Virasoro vacuum construction, minimal-model cofiniteness, and vacuum unitarity

For c∈C put $b_{\mathrm{v}\mathrm{a}\mathrm{c}}=\mathrm{span}\{C,L_n:n\ge -1\}$; it is a Lie subalgebra. Let $C_c$ be its one-dimensional representation with $C=c$ and every $L_n=0$. It is consistent: the central bracket coefficient vanishes for $n=-1,0,1$, so no bracket within $b_{\mathrm{v}\mathrm{a}\mathrm{c}}$ gives a nonzero scalar. Set $V^c=U(\mathrm{Vir}){\otimes }_{U(b_{\mathrm{v}\mathrm{a}\mathrm{c}})}{C}_c$. PBW with the negative block $\{L_{-n}:n\ge 2\}$ proves that the ordered monomials $L_{-n_1}\cdots L_{-n_r}1$, $n_1\ge \cdots \ge n_r\ge 2$, are a basis. In particular 1≠0, $V_0^c=C1$, every homogeneous degree is finite dimensional, and $L_0$ acts by its nonnegative integer degree.

Let $T=L_{-1}$ and $\mathrm{\Omega }(z)={\sum }_{n\in Z}{L}_n{z}^{-n-2}$. It is a field on $V^c$: a positive $L_n$ lowers the degree by n, so kills a fixed homogeneous vector when n exceeds its degree. We have T1=0 and $[T,L_n]=-(n+1)L_{n-1}$, proving [T,Ω(z)]=∂_zΩ(z). For $\omega =L_{-2}1$, repeated brackets give $T^j\omega =j!L_{-j-2}1$, so $\mathrm{\Omega }(z)1=e^{zT}\omega$.

The Virasoro commutator gives the exact distribution identity $[\mathrm{\Omega }(z),\mathrm{\Omega }(w)]={\mathrm{\partial }}_w\mathrm{\Omega }(w)\delta (z,w)+2\mathrm{\Omega }(w){\mathrm{\partial }}_w\delta (z,w)+(c\mathrm{/}12){\mathrm{\partial }}_w^3\delta (z,w)$. For example the coefficients of the central term are $(c\mathrm{/}12)(m+1)m(m-1){\delta }_{m+n,0}$ in the $L_m,L_n$ convention. The identities $(z-w)\delta =0$ and $(z-w){\mathrm{\partial }}_w^j\delta =j{\mathrm{\partial }}_w^{j-1}\delta$ imply $(z-w{)}^4[\mathrm{\Omega }(z),\mathrm{\Omega }(w)]=0$. Thus Ω is local with itself. Its negative field modes are ${\mathrm{\Omega }}_{(m)}=L_{m-1}$, $m<0$, hence are precisely the $L_{-n}$, n≥2, which span $V^c$ on the vacuum by PBW. All four hypotheses of the actual reconstruction lemma are now verified. It supplies the unique vertex-algebra state-field map with Y(ω,z)=Ω(z). The Virasoro relations, $T=L_{-1}$, and the proved nonnegative finite grading verify the local conformal-vertex-algebra definition for $V^c$.

A Virasoro-stable subspace $U\subset V^c$ is stable under every reconstructed state mode. Indeed every state is a linear combination of the negative Ω-mode words above. The field of Ω_{(−k)}b is the corresponding negative field product, obtained inductively by the actual reconstruction proof. Its coefficient on a fixed vector is a finite sum of compositions of Ω-modes and modes of the shorter state's field, by the explicit iterate formula in that supplier. Induction on the state word therefore proves preservation of U. Conversely a vertex ideal is stable under Ω-modes. Thus the two notions coincide. Translation is already an Ω-mode. Every Virasoro-stable subspace is graded, since $L_0$ has distinct integer eigenvalues and polynomial interpolation projects the finite support of any vector to its components. A proper one has no grade-zero vector, since 1 generates $V^c$. The sum J of all proper Virasoro submodules still has zero grade-zero space, so is proper; it is the unique maximal proper submodule. By the preceding paragraph it is a vertex ideal, and $L(c,0)=V^c\mathrm{/}J$ is a simple vertex algebra with the descended conformal vector. Its vacuum remains nonzero, its grade-zero space is C1, and its grades are finite dimensional and nonnegative. Thus it is a simple conformal vertex algebra of CFT type exactly as used by the minimal-model Statement. This includes $c=0$: the simple quotient may be C with zero conformal vector, which the local definition allows. No nonzero-conformal-vector or nontriviality assumption has been silently added.

