Kac determinant for Virasoro Verma modules

The constant vector in ${\mathcal{F}}_t$ is highest weight for the Virasoro action of [F1]. Hence the balanced map $x\otimes v_{h_p(t)}\mapsto x{e}^{t\alpha }$ defines the Verma-to-Fock homomorphism. Let $A_N(p,t)$ be its degree-$N$ matrix, using the same ordering of partitions on the PBW and oscillator-monomial sides. The negative modes act by $L_{-m}=(t+(p-1)(m-1)\mathrm{/}(2p))x_m+(4p{)}^{-1}{\sum }_{j=1}^{m-1}{x}_j{x}_{m-j}+{\sum }_{j\ge 1}j{x}_{m+j}{\mathrm{\partial }}_{x_j}$. Thus the column for $\lambda$ is polynomial in $t$ of degree at most $\mathrm{\ell }(\lambda )$, with leading term $t^{\mathrm{\ell }(\lambda )}{x}_{\lambda }$. Consequently $\det A_N$ is monic of degree $D_N$ in $t$, with coefficients in $\mathbb{C}[p,p^{-1}]$. The pairing pullback is normalized and contravariant, hence by [F3] its matrix is $G_N(c(p),h_p(t))=A_N(p,t{)}^{\mathsf{T}}{H}_N(p)A_N(p,t^{\ast })$, where $H_N$ is diagonal with entries $(-2p{)}^{\mathrm{\ell }(\lambda )}{z}_{\lambda }$. In particular invertibility of the two oscillator maps through a given level implies nondegeneracy of the Verma form through that level.

We prove by induction on $N$ the stronger identity $\det A_N(p,t)={\prod }_{r,s\ge 1,rs\le N}(t-t_{r,s}{)}^{P(N-rs)}$, where $t_{r,s}=(p-1-pr+s)\mathrm{/}(2p)$, for all $p\mathrm{\ne }0$. At $N=0$ the map is the identity. Assume this identity for all smaller levels. It suffices first to fix an integer $p>2N$ and prove the next identity for that $p$.

For $r,s\ge 1$ with $rs\le N$, set $u=(p-1-pr-s)\mathrm{/}(2p)$ and $v=u+r$, so $v^{\ast }=t_{r,s}$ and $h_p(u)-h_p(v)=rs$. In composing two consecutive fields, commuting the annihilation exponential at $z_2$ through the creation exponential at $z_1$ contributes $\exp (-2p{\sum }_{m\ge 1}(z_1\mathrm{/}{z}_2{)}^m\mathrm{/}m)=(1-z_1\mathrm{/}{z}_2{)}^{2p}$. For the positive integer $p>2N$ this is a polynomial, and combining it with $z_2^{2p(t+1)}$ gives $z_2^{2pt}(z_2-z_1{)}^{2p}$. Induction on the number of screenings therefore gives $w=Q^r{e}^{u\alpha }={\mathrm{Res}}_{z_1}\cdots {\mathrm{Res}}_{z_r}\mathrm{\Delta }(z{)}^{2p}{\prod }_a{z}_a^{2pu}\exp ({\sum }_{m\ge 1}{x}_m{\sum }_a{z}_a^m\mathrm{/}m)e^{v\alpha }$, with $\mathrm{\Delta }(z)={\prod }_{a<b}(z_a-z_b)$. Each residue selects a finite homogeneous piece, and $2pu$ and all intermediate charge exponents are integers. The output degree is $rs$. Coefficients of the creation exponential in that degree span all symmetric degree-$rs$ polynomials in the $z_a$, because they are nonzero multiples of power-sum products and Newton's identities generate the symmetric polynomial ring. Apply a linear functional on oscillator degree $rs$ taking this exponential component to $({\prod }_a{z}_a{)}^s$. Its value on $w$ is $\mathrm{CT}(\mathrm{\Delta }(z{)}^{2p}{\prod }_a{z}_a^{-p(r-1)})$. Write $V(z)=\mathrm{\Delta }(z{)}^p={\sum }_{\mu }{c}_{\mu }{z}^{\mu }$. Reciprocal symmetry gives $V(z^{-1})=(-1{)}^{pr(r-1)\mathrm{/}2}V(z){\prod }_a{z}_a^{-p(r-1)}$, so the constant term equals $(-1{)}^{pr(r-1)\mathrm{/}2}{\sum }_{\mu }{c}_{\mu }^2\mathrm{\ne }0$, since the $c_{\mu }$ are real integers and $V\mathrm{\ne }0$. Thus $w\mathrm{\ne }0$ and, by screening commutation, it is singular.

