Диссертация (Isomonodromic deformations and quantum field theory), страница 41

we have two Ramond fermions with two vacuum states, whereasthe r.h.s. corresponds to sum over bosonic modules with half-integer vacuum J0charges. This formula is a simple consequence of the Jacobi triple product identity.Analogously we have similar formula for the bosonization of N S × N S fermions∞YP12q 2n1n∈Z(1 + q 2 +k )2 = Q∞kk=1 (1 − q )k=0(6.248)It is the consequence of Jacobi triple product identity as well.l twisted charged fermionsFor the twisted bosoniφ(z) = −X Jn/l1+ J0 log z + Qn/lnzln6=0196(6.249)6.10. Exotic bosonizationswithJn/l , Jm/l = nδn+m,0[J0 , Q] = 1(6.250)one has for |z| > |w| X −n/l n/lizw1φ+ (z) − J0 log z, φ− (w) − iQ =− log z = − log z 1/l − w1/llnln>0(6.251)whereX Jn/lX Jn/liφ+ (z) = −iφ(z)=−−(6.252)nz n/lnz n/ln>0n<0Define two operators11ψ̂ ∗ (z) = z 2l : eiφ(z) := z 2l eiφ− (z) eiφ+ (z) eQ z J0 /l11ψ̂(z) = z 2l : e−iφ(z) := z 2l e−iφ− (z) e−iφ+ (z) e−Q z −J0 /l(6.253)with the OPE1(zw) 2lψ̂ (z)ψ̂(w) = 1/l: eiφ(z)−iφ(w) :=z − w1/l1 z J0 /l(zw) 2liφ+ (z)−iφ+ (w) iφ− (z)−iφ− (w)e= 1/lez − w1/lw∗(6.254)Then for the modes of their expansionψ̂ ∗ (z) =∗X ψk/lk∈ 12 +Zz,k/lψ̂(z) =X ψk/lz k/l1(6.255)k∈ 2 +Zone gets canonical anticommutation relations{ψa∗ , ψb } = δa+b,0(6.256)Now one can express the l-component fermions in terms of a single twisted boson11ψα∗ (z) = √ z − 2 ψ̂ ∗ (e2πiα z),l11ψα (z) = √ z − 2 ψ̂(e2πiα z),lα ∈ Z/lZ(6.257)and it follows from (6.254), that their OPE is indeedψα∗ (z)ψβ (w)=z→wδαβ+ reg.z−w(6.258)The stress-energy tensor and U (1) current can be extracted from the expansion:Xψα∗ (z + t/2)ψα (z − t/2) =α∈Z/lZ197l+ J(z) + tT (z) + .

. .t(6.259)6. Twist-field representations of W-algebras, exact conformal blocks and character identitiesUsing (6.253), (6.254) and (6.257) one gets for the l.h.s.1−l1(zl1−l+ 2t ) 2l (z − 2t ) 2liφ(e2πiα (z+t/2))−iφ(e2πiα (z−t/2)):=t 1/l : et 1/l(z + 2 ) − (z − 2 )α∈Z/lZX 1l2 − 12πiα=+ t 2 2 eit∂φ(e z) + O(t2 ) =t24l zX(6.260)α∈Z/lZ=l+tXi∂φ(e2πiα z) +α∈Z/lZt l2 − 1 t X−: ∂φ(e2πiα z)2 : +O(t2 )z 2 24l2α∈Z/lZOne finds from hereJ(z) =Xi∂φ(e2πiα z) =α∈Z/lZX Jnz n+1k∈Z(6.261)l2 − 1 1 X : Jn Jk :T (z) =+24lz 2l k+n∈Z z n+k+2which already have expansions over integer powers of z. ThereforeL0 =l2 − 111X+ J02 +J−n Jn24l2ll n>0(6.262)and the character of this module is given byPtr q L0 +rJ0 = ql2 −124ln2q 2l +rnn∈Z∞Q(6.263)n(1 − q l )n=1l charged fermions – standard bosonizationFrom the modes (6.255) of the operators ψ̂(z), ψ̂ ∗ (z) we can construct another lfermions1 X ψ−a+p1 X ψa+p∗√ψ(a) (z) = √ψ(z)=1 ,1(a)(6.264)l p∈Z+ 1 z a+p+ 2l p∈Z+ 1 z −a+p+ 222wherel−1 l−31−l,,...,}2l2l2lThese fermions can be bosonized in terms of l “normal”, untwisted, bosonsa∈{P∗ψ(a)(z)=eiϕ(a),− (z) iϕ(a),+ (z) Q(a) J(a),0eezJ(b),0(−1)b<aP−iϕ(a),− (z) −iϕ(a),+ (z) −Q(a) −J(a),0ψ(a) (z) = eeez(6.265)(−1)J(b),0(6.266)b<awhereJ(a),0 |0i = a|0i198(6.267)BIBLIOGRAPHYComputation of character in this case gives usl−1PPtr q L0 +rPJ(a),0=( 1−l+2kl+nk )2 +r2lq k=0n0 ,...,nl−1∞Ql−1Pnkk=0(6.268)(1 − q n )ln=1One can easily see that equality between (6.263) and (6.268) follows from particularcase of (6.223).Bibliography[AFLT] V.

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