Let $B_N(c)$ be the real PBW Gram matrix on the universal vacuum quotient $V^c=M(c,0)\mathrm{/}U(\mathrm{Vir})L_{-1}1$, using partitions of N with all parts≥2. It is polynomial in real c. If $G_N(c,h)$ is the Verma Gram matrix, then generically in c $\det B_N(c)=2^{-P(N-1)}{\lim }_{h\to 0}\det G_N(c,h)\mathrm{/}(h^{P(N-1)}\det G_{N-1}(c,1))$. To verify this, choose at grade N the PBW block consisting of all descendants of $L_{-1}{v}_h$ and the complementary PBW block with no rightmost $L_{-1}$. All entries pairing a descendant of $L_{-1}{v}_h$ with any vector vanish at $h=0$, because $L_{-1}{v}_0$ is singular and lies in the radical. The first block, divided by 2h, tends to the normalized Verma form of weight1: moving positive modes across the pairing proves contravariance in the limit, the top pairing is 2h, and the order-h errors in the positive-mode highest-vector equations vanish after pairing against the other singular-descendant block. More explicitly $L_1{L}_{-1}{v}_h=2h{v}_h$ and $L_n{L}_{-1}{v}_h=0$ for $n>1$; pairing the exceptional $v_h$ term against a descendant of $L_{-1}{v}_h$ introduces one more factor h by the preceding radical vanishing. Thus the limit is the unique normalized contravariant form on M(c,1). The off-diagonal blocks are O(h). When det $G_{N-1}(c,1)$ is nonzero, the Schur complement of the first block therefore tends to $B_N(c)$, giving the displayed formula. It extends as an identity of rational functions in c and then by polynomial continuation wherever cancellation is needed.

For sufficiently large positive c, $B_N(c)$ is positive definite. Rescale a PBW vector of length l by $c^{-l\mathrm{/}2}$. At c→+∞ only complete central contractions survive. Two distinct partitions have no complete matched central contraction, and the limiting matrix is diagonal with entries ${\prod }_{n\ge 2}{m}_n!((n^3-n)\mathrm{/}12{)}^{m_n}>0$. Every noncentral commutator lowers the power of c after this scaling. Hence the determinant is positive at sufficiently large c.

The Kac factors locate every possible zero of $B_N$. Put $c=1-6(b-b^{-1}{)}^2$ and $h_{r,s}=((rb-s{b}^{-1}{)}^2-(b-b^{-1}{)}^2)\mathrm{/}4$. Apart from the identically zero $h_{1,1}$, a zero at $h=0$ solves $(r-1)b^2=s-1$ or $(r+1)b^2=s+1$. For a reduced positive ratio $b^2=q\mathrm{/}p$, $p,q\ge 2$ coprime, the first possible additional zero occurs at $(r,s)=(p-1,q-1)$, grade $N_0=(p-1)(q-1)$; the other family begins at $(p+1,q+1)$, and multiples have larger grade. Ratios with $p=1$ or $q=1$ give $c\le -2$ and do not occur at positive $c$. At $c=1$ the numerator zeros with $r=s\ge 2$ cancel the denominator zeros with $\mathrm{\mid }r-s\mathrm{\mid }=2$: the numerator factor $h_{r,r}=(r^2-1)(1-c)\mathrm{/}24$ is simple in $c$, while the pair of denominator factors with indices $(s+2,s),(s,s+2)$ has a simple product zero in $c$ (their individual leading terms are opposite nonzero multiples of $\sqrt{1-c}$). Their multiplicities match because $N-1-s(s+2)=N-(s+1{)}^2$. Thus $c=1$ is not an extra zero of $B_N$. For $c>1$, choose $b$ with the stated central charge and put $u=b^2$. A numerator zero at $h=0$ forces $u>0$ by the two equations above. A denominator zero requires $h_{r,s}=1$, equivalently $(r^2-1)u^2-2(rs+1)u+(s^2-1)=0$. For $r>1$ its roots are $(s-1)\mathrm{/}(r+1)$ and $(s+1)\mathrm{/}(r-1)$; after discarding a zero root, every remaining root is positive. For $r=1$ its only possible nonzero root is $(s-1)\mathrm{/}2$, which is positive. Thus every such root is positive, which gives $c=1-6(u-1{)}^2\mathrm{/}u\le 1$, a contradiction. So neither numerator nor denominator has a nontrivial Kac zero for $c>1$, and $B_N$ has no zero there.

