Triple Glue Vertext Part 1 of 1

Triple Glue Vertex data PPP >Ward10 ?git Ward 6 15x 18 5x6 Operator Fki* BVectors Complete Sets typed Misc PPP +DP Sector (n0go) IRREDUCIBLE the n's n-oldway Phil Lucht 1979 WW) Flow Chart for Tensor Problem 0. ItwouljdbenicetoseeexactlywhyI"paused" whenIdidatvariouspoints.Ie,went down various alleys and hit brick.walls. Are those walls still walls? Pause 1: Here the last thing Idid was towrite ahuge table showing 10constraints onthesetof10S, functions including four pairs, 3S'sand#3Afunctions. Table entries aredj,@) andd, thethree dot products. Things areallcompletely analyzed in S,fractions. Atthis point Ididn't know what todoabout the"pairs". Somehow you can't just treat the two members ofeach pair asindependent functions. There are in effect four extra constraints, namely, that each pair isreally apair. Ithink mycurrent feeling onthischargistoexpress thepair“entries usingtheX',operator sothat you have 10operator-constraints on10functions. This iswhere Iplan tocontinue. I stopped because a)the table wssolarge and foreboding; b)Jim swore there were fewer tensors soIthought Ihad counted wrong and was therefore discouraged with the 10-function approach. . Pause 2:Ishowed howtoreduce to6functions from10(albeit notS3symmetric ones). Irewrote the Ward but ran into the same kind ofproblem aswith the pairs above. Here you can extend from 5Ward equations to15byadding inthe cyclics, sothen you have6;effect6x3=18“variables”. Butthisapproach isnottoogoodbecauseyouthen have tosomehow addINtheconstraints (12ofthem) that thecyclic functions really are cyclically related. So15equations in18variables really scared me, soIstopped on this because too foreboding; later Ireturned toit(see below) foranother try. Itthen occurred tomethat Icould replace some ofthe ppproducts with the F-tensors and this might beuseful because Marshall's vertex was inthis form and because these tensors have great orthogonality propoerties tovectors. Icame upwith agood general vertex form, then Iwrote down the Ward but did not Lorentz-tensor parse it. Some ofthe contracted anddouble dotted equation here were quite simple, but still Ididn't know what todowith them: the same 6x3 =18function problem. Toomuch coupling between functions. Ishowed howMarshall's solution really does fitinto this general form, then I Just gave uponthe approach. Pause 3: Here, reelizing that the B-vectors give you avery simply Ward null term, and convinced that finding Ward nulls wastheentire deal, Itried converting toBfrom p. J.gave upwhen Icould not make the three B's add uptozero, hence dould not take overo..independent tensorformsanalysis. Pause4:Ithenreturned tothe6-function earliergeneralpppformsandtheanalysis ©. ofthe 5Ward equations. Idid some elimination and reduce ditall tothree operator equations in6functions; Idecided itwasprogbly better tohave symmetry operators like [e]and[12] sitting inyour equations, rather than having 6x3=16functions. The operator approach seems better because a)itkeeps your number offunctions down to6,not 18 d)noneed toaddinthose extra 12cyclicness constraints. Then Iconsidered the problem ofsolving these operator-like equations. Ifound that the operators appearing tended tobenon-invertable, projection operators. This made life difficult because then along story about howtosolve a“double projection" equation. Ifinally gave uponthat, though theoperator technique should prove ~ useful when Itake upthe thread again elsewhere. Thelast burst: Showed howyoucanslide a“complete basis" into various forms, putitintheF-tensor basis which hastwoWard Null Terms.” Complete analysis ofhow you fit any object into any basis. Then Ward analysis onthegeneral F-tensor form. Result speaks foritself. Method ofWard analysis here wasy this: 0: a)obtainthe5separate equations A,B,C,D,E b)replace Aand Dwith AWD and A-D ¢)fractionate eachequation intotwoequations usingS,analysis. 4)solve byhand theresultant setof10equations in10unknowns. io] Retrospective Comments.// . ie]Look back atthe original Ward Table which was obtained from the Standard Basis. There were 10constraints on14variables. Thus, 4degrees offreédom inthe - solution. Now Iknow where those four degrees are going: . 2gointoapairassocaited withtheF083coefficient function 8,(12), and 1goes into the totally symmetric piece ofthis function and 1goes into $(123), the Ward null BBB coefficients. - Thus, ifIhad known what Iwas doing, once Ishowed those 10constraints were linearly indepeninet, Iwould have known right then that there were four constraints. Iwould have knonw that the two knonw Ward Null Terms ate upfour degrees, so Iwould have known without doing anything else that there were nomore Ward null terms! Toshow linear independence ofthe constraints, Icould have deleted the columns ofpairs and put the 10x10 thing into Reduce and computed the determinant. T£Detf4O then Iknown independent andcanthen addback thefour columns and still will beindependent. Knowing Marshall's solution andknowing theWardNullsfromthestaré, ositwas just amatter ofcouting. BUT: Thewhole question ishowdoyoucount degrees offreedom! Apparanetly you must count "irreducible components" asdegrees 66freedom. Whyshould youcount both members ofafunction pair ?Youknow youcan get onefrom the other byanoperator ofsymmetry. Butasfunctions they are linearly independent. Ie, oneisnot amiltiplce ofthe other. Butyou can't change one without also changing the tther. ~ Sothequestion is: howdoyouknow that ifyousolve allthose equations, thefunctions that aresupposed tobepairs really come outpairs? What guarantees this?? Ithink itisthesymmetry; when youClebsh up,youinput thefact that thefunctions arepairs when youdoyour 12combinations. Soitmust bethat they counddnt come out anyo other way. Similarly, inmyFine] Solution theobjects RyandR,were treated as independent objects, butinthesolution they were indeed pair-like .Again, the Clebshing built this in. le] Bythe way, Icould have had reduce solve these 10equation inten.unknowns (againdetetingthecolums).NO,thatiswrong.Reducecannothandle£k10re)equations in 14unkgwons.. $0Idon't know howIwould have applied those constratins. ~ 7 . Rduce could have solved my6x6problem inthefinal solution, butIdid ] itbyhandsince mostwassosimple. * The read bitch problem is getting those extra degrees out ofthe problem soyou have something you cah solve. The best way todothis ofcdurse was to use the known Ward null terms!It!!!! . @ c*) as lameyo| poe“ta,GOTamdprcgutnsXi| LBreinere2 haMalaifisaun! “a, Satlearamsnekhis 1CouagleleSeta." SgteASCs)os6gece4Sortepeda aCletsledy$029gues) ||b)Peusnloaae. onepssRnaicDede :iOrsBOB)oxargheee £TeenWaddGeneralFram. OFBepepadBeJewsnial : *Sngow. PitPerts=O ares , +AW)Sit) amabaysis@AhapeevedexTome|[Tes sdapemclerToons,| cLtaeLastWedArabs___phestSebs Sone LoTaeas tLDuRobt| a >.Bag.iconsSigeFactorhin: _fa Qe Pawat : ESRagsden enwelHeSorondpage [ees WSCoach .oon jstucs 10duede6) BidvaNolin [Wakifate6Fadee Sra SS) aee /|Teg Silaneseas _™ |“Hey YBoe wc PaDacagsiasA fazPreaate!!]FOfaSboiEyut on4,FOed3c Unde6 [opo8deyaicst ya i) Tee || jeee psBasicDeda -Iebaies |£Chetwidhanye | SgAgen| Pre“athodBp<FadaQos OfSoeRitfertsecEN|| —olGrapgpvedexTomeSayets}; !“thofpduaaleyTon Pauie¥Ruy) . i ResSeahSumy, ng — —_—_ [edSAToataww |oy hailedGed” :5 i. Q, aATeneTas !WideNandWord 45 ajCosbadie5DobleBt, ~CoreesSpFracty sabeVos Wi 7S Talat2Oy_(Sah Pema Paas %oo Pant | pStack& ;Bee | |i |“Rvecins | |iYbref pe ia: &Pomae3 | Tay\o\AN’. STAT 6 QO TsGrongCro}Note ."~Pavin@rehien” Niele. Re-DypineOnnpppBasis,JaProlygeBGS,£8,Fomine NosTae,Basis ookAode atooQuan" oon vrNeralesorgakSee] Moke complete.sddnstuasa, |mdPome» LsOldFrashiwdbean 2Appsfemacsweaot NengGeatexto 9)Vatabastewh. ounhlowed, SYSteeKaewWordNuanlar Sok .8)QuanUatsQUdoakWaddak. fo)3)OntBWorddotkny¥a)Wak? UNSof“De\o~sqvasicona, 2)Woke Chock +Rroomes te : . 9)Shashock: qua“WoidNulla.dedsch OMSe —\(Seeeodande Mu) bRowan Y ie) _ K=(2) Me =UW) ‘ ¥=@~3) yee=0) 25G4) etyS-(33) i Fas anstagt aac a= a B=Gakent 0,=Gayl Gaya! Dee@ayd=(eryae Be(OF -8)08 Sma, XX}... June 2h, 1979 Contents: : : 01.Howtoassociate functions likexy?withUIR'sofSU(2)anddiscrete groups. 2.Interpretation ogthe Great Orthogonality Theorem: a.thenotion ofX;, "projectors" b.example ofS3 group. 3.Left over details onLichtenberg basis functions. 4.Solution toaMatrix Problem. (linearly combining theX,,etetogetorthogonal basis) 5.Theorems relating toprojector construction. fy =X ay= Xe b.How tomake new idempodents thet are notprojection operators. c.How tointerpret the Yand Y"operators ofHammermesh and others. 6.Earlier effort atdecomposing afunction into Irreducible pieces (only 4pieces here) 7.Thetensor problem forself-energy (ie, really only 2/1momenta). : 8.Counting triple glue amplitudes. 9.About multiplication of£(123) byg(123) andclassification ofthe36pieces. fo)10.BasieS;data:. a.definitions oftheX;5projectors, their (vectors), bd.complete Xmultiplication table rec. the group miltiplication table a.The explicit 2x2 representation matrices e.Table showing RXj; multiplications. f.Decomposition ofarbitrary element ofgroup algebra. “g.thereason youreally have six functions formable from £(123), what they do. h,Verification that Hamermesh's Clbbsch table florks for makind symmetric combos. 4.moreexplicit RXftypemultiplications. Jj.how togenerate M-pairs, mixed synmetry 11. The triple-glue tensor problem. 4.classify raw tensor forms into symmetry classes 2.Put 3=-1-2 and count independent tensor forms 3.Write general triple-momenta vertex tensor forms 4.Dothesameforthebuyforms. ra)Summary ofall tensor forms. ae ws What does itmean tosaythat the function f(x,y1z) =x belongs” tothe,Iel re)representation ofSU(2)2 1.First, when Isay that the QMoperator Xisa“vector operator", Imean this: , oaKe ROK RG)=Ky) IfIsandwich this operator between two group elements asshown, Iget alinear combination oftheX'sback, andRy, isthe3-dimensional repoftherotation group inthexyzsense. This iswhat youmean when yousaythat X,belongs tothe IL-1 representation. 2.Aket/jm) certainly "belongs" totherepresentation jwith label mandbythis you mean that: eyRG)iy=ZLDan(Dond IfIwere toclose this with (r/ you would have: a dV .QRKld= Vern <Pline)Soletscallthisfunction bythename fyy(r)- ‘Thenwehave: RG)TAM=2DhanRe 3)TymP)=Z-DeinCQ)Hae@) Afunction whichbehavesunderroations inthiswayisseidtobeafunction belongingtothe"m'" row" oftherepresentation j.Fine anddandy. Ontheother hand, weknow that forany function the following istrue: IN > Pay=RDFM=TCRO) 3.Now,howmightwesearchforfunctions £jq(¥) that"belong" tovarious rep's. That's easy, here isthe answer: Fe(PR)=FOR)Yao ro)ye(E)= )Nim ) AsyouletFrangeoverarbitrary functions ofr,youexhaust ellfunctions Ejq(*)- “Ss we -2- 4.Now for j=0,1,2 Iwill make alist offunctions that belong" torepresentations. lo)js0:AnyF(x?)willdo.Forexample, r,r°,e77allbelongtotherepj=0- JaliHerearetherelevant Yints:s LtF(t)=¢ 4 a: ad a : ESYeSRsuee gf,@)=[x+y] CS+;) es 2 a Yo=Secose > R.@)= =(TR) Bo. Ate 5 .i% You=+JRsned Sal@\= beg) G4) Ofcourse wecan change overall scale. This then isthe usual set offunctions which "belong" tothe L=1 representation: rae ARy =1(x+y) f.@\=+2 = Kye BleLEA. 2 . 0ae) =+Riy) . . 1:2:Again, firetweexpress theYj,'81 Nowtaker°asoverall: LEcvso%* t= Lege 2%\ YasEset 3QE) bigs tyl) | ze. Xa = . aan .SEsiswsee® SE) 2xriy] * E403tee ‘Y No=Ate(Feodo~2) ~(4%)\ee-ac]Fs 2% Yay=-Yu >GFE) alets] ay Pa _ = auu : 2hYayo=Ye ,~(2) (t=)~Ving 32 Thefivefunctions shown ontheright areobtained bylinear combination, soeachof these 5functions maybesaidto"belong" toL=2. Thefunction x-y~ is@congination of/22)and/2,-2),and32°-r?4sentirely/20).Ofcoursebycombiningyoucanget ©)the other obvious functions: =. aeater L ox age -- ~3- Notice ofcourse thatnolinear combination canevergiveyour°.Thatfunction does [o)notbelongtoL=2,itbelongsto1-0. . 5.Bytheway, consider thisdirect product : Fy=jx; +Youareconbining ‘two Le1 objects here, soyou know you must get some 1-0,1 and 2. But there isnoLel because thistensor iseynmetric, soonlyL-0and1-2. Forexample thefunction x” isamixture of1-0 and 1-2 functions. You then have tolinear combine toget functions thatarecompletely 1-2suchasx°-y*. Ietmecorrect this: anyfunction belonging toI=l isproportional toYj, and anysuchfunction mist,change signunder parity. Obviously Fy,obove doesnot.change signunder parity, sonolinear conlgination oftheF,,could giveyouanLal.Forget the symmetry comment. Consider Fyy, =4%, +Inprinciple youcould get1n3,2,1,0. Butparity will now only allow L=3,1. 6.Suppose you aregiven anarbitrary function f(r). How would you "project it" into its irreducible function components? First, for each Lwecan choose aset of bais mfunctions like so: fe) sAw@= eed =BOYLE Ie,foreachdifferent Lyoucanghoose anyFy(r)function youlike,oryoumightset them ellequal to1,orperhaps r‘. Anyway, then here isthedecomposition: = = BSAa <A =Ziloads =Zt Be) Wane 3 Ays Aare Calg) =(se Lace * es > BS=\te. Ann@)2). Having selected asetofbasis functions, wecanthem gettheweights fyq* Thenthe original arbitrary function canbeexpanded interms offunctions which belong toUIRs. an -~h- 7.Construction ofaprojection operator .First consider this: eo| ’ ;RO)AH) =ZSDianayhw@ ey AN Ny 3 =\QiaG Rare O=BHM SagDas) Dia’) a@).BE ya5 =Bde BEbybe ~\=WAH. Le@® Bohan Now look what happens when you integrate against anarbitrary function: Wares .Way RO)FOag) fo) Vk KRxe) > =\dy&KymDateGdRO)AA) =xNonWx’) aya\dsSonn =8. SeAw@)Cuts) vx PS 3 Xu 5 : 3.9.Down RayFR)=Br Lite Ropsat SagDaw(g)RGYFR)aayQc tn(®) S : ie‘OR T= Ged\ayDeeRea) \e] Obviously thisistheprojection operator which, whenapplied toanyfunction, projectsouttheportion which belongs tojm.Youneed toknow thediagonal matrix elements andthe operators, ~_ eS -5- 8.Ifyoutakethepreceing projection operator andsumoverm,yougetanewprojection ©) operator which does this: itprojects out all functions which belong toany row ob the representiation. Soifyou don't care about connecting functions with particular "rows", you can use this: 7 i 8 a Bo ar= yw = Qe .PeGAIA] RA)=cand\tg 26(9)RG)eur os Sw Here wehave identified thetrace oftherepresentation asthecharacter. Idon't know therotation group characters except forspecial cases ofg,soperhaps this isnot too useful for continuous groups. 