Symmetry enriched U(1) quantum spin liquids
SSymmetry enriched U(1) quantum spin liquids
Liujun Zou,
1, 2
Chong Wang, and T. Senthil Department of Physics, Harvard University, Cambridge, MA 02138, USA Department of Physics, Massachusetts Institute of Technology, Cambridge, MA 02139, USA (Dated: May 17, 2018)We classify and characterize three dimensional U (1) quantum spin liquids (deconfined U (1) gaugetheories) with global symmetries. These spin liquids have an emergent gapless photon and emergentelectric/magnetic excitations (which we assume are gapped). We first discuss in great detail the casewith time reversal and SO (3) spin rotational symmetries. We find there are 15 distinct such quantumspin liquids based on the properties of bulk excitations. We show how to interpret them as gaugedsymmetry-protected topological states (SPTs). Some of these states possess fractional response toan external SO (3) gauge field, due to which we dub them “fractional topological paramagnets”. Weidentify 11 other anomalous states that can be grouped into 3 anomaly classes. The classificationis further refined by weakly coupling these quantum spin liquids to bosonic Symmetry ProtectedTopological (SPT) phases with the same symmetry. This refinement does not modify the bulkexcitation structure but modifies universal surface properties. Taking this refinement into account,we find there are 168 distinct such U (1) quantum spin liquids. After this warm-up we provide ageneral framework to classify symmetry enriched U (1) quantum spin liquids for a large class ofsymmetries. As a more complex example, we discuss U (1) quantum spin liquids with time reversaland Z symmetries in detail. Based on the properties of the bulk excitations, we find there are38 distinct such spin liquids that are anomaly-free. There are also 37 anomalous U (1) quantumspin liquids with this symmetry. Finally, we briefly discuss the classification of U (1) quantum spinliquids enriched by some other symmetries. Contents
I. Introduction 2II. U (1) quantum spin liquids enriched by timereversal and SO (3) spin rotational symmetries 4A. Quantum spin liquids with θ = 0 5B. Quantum spin liquids with θ = π SO (3) × T symmetry 7A. Anomalous states with θ = 0 7B. Anomalous state with θ = π π π/
2: Fractional TopologicalParamagnets 10C. Surface states 11V. Combining U (1) quantum spin liquids andbosonic SPTs under symmetry SO (3) × T U (1) quantum spin liquids 14A. Enumerate putative states 15B. Examine the anomalies 15C. Couple the spin liquids with SPTs 16 D. A formal framework 17VII. U (1) quantum spin liquids enriched by Z × T symmetry 17A. Z not acting as a charge conjugation 17B. Z acting as a charge conjugation 19C. Strategy of anomaly-detection 20D. Anomaly classes 21VIII. U (1) quantum spin liquids enriched by someother symmetries 22A. SO ( N ) × T symmetry 22B. SO ( N ) symmetry 23C. Z symmetry 23D. O (2) symmetry 23IX. Discussion 23X. Acknowledgement 24A. Some remarks on time reversal andcharge-conjugation symmetry 24B. An SPT: eCm U (1) (cid:111) T ) × SO (3) 272. Bosonic SPT with symmetry U (1) × T × SO (3) 27 a r X i v : . [ c ond - m a t . s t r- e l ] M a y
3. SPT with symmetry(( U (1) × SU (2)) /Z ) × T of fermions 274. SPT with symmetry(( U (1) (cid:111) T ) × SU (2)) /Z of Kramers singletfermions 28E. Relations between the classification of SO (3) × T symmetric U (1) quantum spinliquids and the classification of some relevantSPTs 29F. Bosonic SPTs with SO (3) × T symmetry 30G. Constraints on the Hall conductances due totime reversal and spin rotational symmetries 30H. Non-edgeability of some Z topological ordersin the presence of nontrivial particles 311. Brief review of the K -matrix theory 312. Non-edgeability of some Z topologicalorders in the presence of nontrivialparticles 32I. Projective representations: the electric(standard), the magnetic (twisted) and thedyonic (mixed) ones 331. Electric (standard) projectiverepresentations 342. Magnetic (twisted) projectiverepresentations 353. Dyonic (mixed) projective representations 35J. Examine the anomalies of Z × T symmetric U (1) quantum spin liquids with θ = 0 361. Constructions of the relevant free-fermionSPTs 372. Anomaly of ( E bT M bT (cid:48) ) − E bT T (cid:48) M b − Z × T symmetric U (1) quantum spin liquidswith θ = π and Z not acting as a chargeconjugation 49L. U (1) quantum spin liquids with some othersymmetries 501. O (2) × T symmetry 502. Z × Z × T symmetry 513. Z × Z symmetry 51References 51 I. Introduction
Symmetry and entanglement both play important rolesin understanding quantum phases of matter. It is by nowwell known that the ground states of quantum many-body systems may be in phases characterized by long-range entanglement between local degrees of freedom. Global symmetry may be realized in interesting waysin such long-range entangled phases. The simplest (andbest understood) cases are gapped topologically orderedquantum phases as exemplified by the fractional quan-tum Hall states. The long-range entanglement in thefractional quantum Hall ground state wavefunctions en-ables gapped quasiparticle excitations showing fractionalstatistics and fractional charge. The fractional statis-tics is a fundamental phenomenon that follows from thetopological order, while the fractional charge describesthe implementation of the global U (1) charge conserva-tion symmetry in this state.Another prototypical class of states that possess long-range entanglement are quantum spin liquid phases of in-sulating magnets. A wide variety of quantum spin liquidshave been described theoretically. Their universal low en-ergy physics is (in most known examples) described bya deconfined emergent gauge theory coupled to matterfields. In the presence of global symmetries it is nec-essary to also specify the symmetry implementation inthis low energy theory. Indeed two phases with the samestructure of long-range entanglement (eg, same low en-ergy gauge theory) can still be sharply distinguished bytheir symmetry implementations. This leads to a sym-metry protected distinction between symmetry unbrokenphases of matter (as is familiar from the theory of topo-logical band insulators).It is useful to distinguish two very broad classes ofspin liquids. The simplest and best understood are onesin which all excitations are gapped. These gapped spinliquids are topologically ordered - they have well definedquasiparticle excitations with non-local ‘statistical’ inter-actions, ground state degeneracies on topologically non-trivial manifolds, etc. Global symmetries can be imple-mented non-trivially in topologically ordered phases. Forinstance a symmetry may be fractionalized. Topologicalphases in the presence of global symmetries have beendubbed “Symmetry Enriched Topological” (SET) mat-ter. Thus symmetry protected distinctions between dif-ferent SET phases may be much more striking than intopological band insulators. Though much of the earlywork on spin liquids dealt with SET phases, it is onlyin the last few years that there has been tremendousand systematic progress in understanding their full struc-ture and classification in two dimensional systems[1–11].Some limited progress has been made for three dimen-sional SET phases as well[12–14]. A different broad classof spin liquid phases have gapless excitations. Theseare much less understood theoretically though they havetremendous experimental relevance.In this paper, we study a particularly simple class ofquantum spin liquids in three spatial dimensions (3D)with an emergent gapless photon excitation. Their lowenergy dynamics is described by a deconfined U (1) gaugetheory. Microscopic models for such phases were de-scribed in Refs 15–21. The emergence of the photon isnecessarily accompanied by the emergence of quasiparti-cles carrying electric and/or magnetic charges that cou-ple to the photon. We will restrict attention to phaseswhere these ‘charged’ matter fields are all gapped . Oneof our main focuses is on the realization of such U (1)quantum spin liquids in 3 D magnets with spin SO (3)and time reversal T symmetries. After warming up withthis example, we will describe a general framework toclassify symmetry enriched U (1) quantum spin liquidswith a large class of symmetries. Then we will applythis framework to the more complicated case where thesymmetry is Z × T . We will also briefly discuss such U (1) quantum spin liquids enriched by some other sym-metries. In previous work by two of us[30] (see also Ref.8) we described the various such phases when time rever-sal is the only global internal symmetry. The extension to SO (3) ×T , Z ×T and other symmetries is non-trivial andrequires some conceptual and technical advances whichwe describe in detail in this paper.For the case with SO (3) × T symmetry, we find thatthere are 15 families of such U (1) quantum spin liquidswhich may be distinguished by the symmetry realizationson the gapped electric/magnetic excitations. We describethe physical properties of these states. We will showthat there are two such quantum spin liquids which havea “fractional” response to a background external SO (3)gauge field. For this reason we dub them “FractionalTopological Paramagnets”. They are closely analogousto the fractional topological insulators discussed theoret-ically.Each of the 15 families is further refined when thequantum spin liquid phase is combined with a Symme-try Protected Topological (SPT) phase of the underlyingspin system protected by the same SO (3) × T symmetry.This does not change the bulk excitation spectrum butmanifests itself in different boundary properties. As de-scribed in our previous work[30] this refinement can benon-trivial: some but not all SPT phases can be “ab-sorbed” by the spin liquid and not lead to a new state ofmatter. Including this refinement we find a total of 168different such U (1) quantum spin liquids with SO (3) × T symmetry.For the case with Z × T symmetry, we find there are38 distinct of such U (1) quantum spin liquids based onthe properties of the bulk fractional excitations. We alsoobtain the classification for such spin liquids with someother symmetries.Studying symmetry enriched U (1) quantum spin liq-uids is of conceptual and practical importance not onlyfor quantum magnetism, but has far reaching connec-tions to many other topics in modern theoretical physics.First as emphasized in previous papers[30], there is a very The problem of gapless matter fields coupled to a (compact) U (1)gauge field is an interesting and extensively studied problem. Forsome representative papers from the condensed matter literaturesee Refs. 22–29. A full classification of such phases with gaplessmatter fields is beyond the reach of currently available theoreticaltools. useful connection to the theory of Symmetry ProtectedTopological (SPTs) insulators of bosons/fermions. It isvery helpful to view these U (1) quantum spin liquids asthe gauged version of some SPTs with a U (1) symme-try, i.e. these quantum spin liquids can be obtained bycoupling the relevant SPTs to a dynamical U (1) gaugefield. There are actually two distinct ways in which thesame U (1) QSL can be viewed as a gauged U (1) SPT -either as a gauged SPT of the electric charge or a gaugedSPT of the magnetic monopole. This leads to a general-ization of the standard electric-magnetic duality of threedimensional Maxwell theory which incorporates the real-ization of global symmetry [30–33]. In the presence of aboundary, this 3 + 1-dimensional “symmetry-enriched”electric-magnetic duality implies interesting and non-trivial dualities between 2 + 1-dimensional quantum fieldtheories[32–35]. This line of thinking has proven to bevery powerful in studying difficult problems in strongly-correlation physics in two space dimensions. Examplesinclude quantum hall systems, especially the half-filledLandau level[31, 36–40], interacting topological insulatorsurfaces[32, 34, 35], quantum electrodynamics in 2 + 1dimensions[33, 41] and a class of Landau-forbidden quan-tum phase transitions known as deconfined quantumcriticality[42]. The lower dimensional dualities are alsointeresting on their own as nontrivial results in 2 + 1 di-mensional quantum field theory[43–46]. Therefore, wediscuss in detail the relation between different symmetryenriched U (1) quantum spin liquids and various SPTs.The rest of the paper is organized as follows. In Sec.II, we enumerate all possible SO (3) × T symmetric U (1)quantum spin liquid states based on the properties oftheir bulk fractional excitations. However, we will findthat 11 of them are anomalous in the sense that thesestates cannot be realized in any three dimensional spinsystem with time reversal and SO (3) spin rotational sym-metries. We will present various ways of understandingthe 15 non-anomalous families. In particular, we de-scribe their physical properties and their construction asgauged SPTs. In Sec. III, we explain why the other 11states are anomalous. In Sec. IV we discuss the topo-logical response of the U (1) spin liquids to an SO (3)probe gauge field, which leads to the notion of “frac-tional topological paramagnets”. In Sec. V, we combinethe non-anomalous U (1) quantum spin liquids with 3Dbosonic SPTs with the same symmetry, and discuss howthe presence of the SPTs further enriches the classifica-tion of the quantum spin liquids. After warming up withthe example of SO (3) × T symmetric U (1) quantum spinliquids, in Sec. VI we describe a general framework toclassify symmetry enriched U (1) quantum spin liquids fora large class of symmetries. We will apply the generalframework to classify Z × T symmetric U (1) quantumspin liquids in Sec. VII, and to classify U (1) quantumspin liquids with some other symmetries in Sec. VIII.Finally, we conclude in Sec. IX. The appendices containsome supplementary details, and some contents there areinteresting and important, albeit rather technical. II. U (1) quantum spin liquids enriched by timereversal and SO (3) spin rotational symmetries We will start by considering systems of interactingspins on a lattice with SO (3) × T symmetries. The mi-croscopic Hilbert space thus has a tensor product struc-ture. Further all local operators in this Hilbert space willtransform under linear representations of the SO (3) × T symmetry ( i.e integer spin and Kramers singlet). Also,these local operators can only create bosonic excitations.Our goal is to classify and characterize U(1) quantumspin liquids that can emerge in such systems with thesimplifying assumption that only the emergent photon isgapless.A first cut understanding of the different possible such U (1) spin liquids is obtained by focusing on the prop-erties of the gapped matter excitations, such as theirstatistics and their quantum numbers under the relevantsymmetries[30]. In three dimensions, the statistics of par-ticles can be either bosonic or fermionic. Under time re-versal symmetry, they can be Kramers doublets or non-Kramers. Under SO (3), they can either be in a linearrepresentation (spin-1) or its projective representation(spin-1/2). Note that any excitation with integer spincan be reduced to spin-1 by binding local excitations ( i.e excitations created by local operators), and half-integerspin excitations can be similarly reduced to ones withminimal spin-1 /
2. Thus the only distinction is betweenlinear and projective realizations of the global symmetry.In the presence of time reversal symmetry, it is helpfulto integrate out the gapped matter fields and considerthe effective theory of the photon field. In general thiseffective theory has the form L eff = L Maxwell + θ π (cid:126)E · (cid:126)B (1)where L Maxwell represents the usual Maxwell Lagrangian,and the second term is of topological character. It iscustomary to define time-reversal transform such that theelectric charge is invariant, namely under time reversal (cid:126)E → (cid:126)E and (cid:126)B → − (cid:126)B . This definition also implies that θ → − θ under time-reversal. It is also known that θ is periodic with a period 4 π if the elementary electriccharge is a boson, or a period 2 π if the elementary electriccharge is a fermion (see Refs. 47 and 48 for argumentsfrom a condensed matter perspective). In all cases, thepossible electric and magnetic charges of excitations forma two-dimensional lattice, and there are only two distinctconfigurations of this charge-monopole lattice, i.e. θ =0 (mod 2 π ) and θ = π (mod 2 π ), as shown in Fig. 1 andFig. 2, respectively. For notational simplicity, we willdenote these two cases by θ = 0 and θ = π , respectively.Notice we take the normalization that the elementaryelectric charge is 1, and the minimal magnetic charge issuch that it emits 2 π flux seen by the elementary charge.It is natural to ask whether time-reversal can act on thecharge-monopole lattice in more complicated ways. Someexamples were discussed in Ref. 43, in which the charge- FIG. 1. Charge-monopole lattice at θ = 0 (mod 2 π ).FIG. 2. Charge-monopole lattice at θ = π (mod 2 π ). monopole lattices undergoes a rotation (also known as S -duality transform) under time-reversal. However, inthose examples the theories can be redefined, through ap-propriate electric-magnetic duality transforms, into theconventional form with the canonical time-reversal trans-form ( (cid:126)E → (cid:126)E and (cid:126)B → − (cid:126)B ). In general, such a re-definition should always be possible if the theory, whilepreserving time-reversal symmetry, has a weakly coupledlimit (with gauge coupling e (cid:28) q e and mag-netic charge q m by ( q e , q m ). When θ = 0, the lattice ofcharge-monopole excitations is generated by the two par-ticles (1 ,
0) which we denote E and (0 ,
1) which we denote M . Then the distinct possibilities for the statistics andquantum numbers of E and M will correspond to dis-tinct U (1) quantum spin liquids. Under time reversal anexcitation with nonzero magnetic charge is transformedto another excitation that differs from the original one bya nonlocal operation. It is then meaningless to discusswhether these excitations are Kramers doublet or not, be-cause T is not a gauge invariant quantity for them[8, 49].On the other hand, all the pure electric charges shouldhave well-defined T , and they are either Kramers singlet( T = 1) or Kramers doublet ( T = − θ = π , time reversal interchanges( ,
1) and ( , − E as the (1 ,
0) excitation, but we denote M as the (0 , A. Quantum spin liquids with θ = 0 We start with phases where θ = 0. Let us consider thedistinct possibilities for the E and M particles. Note thatU(1) quantum spin liquids with both E and M fermionicare anomalous, i.e., they cannot be realized in a strictlythree dimensional bosonic system but they can be re-alized as the surface of some four dimensional bosonicsystems.[49–51] We will therefore restrict to situations inwhich at most one of E and M is a fermion. Consider thecase where E is a boson. Naively then E may have SO (3)spin S = 0 or S = 1 /
2, and may be Kramers singlet ordoublet, while M may be either a boson or fermion, andmay have S = 0 or 1 /
2. This gives 16 distinct possibili-ties. If instead E is a fermion, it may again have S = 0or 1 /
2, and T = ± M must be a boson but mayhave S = 0 or 1 /
2, corresponding to 8 distinct possi-bilites. In total this gives 24 distinct possibilities for the E and M particles which each correspond to a distinctsymmetry enriched U (1) QSL (see Figure 3). Howeverwe will argue below that of these 10 are anomalous ( i.e the symmetry implementation is inconsistent in a strictly3 + 1-D system and is only consistent at the boundary ofa 4 + 1-dimensional SPT phase). We will discard theseso that there are only 14 distinct possibilities for the E and M particles at θ = 0. These will describe 14 distinctfamilies of U (1) QSLs. FIG. 3. Symmetry protected distinctions among symme-try enriched U (1) quantum spin liquids. For example, with SO (3) × T symmetries, two phases, E b M b and E bT M f inthis example, cannot be connected without crossing a phasetransition. When the symmetry is broken, they can be con-nected without crossing a phase transition. In Table I, we list these distinct possible families, andintroduce labels for them that we will use in the rest ofthe paper. The rest of this subsection will explain how toobtain these 14 spin liquids and Sec. III will show thatthe other 10 spin liquids are anomalous.Among the 14 quantum spin liquids, the 6 of them inwhich none of E or M carries spin-1/2 have been de-scribed in detail previously[30] . Below we demonstrate Here we just note that from the point of view of E , E bT M f can be viewed as a bosonic SPT with (( U (1) (cid:111) T ) /Z ) × SO (3)symmetry. On its surface there can be a symmetric Z topologi-cal order, where both e and m carry charge-1/2 under U (1) andspin-1 under SO (3). Interestingly, time reversal exchanges e and m , while their neutral bound state (cid:15) is a Kramers singlet. Thissurface state is labelled as ( eCmC ) T (cid:15) . T E S E S M comments E b M b E b M f eCmC , M: trivial E b M b E: eCm , M: trivial E b M f E: eCmC , M: trivial E bT M b -1 1 1 E: trivial, M: eCmTE bT M f -1 1 1 E: ( eCmC ) T (cid:15) , M: n=2 TSC E b M b eCm E bT M b -1 eCmT E bT M f -1
12 12 E: θ = 2 π , M: n=2 TSC E f M b eCmCE fT M b -1 1 1 E: trivial, M: eCmCTE f M b eCmC E f M b
12 12
E: n=2 TI, M: θ = 2 πE fT M b -1 eCmCT T E S E S M comments E b M b
12 12 anomalous (class I) E f M b anomalous (class I) E fT M b -1 1 anomalous (class II) E b M f E bT M f -1 E bT M b -1
12 12 anomalous (class II) E bT M f -1 1 anomalous (class III) E b M f
12 12 anomalous (class III) E bT M b -1 1 anomalous (class III) E fT M b -1
12 12 anomalous (class III)TABLE I. List of U(1) quantum spin liquids at θ = 0. Thesubscripts “b” and “f” refer to bosonic or fermionic statisticsof the associated particle, respectively. T E = 1 ( T E = − S E and S M refer to the spin of the corresponding particle under SO (3). In this table the spin liquids with both E and M fermions are not listed, because they are known to be anoma-lous. We identified ten more anomalous spin liquids, and theyare divided into three classes. More details can be found inthe main texts. how the other 8 can be constructed. Many of these spinliquids can be obtained simply. Specifically if either E or M is a trivial boson ( i.e has S = 0 and (for E particles) T E = 1), then the corresponding spin liquid is obtainedby gauging a trivial insulator of the other particle. Forinstance, to obtain E b M b , start with a trivial insulatorformed by bosons with S = 1 / U (1)charge that is even under time reversal. Coupling thischarge to a dynamical U (1) gauge field produces a quan-tum spin liquid which is precisely E b M b . If instead wewanted to obtain E b M b , we begin with a trivial insula-tor of a boson with S = 1 / U (1) chargethat is odd under time reversal. Gauging this insulatorproduces E b M b . This kind of construction clearly worksfor 6 of the 8 phases where one of E or M is a trivial bo-son while the other has S = 1 /
2. It is instructive toalso understand these phases from a different ‘dual’ per-spective where we will need to gauge the U (1) symmetryof some SPTs with symmetries that contain a U (1) sub-group. We explain this first below. This will also setthe stage to understand the two interesting remainingcases where neither E nor M is a trivial boson (these are E bT M f and E f M b ).1. E b M b From the point of view of M (that is, viewing M as the gauge charge), E b M b can be viewed asa gauged trivial bosonic insulator with symmetry(( U (1) × SU (2)) /Z ) × T . From the point of viewof E , it can be viewed as a gauged SPT with sym-metry ( U (1) (cid:111) T ) × SO (3), where the microscopicboson is a Kramers singlet. This SPT is denotedby eCm , which means that it can have a surfacetopologically ordered (STO) state with Z topologi-cal order, where the topological sectors, (1 , e, m, (cid:15) ),have e carrying charge-1/2 under the U (1) sym-metry and m carrying spin-1/2 under the SO (3)symmetry. This SPT is discussed in more detailsin Appendix B.2. E b M f From the point of view of M , E b M f can be viewedas a gauged trivial fermionic insulator with sym-metry (( U (1) × SU (2)) /Z ) × T . From the pointof view of E , it can be viewed as a gauged bosonicSPT with symmetry ( U (1) (cid:111) T ) × SO (3), wherethe microscopic boson is a Kramers singlet. We de-note this SPT by eCmC , which can be viewed asa combination of eCm and eCmC , a well-knownSPT with symmetry U (1) ×T or U (1) (cid:111) T .[8, 47, 48]In fact, eCmC is still a nontrivial SPT even if thereis additional SO (3) symmetry that commutes with U (1) × T or U (1) (cid:111) T .3. E b M b From the point of view of E , E b M b can be viewedas a gauged trivial bosonic insulator with (( U (1) (cid:111) T ) × SU (2)) /Z , where the microscopic bosons areKramers singlets. From the point of view of M ,it can be viewed as the gauged eCm , but withsymmetry U (1) × T × SO (3).4. E bT M b From the point of view of E , E bT M b can beviewed as a gauged trivial bosonic insulator with(( U (1) (cid:111) T ) × SU (2)) /Z , where the microscopicbosons are Kramers doublets. From the point ofview of M , it can be viewed as a gauged SPT eCmT under symmetry U (1) × T × SO (3). ThisSPT can be viewed as a combination of eCm and eCmT , another well-known SPT with symmetry U (1) × T .[8, 47] It can be shown that this is stilla nontrivial SPT even if there is additional SO (3)symmetry that commutes with U (1) × T . 5. E f M b From the point of view of E , E f M b can beviewed as a gauged trivial fermionic insulator with(( U (1) (cid:111) T ) × SU (2)) /Z , where the microscopicfermions are Kramers singlet. From the point ofview of M , it can be viewed as a gauged eCmC with symmetry U (1) × T × SO (3).6. E fT M b From the point of view of E , E fT M b can beviewed as a gauged trivial fermionic insulator with(( U (1) (cid:111) T ) × SU (2)) /Z , where the microscopicfermions are Kramers doublets. From the point ofview of M , it can be viewed as a gauged eCmCT under symmetry U (1) × T × SO (3). This SPT canbe viewed as a combination of eCm , eCmC and eCmT .We now turn to the last 2 cases E bT M and E f M b . As both the E and M are non-trivial inthese spin liquids, in both the electric and magneticpictures they should be viewed as gauged SPTs.We state their construction here and describe theirproperties in greater detail later. We will see thatthey should be viewed as “Fractional TopologicalParamagnets”.7. E bT M f From the point of view of M , E bT M f can beviewed as a gauged n = 2 topological supercon-ductor with symmetry (( U (1) × SU (2)) /Z ) × T .From the point of view of E , it can be viewed asa gauged bosonic θ = 2 π SPT under symmetry(( U (1) (cid:111) T ) × SU (2)) /Z , where the microscopicbosons are Kramers doublets.8. E f M b From the point of view of E , E f M b can beviewed as a gauged n = 2 topological insulator offermions with symmetry (( U (1) (cid:111) T ) × SU (2)) /Z ,where the microscopic fermions are Kramers sin-glets. From the point of view of M , it can be viewedas a gauged bosonic θ = 2 π SPT with symmetry(( U (1) × SU (2)) /Z ) × T . The properties of thisSPT is described in Ref. 39. B. Quantum spin liquids with θ = π Now we turn to U (1) quantum spin liquids with θ = π .At θ = π , the charge-monopole lattice is shown in Fig.2. Because time reversal symmetry exchanges ( , , − π mutual braid-ing statistics. This implies that E , the bound state of( ,
1) and ( , − S D comments( E fT M f ) θ E fT M f ) θ anomalous, class IITABLE II. List of U(1) quantum spin liquids at θ = π . S D =1 ( S D = ) represents the case where the ( ,
1) dyon carriesspin-1 (spin-1/2). fermion.[49] Also, because ( − ,
1) dyon is the antiparti-cle of ( , −
1) dyon, it has the same properties as ( , π braiding between ( ,
1) and ( − , M , is also a fermion that carries spin-1 and is non-Kramers. Similar thoughts imply that thestatistics and quantum numbers of ( ,
1) will determinethe statistics and quantum numbers of all gapped exci-tations. So the classification of U (1) spin liquids with θ = π is equivalent to the classification of the statisticsand quantum numbers of the ( ,
1) dyon.It is known that ( ,