For coprime p,q≥2 put $c=1-6(p-q{)}^2\mathrm{/}(pq)$, $N=(p-1)(q-1)$, $D=N\mathrm{/}2$. This proof uses the actual all-parameter Kac determinant and the universal vacuum construction. It proves the original minimal-model C2 clause, including a nonzero pure $L_{-2}$ coefficient; it does not assume a Feigin–Fuchs singular-vector formula or classify modules. Use the vacuum determinant identity proved above: $\det B_n(c)=2^{-P(n-1)}{\lim }_{h\to 0}\det G_n(c,h)\mathrm{/}(h^{P(n-1)}\det G_{n-1}(c,1))$. Here $B_n$ is the Gram matrix on the universal vacuum module in its parts≥2 PBW basis. At b²=q/p, the positive-index solutions of $h_{r,s}(c)=0$, other than (1,1), have (r,s)=(kp−1,kq−1) or (kp+1,kq+1), with k≥1. Their first grade is N, attained only by (p−1,q−1). The positive-index solutions of $h_{r,s}(c)=1$ have (r,s)=(kp+1,kq−1) or (kp−1,kq+1), k≥1; their first grades are (p+1)(q−1) and (p−1)(q+1), both strictly greater than N. These statements follow by solving qr−ps=±(p−q) or ±(p+q), respectively, using gcd(p,q)=1. Thus the determinant formula makes every $B_n$ with $n<N$ nondegenerate and $B_N$ degenerate. The exceptional $h_{1,1}$ factor is exactly the factor $h^{P(n-1)}$ removed before the limit. The derivative of the additional $h=0$ Kac factor at this parameter is nonzero ($b>0$ and b≠1), so the formula has a simple zero at $n=N$, though only existence of a null vector is needed. Choose v≠0 in the radical of $B_N$. Contravariance makes $L_m$ v orthogonal to every vector of grade N−m, for every $m>0$. When N−m≥0, that grade is nondegenerate by the preceding paragraph, so $L_{m}v=0$; when $N-m<0$ it vanishes by the vacuum grading. Hence v is an actual singular vector, not just a vector with determinant zero. The integer N is even: if p and q are both odd, both p−1,q−1 are even; otherwise one of them is even because p,q cannot both be even.

Filter the vacuum PBW space by word length in the modes $L_{-n}$, n≥2. Its associated graded is the polynomial ring $C[x_2,x_3,\dots ]$, where $x_n$ is the symbol of $L_{-n}$; at a fixed conformal degree only finitely many variables and words occur. On the top-length symbol, every positive $L_m$ acts as the derivation $D_m={\sum }_{n\ge m+2}(m+n)x_{n-m}\mathrm{\partial }\mathrm{/}\mathrm{\partial }{x}_n$. Indeed $[L_m,L_{-n}]=(m+n)L_{m-n}$ plus a central term. When n≥m+2 the new negative mode still has index at most−2 and preserves word length. When $n=m+1$ it is $L_{-1}$; moving it to the right until it kills the vacuum replaces two negative factors by one in every surviving commutator. When $n=m$ it is $L_0$ or the scalar central term and lowers length by at least one; when $n<m$ the positive mode must commute through another factor before surviving, again lowering length. Thus there are no other top-length contributions. Lower-length components of v cannot contribute at its top length under any positive mode.

Let f be the nonzero highest-length symbol of v. Singularity gives $D_{m}f=0$ for every m≥1. If $x_R$, $R>2$, is the largest variable occurring in f, then $D_{R-2}f=(2R-2)x_2\mathrm{\partial }f\mathrm{/}\mathrm{\partial }{x}_R$, because no variable of index greater than R occurs. The polynomial ring is an integral domain in characteristic zero, so f is independent of $x_R$, a contradiction. Repeating removes every variable except $x_2$. Homogeneity of v forces $f=a{x}_2^D$ with a≠0. Therefore v has a nonzero coefficient of $L_{-2}^{D1}$, and every other PBW monomial in it has at least one part≥3. This proves the needed coefficient, rather than asserting it from determinant vanishing.

Let L(c,0) be the simple quotient of the constructed universal vacuum module. The vector v is in its proper radical, so its image is zero. Any PBW monomial with some part n≥3 is in C2 of this quotient: order that part leftmost, write $L_{-n}={\omega }_{(-n+1)}$, and use the state derivative formula $(T^{n-3}\omega {)}_{(-2)}=(n-2)!{\omega }_{(-n+1)}$. Thus $v=0$ gives a nonzero multiple of $L_{-2}^{D1}$ equal to a linear combination of C2 vectors. Moreover the span of PBW monomials with a part≥3 is stable under $L_{-2}$: reorder using $[L_{-2},L_{-n}]=(n-2)L_{-n-2}$, whose new index remains at least3. Applying $L_{-2}$ repeatedly to the relation therefore places every pure power $L_{-2}^{j1}$, j≥D, in C2 as well. All remaining PBW monomials already lie in C2 by the same leftmost-part argument. Consequently L(c,0)/C2 is spanned by 1, $L_{-2}1,\dots ,L_{-2}^{D-1}1$ and has dimension at most D. The case $p,q=2,3$ gives $N=2,D=1$ and the one-dimensional trivial quotient, as expected.