9.Extension ofthise ideas toDiscrete Groups. Ican just quote the projection operators etcfrom Tinckham, all obvious once you know about orthogonality. Wehavet R=group element, replaces label gused above. P,=thetransformation operator, replaces R(g) used before. Ie, Pit(r)=£(Rr) 0This operation can bearotation orreflection orwhatever. TR) =element ofrepresentation, calledDo(8)intheabove. Now here are the basic facts. (a)Arbitrary function expanded interms ofbasis functions: f@- ZRH) @s)~~ (b) Transformation rule for basis functions: 2 6) a : feRe@)= 2Mea(®Re@) (2-2) . (c) the projection operators: . * .re : 9, : .P= YZ Om K (2-35) h R %&ON«x@=es)E_K® Pe (3-3) fe)n) ® Here 1;isdimensionality oftherep, andhisorder ofthegroup. Note that (2j+1) does infact ‘appear for the continuous group aswell, but not h. Interpretation ofthe Great Orthogonality Theo @F)_+ Fortintte groups, hereiswhatthetheorem says: a)o>o*WSU, zsDs@®D,.® =&be3en. Now, lets define some elements ofthe group algebra: * v. ” vo ”* 4 is,Kay=LDA Vals)BY Xie=(BVs® R\S CK,Ken=&du)Dyal®) 4i" v ort BS- & =de bsho. Xin=(ZsBml®8)& beiim, Clearly, these two vectors inthe group algebra are orthogonal unless all three sets ofindéces areequal. SeePozzi (2.1)) fordefinition ofgroup algebra scalar product. Think ofthe various Rasunit vectors and the coefficients ascomponents ofvector. Then the line above states precisely that these two vectors are orthognal. 2.Thus,eachUIRofafinitegroupprovidesnvectorswhichare(1)orthognnal to le) each other ,and(2)P orthogonal tothevectors ofalltheother UIR's. Clearly these ere the spanning vectors ofthe regular representation!” 3.Example: Iamnob going toconstruct the 4spanning vectors which are related to - thetwo-dim UIR's ofthegroup $3.First, herearethematrices: que @ “(4He wo. (LZ nn) 7 ve5=CF ws)= apes De@= GI) DBO=ea) DEC =aa] sates Fes ay ‘ ape & -t~Ai\* ei)Were A)droa- (3-* dsCor)=GE, Jes Sot coo la+ Ihavereadthese frompage&Tinckham, butIcould havegottén themfromH'slarge table. Now Iwill construct the 4basis vectors! . . eyKy=6ECR) +2-20) 4£.Gar) ro)io]Cie 8+409 AGa-A 00) ek. 7° ~2- Restate these with shortest form: fe) Gee.2£©®@B.a®ce) ) ;aCroeaOr CE)Me oy :Xeoe ©0-—e&&-L a4 -Y" y(Xa pe00HBRRB8B&Ot) oea)— MUMSb 2Boy dsl \ Youcanseebyinspection that,these fourvectors areorthogonal inthegracup algebra. Lé-we_meitiplyeach-of_these_by1/SQRT3,-they-willbe-erthenormel— Warning:Justbecause twovectors inthegroup algebra arearthogonal, thatdoesnot mean that ifyougroup-miltiply thetwovectors youwill getzerol Youcansort of imagine thegroupmultiplication asacrossproduct. Rjro =AS20% 4h,Now-farget-that-oqrt3-abeve-and considerthis:thefirstobjectabove,X41,may ie)beregarded asa'first row' object. Why? Consider: nw a aX, » = RR=RABIES =ZW) KS- ADyKas)ass 3 xaX * ak oy= ZWD =22 WHT akan. at ae-\ A=MR) 22D O)T =DAR) KG © wt wad =2De(R)Ky nadeARdacawe *[owieSaepaiseth Thus,wecanregardourXj,objectsinthisway Uers wo. me onXy=Xe=ckekiyOuinadQesaryghWEEps. 5+Nowwecanapplypage40ofTinckhan. Probably Tshouldrenormalize theKy,objects sothat they areexactly theobjects ofMnckham (3231). Dee -os" , % SS)) 4=hy $e=Ky. -3- Using Tinckham 3-30 applied here weget: QO Fe oO)Bege=SoSua Ifdiffering TUR's yougetzero, IfthesemeUIR,thenconsider P,operator. Ifthis acts onafunction belonging tothe row k,you get zero unless k=j. Inthat case you getprecisely by. Now translate this toour present notation and situation: YW od) ms) XxXx=SeRyeXt k = HO WyKee=SeKa Viy, Po Ww Ww=fAa Aa =hw Ya Ky=© Wo Ww PKnXa =Xia YeKn =0 B (yaywXu Ya =© Yar Ky =Yar WGA Xu Ya =O YarKee =Kar. This shows that X,;andXj5 tryly areprojection operkors. Notice that, although thegroup algebra vectors X,,andXj,areorthogonal, you donot get zero when you multiply these two objects. Don't make that mistake. Continuing: »‘bnAu=0 XaXv=Ka—Nene XnXa=0 XaWen=Kar XnYar=Xn XeXy=© XuXe =Xa XaXa=0 ~he ExistenceofHermitianAdjointsofGroupAlgebraelements.Letusassumethat fo)wehave ascalar product ofsome sort which isinvariant inthis sense: CH= RRR) Regpesy. Obviously, theoperators Rareunitary with respect tothis scalar product, Rt=RT. Thisserves really todefine R*,since theRT>areknown justfromthegroup data. Now consider anelement ofthe group algebra, + — —x3 A=ZaQR»A=2o®R=Feo®& = * . . .=Ba) R=AXA Papi (uoetabes),& Thus, theHermitian Adjoint ofanelement ofthegroup algebra isprecisely its involution, asdefined byPozzi. Now: Example: here are some Hermitian conjugates: ~ ro)R=G)AREQ) R=Gud =GOGH) S|R=QYGQ= Cy) =Gey) + Xu=Av x Yr =Ka = £16 +8@)-& O48 Gr)=Ky . Ya=Ka. Armed withthisinformation, wecannowinvestigate theorthogonality orlackthereof ofbasis functions. ot -5- Copy1 Copy2 x xOSSD eyoy X22 These results tellusthatwithin eachcopy, Ku%,Xu)=GaXa) =° theaidsfunctions areorthogonal. Ched,ke=CHYarhe)=9 = Andthesetellusthat theties, Git,et)=GokeS) = difgerent rowsareorthogonle Che$ku)>©Kakab=© Qh,YroF)=Gea) =RRA) FO.=ea,Kerf) Thislastresult says, however, thet“thetworow-1 functions ofthetwocopies need not beorthogonal! Theorem: Thefactthattheprojection operators 45areorthogonal inthegroup algebraimpliesthatthefunctionswhichtheyprojectoutofsomecarrierspace ce)are also orthogonal, provided the varrier scalar product is"invariant", inthe carrier space. vaniase.prods y+ Proof:vo ir) PKR KAD =ge DEW) DRO (RHSH)we ORS oe r ; =(Sah) ) =) ) * =!~te2Da(SENDA C5849)air A ik ~dw WEDD. O (15,4) .Kost A a “¥=dt2%Vin(3)DeVAG) Cv) sr aA v * ..S O=42,Kao)OH) LZBLODBLO cS woh.beGulighm , = Buby be: ZeDa (44)ia av A SkimKye Cem, f) A=WwSyd (Kah, £)ve ¢, Ouch. This says youareguaranteed orthogonality only ifpour pesjex projection-like operators areindifferent rows. Ifthey correspond tothe same row, but different copies, ofthesame UIR, youarenotguagenteed orthogonality ofthefunctions, even thouth the projectors are orthogonal inthe group algbra. MingTable(Lad)Lientenberg.‘infornetion,seve e & & &) We) Ce Xn=FS 4CSOT ec yeh 1et 0A= By hoa ws \chy 004s 3X7BYu On,=1O4-|“lLOs%%(Wyte) AYy> 2° a Yu=oOo-4yt 2 3Nau=puAtN We=00OA ae :o1-|“th8 | “Oph=aay ==BY4BKy . hb=Ihreneh=6hr k= -A%e AY BABY =Cut0+0) y= 3%y—3Yu=Cyro~vo~t) %G=BNar~83Vay=Cer(oo71) We Ky +EYe—FBXpABM =Ge1-10) ay=Yt.BYq)BXXu) =FX 3B =BvD) =3 eae=BY+BYu)(Oky~BXu) = =Ay«3h‘a=3(ut Go)=3% NB=(3%+BXv)(B%a—F Xu) =~O+O Oo+0 =O. -J - , OBBB 8XHee. +2BG4-Lyea Solution totheMatrix Problem June 18, 1979 81,Hereisastatement oftheproblem: youhavea4-dimensional vectorspacewhich isspanned byfour vectors called €11€,4€3y€, »These vectors arenotorthogonal nor are they unit vectors. Infact, their dot products are described bythis matrix: NEBR) AAVE ao;eo ales& ={ecafoec)] . fS>])— & = ete “eToty eA fet Ss ocrob Before continuing, here isthe correspondence between these symbols and earlier work: &=\hy =Xalfp BB =CHW =CHG) =Gay ee E=VED=Kal? BE =Gata? =Cuda eS => ~~=@&~\i,>= Xulf) Bis Habe RE 0 s.- _ eeR= thay=Xalf BB =Hac =BR =e 1 Inaddition, thevectors e;andeyspan aninvariant subspace ofthefour-dim space, andthevectors e,,¢, span another invariant subspace. Infact, theaction ofgroup elements onthese two subspaces isthe same, namely: & &&\ (TO\o Va e}= &Re 21a Here,’ (R)isthetwo-dim representation ofS;wehavebeenusing. Onpage2ofearlier notes, Ishowed thattheaction ofgroup element Rintheobjects X,,wassuchthat theaction isexactly thesame oneither "copy" .Innewnotation, e,e, span thefirst copy (subspace), ande3@,, spanthesecond copy. Now here isthe problem: Isitpossible todoa’transformation onthe basis vectors toget some new basis vectors such thet (a)thenewbasisvectorsaremutuallyorthogonal,and fo)(b) the action ofRlike that above but onnew basis isstill inblock form. sO? ~2- 2.Discussion ofthe situation: Asweshall see below, asolution must exist tothe problemoffindinganotthogonal basisinthis4-dimspace.Thisiscompletely obvious, fo) and will beshown tobejust the problem ofdiagonalizing areal-synmetric matrix. The real question is: once you have obtained anorthogonal basis, how doyou know your group action will have block form? Certainly the two invariant subspaces will still exist, but not obvious that the vectors you obtain byobthogonalizing will be the same vectors which, inpairs, span the subspace. Infact, itisnot obvious that you can have block form and orthogonality atthe same time. Lets return tothis discussion atthe end. Itturns out that you can have both atthe same time, though now Idont know why based ongeneral principles, only based onthis calculation. 3.The orthogonality calculation. Here ishow Ipropose totransform the ba’&s vectors: gy &a| [A[B\e & & &o|Dg hy(tase T rs) a)@eee) NAé Al&\fodleA\fAle = =— — & o”™ ele felvaleloPe Wwey Itisourgoaltofindmostgeneral formofthematrix ‘ABCD.YoucanseethatitUsjust theproblem ofdiagonalizing areal, symmetric matrix. Youknow therefore that the matrix M=ABCD mustberealorthogonal sothatweal, Thus, weget: A\®\(st{eh\ -(tt°(aie ely /\ealbt o|ki/\s' > a =woA-|aan]8 3C=ald : ek) =\-(b-by 2A-\SB 4C=Sy These four equations arise just bymultiplying outthemattices shown’. Ihave anti- cipatedthefactthattheeigenvalues areequalinpairs,whichIwillshowinnext ¢>)section. ~ 3. 4:Determination ofEigenvalues. Just outline itasfollows; ° ah 9 cjio hath =are x °bhoO 2.° © 0 bh Wwh. =do-co => xlx-W]ad eo Ber x= Canyerh). x « . Vogiaxtxcd=o aeS eo a xecS fue. SBw&ayeY=e 3he=tly tle | Offhand Idon't know howtodeduce that theeigenvalues areequal inpairs without doingit;theremustbesomefacto aboutthematrig which makesittruebyinspection/. Now wedefine some auxiliaty symbols and state relations tween them: Q@t= 2s. -Gy =o(a-ha) =e (ard.+N) u paw (a~hia-h)+ =o ~ oy * WY)=OE(ea Gyeed oct Inthefirstline,theequality tellsusthatequations 2.andAeonpage2areredundant, soour only conditions sofar onABCD are: s.A= a® 6Q= aD 5.But,wehaveintheabove already assumed thatMwasreal orthogonal. Weused this fact toright multiply ant equation byM.This fact implies furthoer conditions on ABCD as follows: ST,ay ‘AleALIS We) ooAmARe=)Ataalso2 aiogoral) . re)&otaaw=|(cao e) -he So, equations 7.,8.,9. are the conditions that the 2x2 matrices A,B,C,D must satisfy le)inorderthabmatrixMberealorthogonal. Ourgoalnowistofindasetofmatriceswhich satisfy 5,6,7,8,9. Then weare done. Inserting 5and 6into 7,8,9 weget: ~ 1. v. Be= wywee: %.((eaa.nn)88=o) ! oxo, This tells usthat matrices Band Dare “almost real orthogonal". Lets then define new vertions: : 2 aay yy B=B/Fe > Be= A=%08 2 war od DzD/p- ~ DD=[ c=aD Thus, here isour most general solution tothe orthogonalization problem: . g A\B aesAy|PeAsuneAg,KOononbsvhamy n=(S42) = | —= qld ap.A-|e.A- . re) f-A-Ie oad,afeagcmaarie. -(ky©)a,&4). oOAe Avpal pa Here youseethat thematrix ontheright solves theproblem giving orthogonal basis vectors. Then A,andA_separately rotate basis vectors within thetwosubspaces. Obviously these extra arbitrary rotations cannot change the orthogonality. 6.Whatabout theinvariant subspaces? Consider theeffect ofmatrix ontheright above. Weget: f e 2& Tyo ‘ iq &zy.| RISl=nef}[npy|e et«\ee et eae ae ay +- @foo\/A\er APRA&RAresard 0.2GeyettPee etarsole opera opyr x =Aathe|2= o |ATE .7 -5- Thus,youseethatblockformismaintained: vectors e1'andey'spenonesubspace, A endtheother twospantheother subspace. Theeffect ofapplying A,andA_isimplyfo)toconverttoequivalent newtwo-dimensional representation of83+ Iseenoreason tobother with A,andA_. Iwill remain awayre that within each subspace youcanrotate toyour hearts content andgetequivalent represntations. Here then isthe simplest solution: 2, ae a eaka=Pe(4+8) a=A(Cx) a —— as anB~ K(Ls = PRs’) mat ory a2) aeSe -SA =Wy Ones =4-9 = 7.Comment: notice thatthenew"row1"vectors e'ande3"arelinear combinations oftheoriginal row-1 vectors e;endez.Thus, theproblem wassolved byrecongining therow-1 functions seaparately from therow-2 functions. Ithough Ihadtried this! Myother attempt was doomed tofailure, the one where Itried tosolve this byrotatingwithineachCopy.Inowseehowstupidthatapproachwas:ifyouget fe) anorthogonal solution, clearly separate independent rotations within the two subspaces can only lead tooverall orthogonality. Nopossible rotations ortransformations ofanykind within theseparate spaces canever connect youfrom overall orthogonality toour starting non-orthogonla position. 8.Lichtenberg iswrong. Iclaim that thelinear combinations necessary tosolve this problem are non-trivial. Weknow that: de=to[ob +Janeane | wee ae SHY =FCF E-Fed -FGD-4 <i’ be Kye Fee) -Fe GD+FAD -4cd -& Bic= = = HAD+Lad, vhs,eR) tonneSayin isFGxeyxe)FO,%Xs) “? -6- Suppose youaretryingtoanalyze aparticular function £011%%5)+Thenumbers __ayb,¢_ are certain over&ap integrals ofthis function, Ihave assumed that each 0integration hesthesamerangesoIcoulduseasymmetry tosimplify slightly the forms for a,b,c given onlast page. Since thenumbers a,b,c areuncorrelated, (ie, fordifferent functions you will getdifferent a,b,c andthey have noonnstant relation, like sumtozero orsomething),N; isdependent onthefunction f. Now lock atthe general solution: 12> 2.2 > 2»BS=So(BsB) ~CoB +%) ers >e\x a2 3)A= Co(eer&)9Se,(aeB+As) swmdeh AY Iseenowayyoucould select angle 8,sothat thelinear combinations are f-independent. 9.Caveat: However, suppose f(1,2,3) hasthe following very special form: FOa =aGSb@)e@) fe)wherea,b,caremutually orthogonal functions ofasinglevariable. Ifthatisthecase,then allexchange expectation values like (f/(12)/f) vanish andyougetinparticular that number c=0. Inthat case, your original basis ej...¢, wasBready orthogonal and there isthen noneed todoanything. According toLicht's definitions, wehave: erK=2e+3ze, =) b=bey Y=~-e4aBbe, X=-heict “(3 BNE.aaye a)\+33BNeeSaySy) Recall that the most general subspace-transformation which leaves basis vectors orthogonal isthamatrix a8)where theonlycondition isAC+BD=0. Thereason itismoregeneral than arotation isdue tothe extra facts implied bythe group algebra. You see that Licht'schoiceofbasisfunctionsusesjustsuchatransformation onmye,€basis! (oe) Hissecond pair ofvectors areproportional tomye3,+Hemakes noclaims about normalizing his functions. So, heiscorrect, with this caveat that a,b,c are orthogonal functions! !!! SomeTheorems about Constructing Projectors. i Theprem 1:Ifyour group elements Rarerepresented byunitary operators R,andif ‘ yourrepresentation matrices arewitiaythenX,}=X,,.Ie,thenhermitian conjugation with respect totheHilbert Space Racts onisthesame astransposing indices forthe projection operators. Pro6f: Start with definition: —~ x Wowe WE.