arXiv:0812.4408v1 [hep-th] 23 Dec 2008Asolutionmanual forPolchinskisStringTheoryMatthewHeadrickMartinFisherSchoolofPhysicsBrandeisUniversitymph@brandeis.eduAbstractWepresentdetailedsolutionsto81ofthe202problemsinJ.Polchinskistwo-volumetext-bookStringTheory.BRX-TH-604CONTENTS 1Contents0 Preface 41 Chapter1 51.1 Problem1.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Problem1.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Problem1.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Problem1.7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.5 Problem1.9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 Chapter2 112.1 Problem2.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Problem2.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Problem2.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.4 Problem2.7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.5 Problem2.9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.6 Problem2.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.7 Problem2.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.8 Problem2.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.9 Problem2.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 Chapter3 213.1 Problem3.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.2 Problem3.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.3 Problem3.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.4 Problem3.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.5 Problem3.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.6 Problem3.7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.7 Problem3.9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.8 Problem3.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.9 Problem3.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 Chapter4 334.1 Problem4.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 Chapter5 355.1 Problem5.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35CONTENTS 26 Chapter6 406.1 Problem6.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.2 Problem6.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.3 Problem6.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.4 Problem6.7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426.5 Problem6.9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456.6 Problem6.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486.7 Problem6.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 517 Chapter7 527.1 Problem7.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527.2 Problem7.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 547.3 Problem7.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 567.4 Problem7.7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 587.5 Problem7.8. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 597.6 Problem7.9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617.7 Problem7.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 627.8 Problem7.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 627.9 Problem7.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647.10 Problem7.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 668 Chapter8 698.1 Problem8.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 698.2 Problem8.3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 708.3 Problem8.4. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 718.4 Problem8.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 728.5 Problem8.6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 728.6 Problem8.7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 738.7 Problem8.9. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 758.8 Problem8.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 759 AppendixA 789.1 ProblemA.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 789.2 ProblemA.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 809.3 ProblemA.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8010Chapter10 8310.1 Problem10.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8310.2 Problem10.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83CONTENTS 310.3 Problem10.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8410.4 Problem10.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8410.5 Problem10.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8510.6 Problem10.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8510.7 Problem10.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8610.8 Problem10.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8610.9 Problem10.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8711Chapter11 8811.1 Problem11.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8811.2 Problem11.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8811.3 Problem11.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8811.4 Problem11.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9011.5 Problem11.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9111.6 Problem11.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9211.7 Problem11.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9212Chapter13 9412.1 Problem13.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9412.2 Problem13.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9512.3 Problem13.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9612.4 Problem13.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9913Chapter14 10313.1 Problem14.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10313.2 Problem14.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10313.3 Problem14.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10514Chapter15 10814.1 Problem15.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10814.2 Problem15.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10815AppendixB 11015.1 ProblemB.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11015.2 ProblemB.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11215.3 ProblemB.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1130 PREFACE 40 PrefaceThefollowingpagescontaindetailedsolutionsto81of the202problemsinJ. Polchinskistwo-volumetextbookStringTheory[1,2]. Ioriginallywroteupthesesolutionswhileteachingmyselfthesubject,andthenlaterdecidedthattheymaybeofsomeusetoothersdoingthesame. Thesesolutionsaretheworkofmyselfalone,andcarrynoendorsementfromPolchinski.IwouldliketothankR. Britto, S. Minwalla, D. Podolsky, andM. Spradlinforhelpontheseproblems. Thisworkwas donewhileI was agraduatestudentat HarvardUniversity, andwassupportedbyanNSFGraduateResearchFellowship.1 CHAPTER1 51 Chapter11.1 Problem1.1(a) Weworkinthegaugewhere= X0. Non-relativisticmotionmeansXi vi1. ThenSpp= m_d_XX= m_dt_1 v2_dt (12mv2m). (1)(b) Again, weworkinthegauge =X0, andassumeXivi1. DeninguiXi, theinducedmetrichab= aXbXbecomes:hab =_1 +v2uvuv u2_. (2)UsingthefactthatthetransversevelocityofthestringisvT= v uvuu, (3)theNambu-GotoLagrangiancanbewritten:L= 12_d(dethab)1/2= 12_d_u2(1 v2) + (uv)2_1/2 12_d [u[_1 12v2+uv22u2_=_d [u[12v2TLsT, (4)where =12(5)isthemassperunitlengthofthestring,Ls=_d [u[ (6)isitsphysicallength,andT=12= (7)isitstension.1 CHAPTER1 61.2 Problem1.3Itiswellknownthat,theEulercharacteristicofthesurface,isatopological invariant, i.e.doesnot depend onthe metric. Wewill proveby explicitcomputationthat,inparticular, isinvariantunderWeyltransformations,ab= e2(,)ab. (8)Forthiswewillneedthetransformationlawfortheconnectioncoecients,abc= abc +bac+cab dadbc, (9)andforthecurvaturescalar,R= e2(R 2aa). (10)Sincethetangentandnormalvectorsattheboundaryarenormalized,theytransformasta=eta, (11)na=ena. (12)Thecurvatureoftheboundary thustransformsasfollows:k= tanb(atb+ bactc)=e(k tatanbd), (13)wherewehaveused(9), (11), (12), andthefactthatnandtareorthogonal. Iftheboundaryistimelikethentata= 1andwemustusetheuppersign, whereasifitisspacelikethentata=1andwemustusethelowersign. Hencek= e(k +naa). (14)Finally,sinceds = ()1/2dforatimelikeboundaryandds = 1/2dforaspacelikebounday,ds= ds e. (15)Puttingallofthistogether,andapplyingStokestheorem,whichsaysthatforanyvectorva,_Md d ()1/2ava=_Mds nava, (16)wendthetransformationlawfor:=14_Md d ()1/2R +12_Mdsk=14_Md d ()1/2(R 2aa) +12_Mds (k +naa)=. (17)1 CHAPTER1 71.3 Problem1.5Forsimplicity,letusdenea (/2p+l)1/2. Thenwewishtoevaluate

n=1(n ) exp[(n )a]= dd(a)

n=1exp[(n )a]= dd(a)eaea1= dd(a)_1a+ 12+_112 2+22_a +O(a)2_=1(a)2 12_16 +2_+O(a). (18)Asexpected,thecutodependenttermisindependentof;theniteresultis12_16 +2_. (19)1.4 Problem1.7Themodeexpansionsatisfyingtheboundary conditionsisX25(, ) =2

n1n25nexp_incl_sin nl, (20)where the sum runs overthe half-odd-integers,n = 1/2, 1/2, 3/2, 3/2, . . . . Note that there is nop25. Again,HermiticityofX25implies25n= (25n). Using(1.3.18),25(, ) = i2l

n25nexp_incl_sin nl. (21)Wewill nowdeterminethecommutationrelationsamongthe25nfromtheequal timecom-mutationrelations(1.3.24b). Notsurprisingly, theywill comeoutthesameasforthefreestring(1.3.25b). Wehave:i( )=[X25(, ), 25(, )] (22)= il

n,n1n[25n, 25n ] exp_i(n +n)cl_sin nlsin nl.Since theLHSdoes notdepend on ,thecoecientofexp[imc/l]ontheRHSmust vanishform ,= 0:1l