1) must be a boson. [30, 49, 50] Un-der time reversal symmetry, T is not a gauge invariantquantity for ( , SO (3),it can carry either spin-1 or spin-1/2. We will denotethe former by ( E fT M f ) θ and the latter by ( E fT M f ) θ .These states are summarized in Table II B.( E fT M f ) θ has been described in detail in Ref. 30.From the point of view of E , it can be viewed as agauged free fermion topological insulator with symme-try ((( U (1) (cid:111) T )) /Z ) × SO (3), where the microscopicfermions are Kramers doublets. From the point of view of M , it can be viewed as a gauged n = 1 free fermion topo-logical superconductor with symmetry U (1) ×T × SO (3).In Sec. III, we will show that ( E fT M f ) θ is anoma-lous. III. Anomalous quantum spin liquids with SO (3) × T symmetry In the enumeration in Sec. II, 11 states are claimedto be anomalous, where 10 of them have θ = 0 and 1has θ = π . In this section we will provide arguments todemonstrate these anomalies. We start with the 10 with θ = 0. A. Anomalous states with θ = 0 The 10 anomalous quantum spin liquid states at θ = 0are grouped into three classes, such that within each classany one of them can be obtained by coupling another inthe same class and some non-anomalous quantum spinliquids. For illustration, let us demonstrate how to ob-tain E f M b by coupling E b M b and E f M b , a non-anomalous quantum spin liquid. To do this, one can cou-ple E b M b and E f M b , and condense the bound stateof the monopole of E b M b and the anti-monopole of E f M b . This bound state is a trivial boson, so condens-ing it will not break any symmetry. After this condensa-tion, the electric charge of E b M b and that of E f M b will be confined together, and the resulting bound stateis a fermion that carries no nontrivial quantum number.This is precisely E f M b .The above example shows the relation between the twoanomalous quantum spin liquids in class I. We denote thisrelation by E b M b E f M b ←−−−−→ E f M b (2)The relations among the two other classes are listed here:class II: E b M f E bT M f ←−−−→ E bT M fE bT M f ←−−−−−→ E bT M b E f M b ←−−−−→ E fT M b (3)class III: E b M f E bT M f ←−−−−−→ E bT M f E bT M f ←−−−→ E bT M b E f M b ←−−−−→ E fT M b (4)Because of these relations, given that the other 14quantum spin liquids can be realized in strictly three di-mensional bosonic systems, showing that any one of thestates of a certain class is anomalous is sufficient to showthe entire class is anomalous. Below we will show that E b M b , E bT M b and E bT M b are anomalous.States of matter that realize a global symmetry non-anomalously allow a consistent coupling of backgroundgauge fields. In our context a non-anomalous realiza-tion of SO (3) symmetry thus implies that we can consis-tently couple background SO (3) gauge fields. Converselyanomalous states can be detected by finding inconsisten-cies when such background gauge fields are turned on.Let us therefore couple our spin liquids to a probe SO (3) gauge field. Because π ( SO (3)) = Z , there aremonopole configurations of this SO (3) gauge field thatare classified by Z . [52]One explicit expression of a nontrivial SO (3) monopoleconfiguration is: A µ = A µ = 0 , A µ = A U (1) ,µ (5)where A µ = (cid:80) A aµ T a is the Lie algebra valued SO (3)gauge field with T a the generators. A U (1) ,µ is the U (1)gauge field configuration of a U (1) monopole.[52] Oneof the physical consequences of this SO (3) monopole is aBerry phase factor of an excitation going around a closedloop around it: exp (cid:18) i Ω2 S z (cid:19) (6)where Ω is the solid angle of the closed loop with respectto the monopole and S z is the representation of one ofthe generators of SO (3). For spin-1 particles, S z can betaken to be S zS =1 = diag(1 , , − S z can be taken to be S zS =1 / = diag(1 / , − / U (1) chargemoving around a U (1) monopole and using (5).Now consider a Dirac string that ends at thismonopole. According to (6), moving around an infinites-imal loop around the Dirac string, a spin-1 particle willget a unit phase factor, which seems normal. But a spin-1/2 particle will see a phase factor of −
1, which is unphys-ical. To cancel this phase factor, another defect that alsogives a − SO (3) monopole.We will denote such an SO (3) monopole (with this defectincluded) by M SO (3) .Next we argue that the defect trapped at another SO (3) monopole with A µ = A µ = 0 , A µ = − A U (1) ,µ (7)can be essentially the same as the one trapped at theprevious SO (3) monopole, M SO (3) . This is because thisnew SO (3) monopole can be obtained by performing on M SO (3) a π -rotation around any axis on the xy -plane.In the presence of SO (3) symmetry, the defect trappedby it should be the same as that trapped in M SO (3) upto a spin rotation. We denote this SO (3) (anti)monopole(with the same defect included) by M (cid:48) SO (3) . Notice if an M SO (3) and an M (cid:48) SO (3) are fused together, the SO (3)gauge field background will be cancelled, and what re-mains will be an excitation of the original system with-out any SO (3) gauge field. These imply that the de-fect trapped in these SO (3) monopoles can be viewed as“half” of an excitation of the quantum spin liquid.For the quantum spin liquid states E b M b and E bT M b , the Dirac string of a bare SO (3) monopolewill give any excitation with spin-1/2 a − q = q e − q m is odd. Forthese excitations, a ( Q e , Q m ) dyon with odd Q e and Q m will give rise to a phase 2 π ( Q e q m − Q m q e ) around an in-finitesimal loop around the Dirac string, so “half” of sucha ( Q e , Q m ) dyon will give a phase, which is an odd mul-tiple of π , that exactly cancels the − SO (3) monopole. One can also check this − Q e and Q m iseven.According to the argument above, fusing a M SO (3) with M (cid:48) SO (3) here should give rise to an ( Q e , Q m ) dyon,with both Q e and Q m odd integers. That is, M SO (3) × M (cid:48) SO (3) ∼ D ( Q e ,Q m ) (8)However, the ( Q e , Q m ) dyon is a fermion as long as both Q e and Q m are odd,[53] and this is inconsistent: M (cid:48) and M cannot have any nontrivial mutual Berry phase since they differ merely by a continuous gauge rotation, so thebound state of the two cannot be a fermion. Therefore,the above fusion rule is physically impossible. This showsthat all states in class I and class II are anomalous.The anomalies in the states discussed above do not in-volve time reversal symmetry in an essential way, but thisis not the case for E bT M b . For E bT M b , the analogousfusion rule we will obtain is M SO (3) × M (cid:48) SO (3) ∼ D ( Q e ,Q m ) (9)with an odd Q e and an even Q m . As Q m is even, wecan always bind − Q m / U (1) monopoles to M SO (3) and M (cid:48) SO (3) to cancel their magnetic charges. Thus Q m canbe taken to be zero in the above fusion rule. In this case,the time reversal partners of the M SO (3) (and M (cid:48) SO (3) )will differ from itself only by a local operator. This im-plies that they have a well-defined value for T . How-ever this is also seen to be impossible: first note that a( Q e ,
0) dyon with odd Q e is a Kramers doublet in thiscase and all microscopic degrees of freedom are Kramerssinglet. Suppose the fusion rule in Eqn. 9 is possible,then M SO (3) and M (cid:48) SO (3) should satisfy T = −
1. Theargument in Appendix A shows this is impossible unlessthere are microscopic Kramers doublets, which is absentby assumption. Therefore, E bT M b and hence all statesin class III are anomalous. B. Anomalous state with θ = π Now we show ( E fT M f ) θ is also anomalous. Thesimplest way to see this is to first ignore time rever-sal symmetry, then from the point of view of ( ,
1) and( , −
1) dyons, this spin liquid is just E b M b . We haveshown E b M b is anomalous with SO (3) symmetry aloneeven without using time reversal symmetry, this implies( E fT M f ) θ must be anomalous. Another way to see theanomaly is to notice the relation( E fT M f ) θ ( E fT M f ) θ ←−−−−−→ E fT M b (10)This also shows ( E fT M f ) θ is anomalous, and in thepresence of time reversal symmetry its anomaly belongsto class II.A more direct argument similar to the ones used abovegoes as follows. In this case, all ( q e , q m ) dyons with q e anhalf-odd-integer and q m an odd integer carry spin-1/2.This implies the following fusion rule M SO (3) × M (cid:48) SO (3) ∼ D ( Q e ,Q m ) (11)with Q e = 2 n and Q m = 4 m + 2, or Q e = 2 n + 1 and Q m = 4 m , where n and m are integers. One can checkthis dyon must be a fermion, which in turn shows thatthis spin liquid is anomalous. C. Some comments
The above arguments show that the 11 quantum spinliquids cannot be realized in strictly three dimensionsmade of bosons if the symmetry SO (3) × T is present.Careful readers may have noticed that the descendentstates of these anomalous states will still be anomalousif the symmetry is broken down to ( U (1) (cid:111) Z ) × T ∼ = O (2) ×T , where U (1) is the spin rotation around one axis,say, the z axis, and Z is a discrete π -spin rotation aroundan axis perpendicular to the z axis. In this case, we cancouple the system to a U (1) gauge field correspondingto the spin rotational symmetry around the z axis, then M SO (3) and M (cid:48) SO (3) become the monopoles of this U (1)gauge field, and the analogous equations of (8) and (9)still hold. These two monopoles are mapped into eachother by the Z transformation. Because this unitary Z transformation flips both S z , the spin component alongthe z direction, and the field value of the U (1) gauge fieldcorresponding to S z rotational symmetry, there is no mu-tual statistics between these two monopoles. Therefore,all previous arguments still apply. In fact, we conjectureeven if the symmetry is broken down to Z × Z × T , thedescendant states of these anomalous states will still beanomalous . In Appendix C we will show they can berealized as the surface of some four dimensional short-range entangled bosonic systems. In particular, four di-mensional bosonic SPT states with only SO (3) symmetrywere discussed in Ref. 54 using group cohomology, wherethe SPT states have a Z classification. This is consis-tent with our result: the only anomalous U (1) spin liquidwith SO (3) symmetry is E b M b .If these states were not anomalous, they could also beviewed as some gauged SPTs. So their anomalies implythe impossibilities of some SPTs, which is discussed inmore general terms in Sec. VI B. One such example isgiven in Appendix B.We would also like to mention that, although theanomalies of these states are shown by examining the SO (3) monopoles, an alternative argument independentof the SO (3) monopoles is sketched in Sec. VI.In passing, we notice that the fact that “half” of a dyonis confined by itself does not invalidate our arguments. Infact, the phenomenon where a defect is unphysical unlessit traps a confined object is familiar. The most familiarexample may be that in a conventional two dimensionalsuperconductor obtained by condensing charge-1 bosonicchargons from a spin-charge separated described by a Z gauge theory, a π -flux always appears with a vison, whichis confined by itself in the superconducting phase.[55] However, if the symmetry is broken down to U (1) × T , the de-scendants of all the anomalous states will become non-anomalous(see Sec. VIII A). IV. Fractional topological paramagnets
In this section we study the topological response ofthe spin liquid phases to an external SO (3) gauge fieldthat couples with the SO (3) spin degrees of freedom. Inparticular we show that the two phases E bT M f and E f M b exhibit nontrivial fractionalized topological re-sponse, due to which we dub them “fractional topologicalparamagnets”.We start with non-fractionalized (short-range entan-gled) bosonic phases with SO (3) × T symmetry, coupledwith a background SO (3) gauge field A µ . Since the bulkdynamics is trivial by assumption, we can integrate outall the bulk degrees of freedom and ask about the effec-tive response theory for the SO (3) gauge field A . Thesimplest topological response is a theta-term: S Θ = Θ16 π (cid:90) tr SO (3) F ∧ F, (12)where F is the SO (3) field strength. The normalizationis chosen so that if the SO (3) symmetry is broken downto U (1) ∼ SO (2), the term becomes a theta-term for the U (1) gauge field with familiar normalization.It is important to realize that the period of Θ is 4 π forpurely bosonic systems, in contrary to fermionic systemswhere the period is 2 π . In fact a bosonic short-rangeentangled phase with Θ = 2 π is a nontrivial SPT stateprotected by SO (3) × T . The physics behind is whatis known as the “statistical Witten effect”[48]: considerinserting a monopole configuration of A of the form ofEq. (5), we can ask about the SO (3) charge carried bythis monopole. But since the monopole configurationalready breaks the symmetry down to SO (2) ∼ U (1),we can only ask about the U (1) charge it carries. Thestandard Witten effect implies that the monopole carries U (1) charge q s = Θ / π = 1. We can bind a gauge chargeto the monopole to neutralize the gauge charge, but thisconverts the monopole to a fermion[53].The above argument also shows that for short-rangeentangled bosonic phases with SO (3) × T , the mini-mal nontrivial Θ-angle is 2 π since under time-reversalΘ → − Θ. However, it is also known that for long-rangeentangled (fractionalized) phases, time-reversal symme-try could be compatible with smaller Θ-angles[56, 57].This is because the effective period of Θ is reduced due tothe presence of fractionalized excitations. More formally,in the presence of emergent dynamical gauge fields, it ismore appropriate to integrate out only the gapped mat-ter fields and keep the low energy dynamics of the gaugefield explicit. The response theory is then correctly cap-tured by a Θ-term and a dynamical term˜ S Θ = Θ16 π (cid:90) tr SO (3) F ∧ F + S (cid:48) Θ [ a µ , A µ ] , (13)where the second term involves the dynamical gauge field a µ . It is this ˜ S Θ that has a reduced period of Θ. Wewill explain this point in more concrete examples at theend of Sec. IV B. However, to understand the physics,0it suffices to simply study the properties of an SO (3)magnetic monopole (the Witten effect) carefully – we willmainly focus on this approach here.We argue below, in the context of U (1) spin liquids,that the effective period of Θ is reduced to π when spin-1 / π/ π ). A. Triviality of
Θ = π First, we need to show that Θ = π is in some sense triv-ial if (and only if) there are spin-1 / E or M particle). Our argument proceeds by carefullystudying the Witten effect. Consider again a monopoleof A of the form of Eq. (5), denoted by M SO (3) . In gen-eral it could carry both the SO (2) charge q s = Θ / π =1 /
2, and the electric-magnetic charge of the dynami-cal U (1) gauge field ( q e , q m ). We denote this objectas M SO (3) = ( q e , q m , q s , q M ) = ( q e , q m , / , M SO (3) =( q e , − q m , − q s , q M ) = ( q e , − q m , − / ,
1) must also existin the spectrum, and it must have the same statistics with M SO (3) . One can think of ˜ M SO (3) as M SO (3) attachedwith a (0 , q m , ,
0) particle (which implies that this par-ticle should exist in the excitation spectrum). Noticethat if q m = 0, this attachment will change the statisticsof M SO (3) from boson to fermion (or vice versa), and M SO (3) cannot have the same statistics with ˜ M SO (3) – this is precisely why in the absence of fractionaliza-tion, Θ = π cannot be time-reversal symmetric for abosonic system. Now with nonzero ( q m , q e ), the issuecan be cured by another statistical transmutation if2 q e q m + S (0 , q m , , = 1 (mod 2) , (14)where S (0 , q m , , = 0 if the (0 , q m , ,
0) particle is aboson, and S (0 , q m , , = 1 if it is a fermion.Furthermore, any excitations in the (ungauged) U (1)spin liquid ( q (cid:48) e , q (cid:48) m , q (cid:48) s ,
0) should satisfy the general Diracquantization condition with respect to M SO (3) : q e q (cid:48) m − q m q (cid:48) e − q (cid:48) s = 0 (mod 1) . (15)The two conditions Eq. (14) and (15), together withthe existence of (0 , , ,
0) in the spectrum, stronglyconstrains the allowed values of ( q e , q m ) for M SO (3) and the allowed spectra of the U (1) spin liquids. Forexample, the ( E fT M f ) θ spin liquid could satisfy Eq. (14)with q e = 0 , q m = 1, but this choice inevitably violatesEq. (15) with q (cid:48) e = 1 / , q (cid:48) m = 1 , q (cid:48) s = 0. A related phase( E fT M f ) θ could satisfy all conditions since the test par-ticle for Eq. (15) should have q (cid:48) e = 1 / , q (cid:48) m = 1 , q (cid:48) s = 1 /
2– the problem is that this phase is anomalous andcannot be realized in three dimensions on its own.It can be seen after some careful examination, thatamong the anomaly-free U (1) spin liquids, only thosewith either E or M particle (but not both) carry-ing spin-1 / π . These include ( E b M b , E b M f , E b M b , E bT M b , E f M b , E fT M b ).The values of q e and q m for M SO (3) are chosen in thefollowing way: if E particle carries spin-1 /
2, then q e = 1(mod 2) and q m = 1 / M carries spin-1 / q e = 1 / q m = 1 (mod 2). This choiceis needed to satisfy the Dirac quantization conditionEq. (15). It is also easy to check that Eq. (14) is satisfied(for those states without anomaly).It is now easy to see why Θ = π should be consideredtrivial. In those spin liquids where E particles carry spin-1 /
2, we can bind an E particle to M SO (3) . This givesanother, equally legitimate, SO (3) monopole with q s = 0and q e = 0. We still have q m = 1 / / E particle which should be true regardless of what valueΘ takes. Therefore one can equivalently view this phaseas having Θ = 0 (mod 2 π ) (notice that both q s = 0and q e = 0 for the redefined monopole are importantto draw this conclusion). The argument is identical if M particles carry spin-1 / q s = 0 monopole is aboson or a fermion, but this is simply about whetherΘ = 0 or Θ = 2 π (mod 4 π ) – or whether a boson SPTstate has been stacked on top of the U (1) spin liquid. Wetherefore conclude that for a U (1) spin liquid, Θ = π istrivial. B. Θ = π/ : Fractional Topological Paramagnets We now argue that the two U (1) spin liquid phases E bT M f and E f M b effectively have Θ = π/
2, andhence can be called “Fractional Topological Paramag-nets”.Again we consider a monopole M SO (3) . In gen-eral it could carry both an SO (2) charge q s , andthe electric-magnetic charge of the dynamical U (1)gauge field ( q e , q m ). Since both the fundamental elec-tric and magnetic excitations ( E and M ) of the twospin liquids carry spin-1 /
2, according to the argu-ment in Sec. III A we require ( q e , q m ) = (1 / , /
2) for M SO (3) , up to integer shifts. We denote this objectas M SO (3) = ( q e , q m , q s , q M ) = (1 / , / , q s , M SO (3) =( q e , − q m , − q s , q M ) = (1 / , − / , − q s ,
1) must also existin the spectrum. Now take the anti-particle of ˜ M SO (3) and bind it together with M SO (3) , we get an object(0 , , q s , SO (3) gauge field, it must exist in the U (1)spin liquid phase before coupling to A SO (3) . But in thespin liquid phase any particle with q m = 1 and q e = 0must carry spin-1 /
2. Therefore 2 q s = 1 / q s = 1 / / π/ π ).One can also ask whether E bT M f and E f M b arethe only two ( T -invariant) U (1) spin liquids with Θ = π/
2. An argument similar to that in Sec. IV A for Θ = π U (1) spin liquidswith Θ = π/ E f M b as example (the logic will beparallel for the other state). This state can be obtainedfrom E f M b by putting the fermionic E particles intoa topological band. The corresponding surface state for E will have two Dirac cones – one for each spin. It iswell known[58] that this state, when coupled to an SU (2)gauge field, induces a theta-term for the SU (2) gaugefield at Θ SU (2) = π . This implies Θ = π/ SO (3)gauge field.The Witten effect is also easy to study in this picture:an SO (3) monopole M SO (3) is viewed by the spin-1 / E particles as a half-monopole. Therefore it should bindwith a magnetic charge q m = 1 / q m = 1 /
2. The monopole is then viewed as a q m = 1 monopole by the spin-up fermion f ↑ , and a q m = 0 objectviewed by the spin-down fermion f ↓ . Since each fermion(up or down) has one Dirac cone on the surface, similarto the usual topological insulator, the M SO (3) monopolewill trap half of the charge of an f ↑ fermion, which gives q e = 1 / q s = 1 /
4, in agreement with what wasobtained earlier using a direct argument.Alternatively, one can obtain the E f M b state from E b M b by putting the spin-1 / M into a bosonictopological insulating state. The result should be iden-tical, even though the bosonic state is harder to picturedue to the lack of non-interacting limit.We can make the picture slightly more precise by writ-ing down the response theory. We first consider the elec-tric picture, viewing the state as a gauged fermion SPT.This is the more convenient choice if the gauge couplingfor the Maxwell term e is weak. Integrating out thefermion matter field gives (on a general oriented mani-fold Y )˜ S Θ= π/ = π (cid:18) π (cid:90) tr SO (3) F ∧ F + 12 π (cid:90) f ∧ f + 12 · π (cid:90) tr R ∧ R (cid:19) , (16)where f = da is the field strength for the dynamicalgauge field, and R is the Riemann curvature tensor. Thefirst term comes from the Θ SU (2) = π response of thefermion topological band. The second and third termsare the U (1) and gravitational theta-terms induced bythe fermions. The gauge field strength f satisfies thecocycle condition (cid:90) (cid:18) fπ + w T M + w SO (3)2 (cid:19) = 0 (mod 2) , (17)where w T M is the second Stiefel-Whitney class of the tan-gent bundle on Y , w SO (3)2 is the second Stiefel-Whitneyclass of the SO (3) gauge bundle (physically it measuresthe Z -valued SO (3) monopole number and serves as anobstruction to lifting the gauge bundle to an SU (2) one),and the integration is taken on arbitrary 2-cycles on Y .The Maxwell term for f is suppressed in the above equa-tion for simplicity. Physically this cocycle condition sim-ply represents the fact that charge-1 objects under a µ must carry spin-1 / SO (3) symmetry andmust also be a fermion. When w T M is trivial, this re-quires an SO (3) monopole to be accompanied by a half U (1) magnetic-charge, a conclusion we have drawn pre-viously in less formal terms.To show that Eq. (16) is time-reversal invariant, weonly need to show that 2 ˜ S Θ= π/ is trivial (mod 2 π ). Thiswas shown explicitly in Ref. 42 (Sec. VII A therein). Thisalso provides an explicit example, in which Θ = π istrivial in the sense that ˜ S Θ= π = 2 ˜ S Θ= π/ is trivial.Similar result can also be obtained in the magnetic pic-ture (with an inverted Maxwell coupling e ). Integrating out the bosonic ( M ) degrees of freedom gives˜ S (cid:48) Θ= π/ = π (cid:18) π (cid:90) tr F ∧ F − π (cid:90) ˜ f ∧ ˜ f (cid:19) , (18)where the inverted sign of the U (1) theta-term and theabsence of the gravitational term is simply reflecting thefact that for a bosonic integer quantum hall state in twodimensions with U (2) = U (1) × SU (2) / Z symmetry, thespin and charge hall conductance are opposite in sign andthe net thermal hall conductance is zero[59, 60]. Thecocycle condition for the dual field strength is now (cid:90) (cid:32) ˜ fπ + w SO (3)2 (cid:33) = 0 (mod 2) . (19)Following an argument similar to that in Ref. 42(Sec. VII A therein), one can show that ˜ S (cid:48) Θ= π = 2 ˜ S (cid:48) Θ= π/ is trivial (mod 2 π ). Therefore the effective theory in themagnetic picture is also time-reversal invariant. C. Surface states
Perhaps the most striking property of a topological in-sulator is the presence of protected surface states. It isnatural then to ask about the physics at the surface ofthe Fractional Topological Paramagnets. Specifically weconsider an interface between the vacuum and a mate-rial in a Fractional Topological Paramagnet phase. Thegauged SPT point of view then makes it natural that2both E bT M f and E f M b have protected states atsuch an interface.Protected surface states for U (1) quantum spin liq-uids with time reversal were described in Ref. 30. Asdiscussed there, in states where both E and M are non-trivial (i.e not simply a boson transforming trivially un-der the global symmetry) the surface to the vacuum nec-essarily has protected states. Of the 15 families of U (1)quantum spin liquids with SO (3) × T , only E bT M f and E f M b therefore necessarily have protected sur-face states. In both these cases the parent SPTs (ei-ther in the E or M points of view) are such that thesurface exhibits the phenomenon of Symmetry EnforcedGaplessness, ı.e, there is no symmetry preserving gappedsurface even with topological order. Symmetry preserv-ing surfaces are necessarily gapless. For the FractionalTopological Paramagnets a gapless surface state is read-ily described from the fermion point of view. Both statesthen have 2 gapless surface Dirac cones (one for eachspin) that is coupled to the bulk U (1) gauge field. Timereversal acts differently on the surface Dirac fermions inthe two states (the time reversal is inherited from thaton the bulk fermionic quasiparticle). V. Combining U (1) quantum spin liquids andbosonic SPTs under symmetry SO (3) × T We have thus far described the distinct possible re-alizations of symmetry for the bulk excitations of U (1)quantum spin liquids with time reversal and SO (3) spinrotational symmetries. However, strictly speaking, thisis not the complete classification of such spin liquids. Wecan in principle obtain distinct spin liquids with the samesymmetry fractionalization patterns by simply combin-ing spin liquids with SPT states protected by the global SO (3) × T symmetry. This was demonstrated for timereversal invariant U (1) spin liquids in Ref. 30. Furtherit was shown that not all SPTs remain non-trivial whencombined with a spin liquid. In other words some SPTscan “dissolve” into some spin liquids without leading toa distinct state. Determining the distinct spin liquidsthat result when SPTs are combined with spin liquidsis a delicate but unavoidable task that is part of anyclassification of symmetry enriched spin liquids. In thissection we undertake this task for the SO (3) × T sym-metric U (1) QSLs of primary interest in this paper. Wewill show that each of the 15 families of such U (1) spinliquids described so far is further refined to give a totalof 168 distinct phases. We expect that this is the com-plete classification of U (1) QSLs enriched with SO (3) ×T symmetry.Bosonic SPTs with symmetry SO (3) × T are classi-fied by Z . The four root states all admit surface Z topological order { , e, m, (cid:15) } , with different assignmentsof fractional quantum numbers to the anyons e, m, (cid:15) (no-tice that e, m here denote the anyons in the 2 d surfacetopological order, which are not to be confused with E, M T e T m S e S m comments eT mT -1 -1 1 1 efmf e and m are fermions e m
12 12 e mT Z topological orders ofthe four root states of bosonic SPTs with symmetry SO (3) ×T . in earlier sections denoting electric and magnetic chargesin the 3 d bulk U (1) gauge theory). These surface Z topological orders realize symmetries in a way that is im-possible for a purely two dimensional system (see TableIII. More details can be found in Appendix F). The sur-face topological order provides a non-perturbative char-acterization of these SPTs; we therefore label the SPTsthemselves by their surface Z topological orders. Thefour root states generate in total 16 distinct SPTs, andeach can be viewed as a combination of some of the rootstates. For example, if ef mf and e m are taken as tworoot states, weakly coupling them produces a new SPTdenoted by ef mf ⊕ e m . In this example the notationof the state can be simplified because a surface phasetransition can be induced such that the bound state ofthe (cid:15) ’s from the surface ef mf and e m is condensed.This condensation will not change the bulk property, butthe surface now has Z topological order ef mf , whereboth the e and m are spin-1/2 fermions. So for simplicity ef mf ⊕ e m can be denoted by ef mf .Below in Sec. V A we use the same strategy as inRef. 30 to determine if these nontrivial SPTs are trivialor still nontrivial in the presence of the excitations inthe quantum spin liquids. Then in Sec. V B, we applythese results to obtain the enriched classification of U (1)quantum spin liquids combined with SPTs. A. SPTs in the presence of nontrivial excitations
Table IV and table V show whether the nontrivialSPTs are trivial or nontrivial in the presence of frac-tional excitations with all possible statistics and relevantquantum numbers. Below we explain the reasons for theentries of these tables. The notations that will be usedbelow are defined in the captions of these tables.