=(E_DL® RYE . () s\ yt Ww x 4 2XG- B(Z DER) =B(& DI. Ww=( Din(R)R) [hogsssanargunscks.hw f& EAC / =((R ee. 6§o(2D) &) ne:enctanes, 1)=Xu gen. Ihadtoassume theoperators wereunitary inorder togofromRYtoR4.onlythen canyouusetherearrangement theorem. Note: wehave already seen special cases ofthis theorem, likeX})=Xj,for85,ete. Theorem 2+Theobject formed byadding tothestendard projector X,,anarbitrary linear combination ofX; operators (ie,ones forsame rowbutdifferent copies) isanew idempotent. Te: omy a ic NG ~~Ker=Kee+ZeAcXee| =>Ken)=(Kis) Proof: Wemake useofthebasic fact about theprojectors (ie,theGOT): XyXa=Sy,Kb Soyekwast,Sang oh: a3; 1 os rar >2 t (Ke= hey+ZLA:(Soros Kedker)+SZALASKoreas ae iat SE Oa Tyee Ves Nore Kei oie oith is ~ =Yer*EAN =Xe) ag. weyTheorem 3:These newidempotents areorthogonal inthefollowing sense: XfX,,=0. Inother words, (Xf Xsf)=0 Proof: * * : y ~ t(KeYKed)=Lhe+2.AYXie||heeZaAKes| or jaz D8) *: .t=ZK: XKirkss =Ome THS 2K : ta&,AyYorKey=Osmec#5 aras » fe)te BAKA ieYay=0” QE. is— The proof here isreally obvious byinspection: the operators infirst bracket have ras second index; inthe second bracket have sasfirst index. Since r/s, allproducts vanish. Theorem 4:TheHermitien adjoints oftheX,,arealsoidempotent. .oe whet ot Brooft Keefer =freaKeeRie=Yr ep y Comment: Although these newprmjerkmx idempotents Xe"dotheright thing" inthe sense that the chunkes offunctions which they project out are orthogonal, asintheorem 3,theseoperators arenothermitian andaretherefore not“projection Qperators”intheusual seince. Ie,wedonothaveXX, =0forrfe.Ooo, Geek eXie: Unless you have honest~to-goodness projection operators, you cannot "decompose" the identity into sum ofprojectors, even ifthey are indempotents. Exception:Inthespecialcaseofthetwocopiesofthetwo-dimrepofS3rwe re)can consider these two objects tobeacceptible projectors: a 2Rak =He R= ke=0. L >BaNe Pak. te ~3- Thus, the following are acceptible projectors: f=MyFAX f= Ka+Ba Specific Example: Ifyoucompute theprojectors using themethod ofHanmermesh, seenotes above, yougetthis: (isve-te) ; Maat x 4=4 ~os)-rg| =i R=SY=Xr ka(SH) =gleroo-9 oloe -1) a u Ud Gvveoe =tfia3Y= kai Xa©FH\=£Leroy ~O9-O39]akote Thus, weseethat this isaparticular case oftheabove most general form ofthe projectors fortheS,gwo-dim reps. Comment: Inmore complicated cases involving noccurrences ofann-dimensional representation forn-3,l.++. youdonothave thefreedom discussed above. Porexample, re)inne3caseyoucamnotarrangeX,11Xo91X33togetthreeprojectorsintheway done above because there arejust toomany. Ie,ifyouchoose P,andP,asonbottom oflastpage, there isnowaytodefine Byexcept maybe P,=X,,withnomixture. -Probably that isthewayitworks: you can shuffle two ofthebasic. projectors, but all the remaining ones have tobe"basic projectors". Summary: Ingeneral, theee aretwo methods Ihave tocompute theprojectors. (a)IfIknow theactual representation matrices (they aregiven inHammermesh for upthrough S5)Iknow howtocompute the"basic projectors" X,,.ysing thegeneral formila: ie, its alinear combination ofgroup elements where the weights are just the rrentries ofthose matrices. Analternative method is: (b)tise Hammermesh orMessiah method: foreach standard arrangement inagiven pattern, construct the object called Y=QP. The only problem here isthat you have tothen work outthenormalization byhand, even ifyouinclude those Messiah factors. ye “3 Theoren 5:Thesynmetrizers Xx)forma"peir™ ,andsodothesynmetrizers q. le)Proof: -day Qt FreetThemQuindelpndbea oake dD a) .RX=ZLDla)Ke CS = 36o) 3)~ RXX= DAK Avector oftwo objects which behaves this way under the action of Ricalled a"pair". Theorem 6:Anylinear combination oftheabove symmbtrizers ecross copies also forms a"pair", Proof: ~ woeoe RE Res YS, 3 ~> ~ Qa.R= 2bZDe(®)Vid=&Dale) Xe rey z Ow RL-WE —aco. \2 Eroomyba K=KX+X,Ya Ye Kh. Sequeteod : 1im Ha aka wy Ye=aX ree Theorem 7:Ifyouhaveonemember ofageneral pair(theabuve pairwithanya,b), you cangenerate theothermember ofthepairbyapplying either x,orX,*. : Vy 1 2, aSet YaY=aOak)+bOGX) =xh+e=K% as Vv= \ w\ tX= ahm@r)+64) =akax, -K7rae fo)ae - Badewor,yywdoWay7fVw XakXe =O -+: —5- 0BeGMSaa“pu Es,A=RROD —R=RAs) Gre ea pride sumone Osalone i ° xt ae a ° aren ara \ .SY Oe0, xxex ° io) ie) Oy Bad joeeee OCR =hsJhaXo Susman .oe aiaky-v-t wot . | SACU aba tkee 5aap2A POS AVV OL RET cotgeepeterye | Aeee mee Beh Lo eave th ce nn ANE -) (a2) ==(2-e\ 5G22).ee ae ef BeK-B-= a xoNeb=3.Sa egBEC pot4) _- SE AED-oOsthroe %2Chas comes LEED ee a oe Mot _. _— nah ba fi oe Aon eCiaa aWDCaes <i a woo XC NY nn <<oe SeeLS Maer ean.ea uy - eGo etae . — . ©9Ava Ay. L SoSGAKMon)ae eg La, a —-epsain o)__aatte peesarwfao. /wen) _wwtee sefo ereh) =S-bdah ae ee ee Te BEL atcpo¥sinta-ty 5-2es-taEy “Cy. ‘auepyaers penceate a HS_asporS ses lgpews = —.-. a — —- a =G-p=! =.- —acsyC)-G)-(-W=\ ayeven ne pee ee) ieee ee > _BLYaddeyonekmmvache )HAS frete a OS SOSna aeOUDO OOSROO &Note NL at Ss OOee oo osNN \ __3 ek Nsad eedBe Ssies asadad - LY. ° - | es Oe .Qo >) Vion Jol 5 QTV juggs)tt yf $f i Sana aSa [on Aukwhadeeh\dooek To 2Brg. Qedknt oddaangSteTa nent ee ~ we fob mh abl ee iSNndOsean a Wet, 22 eo Vu fo ee re 2. t22 “enToBLS@ EasOd a) enne—2 ee -_ as¥rh4bSoe Oe 8 _u.@. og+)-269Fat ee We a a Sa a a, a ER SO a ee ee a Dos ak SS CNWGNOIE) Fon 50Wereisnysecondverification thattheoixeymmotrizersareLinearlyindependents “1,Gnthegroupalgebraspace),Systematically reducethenetrixtotriangular form If ___athere isazeroonthediagonal anywhere, det~0andsetislindep.Butherethre,|""sgwerenozeros, soallarelinearly independent!! ee ae speengly)bonlesan)NER About Decomposing Functions £(123) intoS,irreducible components. 1.Imagine anobject £(123) such that thesixobjects yougetbypremiting this thing arelinearly independent. Such would bethecase, eg,if£(123) =a(1)b(2)e(3) where a,b,c arethree different functions. Ormore generally, if£(123) issame arbitrary function ofitsarguments that possesses noparticular symmetry. Ifthepieces Rf(123) arelinearly independent, then the6objects youget byapplying thesixsymmetrizers arealso linearly independent. These 6objects are: £,(123) =(1/6) [£(123) +£(213) +£(321) +£(132) 4£(231) +£(312) ] £,(123) =(1/6) [£(123) -£(213) -£(321) -£(132) +£(231) +£(312) } £3(123) =(1/6)[2¢(123) +2¢(213) -£(321) -£(132) -£(231) -£(312)] £3(123) =(6/6)[ =£(321) +£(132) -£(231) +£(312) J £3(123)=(/6)[ ~£6921) +£(132) +£(231) -£(312)J £3(123)=(t/6)[2e(123)-28(213)+£(321)+£(132)-£(231)-£(312)] Le} Thebrackets show pairs offunctions which are"paris" inthesense that they behave under action ofRasabasis fortheM-rep of$3.Here, then, wehavetwolinearly independent pairs. ' Recall that the"group algebra" space isesentially theregular representation space, andbythe"celebrated theorem" this reduces toS+A+2Mreps, soyouexpect tosee two pairs for the M-reps. 2.First special case: suppose £(123) =£(12) only. Forarbitrary function £(12) with nosymmetry this isnorestriction; there are still 6indpendent objects andthe above stuff still lists them, just delote thelast arguement. The6objects areofcourse: £(12) ,£(21), £(13), £(31), £(23), £(32) 3.Symmetry: ifyour £(123) hassome sort ofsymmetry, like thetensor form 143223, youmayfind that thetwopairs aredegenerate; thecopy 2pair mayjust bemultiples ofthecopyonepair.Thatisexactlywhathappensthere.Inthisexamplethesymmetr fe)isthis: £(123) =£(132). -2- 4.SecondSpecialCase:£(123)=£(1)only.Inthiscasethereareclearlyonly fe)three independent objects; the totally symmetryc, and one pair. Infact: £,(1) =(2/3) [£(1) +£(2)+£(3) J £,(1) =0 . £7(2)=(1/6) [(2)+(2)22£03)Jfee=@/6)[£(2)-£(2)] £3(2)=(6/6) [£2)+£(2)=2£(3)J{eo=(3/2)[2(2)~£(2)J Here youseeanexample ofhowthetwopairs become degenerate; thetoppair isrelated tothe lower pair bySQRT(3). 0 ; fe) The tensor problem for self-energy. a1.Consideraselfenergywithmomentap,andp»entering. Ignoretheconstraint that Dy+P220. What isthemost general tensor form involving twomomenta that youcanmake? Answer: this sum: 2 Ayca aot. en Be TeeaR)=eeAgate (pe)) This isonly four terms, easy towrite them out. Now, given four functions and thus anexplicit form ofthe above, what isthe "switched" tensor? Answer: Ha(opt = T By aly RayPsd= +(0bor - -ZeASE) Cry") Here, thenotation Imeans ifiwas1,then I=2,Maybe Ishould write that I=P,,i. You see that the actual momenta ofthe tensor form get switched, aswell asthe arguments ofthe functions. Their indices stay put ofcourse, and the Lorentz indices onthe momenta also switch. Now, rename the summation indices toget hetoe (coe ASAL (a) (Cony\=aAsPafr)(p:)(P§S 2,Wow, what aretheconditions ontheA,,ifyouaretrying toconstruct anantisymmetric tensor? Since nomomentum constraint, each ofthe four tensors inthe sumisindependent, 80you conclude + APD 6920% Par otTH y+ Tet) =0 < +.(otpt =7 (9?92 AY raMi=Z|NahetA Get|Ceyeny ‘y => ta ct AAgate =—Agee) =SAuGd= -An@Aucd=- Anas) Youcanthinkoftherebeingoriginally @functions, AnsGye-AasCa") thesymandantisympartsofeachofthefunctions. Aa(y= ~Aw(uy) These condtions then reduce you to4functions: notice that self-condition removesonlyonefunction, whereasacondition relating two ODaisterent. functions renoves twofunctions, inthesense ofthere being €originally. Sohere then arethe most general antisymmetric (and symmetric )tensors you could have: Bya.¢5ok Ss HyeBe Mieee lo) we) =AG)|Gye" -GG aS. PurBa Arr\Re aANG) |GQGY&(pay(pay aS os +AGG |Guy} as Apghe +Any [Ga*gy] ‘SAS Sorate all. “ToghteT)ald SEASongn Thus,youmightarguethattherearereally4different "tensors" existing inthisobject, each requiring amultiplying scalar function ofsone definite symmetry. 2.Nollets remove theconstraint andletp,+P)-0 .Whatdoesthisdo?Replace Ppwith-p,. The first tensor vanishes identically, the last three tensors are all proportional to oneancther. Butg since p,*=pp,itisnowimpossible tomakeanantisymmetric function 8tomultiply. Thus,therearenoformswhatsoever left!(fortheantisymtensor). For the sym tensor where Sand ASlabels are all switched, you have only one form, which isthe confluence ofthe last three, above. Thus: Pye2 al GF) =© s yp. PyeCgty ot \ye azy TO) =Alea) CHYNCAY +denneGeelyg FOri) 4Yonmin 3.So,tosummarize: wefirst hunted for allthe tensor forms,without the constraint. Wegot some general answers, each having four functions arbitraire ofdefinite symetry. Then weput the constraint back inwith two effects: 1)some ofthe tensor forms identically vanished; 2)itwas nolonger possible tomake functions that were AS. LsTnowwanttoseehowthisgoesforthecaseofa3-vertex. Something tells methat thegroup $3anditsrepresentations might somehow beuseful! A-Review ofthe general construction oftensor forms with two momenta. fo)1,Question: whetarethetensorformsyoucanconstruct fromp,andpy.Ihave already figured this out but here isanother way todoit. 2.Solution: forthe group Sythere are only two representations and soonly two projectors, call them Sand A.Write down all the basic tensor forms and then apply projectors toget forms which are associated with certain representations. Then at the end apply scalar functions. So here we go:8 Sg A ce A arres LCpeepAban Qs"Cay SLexrent ch] ELeheth | AYSaad AyRe(gtec) Certs © oN gie Aween) Ree 3 Aa oe Ride ACAY LP ohms peg aree slate es)ttelee ih Thus,byapplyingourprojectors toalistofbasisfunctions, wehavegeneratedalist fe)ofnew basis functions which have definite symmetry. Namely, wehave: . pe wy Be .seme TPM gee ie JT= ep he LY gtr ke pe garwhe ele attegtet Omasy -— ST bigHepdeyAWe kt tte'), ‘Then here are ‘the most general sym and anti sym tensors wecan constructs aye Spape OS Seyee ASSpyfaWye=Paeoy ¥aRTaeYow aw Ie,youareallowed 4scalar functions andeach must have thecorrect symmetry. Similarly .forTantisym. Count Amplitudes inTriple Glue Coupling byLSCoupling. a1,Imagine thatyouareinarestGameofagluon;itdecaysintotwofinalstate gluons. How many independent amplitudes are there? The pair offinal gluons can have S=0,1,2 . For S=0, toget Jal you can have Ie1 and that all, For S=1, you get Jel with 1-0,1,2. For S=2 you get Jal with L=1,2,3. Itwould seem then that there are 7amplieudes: 8-0 Lal 1 S=1 L-0,1,2 3 S=2 Le1,2,3 3 1 Isthis number further restricted bybose symmetry orparity orsomething %am not thinking of?? Theproblemofrearranging the36objectsoftheform£(123'g(123)intotensors fe) ofdefinite overall symmetry. 1.Here ishowyou doit: first, take each ofthetwo systems and decompose into 6pieces ofdefinite synmetry. Then combine thesystems using direct product to get resultant combined forms ofdefinite symmetry. Namely: (m+B+R+®)x(B+Q+ B+Q) =CmvB)(maxB) +(am+ B)«(fP)-4 Q +» @-4 (+3:RB) (+Ym +4B+4-FR) =60D+6BR +a Note that weget646424 =36objects intheend. Inparticular, youshould beable tomake Bix different totally symmetric tensors inthis way. Two arise inthe obvious way, and 4arise from the mixed symmetry combinations. re sei °i. : /wy)keoOFRI]Oo |shioTs| . eses esee1 Lo \oyfepee MN 1 2e HX XO |%xX86 ¥] oo ° xy x w%/M K 0 © x o°° x x G@ “ 33Expansions intheStandard Basis. Ranie Szdole _ le)1,Herearetheébasisvectorsofthegroupalgebra,andtheirmultiplication table:H 6m = 1 ee | \ \=6KA 6Xs = \ \ \ \ \ \ =6% 6X)=2 2 ~\ ~\ ~\ ~\=6%, bXb= 0 9 -® & BR -B=6K 6XH= 0 o )6-BlhUR HR B= 6% (es 2 2 to yon t+ ;|Xa Xs %, XnKas Xev Owpule ooo =Xs oY Y° } ° ° Au ° o]Xa Ye_ °| Xe°otOO°YouYo| Xu|O oi. Ye 0 o a. | . oe TheseXobjectsaresupposedtoprovideacomplete,altbeitnotorthonormal,basis fe)forthegroup algebra. Thus, anyelements ofthegroup algebra canbeuniquely expanded inthese basis vectors like so: A=Xa asks au%Xy +ae +OyKat eeKer Given the group algebra vector A,how can you deduce the 6coefficients? Here are "some useful facts that follow from theXmultiplication table: AXae KaA= aaa KyA= anKae aaKer A¥s=XsA =asXs YaA=aa%a +2 Ker YuAX =anXu YarA%y =aKa YuAXee=aeKn XaAtn=doKer. le) Forexample, ifyouapply Xyontheleft ofA,andXoontheright ofthat result, yougetaresult which isproportional toXp andthecoefficient will beayo: so this isaway inehich you could compute the 6coefficients, given A. Group Multiplication Table: e a @) ae) (sey e e Ww i. 23 (23 13a i@ | ¢ 32 vey 23 13 i @ 18B 3 e (ar R 23 ! . @) ry (30 Ws e B 12 ' Cia) F123 m xy mu 13u e. fo) (@y)|Br23 12(3 .€ 123 Quorn: |COBB=Cabs) LYCade)=Cee) Cys (eevee) =(Ce) - -3- 2.Action ofgroup elements onbasis symnetrizers: First makechart ofD'(R): my Le aw ~yxfB Ove-= GC? Se- (RY : xt te ay “E283 :Dv(aa)=) v@)=G,“4 ee x H+-& + P28D(os)=(. D> (Heuy Notice that these are instendard order reading down. The general rule for actio ofR onabasis symnetrizer isthis: am. k T ts k vey. RX=ZDQM, ae KE=Xtke For the one-dim reps the group actions are fairly trivial. Using the above matrices we can then generate this complete chart: i é fe) Xs|Xa|Xi Xa. j ‘ ;| i OdX.=-CB)K, elXsi Yl Ka i A A FOP GSE03 Ke GD]Xs5XR, XK ~h rrT iz : Bul yt OSXy= OYXK G3))Xs|-Xa|-2%y- 4K -BxX tke t é toj ays ayy |XeXA|EXASBKL |ERKE KE] =OOK . a en _ i é. | i A fa Q CS)Xi=(132)Xi ary}Xs!Xa[rtRnB |+B-$XeFe aeesOra (32))XoYA[3AyaX|-4BX-4% atee ed an "3, Bytheway,Ithink youcancheck bydoing hermitian conjugation that: right action® bygroup elements mixes copies rather then rows like so: (check ifneeded) le) i 4KER =ZDS@%3 . “he 4.Nowhereisatablewhichwillbelowenableustowriteeachgroupelementas fe)asum ofsymmetrizers. The table isread easily fram table onpage 3: ” Avu cSal & XRK XeRX! XiRY KeRX \ t e Ya ° ° Xe Loy * qe) w ° fo} aK \ \ w a ws EK RX BX thy, vy! \ 4 t @) -th +358KY EeBK Ta \ Pay .