We relate two apparently different bases in the representations of affine Lie algebras of type A: one arising from statistical mechanics, the other from gauge theory. We show that the two are governed by the same combinatorics and therefore can be viewed a
8IGORB.FRENKELANDALISTAIRSAVAGE
Proposition3.0.3.ThemapΦisanisomorphism;foranyX∈IrrL(v,w),thereisauniquefunctiongX∈L(v,w)suchthatforsomeopendensesubsetOofXwehavegX|O=1andforsomeclosedGV-invariantsubsetK L(v,w)ofdimension<dimL(v,w)wehavegX=0outsideX∪K.ThefunctionsgXforX∈IrrΛ(v,w)formabasisofL(v,w).
4.LevelOneRepresentations
WenowseektodescribetheirreduciblecomponentsofNakajima’squivervari-ety.BythecommentmadeinSection2,itsu cestodeterminewhichirreduciblecomponentsofΛ(v,w)arenotkilledbythestabilitycondition.ByDe nition2.0.1andLemma2.0.4,thesearepreciselythoseirreduciblecomponentswhichcontainpointsxsuchthat
(4.0.3)dim(kerxi→i 1∩kerxi+1←i)≤wi i.
We rstconsiderthebasicrepresentationofhighestweightΛ0whereΛ0(αi)=δ0i.Thiscorrespondstow=w0,thevectorwithzerocomponent1andallothercomponentsequaltozero.
4.1.TypeA∞.ConsiderthecasewheregisoftypeA∞.LetYbethesetofallYoungdiagrams,thatis,thesetofallweaklydecreasingsequences[l1,...,ls]ofnon-negativeintegers(lj=0forj>s).ForY=[l1,...,ls]∈Y,letAYbetheset{(1 i,li i)|1≤i≤s}.
Theorem4.1.1.TheirreduciblecomponentsofL(v,w0)arepreciselythoseXf ∞suchthatwheref∈ZV
{(k′,k)|f(k′,k)=1}=AY
forsomeY∈Yandf(k′,k)=0for(k′,k)∈AY.Denotethecomponentcor-respondingtosuchanfbyXY.Thus,Y XYisanatural1-1correspondencebetweenthesetYandtheirreduciblecomponentsof∪vL(v,w0).
′′′′Proof.Considerthetworepresentations(V∞(k1,k1),x∞(k1,k1))and(V∞(k2,k2),x∞(k2,k2))′ofourorientedgraphasdescribedinSection1wherethebasisofV∞(ki,ki)is′{eir|ki≤r≤ki}.LetWbetheconormalbundletotheGV-orbitthroughthe
point
′′′,k)⊕V(k′,k), .x =(xh)h∈ =x∞(k1,k1)⊕x∞(k2,k2)∈EV∞(k11∞22
ByProposition1.2.3,x=(x ,x ¯)=(xh)h∈HisinWifandonlyif
xi+1→ixi+1←i=xi←i 1xi→i 1
foralli.1′2Leteir=0forr<kiorr>ki.Now,xr+1←r(er)=crer+1forsomecr∈Csince
2′xr+1←r(e2r)canhavenoer+1-componentbynilpotency.Supposethatk1≤r+1≤
′k1andcr=0(thatis,xr+1←r(e2r)=0).Thenifr+1>k1,
221xr←r 1(e2r 1)=xr←r 1xr→r 1(er)=xr+1→rxr+1←r(er)=crer=0.
′Inparticular,e2r 1=0andsor 1≥k2.Continuinginthismanner,weseethat
2′′′←k′ 1(e′xk1k1 1)=0andthusk2<k1.1Now,ifr+1≤k2then
22xr+2→r+1xr+2←r+1(e2r+1)=xr+1←rxr+1→r(er+1)=xr+1←r(er)=0.
21Therefore,xr+2←r+1(e2r+1)=0.Butxr+2←r+1(er+1)mustbeamultipleofer+2asabove.Thuswemusthaver+2≤k1andxr+2←r+1(e2r+1)=0.Continuingin
thismannerweseethatk2<k1.RefertoFigure1forillustration.