SPTs with component efmf always enrich the classificationof the quantum spin liquids
When time reversal symmetry is broken on its sur-face, ef mf has surface thermal Hall conductance κ xy =4 (mod) 8 in units of π k B h T . Thus it always enriches To characterize the SPT efmf more formallly, one can considerits response to a change of the background metric. Then this SPT C ˜ C C T C ˜ C C T eT mT × × √ × × × efmf × × × × × × e m × × × × × × e mT × × × √ × × efT mfT × × × × × × eT mT × × × × × × eT mT × × × × × √ ef mf × × × × × × efmf ⊕ e mT × × × × × × e mT × × × × × × efT mfT × × × × × × efT mfT ⊕ e mT × × × × × × eT mT ⊕ e mT × × × × × × ef mf ⊕ e mT × × × × × × ef mf ⊕ eT mT × × × × × × TABLE IV. Triviality of the root states of bosonic SPTs withsymmetry SO (3) × T in the presence of nontrivial bosonic ex-citaions. The rows represent the nontrivial SPT states, andthe columns represent the quantum numbers of the bosonicexcition. C means the elementary boson carries electriccharge 1, and ˜ C means it carries magnetic charge 1. Noticeelectric (magnetic) charge is even (odd) under time reversal. T means the elementary boson is a Kramers doublet, and means it carries spin- . A cross (hook) means the topologicalorder is anomalous (non-anomalous) in the presence of theexcitation from the quantum spin liquid. the classification of the quantum spin liquids[30, 47]. Thesame is true for all SPTs that are obtained by combining ef mf and other root states. Besides ef mf , these in-clude ef T mf T , ef mf , ef mf ⊕ e mT , ef T mf T , ef T mf T ⊕ e mT , ef mf ⊕ e mT and ef mf ⊕ eT mT . SPTs with a component Z topological order where both e and m carry spin-1/2 are anomalous in the presence of thenontrivial excitations These SPTs include e m , eT mT , ef mf , e mT , ef T mf T , eT mT ⊕ e mT , ef mf ⊕ e mT and ef mf ⊕ eT mT . In this case, the SO (3)Θ = 2 π (see Appendix F). But as discussed in Sec.IV, none of the spin liquids have Θ = 2 π . So couplingthese SPTs with spin liquids cannot change Θ from 2 π to 0 (mod 4 π ), and all these surface states will remainanomalous even when coupled to a spin liquid. Below is characterized by a bulk gravitational response term given by π (cid:82) tr R ∧ R , where R is the Riemann curvature tensor. In thisformal language, because none of the U (1) quantum spin liquidsdiscussed has a gravitational response term that can cancel thisone, this SPT cannot be “absorbed” by any of these spin liquids. More formally, this is characterized by
Θ16 π (cid:82) tr SO (3) F ∧ F , aresponse term to a background SO (3) gauge field, where F is the SO (3) field strength and Θ = 2 π for these states. we provide more physical reasoning to demonstrate theiranomalies in the presence of the excitations from the spinliquids.To see their anomalies, we can first assume that such astate can exist in a purely two dimensional system. Thenin the case where the nontrivial excitations carry quan-tum numbers C , ˜ C or C T , we can tunnel an SO (3)monopole through the system, which leaves a flux. Thisis a local process, but due to the spin-1/2 of e and m ,both e and m will see a − (cid:15) has to be generated in the processof tunneling the monopole. As shown in Appendix G,this process will not induce any polarization charge orspin because of the symmetry of the system. Therefore,this local process generates a single neutral and spinless fermion in the system, which is impossible and showsthese states are still anomalous even in the presence ofthe nontrivial excitations.When the quantum numbers of the nontrivial exci-tations are C , ˜ C or C T , tunneling an SO (3)monopole is no longer a local process, but tunneling abound state of an SO (3) monopole and half of a U (1)monopole is still local. Again, this process will generatea single neutral and spinless fermion in the system, whichis impossible and shows the anomalies of these states inthe presence of these nontrivial excitations. eT mT is anomalous in the presence of non-Kramers bosons It is known that the anomaly of eT mT only comes fromtime reversal symmetry, so the presence of bosons withquantum numbers C , ˜ C , C or ˜ C will not removethe anomaly. e mT and eT mT are anomalous in the presence ofnontrivial excitations with quantum numbers C , ˜ C , C T and ˜ C It turns out that e mT and eT mT are also stillanomalous in the presence of nontrivial excitations withquantum numbers C , ˜ C , C T or ˜ C . To see this,for the cases where excitations carry quantum numbers C , ˜ C or C T , we can again tunnel an SO (3) monopolethrough the system. This will leave a flux such that e and (cid:15) see a − m willhave to be generated in the process. As argued in Ap-pendix G, this process cannot induce any polarizationcharge or spin. Because the SO (3) flux background isinvariant under time reversal, such a local process gener-ates a Kramers doublet that carries no other nontrivialquantum number. But there are no such local degrees offreedom in these cases, so this is impossible. Thus these Z topological orders are still anomalous.If the excitations carry quantum number ˜ C , one can4tunnel a bound state of an SO (3) monopole and half of a U (1) monopole. Similar argument shows an m needs tobe produced in the process. Again, because both SO (3)and U (1) commute with T , the flux background left onthe system is time reversal invariant. This again showsthat a local process generates a Kramers doublet with noother nontrivial quantum number and thus it is impossi-ble. ( eT mT, bC T ) , ( e mT, bC ) , ( eT mT, bC T ) , ( e mT, fC T ) , ( eT mT, fC ) and ( eT mT, fC ) arenon-anomalous Denote eT mT in the presence of bosons with quantumnumber C T by ( eT mT, bC T ). It turns out this is non-anomalous. [30] To see this, we can attach a boson withquantum number C T to the e particle, then eT mT willbe relabelled as eC mT . This is a non-anomalous state.To construct it, one can first construct eC T , which isnon-anomalous because the topological order can be con-fined by condensing m without breaking any symmetry.Then putting the (cid:15) into a quantum spin Hall state makesit eC mT . [61, 62]Similarly, with parallel notations, ( e mT, bC ),( eT mT, bC T ), ( e mT, f C T ), ( eT mT, f C ) and( eT mT, f C ) are also non-anomalous. Other entries in table IV and V are anomalous
For other entries in Table IV and Table V, the argu-ments utilized above do not apply. However, they are stillexpected to be anomalous. Below we sketch the logic toshow this, and more details can be found in Appendix H.Suppose any of these Z topological orders is non-anomalous, that is, it can be realized in a purely twodimensional system, it must allow a physical edge, i.e. aboundary that separates this state and the trivial vac-uum. It is believed that the K -matrix formalism candescribe all two dimensional Abelian topological orders,and in particular, K -matrix theory naturally allows aphysical edge.[63] So if no K -matrix description of a Z topological order exists, it should not be edgeable, i.e.it is anomalous. We note the K -matrix formalism hasalready been applied to check edgeability or to classifySPTs and symmetry-enriched topological orders in theliterature.[1, 7, 8, 59]Indeed, in Appendix H we will show all other entriesare not edgeable. This implies they are still anomalous. B. Enriched classification of quantum spin liquidscombined with SPTs
In the previous subsection, we have shown when thenontrivial bosonic SPTs are in an environment with some C ˜ C C T C ˜ C C T eT mT √ × × × × × efmf × × × × × × e m × × × × × × e mT × × × × × √ efT mfT × × × × × × eT mT × × × × × × eT mT × × × √ × × ef mf × × × × × × efmf ⊕ e mT × × × × × × e mT × × × × × × efT mfT × × × × × × efT mfT ⊕ e mT × × × × × × eT mT ⊕ e mT × × × × × × ef mf ⊕ e mT × × × × × × ef mf ⊕ eT mT × × × × × × TABLE V. Trivialness of the root states of bosonic SPTs withsymmetry SO (3) × T in the presence of nontrivial fermionicexcitaions. The rows represent the nontrivial SPT states, andthe columns represent the quantum numbers of the fermionicexcition. C means the elementary fermion carries electriccharge 1, and ˜ C means it carries magnetic charge 1. Noticeelectric (magnetic) charge is even (odd) under time reversal. T means the elementary fermion is a Kramers doublet, and means it carries spin- . A cross (hook) means the topologicalorder is anomalous (non-anomalous) in the presence of theexcitation from the quantum spin liquid. nontrivial particles, which ones are still nontrivial andwhich ones become trivial. In most cases, the nontriv-ial SPTs remain nontrivial. Then each of the quantumspin liquids can become 2 = 16 distinct ones after beingweakly coupled with the bosonic SPTs. In the presenceof the excitations in the quantum spin liquids, the caseswhere nontrivial SPTs become trivial are when eT mT coupled with E bT M b , E bT M f or E f M b , when e mT cou-pled with E b M b or E fT M b and when eT mT coupledwith E bT M b , E bT M f , E f M b or E f M b . For thesequantum spin liquids, each can become 2 = 8 distinctones after weakly coupled with the bosonic SPTs. Allthese SPT-enriched quantum spin liquids are differentfrom each other. Therefore, when weakly coupled withbosonic SPTs with time reversal and SO (3) spin rota-tional symmetries, there are in total 6 ×
16 + 9 × U (1) quantum spin liquids. VI. A general framework to classify symmetryenriched U (1) quantum spin liquids The above discussion on the classification of SO (3) ×T symmetric U (1) quantum spin liquids provides a good ex-ample. In this section we describe a general frameworkto classify symmetry enriched U (1) quantum spin liquids.It involves three steps: enumerating all putative states,examining the anomalies of these states, and couplingthese states to 3D bosonic SPTs with the same symme-5try. This framework is physics-based. After discussingthis framework, we will briefly discuss a supplementaryformal approach to classify such states, which can be po-tentially more useful for thinking about these problemsmore abstractly. A. Enumerate putative states
We begin with the first step: enumerating all putativestates. As discussed earlier, different symmetry enriched U (1) quantum spin liquids are distinguished by the prop-erties of their excitations, and to determine the phase,we need to specify the statistics and symmetry quantumnumbers of the excitations.We start with the simpler case where the symmetry G is unitary and connected, that is, all elements in thesymmetry group are unitary and they can all be contin-uously connected to the identity element. In this case,the symmetry cannot exchange the type of the fractionalexcitations. Also, one can tune θ such that the charge-monopole lattice is of the θ = 0-type, and both E and M are bosons (this is shown more explicitly in the examplesin Sec. VIII B). To fully determine the properties of theexcitations, we just need to specify the symmetry quan-tum numbers of E and M . More precisely, we need to as-sign (projective) representations to E and M , which areclassified by the second group cohomology H ( G, U (1)).While doing this, we also need to keep in mind that E and M are equivalent in this case, so that, for example, E b M b and E b M b are the same SO (3) symmetric phase.When this is done, all putative states will be listed.Next we go to the more complicated case where thesymmetry is G × T (or, more generally, G (cid:111) T ), with G aconnected unitary group. Again, the elements in G willnot change the type of fractional excitations. However,time reversal will necessarily change some types of frac-tional excitations, and we will always take the conventionthat the emergent electric (magnetic) field is even (odd)under time reversal. Then there are two types of charge-monopole lattice, with θ = 0 and θ = π , respectively.Consider the states at θ = 0 first. Then the U (1)quantum spin liquids are classified by the statistics andsymmetry quantum numbers carried by E and M . As forstatistics, the only constraint at this point is just that E and M cannot both be fermions. Below we discuss thesymmetry quantum numbers, or in other words, projec-tive representations.Let us start from the case with θ = 0. Here we needto distinguish two types of projective representations:the electric (standard) one and the magnetic (twisted)one. The electric projective representations are applica-ble to E , and they are classified by H ( G × T , U T (1)),where G × T acts on the U (1) coefficient by taking thecomplex conjugate if the group element is anti-unitary.This is the standard classification of the projective rep-resentations of a group with anti-unitary elements. How-ever, another type of projective representations apply to M , which are classified by another group cohomology H ( G × T , U MT (1)) (see Appendix I for more details).This group cohomology differs from the standard one inthe group action on the U (1) factors, and this differencecomes from the convention that the magnetic (electric)field is odd (even) under time reversal. After assigningstatistics and symmetry quantum numbers to E and M as above, all putative G × T symmetric U (1) quantumspin liquids with θ = 0 will be listed.As for states with θ = π , the properties of all exci-tations are determined by the properties of the ( , ± H ( G × T , U D (1) × U D (1)). After assigning symmetry quantum numbers tothe (cid:0) , ± (cid:1) dyons, all putative G × T symmetric U (1)quantum spin liquids with θ = π will be listed.If the symmetry group is G or G × T , where G is uni-tary but not connected, the elements in G can also per-mute fractional excitations, as we will see in examplesbelow. The putative states in this more complicated sce-nario can be listed in a similar manner as above: one hasto fix the shape of the charge-monopole lattice, specifythe statistics of the relevant excitations, and specify thesymmetry quantum numbers of the relevant excitations.The first two steps are identical as the previous cases, butthe classification of symmetry quantum numbers will bemore complicated in this case, and in this paper we donot attempt to give a mathematical framework for thisstep, although it can be done in a case-by-case manner. B. Examine the anomalies
After enumerating all putative symmetry enriched U (1) quantum spin liquids, we need to examine whichones are anomalous. A general way of doing this is toconsider whether the corresponding SPT of this spin liq-uid can exist. Denote the symmetry of the U (1) quantumspin liquid by G , which is supposed to be a completelygeneral on-site symmetry in this subsection (it can con-tain anti-unitary elements, and its unitary elements donot need to be continuously connected with identity). Bythe corresponding SPT of a spin liquid, we mean an SPTprotected by a U (1) central extension of G , which be-comes this spin liquid once this U (1) symmetry is gauged.Clearly, by definition, as long as this SPT can exist, thespin liquid state must be anomaly-free. If this SPT is in-trinsically inconsistent, then the corresponding spin liq-uid state must be anomalous. To see this, suppose this In the more standard terminology, this U (1) central extensionof G is the projective symmetry group of G , which was firstintroduced in Ref. 64. U (1) gauge field, and the resulting statewill be precisely the corresponding SPT of the spin liquidstate (an example is shown in Figure 4).[31] This leadsto a contradiction to the original assumption that suchSPT is problematic. Therefore, a sufficient and neces-sary condition for a symmetry enriched U (1) quantumspin liquid to be anomaly-free is that its correspondingSPT is consistent. FIG. 4. There is a systematic ungauging procedure that takesa symmetry enriched U (1) quantum spin liquid to its corre-sponding SPT. Consider the time reversal symmetric U (1)quantum spin liquid ( E fT M f ) θ (the upper left system), andwe will try to get its corresponding SPT from the perspec-tive of the electric charge. To do this, we first introducean auxiliary trivial time reversal symmetric insulator madeof fermions that are Kramers doublets, where these fermionsare denoted by c (the lower left system). Next we condensethe bound state of E , the electric charge of the U (1) spin liq-uid, and c † , the holes in the auxiliary trivial insulator. Thisbound state is a boson and a Kramers singlet, so this conden-sation will preserve the time reversal symmetry. The dynami-cal U (1) gauge field in the U (1) gauge theory will be confined,and the resulting state is precisely the Fu-Kane-Mele topolog-ical insulator,[65] which, viewed from the perspective of theelectric charge, is the corresponding SPT of ( E fT M f ) θ (theright system). How do we check whether the corresponding SPT isconsistent? One way is to consider whether it has a con-sistent surface state. This condition - known as “edge-ability” - was defined in Ref. 8. Assuming such SPT isconsistent, one can first condense certain charges on the surface of this SPT and get a surface superfluid. Thenone can try to condense certain vortices to restore thesymmetry on the surface. If the symmetric surface stateis consistent (but possibly anomalous), one can build upthe three dimensional bulk SPT (for example, by a layerconstruction or a Walker-Wang type construction). If theputative symmetric surface state is inconsistent, then thisSPT is inconsistent, because it has an invalid edge state.In summary, a systematic physical way to examinewhether a putative symmetry enriched U (1) quantumspin liquid is anomalous is to check whether its corre-sponding SPT can have a legitimate surface state. If so,this spin liquid state is non-anomalous. Otherwise, it isanomalous. These relations is sketched in Figure 5. FIG. 5. That the G symmetric U (1) quantum spin liquidis anomaly-free is equivalent to that it has a correspondingSPT, which is in turn equivalent to that this SPT can have aconsistent (but possibly anomalous) 2D surface state. This method of anomaly detection applies to any sym-metry enriched U (1) quantum spin liquids, but for someparticular cases, there are more physical ways of doingit by focusing on the spin liquid state itself, instead ofits corresponding SPT. For example, we have used the SO (3) monopole to detect the anomaly of some putative SO (3) symmetric spin liquid states in Sec. III. However,for some more subtle cases, the anomalies are examinedby considering the corresponding SPTs. Some examplesare given in Sec. VII, where Z × T symmetric states arediscussed. C. Couple the spin liquids with SPTs
The above two steps classify symmetry enriched U (1)quantum spin liquids in terms of the properties of thebulk excitations. To complete the classification of thesymmetry enriched U (1) quantum spin liquids, one hasto consider coupling these spin liquids and 3D bosonicSPTs with the same symmetry. In general, when an SPTis coupled with a U (1) spin liquid, the result is still a U (1) spin liquid with the bulk fractional excitations car-rying the same symmetry fractionalization pattern, butthe new state can have a different type of surface com-pared to the original one, due to the nontrivial surfaceof the SPT. Therefore, one has to check if this SPT canbe “absorbed” by the U (1) spin liquid. Physically, thisamounts to checking if the nontrivial surface of the SPTremains nontrivial if it is coupled with the bulk excita-7tions in the spin liquid. Examples of such excercises aregiven in Sec. V. D. A formal framework