& ok way AK <2 x) ay! -2y , A w we ase) 2 +28 YX -+938 X -2y Rememberthatx,andxarerowprojectors,hecneabovetableiseasytosee.Now re)recall the general decompoeition ofany element ofthe group algebra (here take A=R): = ‘ \ al Voy r uiRaRk+RHEHR MRM +eWEE CRE | \ a a u Se Xs MH KH XO e \ \ \ ° } \ @ +) \ \ ° oH ay -\ \ a 38 +8 z &) ~\ { -t tH Hf 4 Ge) \ \ “4 4B 8 -4 (sz) \ \ -4 rtf te ed +l+GA\+Ga] =Yatks 5+Identification ofthe6functions which canbemade from £(123): fo)Backonpage3Ihavewrittenhowthextransform undergroupaction.Thisimplies that theprojections ofafunction will transform inthesame way: kos x ROD) =2LOB)y fr ‘ Que. Se KY, an kes A My=al . 1\‘a+XK\x /A i ¥,=X+ow aNe a3) ‘e] caer woi y a He t-Ya osfiew+ Peoy%Vey Ree ay foo A ae 1D Thus, these erethesixfunctions ofdefinite symmetry that youcanmake from the6permutations off(123). Ofcourse ithappens thatonlyfour ofthemare needed ifyou want topartition f=sum offunctions. Ifyoulook attable onbottom page page andread across thetopasthe sixfunctions, youseethatthefull setofsixfunctions spans the6-dimensional representation ofthegroup $3.Itjusthappens that thefirst rowofthat table involves only the true projectors. -6- 6.Now,ifHammermesh's Clebshc tableiscorrect, herearethefourtotally Weld _aastonett \ Voy an 3 wie uw\B= Vyehe B= Aye tge a Voa Lie q woke BoeRoe Nash ReJa+fg Any choice ofarow-l xrow-1 +row-2 xrow-2 should betotally aynmetric, according tothe Clebsch table. Now lets check .Perhaps here isafaster notation toaccount for all of them : ap tay coy i=4a+ 4Qe Ree yeuee Usetable onpage3toinvestiagate theaction ofvarious group elements onthis FJ: bs ard e:>LAWael +Wie) =7 tyi\ le)a=UNUg) sbE gl=- 7 Gy>LAS eB Vag agllsFEVANALEQrtse ~Gri) Rare Qayhye «0-7 an> L- -VWe -\sl- +t +1 aay> L- 4\b +)4G +e +] Qsya U>~JG~\ +C4-10+-] Bh Tetl\Gs+} +(-YE4 o Thelast three elenents arejust sign changes on(13) cagg/ They allwork. Therefore, I have shown indetail that theClebsch's arecorrect forAimM into theTSrepresentation. Probably then all the other ones are right too. Loe —7- , Action ofgroupelements ontheprojectors. 6 too. \ \ \ \ \ udyX,=7-2(\\48X,)=-Eee] : ie) =-ti(a 2-42-42)=+e(i-t 274-1) \ \ 2 ~ wyY,_~XK Stop. Theeasiest waytovisualize this isthe (13) orRingeneral acts onthe heading and rearranges the headings. When you unshuffle, you find that the integers havesimply beenrearranged! Forexample, X,*hastwo2'sandfour-1's. Then(13)x} alsohastwo2'sandfour~1's, Infactanyobject like'x,'R orRX,’mustcontain two 2'e and four -1's. That set ofintegers isinthis sense "invariant" under group multiplication oneither side. However, there are 30ways tohave two 2's and four -1's. Ofthese, only 6are accessible, sotospecak. Eg: \ KrP52 aL-\ —\ —\ -\y=CRe=e)=e GG OO @) Gey (sz) fe)> © G GO esBEIrI2-) AV 2-12% (Wr?(~&(2)(iv) e (3) ty=pCE ) (By @)=lov a-tp (nd\>as)us)(2)(wy)eSa)> ) ry>(WY (v) Gd a QB @&2(H17712)(sx)>Wy ( ) ’) 2 fas) AY aKe (2a~v~v~v-4) 4 4 VOD X= |HrAAA-Y =Kop ‘ . (8)KsGl e-bay) =Yo) \@KsGrvatz)=Qe ie) Ge)Y= GIB ANY SYD \ Le )Kyz Gert an-vay = +d _— —3- Geomnadr cota nck itt rs) CabvedesSs (abedte)§Sdwad, Sedalle also soy2! axX(2)=Cr-1-e1-!) . x(a)=(rite -12) MWA =(tv 22-t) x(av)=(-4-8z~i-tt) Xfrd=Grote te-8) rd Caemare)=x=(2)X\=xX(az) | Gta-at)=adyt=ce! | |G11 =@X\= 0), i Crta-v-v2) =XbCs)=X(432) (eis) 22-1) =Xi) =X)es) aNor, CwLr 4 ° ° -\ \ \ -\ Ky > fo)Afastwaytodothesethings:reorderintegers according toneworderofgroupelements inthe group multiplication table. ws -4- Forexample, letsredotheX,*thininthiswey: ie)OX,=Vw=-V~V-V=h dX =Ay 2-1 2-\ Vv Y= okedi What Iam doing isthis: take 22-1-1-1-1 andimagine itasaheading on the group mult table. Then look ateach row tosee how they are reordered. So: X= gosy desir Pay =(oonary |e(oom tiny=YXCe) CNH=GVotkov PPCrresavey= ¥(3) | ee (-1ond (i100-\)= &@a)=h Beant =to-oo) (\to-10) =ney(32) aKG =)|ont0) (soe =i)=HOe)ete| 6X=@oAa) Ooee(ool mtyx) Coot-\~V 1)=X(12) [onRe (4-18101) Gio veo)=XU3) adXsCh otoY Crave ot)=O(28) hoykee(bb Osh 0-0e' (ve 0) =Ke) Ose Git 10 4y= (Er Loos ve(23). ; fo Ways togenerate functions which sre"pairs" formixed symmetry. Le)1.Ingeneral,weknowhowtogetasetofpartnersforarepofanydimensionforarbitrary groupS,.Youactonafunction withthesetofprojectors x8forany fixed copy index k,letting igoover the rows. 2.Any linear combination across the copies also works. For exemple, consider; Bar .he =CKKE k7 k =RTL=BARK = LSAZ DOKK 3 >ELYd: eGs Since any such linear combination gives aset ofacceptible "irreducible functions” orsymmetrizers, you might just aswell choose the simplest ones. Notice ofcourse fo)thatyouhavetoaddthecopyelements withthesameweightsforalli. 3.Example: ft a 3 wa Bess os \ 2We22 ~\ oN -\ ~\ \KEK - \ a = -& ~ = + ye Oo8 -R BB -f S38 co :Yee 22 \ \ ~\ ~\ Ts 2 2 72 ° ° -2 7 Aea fe)So,contrary tomyfirst impression, youcannot define 7,andT,here suchthat each oneinvolves only 4group elements. NordoTyand Ypgive youanacceptible pair. hh.Lets check outYandY'andshow (Ithink) that they dont generate anacceptible pair: A ‘ .syY=w=y+BK, [cwtives poor) u i} ays f=Ki-BX 1 1 a \ 1 wd. RR= RRB =AO BZMOK> = > = =yan skde! So,although these P,andP,arerowandcolumn projectors, they dont transform inthe standard way and thus they wont generate acorrect partner. You need a"correct pair” ifyou are going toutilize standard Clebsch coefficients. 5.SoIseem tobestuck with these operators asmypair generators: \ aL= awa Ko Gr aaanyt 8 —— i + % =©oBBB - &.Example:_ Thesimplest non-trivial case ofgenerating apair offunctions like this istoassume that (123) =f(1) only. Then you find: ‘ J=BO)+l@Q-24 .f= B(&O-f@). Chmment: Ifyouaretrying tomake apair offunctions, youcanalways premiltiply onthe right bysome fixed group element. See later. aaa aie Be Trreducible Compoenents ofTensor Forms. 81.Letsconsider twomomentathesame.Inparticular, take 171723=p."p,¥2p.¥3. This isanexample ofa£(123) object. Icanform sixirreducible objects byapplying nysixbasic Kj,symmetrizers. Letsdoit: First, compute the action ofall group elements onthis £(123) object. You should doitintwo sages orhowever you like. When Ileave the superscripts off, itmeans thpy havebeen ordered in“standard order" meaning wyuju,. SoIfind: = ws ws we WL Ae 22 233 BY 32a 3 Here bythe way you see the six tensor forms that get mixed together. The complete set of"two momenta the same" raw tensor forms really has 18elements; its like three separate functions f(123) which donot mix. Lets keep tothis setof6fornow. Now use table onpage 1of"expansions", Set the root threes equal to1,Idont see any need tokeep them, and ignore the overall factor of6's. Then here are the six functions: CUDA =CWE =Ze 23B= AH 827431. Ms =LUT FU 4233: +1B143224.213} \ GNDy =(CQL AVAL 2B 191-322-313] os cue = T=B413)a3rr-a3IG \ CI), = (+ ~3n433)G .UeALVUTAIWw AA.$7S_41B1 ~32v_--313] The only work isgetting the table headings. The coefficients here are just. those ofthe symmetrizers. So,wecould replicate this information like so: a mh » et -\-(yy ru We 423 ws U3 (83 Wa cant 233 (3) rr 313 cory ~\ -\ ~\ \ \ (ry \ als Ail) \ \ \ \ \ \ Gut _ arya zu -\ “\ ~ \ aut (wy se |eeeeo° (au . CHa, ~\ + -\ +) x3 cry. ad. 2 -2 +\ ty = -\ Tohandle theother 12functions inthis calss, wesimply change thelabels on the table; the entries are just the same. The facts weneed are: . a as ies wee re)2 We2233A crxaas vat (UW -(12 ~BW ~~43 US \33 val) uM aro BW aa 33\ Thaveshown inredhowyouwould change theheadings togetthesecond setof6 irreducible tensors. Conclusion: Having doneverylittle work, Ihaveclassified the19rawtensor forms which havetwomomenta thesameinto18irreducible combinations. Muchfaster than the last time idid this!!! .. r" 73> 2.Now lets dothe same thing for "all momenta different". Again, its just aquestion 0ofgettingtheheadings. Wehave: : <aaveas aAwwst \ BLBBB fareants 32|LW UoBUBUA» Vies Biaes WIT B_— BBY /* ; UZ] 23 air} se ; : B2i|32h : a . Maybe Idon't have todothese last three. Iamstarting asusual with asetof6raw - tensors,ifIget6tensorsofdefinitesymmetryIknowIamdone. fe) Solets take each row asaheading and see where itleads: (By, =0 OM =© (ai, £afer wed Gas£23" (se,&ag[trarase0| ys £sparred 7 qey =0 (ao)£Larsnnrael (si!=© ae=o ante cadds Qay=© Cd.=© sult Wwe fay=0 Gai=o asurt 3Teese) asOo Since wedoget6tensors, weknowwearegolden. Also, notice thatit.isallright for thesame tensore toappear intwoplaces asshown: same rowidea. 6 Los a : Next, lets do"all momenta the same" : loi =awacSeedeWA NY UL BBW BD. eahelabe Ue aus QW,=0 aye2es (uy©almaerassss] any,Swed 2ow (vy)|twee %333) O)L=sfw-ruy Finally, lets combine one momenta with akronecker delta. fo) _ \.a a Ta i ep we Se Bs te ge ws ss xa=zePeal st ={sax Weshould generate 4tensors here.Whatarethey? L Sy =SQ-d' +sSP@aAv4 8"G-Y Se =Sara «shar caer" BY =BEA Saayl= 8"Gar” 3a3r ? Ove =fsa-svt+ 8% eae ‘ \ BOE =[Soest eHy'y® « \3zr fol WYSE =2Sys Sook Es) foe . -5- The second header gives these tensors: Oae w= Oo (SS =alstare Ths TY &2), =Lat2:asx‘se] Se =© Se =e Hvis oe>=ro) \ 0 ee 477 ae ee Oo=teadoy ee oo A 1 ANzie‘eel22 wa] rs [how cachofthe27monentaobjects Sn re . _|__ reduces toalinear combination ee \ Pete nesee ofthe remaining 8independent ie ; tensor forms. ns (2 ee xen ...- -$0.g0boyoursetofinterestyen |Ct del t a_setinvolving 6objects. Then BL TH lookupeachentryinthistable AB PRA : _tunbers ingtablewith6rows. —:a_[1_____Thenaddthese6rowsinthedifferent_a j.-.—- Ways toget_your results, Te,add wejp, therowsasifyouwereaddingthe_ Aa|beeeet _Sixobjectsaccordingtoyourrulesa.-Ml }| toratuctngrmecutineengects—- | Lt. i se . a a a a al |=ft . a a a) Ceee oe —— wee weee BU fe Teee. 2ee RR a _BrL~=aan ceei lowly) Po woe | ne oe en a — ++ | aen oo on Ea eoaa [+ Ic ! ¥ _ ee 7 +t[st ne 2 fom|A[==no|v _ (ee et | ! Yo eee _ eee OrCal _ : ee ee OO _Lao [==i ~ TE Ree = oeOw a 24) +Ly. LUNY. H|anteeto x a =mFES eea H= i+ 5 on ee! | ee| —__ =ZNoite 0 bi2 __ f LT | a Ma x . ( & eoBLoyaiethay(lpn} ac] yy==] -\ Jae_———..rl.a:=ti (sre) eeee _.watAA|itt+++looutaa /@)s=Joo153!]o0oof, aabert] =uly” See. a | - ae wae et |Popo te — edeeee 2s | J(unv=. 00,00 Joooor (ue Patazffoo [-U-t fea PL _ wr tb sease Pe_ es ee < ay 2 ' [oe a we ' oa foo oy i—— Ed : ¢ rereoie ae ny aes|eeee Swi 10000footeo -2 2looRe, = — | De | a 1 ‘ II tt ee tt - en eeeeee|e | It aJ ee eee|ee tof !| fo) ‘a eeee|| _ TT tt ee || a ae | ee ee a ae pe _ oe om—‘trap al_|J.ee yas= oAANT Uiolei=2@aifshSD—— re eeree a py WE Se a eee favs O-IA-2 tol ava Raa af ‘jt ' \ 2 HH+ z, 1 R=-asf-|lloon0We=tlI-y\ fioon2p oa ~—Lo Jj'{ 1a ro to Op aneae Cl.=~tLe 00,600, 00. HK A, == ; ee —L(X)s_= a nn __ABdDI\= os ZeNol oe ~—— he D4, Hre aaayiaA f2~ ee a .—Aaa lena e0Vo . ei + ot.Tottae fretFOB7aoloo>|a ee poolsbeenLamia ~ub Lt H : _Au. ~(lotSt 0Ye(To40 00To-WO. . i fo ee Se rsgee A gltrwiiee zz eeKy,—_ — AeLUDA =*22, 02,-AZY launeg. _- a meery—.- VIA=o\,00VUVO.= wae a Ses Steet en (ns =|00,33,00,9947 cee Ras 80OGODOO waebas=es,09-110 Jeeeee SS S2veo Ao (QO. Yaa =oy eene _.)— Ketype- 32;2524S _- — Ce nn nnQuy 24-Bt,00VL =_33.9390a —{[Q® Zlwde=2ta\90,=10.. .ee weea heey ee on wee yh —Lae eeteeOWN =10,99,oaoh Summary to this Point. 61.IfiguredoutthatXyyX12XoXoreallyarethecorrectprojector objectsto use toget functions which belong tothe two M-reps . 2.Next, Itook the 27triple-momenta tensor forms and Icompletely classified them according totheir symmetry. These 27forms grouped like so: allmomenta different twomomenta _same three momenta same dval A 1 3 O° 4 s 3 3 | ? x, { 6 ( 3 % \ 6 \ & a) 3.Next, Iclassified the9forms involving Buy" _4.Next,Ireplaced3=-1-2andfiguredouthowmanytensorswereleftandinwhat le)catagories. Here arethose results: Weknow there are&rawtensors available, this means weexpect toget8linear combinations grouped bysymmetry. Itturns outthat there aretkrumxtax twoforms ineach ofthefour groups: Wy An AAA \aa au Qe 221 22% tay 2 2 \ -\ ° ° -2 “2° ziie)a taro t © © -\ \ ~\ o =@, Sy ° ° \ \ fo) ° ° Oo=(i\s Sn 0 HN ° \ \ \ ®=(as)s Ly : 1 4 ° \ \ 1 -2 a \ o =Qe, ne . a \\\\ °o\=ae), Ky- = ‘ L ° Loom \ ° ° V8 =ep, 7 . oh \ ° ° ° ° ° oONe ad. Choosing8BasicTensorsfortripleglue.(3-cmaate, ) 1.Once ithas been established that there are precisely two tensors ineach catagory, you can choose the pairs within each symmetry catagory inany way you want, providing you don't accidenthlly pick the same one orproviding one diesnt vanish. Iwill try tochoose these forms which are easiest towrite down. Sohere ismylatest choice: Ws =123 Gis =23D Qa=WaT aQa =2~22-233 ~49)4922 4312, (3a, £y= ABLABU- 2-213 a)tT ce re)Cea,=(3a3an)sJ3 | ge cor WW) IM22222-3383 a Ve Guy, =(A-aaa)JS 1a Comments: Notice that,concerning the"pairs" offunctions forthemixedrep,the pairs really dohave tobepairs inthis sense: they transform among themselves exactly dsprescribed bytheDmatrices. ThismeansthattheyhavetobeX~4Projections ofthesame object. Youcannot justtake anarbitrary "first row”function andexpect. ittobeapair tosome unrelated "second rowfunction". They have tobeanhonest pair. Also, inorder totransform correctly they have tohave relatively correct normalization! Ihavemadesurethat, intheprojection process, Ihavenotdropped anyfactors of2or3orsigns (red N's). Ofcourse this normalization question does not arise for the one dim reps AadS. + : : -~2- 2. General form of the vertex so far: A St . ShFaas =2.@ada +Feard, . AN AL . aPads +Pocaidss aL \ mt \ ~~ ~z|iGx),t(9,| m2.‘ay aoaAW-Aue e[Hoo,-Poa] Ithink you could argue that there are infact six arbitrary functions involved here. For example, you could say: S\ AY Co)P= %ACD Pe XeLow SL ALPe Xsaw $F XQAad nN \ aL \ ‘=Kyde f=xLae)ma ‘ . ne 1Fn Kedoled re xfae). . Initsmostgeneral form,thisamplitude involves sixdifferent andunrelated functions! Thaveshowapossible waytogenerate function-pairs usingX41andX1acting on arbitrary function. Couldalsohaveusedtheotherpair“ofprojectors, nomatter. o- 3. 3.Nowletsexaminethemixed-terms inLightofresultofnextsection:Ie,suppose fo)werestrict ourinterest tofunctions ofasingle permutative variable like £(1) instead of£(123). Then weatonce get the basis functions asshown innext section. Putting these functions inweget: Lod) =£03 Fd) =go (ardor) = Lid]($@-4@)) +(ath (£@ -F0) +(23) 4) ~F@) GandWaa) = LAD(g~-9@) +Le)(qe) ~90)) +TY 3)~3@) You can check thet each ofthese terms iscyclically symmetric. That must bethe 0case since each term should betotally antisymmetric. 