A level-three Virasoro null relation restricts primary channels

The Virasoro bracket [F1] gives $[L(0),L(-r)]=rL(-r)$ for $r\ge 1$. In each algebraically generated highest-weight module, reorder any finite mode word by $XY=YX+[X,Y]$, moving negative modes left, zero and central modes next, and positive modes right; induction on word length for bracket terms and inversion count for swaps terminates. Positive modes then kill the top vector, and $L(0)$ and the central generator act by scalars. Thus negative-mode words span, each has weight equal to the top weight plus the sum of its mode indices, and only the empty word has grade zero. Distinct eigenspaces are linearly independent, so this proves the algebraic gradings and top lines stated above. Let $u_2,u_3$ be the top vectors of $B,C$. If $I(v,z)u_2=0$, then for each $m\ge 1$ the commutator formula with $n=-m$ expresses $I(v,z)L_B(-m)u_2$ through $L_C(-m)I(v,z)u_2$, $I(v,z)u_2$, and ${\mathrm{\partial }}_{z}I(v,z)u_2$: because $v$ is primary, the only terms in the commutator are those involving $L(-1)v$ and $L(0)v$. All vanish. Induction on negative-mode words shows $I(v,z)$ vanishes on $B$; the derivative identity and descendant-residue identities then give $I(w,z)=0$ for every Virasoro descendant $w$ of $v$, so $I=0$ on $A$. Thus for nonzero $I$ we have $I(v,z)u_2\mathrm{\ne }0$. The grading condition implies that the coefficient of $z^{h_3-h_2-d+\mathrm{\ell }}$ lies in $C[h_3+\mathrm{\ell }]$. Lower truncation gives a least target grade with nonzero coefficient. For each $n>0$, the primary commutator $[L(n),I(v,z)]=z^n(z{\mathrm{\partial }}_z+(n+1)d)I(v,z)$ shows coefficientwise that $L_C(n)$ sends this coefficient to a lower-grade coefficient, which is zero. Hence it is singular. A positive-grade singular vector in simple $C$ would generate a proper submodule, because its descendants have weights strictly above the highest weight; thus the least nonzero coefficient is in grade zero. Let $u_3^{\ast }$ be the linear functional equal to $1$ on $u_3$ and zero on all positive grades. The $L(0)$ identity then gives $g(z):=u_3^{\ast }(I(v,z)u_2)=C_0{z}^a,C_0\mathrm{\ne }0,a=h_3-d-h_2\mathrm{.}$

For any $w\in A$ put $g_w(z)=⟨u_3^{\ast },I(w,z)u_2⟩$. In the $k=2$ descendant-residue identity, the target-side expansion is $(y-z{)}^{-1}={\sum }_{j\ge 0}{z}^j{y}^{-j-1}$; after multiplication by $T_C(y)$ its residue contains only $L_C(-2-j)$, whose pairing with $u_3^{\ast }$ is zero because these modes raise target weight. The source-side expansion is $(y-z{)}^{-1}=-{\sum }_{j\ge 0}{y}^j{z}^{-j-1}$; the outside minus sign leaves the $L_B(-1)$ and $L_B(0)$ terms. In the $k=3$ identity the expansions are ${\sum }_{j\ge 0}(j+1)z^j{y}^{-j-2}$ and ${\sum }_{j\ge 0}(j+1)y^j{z}^{-j-2}$; the target residue again contains only negative modes, and the source residue has $L_B(-1)$ and $L_B(0)$ with coefficients $-z^{-2}$ and $-2z^{-3}$. Positive source modes annihilate $u_2$. The $n=-1$ commutator and derivative identity yield $⟨u_3^{\ast },I(w,z)L_B(-1)u_2⟩=-g_w^{\mathrm{\prime }}(z)$, while $L_B(0)u_2=h_2{u}_2$. Therefore $⟨u_3^{\ast },I(L(-2)w,z)u_2⟩=h_2{z}^{-2}{g}_w(z)-z^{-1}{g}_w^{\mathrm{\prime }}(z),⟨u_3^{\ast },I(L(-3)w,z)u_2⟩=-2h_2{z}^{-3}{g}_w(z)+z^{-2}{g}_w^{\mathrm{\prime }}(z)\mathrm{.}$ These are coefficientwise residue calculations; no contour deformation or analytic continuation is used.

Apply the derivative identity and step 2.1 to the three descendants in the null relation, using $g(z)=C_0{z}^a$ from step 1.1. Their top coefficients are respectively $a(a-1)(a-2)C_0{z}^{a-3}$, $(h_{2}a-a(a-1))C_0{z}^{a-3}$, and $(a-2h_2)C_0{z}^{a-3}$. Since the null state is zero and $C_0\mathrm{\ne }0$, its image coefficient gives $a(a-1)(a-2)-4p(h_{2}a-a(a-1))+2p(2p-1)(a-2h_2)=0.$ Substitute $4p{h}_2=x^2-(p-1{)}^2$ and expand; the left side factors as $(a-(1-2p))(a-(1-p+x))(a-(1-p-x))\mathrm{.}$ Thus $a$ is one of those three roots. Since $h_3=h_2+d+a$ and $d=2p-1$, the roots yield $h_3=h_2$, $h_3=((x+2p{)}^2-(p-1{)}^2)\mathrm{/}(4p)$, or $h_3=((x-2p{)}^2-(p-1{)}^2)\mathrm{/}(4p)$. If $h_2=h_{r,s}$, take $x=r-sp$; replacing $x$ by $x\pm 2p$ changes $s$ to $s\mp 2$. This proves only the stated necessary restriction.