We would like to close this section by briefly discussinga more formal approach to classify symmetry enriched U (1) quantum spin liquids. In this formal approach, theproblem amounts to classifying the action, or more pre-cisely, the universal part of the partition function, of the U (1) gauge theories. To encode the information aboutsymmetries, in this action the U (1) gauge field shouldbe coupled to a background gauge field corresponding tothe symmetry and a background spacetime metric. If theglobal symmetry includes time reversal the equivalent ofcoupling a background gauge field is to place the theoryon an unorientable space-time manifold.Note that we are considering spin liquids that arise ina UV system made out of bosons. To impose this re-striction directly in the low energy continuum theory wedemand that the low energy theory can be consistentlyformulated on an arbitrary non-spin space-time manifold.On an orientable manifold, this is achieved by requiringthat the emergent gauge field be either an ordinary U (1)gauge field (when the emergent electric charge E is aboson) or that it is a Spin c connection (when E is afermion). On an unorientable manifold, there is a gener-alization of a Spin c connection known as Pin c ± connec-tions - the ± sign correspond to the two possibilities that E is non-Kramers or Kramers under time reversal (moredetail is in Ref. 32). We will not make an explicit dis-tinction in the schematic discussion below between thesedifferent kinds of U (1) connections.Denote the U (1) gauge field by a , the gauge field cor-responding to the global symmetry by b , and the back-ground metric by g . In general, the action can be writtenin a form S [ a ; b, g ] = S U(1) [ a ] + S SPT [ b, g ] + S mixed [ a ; b, g ] (20)The first term contains the Maxwell action and the θ term of a U (1) gauge field, and it is present in generalfor a U (1) quantum spin liquid and are independent ofsymmetries. The third term, S SPT [ b, g ], depends onlyon b and g . This term physically describes 3D bosonicSPTs with the same symmetry as the U (1) spin liquid,and adding it into the action means coupling a U (1) spinliquid and a bosonic SPT with the same symmetry. Asdiscussed before, this will potentially change the system A Spin c connection differs from an ordinary U (1) gauge fieldthrough a modification of its flux quantization condition: the cur-vature F of a Spin c connection satisfies (cid:82) F π = (cid:82) w TM ( mod Z )on oriented 2-cycles where w TM is the second Stieffel-Whitneyclass of the tangle bundle of the manifold. For more detail seeRefs. 32 and 43 and references therein. into a different U (1) spin liquid. In order to see if such anSPT can be “absorbed” into a U (1) spin liquid, one needsto check if the universal part of the partition function willchange due to the presence of this term. The last term, S mixed [ a ; b, g ], only involves terms that couple a with b and/or g . This term encodes the information about sym-metry fractionalization on the bulk excitations.In general, such an action is constrained by gauge in-variance. In addition, certain constraints on these fieldsmay apply analogous to the modification of the flux quan-tization condition for Spin c connections when E is afermion. For example, for fractional topological para-magnets, there is a constraint on such fields given by(17). To classify symmetry enriched U (1) quantum spinliquids, one can first write down all possible such actionsand then classify the resulting universal part of the par-tition function. We leave this for future work. VII. U (1) quantum spin liquids enriched by Z × T symmetry In this section we apply the above general frameworkto classify U (1) quantum spin liquids enriched by Z × T symmetry. This symmetry can be relevant for experi-mental candidates of quantum spin liquids made of non-Kramers quantum spins, i.e. for example, two-level sys-tems made of m z = ± S z and Z acts as a π spin rotationaround the x axis. Below we will first list all putativestates, including the anomalous ones. Then we will ex-amine the anomalies of these states. We will leave theproblem of coupling these spin liquids with SPTs for fu-ture work.It turns out there are two types of Z actions thatdeserve separate discussions. In the first type, the Z symmetry does not change one type of fractional exci-tation to another. More precisely, the electric chargeand magnetic monopole will both retain their charac-ters under this type of Z action. In the second type,the Z symmetry changes the fractional excitations. Inparticular, it can change the electric charge into the anti-electric charge, and at the same time change the magneticmonopole into the anti-magnetic monopole. This type of Z action is physically a charge conjugation. One maywonder whether it is possible to change an electric chargeinto a magnetic monopole, but Ref. 66 pointed out thisis impossible in a strictly 3D system.Below we will discuss these two types of Z actions inturn. A. Z not acting as a charge conjugation We start from the case where Z does not act as acharge conjugation, that is, it does not change a type offractional excitation to another type.We will begin with the simpler case that has θ = 0. In8 T E T (cid:48) E [ T , Z ] M E b M b E bT M b − E bT (cid:48) M b − E bTT (cid:48) M b − − E b M b − − E f M b E fT M b − E fT (cid:48) M b − E fTT (cid:48) M b − E fTT (cid:48) M b − − − − E b M f E bT M f − E bT (cid:48) M f − E bTT (cid:48) M f − − E b M f − − TABLE VI. List of non-anomalous Z × T symmetric U (1)quantum spin liquids that have θ = 0 and have Z not actingas a charge conjugation. All these states are anomaly-free. T E = 1 ( T E = −
1) represents the case where E is a Kramerssinglet (doublet) under T . T (cid:48) E = 1 ( T (cid:48) E = −
1) representsthe case where E is a Kramers singlet (doublet) under T (cid:48) .[ T , Z ] M = + ([ T , Z ] M = − ) represents the case where Z and T commute (anti-commute) on M . this case, to classify the quantum numbers of the electriccharge, it is appropriate to look at the projective repre-sentations of Z × T , which are classified by Z , wherethe nontrivial projective representations can be viewed asbeing a Kramers doublet under the original time rever-sal and/or under a new anti-unitary symmetry T (cid:48) , whosegenerator is the product of the generator of Z and thegenerator of T . Although it is not meaningful to talkabout whether the magnetic monopole is a Kramers sin-glet or doublet under T or T (cid:48) , there are still two typesof quantum numbers of the magnetic monopole under Z × T : on the monopole the Z and T can either com-mute or anti-commute. This relation between Z and T is gauge invariant for the monopole, but not gaugeinvariant for the charge.Therefore, we can make a list of putative U (1) quan-tum spin liquids with this type of symmetry, and thereare 3 × × θ = π , we note thatone point deserves immediate clarification. That is, one If T and Z commute (anti-commute), T (cid:48) defined above will alsocommute (anti-commute) with Z . T E T (cid:48) E [ T , Z ] M anomaly class E bT (cid:48) M f − − − class a E fT M b − − − class a E bT (cid:48) M b − − − class a E fT (cid:48) M b − − − class b E bT M f − − − class b E bT M b − − − class b E bTT (cid:48) M f − − − − class c E bTT (cid:48) M b − − − − class c E f M b − − class cTABLE VII. List of anomalous Z × T symmetric U (1) quan-tum spin liquids that have θ = 0 and have Z not actingas a charge conjugation. All these states are anomaly-free. T E = 1 ( T E = −
1) represents the case where E is a Kramerssinglet (doublet) under T . T (cid:48) E = 1 ( T (cid:48) E = −
1) representsthe case where E is a Kramers singlet (doublet) under T (cid:48) .[ T , Z ] M = + ([ T , Z ] M = − ) represents the case where Z and T commute (anti-commute) on M . The last column in-dicates the anomaly classes. may wonder, for example, whether E bT M b and E bT (cid:48) M b are truly distinct, since they are related to each other byrelabelling T ↔ T (cid:48) and E ↔ M . At the first glance,these two states indeed seem to have identical physicalproperties when examined on their own. However, oncethe definitions of T and T (cid:48) are fixed, these states aredistinct. One physical way to see this is to consider thetwo states at the same time, clearly without breakingeither T or T (cid:48) , one state cannot be connected to anotherwithout encountering a phase transition. Therefore, allthese 24 states are truly distinct.Now we turn to the states with θ = π . In this case,the quantum number of the (cid:0) , (cid:1) dyon determines thequantum numbers of all other dyons. However, the (cid:0) , (cid:1) dyon does not have any projective representation of the Z ×T symmetry, so there is only one state: ( E fT T (cid:48) M f ) θ ,as described in Table VIII. The electric charge has tobe Kramers doublet under both T and T (cid:48) , because itis a bound state of the (cid:0) , (cid:1) and (cid:0) , − (cid:1) dyons, whichhave π mutual braiding and are exchanged under both T and T (cid:48) . Naively, the M particle (the (0 ,
2) dyon in thiscontext) can either have Z and T commuting or anti-commuting. But it turns out the latter possibility canbe ruled out, as shown in Appendix K. This ( E fT T (cid:48) M f ) θ state can be viewed as a descendant of the SO (3) × T symmetric ( E fT M f ) θ , so it is anomaly-free. T E T (cid:48) E [ T , Z ] M ( E fTT (cid:48) M f ) θ − − Z × T symmetric U (1) quantum spin liq-uids that have θ = π and have Z not acting as a chargeconjugation. This state is anomaly-free. T E = 1 ( T E = − E is a Kramers singlet (doublet)under T . T (cid:48) E = 1 ( T (cid:48) E = −
1) represents the case where E is a Kramers singlet (doublet) under T (cid:48) . [ T , Z ] M = +([ T , Z ] M = − ) represents the case where Z and T commute(anti-commute) on M . Z does not change one type of frac-tional excitation into another type, there are 16 distinctanomaly-free Z × T symmetric U (1) quantum spin liq-uids. B. Z acting as a charge conjugation Now we turn to the more complicated case where the Z symmetry acts as a charge conjugation. Let us firstpause to lay out the principle of organizing these states.Let us focus on the case with θ = 0 for the moment.In this case, it is meaningful to discuss whether E is aKramers doublet under the original time reversal T , andwhether M is a Kramers doublet under T (cid:48) . Also, noticenow it is also meaningful to ask whether Z squares to+1 or − E and M (see Appendix A for moredetails). We will use ( · · · ) − to indicate that Z acts as acharge conjugation, and a subscript Z to represent thatcertain excitation has Z squaring to −
1. For example,( E fT M bT (cid:48) Z ) − means Z flips both the electric charge andmagnetic charge, and E is a fermionic Kramers doubletunder T , while M is a boson where Z squares to − M is also a Kramers doublet under T (cid:48) .With this notation, we can list all 3 × × = 48possible distinct states with θ = 0 and Z acting as acharge conjugation, and they are shown in Table IX andTable X. T E Z E T (cid:48) M Z M ( E b M b ) − E bZ M b ) − − E bT M b ) − − E bTZ M b ) − − − E b M bZ ) − − E b M bT (cid:48) ) − − E b M bT (cid:48) Z ) − − − E f M b ) − E fZ M b ) − − E fT M b ) − − E fTZ M b ) − − − E b M f ) − E b M fZ ) − − E b M fT (cid:48) ) − − E b M fT (cid:48) Z ) − − − E fT M bT (cid:48) ) − − − E bT M fT (cid:48) ) − − − E fZ M bT (cid:48) Z ) − − − − E bTZ M fZ ) − − − − E fTZ M bZ ) − − − − E bZ M fT (cid:48) Z ) − − − − Z ×T symmetric U (1) quan-tum spin liquids that have θ = 0 and have Z acting as acharge conjugation. T E = 1 ( T E = −
1) represents the casewhere E is a Kramers singlet (doublet) under T . T (cid:48) M = 1( T (cid:48) M = −
1) represents the case where M is a Kramers singlet(doublet) under T (cid:48) . Z E,M represents the result of acting thecharge conjugation twice on E and M , respectively. T E Z E T (cid:48) M Z M anomaly class( E bZ M bZ ) − − − E bTZ M bT (cid:48) Z ) − − − − − E fT M bZ ) − − − E bZ M fT (cid:48) ) − − − E fT M bT (cid:48) Z ) − − − − E bTZ M fT (cid:48) ) − − − − E bTZ M bZ ) − − − − E f M bZ ) − − E bTZ M f ) − − − E bZ M bT (cid:48) Z ) − − − − E bZ M f ) − − E f M bT (cid:48) Z ) − − − E bZ M bT (cid:48) ) − − − E fTZ M bT (cid:48) ) − − − − E fTZ M bT (cid:48) Z ) − − − − − E bT M bT (cid:48) Z ) − − − − E bT M fZ ) − − − E bZ M fZ ) − − − E bT M bZ ) − − − E bT M fT (cid:48) Z ) − − − − E bTZ M fT (cid:48) Z ) − − − − − E bTZ M bT (cid:48) ) − − − − E fZ M bT (cid:48) ) − − − E fZ M bZ ) − − − E bT M bT (cid:48) ) − − − E bT M f ) − − E f M bT (cid:48) ) − − Z ×T symmetric U (1) quantumspin liquids that have θ = 0 and have Z acting as a chargeconjugation at θ = π . T E = 1 ( T E = −
1) represents the casewhere E is a Kramers singlet (doublet) under T . T (cid:48) M = 1( T (cid:48) M = −
1) represents the case where M is a Kramers singlet(doublet) under T (cid:48) . Z E,M represents the result of acting thecharge conjugation twice on E and M , respectively. The lastcolumn lists the anomaly classes. Similarly, for states with θ = π and Z actingas a charge conjugation, there are only two states:( E fT M fT (cid:48) ) θ − and ( E fT M fT (cid:48) ) θ − Z . In both states, Z takes the ( ,
1) dyon into the ( − , −
1) dyon. Becausetime reversal takes ( ,
1) into ( , − T (cid:48) takes ( ,
1) to ( − , M , the boundstate of ( ,
1) and ( − , T (cid:48) . [30, 49] The difference in these two states is that Z squares to +1 ( −
1) on the (cid:0) , (cid:1) dyon in the former(latter). In fact, the former state is just the time rever-sal symmetric ( E fT M f ) θ further equipped with a chargeconjugation symmetry, so it must be anomaly-free.So without examining anomalies, there are in total 50possible distinct Z × T symmetric U (1) quantum spinliquids where Z acts as a charge conjugation. It turnsout that, together with the anomaly-free ( E fT M fT (cid:48) ) θ − ,22 of these states are free of anomaly. The other 28states are all anomalous, and there are 6 anomaly classes.The strategy to show the anomalies will be given in Sec.VII C, and the arguments for this anomaly-detection willbe completed in Appendix J.0 T E Z E T (cid:48) M Z M Z D comments( E fT M fT (cid:48) ) θ − − − E fT M fT (cid:48) ) θ − Z − − − Z × T symmetric U (1) quantum spinliquids that have θ = π and have Z acting as a charge con-jugation. T E = 1 ( T E = −
1) represents the case where E isa Kramers singlet (doublet) under T . T (cid:48) M = 1 ( T (cid:48) M = − M is a Kramers singlet (doublet)under T (cid:48) . Z E,M,D represents the result of acting the chargeconjugation twice on E , M and the (cid:0) , (cid:1) dyon, respectively. Therefore, combined with the 16 states where Z doesnot permute any excitation, there are in total 38 distinctnon-anomalous Z ×T symmetric U (1) spin liquid states,and they can be found in Table VI, Table VIII, Table IXand Table XI.We note that models that discuss Z × T symmetric U (1) quantum spin liquids have been proposed in the lit-erature, and a prototypical set includes but is not limitedto Ref. 19–21. In these models, the Z × T symmetric U (1) quantum spin liquid states are ( E b M b ) − . C. Strategy of anomaly-detection
In this subsection we lay out the strategy to show theanomaly of the other 36 states. It turns out to be eas-ier to first show that ( E bZ M bZ ) − is anomalous with the Z symmetry (independent of time reversal), and thiswill be done later in this section. This immediately im-plies that ( E f M bZ ) − and ( E bZ M f ) − are also anomalouswith Z symmetry, because these states can be related to( E bZ M bZ ) − by tuning θ by 2 π . It also immediately im-plies that ( E bT Z M bZ ) − , ( E bZ M bT (cid:48) Z ) − , ( E bT Z M bT (cid:48) Z ) − ,( E fT M bZ ) − , ( E bZ M fT (cid:48) ) − are anomalous with Z × T symmetry, because breaking T will make them one of( E bZ M bZ ) − , ( E f M bZ ) − and ( E bZ M f ) − . Furthermore,this means ( E fT M fT (cid:48) ) θ − Z is anomalous, because evenif the time reversal symmetry is broken, this state issmoothly connected to the anomalous ( E bZ M bZ ) − .Next, using a generalization of the method for show-ing the anomaly of ( E bZ M bZ ) − , we show ( E bT M bT (cid:48) ) − and E bT T (cid:48) M b − are anomalous with Z × T symmetryin Appendix J. It turns out this is enough to show theremaining states are all anomalous. More precisely,1. showing that ( E bT M bT (cid:48) ) − is anomalous is suffi-cient to show that the other entries in Table X areanomalous.2. showing that E bT T (cid:48) M b − is anomalous is sufficientto show that the rest entries in Table VII areanomalous.To see the first claim, let us consider ( E bT M bZ ) − and( E bZ M bT (cid:48) ) − . These two states must be simultaneouslyanomalous or anomaly-free, because they are related to each other by the relabelling T ↔ T (cid:48) and E ↔ M . Sup-pose they are anomaly-free, then by combining them withthe states that will be constructed in Appendix J 1, wewill get ( E bT M bT (cid:48) ) − , which is in contradiction with that( E bT M bT (cid:48) ) − is anomalous. This means if ( E bT M bT (cid:48) ) − isanomalous, then ( E bT M bZ ) − and ( E bZ M bT (cid:48) ) − will alsobe anomalous. Combining these three anomalous stateswith the anomaly-free states constructed in AppendixJ 1, one can show all other entries in Table X are alsoanomalous.To see the second claim, consider E fT M b − and E fT (cid:48) M b − . These two states must also simultaneouslybe anomalous or anomaly-free, because they are relatedto each other by the relabelling T ↔ T (cid:48) and E ↔ M .Suppose they are anomaly-free. By combining them andthe anomaly-free states constructed in Appendix J 1, wecan get E bT T (cid:48) M b − , which contradicts that E bT T (cid:48) M b − is anomalous. This means if E bT T (cid:48) M b − is anomalous,then E fT M b − and E fT (cid:48) M b − must also be anomalous.Combining these anomalous states with the anomaly-free states constructed in Appendix J 1, one can showall other entries in Table VII are also anomalous. Anomaly of ( E bZ M bZ ) − In the spirit of Sec. VI B, here we will show that( E bZ M bZ ) − is anomalous with a Z charge-conjugationsymmetry (independent of time reversal), by showing itscorresponding SPT has an inconsistent surface.To show the anomaly of ( E bZ M bZ ) − , we will considerfrom the perspective of E bZ , and suppose there is an SPTmade of E bZ that after gauging becomes ( E bZ M bZ ) − . Onthe surface of this SPT, we first condense the bound stateof two E bZ , and this makes the surface a superfluid witha Z symmetry. There will be various vortices, and the4 π vortex is the minimal trivial boson. So we can thencondense the 4 π vortices to restore the U (1) symmetry,and this gives a symmetric gapped surface state wherethe U (1) charge of the excitations is quantized in unitsof 1 / { , M bZ , X, N I } × { , E bZ } , where M bZ is the remnantof the strength-1 monopole, X carries half charge un-der U (1), and N I ’s are neutral. In general, there canbe many flavors of N I , but only the case with a single X needs to be considered, because other X ’s can be re-lated to a single one by attaching certain N I . Notice inthis notation the inverses and bound states of these ex-citations are understood to be implicitly displayed. Wewould like to check whether such a surface is consistent.That is, it has consistent braiding, fusion and symmetrytransformation rules.As for braiding, we know E bZ is local, M bZ and X have mutual π statistics, and M bZ has no mutual statis-tics with N I . X and N I can have complicated braidingthough, and it can even be non-Abelian.1As for fusion, we have M bZ × M bZ = 1 (21) X × X = E bZ + E bZ N + E bZ N M bZ (22)and N I × N J = N k + M bZ N k (23)(21) comes from that this surface is obtained by con-densing M bZ , and (22) and (23) are obtained under theconstraint due to charge conservation. Notice in (22)the fusion product cannot be E bZ M bZ , because this willbe inconsistent with the general condition that a parti-cle and its anti-particle should have the same topologicalspin. Also, all potential fusion multiplicities are sup-pressed, and they turn out to be unimportant for ourdiscussion.Now if we are willing to break the U (1) symmetryon the surface by condensing E bZ M bZ , X will be con-fined and N I ’s will remain, and we will be left with { , M bZ , N I } that has a Z symmetry. Notice that { , M bZ , N I } is closed under fusion and braiding, andit is known that in three dimensions there is no bosonicSPT protected by Z symmetry. This means N I can befurther confined (without breaking the Z symmetry),and we are left with { , M bZ } . In other words, N I ’s canbe viewed as emergent particles of a system made of M bZ in the presence of the Z symmetry but in the absence ofthe U (1) symmetry. However, because neither M bZ nor N I carries a U (1) charge, even in the presence of U (1)symmetry N I can still be viewed as emergent particles ofa system made of M bZ .So we can get rid of N I and be left with { , M bZ , X } ×{ , E bZ } . Now the fusion of X must be X × X = E bZ (24)The only possible consistent topological order of thisstate is a Z topological order (or its twisted version,the double-semion theory).Let us turn to symmetry assignment, and we will par-ticularly consider how charge-conjugation acts on X . No-tice when defining the charge-conjugation action on X ,there is an ambiguity due to our freedom to multiply itby a gauge transformation. But because this topologi-cal order is a Z gauge theory, the action of the globalcharge-conjugation symmetry twice should have an un-ambiguous result on X . So in order to be consistentwith the above fusion rule, X must go to ± iX upon act-ing with charge-conjugation twice. Below we show thisis impossible.Suppose the action of charge-conjugation on X is im-plemented by a generic matrix CX i → C ij X † j , X † i → C ∗ ij X j (25) This is a boson in this surface topological order.