4.General Case formixed terms: Ifyoujust multiply things outkeeping arbitrary functions £5andf%here iswhat youget: : : i w mixedterm#1=|{32"](45(123)+.213)~¥,(132) ~(s®))Aeyk mixedterm=LETH(FA) +Xcad)-£0139 ()\ xyoke wots a 5.Here then ismymost general form except fortheGyterms: Jipys Bey KeMefy]>Way =Leeet- itt) &fox beme ye He my+WR CR see] EBAes) MMe dy +Lee et]BAG Ay He Me MA thy,+eBeReRLRAG ae Mey aAL¢PselkFaz)~oanfwaok AyAeAS fo)*{Ueee|Rfae+cate Wha Roe |4Gt) Cs)4C28)©Cee)+(132) Pa=e=G2)~08)=(2d)4Ces)+(182) t foe e2) ~(3) ~(32) =(DK, Thus, there aresixindependent, unrelated scalar functions which Ihave called fyeeefge Notice that thesymmetrizers donotactonthefunction names like the1offy. The above form isthe most general possible form the triple vertex could have (Iam assuméng acolortensoroftheusualCayagestotheaboveisreallythemostgeneralantisymmetric tensor you can make using triple momentum terms. NextstepistoaddtheBuytermsandthenthinkaboutimplicationsofthe ro)Ward Ideneity. tems Aw aorta [ersani-2.28]- (oo-01~1aE 6-Wwealf Gray #. 2h TaN =UsqG2-t° 202) Ce nyest)= fa. . +Bal(2Mere0-20) +Wi&©2-t2-2) 2: 2L)%F.EEa ahhOurQh SedeLahseytet aanay )RAYA ode. O & 4 At<-- ; 5)TRAYS +ye... . 6 8 : OY) |-\ \ v3) —2 \4 GPs, 11A °047 @vy(e eo'24i) BYMe—-o--- tr —-BE *):.a2-“2ve\S/S Se -y-4 tO o -2 So.ele)<2oO8B\ari GW ery i(se) (S38)« Sey; Cs) - | an 2 © . anty =RCD TG\NS O BYE =e da 4BFGDa B70 Condouue uta*gous’: .pod (a La woos8, a 4\28(DAP CIVUS] (27a1Aa) aan 46)0S+Sea]B(0on ie ne = BY eee) =SUD X aeSeay (22421-2) +E LOK +8G4)"(2e-24{-2-4)) soa L(BDX =3s (WW-e-e-2) &oyae bytheotherprojectors atand¥,+Wouldyougetaatstoreitnewmeemet [asro s-3asc-a¥) (o©sthVARyoe*[3°34Sea)Becma(465 aenN1 AS ty =BO wey -=Seo? (ooarin) ory, - ! santea “OL(Vrvehe)SW ~~) i, +8&Koriesswart—IsCRY fo) Si.(pt 29bay 2 ari +33a) rere K aes :ae oo t.BSarPL’ egcc =SY (Go110-8) +egelee BONKS+ye os Ws com ae wsQe} =[wpastt— 2.3°8 Wysyt 2 PB LaeBOM [Elwes Itseems tomethat these guys combine just thewaythe[123] triple momenta forms did. Replace things inthe obvious way. Result will be: <a) 23\ 0CT)Pahaeener. o& (OB)XAHAege. Soherearetheresults: TheBuy“mixed” terms looklikethis: [e* Sn$ .GDA +ye]DQ cquend. Lt a a ty! .\ \ [S°U-2F Kw ote]. ¥-yf Iamstill not sure whether ornot tocount both ofthese possiblities separately, oraretheyessentiallythesame?Therereallyaretwodifferentwaystoproject fo)afirst-row function out ofanarbitrary function. Foranyfunction fyouuseinthefirstform,IcanuseX',fformy£"and the second form will duplicate it. Alternatively, ifyou give meanf', and use thesecond form, Icanuse£=X*,f'andduplicate yourresult fromthefirst form, Soifforf*isallowed tobearbitrary, then really one form will do. Ararows Fite ;[33Kak©eye| . Ye Comment: Recall that thematrix for 12isD(12) =dieg(1,-1). This tells you that if you multiply any "first row" object by(12), youget same thing back times +1,whereas ifyou take a"second row" object and multiply (onthe left) by(12), you get same thing back but with aminus sign. Inparticular, wehave: t i (ayX= +@xX) é c Ce)Xo =Ke Wecanusethis observation toverify that ourg..,tensor forms areinfact totally antisymetre. Asshownabove,theyareclearlycyclicelly symmetric, soallyou le)have todoisshow that thing negates under (12). Lets try itfor first form: mA Meat goXt«~oF TEars 4=Meany axt = Ky Sothe first term is12antisymmetric. You dont have toeven look atthe other terms because they will automatically be12antisymmetric. Here isaproof: AFGrd)«eaig=Leeasytcad)$a) & , apik=ay[e+Oed)+33]Ge)CryHAs), Yt -$02) ==>|Ge)Let(ra)+08]Ce)Fa] Butrecallthatanyclassisjustrearrangedbyconjugationwithanygroupelement.In re)this case theclass containing thecyclic andanticyclic permatotors hasitselements switched. QED. Soshowing thatthefirst function £(123) was12anti symissufficient together with thecyclic symmetry toshowthat theentire thing is12antisymmetric. Butof course this implies anti symonanyother pair, etc. Bg: >)PF rag=ONwd} f=GOPt= -Ieu, Similarly, itisclear that thesecond form isalso antisynmetric. Restate the two mixed forms: myBoe ot q3aryxyP+ayer¢ : W338 : T)SK 4%ayef .. O28 <2lay ro)Cty 4eqaaty = CyIA 2 weLemar eycce§=Q's)s Pat 3 _— Sohereisourmostgeneral tensor formforvertex withonlybuyterms: yp As PPray=LCreas ga} BJag +COMP ~ogee) Pa4,Ge) ay My? ‘ *;5Ces) Xx4,12)%ogee} ype ais!TOT SGAG=gah Ofcourse wecan replace the X's with slightly simpler permiters asfollows: 2 2 ~ KA=Yeah =Orneeoh . of ve) Pa e©sOd-09-32) =fy \ \ 5 Sf aaaPav @=th+s) ~C3e) =Pa Actually, I'd rather dothis inaway that yiels the forward cyclic and not the reverse cyclic, just toavoid later confucion. So: 2 I WAL=KCwuroaay he=Cuears)h est)— (sy-Cea) =Ra \ \ YA,=Ketaray fh.=press oy a eit)£023)~(1203)=Paw O%FadiWanteSodYoYornaemensneers: Aloruns . Guys =[+Weeaa4zeeear 4er43is] [241.22 20—. 2.Oa =[eee aiaaa $13. tear 4313] 22.Fr)99~2-0_oS =LEU 2a2agp BE-322-3 ||ow 22Jo. -——. Loy =BL meee tree488 ©\-\\ ag~\0) _Scans —.Blows ris+3=38|Beetay-er-bv a.bombs Vee=vey aseti nae=318Jfd SU Gras =Liar+vr+3234a422a+123)|00320099 ~——_Garas.Dai = es aesat233tsDpee oe ay sSa2er ate +3=tue=223=338) |od00-00 99__Tonal $-—___..... BY23 =1+:138-|B000000 90) agona. —BL -33+3+e=8]ifark22.202d -board. =Baer =eae aedte tes=153Jit. ae (ns =+a +a24we at 2e¥3a)Jforuwwie On =Caan ee Sa atee 48) OA332]—joe. Extau4Veta >332.=3sase=38) O-\ 422-10 —.S@mp sf BT=332tau=ae,433) How19010)fen sn BL=3xau+e=35)VGmtee2)SE =Pty Ue BW 28e351_7]Patsyoowev G2) s-s6]\23} —~-- | 0069020 --Wssdhhazs seeag} LL Hot14 40)——fod2batsse=2a_.ee HLan2arto —-Usos SBlwe-aedo oe o-\ \-L00 10 __G@)s Sized oe “30-100sts=10)NA sBTaa- sz Bho00 9) ——Us22st 222aaa) oasts=11-19)mySOUEL xeeesey Lr 22 wh= We=222) — 34lo op_o-\) .. ~t- ~ SP =Ista 4Stra aSays ieeetere 0es A=[AtO-W_as @- 4eat) fa aese— FSR = axtQen’ -Saas =Gent fae eetaoTE =2BLES Ges)x3Gent] _Lb(co 102). Ses BES? a = (eo wadfee =('-o >BeGass! ~aGry]fare ty BS).=Sas ate Fan=Lasts atyo ete 0 fa 420on. (e's, =BiesPta sey O°_-260%) __Ohere are36entitiesinthislist.Thereare9S,5Aand11pairs,too.______If wenowlet3-1-2,thereareonly8independent, tensorformsdnvevlving three___ ___nonentayandthereareonly6involving onemomentaandad,.,«Thus,eachoftheabove_ ____forms canberepresented bya@or6component, vectorwithorderdefinedhere: _triple mowenta: Ladd 21212, aan app cy ___IntheaboveListing, thevectorforeachformislistedattheright.= Nowtherecanbeatmost8independent formsinthetriplepgroup,and6in __todyBFOUP.Hereorewychoices,including theirvectors: ——@s)s =Wea olooto CAH ANB A 99WMO ~—GRYA=W2>22a23amVagree3ia|WWI08-2 Suawataat-2.218 otaeoX -—S00r= Tele aed| otoetoxFind) AAa222naga teeeeX —One Tw ez to00ooKT SP=SAAS G “10+1 _ Btaa SOWad?es)Vat A=f awa Open:Rhys?(ea3%ehoo122% . AG Es2G se Gt|aeee_5ahs2arote e fstjoop—@ Mes BowNett] ooto oy s—S—S no ee re -~3- . 3.Wecanchecktheindependence ofthe8vectors inthefirstset,andthe6vectors 0 “inthesecond set, merely bymakingsurethatdetJo.Te,bringeachdettotriangular "formandmakesireno‘seros onthediagonal. __ Birstsete wee : a - eo ee <=en Lobee ees Oa Xx9 ee See 6 eT soBO aLeJe Se SS ‘ we nr” ara 36#0.Se e8. cee i ae a *£ =, Qo We|MefoA-¥Yo eeeep OLYakk 7 ae—-+}be ANS > Thisisddifficult dettoget;should bemoresystematic andbhowwhatneodS-not ee° “beconsidered. Bettértoletreduce dothisstuff! 6.4;Actually thereisnoneedtodoallthis;youneedonlychecklinearindependence. ———Within eachrowofeachrep._Asynmetric function canneverbewritten asasum.——.. offunctionsofothersymmetries. Thisiseasytocheckandyouconclude that. —..theabovetensors areindeedindependent. ee - |____5. Whattodowiththemixedreppairs. Ifyouhavetwopaireoucanmeketaobject_ likeso(sedtheClebsh's)? —Geo. — afes slon ee _ fi hs=hatn _Sowetaleeachofourtensorformpaireandcombinewithanarbitrary function pair_ togetfull list ofacceptible triple glue vertex tensor forms with functions: _ ae nies Coe Cad,ha.weeeeee Pl AaSoe ee . 0 (asfa. CaMatspes weeeeee Gad LK OY Gdn4s \OSA -ONG .Gay) =O32) ee . wee |Gy Fer Ge yk fee Thesethenaroyourtenbasicforms | __.3sTtmightbenicertowriteeachformassonething pluscyclic.This1sdoneasfollows:_|on forexamples ©TRYST iSSaCC600 2B®2-9)!Fientbe)ee ee os2TS SDWON]SSG OX . pe ee SEBO KEseyeQT wee. As =02-1 ze2g=ASEH v a -Os. 5.02 BR_00 0FY Mo ne -Ws =OfUt jo SL Y . 2ee-—Wrs)s. =99~€6_00 99==28) Se-38D speonASLAstaie=Se ae ..—adshe 0-\oe=i-02SAGawe =Ouse ontmintthteSk -ORAS22eo eA 2a=22stooeteA ——— Gua=2e)_to) 3:3ea=ABADooOa)afpsy oovos A oytieesOily) —~OO) =.ow tte es ey OME GAN UAHA" ss -Y _——— ())_S 90.00 oopon KO—— Gaps Ger-229ae)8 Lso-Go Lov ee a Sey tsSB —--G3QLFom\cLaeto-—. du)=3%e222 23 4-380 YO. — Summary Combining theTriple Momenta formswiththeBuySingle momenta forms: Oo1.First,hereareallthe"mixedsymmetry"forms;ie,theformswhichareofcourse totally antisymmetric, but which got that way vie the mxed-symmetry reps interacting with each other: vos) Ot 2TRYST LA+cgaic LN ea ros Jr. BERK +yee seyyy! a 73 (SSK aqae Ad,2 . ye(OS) &, aqebi Notice that thefirst three terms have a12symmetric bracket, whereas thelast has a12anti sym bracket. That iswhy its Xprojector isdifferent from the others. 2.Now the symmetrics and antisymmetrics: ce} yee gesfs, \2tat- st] x,& Les as 46 LOREAL vege Xe cs 2yo{aQ-2)+agtie|XsYy Lea -&WIRY Vax a3 14 [U8Vs B12]XAd, A? . po.\S%ara eeetic] Yate 3.Thus, there are10independent tensors forms!! Ifyoulike, youcanreplace the mixed projectors asfollows: ‘ \ a Yio We e-cytay~asy =YX:(x32) a Yr feseeGy) -() =ren(32) k heck list of tensor forms. aS1,Piret,letsreviewhowtheyarrived. Priortogoingonmasmmomentum conservation we had: MA+78 +8paire =19tensors offormpyp5p, 1A+25 ¥3pairs =_6tensors ofdj5p,form 25 total Next weset 3=-1-2 and then counted only the independent tensor forms. Wegot: 2h+28+2pairs =6tensors offormpipsp,1A+1S+2pairs =_4tensors offormd,,p otc atk The main problem here was counting and constructing the tensors. Once you know there are ten ofthem, ifyou have alist you can check them for antisymmetry. They have already been checked for independence. This check isjust tomake sure notyposs. 2.Solets inspect the list. Interms that are +cyclic, Ihave kown that you need only show 12anti sym inthe exposted term. For the first four terms inthe list, this inspection iseasyonceyouknwotheeffect of(12)operation onthe5objects: second row objects change sign, first row objects don't. That completes the verification’ BHthatterms1through4aresymmetric! Justlookatthem.Term 5isobvious from inspection. Term 6isalso obvious (cyclic commutes withX,andX,).Term 7obvious. Term 8:the [...] isclearly totally symmetric: itis12symandcylcic sym. Term 9:again, bracket iscyclically sym and 12sym Term 10: cyclic sojust check the 12, locks find. 3.Ihave now explicitly checked that my10tensor forms are completely antisymmetric. Ibelieve this isthe complete set ofindependent forms. Ward 10 July 1,1979 Contents. ce)Comment: Thisbindercontains anattempttostartwiththegeneraltensorformofthe triple glue vertex andseehowtheWard identity restricts thet form. However, Ithink this isell obselete because Ireelly assumed too many terms; there ere infact fewer then tenindependent terms, right now(Jhly 1)Iamnotsure howmany, andhowthey should bewritten. SoIhave togoback and see, 1.This section leads uptothe Table atthe end ofthe section. This teble has 5 columns which arise bybalenceing the tensor terms inthe Ward identity. seeac -sazazaes Thee rows correspond tothe supposed 10terms inthe vertex. 2.Itseemed reasonable totake eech colum and project itinto its symmetry components; thus each colum might actually by4identities rather than one. Inorder toproject each colum-identity into its synmetry frections, Ihad todoacertain Clebsching ofproducts ofobgJets. Inmyfirst version ofthis calculation Ididthet wrong. This section contains residual pages from that wrong calculation. . 3.Here Icorrect this Clebsh error and redo the compuetion .The final result here Se)isatablewhichliststhe10constraints imposedbytheWardidentity. Ihavenever done enything with this table because Iamebout tosimplify my"most generel form" of the vertex. Constraints ontheGenerel Vertex Imposed bytheWardIdentity. | D 1.Iamnotyetreadytodothisingenerality. RightnowIhavenotyetfigured out the n-dependent terms inthe vertex. Later Iwill have toadd them. For now, Iam going toignore n-dpendent terms inthevertes. Thus, non-dependent (gauge variant, ifyou like) terms will begenerated bytheLHS ofthe Ward. Therefore, ontheRHS Iwill setthe function ¢=0(which isthe simple ansatz), andthen there will also beno nterms onthe RHS ofthe Ward. Iwill then belence the Lorentz tensor forms and the symmetry "fractions" onboth sides. Ifall goes well, Iwill obtain the foldowing object: the Most General Form ofthe Vertex that contains non-terms andwhich isconsistent with the simple ensatz. Ipresume that Iwill get something Like Marshall's result plus aset ofWard Null terms. However, Idon't even want to guess orlook atananswer: Iwant the machinery togrind out the result. Ifthis works out all right, Iwill reinsert the function c,figure out all the ntensors, andre-balance thewhole thing. This isalonger problem, Iprefer todothe shorter onefirst. Ofcourse information from the shorter problem will be helpful inthe full version ofthe problem. 2.ThereareofcoursethreeseparateWardidentitiesonthevertex,oneforeach co)monentum.° Butifyouarrnage tohave thevertex satisfy oneofthese Wards, itwill satisfy theothers since itiscyclically symmetric. Ie,theWard inmost general form nowisC-symmetrie for any choice ofthe 10arbitrary functions. Thefact that oneoftheWard's puts restrictions onthose 10functions does notdestroy cyclicness. 3.Inpassing, onewonders ifthere areanyother constraints onthetriple-glue vertex. Bose symmetry and Ward identity inexial gauge have been considered. What isthe effect ofcharge conjugation andparity? Also: what about gauge invariance? Ie,youcould dosome kind ofWigner Eckart analysis with gluons being octet members and soon.I predsume that all this iscontained inthe Ward identity, but should check that! Onthis Wigner thing: howmany times does theadjoint appear inadjoint xadjoint? Iets continue this line omeother time. Ithink itisimportant toknow howmany amplitudes you have, and soon. ‘ akeo ot 4.SohereistheWardidentityanditsRHS: \Boa Va Vom@30=ToS 5are [ors lr]oo+a@Q_ a) movie=RUSWah=~lanst {eo) +Lars —~LE\b@) . —Q- a wti" re) RWW =BVT b(d2T 43[cearecey ~CiLOS] Le Now lets attack the ten general terms, one atatime: Term1:Inorder todot2?intothisthing, youreally havetowrite outthe cyclic terms completely. Suppose Pisaprojector ofsome sort acting onafunction f.Ifyou see Pf appearing inthe first term, then you have toremember that cylcic implies action onthis vector (Pf), itbecomes (123)P finthenext term Thus, here iswhat wewant: 3 St37h +34 \ 3h 2 ‘ey|LVS 2oa f+[FazWoW Mygeneral procedure willbethis: dotinthe3°andleave dotproducts inrawform. Butforuncontracted momenta always replace 3--1-2 andthen group things ina standard order interms ofthe H€X 5possible tensor forms. That standard order will be: fo) hy? ph? 2h? atgt a? Forexample,/nereiswnaboveneon8ratih):> v y . rane L-GyGat F-@-dos) -aumt [Gotan cg{pt>\X,4\ Thenotation here isthis: each curly bracket isacoefficient ofoneofthestandard order tensors shown: each bracket really contains thefactor wayover ontheright, noneed towrite it4times. Thered checks mean that this version ofthe calculation agrees with myprevious, independent calculation. Sofar90good. Term 2: 3 es za BAe \3?L3SSTO+ [OVSae[2eTuae3sKh - ’ z So ev>{Ga+aday Tat[6NZGD+G2)OEFPSZS XLL fo)Again,redchecksindicate agreement withlastcalculation. Sofarsogoodagain... oo. -3- Term 3: 3923% \ fe)3iLSeas" ylGe)+82(ragmY =avl Qe) Obviously thenumbers 1thru5refer 42 =Urv)-as) \ totensor forms instanrd order. Easier 4200 oX24, tolistthemoffthisnay.Agrees. 