The source charges $u$ and $u^{\ast }=(p-1+pr+s)\mathrm{/}(2p)$ avoid every root $t_{a,b}$ with $ab<N$. Indeed $u=t_{a,b}$ would require $p(a-r)=s+b$, impossible because a positive left side is at least $p>2N$ while $s+b\le 2N$; if $a\le r$ its sign is already impossible. Likewise $u^{\ast }=t_{a,b}$ would require $p(r+a)=b-s$, impossible by the same bounds. The induction hypothesis therefore makes both source Fock maps invertible at all levels at most $N-rs<N$. By step 1.1 the source Verma form is nondegenerate in that range. The homomorphism $Q^r\circ A(p,u)$ sends the source top to $w\mathrm{\ne }0$ and is injective through relative level $N-rs$: a nonzero homogeneous kernel vector of least degree would be singular, since every positive mode enters a lower kernel degree. A positive-level singular vector in a Verma module pairs to zero with every PBW word of the same level by transposing its nonempty negative-mode word; it lies in the radical, contradicting the established nondegeneracy. Thus the descendants of $w$ at target degree $N$ have dimension exactly $P(N-rs)$.

These descendants annihilate the image of $A_N(p,t_{r,s})$ under the nondegenerate pairing of ${\mathcal{F}}_{t_{r,s}}$ and ${\mathcal{F}}_v$. Indeed the submodule generated by $w$ has degrees at least $rs>0$, whereas transposing a degree-$N$ negative PBW word applies positive modes of total degree $N$ to a degree-$N$ vector in that submodule, landing in its zero degree-zero space. The corank of $A_N(p,t_{r,s})$ is therefore at least $P(N-rs)$. For a square polynomial matrix of corank at least $k$ at $t=t_0$, constant column operations place $k$ independent kernel vectors in columns which vanish at $t_0$. Those columns are divisible by $t-t_0$, so its determinant is divisible by $(t-t_0{)}^k$. This proves the required multiplicity for the root $t_{r,s}$.

At $p>2N$ all roots $t_{r,s}$, $rs\le N$, are distinct: equality implies $p(r-r^{\mathrm{\prime }})=s-s^{\mathrm{\prime }}$, and the right side has absolute value less than $N$ while a nonzero left side has absolute value at least $p$. Their multiplicities sum to $D_N$, since $P(N-rs)$ counts partitions with at least $s$ parts equal to $r$, and summing first over $s$ then over $r$ counts all partition lengths. Thus the product of all factors supplied by step 5.1 has the same degree and leading coefficient as the monic determinant in step 1.1, proving equality for this fixed integer $p>2N$. Each coefficient of the difference of the two sides is a Laurent polynomial in $p$ vanishing at infinitely many such integers. After multiplication by a power of $p$ it is a polynomial with infinitely many roots, hence zero. This proves the oscillator determinant identity for every $p\mathrm{\ne }0$ and completes the induction, including at collided roots.

Taking determinants in the pairing identity of step 1.1 and using step 6.1 gives the claimed Gram determinant. Explicitly, $\det H_N=(-2p{)}^{D_N}{\prod }_{\lambda }{z}_{\lambda }$, $t^{\ast }-t_{r,s}=-(t-t_{r,s}^{\ast })$, and $p(t-t_{r,s})(t-t_{r,s}^{\ast })=h_p(t)-((pr-s{)}^2-(p-1{)}^2)\mathrm{/}(4p)$. The reflection sign cancels the sign in $\det H_N$, and the powers of $p$ cancel, leaving ${\alpha }_N=2^{D_N}{\prod }_{\lambda }{z}_{\lambda }$. For fixed $p\mathrm{\ne }0$, the quadratic $h_p(t)$ ranges over all complex $h$, so the identity holds for every $h$. Put $p=b^2$ to obtain the statement. Conversely every $c$ has a nonzero solution of $c=13-6(p+p^{-1})$; swapping $p$ with $p^{-1}$ exchanges $r,s$ in the ordered product. Since the Gram entries are polynomials in $c,h$ by their commutator definition, this is the asserted polynomial identity at all parameters.