Notice the indices label different components of X thatdiffer by some local operations. This implies that actingcharge-conjugation twice on X gives X i → ( CC ∗ ) ij X j (26)Now consider the operator X i M ij X j with an arbitrarymatrix M , which is a charge-1 boson, so the charge-conjugation acting on it twice gives −
1. This requires( CC ∗ ) T M ( CC ∗ ) = − M (27)Because M is arbitrary, this is possible only if CC ∗ = ± i , which confirms the previous statement that X →± iX upon acted by Z twice. However, no matrix C can possibly satisfy CC ∗ = ± i . To see this, suppose CC ∗ = ± i , then ( CC ∗ ) = −
1. On the other hand, C ∗ C = ∓ i and ( CC ∗ ) = CC ∗ CC ∗ = C ( C ∗ C ) C ∗ = 1,which contradicts with the previous result.The above contradiction shows that there cannot evenbe any X . So the surface is just { , E bZ , M bZ } , whereeverything is local. This means that there is a chargeneutral excitation that has Z squaring to −
1, whichcontradicts the original assumption. Therefore, this SPTcannot exist, and furthermore, ( E bZ M bZ ) − is anomalouswith a Z charge-conjugation symmetry. Notice from theabove argument we see the anomaly of ( E bZ M bZ ) − is in-dependent of time reversal.Note this argument can be easily modified to show that E b M b is anomalous with SO (3) symmetry, by chang-ing every Z symmetry by SO (3) symmetry, and chang-ing every excitation with charge-conjugation squaring to − SO (3) monopole to show the anomaly. D. Anomaly classes
Before finishing this section, we make some brief re-marks on the anomaly classes of these anomalous spinliquid states. Here by an anomaly class, we mean a groupof anomalous states which can be turned into each otherby coupling it with a state that is anomaly-free. Becausethe anomalous spin liquid states can in principle be re-alized on the surface of some 4 + 1-d Z × T symmetricbosonic SPTs, analysing the anomaly classes of these 37anomalous spin liquid states gives some information onthe properties of these SPTs.We first discuss the anomalous spin liquid states with Z acting as a charge conjugation. It is straightforwardto check that within each of the following 6 groups ofanomalous states, all states have the same anomaly:1. ( E bZ M bZ ) − , ( E bT Z M bT (cid:48) Z ) − , ( E fT M bZ ) − ,( E bZ M fT (cid:48) ) − , ( E fT M bT (cid:48) Z ) − , ( E bT Z M fT (cid:48) ) − ,( E fT M fT (cid:48) ) θ − Z .2. ( E bT Z M bZ ) − , ( E f M bZ ) − , ( E bT Z M f ) − .3. ( E bZ M bT (cid:48) Z ) − , ( E bZ M f ) − , ( E f M bT (cid:48) Z ) − .24. ( E bZ M bT (cid:48) ) − , ( E fT Z M bT (cid:48) ) − , ( E fT Z M bT (cid:48) Z ) − ,( E bT M bT (cid:48) Z ) − , ( E bT M fZ ) − , ( E bZ M fZ ) − .5. ( E bT M bZ ) − , ( E bT M fT (cid:48) Z ) − , ( E bT Z M fT (cid:48) Z ) − ,( E bT Z M bT (cid:48) ) − , ( E fZ M bT (cid:48) ) − , ( E fZ M bZ ) − .6. ( E bT M bT (cid:48) ) − , ( E f M bT (cid:48) ) − , ( E bT M f ) − .It is clear that combining group 1 and group 4 results ingroup 3, combining group 1 and group 5 results in group2, and combining group 4 and group 5 results in group6. This implies that the 4D bosonic SPTs with Z × T symmetry at least have a classification of Z . Group co-homology gives precisely the same classification,[54] andour results suggest that the surface states of these SPTscan be the above anomalous U (1) gauge theories.Notice there is another SPT that is beyond groupcohomology, and that SPT is protected purely by Z symmetry.[67, 68] One physical realization of the bulkof this SPT is to consider a decorated domain wall con-struction, where on each Z domain wall we place an ef mf state. [67] The surface properties of this beyond-group-cohomology SPT is unclear at this point, and itmay be interesting to work it out.Taking all these together, we propose that the com-plete classification of 4D bosonic SPTs with Z × T sym-metry is Z . This agrees with the classification of 4Dbosonic SPTs with Z × Z P symmetry, where Z P is areflection symmetry that results in a trivial action whenacted twice.[69]Next we discuss the anomalous spin liquid states with Z not acting as a charge conjugation, whose anomalyclasses can be organized asa. E bT (cid:48) M f − , E fT M b − , E bT (cid:48) M b − .b. E bT M f − , E fT (cid:48) M b − , E bT M b − .c. E bT T (cid:48) M b − , E bT T (cid:48) M f − , E f M b − .First notice all these states are anomalous with the full Z × T symmetry. If T is broken, all these states shouldbe non-anomalous Z symmetric states. Second, class aand class b differ by the relabelling T ↔ T (cid:48) and E ↔ M ,and class c can be obtained by combining states in class aand class b. Notice before that class 2 and class 3 differby this relabelling, so do class 4 and class 5, and thatclass 6 can be obtained by combining class 2 and class 3,or by combining class 4 and class 5. This suggests statesin class c here are of the same anomaly class as states inclass 6 as above. We do not attempt to completely settledown the relation among these anomaly classes in thispaper. VIII. U (1) quantum spin liquids enriched by someother symmetries In the spirit of the general framework in Sec. VI, inthis section we briefly discuss U (1) quantum spin liq-uids with some other symmetries. Part of the motivation comes from the existing lattice models that realize U (1)quantum spin liquids with O (2) × T = ( U (1) (cid:111) Z ) × T symmetry.[15–17, 70] In all these models, the improper Z rotation of the O (2) symmetry acts as a charge con-jugation. Ref. 15–17 studied a couple of different latticemodels that realize ( E b M b ) − , where θ = 0 and both E and M are bosons, and M carries half charge under the U (1) subgroup of O (2). Ref. 70 constructed two othermodels of O (2) ×T symmetric U (1) quantum spin liquids,where one of them has a bosonic monopole and the otherhas a fermionic monopole. These two states are ( E b M b ) − and ( E b M f ) − , respectively. Notice, for simplicity, in thissection we will not consider the refined classification thatconsiders coupling these spin liquids with SPTs. A. SO ( N ) × T symmetry We can generalize our classification of SO (3) × T sym-metric U (1) quantum spin liquids to SO ( N ) ×T symmet-ric U (1) quantum spin liquids, with the integer N > SO ( N ) × T have thesame classification as those of SO (3) × T : there is aspinor representation of SO ( N ) and a Kramers doubletrepresentation of time reversal. Therefore, the enumer-ation of states with this symmetry goes in a parallelway as those with SO (3) × T symmetry, and all non-anomalous states with SO (3) × T can be generalized totheir SO ( N ) ×T analogs. Furthermore, π ( SO ( N )) = Z and the monopole structure of an SO ( N ) gauge field issimilar to that of an SO (3) gauge field, so the gener-alization of the anomalous states in the SO (3) × T casewill still be anomalous. Therefore, we conclude that with SO ( N ) × T symmetry there will also be 15 distinct non-anomalous U (1) quantum spin liquids, and they havesimilar properties as those with only SO (3) × T sym-metry in terms of the bulk fractional excitations.For the special case of SO (2) × T ∼ = U (1) × T , its pro-jective representations on E are classified by Z . One ofthe nontrivial root projective representation correspondsto Kramers doublet of time reversal, while the other cor-responds to half-charge of SO (2), which is protected bytime reversal here. Therefore, although SO (2) has noprojective representation on its own, the projective rep-resentations of SO (2) × T on E can still be viewed asdescendants of those of SO (3) × T .If θ = 0, there is no projective representation on M . So there are 3 × = 12 putative states with θ = 0, which can all be viewed as descendant states of SO (3) × T symmetric states when the symmetry is re-duced to SO (2) ×T . Notice some distinct SO (3) ×T sym-metric states have the same SO (2) ×T symmetric descen-dant, because there is no fractional quantum number onmonopoles anymore. The descendants of the 15 anomaly-free states of course remain anomaly-free. By inspectingthe anomaly classes of anomalous SO (3) × T symmetricstates listed in Sec. III, we see in each anomaly classthere is at least one state that has a trivial monopole3when the symmetry is reduced to U (1) × T , which meansall these anomalous states will become anomaly-free. Soall these 12 SO (2) × T symmetric states with θ = 0 areanomaly-free.If θ = π , the properties of the (cid:0) , ± (cid:1) dyons will de-termine the phase, which does not have any projectiverepresentation of this symmetry. So it contributes oneanomaly-free state, which can be viewed as a descendantof the SO (3) × T symmetric ( E fT M f ) θ state when thesymmetry is broken down to U (1) × T .Therefore, there are in total 13 distinct anomaly-free U (1) quantum spin liquids with U (1) × T symmetry. B. SO ( N ) symmetry In the following few subsections we will consider thecase in the absence of time reversal symmetry. In thissubsection we start by discussing U (1) quantum spin liq-uids with only SO (3) spin rotation symmetry, which canbe realized in systems with a spin chirality term in theHamiltonian.As for the elementary excitations, we focus on (1 , q e , q e a real number. It is not hard to seethere must be bosons for some q e , and we will considersuch a bosonic ( q e ,
1) excitation. Due to the absence oftime reversal symmetry, θ in (1) can be tuned contin-uously, so that the ( q e ,
1) particle chosen above can betuned to (0 ,
1) by Witten effect. That is to say, now wefix (0 ,
1) to be a boson. Similarly, we can make (1 , E = (1 ,
0) and M = (0 ,
1) under SO (3), we can have E b M b , E b M b and E b M b , andthese exhaust all (including anomalous) SO (3) symmet-ric U (1) spin liquids. E b M b and E b M b are clearly notanomalous, and their description as a gauged SPT is sim-ilar to their cousins with SO (3) × T symmetry. Also, asargued in Sec. III, E b M b is anomalous even with only SO (3) symmetry.Therefore, with only SO (3) symmetry, there are onlytwo (anomaly-free) distinct possible symmetry realiza-tions in the U (1) quantum spin liquids: E b M b and E b M b . This concludes the classification of U (1) quan-tum spin liquids with SO (3) symmetry.Similar reasoning as above can be applied to U (1)quantum spin liquids with SO ( N ) symmetry with N > SO ( N ) groups have one nontrivial projec-tive representation, the spinor representation. Also, asmentioned above, the monopole properties of an SO ( N )gauge field is similar to that of an SO (3) gauge field.Therefore, the arguments for the enumeration of all the states, construction of the non-anomalous states and ex-amination of the anomalous states are parallel to that of SO (3), and gives only 2 distinct SO ( N ) symmetric U (1)quantum spin liquids.For the special case of SO (2) ∼ = U (1), because of theabsence of any nontrivial projective representation of thissymmetry, there will be only a single type of symmetric U (1) quantum spin liquid. C. Z symmetry For the case with Z symmetry, as above, all statescan be symmetrically tuned so that it has the θ = 0-typeof charge-monopole lattice with both E and M bosonic.Also notice there is no projective representation of Z ,so nontrivial states must have Z acting as charge conju-gation. Then there can be E b M b , ( E b M b ) − , ( E bZ M b ) − and ( E bZ M bZ ) − . The first three states can clearly berealized, but as shown before, ( E bZ M bZ ) − is anomalouswith a Z symmetry. So there are 3 distinct Z sym-metric U (1) quantum spin liquids: E b M b , ( E b M b ) − and( E bZ M b ) − . D. O (2) symmetry Similar considerations can be applied to the case with O (2) symmetry. There will still be two spin liquid stateswhere the improper Z component does not act as acharge conjugation, and these are the descendants of E b M b and E b M b with SO (3) × T symmetry. As shownin Sec. III, the descendant of E b M b is still anomalouswith O (2) symmetry. Again, for a complete classifica-tion, states where Z acts as a charge conjugation needto be taken into account. Unlike the case with O (2) × T symmetry, the fractional excitations always have inte-ger charges under the U (1) subgroup of O (2), if the im-proper Z component acts as a charge conjugation. Sothese states include ( E b M b ) − , ( E bZ M b ) − , ( E bZ M bZ ) − ,and the first two are anomaly-free, while the last one isanomalous. Therefore, there are in total 4 distinct non-anomalous O (2) symmetric U (1) quantum spin liquids: E b M b , E b M b , ( E b M b ) − and ( E bZ M b ) − . IX. Discussion
In this paper we have classified and characterized 3Dsymmetry enriched U (1) quantum spin liquids. One ofour focuses is on such spin liquids with time reversal and SO (3) spin rotational symmetries. 26 states were enu-merated based on the properties of the bulk fractionalexcitations, among which only 15 can be realized in 3Dlattice systems. We explain in details how to view thesequantum spin liquids as gauged version of some SPTs.The other 11 are shown to be anomalous, i.e. they can-4not be realized in a 3D bosonic system with these sym-metries. In Appendix C, they are constructed on thesurface of some 4D bosonic short-range entangled states.The anomalies of the anomalous states become clearwhen the properties of the SO (3) monopoles are exam-ined. Although checking the topological defects of thegauge field that corresponds to certain symmetry hasbeen widely applied to detect anomalies, to the best ofour knowledge, the properties of SO (3) monopoles havenot been investigated in previous studies. We expect itto be helpful in studying other problems that involves SO (3) symmetries.When combined with bosonic SPTs with time reversaland SO (3) spin rotational symmetry, we find a furtherrefined classification which shows there are 168 different U (1) quantum spin liquids.After warming up with the example of SO (3) ×T sym-metric U (1) quantum spin liquids, we have described ageneral framework to classify such spin liquid states witha general symmetry. This approach is again physics-based, and it has the advantage of providing us withintuition both on the classification and the physical char-acterization. However, it is not always easy to implementthis framework, and so it is desirable to find a simpler sys-tematic way to do the classification. The field theoreticformal approach discussed in Sec. VI may be potentiallyhelpful. Another possibly helpful formal approach is togeneralize the categorical theory that is used to study2 d SETs to U (1) quantum spin liquids.[9, 11] This maybe possible because in both cases the excitations are allparticle like, although there are infinitely many types offractional excitations in a U (1) quantum spin liquid.In the spirit of this general framework, we have alsodiscussed U (1) quantum spin liquids with some othersymmetries, and found some very rich structures. Inparticular, we discussed U (1) spin liquids with Z × T symmetry in great detail. Based on the properties ofthe bulk fractional excitations, there are 38 such Z × T symmetric states that are free of anomaly. The anoma-lies of the other 37 such Z × T symmetric states aredetected based on the method in the general framework.The study of Z × T symmetric U (1) quantum spin liq-uids have some implications on some SPTs, as discussedin Appendix J.Besides looking for a simper systematic classificationof these symmetry enriched U (1) quantum spin liquids,the other most important open questions are of coursewhich microscopic models and experimental systems re-alizing these different symmetry enriched U (1) quantumspin liquids, and how to detect and distinguish them nu-merically or experimentally. One particular interestingtheoretical aspect of this question is how lattice symme-tries interplay with these quantum spin liquids. Theseare beyond the scope of the current paper and are worthfurther investigating in the future, and we note some re-cent progress in this aspect.[71–74]Another interesting theoretical challenge is to clas-sify and characterize symmetry enriched gapped quan- tum spin liquids, some of which can be obtained by con-densing some excitations in the U (1) quantum spin liq-uids. One complication in this problem is that there usu-ally exists loop-like excitations, whose properties are notcompletely understood to date. Because some of thesegapped quantum spin liquids are descendants of the U (1)quantum spin liquids, relating the properties of loop-likeexcitations in the former to the properties of the particle-like excitations in the latter may shed light on this prob-lem. X. Acknowledgement
We acknowledge fruitful discussions with MaissamBarkeshli, Meng Cheng, Meng Guo, Max Metlistki, YangQi, Chenjie Wang and Boyu Zhang. T. S. was supportedby NSF grant DMR-1608505, and partially through a Si-mons Investigator Award from the Simons Foundation.L. Z. acknowledges support by the 2016 Boulder Sum-mer School for Condensed Matter and Materials Physics- where a part of this work was done - through NSF grantDMR-13001648. C. W. was supported by the HarvardSociety of Fellows.
A. Some remarks on time reversal andcharge-conjugation symmetry
In this appendix we discuss some general propertiesof time reversal and charge-conjugation symmetry. Inparticular, we would like to check the values of T and C on a fractionalized excitation. By abuse of notation,we will denote this excitation by E in general, whichincludes but does not limit to the case where E is theelectric charge of a U (1) quantum spin liquid. In general, E can be a multicomponent object, and let us denote the i th component as E i .We first consider time-reversal symmetry T . We as-sume the anti-unitary time reversal symmetry acts on E i as T : E i → U ij E j (A1)with U a generic matrix. That is, time reversal changes E to something that differs from it only by a local oper-ator. We also assume E † i M ij E j is always a local object,for any matrix M . Notice in the above notation, somecomponents of E can be a bound state of the fractional-ized excitation E and some local particles.Straightforward algebra indicates that acting time re-versal twice, the original E becomes E i → ( U ∗ U ) ij E j (A2)and the local object becomes E † i M ij E j → E † i (( U ∗ U ) † M ( U ∗ U )) ij E j (A3)Suppose the system is made of Kramers singlets. Thatis, E † i M ij E j is invariant upon acted by time reversal5twice for any M , which is equivalent to that( U ∗ U ) † M ( U ∗ U ) = M (A4)for any matrix M . This is possible only if U ∗ U = e iφ I ,where I is the identity matrix. Because ( U ∗ U ) = e iφ I and U ∗ U U ∗ U = U ∗ ( U U ∗ ) U = I . This implies e iφ = ± T acting on such fractionalized excitations mustgive ± T is well-defined, T canonly be ±
1. However, if there are microscopic Kramersdoublets in the system, it is possible to have T = − T = − T does not have tobe well-defined.Now T = − T = ± i ) can happen if time reversalchanges the relevant excitation by a nonlocal operation,and a typical example for this case is ( eCmC ) T (cid:15) men-tioned in the main text, where under time reversal e and m are exchanged. It can also happen if time reversalchanges this excitation by a local operation. In this case,time reversal has to attach a local Kramers doublet tothe relevant nontrivial excitation. To see this, withoutloss of generality, let us assume the time reversal partnerof E is F , i.e., E → F under time reversal. To have T = ± i for E , we need F → ± iE under time reversal.These also imply T = ∓ i for F , so E and F defer by alocal Kramers doublet.We now consider charge-conjugation symmetry C . Ingeneral it is a unitary symmetry acting on E i as C : E i → V ij E † j , (A5)where V is a matrix. Taking the hermitian conjugate ofthe above equation, we have C : E † i → V ∗ ij E j . (A6)So C acting twice should give C : E i → ( V V ∗ ) ij E j . (A7)Again we require the local operator E † i M i jE j to have C = 1, for any matrix M . Following the same logicas we did for time-reversal symmetry, we conclude that C = ± E .Also notice that the value of C is invariant under a U (1) gauge transform U θ , namely ( U θ C ) = C . Thismeans that the value C = ± C should be enough forour purpose. In the context of gapped topological orders(for example in Z N gauge theories), the mathematicallymore precise meaning of C on fractionalized excitationshas been discussed in Ref. [9]. B. An SPT: eCm In this appendix we describe a 3D SPT, eCm , undersymmetry U (1) × SO (3), U (1) × T × SO (3) or ( U (1) (cid:111) T ) × SO (3). The defining property of this SPT is that itcan have a symmetric surface Z topological order withexcitations { , e, m, (cid:15) } , where e carries charge-1/2 under U (1) and m carries spin-1/2 under SO (3). We beginby giving field theoretic descriptions of this surface statewith the above symmetries.If the symmetry is simply U (1) × SO (3), a field the-ory for this surface can be described by the followingLagrangian: L = (cid:88) s = ± | ( ∂ µ − ia µ ) z s | + V ( | z | )+ 14 e ( (cid:15) µνλ ∂ ν a λ ) − π Ada (B1)where Ada is a shorthand for (cid:15) µνλ A µ ∂ ν a λ . z s is a two-component complex field that transforms as a doubletunder SO (3), a µ is a non-compact U (1) gauge field, and A is a background U (1) gauge field corresponding to theglobal U (1) symmetry.Condensing a bound state of two z ’s in the singletchannel (i.e. letting (cid:104) z ∂ x z − z ∂ x z (cid:105) (cid:54) = 0 for exam-ple) gives us the above surface topological order, wherethe uncondensed single z will be identified as m thatcarries spin-1/2. After this condensation, the flux of a isquantized in unit of π , and the last term in the above La-grangian implies that this π -flux carries charge-1/2 underthe global U (1). This π -flux can be identified as e . Theresulting state is precisely eCm .If the symmetry is ( U (1) (cid:111) T ) × SO (3), the surfacetheory of eCm can still be described by the field theorygiven by (B1), but now the spin operator is representedas (cid:126)S ∼ z † (cid:126)σi∂ t z and under time reversal z → z † , (cid:126)a → (cid:126)a, a → − a (B2)In order to obtain eCm , we need to make (cid:104) z ∂ x z − z ∂ x z (cid:105) = (cid:104) z † ∂ x z † − z † ∂ x z † (cid:105) (cid:54) = 0.If the symmetry is U (1) × T × SO (3), the surface stateof eCm can be described by a field theory similar to(B1): L = (cid:88) s,α = ± | ( ∂ µ − ia µ ) z sα | + V ( | z | )+ 14 e ( (cid:15) µνλ ∂ ν a λ ) − π Ada (B3)Notice now each component of z s contains two compo-nents, z sα with α = ± , and the generators of spin rota-tions become (cid:126)S = z † sα (cid:126)τ ss (cid:48) z s (cid:48) α . Under time reversal z sα → ( σ ) αα (cid:48) ( τ ) ss (cid:48) z s (cid:48) α (cid:48) , (cid:126)a → − (cid:126)a, a → a (B4)where σ and τ are the standard Pauli matrices. Thereason to give more components to z is to make it not aKramers doublet. To get eCm , we can also condensethe bound state of two z ’s in the singlet channel, that is,let (cid:104) z T σ τ ∂ x z (cid:105) (cid:54) = 0. Similar argument as above impliesthe resulting state is eCm .To see that this state as a strictly 2D system is anoma-lous, consider tunneling a U (1) monopole through this 2D6system. This will leave a 2 π flux on the system. For sucha local process, no excitations far away should be ableto tell the existence of this 2 π -flux. But e carries halfcharge under U (1), it will pick up a nontrivial phase fac-tor upon circling around this 2 π -flux, regardless how farit locates away from it. To cancel this phase factor, an m needs to be present at the 2 π -flux. Because m carriesspin-1/2, this then implies tunneling a monopole leaves aspin-1/2 on the surface. This is not possible for a strictly2D system with symmetry U (1) × SO (3). Notice thattime reversal symmetry is not involved in the anomaly,so this surface is still anomalous even if the symmetry is U (1) × SO (3).To visualize this SPT, the simplest way is to do a layerconstruction similar to that used in Ref. 8. Because simi-lar method will be used in Appendix C to construct 4+1-d systems whose surfaces realize the anomalous quantumspin liquids, we do not explicitly display it for eCm here.Notice when the symmetry is ( U (1) (cid:111) T ) × SO (3),to realize eCm , we have assumed that the microscopicbosons are Kramers singlet. Below we argue that for mi-croscopic Kramers doublet charged bosons, eCm can-not be realized. This fact is important, because otherwisegauging eCm in such a system would lead to E bT M b ,which is argued to be anomalous in Sec. III.Suppose eCm can be realized in a system made ofKramers doublet charged bosons. Fusing two e ’s givesa charge-1 local particle, which must be a Kramers dou-blet by assumption. Given the excitation content of thistheory and that time reversal keeps the U (1) charge, thetime reversal action on e can always be represented as e i → U ij e j (B5)Now notice e i M ij e j is a local charge-1 operator for anymatrix M , so this operator must be a Kramers doublet,which implies that( U ∗ U ) T M ( U ∗ U ) = − M (B6)This is possible only if U ∗ U = ± i . As shown in AppendixA, no matrix U can have this property. This implies that eCm cannot be realized in a system made of Kramersdoublet charged bosons. C. Anomalous spin liquids as surface states ofsome -d systems
In this appendix we will show the anomalous spin liq-uids can be obtained on the surface of some 4 + 1-d sys-tems. The simplest way to construct these 4+1-d surfacestates is the following layer construction, which has beenwidely used to construct topological states[8, 13, 50, 75].For example, to construct a 4 + 1-d system whose sur-face can realize E bT M b , one can start by stacking alter-nating layers of non-anomalous spin liquids E bT M b and E b M b (see Fig. 6). Then on the i th, i + 1th and i + 2th FIG. 6. Layer construction of the 4+1-d system whose surfacerealize an anomalous spin liquid. layers, one can condense the bound state of E i , M i +1 and E i +2 with the subscript indicating the layer index. Thisbound state, B i = E i M i +1 E i +2 , is a trivial boson, and B i ’s with different i ’s commute, so they can be simultane-ously condensed without breaking any symmetry. Afterthis condensation, the gauge field in the 4 + 1-d bulk willbe confined (or Higgsed) and this bulk becomes short-range entangled, but on the surface some nontrivial ex-citations still survive. These survivors are E and M † E on the top surface, and E N and E N − M † N on the bottomsurface. Now the top (bottom) surface realizes E bT M b ,and E ( E N − M † N ) and M † E ( E N ) can be viewed asthe electric charge and magnetic monopole, respectively.The 4 + 1-d system constructed here is an SPT undersymmetry SO (3) × T because its surface, E bT M b , isanomalous.To obtain 4 + 1-d systems whose surface realize allother anomalous spin liquids, one only needs to replaceeach layer by the appropriate non-anomalous spin liquidand condense the proper bound states.For some spin liquids, there is a more isotropic con-struction of the corresponding 4 + 1-d systems by using anon-linear Sigma model (NLSM) with appropriate topo-logical terms and anisotropies, similar to that used in Ref.76. For example, to construct the corresponding 4 + 1-dbulk of E bT M b , consider a 4 + 1-d O (6) NLSM with atheta-term at θ = 2 π . Its surface theory is a 3 + 1-d six-component NLSM with a Wess-Zumino-Witten (WZW)term at level-1, with Lagrangian L = 1 g ( ∂ µ n a ) + 2 πi Ω (cid:90) du(cid:15) abcdef n a ∂ u n b ∂ x n c ∂ y n d ∂ z n e ∂ τ n f (C1)with Ω the surface area of a five-dimensional unit sphere.The six-component vector transforms under time reversalas n , , → − n , , n , , → n , , (C2)This theory is invariant under O (6) × T .To see how E bT M b can be accessed by the abovetheory, let us first add some SO (3) × SO (3) anisotropy,7such that the first (second) three components transformas a vector under the first (second) SO (3). Consider theweak coupling limit of the theory where both the first andthe second three components are ordered. Now disorderthe second three components by proliferating its hedge-hog defects. In this way, the second three componentsthemselves form a trivial state that preserves the second SO (3) symmetry. Due to the WZW term, the hedge-hog defects of the first three components carry spin-1/2under the second SO (3), and it will be identified as themagnetic monopole of E bT M b later.Now disorder the first three components by proliferat-ing the spin wave excitations while keeping its hedgehogdefects gapped. In this way we will get a U (1) spin liquid.One way to see this is to write the first three componentsin the CP representation, n a = z † α σ αβa z β , for a = 1 , , z is a two-component complex spinon field with | z | + | z | = 1, and σ ’s are the standard Pauli matrices.Under the previous defined time reversal and the first SO (3), the spinon field transforms as a Kramers doubletand SU (2) doublet. This spinon will be identified as theelectric charge of E bT M b later.The Lagrangian that only involves the first three com-ponents can now be written as the following gauge theory L (cid:48) = | ( ∂ µ − ia µ ) z | + V ( | z | ) + 14 e ( (cid:15) µνλ ∂ ν a λ ) (C4)where a µ is an emergent U (1) gauge field due to the U (1) gauge redundancy in (C3), i.e., n a is invariant when z → ze iθ for any real θ . The ordered state of the firstthree components corresponds to the Higgs phase of the U (1) gauge theory, and proliferating spin wave excita-tions corresponds to making spinons gapped and give riseto a U (1) spin liquid, where the gapped spinons are theelectric charge. The monopole of this U (1) spin liquid,which is the source of magnetic flux, should be identi-fied as the un-proliferated hedgehog defect, which is thesource of the skyrmions.Finally, adding a weak anisotropy to collapse the SO (3) × SO (3) symmetry to its diagonal SO (3) subgroup,we get E bT M b with SO (3) × T symmetry, where theelectric charge is a Kramers doublet and SU (2) doublet,and the magnetic monopole is an SU (2) doublet.The construction above gives a 4 + 1-d system whosesurface realizes E bT M b . If time reversal is ignored, theabove construction gives the 4+1-d system whose surfacerealizes E b M b . The 4 + 1-d system whose surface canrealize E bT M b can be obtained similarly, where time re-versal acts in the same way as before, while under SO (3)only the last three components transform as a vector andthe first three components do not transform.Notice in the construction based on NLSMs, even if SO (3) is broken to Z × Z , all components still transformnontrivially under this symmetry. Then it is believedthat the constructed 4 + 1-d states are still nontrivial SPTs. This motivates us to conjecture that even if thesymmetry is broken to Z × Z × T , the descendantsof the anomalous states remain anomalous because theystill live on the surface of some SPTs. D. Classification of some SPTs
In this appendix we classify some SPTs which, oncegauged, can become some of the U (1) quantum spin liq-uids studied in the main text.
1. Bosonic SPT with symmetry ( U (1) (cid:111) T ) × SO (3) We start with bosonic SPT with symmetry ( U (1) (cid:111) T ) × SO (3), where the microscopic boson is a Kramerssinglet. Without SO (3) symmetry, the classification ofthis SPT is well established. They are classified by Z , where the three root states are eCmC , eT mT and ef mf .[47] With SO (3) symmetry, Appendix B showsthere is another root state: eCm . In fact, there twomore root states: e mT and e m . That these twoare nontrivial SPTs can be inferred from the classifica-tion of bosonic SPT with symmetry U (1) × T . Indeed,once SO (3) is broken to U (1), e mT and e m become eCmT and eCmC , respectively.Therefore, we propose the classification of these SPTsis Z . Notice that among the six root states, only two ofthem need protection from the U (1) symmetry: eCmC and eCm .
2. Bosonic SPT with symmetry U (1) × T × SO (3) If the symmetry is U (1) ×T × SO (3), the understandingof bosonic SPTs with symmetry U (1) × T implies thereis one more root state: eCmT . This state is protectedby both U (1) and time reversal.Therefore, we propose the classification of these SPTsis Z . The properties of the surface Z topologically or-dered states of the root states are summarized in TableXII.