74 =(30) 76. (33) Term At 3 Ks BwSES Ses Gos OTengo YK SAV ~UB)+2-132) Yr =(2d) +032) 2 va-rUwy23}YL fo)YX AUCW) *&CS) 4S. LQay-@3)} Term 5: va \SPATE AAAs YA = 71. G2) v% 0 2% G)=GA) Yhe “4% -@a)y vs 0 Term 6 SLVte-1ths2s-sek4BTS OK rol“Ww~29 72. Ga)~(32)¥3. 0 Xt AGED) vs. 2° a -h- Only four more terms togo. Sofar noerrors informer calculationt.... Term7: oO aSMt,3Casweet oBLshary 4S°E-V4 SG 33eh 4 “\ +h “uO 4,Eco) sh,Yq AN . 5.Gy-(22) me“ M 72 (3.3) . 4Term8: 3B0Yahy3N23 “% @VNTR S “5.0 Term 9: ag 23. JaaBLS CsSTAR xy =a4 ~Q3) va0 le) 3,-W3)-@3) =+@3) Yay y—03) “sO Aha!Anerrorintheprevious calculation. Ifoundtheerror,thisisthecorrect | result forthisterm. Ihavedonethisonethreetimestomakesure.Notmuchtodo,really. Term 10: — a. a at SisSeSladThaxpho at vez -2 73.6 Kabe - wy aN vs. (33) . NowthatIhavedoneeachtermtwice,I'llthrowouttheoriginalroughcalculations fo)and retain only this. Original summary sheet follows: o as & ¢ S ve a ae tat ae ar ait Sohn. 1,[-aaeySeopa pay|-Gomy|6°|Xf 2a WGay) Gay Gy Gassaey or| i op | «|= +2-(ey|-23)4(927|-20e)eecisey]~2-CeH(ieey] Toren) |xFY —S..Gee? _|(eae -@yr 4: - i {4 _ | _ one teyieay 6|se@ey|ob TyR- —. j — —} i—| |—— -4!—.ileowtbits-3ay]Yshy ale Gy|} b° 67|yates 4ayTt6-<42)-=~|Oesce | eeee = Oo) G8) whe a GG ee LT | py , a a, SL a tt fat ate TS A GORY Oth-e088GI ene|eo| Tew ‘oakésoutte aDet"GHRA DRAGOE OT =e—Ys) |e eas Tees | ee oeSa|SsealSake 8fe a Summary ofInformation Deleted atthis Point: fe)1.IdeathateachWardequation isavectorequation inthe(s,1,2,a) sense. 2.Idea that R[fgJ} =Rf-Rg where R=group element, but not true for Rreplaced byX,some element ofthe group algebra. 3.Explanation ofwhy only a,2 components appear inmost Ward equations. (Obvious) 4.Expansion ofWard table terms into symmetry fractions. Did this wrong originally because had Clebsching wrong, later repaired. 5.Derivation offacts stated on¥Main Data Sheet" see below. 6,Ingeneral, this section contained thework needed tosetup the 10-equation Ward Chart which Iheve retained below. Iamquite sure this chart iscorrect, but Inever used itfor anything. Later ofcourse Isolved the entire problem very quickly using the cyclic operator method ofsolution. That ismymain reason for dumping stuff out ofthis section. MainDataSheetforAnalyzing Component, TermsintoSymmetry Fractions, 4QY 1.General clbsch Forma: Sas RESEND| Fre = ! | SLBGT +URGSEG)+2(ROrHd]|n aw eeALR GsGsshGhGe +HKG) +4(EGEG)|| |ST[BGAEGHEG HG.TEREARS) +4(RGAEGY] x ne atisTRGARG 0.+ERGAG) +A(RG-RBS] 2.Diagonal dot products: . nnne fae PON=aSeata et SG |A A “— oeQMA=eStat-eto .e) GY= @§-2et—— &=Re 3.Off-diagonal dot products: Syahasa yn, 4 BE @d= s+dt44.2 Lekha BeEUD= dS4dt-a2 w FEGD AS +2A,F a= Ay 4.Relations implied bymomentum conservation: QN+@D4@3)= 0 — $ds+&+dszo >Tee-2ds | q -WFe taze Sad . 2 Oo- Gta so =A, 3ree et “Gays-2ds$4dt44,2 ' coi (QV =-2a8 +4%-3,2 - GNF -~2as$ ~247 —CreoColer;sumdov\MainQa" dkDs=FLOP+oy] e]Q)=ELA+@-2@q Oi= £[O)-@] @=Ws$*OT-2 : @= Os$-20,7 |1" 8-208 —|OM=O=OO=ROL TT |GerWO= KO)=BOL|G,=%OQ=WO=-FO | 9° | @= KO. =KG) =RO), a,=F@c=KO=-KO, | @=KOr= Keys 0 lo; .CERO) @)ofow««6 aee ye |:: Ay S$) 44 Sy os ogop lop og A,Se ofBR egRf yg A.RA BedRaEda fk sES Alcsh tNSp th AKG BL ek deta, “RA -2Hd og \ sin Q,aa tha ~4 6g tg 1‘ : t GQ|tBde ~ded aee ¢2G fl : . : i { ;a a i \|Hy,‘ CQ. the <4 (spa|~a) 4Leog og EE#go#|g|sha,[ted keafd ee es '5\toa-Lt — : i i i 1 1re Apig ji¢go i a a ce a ee era a a a er ee |¢¢Bide gpg igog4 oa ¢ | |; gpfgtd fg |2an be 6gg lt ¥ y4go |g|gp|2|gpo~2a24,ene.| 1 : ; : nr ns | i|! ||ti an i Ward 6 Status Report. ’ i fo)1,Ididindeed find 14“linearly independent tensor forms". There were 3synmetrics, { 3antisymmetric, and4mixed "pairs offorms. Ittodk alotofwork formetobe ' convinced that this was the complete set ofindppendent tensor forms. 2.Having found these "tensor forms", Ithen considered how toconstruct "vertex terms". Idid this inanobvious way and Ithought Ihad thereby constructed 10independent “vertex terms". BUT, Ithen discovered that some ofthese terms contained others sothat some cotld be removed from the set. 3.Inow wish toconsider the "most general form" problem interms ofthese "vertex terms". This isreally kaanew problem inreduction. Ithink Iknow that Ihave generated “vertex terms" from mycomplete set of“tensor forms" inellpossible ways. Ithink, therefore, that nothing has been omitted. Itiseproblem now ofreducin this set to its minimum somehow. Material Deleted From this Section. fe)1.Thiswas‘whereIreducedmylistof10tensorformstoalistof6forms using the "extension idea". This led tothe Most General Form ofthe vertex given below assix terms (non-n ofcourse), Ithen took that xixk form ofthe vertex and constructed anew Ward table, see below. Atthis point Itried to"solve" this Ward table but Ifailed because Iwas not yet using the cyl,ci-operator method. Ihave delted pages ofhow this went since Inolonger care about it. 8 8 MostGeneral FormofTriple GlueVertex Prior toWardIdentity:rRestrictions.= fe) 1.Howmany totally antisymmetric, linearly independent tensor forms canbemade from py,Pyandg,, ?Theanswer is10. Here they are: vagy pAaLIAL (ery A Veewae wns P+ofzyfosd ba tua psLAS V~srr] ces) ane L .L113 Fay +que| vad pr %{332 FGd +ope] Ss Meas 6BLQIL- ett). eyeric|X(23) Wa,a Ss 6ea(-2Y«eae[ft(za) 2s A FAGBxyacl $Gu) R as. &(S"Qwy Bay 4ejaic4 “sy wis A 3LSsary xegchic| However, Inowseethatsomeofthese“vertex forms" arecontained inotherstfAlthough nyoriginal tensor forms were truly linearly independentg, when Icombine them with functions tomake overall antisymmetric, some vertex forms become contained inothers. Inparticular, itappearsthattheformsarisingthroughthe“mixedsymmetry" “route are somehow more general theh the other forms. 1.Consider form 1.above. Asaspecial case ofthis form wemight consider fytobeatotally antisymmetric function, instead ofjust 12Anti-sym. Then youget: 53) pat att gTA be[NS4FaleP3T)FOy) fe)A . . =s+ iaasan) fav) A=Gad. Pq) a a But this symmetric tensor isalinear combinations offorms 8and 9. Thus, weshould rs)ruleouteither8or9asbeinganindependent form.2.Lets dothe same thing here. Wethen get: A * (BBs UVFras) -cays{U0s)Qs =sds Here weget essentially the same form as before, sowedon't have toknogk outtwo. Only 8or9mst go. Lets kmock out 9since itismore complicated. 10. This case iscontained in3.bysame idea, sokill 10. 7.This case iscontained in4sokill 7. Soour list has now shrink tothe following: . Vou dy QAesLt dyficas) 1b3a3Ss y fo)sLava. 42)0qa) 6LOAVD~ BEL) eenUNE 1a) ~Ur egcle 6G vas pA is(a3 aary|weogele 22 QA2LS3s Pay) +qac . w a9ssLSaranonlAeoch easaLehaRag] «see. Lets see iffurther reductions can't befound. Can Imatte 1and 2coincide?? . : -~4- 3.8 LATESAG) 4ESAT Peay 4ListSanh =?= 0- Lassa) +LUv@al +reekey) 2 Reduce both sides toPyPo tensor forms andseeifthefunctions canberelated. uss Moy v3.4 4ediaxat ayaa =Maya erd+ far rr esFaery =Fr)Lanearry =F@DLies 22)&~FN) Lies+221] . 2 3 RES==CDF (HL)ge) +WegCu)+TW4) aul aes)~3(ry4ereaay— aw) aawe)(ure\UA+WHR*zr+2ri| : E3sciateDoansquolaeonasds gy@) =a) LQ =gle) a)a)(vy) Fd=g(r)F@N= 4ar) Stop, Have toconsider howyoumight make afunction like g(23), forexample. Ascalarfunction cenonlydependonp,sPpsPy-Py+ ALLotherscalars like22.3 can bereduced tothese. So; ' th(12) =F°(p,2-p.") mostgeneral form,oddparity function ofoneargument. Aspeciel case ofthis generel form would be O(py779") +1(D375) +O°(P57-7,”) which istotally antisymmetric. Comments about the Vertex Form: ie)1.Inowseethatinfactmy10vertextermsdonotoverlapasIthoughtearlier. Toseehowthis works, lets classify thespace offunctions into 6orfour pieces like so: _. : x tof3°~N Ss - Qe:y 1 4 Sse an) Yceramungn.ct “ee o)xmy .* \ This issupposed toillustrate that the totally symmetric functions together with the row-1 functionss comprise the set of12symmetric functions, whereas the lower half ofthe diagram represents the 12- antisymmetric functions. “Now, mytensor vertes term #4represents combinations ofthat tensor form only with row-1 functions, whereas myvertex term #7 represents the same fe)tensorformcombinedwithtotallysymmetricfunctions. Youcouldcombinethese two “vertex terms" into one byjust saying you are going tocombine with an arbitrary (12)-symmetric function. But maybe there issome advantage tokeeping these catagories separated. ‘The set offunctions which are (12)-symmetric isnot really anirreducible set relative to$3;itistheunion oftwosuch sets. 2.Similarly, consider tensor form 1,IfIwere toextend the row-2 function there toinclude the class ofother 12-antisymmetric functions, that would bring inthe totally antisymmetric functions ,then term 1overlaps with &and 9. 3.Sohere isthe major result here. {12-symmetric functions] =[row-1 functions] [totally Sfunctions] : [12-antisym functions] =[row-2 functions] [totally Afunctions] iwee epee So ee eb ie] Extensions andRemovel ofsome Terms. 1.Ihave chown that the extension ofterm 4,equals term 7,soifyou use the extended term 4then you can eliminate term 7altogether. Similarly, the extension ofterm 3equals term 10, soyou can drop 10 and use an extended term 3. 2.The extension ofterm 1gives alinear combination of8and 9.Thus, ifyou use anextended term 1,you can drop &or9,Specifically, the extension ofterm 1gives theform (132), =combination of(123), and(231), . 3.Interestingly, theextension ofterm2alsogives (132),. Thus, ifyouuse anextended term 1andanéxtended term 2,youhave ineffect “wovered" (132), twice. Perhaps itwould bebetter totake adifferent pair atthe start toremove this double overlap? JERE fa)4.OK,Ihaverelisted thechoiceswehaveforvarioustensorsinthepppclasses. AndIhave looked attheir "reductions" when you "extend". (see attached page). Bg,ifyoustartwith(112); asyourchosen "pair", thenwhenyouextned this thing youpick up(112), .Thething youpick upisalways intheclass you started with. Sohere isanice proposed list: =o 28,781cr et \Cuayy a’9) Wdudz 2wdepadich S%,FE0y oit aundspdact A:Qe (GaCups Zirondapunditpad. ane . Ay(Cea)sy 38 as / f 7 -2- _ 0,ifweinclude extensions, ourlistcontains onlysixentries. Heretheyare:O—————— tei)convenfiendAol, |oOaoO e2)Qua~221)S02) _vgckic VY. SFR: . ;}3)(@aryair)as yryr i) ZW BW ea sy=342) aAeyweeghicSe ok jreCane |68o-asa egele | le) 4 |av|af! a/ FQ. A &4qq D oom . _) RGD Sr)|-GOA. |-@94e9 ||) |+Gs)4)=eo¥e) 63.36)=B04a) . 2). 3.9kww)|-GaQe)[+39Ke)~Gx)aee ee | -B DAB)|eG {@_1-G ADA) 1+BML) | we ~~) (arse) [~C ~(ea Xs =Va) S08) [~0s)S3'aee eee — ©) eaSes) 0s) Sie)(=O)Bis) |nn a ODM) et 3) |=e (> esA esCD)SE :i@a ee ~$0) - a 9 |Se) |=es) |28s) 2es) 1G)-@diyeee +24G)_ | AAG! +A SO) Lo _ RUSsoo.=2)[@are\-nyuy] -akaerayaneatA Decae BIodA |” : ee ee Ge a a een :ae ee ie es eee a a 15x 18 Aplan for solving the Ward Identity fo) 1,Theproblemwasthis:howdoyouconveytoyourequation systemthatf(12)issomehow related tof(23) andsoon? Inowseethat thewaytodothis istotake each equation and state its cyclic and reverse cyclic duplicate equations. In effect this says you are forcing the other two Ward Identities, but that isjust eninterpretation, Infactyoucanapply anyRofS,toanyequation togetanew q equation, sothe question isnow: how many different equations are there? 2. Here are some facts we do know: (12)Be-B (12)C=-c¢ (12)B=-8 Thus, read as" application ofpermutation (12) toequation "BMyields -B.Inother words, does not yield anew equation. Same for Cand E. Wealso know: (12) A=-D 3.Conjecture: Ibet that the two cyclics oneach ofthe above equations exhausts the entire set ofequations, Lets try itand seer | (123)B =BYdef. lo} (132)B=BDdef.andsoonfortheohters. @b&b=8 adb=-B a“ .CaB=HSVNB=-Gere=— 8|soWCB enmgtalewheaebie @G=WHAOB=-(B=-C Senne donSEet et > (NBEB a (db=8 @A=A @v=d GA =-0 C\D =-A . + 1 _ $F GAA=rast =08)0=-8 (aB:-A LsAKAD SFA= GAMWNA=—13)VD~-V @p+-A =Yulach G@uayA= (woe csdA~ A QAld=T _ Thus, thecomplete setofequations includes ¥3+3+343+3 =15equations! Imean there ©)_38nototherequation youcangetbydoingapermutation. Soletsmakealist. -~2- 4sSuggested notation for the dot products: fe) (1.2)=4x (1.1)=2 (2.3) =y (2.2) =-y-x (3.1) =2 (3.3) =-z-y : Note that x,y,z are independent since there are three independent dot products. Note also that these things "cyclic" inthe obvious way. 5.Well, Ihave made thetable andthere are 15constraints on14functions. But the function sets are sostrongly coupled Idon't know what todo! Inprinciple this sytem could besolved, but practically itistoomuch ofamess. 6.What about this idea offinding asolution tothe Ward identity and then adding null terms tothat solution. How doyou know there isnot some other non-trivial sol'n? Fact: Any two solutions ofthe Ward mst differ byaWard null term. Ifyou have asolution andifyouhave exhausted allpossible Ward null terms, then youhave exhausted all solutions! fe)7.Ithinkitwouldhelptohaveamoreintelligent setofstartingtensorforms! '. | . i : Lona’ [aa |war’ “ALge’ |“toa | 6Aly_(23).mconrsoS=|_#— -|--A@ Gn eR nn ns * a-AGLI @aty =Pef -fm. ;1 1 Se |@3)G@-00f aT sa)levayfSee Le|tyre ; SG) 16)-G8) 1 GDffteey | Px_..AWRay eD)Lara} coy[sep —-. Adrd_,=@3 =G" 27) = 7 ~*~ a a eens | a—Saf+@ay_|+Ga)|eWO|-¥ee ~A@L | a | es - ee oeSO a | 2 aOso| oa Is — _—S(a)) +24Nf Le x = a~..”=icn |_=-..WOfat :af edOs ee _— Wo a — {=) ee ___9):P| = : esa [oso [so ee se Te —oo . _ | eeseee . Cramers Solution toaCyclic System okB02)baoben.fesomeobenFQ23). OgePanIYteckedviueeyehe, wba avaot? =F GrowSheayatiin lereames ie&BEEytoatfgt{=fp Poot oat UL cas F Oo co lefR|anae cae aa aronot DngatSAbycypher. Ofcoumschataorice aAdorm dak=0, fo} Soe atesercenseeny oeeeeeneceeepeeTAROA) ®;8.8]cs.AM:43%| DoseGetors Aw) 2x|~bagey S08 4bee i ix1an KE Se) D2: Ayeri“4 Pog nary SQ) Zay boy foe fe ety Ke AW) 4De eG ' Tey eR SQe) y|2PK Fo neayfboeta Ace : ~\ portob sk : --‘ i’ A,@) a i~\:“Vo: sy Se) cn re : '\~lo, 1eee nnn Seeeenee nee er) : XC) ee Pata|lLbet Oc ;i|iai| | H i H {' i : : ‘ i | \ | Sopp nee ~ + ~ _ +44joe |4| awe+44%|2*:|:a =: \ foo anaes IEeo be_.wee ¥~4e%|-eO |yey‘ ye yee [ate x | ie idyea heed;-z ~*:i \eo Ea a oo ~loo -ey : jo ~k-2 |- ~yo ~ Lope : F Po te. hy ee :: |+4~U| ::4-*: a ! : \ -xe | “%ee ° L«t ! ytxSma : ‘ : {ot [Bey ine =O} =0,;=0 550 '>501=0 =0 ,to, i ! | j ‘ : ' i ; ' \ 5x6 Operator Material Deleted from this Section. Oc stneaee 1.Back insection called."Ward-6" Imade aWard-Table using thelist of . ‘6terms forthevertex. Here’ isoneofmyattémpts tosolve that Table. Thiswasthefirst, timeIthought abut using operator equations, ‘but‘myimplementation wasnogood. Later ILearned howtodoit. SoIamdumpting this entire section except the page showing howtoinvert operators involving (e)end(12)). * ie) 8 : i Recenviderthe-B-equettor co) Thinkofthisas£(S,45,*, Sp»8,*) =0.Afunction offourvariables. However, wehavesomeother implicit ‘conditions, nameyf S,*=(123)8,.