n1n[25n, 25mn] sin nlsin (n m)l= ( )m,0. (23)1 CHAPTER1 8Multiplyingbothsidesbysin[n/l]andintegratingovernowyields,12n_[25n, 25mn] sin (n m)l+ [25n+m, 25n] sin (n +m)l_(24)= sin nlm,0,or,[25n, 25mn] = nm,0, (25)asadvertised.ThepartoftheHamiltonian(1.3.19)contributedbytheX25oscillatorsisl4p+_l0d_2_25_2+12_X25_2_=14p+l

n,n25n25n exp_i(n +n)cl_

_l0d_sin nlsin nl+ cos nlcos nl_=14p+

n25n25n=14p+

n=1/2_25n25n +25n25n_=12p+

n=1/2_25n25n+n2_=12p+__

n=1/225n25n+148__, (26)wherewehaveused(19)and(25). Thusthemassspectrum(1.3.36)becomesm2=2p+H pipi(i = 2, . . . , 24)=1_N 1516_, (27)wherethelevelspectrumisgivenintermsoftheoccupationnumbersbyN=24

i=2

n=1nNin +

n=1/2nN25,n. (28)Thegroundstateisstillatachyon,m2= 1516. (29)1 CHAPTER1 9TherstexcitedstatehasthelowestX25oscillatorexcited(N25,1/2= 1),andisalsotachyonic:m2= 716. (30)Therearenomasslessstates,asthesecondexcitedstateisalreadymassive:m2=116. (31)Thisstateis24-folddegenerate,asitcanbereachedeitherbyNi,1= 1orbyN25,1/2= 2. Thusitis a massivevectorwithrespect to theSO(24,1)Lorentzsymmetry preserved by the D-brane. Thethirdexcitedstate,withm2=916, (32)is25-folddegenerateandcorrespondstoavectorplusascalarontheD-braneitcanbereachedbyN25,1/2= 1,byN25,1/2= 3,orbyNi,1= 1,N25,1/2= 1.1.5 Problem1.9Themodeexpansionfor X25respectingtheboundaryconditions is essentiallythesameas themodeexpansion(1.4.4), theonlydierencesbeingthatthersttwotermsarenolongerallowed,andtheoscillatorlabeln,ratherthanrunningoverthenon-zerointegers, mustnowrunoverthehalf-odd-integersasitdidinProblem1.7:X25(, ) = (33)i_2

n_25nnexp_2in( +c)l_+ 25nnexp_2in( c)l__.Thecanonicalcommutatorsarethesameasfortheuntwistedclosedstring,(1.4.6c)and(1.4.6d),[25m, 25n]=mm,n, (34)[ 25m, 25n]=mm,n, (35)asarethemassformula(1.4.8),m2=2(N+N+A+A), (36)thegeneratorof-translations(1.4.10),P= 2l(N N), (37)and(therefore)thelevel-matchingcondition(1.4.11),N=N. (38)1 CHAPTER1 10However,theleveloperatorNisnowslightlydierent,N=24

i=2

n=1inin +

n=1/225n25n; (39)infact, itisthesameasthelevel operator fortheopenstringonaD24-braneof Problem1.7.Theleft-movinglevelspectrumisthereforegivenby(28),andsimilarlyfortheright-movingleveloperatorN. Thezero-pointconstantsarealsothesameasinProblem1.7:A =A=12__24

i=2

n=1n +

n=1/2n__= 1516. (40)Atagivenlevel N=N, theoccupationnumbersNinandNinmaybechosenindependently, solongasbothsetssatisfy(28). Thereforethespectrumatthatlevel will consistoftheproductoftwocopiesoftheD-braneopenstringspectrum,andthemass-squaredofthatlevel(36)willbe4times the open string mass-squared (27). We will have tachyons at levels N= 0 and N= 1/2, withm2= 154(41)andm2= 74, (42)respectively. Thelowestnon-tachyonicstateswillagainbeatlevel N= 1: asecondrankSO(24)tensorwithm2=14, (43)which can be decomposed into a scalar, an antisymmetric tensor, and a traceless symmetric tensor.2 CHAPTER2 112 Chapter22.1 Problem2.1(a) Forholomorphictestfunctionsf(z),_Rd2z ln [z[2f(z)=_Rd2z1zf(z)= i_Rdz1zf(z)=2f(0). (1)Forantiholomorphictestfunctionsf( z),_Rd2z ln [z[2f( z)=_Rd2z 1 zf( z)=i_Rd z1 zf( z)=2f(0). (2)(b) Weregulateln[z[2byreplacingitwithln([z[2+ ). Thisleadtoregularizationsalsoof1/ zand1/z: ln([z[2+) = z[z[2+= z[z[2+=([z[2+)2. (3)Working in polar coordinates, consider a general test function f(r, ), and dene g(r2) _d f(r, ).Then,assumingthatgissucientlywellbehavedatzeroandinnity,_d2z([z[2+)2f(z, z)=_0du(u +)2g(u)=_u +g(u) + ln(u +)g(u)_0_0du ln(u +)g(u).=g(0)=2f(0). (4)2 CHAPTER2 122.2 Problem2.3(a) Theleadingbehavioroftheexpectationvalueasz1 z2is_n

i=1: eikiX(zi, zi):_=iCX(2)DD_n

i=1ki_n

i,j=1[zij[kikj= [z12[k1k2iCX(2)DD(k1 +k2 +n

i=3ki)

n

i=3_[z1i[k1ki[z2i[k2ki_n

i,j=3[zij[kikj [z12[k1k2iCX(2)DD(k1 +k2 +n

i=3ki)n

i=3[z2i[(k1+k2)kin

i,j=3[zij[kikj= [z12[k1k2_: ei(k1+k2)X(z2, z2):n

i=3: eikiX(zi, zi):_, (5)inagreementwith(2.2.14).(b) Thezi-dependenceoftheexpectationvalueisgivenby[z23[k2k3[z12[k1k2[z13[k1k3= [z23[k2k3[z12[k1k2[z23[k1k31 +z12z23k1k3(6)= [z23[(k1+k2)k3[z12[k1k2_