3. SPT with symmetry (( U (1) × SU (2)) /Z ) × T offermions For fermionic SPT with symmetry(( U (1) × SU (2)) /Z ) × T , free fermion band theorygives a Z classification, and each state can be labelledby an integer k , which is basically the number of pairsof massless Dirac fermions on the surface.The root state can have a surface state with two mass-less Dirac fermions with the following Hamiltonian H = ψ † ( − i∂ x σ x − i∂ y σ z ) ⊗ τ ψ (D1)8 q e q m T e T m S e S m comments eCmC
12 12 eT mT efmf eCm
12 12 e m
12 12 e mT eCmT Z topological ordered states of SPTsunder symmetry U (1) × T × SO (3). The topological sectorsare denoted by { , e, m, (cid:15) } . q e and q m represents the charge of e and m under U (1), T e and T m represents the Kramersnessof e and m under time reversal, and S e and S m representsthe spin of e and m , respectively. If the symmetry is ( U (1) (cid:111) T ) × SO (3), eCmT will be absent and all other six root statesremain. where σ and τ are the standard Pauli matrices with σ = τ = I , and σ acts on the internal indices of the Diracfermions and τ acts on the spin indices. Under U (1), ψ → ψe iθ (D2)Under SU (2), ψ → σ ⊗ U ψ (D3)with U an SU (2) matrix. And under time reversal ψ → iσ y ⊗ τ ψ † (D4)Notice the inverse of this root state, i.e. the state thatcan trivialize the root state when coupled together, canhave the same surface Hamiltonian as this root state ex-cept that time reversal acts as ψ → − iσ ⊗ τ ψ † . Thismeans the state labelled by k and that labelled by − k are identical after gauging the U (1), because the afore-mentioned sign difference in the time reversal action canbe eliminated by a U (1) gauge transformation. [30]When the U (1) symmetry is gauged, the monopole ofthe corresponding U (1) gauge field is a Kramers doubletthat carries spin-1/2.[77, 78] Because the method thatleads to this result will be used extensively later, it ishelpful to review it here.Since the surface is described by two free Diracfermions, which is a conformal field theory, one can usestate-operator correspondence to study the properties ofmonopoles, by imagining putting the surface on a spherewith 2 π flux threading out. Guaranteed by the indextheorem, each Dirac fermion will contribute a zero modein the background of the 2 π flux, in our case denoted by f and f , respectively. We also denote the flux back-ground with both zero modes empty by | (cid:105) . Because thetime reversal symmetry flips the U (1) charge here, thephysical gauge invariant states must have one of the zeromodes being occupied. That is, it should be f † | (cid:105) or f † | (cid:105) , which are bosonic. In light of state-operator corre-spondence, these two states correspond to two differentcharge-neutral monopole operators, denoted by M and M , respectively. Also, | (cid:105) corresponds to the operator of the ( − ,
1) dyon, and f † f † | (cid:105) corresponds to the op-erator of (1 ,
1) dyon. Then the quantum numbers of themonopole can be read off from the properties of thesestates.For example, for the surface theory described above,because the two Dirac fermions transform as spin-1/2under SU (2), the monopoles M ∼ f † | (cid:105) and M ∼ f † | (cid:105) also transform as spin-1/2. As for time reversal, thesetwo states transform as M ∼ f † | (cid:105) → f f † f † | (cid:105) = f † | (cid:105) ∼ M M ∼ f † | (cid:105) → f f † f † | (cid:105) = − f † | (cid:105) ∼ − M (D5)This means the monopoles are Kramers doublet undertime reversal. Therefore, after gauging the U (1) symme-try, this state becomes E bT M f .Now we turn to the classification of such fermionicSPTs. Upon adding interactions, the free fermion clas-sification collapses to Z .[77] It can be shown that thestate with 8 massless Dirac fermions on the surface istrivial, and the state with 4 massless Dirac fermions onthe surface is equivalent to eT mT , a bosonic SPT withsymmetry SO (3) ×T . There can be interacting SPTs be-yond band theory, which can be viewed as bosonic SPTswith symmetry SO (3) × T . They are classified by Z .One of the root states of these bosonic SPTs coincidewith a free fermion SPT that can have 4 massless Diracfermions on the surface, so we propose the complete clas-sification is Z × Z .
4. SPT with symmetry (( U (1) (cid:111) T ) × SU (2)) /Z ofKramers singlet fermions Consider fermionic SPT with symmetry (( U (1) (cid:111) T ) × SU (2)) /Z and assume T = 1 for these fermions. Freefermion band theory gives a Z classification. The rootstate can have a surface state with two massless Diracfermions, described by the same Hamiltonian as (D1),with the only difference that under time reversal ψ → σ y ⊗ τ y ψ (D6)When the U (1) symmetry of these Dirac fermions isgauged, the monopole of the corresponding U (1) gaugefield carries spin-1/2. [39] This can also be seen by usingthe method of state-operator correspondence reviewedabove. Notice in this case the time reversal symmetrydoes not flip the U (1) charge, so it is convenient to mo-mentarily suppose equipping the system with a furthercharge conjugation symmetry. We will determine theproperties of the monopoles in the presence of this furthersymmetry first, and then break this symmetry. Becausethe properties of the monopoles are described by somediscrete data, breaking this symmetry will not changeany of them.Again, each Dirac fermion will contribute a zero modeto the 2 π flux background, and the neutral bosonicmonopoles correspond to the two states with one zero9mode occupied: M , ∼ f † , | (cid:105) . Because the two Diracfermions carry spin-1/2 under SU (2), as above, themonopoles must also carry spin-1/2. It is not meaning-ful to discuss whether monopoles are Kramers doubletor not under time reversal, so this finishes determiningthe properties of the monopoles. From this discussion,we see that after gauging the U (1) symmetry this statebecomes E f M b .Although it is not meaningful to discuss whethermonopoles are Kramers doublet or not under time re-versal, it is interesting and helpful to determine howmonopoles transform under time reversal. To be con-sistent with that time reversal and SU (2) commute onthe monopole, under time reversal the monopole oper-ators must transform as M , → M † , (with a possiblephase ambiguity).How do we understand this time reversal action onmonopoles from the point of view of state-operator cor-respondence? This is a little tricky because in this casethe 2 π flux will be turned into a − π flux under timereversal, which also has two zero modes, denoted by ˜ f , ,where ˜ f , is contributed by ψ , , respectively. In partic-ular, denote | ˜0 (cid:105) as the state with − π flux backgroundand neither zero mode occupied. This is the time rever-sal partner of | (cid:105) , so it corresponds to the ( − , −
1) dyon.Similarly, ˜ f † ˜ f † | ˜0 (cid:105) is the time reversal partner of f † f † | (cid:105) ,so it corresponds to the (1 , −
1) dyon.Under time reversal, f † | (cid:105) → ˜ f † | ˜0 (cid:105) f † | (cid:105) → − ˜ f † | ˜0 (cid:105) (D7)where an unimportant U (1) phase factor has been sup-pressed. For this to be compatible with that M , → M † , under time reversal, we must identify (with anunimportant phase factor) M † ∼ ˜ f † | ˜0 (cid:105) , M † ∼ − ˜ f † | ˜0 (cid:105) (D8)To the best of our knowledge, this identification of theHermitian conjugate of the monopoles in the context ofstate-operator correspondence has not been given before,and it will be used later. We remark that this identifica-tion is true as long as the long-distance conformal fieldtheory is described by two massless Dirac fermions, andit should be independent of the microscopic symmetriesof the system, although we obtained it by considering asystem with a particular symmetry.Now we return to the classification of these fermionicSPTs. Upon adding interaction, the nontrivial state isstable. There are also SPTs beyond band theory withroot state that can be viewed as bosonic SPTs with sym-metry SO (3) × T , and they can be classified by Z . Oneof the four root states becomes trivial in the presence offermions with this symmetry (see Table V), therefore, wepropose that the total classification is Z . E. Relations between the classification of SO (3) × T symmetric U (1) quantum spin liquids and theclassification of some relevant SPTs In the main text SO (3) × T symmetric U (1) quantumspin liquids can be classified into 15 phases, as summa-rized in Table I and Table II. In particular, how they canbe viewed as gauged SPTs are also discussed. It is help-ful to understand the relation between the classificationof the U (1) quantum spin liquids and the classificationof the relevant SPTs. Below we give some examples.From the point of view of E , E b M b , E b M f , E b M b and E b M f can all be regarded as the gauged insulatorsof Kramers singlet bosons with symmetry ( U (1) (cid:111) T ) × SO (3). In Appendix D we propose that the bosonic SPTsunder this symmetry are classified by Z , where only twoof the six root states requires protection from the U (1)symmetry. It is not hard to see, after gauging this U (1)symmetry, the Z subset of SPTs coming from these tworoot states become precisely the above four quantum spinliquids.With the same symmetry as above, if the bosons areKramers doublet, Appendix B shows that only one of thetwo root states survives. Gauging the trivial insulatorand the nontrivial SPT from the other root state leadsto E bT M b and E bT M f , respectively.From the point of view of M , E b M b , E bT M b , E b M b , E bT M b , E f M b , E fT M b , E f M b and E fT M b can bethought of as the gauged bosonic insulators with symme-try U (1) × T × SO (3). In Appendix D we propose thatthe bosonic SPTs under this symmetry are classified by Z , where only three of the root states requires protectionfrom the U (1) symmetry. Gauging the Z subset of theSPTs generated by these three root states gives preciselythe above eight quantum spin liquids.From the point of view of M , both E b M f and E bT M f can be viewed as a gauged topo-logical superconductor of fermions with symmetry(( U (1) × SU (2)) /Z ) × T . In Appendix D we proposethat the topological superconductors with this symme-try are classified by Z × Z , where the first Z factorrepresents those can be realized with free fermions, andthe last Z factor corresponds to interacting topologicalsuperconductors beyond band theory. For states thatcan be realized by band theory, the nontrivial topologi-cal superconductors can have 2 k massless Dirac fermionson the surface, where k = 0 , , , k leads to E b M f (up to a bosonicSPT eT mT ) and gauging states with odd k leads to E bT M f . For states beyond band theory, gauging themresults in the same quantum spin liquid as their corre-sponding state within band theory up to a bosonic SPTwith symmetry SO (3) × T .From the point of view of E , both E f M b and E f M b can be viewed as gauged topological insulatorsof fermions with symmetry (( U (1) (cid:111) T ) × SU (2)) /Z ,where the microscopic fermions are Kramers singlets. In0Appendix D we propose that these topological insulatorsare classified by Z , where the first Z factor correspondsto those realizable by free fermions, and the nontrivialstate can have 2 massless Dirac fermions on the surface.Gauging the trivial state leads to E f M b and gaugingthe nontrivial state leads to E f M b . Gauing the statesbeyond band theory gives the same quantum spin liq-uids as their corresponding free fermion cousins up to abosonic SPT under symmetry SO (3) × T .The above examples show that gauging different SPTsmay results in the same quantum spin liquid, and no onesingle class of SPTs will lead to all quantum spin liquidsafter gauging, so the classification of these quantum spinliquids does not form a simple group structure, whilethe classification of SPTs does. As mentioned in theintroduction, viewing a single quantum spin liquid as twodifferent gauged SPTs leads to some helpful dualities onthe surface theories of these SPTs, which can be inferredfrom the above discussion.Also, if two different quantum spin liquids can beviewed as two different SPTs with the same microscopicconstitutes and symmetry, the quantum phase transitionbetween them can also be viewed as the gauged versionof the quantum phase transition between the two corre-sponding SPTs. For example, the quantum phase transi-tion between E f M b and E f M b can be viewed as thegauged version of the quantum phase transition betweenthe trivial and nontrivial fermionic insulators with sym-metry (( U (1) (cid:111) T ) × SU (2)) /Z . We will not go into thedetails in this paper. F. Bosonic SPTs with SO (3) × T symmetry In this appendix we discuss bosonic SPTs with symme-try SO (3) × T , with the assumption that the microscopicdegrees of freedom are non-Kramers bosons with spin-1. Group cohomology gives classification Z ,[54] but itmisses one root state[47], and the complete classificationshould be Z . The four root states all have anomaloussurface Z topological orders, where symmetries are re-alized in a way impossible in a purely two dimensionalsystem (see Table III). Among the four root states, eT mT and ef mf are protected by time reversal alone. Belowwe review the anomalies of e m and e mT .Suppose e m is realizable in a purely two dimensionalsystem, then tunneling an SO (3) monopole through itleaves a π -flux seen by both e and m . Because such alocal process should not have nonlocal observable effect,an (cid:15) , the fermionic bound state of e and m , must betrapped at this π -flux. Due to time reversal symmetry,there is no polarization spin in this process and this fluxis just a fermion. Therefore, a local process generates afermion in this 2D system, which is impossible. Noticethis anomaly is just the SO (3) version of the anomalyof eCmC under symmetry U (1) × T . In fact, when thesymmetry SO (3) × T is broken down to U (1) × T , the descendant state of e m is precisely eCmC , which hasa U (1) theta angle to be 2 π . This also implies the SO (3)Θ = 2 π for e m .As for e mT , tunneling an SO (3) monopole leaves a π -flux seen by e and (cid:15) , so this process must trap an m .Because SO (3) commutes with T , the SO (3) flux is in-variant under time reversal, and a Kramers doublet isgenerated by this local process. This contradicts the as-sumption that there is no local Kramers doublet parti-cles. Again, this anomaly is the SO (3) version of theanomaly of eCmT under symmetry U (1) × T , and thedescendant state of e mT is just eCmT when the sym-metry SO (3) × T is broken down to U (1) × T . G. Constraints on the Hall conductances due totime reversal and spin rotational symmetries
Suppose in addition to a U (1) c charge conservationsymmetry, a two dimensional gapped system also hastime reversal symmetry and spin rotational symmetry.One can also consider the spin quantum Hall conduc-tance, σ sxy , which relates the spin current due to a gra-dient Zeeman field.[79] This can be formally viewed asthe response of the system to a probe gauge field, A s ,which corresponds to the S z rotation symmetry, U (1) s .There can also be quantum spin Hall conductance, σ csxy ,which relates the spin current and the electric field of thegauge field A c , the gauge field corresponding to symme-try U (1) c .[80] This appendix discusses the constraints onthese Hall conductances due to time reversal and spinrotational symmetries. The results are useful in deter-mining what polarization charge or spin will be gener-ated when flux is inserted in the system, or equivalently,when a monopole tunnels through the system.To this end, we first reorganize the charge conservationand S z rotation symmetries in terms of two other U (1)symmetries, denoted by U (1) ↑ and U (1) ↓ . These two U (1) symmetries can be viewed as separate conservationsof spin-up and spin-down particles. The correspondinggauge fields and charges of these two symmetries are re-lated to A c and A s by A ↑ = A c + A s , A ↓ = A c − A s Q ↑ = Q c + Q s , Q ↓ = Q c − Q s Q and Q can independently take any integers, Q c and Q s have to either be both even or both odd.Now we can discuss the Hall conductances due to cou-pling to A ↑ and A ↓ . The general Hall response theoryreads L = 14 π ( σ ↑ xy A ↑ dA ↑ + σ ↓ xy A ↓ dA ↓ + 2 σ ↑↓ xy A ↑ dA ↓ )(G2)where AdB is a shorthand for (cid:15) µνλ A µ ∂ ν B λ . Using (G1),1we get σ cxy = σ ↑ xy + σ ↓ xy + 2 σ ↑↓ xy σ sxy = σ ↑ xy + σ ↓ xy − σ ↑↓ xy σ csxy = σ ↑ xy − σ ↓ xy (G3)Clearly any element in the spin rotational symmetrythat takes spin-up to spin-down (such as rotation around x -axis by π ) requires σ ↑ xy = σ ↓ xy . Below we study theconstraints from time reversal symmetry. Notice thatthe U (1) s charge is always odd under time reversal, butthe U (1) c charge can either be time reversal even or odd,and we discuss these two cases separately.We start from the case where the U (1) c charge is evenunder time reversal, which means A , → A , , (cid:126)A ↑ , ↓ → − (cid:126)A ↓ , ↑ (G4)For the response theory to be time reversal symmetric,we need σ ↑ xy = − σ ↓ xy , σ ↑↓ xy = 0 (G5)For the case where the U (1) c is odd under time rever-sal, time reversal transformation takes A ↑ , ↓ → − A ↑ , ↓ , (cid:126)A ↑ , ↓ → (cid:126)A ↑ , ↓ (G6)For the response theory to be time reversal symmetric,we need σ ↑ xy = σ ↓ xy = σ ↑↓ xy = 0 (G7)From these constraints and (G3) one can easily ob-tain the constraints on σ cxy , σ sxy and σ csxy . We note allthese constraints can also be obtained simply by apply-ing Laughlin’s flux insertion argument.We notice that all these Hall conductance vanishes ifthe system has both time reversal symmetry and spin ro-tational symmetry that contains at least O (2) (cid:39) U (1) s (cid:111) Z , where U (1) s is a rotational symmetry around one axisand Z is the π -rotation around another axis perpendic-ular to the previous one. This implies that inserting fluxor tunneling a monopole through such two dimensionalsystems will not lead to any polarization charge or spin. H. Non-edgeability of some Z topological ordersin the presence of nontrivial particles In the Sec. V A we claimed some Z topological ordersare not edgeable even in the presence of some nontrivialparticles, i.e. they do not allow for a physical edge sep-arating it and the trivial vacuum. In this appendix, wewill justify this claim by showing that these Z topolog-ical orders allow no K -matrix theory to describe them.