° Yorany "form "assumed byS;,thistells youwhat$,*is.Thus, it saysF(S,,5,*)=0 ! This isasmich a"condition" asanyother condition/. Sowemight insert both these auxiliary conditions intoouroriginal equation toget £(S;1(123)8, .8,,(123)s,) =0. Solywaning,cqanadsapa / ae | —_—_ GroupAlbegraProblem:doelementsofagroupalgebrahaveinverses? Le]1.The answer is: itdepends what elements you are looking at; not all have inverses. Here isthe poop: A=ZAacmR . R : . Bik B=ZtOS ach tok BASE | SoYork bE)sudGack 57,a(R)LS)SR=57aSHLOKR RS &S =AZoaGare@ yr~~2|atacoea@|e= € 5 & Qe wewok: _ ZLa(seyeS’)=Sez ; g =\ . fe)a=aR) b(R)—Ssge_ a . .” . a a. a(R)af)a(@) --. aRh) ‘ce $;©FARR) aCGeh) ACRR) ~~~afRee)|WIM’)]=fS (fe)allem) =- : | aR)alee)as)==.af) Regunsa: a(R’)‘ay—alf})~~ £0alt) ala’) — Soinorder forelement Aofthegroup albegra tohave aninverse, this determinant must notvanish. Notice that thearguments ineach rowarearearrangement ofthe first row, sothis islike aregular reprepresentation. This says that theh veagrnangements ofthevector a(R) mustbelinearly independent! Obviously this worksfora(R)=dg,g+thisjustverifiesthattheelementsofthegrouphave fe) inverses. Sogiven a“vector "A(R), howdoyouknow ititshrearrangements are linearly independent! ? Think ofthecoefficients a(R) asavector inh-dimensional space. The conditions thatdet=0isoneconditiononh-mimbers(ie,onthea(R)).Only°these hnumbers appear inthedetbecause the rows are just rearrangements. Therefore, the "surface" upon which det-0 isone dimensional down-from thewhile space (codimension 1?).Thus, weexpect that det=0 onsome h-1 dimensional manifold within the h-dimensional space. Forexample, ifh=2 weknow that det=0 obly ifa(E)=a(R). Ie,onaline at45°through theorigin intheplane. Every “vector” inthatplane is"invertible" except vectors onthe line. So"most" vectors are invertible. ‘Jfh=3, again there will beaphafiec onwhich vectors arenon-invertible . There will besome two-dimensional surface onwhich vectors vanish §notaplene because thedet=0 condition isnon-linear). Infact, ifthevector is(x,y,z), you know that the3rearrangements arejust thecyclics soyougetthis. surface: Pry 4232-0 Some kind ofsurface in3-space. Obviously ifyouchoose (x,y,z) =(1,1,1) yougetnon-invertibility. This obviously extends tothe general case. Bytheway, note that forn=4,5.. andsoon’thearrangements arenotjustthecyclicsbecauseyoucanarrangethingsmovingeverything inotherways. cv) Eg,with n=4 youcanreaarange like this (abcd) andalso like this (1b)(cd). So you don't know without naming the group which rearrangements occur inthe dt: So theinvertibility ofa4-dim vector depends onwhich 4~dim group youareusing. ‘ Conclusion: Forarbitrary vectors inagroup algebra, theprobability is overwhelming that thevector isinvertible, because thechances arealways great that avector does notlie onasurface within aspace. Ofcourse projection operators correpond togroup algebra vectors which arenotinvertible, asdooperators where all coefficients are the same. AGroup Algebra Problem in$2. 0 Motivation: Ithinkthishassomething todowithsolving myconstraint equations for the Ward Problem. 1.Consider this equation: ( a(12) [e] +b(12) [12] )F(12) =G(12) Here a,b,F,@ areallfunctions oftwoarguments, ex,a(x,,X,),.and [...] denotes apermutation group element. The problem here is,given a,b,G, how doyou-solve for F(12) ?! Ineffect, then, wewant toinvert the Bracket ( ... _)onthe left. What isthis object? Itisanelement ofthe group algebra but with coefficients which are functions of12, not just constants. Eariler weshowed that elements ofgroup algebras (with constant doefficitns) are ingeneral invertible. Here things are alittle different because the coefficients here are functions which feel the action ofgroup elements. However, Ifeel that the general result isthe same: generally there will exist aninverse. For this particular problem, wecan explicitly construct aninverse like so: (ALY +REx) CayCelebayLD)=fe A(060)fed+boustd) +B(9Gty4ctel)=cal. ~»Aaw)y+®baiy= 4 a(a)KOYVYVA) 2(1 Abe) +R&(a) =0 bay)a) A® o 3(A)a(2)POD Aaay) aAS dk\-bGe} aftr, j\o A\way) .. aA=aai/a enn \se\@) B=-w(e)/a =[waryay|=oGaacar)= bev)bea) = AQG2)\ =aiy/ace) Soherebelow isthegeneral resilt showing the&Git)=~bowfare) explicitconstruction oftheinverse.Ofcourse=~ ifthedethappenstovanish,youaresunk. le) ALY=alte)a(er\—Lolre) b(21) = (acer)Te]~bO2)Ld(atateLab(2)te)=At)Ce]. Se,solukindoquashospratdoe ~ 9 -\ F@)=262)[o@)t1-bary cyG(r), . 3.Suppose the‘determiriant ‘does vanich. Then what doyou have? You cannot invertg. Consider: . . . . 1. (wTe\+bay,ba)=(aera+bx)ws)(atTelsbay(a) =ale)LaosTel+bo)wa).veal,aCe)Tre)4b(20)tel) =face),+WOb@)] Tea+[acer) 4bey4e\Ley . =a(irya(u) a=(aGesate)) (soot9+(2)wy) <Taps2.nal aly) Le+Ge)Ue] hoa \|eeneee Ae+ ialit)+a(z') ree t \ So, ifdet=0, then our operator isprojectionlike .Soconsider PF-G where Pis aprojectionlike operator. Either Gisorisnot inthe range ofP.Ifitisnot, then there isnoFthat makes PF=G .Soassume that Gisinrange. Then wecan"solve" the equation for Flike so: P?.aP forprojection like operator. PP=G withGin range implies F=G/a+Fiz where’ Fy, isanarbitrary function lying inthenullspace ofP. FkI® Material Deleted From this SBction. CO)1.once1gottheformgivenbelowwithsomeF'sinit(laterIbacktracked and pulled theF'sbckoutofthethird andfourth terms!), Iapplied 3°andtried to"solve" the remaining equations. Iconsidered the double-dot and trace ofGAMMA equations instead ofthe tensor Ward table equations; ie, Inever constructed a Ward Table for these six terms. But Icould notsolve, sopages now axed. 2,Also axed: pages showing that Marshall's Vertex does infact satisfy the Ward. This proof iscontained implicitly inReduce program Ilater did with the nterms back inj there isnoquestion but that-Marshall's vertex wrks; Idid itbyhand alonb time ago anyway. t Le] 8 NewApproach duly3,1979 ie) 1/.Ihaveobtained general tensor formforthevertex. NowIamconsidering the general Ward problem. Ihave written down amethod ofsolution which, although well- Gefined, isnotvery practical. Nowisthetime tostart making useofother people work onthis subject. Eg,itwill probably help ifIpick mylist oftensor forms inaclever way, compared toanarbitrary way. Like choosing good coordinates! 2.Soforopeners, lets seehowthe"Marshall Vertex" fits into mylist oftensor vertex terms. Start byrewriting the RHS Ward like so: RWs= gylBWy|-3@[8"ar| . ay Cee -\ a, weegQ= -bOUD=-G) BOQ Db= 20) ¥ -Here youseecertain rank-2 tensors that occur "naturally" ontheRHSWard. Perhaps theseshouldreplacelessnaturaltensors,likebuyallbyitsdle. fo)Infact, here are some tensors that always seem tobeuseful: Py, m artes 0 myeRe.Veearea oRAy-\3-at % CeesBa) cay Faxcrampdt 0 « RWS=90)Ay-g@)Aa a Ay) Aye, De AyBeAje=LesGenes y=—>} GH Pye MsY‘. A A Al i= Pe'— (RRP HeSn . SS -2- e ea ant otPGOT le)Ag=[3-gi| tARS k-lest‘ a+. syGa Lge wa |Ay [83a =h eo ghos ‘ ~ |e la &Ay Ay4=° 4> 3GO SR] theeAg=Ag (An=0 im=O Weshouldreallybecomparing thistensoraywiththetensor xh st?which itcan beregarded asreplacing. Ie, products oftwo momenta. Itseems reasonable thatuseofAwillsimplify things because Aisorthogonal toi}and3%,whereas thetensor j4i7 isnotorthoggnal toanysingle momenta. Thisorthogonality will beuseful because ward identity involves dot with single momenta!!! 3.Sohow can werecast our "complete list ofvertex terms" using the tensor Aabove??? CO)onideaistotaketheoriginal listandreplaceJi”everywhere itocoursby roa this:/ ~. ¢ ghia ee wy)CasGyls®Ag) ifrtotalaSETAETirat-tourtonneofCC.generalfomula,thengroupallthe deltas together into the last term, here iswhat Iget: ws rar J=LODATH@ONTAGD 4& | wos ry ¥~LQDAE- GdARLES) ««. as uy 3TearAplsbaAQTEAG +e 3 ah3: Y=2@dAal —GdAQey SCA *¢ care‘ Jo +V™(sd Ale) +c. ar i yOBOY SU) 5 . fo)te -3- Contemplate this new list: all triple momenta occurrences are now replaced with Apoccurrences. Thefunctions areallthesameexceptthelasttwofunctions 8 are linear combinations ofthe old set, modified bydot products toget the right symmetry. Each curly bracket has adefinite 12symmetry either Aor$and this can beseen byeyeball. Note also that Ihadtofake upterms 2and3byartifically adding cyclic. Then when you convert toA's, you now really need the "+cyclic" inthese terms because the delta parts are not done cocrectly ifyou leave that off. 4.Here isaminor technical detail: goback andlook atterms 3abd,inthe original deal. LetA(123) beanarbitrary TAfunction. Iclacim that anysuch function can bedecomposed inthe following way: AQ)=LAG) +ACs)+AQ@)] “Naw, a“ =|SAU) +SAC)¥4Away] Bytheway, thenotation A(12) means afunction of1,2,3 which isanti symin12. Therefore, Ican replace terms ..wait. Suppose you write term 3as stuff +cyclic. Then the original function nolonger need beTA: anarbitrary A(12) will do, because the A(123) sogenerated bythe sum will still beTA. \ The question is: are these "classes" equal ordoes one contain the other: a) [2314312] a(123) b) (2314312) A(12) +cyclic =[2314312](A(12)+A(23)+A(31)) Can anarbitrary A(123) begenerated from A(12) functions? Yes because you can certainly take A(12)=A(123). Onther other hand, given A(123), canyou@ind A(12)? Same solution. Sothese waysofwriting things arethesame. Sowecanadd+cyclic and get reid ofA(123) infavor ofA(12). Thus, Iwill nowrewrite thelast like so: wos ais; i VLOYALT+@PARPIA GO©egate |cS ory 3 . | 2»)[@DA2- @aAg WS\@ +ogee 23 aALF a»[@OAgUS OAR] AQ) 4ae wa ay 4)LRAal= CYAV?|SG) <=sae we 3 18) LOTQaeyTAsGd 2yee | Ne 3bog Leary jSaG2)+zeget1 Iclaim this isstill the most general form for writing the vertex, pricir tothe Ward restrictions. Now there are 6functions that are undetermined: three are A,and three are S,inonly one pair ofvariables. All terms are +cyclic and must beso. The symmetry orantisynmetry ofeach bracket can beseen easily, sono question but that all these tensor terms are TA. The important fact isthat this is all there are! : © 5.xwonderifTshouldattempt toreplace thesinglemomenta vectors withmy fancy Bvectors ?Maybe first Iwill try toapply the Wards tothe above! 6.Pauseandmakeobservation: itwouldcertainly benicerifwehadagtype objects appearing instesd ofthoseA};.Thereasonisthat,intheWard,theA12would beannihilated byeither 1)orby2! Iamfurther reinforced inthils viewbythefactthattheMarshall vertex doescontain theAj3object. Soletsgo back and see if this cannot be worked out. Better Choice ofOriginal Tensor forms. 1.Based onthe suggestion atthe end ofthe last section, Iamgoing torechoose12 ie] mytensors totry andmaximize the occurrence ofA;5 type objects. Note that all indices cyclicize sothis object retains its high degree oforthogonality under cyclization. Secondly, Iwill try tomake the next worse objects have the form a2orthelike sothat atleast they vanish ononeside. Myidea isthat somewhere13_ . 1 woul later onthis will allow projection ofequations; eg, apply 1”and euch aterm would vanish. SoIwent back tothe original list and here ismynew choice set 5endCarsfa i basa f ~O%(ary, =CHIP AZ) Hotcgane Se ~> Ab ,rs Ginards f 2 tase . o,GY GP-Lit)fsey SAA3Ab ~k sasAs ‘f. AyGata=CGV31%) ~yAb,As ype ts “% -Seh5, G3=Gt+Sie) ~yAL\AS Howwerethesechosen?‘Themixedpairechosenfromobviousdesiretogettheal? ©) crsects. these arethen “extended” aswasdiscussed earlier. Ithen hadtopick onemore Atensor. Mydesire toremove theS(123) infavor ofS(12) byprocedure mentioned above forces metotakeonofthe“short forms" and(231)4istheonly oneavailable. Luckily thisAtensor isindependent ofthe(211), which comes out of the extension. Then Ihadtochoose anStensor. Again, Iwanted ashort form, butnowthere aretwoshort forms. Theultra short (123), waschosen against because itmade aAl?typeobject which diesagainst no 1or22or3?vector. 21 Now lets make an“original table" like sousing the above forms: Pry CA4LR)AYR) ogehee a3. AR LYSar) &eyelte V2 123°TSPABR) ALCO +egckic a(tayasyJA(te) “4 as423 re)CAN=BD) SLQQ) aetee Jn(23\-3\Y) S(ie) a 33(+2) Asad +my.nmSAV G(ey +ve: -2=. 2.Now Iwant toreplace the first pair ofmomenta with Aobjects inthe first 0.fourlines.FirstIwanttoargueclearlythatthedeltatermssocreatedcanZorsurebeabsorbed intothelast twolines by@simple redefinition oftheAg andSjfunctions. Thedelta's picked upfrom thefirst twoterms ereOKbecause firsttermgives(1.2)(d!)(142)? A,(12) whichclearly canbeabsorbed bythe second last term. Etc. Sothe third and fourth terms are the only ones that worry me so lets do them out: Vvwe wu wu R= TBLAL =NSH DAK a4leyoy@roiedalAQ?) aes ye ,=ELQD[ HAPs BERG CVS Ale Ry =Sap LE@agsrGapacy foSorry Aly) seashQaParas LOOP Aue) [=Say Se) Thisshowsthattheterm-3deltasdogetabsorbed,andterm-/,isthenobvious,just fe)afew sign changes and Agoes toSetc. Solets towrewrite our general form in terms of the A's: ~cen 32AC«he ab abab LeaAn(Pe?YAUe)&gece RI=(eS —Fax . we ;LEAL (OS Gt)ahe Ws A ea Ane(18)Aga]Acie*< 3 a \=G3)AgeOAT Suu)xc. mw 3 = GaAs Aca ye3 G2)AnG-9SQ) wD a p83 0 LQdARP& (0.3)Aa2AsGe)3 y LOAARE=~QA)ARV) Bele) an 3s” -3- Iamnotyetsure what isthebest waytodisplay these terms. Icould give fa) themallthesamedimension, PerhapsIshouldredefine myAobjecttoabsorball those dot products. Lets try it: b ob & a ae ob(eS)Ai=Ce]T=ae=|{&OTH VRSAy Ce) &=&dD AB, aFl.o Paws Fad ~~ Fo=an bole=FSA)oy]|wh5 Fay sar)+sy wma3Rs Cals oere=33Fy ‘LPRVA RITA) «3: aacue =ODED-GC) \Fat~FET) Se)aye 3.at \\ DFay=23-92 1% 3 |SSAW)*oy petsoto SOS ; : . 2 |C2YSaQ)49g i SPUD GDI This looks much nices, Iamready tostart dotting fortheWard. Deatail: Notice that Fislinear ineither ofits momenta. Itfollows then that: ob a” ive«FagetBS=o\ Show how Marshall's Vertex fits into myForm. lo a Y (RQ, a .=3M 2, 9@ mot A xe 3 PoweSlyBSCasa] +(eats)[helta“yeeVv. n=SLES] +GO") REGay?+ye/=) 1=STLSE)EEAHHI~ELH -G-0F]a , J =Sy Eb-Hey)"SECogs(HOH), J < 4 monoy=+6 "+WO SiGe) +ae +6 ~6n . 0 +Ses AG +ye =Sy SMa «5¢ SShoe)=[oadbo]/ (a2) ~(). Nop=seve) 7 Boe= =LEO] 7 . fo} BVectors Material Deleted from this Section. 