k=0(12k1 k3 + 1)k! (12k1 k3k + 1)_z12z23_k_

_

k=0(12k1 k3 + 1)k! (12k1 k3k + 1)_ z12 z23_k_.Theradius ofconvergenceofapowerseriesisgivenbythelimitask of [ak/ak+1[,wheretheakarethecoecientsoftheseries. Inthiscase,forbothoftheabovepowerseries,R= limk(k + 1)! (12k1 k3k)k! (12k1 k3k + 1)z23= [z23[. (7)(c) Considertheinteriorofthedashedlineingure2.1,thatis,thesetofpointsz1satisfying[z12[ < [z23[. (8)2 CHAPTER2 13Byequation(2.1.23),theexpectationvalue: X(z1, z1)X(z2, z2) : /(z3, z3)B(z4, z4)) (9)is a harmonic functionof z1withinthis region. It cantherefore be writtenas the sumof aholomorphic andanantiholomorphic function(this statement is true inanysimplyconnectedregion). The Taylorexpansion of afunctionthat is holomorphic onanopen disk (about the centerof the disk), converges on the disk; similarly for an antiholomorphic function. Hence the two TaylorseriesontheRHSof(2.2.4)mustconvergeonthedisk.2.3 Problem2.5Underthevariationoftheelds() () +(),thevariationoftheLagrangianisL =L +L(a)a. (10)TheLagrangianequationsofmotion(Euler-Lagrangeequations)arederivedbyassumingthattheactionisstationaryunderanarbitraryvariation()thatvanishesatinnity:0=S=_dd L=_dd_L +L(a)a_=_dd_LaL(a)_(11)impliesLaL(a)= 0. (12)Insteadof assuming thatvanishes atinnity, letus assume thatit is asymmetry. In this case,the variation of the Lagrangian (10) must be a total derivative to insure that the action on boundedregionsvariesonlybyasurfaceterm,therebynotaectingtheequationsofmotion:L = a/a; (13)/aisassumedtobealocal functionoftheeldsandtheirderivatives, althoughitisnotobvioushowtoprovethatthiscanalwaysbearranged. Using(10),(12),and(13),aja=2ia_L(a)1/a_=2i_L +L(a)aL_=0. (14)2 CHAPTER2 14If we now vary the elds by ()(), where is a symmetry as before but is an arbitraryfunction,thenthevariationoftheactionwillbeL=L(a)a() +L=_L(a)a +L_ +L(a)a. (15)Equation(13)mustbesatisedinthecase()isidentically1,sothefactorinparenthesesmustequala/a:S=_dd_a/a +L(a)a_=_dd_/a+L(a)_a=2i_dd jaa, (16)where we have integrated by parts, assuming that falls oat innity. Since exp(S) =exp(S)S, this agrees with(2.3.4) for thecaseof at space, ignoringthetransformationofthemeasure.2.4 Problem2.7(a) X:T(z)X(0, 0) = 1: X(z)X(z) : X(0, 0) 1zX(z) 1zX(0)T( z)X(0, 0) = 1:X( z)X( z) : X(0, 0) 1 zX( z) 1 zX(0) (17)X:T(z)X(0) 1z2X(z) 1z2X(0) +1z2X(0)T( z)X(0) 0 (18)X:T(z)X(0) 0T( z)X(0) 1 z2X( z) 1 z2X(0) +1 z2X(0) (19)2X:T(z)2X(0) 2z3X(z) 2z3X(0) +2z22X(0) +1z3X(0)T( z)2X(0) 0 (20)2 CHAPTER2 15: eikX::T(z) : eikX(0,0): k24z2: eikX(0,0): +1zik: X(z)eikX(0,0):k24z2: eikX(0,0): +1zik: X(0)eikX(0,0):T( z) : eikX(0,0): k24 z2: eikX(0,0): +1 zik:X( z)eikX(0,0):k24 z2: eikX(0,0): +1 zik: X(0)eikX(0,0): (21)(b) Inthelineardilatontheory,theenergy-momentumtensorisT= 1: XX: +V2X,T= 1:XX: +V2X, (22)soitsucestocalculatetheOPEsofthevariousoperatorswiththetermsV2XandV2Xandaddthemtotheresultsfoundinpart(a).X:V2X(z)X(0, 0) V2z2V 2X( z)X(0, 0) V2 z2(23)NotonlyisXisnotatensoranymore,butitdoesnotevenhavewell-denedweights,becauseitisnotaneigenstateofrigidtransformations.X:V2X(z)X(0) Vz3V 2X( z)X(0) 0 (24)SoXstillhasweights(1,0),butitisnolongeratensoroperator.X:V2X(z)X(0) 0V 2X( z)X(0) V z3(25)Similarly,Xstillhasweights(0,1),butisnolongeratensor.2X:V2X(z)2X(0) 3Vz4V 2X( z)2X(0) 0 (26)2 CHAPTER2 16Nothingchangesfromthescalartheory: theweightsarestill(2,0),and2Xisstillnotatensor.: eikX::V2X(z) : eikX(0,0):iVk2z2: eikX(0,0):V 2X( z) : eikX(0,0):iVk2 z2: eikX(0,0): (27)Thus: eikX:isstillatensor,but,curiously,itsweightsarenowcomplex:_4(k2+ 2iVk), 4 (k2+ 2iVk)_. (28)2.5 Problem2.9Since we are interestedinndingthe central charges of these theories, it is onlynecessarytocalculate the 1/z4terms inthe TT OPEs, the rest of the OPEbeingdeterminedbygeneralconsiderations as in equation (2.4.25). In the following, we will therefore drop all terms less singularthan1/z4. ForthelineardilatonCFT,T(z)T(0)=12: X(z)X(z) :: X(0)X(0) :2V: X(z)X(z) : 2X(0)2V2X(z) : X(0)X(0) : +VV2X(z)2X(0)D2z4+3V2z4+O_1z2_, (29)soc = D + 6V2. (30)Similarly,T( z)T(0)=12:X( z)X( z) ::X(0)X(0) :2V:X( z)X( z) :2X(0)2V2X( z) :X(0)X(0) : +VV 2X( z)2X(0)D2 z4+3V2 z4+O_1 z2_, (31)so c = D + 6V2. (32)2 CHAPTER2 17Forthebcsystem,T(z)T(0)=(1 )2: b(z)c(z) :: b(0)c(0):(1 ) : b(z)c(z) :: b(0)c(0) :(1 ) : b(z)c(z) :: b(0)c(0) :+2: b(z)c(z) :: b(0)c(0) : 62+ 6 1z4+O_1z2_, (33)soc = 122+ 12 2. (34)OfcourseT( z)T(0) = 0,so c = 0.Thesystemhasthesameenergy-momentumtensorandalmostthesameOPEsasthebcsystem. While(z)(0) 1/z as inthebcsystem, now(z)(0) 1/z. Eachtermin(33)involvedoneb(z)c(0) contractionandonec(z)b(0) contraction, sothecentral chargeof thesystemisminusthatofthebcsystem:c = 12212 + 2. (35)Ofcourse c = 0still.2.6 Problem2.11Assumewithoutlossofgeneralitythatm >1; form= 0andm = 1thecentral chargetermin(2.6.19)vanishes,whilem < 1isequivalenttom> 1. ThenLmannihilates [0; 0), asdoall butm1ofthetermsinthemodeexpansion(2.7.6)ofLm:Lm[0; 0) =12m1

n=1nm(n)[0; 0). (36)HencetheLHSof(2.6.19), whenappliedto [0; 0),yields,[Lm, Lm][0; 0)=LmLm[0; 0) LmLm[0; 0)=14

n=m1

n=1mn n nm(n)[0; 0)=14m1

n=1

n=_(mn)nnn + (mn)nmn,n_[0; 0)=D2m1

n=1n(mn)[0; 0)=D12m(m21)[0; 0). (37)2 CHAPTER2 18Meanwhile,theRHSof(2.6.19)appliedtothesamestateyields,_2mL0 +c12(m3m)_[0; 0) =c12(m3m)[0; 0), (38)soc = D. (39)2.7 Problem2.13(a) Using(2.7.16)and(2.7.17),b(z)c(z) =