1. Brief review of the K -matrix theory We begin with a brief review of some general aspectsof the K -matrix theory. For more details, see Ref. 63and Ref. 1, 7, and 59.The Lagrangian of a K -matrix theory of a system thatcouples to an external U (1) gauge field A c is given by L = K IJ π a I da J − q Ic π A c da I (H1)with K a symmetric invertible matrix with all entriesintegers, and q c a vector with all entries integers.An excitation of this theory can be represented by anintegral excitation vector, l . The charge of this excitationunder the external gauge field A c is l T K − q c (H2)And this excitation has self-statistics angle πl T K − l (H3)For two excitations represented by excitation vectors l and l , respectively, the mutual braiding angle betweenthem is 2 πl T K − l (H4)A simple example is that K = (cid:18) (cid:19) (H5)which represents Z topological order, where e can betaken to be represented by excitation vector (1 , T and m can be taken to be represented by (0 , T .The 2+1-d bulk theory (H1) allows the followingboundary theory: L = 14 π ( K IJ ∂ t φ I ∂ x φ J − V IJ ∂ x φ I ∂ x φ J ) (H6)where φ I are bosonic fields such that e il I φ I is the anni-hilation operator of excitation l on the boundary. Thesebosonic fields satisfy Kac-Moody algebra[ φ I ( x ) , ∂ y φ J ( y )] = 2 πi ( K − ) IJ δ ( x − y ) (H7) V IJ is called the velocity matrix that gives the velocitiesof these bosonic fields.The above summarizes the topological properties of the K -matrix theory, (H1). Below we review the symmetryactions on this theory.In general symmetries act on the gauge fields a I as amatrix. For example, we denote the time reversal actionas a I → T IJ a J (H8)with T an integral matrix. Notice the above equationonly gives the transformation of the spatial componentsof the gauge fields, and the temporal components should2have a minus sign in front due to the anti-unitary natureof time reversal symmetry.It is important to notice the above does not fully spec-ify the symmetry action, and to that end, one needs tospecify how the boundary bosonic fields transform. Ingeneral, they transform as φ I → T IJ φ J + t I (H9)with t I a real vector. [1]To make the bulk theory (H1) invariant under anti-unitary time reversal symmetry, we need K → T T KT = − K (H10)If the U (1) charge is even under time reversal, we furtherrequire q c → T T q c = q c (H11)while if the U (1) charge is odd under time reversal werequire q c → T T q c = − q c (H12)
2. Non-edgeability of some Z topological orders inthe presence of nontrivial particles Now by showing some Z topological orders even inthe presence of nontrivial particles do not allow a K -matrix theory description, we show their non-edgeabilitybecause K -matrix theories are supposed to capture alltwo dimensional Abelian states.Here we list the Z topological orders of interests. Wedenote eT mT in the presence of bosons with quantumnumbers C T by ( eT mT, bC T ), and eT mT in thepresence of fermions with quantum number ˜ C by( eT mT, f ˜ C ). Besides these two, we will also con-sider ( e mT, bC T ), ( eT mT, bC ), ( eT mT, f C T ),( eT mT, f C ), ( eT mT, f ˜ C ), ( eT mT, f C T ),( e mT, f C ) and ( eT mT, f C T ), with similarnotations as before.We immediately have two main difficulties in showingtheir non-edgeability. First, in some cases the nontriv-ial particles carry spin-1/2 and we need to incorporate SU (2) symmetry in the K -matrix theory, but continu-ous non-Abelian symmetries are usually not manifest ina K -matrix theory and dealing with them directly is gen-erally difficult. To resolve this difficulty, we will insteadjust show that the descendants of the relevant states arestill not edgeable when the SU (2) symmetry is brokendown to U (1), which is sufficient to show the originalstates are non-edgeable with the full SU (2) symmetry.To distinguish this U (1) from the original charge U (1),we will denote the charge U (1) by U (1) c , and this U (1) by U (1) s . Unit charge under U (1) s will be denoted by C (cid:48) .Therefore, for example, we will consider ( eT mT, f C C (cid:48) )instead of ( eT mT, f C ). Now (H1) needs to be mod-ified to include the coupling to A s , the external gauge field corresponding to U (1) s L = K IJ π a I da J − q Ic π A c da I − q Is π A s da I (H13)Notice the charge of U (1) s is always odd under time re-versal, so time reversal symmetry requires that q s → T T q s = − q s (H14)The second difficulty is that in general the state thatwe are interested in may be described by a K -matrix witha large dimension, but dealing with a large-dimensional K -matrix is daunting. However, the following observa-tion suggests we actually only need to consider a 2 × K -matrix.Notice all these Z topological orders come from eT mT , e mT and eT mT . For both cases, the non-trivial topological quasiparticles only need a single com-ponent to describe them. This is because whenever thereare two components of them, one can condense somebound states of them that are singlets under all symme-tries. This will not change the topological order or thesymmetry of the system, but only one component will beleft over.[8] More concretely, this means to describe theputative Z topological orders that we are interested in,we should always be able to write the K -matrix as K = 2 σ x ⊕ L (H15)where L is an invertible symmetric integral matrix thatcan be large in dimension, and L describes only local exci-tations. For bosons, L can be written as σ x ⊕ σ x ⊕· · ·⊕ σ x ,while for fermions, L can be written as σ z ⊕ σ z ⊕ σ x ⊕· · · .In this form, any excitation with an excitation vector ofthe form (1 , , · · · ) T can be identified as e , and all excita-tions with an excitation vector of the form (0 , , · · · ) T canbe viewed as m , where the “ · · · ” can be nonzero. At thismoment, the nontrivial topological quasiparticle that weare after, for example, the Kramers doublet e particle in eT mT , can still be represented by (1 , , · · · ) T with “ · · · ”nonzero. But we can always bind proper local excitationsto this excitation so that the excitation vector becomes(1 , , , , · · · ) T with “ · · · ” all zeros.The argument above shows that, in order to showthe non-edgeability of those Z topological orders, itis sufficient to show that no 2 × K -matrix can de-scribe the topological quasiparticles with the correspond-ing quantum numbers, up to binding local excitations.Let us demonstrate this via the following example. In( eT mT, bC T C (cid:48) ), eT mT can be relabelled as, for ex-ample, eC C (cid:48) mT . The above statement means that, inorder to show that ( eT mT, bC T C (cid:48) ) is not edgeable, itis sufficient to show that none of eT mT , eC C (cid:48) mT andall other states related to these by binding a local exci-tation made up of bC T C (cid:48) can be realized by a 2 × K -matrix.Because time reversal should not convert a local exci-tation into a nonlocal one, we expect that the matrix T can be written in the following form T = (cid:18) T T T (cid:19) (H16)3where T is a 2 × K and T into (H10), we see time reversal symmetryrequires that T T σ x T = − σ x (H17)It is easy to show the only solutions are T = ± σ z or T = ± (cid:15) , where (cid:15) = iσ y .Notice in all the cases we consider, the quantum num-bers of e and m are always nontrivial. If T = ± (cid:15) , then T = −
1. This does not allow Kramers doublet struc-ture, and it also does not allow nonzero q c and q s thatsatisfy (H11) or (H12) and (H14). So this choice of T can never work.So we can focus on the case with T = ± σ z . Withoutloss of generality, we take T = − σ z . Notice now e ,represented by the excitation vector (1 , T , is always aKramers singlet independent of t . If the second entry of t is π/ m , represented by excitation vector (0 , T , is aKramers doublet.With this choice of T , in order to satisfy (H11) or(H12) and (H14), q c and q s can only be taken as q c = (0 , q ) T q s = ( q , T (H18)when the charge under A c is even under time reversal, or q c = ( q , T q s = ( q , T (H19)when the charge under A c is odd under time reversal.In the first case, e carries charge q / A c and zerocharge under A s . In the second case, e carries zero chargeunder both A c and A s . eT mT is not edgeable in the presence of bC T C (cid:48) The above discussions immediately imply that eT mT is not edgeable in the presence of bC T C (cid:48) . This is be-cause e cannot be a Kramers doublet, an odd number of bC T C (cid:48) s have to be attached to it to cancel its Kramer-sness. Then e carries nonzero charge under A s , which isin contradiction with e always carrying zero charge under A s . So ( eT mT, bC T ) is not edgable. eC (cid:48) mT is not edgeable in the presence of bC T C (cid:48) Here e is not a Kramers doublet but it carries nonzerocharge under A s . To cancel this charge, an odd numberof bC T C (cid:48) s have to be attached to e , which makes ita Kramers doublet. This is again impossible as arguedabove. One can also try to switch the label between e and m , then it becomes eT mC (cid:48) . The argument above im-plies this is inconsistent even in the presence of bC T C (cid:48) .So ( e mT, eC T ) is not edgeable. eT C (cid:48) mT is not edgeable in the presence of bC C (cid:48) Here no matter how many bC C (cid:48) ’s are attached,the Z topological order always has both e and m be-ing Kramers doublet. This cannot be realized. So( eT mT, bC ) is not edgeable. eT mT is not edgeable in the presence of f ˜ C or f ˜ C C (cid:48) Here e is a Kramers doublet, so an odd number of f ˜ C s or f ˜ C C (cid:48) need to be attached to it, which makes itbecome (cid:15) and the new e carry ˜ C . This is in contradictionwith e carrying zero charge under A c in this case. So( eT mT, f C ) and ( eT mT, f C ) are not edgeable. eT mT and eT C (cid:48) mT are not edgeable in the presence of fC T or fC T C (cid:48) Here e and m are Kramers doublets, and attaching anynumber of f C T or f C T C (cid:48) always leaves both e and m Kramers doublets, so ( eT mT, f C T ), ( eT mT, f C T )and ( eT mT, f C T ) are not edgeable. eT mT and eC (cid:48) mT are not edgeable in the presence of fC C (cid:48) Here an odd number of f C C (cid:48) s need to be attachedto e , which converts it to (cid:15) and make the new e carrynonzero charge under A c and A s . But e cannot carrynonzero charge under both A c and A s . So ( eT mT, f C )and ( e mT, f C ) are not edgeable.In summary, none of the Z topological orders is edge-able in the presence of the relevant nontrivial excitations.This implies they are all still anomalous. I. Projective representations: the electric(standard), the magnetic (twisted) and thedyonic (mixed) ones
In this appendix we discuss in detail various projec-tive representations of a symmetry group G : the elec-tric (standard), the magnetic (twisted) and the dyonic(mixed) ones. We always assume this group G can con-tain time reversal, but besides, for simplicity, all otherelements form a connected unitary group. That is, theseelements are all unitary and they can all be continuouslyconnected to the identity element. We will see althoughall these projective representations are classified by somegroup cohomologies, but different cases are classified bydifferent group cohomologies.4
1. Electric (standard) projective representations
We begin with the familiar case of standard projectiverepresentations. Although our results will be identical tothe ones in textbooks, we will use a different formulationthat is more appropriate for our purposes and easier togeneralize to twisted projective representations.Suppose there is a symmetry G , which can in prin-ciple contain anti-unitary element. If all elements of G only change an excitation by a local operation, then it isappropriate to discuss the standard projective represen-tations of G on this excitation.A prototypical example of this case is that the relevantexcitation is the electric charge E of a U (1) quantum spinliquid. In general, the action of an element g ∈ G on E can be written as E i → U ( g ) ij E j (I1)Here different components of E i differ from each otherby a local operation, and U ( g ) is a matrix representationof g .Because E ∗ i A ij E j is a local operator for any matrix A , this operator is supposed to transform in the linearrepresentation of G . For g , g ∈ G , acting g followedby g on this operator gives E † (cid:16) U ( g ) s ( g ) U ( g ) (cid:17) † · A s ( g g ) · (cid:16) U ( g ) s ( g ) U ( g ) (cid:17) E (I2)where now E represents a column vector with compo-nents E i , and, for an arbitrary matrix M , M s ( g ) = M if g is unitary, while M s ( g ) = M ∗ if g is anti-unitary. Forthe special case where M is just a phase factor, s ( g ) = 1( s ( g ) = −
1) if g is unitary (anti-unitary).The above result should be identical to the one ob-tained by acting g g on this local operator directly: E † · U † ( g g ) · A s ( g g ) · U ( g g ) · E (I3)For these two results to be identical for an arbitrary ma-trix A , we must have U ( g ) s ( g ) U ( g ) = ω ( g , g ) U ( g g ) (I4)where ω ( g , g ) is a phase factor.The above equation can be written in the followingequivalent way: U ( g g ) = ω ( g , g ) − U ( g ) s ( g ) U ( g ) (I5)For any g , g , g ∈ G , applying this equation to U ( g g g ) yields U ( g g g )= ω ( g , g g ) − U ( g g ) s ( g ) U ( g )= ω ( g , g g ) − ω ( g , g ) − s ( g ) U ( g ) s ( g g ) U ( g ) s ( g ) U ( g )= ω ( g g , g ) − U ( g ) s ( g g ) U ( g g )= ω ( g g , g ) − ω ( g , g ) − U ( g ) s ( g g ) U ( g ) s ( g ) U ( g ) This implies the following associativity condition for thephase factor ω ’s: ω ( g g , g ) ω ( g , g ) = ω ( g , g g ) ω ( g , g ) s ( g ) (I6)There is a gauge freedom for the phase factor ω ’s. Tosee this, notice the symmetry action of g ∈ G on E canbe modified by a gauge transformation: E i → λ ( g ) U ( g ) ij E j ≡ ˜ U ( g ) ij E j (I7)where λ ( g ) is a U (1) phase factor. The action of g ∈ G on any local operator will be the same, which means˜ U ( g ) is an equally good representation of g . Under thistransformation, it is straightforward to check that˜ U ( g ) s ( g ) ˜ U ( g ) = ˜ ω ( g , g ) ˜ U ( g g ) (I8)where ˜ ω ( g , g ) = ω ( g , g ) · λ ( g ) λ ( g ) s ( g ) λ ( g g ) (I9)The factor systems ω and ˜ ω related in this way shouldbe regarded to be in the same class, because they give riseto identical results in any local operator. It is straight-forward to check that the relation (I9) is an equivalencerelation, that is, it is reflexive, symmetric and transitive.Furthermore, it is clear if ω and ω are the two classesof factor systems corresponding to representations U ( g )and U ( g ), respectively, ω · ω will be the factor systemof the representation U ( g ) · U ( g ). This defines a multi-plication operation among the classes of factor systems.With this multiplication, the classes of factor systemsform an Abelian group, where the trivial element is theclass of factor systems of a linear representation. In fact,this Abelian group form a structure of group cohomol-ogy H ( G, U T (1)).[54] In this group cohomology, the n -cochains ω n ( g , g , · · · , g n ) take value as a phase factor,the 1-coboundary operation is defined as d ω ( g , g ) = ω ( g ) ω ( g ) s ( g ) ω ( g g ) (I10)and the 2-coboundary operation is defined as d ω ( g , g , g ) = ω ( g , g ) s ( g ) ω ( g , g g ) ω ( g g , g ) ω ( g , g ) (I11)It is straightforward to check that d d = 1. Also,the solutions to the associativity condition (I6) are 2-cocycles, and different solutions are identified up to a1-coboundary. Therefore, the classes of factor systems,or the (standard) projective representations, are indeedclassified by this cohomology. Below we will see thetwisted and mixed projective representations are alsoclassified by some group cohomologies, which are how-ever different from this H ( G, U T (1)).5
2. Magnetic (twisted) projective representations
Next we turn to twisted projective representations. Aprototypical example where it is appropriate to considertwisted projective representations is, when the symmetryincludes time reversal, to consider the fractional quantumnumbers on the magnetic monopole, M .Suppose g ∈ G is unitary, its action on M can berepresented as M i → U ( g ) ij M j (I12)Suppose g ∈ G is anti-unitary, its action on M can berepresented as M i → U ( g ) ij M ∗ j (I13)Again, because M ∗ i A ij M j is a local operator for anymatrix A , it is supposed to transform in the linear rep-resentation of G . Similar to the analysis in the previouscase, this implies, for g , g ∈ G , U ( g ) s ( g ) U ( g ) s ( g ) = ω ( g , g ) U ( g g ) (I14)where ω ( g , g ) is a phase factor.Similar as standard projective representations, thesephase factors need to satisfy an associativity condition: ω ( g g , g ) ω ( g , g ) s ( g ) = ω ( g , g g ) ω ( g , g ) s ( g ) (I15)Further, there is also a gauge freedom that leads to thefollowing equivalence relation ω ( g , g ) ∼ ˜ ω ( g , g )= ω ( g , g ) · λ ( g ) s ( g ) λ ( g ) s ( g ) λ ( g g ) (I16)where λ ’s are phase factors.Just as in the case of the standard projective repre-sentations, the classes of factor systems of a twisted pro-jective representation also form an Abelian group, whosemultiplication, trivial element, and inverse element aredefined in parallel as in the case of the standard pro-jective representation. This group is also described bya cohomology, denoted by H ( G, U MT (1)). This coho-mology is different from the previous one, H ( G, U T (1)),in the coboundary operations. In this cohomology, the n -cochains ω n ( g , g , · · · , g n ) still take values as phasefactors, the 1-coboundary operation is defined as d ω ( g , g ) = ω ( g ) s ( g ) ω ( g ) s ( g ) ω ( g g ) (I17)and the 2-coboundary operation is defined as d ω ( g , g , g ) = ω ( g , g g ) ω ( g , g ) s ( g ) ω ( g g , g ) ω ( g , g ) s ( g ) (I18)It is straightforward to check d d = 1. Again, thesolutions of the associativity condition are 2-cocycles, and they are identified up to a 1-coboundary. There-fore, twisted projective representations are classified by H ( G, U MT (1)).Interestingly, for the cases with G = 1, G = Z and G = U (1), H (cid:0) G × T , U MT (1) (cid:1) = H ( G × Z , U (1)),where the latter is the standard group cohomology with G × Z acting trivially on the U (1) coefficient. It will beinteresting to know if this relation is always true.
3. Dyonic (mixed) projective representations
In the case of a G symmetric U (1) quantum spin liq-uid at θ = π , the property of the phase is determined bythe (cid:0) , (cid:1) and (cid:0) , − (cid:1) dyons. Denote these two dyonsby D (+) and D ( − ) , respectively. The symmetry quan-tum numbers of these two dyons are given by the dyonic(mixed) projective representations.Again, assume the only part of the symmetry that canchange the type of fractional excitations is time reversal,the action of g ∈ G on D (+) and D ( − ) can be written as D (+) i → U + ( g ) ij D (+) , D ( − ) i → U − ( g ) ij D ( − ) (I19)if g is unitary, and D (+) i → U + ( g ) ij D ( − ) , D ( − ) i → U − ( g ) ij D (+) (I20)if g is anti-unitary.Now using that D (+) ∗ i A ij D (+) j and D ( − ) ∗ i A ij D ( − ) j arelocal operators for any matrix A , we get U s ( g ) i ( g ) U s ( g ) · i ( g ) = ω i ( g , g ) U i ( g g ) (I21)where i = ± , and ω i ( g , g ) is a phase factor.In this case the associativity condition becomes ω i ( g , g g ) ω i ( g , g ) s ( g ) = ω i ( g g , g ) ω s ( g ) · i ( g , g ) (I22)And the equivalence relation becomes ω i ( g , g ) ∼ ˜ ω i ( g , g )= ω i ( g , g ) · λ s ( g ) · i ( g ) λ i ( g ) s ( g ) λ i ( g g ) (I23)Similar as the twisted projective representations, themixed projective representations also form an Abeliangroup and are also classified by a group cohomology, de-noted by H (cid:0) G, U D (1) × U D (1) (cid:1) . Here the n -cochains ω i,n ( g , g , · · · , g n ) take values as a phase factor (for i = ± separately), the 1-coboundary operation is definedas d ω i, ( g , g ) = ω s ( g ) · i, ( g ) ω i, ( g ) s ( g ) ω i, ( g g ) (I24)and the 2-coboundary operation is defined as d ω i, ( g , g , g ) = ω i, ( g , g g ) ω i, ( g , g ) s ( g ) ω i, ( g g , g ) ω s ( g ) · i, ( g , g )(I25)6 T E T (cid:48) E [ T , Z ] M E b M b E bT M b − E bT (cid:48) M b − E bTT (cid:48) M b − − E b M b − − E f M b E fT M b − E fT (cid:48) M b − E fTT (cid:48) M b − E fTT (cid:48) M b − − − − E b M f E bT M f − E bT (cid:48) M f − E bTT (cid:48) M f − − E b M f − − TABLE XIII. List of non-anomalous Z × T symmetric U (1)quantum spin liquids that have θ = 0 and have Z not actingas a charge conjugation. All these states are anomaly-free. T E = 1 ( T E = −
1) represents the case where E is a Kramerssinglet (doublet) under T . T (cid:48) E = 1 ( T (cid:48) E = −
1) representsthe case where E is a Kramers singlet (doublet) under T (cid:48) .[ T , Z ] M = + ([ T , Z ] M = − ) represents the case where Z and T commute (anti-commute) on M . It is straightforward to check d d = 1. Clearly, the so-lutions of the associativity condition are 2-cocycles, anddifferent solutions are identified up to a 1-coboundary.So the mixed projective representations are classified by H ( G, U D (1) × U D (1)). J. Examine the anomalies of Z × T symmetric U (1) quantum spin liquids with θ = 0 In this appendix we will give more details of theanomaly-detection of the 72 different putative U (1) quan-tum spin liquids with Z × T symmetry that have θ = 0.Among these states, Z does not act as a charge conjuga-tion for 24 of them and acts as a charge conjugation forthe other 48. These 72 states are all listed in Sec. VII,and they are copied in Table XIII, Table XIV, Table XVand Table XVI for convenience.We will first show that the 15 states in Table XIIIand the 21 states in Table XV are anomaly-free, andgive their constructions. Then we will show that the 9states in Table XIV and the 27 states in Table XVI areanomalous.Among the 36 anomaly-free states mentioned above,26 of them have at least one of E and M being a triv-ial boson. These states clearly do not suffer from anyanomaly, and they can be viewed as some gauged trivialinsulators. The other 10 states, E bT M f , E bT (cid:48) M f , E bT T (cid:48) M f , E fT T (cid:48) M b − , ( E fT M bT (cid:48) ) − , ( E bT M fT (cid:48) ) − , ( E fZ M bT (cid:48) Z ) − , ( E bT Z M fZ ) − , ( E fT Z M bZ ) − , ( E bZ M fT (cid:48) Z ) − , (J1)can be viewed as gauged free-fermion SPTs, which will T E T (cid:48) E [ T , Z ] M anomaly class E bT (cid:48) M f − − − class a E fT M b − − − class a E bT (cid:48) M b − − − class a E fT (cid:48) M b − − − class b E bT M f − − − class b E bT M b − − − class b E bTT (cid:48) M f − − − − class c E bTT (cid:48) M b − − − − class c E f M b − − class cTABLE XIV. List of anomalous Z ×T symmetric U (1) quan-tum spin liquids that have θ = 0 and have Z not actingas a charge conjugation. All these states are anomaly-free. T E = 1 ( T E = −
1) represents the case where E is a Kramerssinglet (doublet) under T . T (cid:48) E = 1 ( T (cid:48) E = −
1) representsthe case where E is a Kramers singlet (doublet) under T (cid:48) .[ T , Z ] M = + ([ T , Z ] M = − ) represents the case where Z and T commute (anti-commute) on M . The last column in-dicates the anomaly classes. T E Z E T (cid:48) M Z M ( E b M b ) − E bZ M b ) − − E bT M b ) − − E bTZ M b ) − − − E b M bZ ) − − E b M bT (cid:48) ) − − E b M bT (cid:48) Z ) − − − E f M b ) − E fZ M b ) − − E fT M b ) − − E fTZ M b ) − − − E b M f ) − E b M fZ ) − − E b M fT (cid:48) ) − − E b M fT (cid:48) Z ) − − − E fT M bT (cid:48) ) − − − E bT M fT (cid:48) ) − − − E fZ M bT (cid:48) Z ) − − − − E bTZ M fZ ) − − − − E fTZ M bZ ) − − − − E bZ M fT (cid:48) Z ) − − − − Z × T symmetric U (1)quantum spin liquids that have θ = 0 and have Z acting asa charge conjugation. T E = 1 ( T E = −
1) represents the casewhere E is a Kramers singlet (doublet) under T . T (cid:48) M = 1( T (cid:48) M = −
1) represents the case where M is a Kramers singlet(doublet) under T (cid:48) . Z E,M represents the result of acting thecharge conjugation twice on E and M , respectively. be constructed below. To show that all other states areanomalous, as discussed in Sec. VII C, it is sufficient toshow that ( E bT M bT (cid:48) ) − and E bT T (cid:48) M b − are anomalous.The rest of this appendix is organized as follows. InAppendix J 1 we will construct the relevant free-fermionSPTs, which after gauging give rise to states in (J1).Then we will show that ( E bT M bT (cid:48) ) − is anomalous in Ap-pendix J 2, and that E bT T (cid:48) M b − is anomalous in Appendix7 T E Z E T (cid:48) M Z M anomaly class( E bZ M bZ ) − − − E bTZ M bT (cid:48) Z ) − − − − − E fT M bZ ) − − − E bZ M fT (cid:48) ) − − − E fT M bT (cid:48) Z ) − − − − E bTZ M fT (cid:48) ) − − − − E bTZ M bZ ) − − − − E f M bZ ) − − E bTZ M f ) − − − E bZ M bT (cid:48) Z ) − − − − E bZ M f ) − − E f M bT (cid:48) Z ) − − − E bZ M bT (cid:48) ) − − − E fTZ M bT (cid:48) ) − − − − E fTZ M bT (cid:48) Z ) − − − − − E bT M bT (cid:48) Z ) − − − − E bT M fZ ) − − − E bZ M fZ ) − − − E bT M bZ ) − − − E bT M fT (cid:48) Z ) − − − − E bTZ M fT (cid:48) Z ) − − − − − E bTZ M bT (cid:48) ) − − − − E fZ M bT (cid:48) ) − − − E fZ M bZ ) − − − E bT M bT (cid:48) ) − − − E bT M f ) − − E f M bT (cid:48) ) − − Z ×T symmetric U (1) quan-tum spin liquids that have θ = 0 and have Z acting as acharge conjugation at θ = π . T E = 1 ( T E = −
1) representsthe case where E is a Kramers singlet (doublet) under T . T (cid:48) M = 1 ( T (cid:48) M = −
1) represents the case where M is a Kramerssinglet (doublet) under T (cid:48) . Z E,M represents the result of act-ing the charge conjugation twice on E and M , respectively.The last column lists the anomaly classes. J 3.
1. Constructions of the relevant free-fermion SPTs
This subsection gives the construction of the free-fermion SPTs corresponding to states in (J1). All thesefree-fermion topological insulators have two Dirac coneson the surface, and the surface Hamiltonian can be writ-ten as H = (cid:88) i =1 ψ † i ( − i∂ x σ x − i∂ y σ z ) ψ i (J2)The differences among these states are in the symmetryassignments. Denote ψ = ( ψ , ψ ) T , in all cases there isa U (1) symmetry under which ψ → e iθ ψ . We will alsoassign time reversal and Z symmetries to these states,such that there is no symmetry-allowed fermion bilinearterm that can open a gap on the surface. Then we willshow the bosonic monopoles of these topological insu-lators have the desired nontrivial properties, using the method in Ref. 78 (reviewed in Appendix D).Let us start with the case where Z does not act asa charge conjugation, and give the construction of free-fermion SPTs corresponding to E bT M f and E fT T (cid:48) M b − .For the corresponding SPT of E bT M f , let the symmetriesbe assigned as T : ψ → σ y ψ † Z : ψ → τ y ψ T (cid:48) : ψ → σ y τ y ψ † (J3)Clearly the action of T and Z commute on thefermion, so after gauging the fermion will become the M f . Now we check the symmetry quantum number ofthe E , which is the monopole of ψ . Using state-operatorcorrespondence, this is equivalent to checking the prop-erties of the two zero-energy state in the presence of a 2 π flux background with one of the two zero modes being oc-cupied. Denote these zero modes by f and f , which arerelated to ψ and ψ , respectively. And denote the statewith a 2 π flux background and none of the zero modesbeing occupied by | (cid:105) . Because T will flip the charge butkeep the flux, under T , f † | (cid:105) → f f † f † | (cid:105) = f † | (cid:105) f † | (cid:105) → f f † f † | (cid:105) = − f † | (cid:105) (J4)Notice the above transformations can be modified byan unimportant phase factor. This means E will be aKramers doublet under T . In fact, here the particle-hole-like T is enough to protect the Dirac cones, and itis shown this is sufficient to show that E is a Kramersdoublet under T .[77] Under T (cid:48) , f † | (cid:105) → f f † f † | (cid:105) = − f † | (cid:105) f † | (cid:105) → − f f † f † | (cid:105) = − f † | (cid:105) (J5)This means E will be a Kramers singlet under T (cid:48) . Thisis consistent with that T (cid:48) is not enough to protect theDirac cones. Therefore, after gauging this state indeedbecomes E bT M f .From this, E bT (cid:48) M f can be constructed similarly, and E bT T (cid:48) M f can be obtained by combining E bT M f and E bT (cid:48) M f . To obtain E fT T (cid:48) M b − , let the symmetries beassigned as T : ψ → σ y ψZ : ψ → τ x ψ (J6)One can show that this state becomes E fT T (cid:48) M b − aftergauging by using state-operator correspondence, but analternative point of view can be obtained by consideringthis state as a descendant of the corresponding SPT ofthe SO (3) × T symmetric E f M b , which has been de-scribed in details in Appendix D. To see it, denote thethree generators of SO (3) by S x , S y and S z , and denotethe generator of T by t . Now break the SO (3) × T to Z × ˜ T , where the Z is generated by exp ( iS x π ), and8˜ T is generated by exp ( iS y π ) · t . It is straightforward tocheck that the descendant state is E f ˜ T ˜ T (cid:48) M b − Now we turn to states with Z acting as a charge con-jugation. Let us start with the example correspondingto ( E fT Z M bZ ) − . This state is actually the gauged ver-sion of the free-fermion topological insulator in class CII,which has been discussed in Ref. 81 (but using a dif-ferent notation as here). The time reversal and chargeconjugation symmetries are assigned as T : ψ → σ y ψZ : ψ → τ y ψ † (J7)Now let us first examine the T (cid:48) action on the two statescorresponding to monopoles, whose action on ψ is T (cid:48) : ψ → σ y τ y ψ † (J8)This is the same T (cid:48) action as in the corresponding SPTof E bT M f , so f † | (cid:105) and f † | (cid:105) correspond to Kramers sin-glets under T (cid:48) .Next let us examine the Z action on f † | (cid:105) and f † | (cid:105) .Notice the 2 π flux background is converted into a − π flux background, which also has two zero modes. Usingthe method in Ref. 81, we argue that for such systemswith two symmetry-protected Dirac cones, the value ofcharge-conjugation squared on the neutral monopole isthe same as the value of charge-conjugation squared onthe Dirac fermions. The simplest way to see this is tonotice that these states have θ = 2 π . For the state cor-responding to a trivial insulator, which has θ = 0, themonopole has trivial quantum number, that is, the valueof charge-conjugation squared is 1. Then one can tune θ by 2 π to get a state corresponding to the topologi-cal insulator. In intermediate process of tuning θ , thetime reversal symmetry is generically broken. But theexistence of such nontrivial topological insulator impliesat the end the system will have time reversal symmetrywhen θ becomes 2 π . On the other hand, this process willnot break the charge conjugation symmetry. Then ac-cording to the Witten effect,[82] the original (1 ,
1) dyonnow becomes the (0 ,
1) monopole, and this new monopolehas the value of charge-conjugation squared to be − θ ofthe U (1) gauge theory. It can actually also be obtaineddirectly by using state-operator correspondence. Recallit has been shown in Appendix D that M † ∼ f † | ˜0 (cid:105) and M † ∼ − f † | ˜0 (cid:105) , where | ˜0 (cid:105) is the state with − π flux back-ground and neither zero mode being occupied. Undercharge conjugation, both charge and flux will be occu-pied. So under a convention of phase factors we canchoose | (cid:105) → f † f † | ˜0 (cid:105) under charge conjugation, then themonopole operators transform as M ∼ f † | (cid:105) → ˜ f ˜ f † ˜ f † | ˜0 (cid:105) = − ˜ f † | ˜0 (cid:105) ∼ M † M ∼ f † | (cid:105) → − ˜ f ˜ f † ˜ f † | ˜0 (cid:105) = − ˜ f † | ˜0 (cid:105) ∼ − M † (J9)Again, unimportant U (1) phase factors have been sup-pressed. The above transformation shows that the value of charge-conjugation squared is indeed − E fT Z M bZ ) − .Next we turn to the free-fermion SPT corresponding to( E fT M bT (cid:48) ) − , where the assignments of the time reversaland charge conjugation symmetries are T : ψ → σ y ψZ : ψ → ψ † (J10)Now we check the whether these monopoles are Kramersdoublets under T (cid:48) , whose action on ψ is T (cid:48) : ψ → σ y ψ † (J11)This is the same as the T action in the SPT correspond-ing to E bT M f , so here the monopole must be a Kramersdoublet under T (cid:48) .As for the value of charge-conjugation squared on themonopole, f † | (cid:105) → f ˜ f † ˜ f † | ˜0 (cid:105) ∼ ˜ f † | ˜0 (cid:105) ∼ M † f † | (cid:105) → f ˜ f † ˜ f † | ˜0 (cid:105) ∼ − ˜ f † | ˜0 (cid:105) ∼ M † (J12)so the value of charge-conjugation squared is 1 for themonopoles. Therefore, after gauging this state becomes( E fT M bT (cid:48) ) − .Finally, for the free-fermion SPT corresponding to( E fZ M bT (cid:48) Z ) − , the assignments of the time reversal andcharge conjugation symmetries are T : ψ → σ y τ y ψZ : ψ → τ y ψ † (J13)This state has the same Z action as the one giving rise to( E fT Z M bZ ) − , and the same T (cid:48) action as the one givingrise to ( E fT M bT (cid:48) ) − . In light of the previous discussion,the monopole should have charge-conjugation squared tobe − T (cid:48) . Therefore,after gauging this state becomes ( E fZ M bT (cid:48) Z ) − .By now the constructions of free-fermion SPTs corre-sponding to states in (J1) are given. Before we finishthis subsection, we make some remarks on free-fermiontopological insulators with time reversal and a unitary Z symmetry, which may or may not act as a chargeconjugation. In all these free-fermion topological insu-lators, the surface will always have an even number ofDirac cones in order to have θ = 0. If it has 4 Diraccones on the surface, the corresponding U (1) quantumspin liquid in general has a trivial monopole. In order toget a U (1) quantum spin liquid with nontrivial monopole,the corresponding free-fermion SPT should have only 2surface Dirac cones. We have actually exhausted all pos-sible free-fermion topological insulators with two surfaceDirac cones, and they only give the 10 U (1) quantumspin liquids in (J1) that have nontrivial monopole. Onthe other hand, if a nontrivial fermionic topological insu-lator is equivalent to a bosonic SPT, the monopole mustalso be trivial.[49, 77] Thus, if a state with fermionic9charge and nontrivial monopole is anomaly-free and dis-tinct from the above three (such as ( E f M bT (cid:48) ) − ), it im-plies the existence of an intrinsically interacting fermionicSPT, which is a nontrivial fermionic SPT that cannotbe realized by free-fermions and is not equivalent to abosonic SPT.[83, 84] These SPTs are very interesting,but in the discussion below we will argue that no otherspin liquid state with fermionic charge and nontrivialmonopole is anomaly-free, which means no such intrinsi-cally interacting fermionic SPT can be found with U (1),time reversal and Z symmetries (even if the fermionsare allowed to transform projectively under these sym-metries).