81,MyideawastoreplacepwithB‘everywhere throughout the‘tensoranalysis. * Iinvented some B-vectors (later definition changed!) which hadthese properties: a) (12)B, =B (12)B, =By i . Theidea, forexample, would havebeentoJustreplace 123tensor formwithB,B,B, etc, and just take over all that stuff. - Butthen Ifound thatthese B'scould not,bemaitetosumtoderoaoI could not just "take. over" allthat tensor foims stuff. . Inretfospect, Idont think itreally very helpful toput..eyerything into such B-vectors anyway. Sothis isjust anidea that never gotanywhkre. a “ ~ ~ayb=\é_3|Cary~O9) . i]vy : : Cy)sy) \ : re)(uncehk OY . 4 &weet. . Complete Sets Point ofDeparture. le)1.AtthemomentIamfavoringtheF-tensorwayofdoingthingsonseveral.grounds: a)Ithink the Ward Table will besimple because F's are soorthogonal b)Marshall's solution fits easily into this form c)there are only 6functions toplay with, not 10. 2.Once the Ward Table isconstructed, there will beonly four independent operator equations insix functions. This sounds like atractable problem. The plen isto use the operator-inversion method wherever possible and just solve the whole thing. 3.Thetwoequations obtained fromthedouble-dot 273%¢!*3 andthecontraction equation G!? mustbethree ofthefourequations. Iguess theother iscontained in1°33G!?3. 1think itisprobably better tojustgoahead anddecomposed things into therawtensor forms like 172” endsoon. Lets see what our four equations look like, compare toearlier formilation tosee Afthis F-tensor deal does ordoes not simplify things. Bythe way, donot compute AsD type things; this isreally [e]- [12] onAandthat isaprojector. Well, I guessitsOKtotekesymandantisymperts,butletsgetthestraighttablefirst. fe)Ithink Iwill first check the Ward identity tomake sure it4sright before doing this table. 8 Material Deleted from thi i 1,Here for the first time Iconsidered the problem ofhow you "fit" agiven tensor into aclaimed "complete set". Iconcluded that you should take your candidate tensor, break itdown into its "irreducible components" using some kind ofLookup table, and then fit those irreducibles into your extended form X- 2.Imade atable using reduce toanalyze ppp forms into irreducible fomponents. However, later Imade amore comprehensive table sothis one has been deleted. ‘The names ofthe Standard Basis tensors were later changed toconform more to the F-tensor basis. 3.Deleted: sheet showing irreducible basis for ppp sector. 4,Analysis ofBBB into irredubiels. See later work inppp section onthis. 5.Connectionbetweenirreducible besisandF-tensor'basis.SinceIhavechanged [e)myirreducible basis to more useful form, Inolonger care about this connection, so axe. Question: What does itmean for_abunch ofVertex Terms toform a"Complete Set" fe} 1.Consider thisso-called complete set: \(2114212)A, (12)+cyclic Pvi 2.(211-212)8, (12)+cyclic2,(2314312)A,(12) +cyclic\y4.(231-312)8,(12) +cyclic 6 Inthe ppp sector, weclaim this isacomplete set ofvertex terms. What wemean, of course, orare claiming, isthat there isnototally antisymmetric ppp-related vertex term which cannot bewritten ordecomposed into the above forms. 2.Example: Consider thevertex term A(123)123. Howdoes this "fit" into the above complete set? Ie,tell meexactly what functions A,S, andsoon"abosorb™ this new vertex term. Todothis, Ithink Ifirst have to"undo" the complete set and write itin adifferent way: Ie, Ihave toundo the extension" and say that the set ist (211), (123) ‘ 1 1La 0(211); Fy(a23) =(211)3 Rca) (211), $3(123) * 2Wy 20(211)5 Fy(123) -(211)7 (123) 3.(231), (123) 4.(231), 8,(123) Icouldfindexpressions forA,(12) intermsofA3(123) andFP;andsoon,Lets come back later to do this. Nowletsconsiderthetermunderthespotlight: sincel23=(123),we look uptofind: 4 (211),=8(231),-Es,+(1/3)8)\3 (123)=m28,7 Therefererores Yai)g +(231)g =1/3), =(-1/4) (123), % -L\ ‘Thus: (123),Sean, +(231),] Thereforethistermcanbe"ebsorbed:"bysetting: A,(123)=-6A(123) re) 4,(123)=p02) ot —L- “enogooat: , ao) 0) 3) 0Cy FPeHawy,PO=(aeaaare]oye ey : =AA@= aha) Bao: . AMABWUAG (= N-e342) [SAG2s\\ Aeqelee =>ACY =SA(ea) . -_@> A\G2)=QVary+Age) Mert: : — ausAgr’) =foxesar)AgWe)+~lee\ >6AGQe)=Agtirs) 4 fo)Anke) =athy (\) AWAUA. >9Ae)=-—GAG)AaGd =-2& AUB) da = 0 Sothis shows exactly howournewterm "registers" inthecomplete set. Inretrospect, obvious thatonlytheAyandAgtermsareactive herebecuase thosetermsarewhere the totally symmetric tensor forms reside. . endofexample 3.Could Ihave done this example insome ofher way? Mymethod above was tofirst express the complete setinterms of"irreducible objects". Since my"new" tensor vertex term was itself "irreducible", itwas easy tosee how itfit in. IfIsithereandtryto"absorb"thisnewtermdirectlyIfinditnearly fe)impossible todo.Ireally havetogoviatheirreducible channel. Eg,youseethat thenewterm123contains only121and122pieces, soitatfirst seemsunlikely thattheAogetsinvolved since : -3- itcontains neither ofthese desired pieces. Itturns out that you need this fe)A,termtocanceloffpiecesoftheA,termtomakeitwork. Soitisincredibly easier togothrough the IRchannel!!!!! 4.Example: Nowsuppose someone hands youthis: (121+122) Ag(12) +cyclic. How can wefit this one into our complete set? This example isalittle harder because this object itself isnot irreducible. Itwould probably bed nice tobeable todecompose the raw tensor forms like 112 into their ireeducible pieces. Have Iever done that? Qisromghe (asohdseen! C+ VR)=123 =-$(\23)5 a S =~ Es Ase) +AS, +Acy\ oe) =SPalade sty. =[aansecanyt ie Qa SH BWsHAW gaawadedsohe, =Ges) =FLASK ~ee, AsGe)=3[Aatelxor| Generel Procedure: Take anycandidatge totally antisymmetric tensor. Ttisalinear combination offunctions times rawtensor forms. Analyze each tensor form into itsirreducible components. Thenyouknowtheamount of$144, eteinthisvertex term. Then expand back out toget todesired basis. SymmetryAnalysisTableforBasicrawtensorforms:_ 81.This table resulted from simply inverting the 8x8 matrix asdescribed onlast fewpages.Wefind: H | : |zlelete leleslel aeSs St ACTA GO+02} Tra|atGus |ds|ada [Fal Got[FAONFL Cat|GOL = ane ———- +: } - {wit |e] epo fe fot tie!—- ceSee ee Saano—j-———-\at t ° + fo) ° oyt -f— LoofSopp a aas ! a —1 at \er6 ° | ° ° ° |e =' eenSe|-: wef feepee atcaos .~ ~4 } ary 8 t = z |e ne ss~Lawt \ On|Seif 4 fA tty aL a 4 + a4 j wu¢ s ¢ o é ° ¢ ° | —* + —_+t ~ wa}© 3. ° ° x ry + $ This stuff was read directly from Reduce output, sohopeffilly noeyfrors. Note that you read this table across, not down. (Theorem: you cannot invert amatrix bylooking atit). BxampleofTableUseage: r{ook“9 GAS=6123=COTA —Cs) [ES]=By, 4 ASpillover Theorem fortheManiupulation ofComplete Tensor Sets. (?)1,Supposeyoustartoffwithwhatyouknowisacompletetensorset,eb, Tfy+Tyfy tover +Ty Orfor that matter, why just treat them assingle objects, tensor vertex terms: T14+12+3 +....4T™ You know that any tensor ofcertain criteria cen bedissected into these forms. Now someone comes along and hands you some new form call itFl. You analyze Fland you find that: FleaM+dB+e% You can replace any ofthose forms which appears inF1byF1; eg, ifyou replace T2byFl, youhave added anew form (namely, £1), delted oneform (namely T2), and you have "spillover" into the ober forms T3and T4. Thus, your new "List" is: fo) TL+Fl+ TM+....4™ There isnoneed toindicate that T3and Thwere modified byspillover since they are “arbitrary amounts ofthe tensor form indicated" anyway. So, inshort, because 2appeared inthe list ofFl, you could simply replace 12byFlinthe list. 2,Now somehoands you another form, call itF2. You analyze F2wrt the original set and you find: {24 TL+12+1)+T78 with some coefficients Now you have tobealittle careful. Can you simply replace say Tywith F2? In the new, once-modified list? what you really should doisfirst eliminate T2 from the F2list, then re’ask the question. Eg, suppose upon elimination of12 you find that: FQ =tl +18+ Fl Then obviously you cannot replace 1byF2 since Thnolonger appears inthe list. 3.Moral: Fro each new substitution, you have toanalyze inthe n-lth basis. QO.Steurlh teFea boi wee ee weee Oemarnw sel me. QnFz2) Vile) ve Te - cee a - CARY AMMD AE TEL we eit) RA. WY oe we ~Ounadine BBBSoDan te VSTTT Que,nsdocnssamyTMoy888,2ATHT2gpsSplat cee ee te) kwSte TRCe pat) xArs]ee Te(ABR XKSGU OB aoeWeil,ecaadin O28)5AWS)aaFLGael _ Spe 28S ELS2AG 8 a SAWS =AS) eedhwee STAN = 38h) 6A) COC _. AAW) =SA(iw)=-2AU8) a “Qa FeSTATS ondesyneSo pgaTSky PLyve $$ agp ~ ee Yun Gk: usaaAliyee CS oe tS Gade TRA) oe _. BRB SO Os a = \w~c A WMAAYS2(N*e TA FA) BS) 8 —Q- OF ate. an an ©Nagle DaherWichemeds 7 So, .Whsomneglect TVaERDesedan ohTSTe = en aCe ea =[avigeey—~- - 2Reb 6 NGA wee SOY Ss_ Orr Aare) FE eeoe AS SGD) EE @Man,—cena ~Ue) Gy=eee SeCce) cake en OETA (0 72 BRO see Base) a nce a oo BS —3p—-Qua =Bah Solec, ae —3- —LbReathsdcamsanvonend OakhaJaltows . ._-ceeeHehomegaStDafey,weewefeee RB Sate eZ a BER Se). ad aa Sf SPS AW) ae aoeravers Qaoaziayss sFias* aac | os ost 2 — Se oO Lost Ward ~™Se)=G)— Lhe. Projection Analysis ofS(12)andA(12)typeobjects. Bre fe) 1.Thiskeepsrecurring andIdon'thavethefullthingdownsoletsjust doitnow. Remember that these are really S(12,3) eg. Turn tothat page where the general function f(123) isdecomposed. Then you get: S@ep =E]8a2)+ Wer+ 8@] SQ%,~, =0 Sd, =4asaedy- S@-S@S] =MA, =—2S, MWA, =0 ~~S@)g =Bs S@a=0 DB SA=7ELAM)-S@)-sey] =seed, S@_= £[SE~809). =sedi SoGosh: g to S@) 2), Se), S@), ia2 sen 3s ST,-SMB). SO S02) Ss BSA of “al * Awe Sea S(SG)* ewe] ‘ KB =~Elasad-ses) say] oS SO.=+[scs)-sc). y - 4 Le)AQ),=0J AC) =(ACS) ~AUA)-AR- ABD+AGOFAGEY =STAR +eye]V=pede AGB,=ELA LTAGD= AA)=AG)~AG)=AMY] Y =FLAC-AB]Y =ACD,Y ACY=[LAR~LAW)+Aah+AG)~Ala)—Alny] v =CLAGS+AG)=ZAC]Ye“SEPALe)-A()ABSAQ’), AG,=EL2Aers2NB9—AGa)~AK)~Ale)~Reh re)~$[A@)-Aeal’s ~+TA@d-Awl =—ACY,J ACY =Gaal rate) 4£0y+acl<Any—Hea)J 7 =FLAGS +Alay-2AGY] =Alma 4 re AQ) A. A, ALY AG) A. =AQD, ADL” AW) A(23)a BG~-2A3), * sreA@)a=SFTAGS)eg]¥ ie] A(@)= ELAMay-aeryy y A(B)e= -FRAGA AGS)=ACa| Los 4 fe)feSH=LO,=B[-sew+19)+8Es)~8cry|=o7 KMWA= WAL=GY-sey+94~te)+ay) =&@B]sod-s@)] =Se, KKB)=SE=&[-se0)4S(82)4S@\)~sar) |=HEHL2909-Se-se)] =SB, Yix(28)= BY]-S()+S3e)~S(3) 4Spe) : =&[sGY-S@)) =sC%, Xs)=sey=£\—SUS) +S@1) 4S5Q2)~S@s)| 0 >&ETesuy sud~se] =SRL Kai) =slay=ro=n)+st)-s(f+s(z3)|—&se)~s(3)).=SBN, Snr @,=-A5@, - SM, =© 4. SA,=JzSQ), “ . = +Ksed, a SG),=+ESO), ie) Gai),=-&s@, sswtOS,ke, oe® ~4_ J YEAGd=&[ade)~2rBeayWwe2is[AGD-aGi]=&A=A®, Oyinmo= Eloj-o =aw, KA@= LL-ACHSAGD) =ABLE =BEAnced,=“Lacey AB, AG) =LeAly—AS)~ Alay)=~ACS),=AC), KAwy= LLf-aersacn\ >BS[atedats]=3AG),=AB), KeA@~ AL-28 04st43(s)}=BEY] =RAG,= At, Se: ; _ AQ),=SAQA ~~ AG), +0 y 0 A@e= -KALD vo>. N®, =-BACL v AB,= -BACD, ¥ t 1AGS= BA. 4 This list isrelated tothe S-function list inasimple way: replace partner labels 1interchange 2, and then change all signs. The fact that itissoregular suggests tomethat both lists are correct, nomistakes. XPAW)=FAC,=KAM, =-aXVA@h [dreds HH,| je} YWAG= -BAG~>,Aled), 3|XA@3)=-BAQ, TLS TENA =WDA =SGesjawjyeye 0 oo k( & e >| € tao on es st | _ ‘oA Ve 2h i22 L O Ff Yo 5 nr eeeépo |. —@®|po+69)S00y7]_ppa awee eee ee - to |---| og foun Z| ig we foot ee a neeraa eeane”Aan9S,Gi)-$@sevsifRlas 23)S1)-25,@)|anySA]_Thisfableisjust[phenomenallyforesimplethananythingIeee ~|___The udeofthe123|tensorsimplifies row(3)dlot(comparel toearlieruse| __....o£.(2314312)whichspreaditselPoutintofoisons!eaters ou2the Ayd null terms fis obvious. $0lets get on _ ee —— . +. 2 pay - ; - ore) ce ws EdSC) a aaa — om |. — — _ ! ' | poiAro |A-0 & <|5|+22 \L-2 An 2) |3 Oo aSyan #sdipa ee{ ~~ fee .waeee ®|pa= OARS6bg a 2 ty ae et ws Coe ©.|-AeO-KGI7[=AGSAL)/|-A@-Ai gefemacy | H | ne Serene esaane ane —+— . ESEEEEL EEEETEEEEEEEREEEEEE PEeEERREEEE Geena —|A a a Material Deleted from this Section . 8 1,This wasmyfirst solution to.thenon-n sector Ward Table. problem. Ifractionated theterms oftheWard table andgotthe10equations stated below. Isolved. these equations andgotMarshall's Vertex exactly. Ihave deleted allthe‘ffactionation work that wasneeded togetthese tenequations since thecyleic operator method ismich more effieient (and starts ‘from thesame Ward table given inthis section) fe) : fo)\ -3- [email protected]: “— ie] 1)resolve thepofi-nproblem withtheslight #odification ‘ined _ - _2)writeapeStgeneral formforthen-terffs vertex oe ee _3) balapé the n-terms vertex agains}fhe n-terms RHS, oo . _a.betasotaboutthetirsttag-Since itsoonsgatlest,. Thiswillbequiteeasy. _—/@ -2R=-6% =+28, -aal ,SSt _©..~28 +aT =+28 -2G ee ee TO abet ee “ALl®-24,.0-28,=90 __\ba i J@ =24V=2R -2%- =O — —4@__.- ald S\-AS| so »@.. ~a\dS\sd Salt AdSs=NTso ——@D -~¥Gy) Sy+Coytae=2ay) Sy —___ =Dds8.+2da2d ea ReFdR=OL. —-@®. ~*(2-3) S\4xGrg) SyGyana2ay) Sa_— co DAReEIT 24 24VeGddR TSAR Thesearetheraw10equations withnosubstitutions made,al].termspresent. Youcan _....gee theextent ofthechange shown inunderline. Wecould justfeedthisintoReduce . ——-but_might aswellsimplify byhandfirsts 0 ee Comments: Lets think ofthese equations. Weknow there are really only four equations sittingthere,wecanignorethecolumD.,Iwouldfeelmuchbetter fe) ifeachofthese fourequations weresubmitted toacomplete 8,symmetry feactionation analysis. Remember that S(12) means afunction ofthree variables, Imight have written itlike (12,3). Solets analyze every term into symmetry pieces. Queuaedads oaowbene dX Cds =blxtyee] v +@d,= +Lasy-2xl v y=dyswae*d.2 =9)) ZadSadt—4e2 =a3) a= @r= tly-ay x=ds$~2ah =Gs) Sogy=2d/ ga)=Qayd v =(ya » n Cara) =(24dS4(AQTKH fe)Grz) =AS +WTV ;x3) =Qde(2ard)2 =Heuyt . Wide ney =WS-asliad] =$14 \s ~ ait] |=2Caares) +3(Caneay +YQ) )$2[dsnay+(C2424) +/°7) / +a BGoCOR ESN I ae / vw ~.: =2Ladd-2hb 4-24he]=2[add2A] i ~~ Aba nn +&[+PadsUHI, sétb(ARTY {says ..onss /ws =_ A =al@dd- dd] =Qala =ead % { Sothisgroupingofdotproductsisentirely2-like.Seeappendedsheetforsoir fe)way toconclude the same thing. oe 8 ee -- aoe Y.moe a ~(o— _ Sohere comes the bobtom line: _ - “Question: what isthe formofthemostgeneral solution totheWardIdentity? _ “Answert first, youcanhaveanyWardNulltermsyouwant. Thenifyou impose the Ward onthe remaining terms asparametrized inthe purpole box, here - iswhat you find: oo —. Addy=-8 = ET) -be a - SaQ2= BorBy = eLb@aey G a Se eee a2 0%= -B/d0=sh@Q=eo — ee a ean -2 et 5SUB) Seeeagttiegy —— — Oo RG0. But ‘thesolutign Ihavegenerated isprecisely Marshall's solution! : Conclusion: Assuming thesimpleansatz thate=0soonlyonefunction bonthea RHS oftheWard identity, and assuming that there are non-terms among the tensorforms,wearenowsurethat:—Themostgeneral formofasolution tothe|_WardIdentity isMarshall's solutions plusarbitrary linearconbinations ofthe| two Wafd Null Terms. ! ft! / Conversion ofMarshall's Solution toMyVariables OoSosaindakfrshade rYorbnsd "y = Oey S02) «aye: +BeAry *«ose. ~Sar SaQe)©ee. ern S00)=~pase|(@2~Ayy A\@)= tTve)—won) 7 Sa.Q2)=ELE@ 4¥6)|% O8== oe -8¥ ; SaGO= LOLA wYE BehB,% S42) ==2 heehee ae ee Se enea Poe ~ = S=-B&/k, =SS~28,Nee Oe ase LE 02dSe+h,N=-BYLe ’.3HYIfea(ale tAbGAVE, ATE