m,m=bmcmzm+zm+1=

m,m=bmcmzm+zm+1

m=01zm+zm+1=b(z)c(z) _zz_11z z. (40)With(2.5.7),: b(z)c(z) : b(z)c(z)=1z z__zz_11_. (41)(b) Bytakingthelimitof(41)asz z,wend,: b(z)c(z) : b(z)c(z)=1 z. (42)Using(2.8.14)wehave,Ng=Qg +12=12i_dz jz +12= 12i_dz : b(z)c(z) : +12= 12i_dzb(z)c(z) 12. (43)(c) Ifwere-writetheexpansion(2.7.16)ofb(z)inthew-frameusingthetensortransformationlaw(2.4.15), wend,b(w)=(zw)b(z)=(iz)

m=bmzm+=ei/2

m=eimwbm. (44)2 CHAPTER2 19Similarly,c(w) = ei(1)/2

m=eimwcm. (45)Hence,ignoringordering,jw(w)= b(w)c(w)=i

m,m=ei(m+m)wbmcm , (46)andNg= 12i_20dwjw=

m=bmcm=

m=bmcm

m=01. (47)Theorderingconstant isthusdeterminedbythevalueof thesecondinnitesum. If wewrite,moregenerally,

m=0a, thenwemustregulatethesuminsuchawaythatthedivergentpartisindependentofa. Forinstance,

m=0aea=a1 ea=1+a2+O(); (48)the-independentpartisa/2,sotheorderingconstantin(47)equals 1/2.2.8 Problem2.15ToapplythedoublingtricktotheeldX(z, z),denefor z< 0,X(z, z) X(z, z). (49)ThenmX(z) =mX( z), (50)sothatinparticularforzontherealline,mX(z) =mX( z), (51)2 CHAPTER2 20ascanalsobeseenfrom themodeexpansion(2.7.26). Themodes maredened asintegralsoverasemi-circleofX(z) +X( z),butwiththedoublingtricktheintegralcanbeextendedtothefullcircle:m=_ 2_dz2zmX(z) = _ 2_d z2 zmX( z). (52)At this point the derivation proceeds in exactly the same manner as for the closed string treatedinthetext. Withnooperatorattheorigin,theeldsareholomorphicinsidethecontour,sowithmpositive, thecontour integrals (52) vanish, andthestatecorrespondingtotheunit operatorinsertedattheoriginmustbethegroundstate [0; 0):1(0, 0)= [0; 0). (53)The state m[0; 0) (m positive) is given by evaluating the integrals (52), with the elds holomorphicinsidethecontours:m[0; 0)=_ 2_1/2i(m1)!mX(0) =_ 2_1/2i(m1)!mX(0). (54)Similarly,usingthemodeexpansion(2.7.26),weseethatX(0, 0)[0; 0)= x[0; 0),sox[0; 0)= X(0, 0). (55)Asintheclosedstringcase, thesamecorrespondenceapplieswhentheseoperatorsactonstatesotherthanthegroundstate, aslongaswenormal ordertheresultinglocal operator. Theresultisthereforeexactlythesameas(2.8.7a)and(2.8.8)inthetext; forexample, (2.8.9)continuestohold.2.9 Problem2.17Take the matrix element of (2.6.19)between 1[ and [1),with n = m and m > 1. The LHS yields,1[[Lm, Lm][1)= 1[LmLm[1)= |Lm[1)|2, (56)using(2.9.9). Alsoby(2.9.9), L0[1) = 0,soontheRHSweareleftwiththetermc12(m3m)1[1). (57)Hencec =12m3m|Lm[1)|21[1) 0. (58)3 CHAPTER3 213 Chapter33.1 Problem3.1(a) ThedenitionofthegeodesiccurvaturekofaboundarygiveninProblem1.3isk= nbtaatb, (1)where tais the unit tangent vector to the boundary and nbis the outward directed unit normal. Fora at unit disk, R vanishes, while the geodesic curvature of the boundary is 1 (since taatb= nb).Hence =12_20d= 1. (2)Fortheunithemisphere,ontheotherhand,theboundary isageodesic,whileR = 2. Hence =14_d2 g1/22 = 1, (3)inagreementwith(2).(b) If we cut a surface along a closed curve, the two new boundaries will have oppositely directednormals, so their contributions to the Euler number of the surface will cancel, leaving it unchanged.TheEulernumberoftheunitsphereis =14_d2 g1/22 = 2. (4)If wecutthespherealongbsmall circles, wewill beleftwithbdisksandaspherewithbholes.TheEulernumberofthedisksisb(frompart(a)),sotheEulernumber ofthespherewith bholesis = 2 b. (5)(c) AnitecylinderhasEulernumber0,sincewecanputonitagloballyatmetricforwhichtheboundariesaregeodesics. Ifweremovefromasphereb + 2gholes, andthenjointo2goftheholesgcylinders,theresultwillbeaspherewithbholesandghandles;itsEulernumberwillbe = 2 b 2g. (6)3.2 Problem3.2(a) This is easiesttoshow incomplexcoordinates,where gzz= g z z= 0. Contractingtwoindicesof asymmetrictensorwithlower indicesbygabwill pickoutthecomponentswhereoneof theindicesiszandtheother z. Ifthetensoristracelessthenallsuchcomponentsmustvanish. Theonlynon-vanishingcomponents arethereforetheonewithall z indices andtheonewithall zindices.3 CHAPTER3 22(b) Letva1anbeatracelesssymmetrictensor. DenePnby(Pnv)a1an+1 (a1va2an+1)nn + 1g(a1a2[b[vba3an+1). (7)Thistensorissymmetricbyconstruction, anditiseasytoseethatitisalsotraceless. Indeed,contractingwithga1a2,thersttermbecomesga1a2(a1va2an+1)=2n + 1bvba3an+1, (8)wherewehaveusedthesymmetryandtracelessnessofv,andthesecondcancelstherst:ga1a2g(a1a2[b[vba3an+1)=2n(n + 1)ga1a2ga1a2bvba3an+1+2(n 1)n(n + 1)ga1a2ga1a3bvba2a4an+1=2nbvba3an+1. (9)(c) Forua1an+1atracelesssymmetrictensor,denePTnby(PTn u)a1an buba1an. (10)Thisinheritsthesymmetryandtracelessnessofu.(d)(u, Pnv)=_d2 g1/2ua1an+1(Pnv)a1an+1=_d2 g1/2ua1an+1_a1va2an+1 nn + 1ga1a2bvba3an+1_= _d2 g1/2a1ua1an+1va2an+1=_d2 g1/2(PTn u)a2an+1va2an+1=(PTn u, v) (11)3.3 Problem3.3(a) Theconformal gaugemetricincomplexcoordinatesisgz z=g zz=e2/2, gzz=g z z=0.Connectioncoecientsarequicklycalculated:zzz=12gz z(zgz z +zg zz zgzz)=2, (12) z z z=2, (13)3 CHAPTER3 23allothercoecientsvanishing.Thisleadstothefollowingsimplicationintheformulaforthecovariantderivative:zTa1amb1bn=zTa1amb1bn+m

