Affine W algebra by quantum Drinfeld-Sokolov reduction

Use $b_a={\psi }_a$ and $c^a={\psi }_a^{\ast }$. Their singular contractions are $b_a(z)c^b(w)\sim {\delta }_a^b\mathrm{/}(z-w)$ and $c^b(z)b_a(w)\sim {\delta }_a^b\mathrm{/}(z-w)$, with zero b–b and c–c contractions. These follow directly by splitting the displayed Clifford modes into the given annihilation and creation parts. Currents commute with the ghost modes. Let $j_a=J^{u_a}-{\chi }_a$ and $f_{ab}^s$ be the structure constants. Since $(m\mathrm{\mid }m)=0$ and $f_{ab}^s{\chi }_s=0$, their OPE is $j_a(z)j_b(w)\sim f_{ab}^s{j}_s(w)\mathrm{/}(z-w)$, with no double pole. The current is $Q=Q_1+Q_2$ with $Q_1=:j_a{c}^a:$ and $Q_2=-(1\mathrm{/}2)f_{ab}^s:c^a{c}^b{b}_s:$. Normal ordering may be defined by moving the annihilation modes to the right, with a minus for each swap of two odd modes. Expanding a product of two such ordered monomials and commuting annihilation modes through the opposite monomial gives precisely the sum over cross-contractions; this follows by induction on the number of modes moved, since the only nonzero cross-commutators are the displayed scalar contractions or the current bracket. Thus the following is a finite Wick calculation justified from the declared mode relations.

The single-pole $Q_1{Q}_1$ contribution is $f_{ab}^s:j_s{c}^a{c}^b:\mathrm{/}(z-w)$. In the sum $Q_1{Q}_2+Q_2{Q}_1$, contracting the one b against either c of $Q_1$ gives its negative, $-f_{ab}^s:j_s{c}^a{c}^b:\mathrm{/}(z-w)$. There are no double contractions in these mixed terms because the j current has no ghost contraction. If J rather than j is used to perform the calculation, the uncancelled character term is $f_{ab}^s{\chi }_s:c^a{c}^b:\mathrm{/}(z-w)$, identically zero. The affine double-contraction term is $k(u_a\mathrm{\mid }{u}_b)c^a(z)c^b(w)\mathrm{/}(z-w{)}^2$, also identically zero.

For $Q_2{Q}_2$, a single contraction leaves three c's and one b. Anticommuting the three c's to a common order antisymmetrizes its coefficient in a,b,c; that coefficient is a scalar multiple of $f_{ab}^r{f}_{rc}^s+f_{bc}^r{f}_{ra}^s+f_{ca}^r{f}_{rb}^s$, which is zero by Lie Jacobi. This is zero as a tensor, so any overall Wick sign or common nonzero normalization does not affect its vanishing. The only possible double contractions use the sole b of each factor against a c in the other. The remaining coefficient is a scalar multiple of $f_{ar}^s{f}_{bs}^r=t{r}_m(ad{u}_{a}ad{u}_b)$, multiplying $c^a(z)c^b(w)\mathrm{/}(z-w{)}^2$. This tensor vanishes identically: choose a homogeneous basis of the positive-graded m; $ad(u_a)$ and $ad(u_b)$ strictly raise degree whenever the u's are homogeneous, so their product has zero diagonal and trace. Bilinearity gives the result for every basis, including the arbitrary one in the original definition. There cannot be three contractions because each monomial contains only one b and all c–c contractions are zero. Consequently there are no other poles.

Thus the entire singular part of Q(z)Q(w) is zero. In particular $Q_{(0)}Q=0$. The super mode commutator, obtained by taking the two residues of signed Jacobi, gives $[Q_{(0)},Q_{(0)}{]}_{+}=(Q_{(0)}Q{)}_{(0)}=0$. Over C this is $2Q_{(0)}^2=0$, hence $d^2=0$. This proves the exact missing condition for the actual charged-ghost construction, for every level k and all the supplied good-grading/isotropic data. It proves neither nonzero cohomology nor its PBW identification.

Taking the residue in signed Jacobi shows that d=Q_(0) is a signed derivation of every state product. Therefore ker d is product stable and im d is an ideal in ker d, so the cohomology inherits the vertex-superalgebra products and its degree-zero part defines the stated universal W algebra. The square-zero condition in the unchanged Definition is now proved above.