhoods of the P i s and fintely many other points on C p,q. For c near 1, the matrix M N (c, h) is positive definite on the c > 1 side of C p,q, invertible off C p,q and has one dimensional kernel on C p,q. On the other hand detm N < 0 on the c < 1 side of C p,q. By minimax, at most one eiegenvalue of A(z) can change sign crossing C p,q. The determinant condition therefore implies that it is the lowest eigenvalue of A that changes sign. (This evidently corresponds to the lowest eigenvalue of A 2.) Thus λ(z) < 0 on the c < 1 side of C p,q. Since λ(z) is real and non zero on the c < 1 side of C p,q, the last assertion follows. This last corollary completes the proof of the unitarity theorem. 10