i=1aizcTa1camb1bnn

j=1czbjTa1amb1cbn=__ + 2m

i=1aiz2n

j=1zbj__Ta1amb1bn; (14)inotherwords, itcountsthedierencebetweenthenumber ofupper zindices andlowerzindices,while zindicesdonotenter. Similarly, zTa1amb1bn=__ + 2m

i=1ai z2n

j=1 zbj__Ta1amb1bn. (15)Inparticular,thecovariantderivativewithrespecttozofatensorwithonly zindicesisequaltoitsregularderivative,andviceversa:zT z z z z=T z z z z, zTzzzz=Tzzzz. (16)(b) As showninproblem3.2(a), theonlynon-vanishingcomponents of atraceless symmetrictensorwithloweredindiceshaveallthemzorallofthem z. If visann-index tracelesssymmetrictensor,thenPnvwillbean(n + 1)-indextracelesssymmetrictensor,andwillthereforehaveonlytwonon-zerocomponents:(Pnv)zz= zvzz=_12e2_nzv z z=_12e2_nv z z=( 2n)vzz; (17)(Pnv) z z=( 2n)v z z. (18)Similarly,if u is an (n+1)-index tracelesssymmetric tensor, thenPTn u will be an n-index tracelesssymmetrictensor,andwillhaveonlytwonon-zerocomponents:(PTn u)zz= bubzz= 2e2zu zzz2e2 zuzzz= _12e2_n1u z z z2e2uzz= 2e2uzz; (19)(PTn u) z z= 2e2u z z. (20)3 CHAPTER3 243.4 Problem3.4TheFaddeev-Popovdeterminantisdenedby,FP() __[d] _FA()__1. (21)Bythe gaugeinvarianceof themeasure [d] onthe gaugegroup,this is agauge-invariantfunction.Itcanbeusedtore-expressthegauge-invariant formulationof thepathintegral, witharbitrarygauge-invariantinsertionsf(),inagauge-xedway:1V_[d] eS1()f()=1V_[d] eS1()FP()_[d] _FA()_f()=1V_[d d] eS1()FP()_FA()_f()=_[d] eS1()FP()_FA()_f(). (22)Inthesecondequalityweusedthegaugeinvarianceof[d] eS1()andf(),andinthethirdlinewerenamedthevariableofintegration,.Inthelastlineof(22), FPisevaluatedonlyforonthegaugeslice,soitsucestondanexpressionforitthatisvalidthere. Letbeonthegaugeslice(soFA()=0), parametrizethegaugegroupneartheidentitybycoordinatesB,anddeneBFA() FA()B=0=FAiiB=0. (23)If theFAareproperlybehaved(i.e. if theyhavenon-zeroandlinearlyindependentgradientsat),andiftherearenogaugetransformationsthatleavexed,thenBFAwillbeanon-singularsquarematrix. IfwechoosethecoordinatesBsuchthat[d]=[dB] locally, thentheFaddeev-Popov determinant is precisely the determinant of BFA, and can be represented as a path integraloverghostelds:FP()=__[dB] _FA()__1=__[dB] _BFA()B__1=det_BFA()_=_[dbAdcB] ebABFA()cB. (24)Finally,we canexpress the delta function appearing in the gauge-xedpath integral (22)as a pathintegralitself:_FA()_ =_[dBA]eiBAFA(). (25)3 CHAPTER3 25Puttingitalltogether,weobtain(4.2.3):_[ddbAdcBdBA] eS1()bABFA()cB+iBAFA()f(). (26)3.5 Problem3.5For eacheldconguration, thereisauniquegauge-equivalent congurationFinthegaugeslicedenedbytheFA,andauniquegaugetransformationF()thattakesFto: =F()F. (27)For nearF, F() will be near the identity andcanbe parametrizedbyBF(), the samecoordinatesusedinthepreviousproblem. ForsuchwehaveFA() = BFA(F)BF(), (28)andwecanwritethefactorFFP()(FA())appearinginthegauge-xedpathintegral (22)intermsofBF():FFP()_FA()_=FFP(F)_FA()_=det_BFA(F)__BFA(F)BF()_=_BF()_. (29)DeningG()inthesameway,wehave,G_1GF()_ = F() . (30)Dening 1GF(), (31)itfollowsfrom(29)thatFFP()_FA()_ = GFP()_GA()_. (32)Itisnowstraightforwardtoprovethatthegauge-xedpathintegral isindependentofthechoiceofgauge:_[d] eS()FFP()_FA()_f()=_[d]eS()GFP()_GA()_f()=_[d] eS()GFP()_GA()_f().(33)Intherstlinewesimultaneouslyused(32)andthegaugeinvarianceof themeasure[d]eS()andtheinsertionf();inthesecondlinewerenamedthevariableofintegrationfromto.3 CHAPTER3 263.6 Problem3.7Let us begin by expressing (3.4.19)in momentum space,toknow what were aiming for. The Ricciscalar,tolowestorderinthemetricperturbationhab= gabab,isR (abab2)hab. (34)Inmomentumspace,theGreensfunctiondenedby(3.4.20)isG(p) 1p2(35)(againtolowestorderinhab),sotheexponentof(3.4.19)isa18_d2p(2)2hab(p)hcd(p)_papbpcpdp22abpcpd +abcdp2_. (36)Torstorderinhab,thePolyakovaction(3.2.3a)isSX=12_d2_aXaX + (12habhab)aXbX_, (37)whereh haa(wehaveset 2to1). We will usedimensional regularization, whichbreaksconformal invariance because the graviton trace couples to Xwhen d ,= 2. The traceless part of habinddimensions ishab= habh/d. This leavesacouplingbetweenhandaXaXwithcoecient1/2 1/d. Themomentum-spacevertexforhabishab(p)ka(kb +pb), (38)whilethatforhisd 22dh(p)k(k +p). (39)There are three one-loop diagrams with two external gravitons, depending on whether the gravitonsaretracelessortrace.We begin by dispensing with the hh diagram. In dimensional regularization, divergences in loopintegralsshowupaspolesinthedplane. Arisingastheydointheformof agammafunction,thesearealwayssimplepoles. Butthediagramismultipliedbytwofactorsofd 2fromthetwohvertices,soitvanisheswhenwetakedto2.The hhabdiagram is multiplied by only one factor of d2, so part of it (the divergent part thatwouldnormallybesubtractedo)mightsurvive. Itisequaltod 24d_ddp(2)dhab(p)h(p)_ddk(2)dka(kb +pb)k(k +p)k2(k +p)2. (40)Thekintegralcanbeevaluatedbytheusualtricks:_10dx_ddk(2)dka(kb +pb)k(k +p)(k2+ 2xpk +xp2)2(41)=_10dx_ddq(2)d(qaxpa)(qb + (1 x)pb)(q xp)(q + (1 x)p)(q2+x(1 x)p2)2.3 CHAPTER3 27Discardingtermsthatvanishduetothetracelessnessof haborthatareniteinthelimitd 2yieldspapb_10dx(12 3x 3x2)_ddq(2)dq2(q2+x(1 x)p2)2. (42)The divergent part of the q integral is independent of x, and the x integral vanishes, so this diagramvanishesaswell.We are left withjust thehabhcddiagram, which(includingasymmetryfactor of 4for theidenticalverticesandidenticalpropagators)equals14_ddp(2)dhab(p)hcd(p)_ddk(2)dka(kb +pb)kc(kd +pd)k2(k +p)2. (43)The usual tricks,plus the symmetry and tracelessness of hab, allow us to write the kintegralin thefollowingway:_10dx_ddq(2)d2d(d+2)acbdq4+1d(1 2x)2acpbpdq2+x2(1 x)2papbpcpd(q2+x(1 x)p2)2. (44)The q4andq2terms inthe numeratorgiverise todivergentintegrals. Integratingtheseterms overqyields18_10dx(1 d2)_x(1 x)p24_d/21(45)