Set $c_m=1-6\mathrm{/}[m(m+1)]$, m≥2. Fix $c_m<c<c_{m+1}$ and $N=m(m+1)=A$. There is precisely one possible determinant zero above $c_m$, namely $c_{m+1}$, and it is simple. Here is the elementary bound excluding every other rational root. If $c_{p,q}>c_m$, with $d=\mathrm{\mid }p-q\mathrm{\mid }$, then $pq>d^{2}A$. The increasing function x−sqrt(d²+4x)+1 gives (p−1)(q−1)>d²A−d(2m+1)+1. For d≥2 this is greater than A: $(d^2-1)m(m+1)-d(2m+1)+1>0$ for m≥2, starting already at $d=2$ and increasing thereafter. If $d=1$, write $q=p+1$. Then $c_{p,p+1}>c_m$ implies p≥m+1, while (p−1)p≤A implies p≤m+1. Thus only $p=m+1,q=m+2$ occurs, with $N_0=A$. At that root the additional Kac factor has multiplicity P(0)=1. The denominator det $G_{N-1}(c,1)$ is nonzero there: the equation $h_{r,s}=1$ gives qr−ps=±(p+q), whose first positive solutions are (p+1,q−1) and (p−1,q+1), both of product greater than N−1. There are no other $h=0$ numerator factors at grade≤N there. Hence det $B_N$ has one simple zero at $c_{m+1}$ and no other zero in $(c_m,+\mathrm{\infty })$. It is positive for large c, so it is negative on $(c_m,c_{m+1})$.

A positive invariant Hermitian form on the simple vacuum quotient pulls back, after normalizing the vacuum norm to1, to the canonical vacuum Gram form, because adjointness determines each PBW pairing. Every $B_N$ would therefore be positive semidefinite. The preceding negative determinant excludes every open interval $(c_m,c_{m+1})$. Negative c is already excluded by $\mathrm{\mid }\mathrm{\mid }{L}_{-2}1\mathrm{\mid }{\mathrm{\mid }}^2=c\mathrm{/}2$. Since $c_m$ increases from0 to1, the only allowed c in [0,1) are the $c_m$. This proves precisely the necessary vacuum restriction used below, without an unproved FQS curve argument.

Construct the algebraic NS Fock space on two complex fermion species $p_r^a,q_r^a$, $a=1,2$, r∈Z+1/2, with $\{p_r^a,q_s^b\}={\delta }_{a}b{\delta }_{r+s,0}$, the other anticommutators zero, and all positive modes killing a vacuum of norm1. Fix the order of negative generators by increasing positive frequency, then by species and by p before q. Ordered negative-mode exterior monomials are an orthonormal basis, with $(p_r^a{)}^{†}=q_{-r}^a$, $(q_r^a{)}^{†}=p_{-r}^a$. This defines a positive Hermitian space directly. Give each negative generator $p_{-r}^a,q_{-r}^a$, $r>0$, degree r, the vacuum degree0, and each exterior monomial the sum of its generator degrees. This is a grading by nonnegative half-integers. Each degree E has finitely many monomials, because only the four generators at each positive half-integer frequency r≤E can occur and each occurs at most once. CAR makes every mode $p_r^a,q_r^a$ homogeneous of degree−r: negative modes create, and positive modes contract the paired negative generator. Consequently each normally ordered summand of $J_n^{ab}$ below has degree−n; so does its locally finite sum. In particular $J_n^{ab}$ kills degree E whenever $n>E$. Define $J_n^{ab}={\sum }_r:p_r^a{q}_{n-r}^b$:, with creators left and the fermionic sign on interchange. Each sum is locally finite on a finite exterior monomial. Direct CAR commutation gives $[J_n^{ab},p_s^c]={\delta }_{b}c{p}_{n+s}^a$ and $[J_n^{ab},q_s^c]=-{\delta }_{a}c{q}_{n+s}^b$. Therefore the current-current commutator differs from δ_bc $J_{m+n}^{ad}-{\delta }_{a}d{J}_{m+n}^{cb}$ by an operator commuting with all fermions, hence a scalar. The joint kernel of the positive modes is exactly the vacuum line: the annihilator paired with a negative generator contracts it from each ordered exterior word containing it, with sign ±1; distinct words containing that generator give distinct shortened words. Thus annihilation forces every coefficient of a word containing that generator to vanish, and applying this to every generator leaves only the empty word. A commuting operator sends the vacuum into this joint kernel, so its value there is scalar; commuting with all creators fixes that same scalar on the entire exterior basis. Energy forces that scalar to vanish unless $m+n=0$. For $m>0$ its vacuum matrix coefficient counts the m half-integers between−m and0 in the two creating modes, giving mδ_adδ_bc. Antisymmetry gives all other m. Hence $[J_m^{ab},J_n^{cd}]={\delta }_{b}c{J}_{m+n}^{ad}-{\delta }_{a}d{J}_{m+n}^{cb}+m{\delta }_{a}d{\delta }_{b}c{\delta }_{m+n,0}$, $(J_n^{ab}{)}^{†}=J_{-n}^{ba}$.