_2dx(1 x)acbdp2+ (1 2x)2acpbpd_.Thedivergentpartofthisisacbdp22acpbpd24(d 2). (46)However, itisafactthatthesymmetricpartof theproductof twosymmetric, traceless, 22matrices is proportional to the identity matrix, so the two terms in the numerator are actually equalaftermultiplyingbyhab(p)hcd(p)weseethatdimensionalregularizationhasalreadydiscardedthedivergenceforus. Thenitepartof(45)is(usingthistrickasecondtime)acbdp28_10dx__ ln_x(1 x)p24__(12 3x + 3x2) x(1 x)_. (47)Amazingly, this alsovanishes uponperformingthexintegral. It remains onlytoperformtheintegralforthelastterminthenumeratorof(44),whichisconvergentatd = 2:_10dx_d2q(2)2x2(1 x)2papbpcpd(q2+x(1 x)p2)2=papbpcpd4p2_10dxx(1 x) =papbpcpd24p2. (48)Pluggingthisbackinto(43),wendforthe2-gravitoncontributiontothevacuumamplitude,196_d2p(2)2hab(p)hcd(p)_papbpcpdp2abpcpd +14abcdp2_. (49)3 CHAPTER3 28This result does not agree with (36), and is furthermore quite peculiar. It is Weyl invariant (sincethe trace h decoupled), but not di invariant. It therefore appears that, instead of a Weyl anomaly,we have discoveredagravitational anomaly. However, just because dimensional regularizationhas (rather amazingly) thrownawaythe divergent parts of the loopintegrals for us, does notmeanthatrenormalizationbecomesunnecessary. Wemuststillchooserenormalizationconditions,andintroducecounterterms tosatisfythem. Inthiscase, wewill imposediinvariance, whichismoreimportantthanWeyl invariancewithoutit, itwouldbeimpossibletocouplethisCFTconsistentlytogravity. Localityinrealspacedemandsthatthecountertermsbeofthesameformasthelasttwotermsintheparenthesesin(49). Wearethereforefreetoadjustthecoecientsof thesetwotermsinorder toachievediinvariance. Since(36) ismanifestlydiinvariant, itisclearlythedesiredexpression, witha1takingthevalue 1/12. (Itisworthpointingoutthatthereisnolocal countertermquadraticinhabthatonecouldaddthatisdiinvariantbyitself,andthatwouldthereforehavetobexedbysomeadditional renormalizationcondition. Thisisbecause di-invariant quantities are constructed out of the Ricci scalar, and_d2R2has the wrongdimension.)3.7 Problem3.9Fixcoordinatessuchthattheboundaryliesat2= 0. Followingtheprescriptionofproblem2.10for normal orderingoperators inthepresenceof aboundary, weincludeinthecontractiontheimageterm:b(, ) = (, ) + (, ), (50)where1= 1,2= 2. Ifandbothlieontheboundary, thenthecontractioniseectivelydoubled:b(1, 1) = 2_(1, 2= 0), (1, 2= 0)_. (51)If Tis a boundary operator, thenthe 2and2integrationsin the denition(3.6.5)of [T]rcanbedonetrivially:[T]r= exp_12_d1d1 b(1, 1)X(1, 2= 0)X(1, 2= 0)_T. (52)Equation(3.6.7)becomesW[T]r= [WT]r +12_d1d1Wb(1, 1)X(1)X(1)[T]r. (53)Thetachyonvertexoperator(3.6.25)isV0= go_2=0d1g1/211(1)[eikX(1)]r, (54)3 CHAPTER3 29anditsWeylvariation(53)isWV0=go_d1g1/211(1) ((1) +W) [eikX(1)]r=go_d1g1/211(1)_(1) k22Wb(1, 1)_[eikX(1)]r=(1 k2)go_d1g1/211(1)(1)[eikX(1)]r, (55)wherewehaveused(3.6.11)inthelastequality:Wb(1, 1) = 2W(1, 1) = 2(1). (56)Weylinvariancethusrequiresk2=1. (57)Thephotonvertexoperator(3.6.26)isV1= igo2e_2=0d1 [1X(1)eikX(1)]r. (58)Thespacetimegaugeequivalence,V1(k, e) = V1(k, e +k), (59)isclearfromthefactthatk1XeikXisatotal derivative. Theexpression(58)hasnoexplicitmetricdependence,sothevariationofV1comesentirelyfromthevariationoftherenormalizationcontraction:W[1X(1)eikX(1)]r=12_d1d1 Wb(1, 1)X(1)X(1)[1X(1)eikX(1)]r=ik1Wb(1, 1)1=1[eikX(1)]rk22Wb(1, 1)[1X(1)eikX(1)]r=ik1(1)[eikX(1)]rk2(1)[1X(1)eikX(1)]r, (60)whereinthelastequalitywehaveused(56)and(3.6.15a):1Wb(1, 1)1=1= 2 1W(1, 1)1=1= 1(1). (61)IntegrationbypartsyieldsWV1= i_2 go(ekkk2e)_d1(1)[1X(1)eikX(1)]r. (62)Forthis quantitytovanish forarbitrary (1) requires the vectorekk k2etovanish. This willhappenifeandkarecollinear,butby(59)V1vanishesinthiscase. Theotherpossibilityisk2= 0, ek= 0. (63)3 CHAPTER3 303.8 Problem3.11SinceweareinterestedintheH2term,letus assumeGtobe constant, tovanish,and BtobelinearinX,implyingthatH= 3[B](64)isconstant. Withthesesimplications,thesigmamodelactionbecomesS=14_d2 g1/2_GgabaXbX+iBabXaXbX_=14_d2 g1/2_GgabaXbX+i3HabXaXbX_.(65)InthesecondlinewehaveusedthefactthatabXaXbXistotallyantisymmetricin, , (uptointegrationbyparts)toantisymmetrizeB.Workinginconformalgaugeontheworldsheetandtransformingtocomplexcoordinates,g1/2gabaXbX=4X(X), (66)g1/2abaXbX= 4iX[X], (67)d2=12d2z, (68)theactionbecomesS=Sf+Si, (69)Sf=12G_d2z XX, (70)Si=16H_d2z XXX, (71)where wehavesplititintotheactionforafreeCFT andaninteractionterm. Thepathintegralisnow. . . )= eSi. . . )f= . . . )f Si. . . )f+12S2i. . . )f+, (72)where )fis thepathintegralcalculatedusing onlythefreeaction(70). The Weylvariationoftherst term givesrisetotheD26Weylanomalycalculatedinsection3.4,while thatofthesecondgives rise to the term in Bthat is linear in H(3.7.13b). It is the Weyl variationof the third term,quadraticinH,thatweareinterestedin,andinparticularthepartproportionalto_d2z : XX: . . . )f, (73)3 CHAPTER3 31whosecoecientgivestheH2terminG. Thisthirdtermis12S2i. . . )f=12(6)2HH (74)

_d2zd2z: X(z, z)X(z)X( z) :: X(z, z)X(z)X( z) : . . . )f,where we have normal-ordered the interaction vertices. The Weyl variation of this integral will comefrom the singular part of the OPE when z and z approach each other. Terms in the OPE containingexactlytwoXelds(whichwill yield(73)afterthezintegrationisperformed)areobtainedbyperforming twocross-contractions. There are18dierentpairsof cross-contractionsone canapplyto the integrand of (74),but, since they can all be obtained from each other by integration by partsandpermutingtheindices, , ,theyallgivethesameresult. Thecontractionderivedfromthefreeaction(70)isX(z, z)X(z, z) =: X(z, z)X(z, z) : 2 Gln[z z[2, (75)so,pickingarepresentativepairofcross-contractions,thepartof(74)weareinterestedinis182(6)2HH

_d2zd2z_2_G ln [z z[2_2_G ln [z z[2: X(z)X( z) : . . . )f= 1162HH_d2zd2z1[zz[2: X(z)X( z) : . . . )f. (76)TheWeyl variationof thistermcomesfromcuttingothelogarithmicallydivergentintegral of[zz[2nearz= z,sowecandropthelesssingulartermscomingfromtheTaylorexpansionofX( z):1162HH_d2z : X(z)X( z) : . . . )f_d2z1[zz[2. (77)Thedi-invariantdistancebetweenzandzis(forshort distances)e(z)[zz[,soadi-invariantcutowouldbeat [z z[ =e(z). TheWeyl-dependentpartof thesecondintegral of (77)isthen_d2z1[zz[2 2 ln(e(z)) = 2 ln + 2(z), (78)andtheWeylvariationof(76)is18HH_d2z (z): X(z)X( z) : . . . )f. (79)Using(66)and(68), andthefactthatthedierencebetween )and )finvolveshigherpowersofH(see(72))whichwecanneglect,wecanwritethisas116HH_d2 g1/2gab: aXbX: . . . ). (80)3 CHAPTER3 32Thisisoftheformof(3.4.6),withTaa=18HHgabaXbX(81)beingthecontributionof thistermtothestresstensor. Accordingto(3.7.12), Taainturncon-tributesthefollowingtermtoG:4HH. (82)3.9 Problem3.13If the dilaton is constant and D = d+3, then the equations of motion (3.7.15) become, to leadingorderin,R 14HH=0, (83)H=0, (84)d 2314HH=0. (85)Lettingi, j, kbe indices onthe 3-sphereand, , be indices onthe atd-dimensionalspacetime,weapplytheansatzHijk= hijk, (86)where h is a constantand is the volume form on the sphere, with all other components vanishing.(NotethatthisformforHcannotbeobtainedastheexteriorderivativeof anon-singulargaugeeldB; BmusthaveaDirac-typesingularitysomewhereonthesphere.) Equation(84)isthenimmediatelysatised, becausethevolumeformisalwayscovariantlyconstantonamanifold, soiHijk=0, andallothercomponentsvanishtrivially. Sinceijkijk= 6, equation(85)xeshintermsofd:h2=2(d 23)3, (87)implyingthattherearesolutionsonlyford>23. TheRicci tensorona3-sphereof radiusrisgivenbyRij=2r2Gij. (88)Similarly,ikljkl= 2Gij. (89)Most components of equation(83) vanishtrivially, butthosefor whichbothindicesareonthespherexrintermsofh:r2=4h2=6d 23. (90)4 CHAPTER4 334 Chapter44.1 Problem4.1Tobegin, letusrecall thespectrumoftheopenstringatlevel N=2inlight-conequantization.InrepresentationsofSO(D 2),wehadasymmetricrank2tensor,fiji1j1[0; k), (1)andavector,eii2[0; k). (2)Together,theymakeupthetracelesssymmetricrank 2tensorrepresentationofSO(D1),whosedimensionisD(D 1)/2 1. Thisiswhatweexpecttond.IntheOCQ,thegeneralstateatlevel2is[f, e; k) =_f11 +e2_[0; k), (3)atotalofD(D + 1)/2 +Dstates. Itsnormise, f; k[e, f; k)= 0; k[_f11+e2_ _f11 +e2_[0; k)=2_ff+ee_0; k[0; k). (4)ThetermsinthemodeexpansionoftheVirasorogeneratorrelevanthereareasfollows:L0=p2+1 1 +2 2 + (5)L1=2p1 +1 2 + (6)L2=2p2 +121 1 + (7)L1=2p1 +2 1 + (8)L2=2p2 +121 1 +. (9)Asinthecasesofthetachyonandphoton,theL0conditionyieldsthemass-shellcondition:0=(L01)[f, e; k)=(k2+ 1)[f, e; k), (10)orm2=1/, thesameasinthelight-conequantization. Sincetheparticleismassive, wecangotoitsrestframeforsimplicity: k0=1/, ki=0. TheL1conditionxeseintermsof f,removingDdegreesoffreedom:0=L1[f, e; k)=2_2fk+e_1[0; k), (11)4 CHAPTER4 34implyinge=2f0. (12)TheL2conditionaddsonemoreconstraint:0=L2[f, e; k)=_22ke+f_[0; k). (13)Using(12),thisimpliesfii= 5f00, (14)wherefiiisthetraceonthespacelikepartoff.ThereareD + 1independentspurious statesatthislevel:[g, ; k)=_L1g1 +L2_[0; k) (15)=_2g(k) +2_11[0; k) +_g +2k_2[0; k).Thesestatesarephysicalandthereforenullforg0= = 0. RemovingtheseD1statesfromthespectrum leavesD(D1)/2 states,the extra one with respect to the light-conequantizationbeingtheSO(D1)scalar,fij= fij, f00=D 15f, e0=2(D 1)5f, (16)withall othercomponentszero. (Stateswithvanishingf00mustbetracelessby(14), andthisistheunique statesatisfying(12)and(14)thatisorthogonaltoallofthese.) Thenorm ofthisstateisproportionaltoff+ee=(D 1)(26 D)f225, (17)positiveforD26. InthecaseD=26, thisstateisspurious, corre-spondingto(15) with=2f, g0=32f. Removingitfromthespectrumleaves uswiththestatesfij, fii=0, e=0preciselythetracelesssymmetricrank2tensorofSO(25)wefoundinthelight-conequantization.5 CHAPTER5 355 Chapter55.1 Problem5.1(a) Ourstartingpointisthefollowingformalexpressionforthepathintegral:Z(X0, X1) =_X(0)=X0X(1)=X1[dXde]Vdiexp (Sm[X, e]) , (1)wheretheactionforthemattereldsXisSm[X, e] =12_10d e_e1Xe1X +m2_(2)(where d/d). We have xed the coordinate range for to be [0,1]. Coordinate dieomorphisms: [0, 1] [0, 1],underwhichtheXarescalars,X() = X(), (3)andtheeinbeineisaco-vector,e() = e()dd, (4)leavetheaction(2)invariant. Vdiisthevolumeofthisgroupofdieomorphisms. Theeintegralin(1)runs overpositivefunctionson[0,1],andtheintegrall _10d e (5)isdieomorphisminvariantandthereforeamodulus;themodulispaceis(0, ).In ordertomakesense of the functionalintegralsin(1)wewill needtodene aninner producton the space of functions on [0,1],which will induce measures on the relevant function spaces. Thisinnerproductwilldependontheeinbeineinawaythatisuniquelydeterminedbythefollowingtwoconstraints: (1)theinnerproductmustbedieomorphisminvariant; (2)itmustdependone()onlylocally,inotherwords,itmustbeoftheform(f, g)e=_10d h(e())f()g(), (6)forsomefunctionh. Aswewill see, theseconditionswill benecessarytoallowustoregularizetheinniteproductsthat will ariseincarryingout thefunctional integrals in(1), andthentorenormalizethembyintroducingacounter-termaction, inawaythatrespectsthesymmetriesoftheaction(2). Forfandgscalars,theinnerproductsatisfyingthesetwoconditionsis(f, g)e _10d efg. (7)5 CHAPTER5 36Wecanexpressthematteraction(2)usingthisinnerproduct:Sm[X, e] =12(e1X, e1X)e +lm22. (8)We now wish to express the path integral (1) in a slightly less formal way by choosing a ducialeinbeinelfor eachpoint l inthemoduli space, andreplacingtheintegral over einbeins byanintegraloverthemodulispacetimesaFaddeev-PopovdeterminantFP[el]. DeningFPby1 = FP[e]_0dl_[d] [e el], (9)weindeedhave,bytheusualsequenceofformalmanipulations,Z(X0, X1) =_0dl_X(0)=X0X(1)=X1[dX] FP[el] exp (Sm[X, el]) . (10)To calculate the Faddeev-Popov determinant (9) at the point e = el, we expand e about elfor smalldieomorphismsandsmallchangesinthemodulus:elel+l= deldll, (11)whereisascalarfunctionparametrizingsmall dieomorphisms: =+ e1; torespectthexedcoordinaterange,mustvanishat0and1. Sincethechange(11)is,likee,aco-vector, wewillforsimplicitymultiplyitbye1linordertohaveascalar, andthenbringintoplayourinnerproduct(7)inordertoexpressthedeltafunctionalin(9)asanintegraloverscalarfunctions:1FP[el] =_dl[dd]exp_2i(, e1l e1ldeldll)el_(12)The integralisinvertedby replacingthe bosonicvariablesl,,andbyGrassmanvariables,c,andb:FP[el]=_d[dcdb]exp_14(b, e1lc e1ldeldl)el_=_[dcdb]14(b, e1ldeldl)el exp_14(b, e1lc)el_. (13)Wecannowwritethepathintegral(10)inamoreexplicitform:Z(X0, X1) =_0dl_X(0)=X0X(1)=X1[dX]_c(0)=c(1)=0[dcdb]14(b, e1ldeldl)el(14)exp (Sg[b, c, el] Sm[X, el]) ,whereSg[b, c, el] = 14(b, e1lc)el. (15)5 CHAPTER5 37(b) Atthispointitbecomesconvenienttoworkinaspecicgauge,thesimplestbeingel() = l. (16)Thentheinnerproduct(7)becomessimply(f, g)l= l_10d fg. (17)InordertoevaluatetheFaddeev-Popovdeterminant(13), letusdecomposebandcintonor-malizedeigenfunctionsoftheoperator = (e1l)2= l22: (18)b()=b0l+_2l