Take $E=J^{12}$, $F=J^{21}$, $H=J^{11}-J^{22}$. These are restricted sl2 currents at level1, with normalized form B(E,F)=1,B(H,H)=2 and adjoints $E_n^{†}=F_{-n}$, $H_n^{†}=H_{-n}$. Every nonnegative current mode kills the vacuum. Tensor l+1 copies of this explicitly positive Fock representation. The sum of currents in the first l factors has level l; the last factor has level1; their diagonal has level l+1. The affine brackets follow by adding independent-factor brackets. All current actions are restricted and homogeneous for the finite-piece fermion energy grading.

Apply the published Sugawara current-action and Virasoro theorems to these three actions. For sl2 the adjoint Casimir for B is4, so $h^{\vee }=2$; this follows immediately from the basis E,F,H and their three brackets. Their stress tensors have central charges 3l/(l+2), 1 and 3(l+1)/(l+3). The Sugawara modes are Hermitian in the stated adjoint convention: the quadratic Casimir sum is H H/2+EF+FE, which is fixed by adjoint and index reversal; reindexing the finite-on-vectors normal-order sums gives $L_n^{†}=L_{-n}$. Put $C_n=L_n^{(l)}+L_n^{(1)}-L_n^{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(l+1)}$. The published current-action formula gives $[C_m,J_n^{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}}]=0$. Since the denominator Sugawara modes are locally finite quadratic expressions in those currents, $C_m$ commutes with every denominator Sugawara mode as well. Subtracting the two Virasoro brackets therefore proves that $C_n$ is Virasoro with charge 3l/(l+2)+1−3(l+1)/(l+3)=1−6/[(l+2)(l+3)].

No branching or integrable-module classification is used. These $C_n$ inherit $C_n^{†}=C_{-n}$ on the same positive Fock space. The vacuum is killed by $C_n$ for n≥−1: each Sugawara term has this property because every nonnegative current kills the vacuum and two strictly negative integer indices cannot sum to−1 or a larger integer. Let S be the cyclic Virasoro module on that vacuum under $C_n$. It has finite-dimensional nonnegative integer graded pieces: every $C_n$ has fermion energy degree−n, and $C_0$ acts by grade on its cyclic span, since $C_0$ kills the vacuum and $[C_0,C_{-n}]=n{C}_{-n}$. Thus a word of total index N has both fermion degree N and $C_0$ eigenvalue N, including the empty word at $N=0$. Thus its inherited positive form is an actual vacuum invariant Hermitian form. It is simple. Indeed any proper submodule is graded by interpolation in $C_0$, has zero vacuum grade, and has a grade-by-grade orthogonal complement stable under every $C_n$ by the adjoint relation. That complement contains the vacuum and hence all its cyclic span S, so the submodule is zero. The universal vacuum construction therefore identifies S with L(c,0) at $c=1-6\mathrm{/}[(l+2)(l+3)]$. Taking $l=m-2$ gives every $c_m$ for m≥3. For $m=2$ use the one-dimensional trivial Virasoro representation at $c=0$, whose vacuum norm is positive. This proves sufficiency for every $c_m$, including the degenerate minimal model.

For coprime p,q≥2 set $d=\mathrm{\mid }p-q\mathrm{\mid }$ and $c=c_{p,q}$. If $c<0$ there is no positive form by the $L_{-2}$ norm. If $c=0$, pq=6d² and gcd(d,pq)=1 force $d=1$. If $0<c<1$, the necessary restriction gives $c=c_m$, so pq=d²m(m+1). Again gcd(d,pq)=1 forces $d=1$. Conversely $d=1$ means {p,q}={m,m+1} for some m≥2, and the explicit construction above supplies the positive form. Thus the original vacuum unitarity clause holds exactly for |p−q|=1, independently of the still-separate general minimal-model Zhu polynomial/classification/cofiniteness proof.