j=1bj cos(j), (19)c()=_2l

j=1cj sin(j), (20)witheigenvaluesj=2j2l2. (21)Theghostaction(15)becomesSg(bj, cj, l) = 14l

j=1jbjcj. (22)The zeromode b0does not enterintothe action,but itissingledoutby theinsertionappearing infrontoftheexponentialin(13):14(b, e1ldeldl)el=b04l. (23)TheFaddeev-Popovdeterminantis,nally,FP(l)=_

j=0dbj

j=1dcjb04lexp__14l

j=1jbjcj__=14l

j=1j4l=14ldet_162_1/2, (24)theprimeonthedeterminantdenotingomissionofthezeroeigenvalue.5 CHAPTER5 38(c) LetusdecomposeX()intoapartwhichobeystheclassicalequationsofmotion,Xcl() = X0 + (X1X0), (25)plusquantumuctuations; theuctuationsvanishat0and1, andcanthereforebedecomposedintothesamenormalizedeigenfunctionsofascwas(20):X() = Xcl() +_2l

j=1xjsin(j). (26)Thematteraction(8)becomesSm(X0, X1, xj) =(X1X0)22l+2l2

j=1j2x2j+lm22, (27)andthematterpartofthepathintegral(10)_X(0)=X0X(1)=X1[dX]exp (Sm[X, el])=exp_(X1X0)22llm22__D

=1

j=1dxjexp__2l2

j=1j2x2j__=exp_(X1X0)22llm22_det__D/2, (28)wherewehaveconvenientlychosentoworkinaEuclideanspacetimeinordertomakeall of theGaussianintegralsconvergent.(d) Putting together the results (10), (24),and (28),and dropping the irrelevant constant factorsmultiplyingtheoperatorintheinnite-dimensionaldeterminants,wehave:Z(X0, X1) =_0dl14lexp_(X1X0)22llm22__det_(1D)/2. (29)We will regularize the determinant of in the same way as it is done in Appendix A.1,by dividingbythedeterminantoftheoperator + 2:detdet( + 2)=

j=12j22j2+ 2l2=lsinhl2l exp (l) , (30)5 CHAPTER5 39wherethelastlineistheasymptoticexpansionforlarge. Thepathintegral(29)becomesZ(X0, X1) (31)=14(2)(D1)/2_0dl lD/2exp_(X1X0)22ll(m2(D 1))2_.The inversedivergenceduetothefactorof(1D)/2infrontoftheintegralcanbe dealtwithbyaeldrenormalization,butsincewewillnotconcernourselveswiththeoverallnormalizationofthepathintegral wewill simplydropall of thefactorsthatappearinfront. Thedivergencecomingfromthetermintheexponentcanbecancelledbya(dieomorphisminvariant)countertermintheaction,Sct=_10d eA = lA (32)Themassmisrenormalizedbywhatisleftoverafterthecancellationofinnities,m2phys= m2(D 1) 2A, (33)but for simplicity we will assume that a renormalization condition has been chosen that sets mphys=m.Wecannowproceedtotheintegrationovermodulispace:Z(X0, X1) =_0dl lD/2exp_(X1X0)22llm22_. (34)Theintegralismosteasilydoneafterpassingtomomentumspace:Z(k) _dDXexp (ikX) Z(0, X)=_0dl lD/2exp_lm22__dDXexp_ikX X22l_=_2_D/2_0dlexp_l(k2+m2)2_=_2_D/22k2+m2; (35)neglectingtheconstantfactors,thisispreciselythemomentumspacescalarpropagator.6 CHAPTER6 406 Chapter66.1 Problem6.1Intermsofu = 1/z,(6.2.31)isd(

ki)

i