2. Anomaly of ( E bT M bT (cid:48) ) − In this subsection, by using the same logic as before,we will examine the anomaly of ( E bT M bT (cid:48) ) − . That is, wewill consider the corresponding SPT from the perspectiveof M bT (cid:48) , and check whether it is possible to have a con-sistent surface topological order. However, unlike in thecase of ( E bZ M bZ ) − , where we can reach the conclusion byquite general arguments, here we need to examine somerather detailed properties of the surface states.Again, we will first condense the bound state of two M bT (cid:48) on the surface, which reduces the surface symmetryto T (cid:48) × Z . We would like to point out that there are twopossibilities for the surface at this point: the surface caneither be a simple superfluid, or the surface superfluidhas to coexist with another anomalous topological order.The latter happens if the bulk is still a nontrivial SPTeven if the bulk symmetry is broken down to T (cid:48) × Z ,in which case there must be another anomalous surfacetopological order of an SPT with T (cid:48) × Z symmetry, ifthis symmetry is to be preserved.For the case of a simple superfluid surface, we canshow there is inconsistency of the surface topological or-der. As for the more complicated case where the sur-face superfluid has to coexist with another anomaloustopological order, we need the properties of the surfacesof 3D bosonic SPTs with T (cid:48) × Z symmetry, which, tothe best of our knowledge, are lacking in the literature.So we will first discuss the classification of such SPTs,and then show that there will still be some inconsis-tency even for the more complicated case. This leads usto concluding that there is no such corresponding SPTsthat can become ( E bT M bT (cid:48) ) − after gauging, which means( E bT M bT (cid:48) ) − is anomalous.Before the detailed discussion on this problem, let usfirst collect a few useful tools that will be applied repeat-edly below.1. In a topological order, a particle always has thesame topological spin as its anti-particle. That is,denote a − as the anti-particle of a , then θ a = θ a − (J14) 2. Suppose a and b are two anyons in a topological or-der. Suppose c is a possible fusion product of a and b , that is, a × b = N cab c + · · · with N cab the fusionmultiplicity. Then ( R abc ) = θ c θ a θ b , where ( R abc ) isthe mutual braiding between a and b when theirfusion product is fixed to be c , and θ a is the topo-logical spin of a . In the case of Abelian topologicalorder, the mutual braiding between a and b canbe simply denoted as θ a,b , and the above formulabecomes θ a,b = θ c θ a θ b (J15)3. Braiding and fusion commute in a topological or-der. For example, in an Abelian topological order, θ ab,c = θ a,c θ b,c (J16)4. If the time reversal partner of an anyon a is b , and c = a × b is the bound state of a and b , then theKramers parity of c is determined by T c = θ c (J17) Simple superfluid
We start our discussion with the case of a simple super-fluid surface. This superfluid has vortices with vorticityquantized in units of π , and the minimal trivial vortexis the 4 π vortex. We will condense these 4 π vortices torestore the full symmetry of the surface and get a sym-metric gapped surface topological order, where the U (1)charge is quantized in units of 1/2. The π vortices andthe 2 π vortices will remain gapped, and we will denotethe π vortex by v and the 2 π vortex by E bT .The vortex condensation above will in general generatea charge-1/2 boson, which we denote by X . Physically, X is the 2 π vortex of the 4 π vortices. This X should bean Abelian boson. To see this, let us go to the energyscale below which we can consider only the 4 π vorticesthat are to be condensed. Limiting ourselves below thisenergy scale should not change the topological data of X .Although the π and 2 π vortices are nontrivial, below thisenergy scale they do not play any role. Then because the4 π vortices are trivial bosons, X , the 2 π vortex of the 4 π vortices, is expected to be a simple Abelian boson.The bound state of two X ’s can be combined with M bT (cid:48) to generate a charge-neutral bosonic excitation, whichwill be denoted by N . The bound state of two N ’s havetrivial braiding with all other excitations, so this shouldbe viewed as a local excitation. Therefore, the particlecontents of the surface theory can be written as { , X, N, X − , v, E bT , v − } × { , M bT (cid:48) } (J18)The various bound states of these excitations are under-stood to be implicitly displaced. Also, X − represents0the excitation that can fuse with X into the trivial vac-uum, 1, which does not carry any quantum number, andit should be distinguished from XM † bT (cid:48) .Below we will determine the braiding, fusion and sym-metry assignments of these excitations. Without loss ofgenerality, we will always take v to be neutral, because itscharge can always be cancelled by binding it with certainamount of X and M bT (cid:48) .We start with braiding. For self-braiding, the onlyuncertain part is about v : it can either be Abelian ornon-Abelian. Now we turn to mutual braiding. The mu-tual braiding within the charge sector (built up with X and N ) is always trivial. For the vortex sector (built upwith v and E bT ), the braiding between v and E bT is triv-ial because v is neutral and E bT is the remnant of the2 π vortex, and the braiding between v and v − is to bedetermined.The mutual braiding between the charge sector and thevortex sector can be determined in the following way. Be-cause condensing X will make the surface back into thesimple superfluid, we can view X as something that iscondensed in the superfluid phase. From the Meissnereffect we know the vortices come with certain fluxes inthe superfluid phase, and this combined object of vor-tices and fluxes should be local with respect to the X condensate. Therefore, the mutual braiding between thevortices themselves with the X condensate is the conju-gate of the charge-flux Aharonov-Bohm phase. This tellsus θ X,v = − i, θ X,E bT = θ N,v = − , θ N,E bT = 1 (J19)Notice the third relation comes from the identification N = X M † bZ and that X is condensed in the superfluid( N is not condensed in the superfluid, so we cannot say θ v,N = 1 because N is neutral). The mutual braidinglisted here will be used repeatedly below.Now we turn to fusion. Most fusion rules can be de-termined by the charge and vorticity assignment: X × X = N M bZ , N × N = 1 , E bT × E bT = 1(J20)However, there is some flexibility for v . For example,even if the v is Abelian, we can have either v × v = E bT or v × v = E bT × N . Of course when v is non-Abelian,we must have v × v = E bT + E bT N (with potential fusionmultiplicities suppressed). Because N is a boson that islocal with respect to E bT , N must have trivial braidingwith v in this non-Abelian case, otherwise v and its anti-particle would have opposite topological spins, which vi-olates (J14) and is thus disallowed. However, θ v,N = − v cannot be non-Abelian. Furthermore,for the same reason, the fusion rule for v has to be v × v = E bT (J21) In contrast, if the local particle is a fermion, v can still be non-Abelian. Finally we discuss the symmetry assignment. The U (1)charges of these excitations are clear: X carries halfcharge, M bT (cid:48) carries unit charge, and other excitationsare neutral. The assignment of the T and Z symmetriesis constrained by some general rules. First, T shouldconjugate the topological spins of the excitations, and Z should keep their topological spins. Second, the be-havior of charge and vorticity under various symmetriesis fixed. For example, because the charge flips and thevorticity does not change under T , T will take v to ei-ther v or vN . Because of the fusion rule, v × v = E bT , v cannot become v under time reversal. That is, v shouldgo to vN .Putting all these constraints together, there is actuallynot too much freedom for this topological order. Onechoice is the Z gauge theory listed in Table XVII, andthe only thing that one can modify on top of this stateis to change the values of Z for X and v , and the valueof T (cid:48) for X . Notice in all these cases, N is a Kramersdoublet under T (cid:48) . X N v E bT U (1) T X − N vN E bT T − − Z X − N v − E bT Z ± ± T (cid:48) X N v − N E bT T (cid:48) ± − ± E bT M bT (cid:48) ) − . The first row lists all nontriv-ial excitations, from which the symmetry assignments on alltheir bound states can be inferred. The second row lists thecharges of these excitations under U (1). The third row liststhe time reversal partners of these excitations. The fourthrow lists the values of T of these excitations, with emptyentries representing that T is not well-defined. The fifth rowlists the Z partners of these excitations. The sixth row liststhe values of Z of these excitations. The seventh row liststhe T (cid:48) partners of these excitations. And the last row liststhe values of T (cid:48) of these excitations. The above theory is actually inconsistent. To see this,notice that since the T partner of v is vN and θ v,N = − v must be a semion or anti-semion, so that T can con-jugate the topological spin of v . This means the boundstate of v and its T (cid:48) partner, v − N , is a boson, so thisbound state should be a Kramers singlet under T (cid:48) accord-ing to (J17). However, as discussed above, this boundstate is v × v − N = N = X M † bT (cid:48) , which is a Kramersdoublet under T (cid:48) . This contradiction shows that the sim-ple superfluid surface is inconsistent. To fit into the usual notation of a Z gauge theory, one can takethe Z charge to be X , and take the Z flux to be Xv . Superfluid coexisting with another anomalous topologicalorder
Now we turn to the case where the surface superfluidhas to exist with another anomalous topological order.As discussed earlier, this happens if the bulk remainsto be a nontrivial SPT when the bulk symmetry is alsoreduced to T (cid:48) × Z . We will call such SPTs the reducedbulk SPTs. To complete the discussion, we need theproperties of 3D bosonic SPTs with this symmetry,which will be discussed below.
3D bosonic SPTs with T (cid:48) × Z symmetry Notice there exists local Kramers doublet under T (cid:48) ,so more precisely, the symmetry group of the surface su-perfluid should be denoted by Z T (cid:48) × Z . From groupcohomology, the classification of such SPTs is Z , andthere should still be another SPT whose surface is ef mf ,and this SPT is beyond group cohomology. So we pro-pose that the complete classifications of such SPTs is Z .This proposal is further supported by the classificationof 3D bosonic SPTs with Z P × Z symmetry, where Z P is a reflection symmetry that results in a trivial actionwhen acted four times. SPTs with Z × Z T are believedto have the same classification of SPTs with Z × Z P ,where the latter are classified by Z .[69]What are the surface topological orders of the otherthree root states? We show that they can all be Z topological orders, and they are denoted by ( eT (cid:48) i mT (cid:48) i ) T (cid:48) , eZmZ and eT (cid:48) i mZ . Below we explain the properties ofthese states.The first state, ( eT (cid:48) i mT (cid:48) i ) T (cid:48) , is protected by T (cid:48) alone,and in this state e and m are exchanged under T (cid:48) . Fur-thermore, T (cid:48) acting on e or m four times gives − T (cid:48) i ). The action of Z is trivial on both e and m . In fact, this state is the descendant of ( eCmC ) T (cid:48) (cid:15) when e is condensed without breaking time reversal, and( eCmC ) T (cid:48) (cid:15) is a surface state of the bosonic topologicalinsulator made of Kramers bosons.To justify that this is a legitimate surface state of anSPT protected by T (cid:48) , we need to show this descendantstate is still a nontrivial SPT. That is, the bosonic topo-logical insulator made of Kramers bosons is still a non-trivial SPT when double charge is condensed withoutbreaking time reversal. This can be seen by checkingthe time reversal domain wall of this state. Considerbreaking T (cid:48) in two opposite ways in the two sides of a2D domain wall, while keeping a unitary Z symmetryintact through the entire system (this unitary Z is justthe symmetry generated by acting the generator of T (cid:48) twice). Notice before the double charges are condensed,the time reversal domain wall of this bosonic topologicalinsulator is the elementary bosonic integer quantum Hallstate,[60] because it has σ xy = 2 e /h .[47] When the dou-ble charge is condensed, this bosonic integer quantumHall state becomes the Levin-Gu state.[85, 86] That isto say, the time reversal domain wall of this descendant state is a Levin-Gu state. But this cannot happen unlessthe original T (cid:48) symmetric system is a nontrivial SPT. The above discussion establishes that there is a 3Dbosonic SPT protected by T (cid:48) , and its surface can be( eT (cid:48) i mT (cid:48) i ) T (cid:48) . For our purposes, it will be useful to thinkabout what state this SPT becomes if the symmetry is en-hanced to the full (( U (1) (cid:111) Z ) × T ) /Z = ( O (2) × T ) /Z symmetry of M bT (cid:48) , which can be obtained by consid-ering what the surface topological order becomes whenthe symmetry is enhanced. Because the π rotation of the U (1) is locked with acting T (cid:48) twice, e and m should carryhalf charge under U (1). The entire symmetry assignmentof this surface state is shown in Table XVIII. e m (cid:15) ≡ em † U (1)
12 12 T m † e † (cid:15)T − Z e † m † (cid:15)Z T (cid:48) m e (cid:15)T (cid:48) ± i ± i E fT M bT (cid:48) ) − . The firstrow lists all nontrivial excitations, from which the symmetryassignments on all their bound states can be inferred. The sec-ond row lists the charges of these excitations under U (1). Thethird row lists the time reversal partners of these excitations.The fourth row lists the values of T of these excitations, withempty entries representing that T is not well-defined. Thefifth row lists the Z partners of these excitations. The sixthrow lists the values of Z of these excitations. The seventhrow lists the T (cid:48) partners of these excitations. And the last rowlists the values of T (cid:48) of these excitations, with ± i standingfor that T (cid:48) = − It is straightforward to check that this state can bethe surface topological order of the corresponding SPTof ( E fT M bT (cid:48) ) − (viewed from the perspective of M bT (cid:48) ).This observation implies that if the reduced bulk SPTis ( eT (cid:48) i mT (cid:48) i ) T (cid:48) , we can reduce the surface state intoa simple superfluid by coupling the original SPT tothe corresponding SPT of ( E fT M bT (cid:48) ) − . Because cou-pling ( E bT M bT (cid:48) ) − and ( E fT M bT (cid:48) ) − should result in( E f M bT (cid:48) ) − , if we can show in the scenario of a sim-ple superfluid surface, no SPT made of M bT (cid:48) can be-come ( E f M bT (cid:48) ) − after gauging, it is sufficient to showno SPT with ( eT (cid:48) i mT (cid:48) i ) T (cid:48) reduced bulk SPT can become( E bT M bT (cid:48) ) − after gauging.Next we turn to explaining the properties of eZmZ .In this surface Z topological order, the Z symmetryacting on e or m twice gives a − T (cid:48) More precisely, this is because the Levin-Gu state is the rootstate of 2D Z SPTs, which means there is no 2D Z symmetricshort-range entangled bosonic state that becomes the Levin-Gustate when it is stacked with its time reversal partner. e and m . Again, we need to show thatthe bulk with this surface topological order is a nontrivialSPT with T (cid:48) × Z symmetry, or equivalently, that eZmZ is anomalous with T (cid:48) × Z symmetry.The way to understand the anomaly of eZmZ is torelate it to eCmC , the surface state of a nontrivial SPTwith T (cid:48) × U (1) symmetry. Notice this symmetry is not( U (1) (cid:111) T ) /Z , the symmetry of charged Kramers bosons.In particular, the π rotation of the U (1) here is not lockedwith acting time reversal twice, as in the latter symme-try. There is a bosonic topological insulator at θ = 2 π with T (cid:48) × U (1) symmetry, and this is independent of thepresence of local bosons that are Kramers doublets under T (cid:48) . The surface state of this bosonic topological insulatoris eCmC , which means a Z topological order with both e and m carrying half charge under U (1), and T (cid:48) actstrivially on e and m . Again, the time reversal domainwall of this bosonic topological insulator will have thecharacter of an elementary bosonic integer quantum Hallstate. Breaking the U (1) symmetry down to Z resultsin the eZmZ state, and as before, the time reversal do-main wall of this descendant 3D state will be a Levin-Gustate, which is disallowed unless the 3D bulk is a nontriv-ial SPT. This shows that eZmZ is still anomalous with T (cid:48) × Z symmetry, and we also see T (cid:48) plays a role inprotecting this state, even though it appears to act on e and m trivially. Again, it will be useful for us to understand what thisstate becomes when the symmetry is enhanced to thefull ( O (2) × T ) /Z symmetry of M bT (cid:48) . Because only the Z acts nontrivially in this state, when the symmetry isenhanced, U (1) and T (cid:48) should still act trivially. Thismeans there is a corresponding SPT of ( E b M bT (cid:48) ) − thatbecomes eZmZ when the symmetry is broken down to T (cid:48) × Z . Therefore, when the reduced bulk SPT is eZmZ ,we can always cancel the anomaly of eZmZ by couplingthe original putative SPT to this corresponding SPT of( E b M bT (cid:48) ) − .Lastly, we turn to discuss eT (cid:48) i mZ , which is a Z topo-logical order with e having T (cid:48) = − m having Z squaring to −
1. As argued in Appendix A, in order for e to have T (cid:48) = − T (cid:48) should attach M bT (cid:48) , a localKramers doublet under T (cid:48) , to e . This surface state isanomalous because when two Z fluxes are inserted inthe system, both m and (cid:15) will see a nontrivial phase fac-tor when moving around it. But this is a local process,so this − Z flux when it is inserted, andthis nucleated object must be e . That is to say, a localprocess produces an object with T = −
1, which is ab-sent in the system by assumption. Therefore, this stateis anomalous. We point out that in order to establish that eZmZ is anomalous,it is actually important that T (cid:48) acts trivially on both e and m , sothat this eZmZ state can be viewed as the descendant of eCmC after the U (1) is broken down to Z . What does eT (cid:48) i mZ become when the symmetry isenhanced to the full ( O (2) × T ) /Z symmetry? Itturns out the symmetry cannot be enhanced to the full( O (2) × T ) /Z symmetry. In other words, there is nobosonic SPT with ( O (2) × T ) /Z symmetry whose de-scendant can be eT (cid:48) i mZ . However, we still know that,because the π rotation of the U (1) is locked with acting T (cid:48) twice, when the 4 π vortices are condensed and the fullsymmetry is recovered on the surface, e should carry halfcharge of U (1), and m carries integer charge but has Z squaring to − ef mf , when the full symmetry is recovered,none of e and m get nontrivial action under the symme-try, and its anomaly can be cancelled by just coupling itwith another ef mf state with the full symmetry, whichafter gauging becomes ( E b M bT (cid:48) ) − .The above discussion implies that, in the scenariowhere the surface superfluid has to coexist with anotheranomalous topological order, in order to show there is noSPT made of M bT (cid:48) that can become ( E bT M bT (cid:48) ) − aftergauging, it is sufficient to show:1. In the scenario of a simple superfluid surface,there is no SPT made of M bT (cid:48) that can become( E f M bT (cid:48) ) − .2. In the scenario of eT (cid:48) i mZ reduced bulk SPT,there is no SPT made of M bT (cid:48) that can become( E bT M bT (cid:48) ) − .3. In the scenario of eT (cid:48) i mZ reduced bulk SPT,there is no SPT made of M bT (cid:48) that can become( E f M bT (cid:48) ) − .Below we will show these three statements in turn. Simple superfluid surface for ( E f M bT (cid:48) ) − We start from the first statement. The similar argu-ment as before implies that the surface topological orderin this case will be a Z topological order: { , X, N, X − , v, E f , v − } × { , M bT (cid:48) } (J22)with the symbols standing for parallel excitations as be-fore, and a difference is that here E f is a fermion withno symmetry fractionalization in terms of T and Z .In this case, most of the topological data (braidingand fusion) will be the same as in the simple superfluidfor ( E bT M bT (cid:48) ) − . The only difference is that, because E f is a fermion, the fusion product of two v ’s should be E f N , otherwise the antiparticle of v will have a differenttopological spin from v . That is, v × v = E f N (J23)Because E f N is still a fermion, v has topological spin θ v = exp (cid:0) i (cid:0) π + πn (cid:1)(cid:1) with n an integer. In order for T to conjugate θ v , the T partner of v has to be vX or3 vX − . This then changes the charge of v , which is notlegitimate.This establishes the first statement: in the scenarioof a simple superfluid surface, there is no SPT made of M bT (cid:48) that can become ( E f M bT (cid:48) ) − . eT (cid:48) i mZ reduced bulk SPT for ( E bT M bT (cid:48) ) − Now we turn to the second statement, where the sur-face superfluid has to coexist with the anomalous topo-logical order, eT (cid:48) i mZ . In the superfluid phase of the sur-face, the vortices and anyons of this anomalous topologi-cal order are distinct. One of their distinctions is that thevorticies carry (logarithmically) expensive energy cost,while the anyons have a finite energy gap. The fusionrules of e and m in the superfluid phase is that e × e = M bT (cid:48) , m × m = 1 (J24)These seemingly innocuous fusion rules actually deservefurther clarification. Purely in terms of topological sec-tors, there is no difference in M bT (cid:48) and 1 in the righthand sides of these fusion rules, because they can beturned into each other by binding an M † bT (cid:48) , a local ex-citation. However, in terms of symmetry quantum num-bers, M bT (cid:48) and 1 are of course different. It makes senseto talk about fusing two e particles or two m particlesthat have the same symmetry quantum number (withthe difference due to attaching an M bZ resolved), andthe above equations should be interpreted as the fusionrules of fusing two identical e ’s or m ’s. This distinction isimportant when we try to determine the fusion rules of e and m after the surface gets into the symmetric topolog-ically ordered phase, where the right hand sides of thesefusion rules can potentially be modified by multiplyingsomething condensed in the superfluid phase.When 4 π vortices are condensed and the full symme-try is restored, e will remain deconfined because it willcarry half charge under U (1) ( m will of course also re-main deconfined because it is neutral). Then the surfacetopological order can be written as { , X, N, X − , v, E bT , v − , e, m } × { , M bT (cid:48) } (J25)where the symbols have parallel meaning as in the case ofa simple superfluid. As before, the various bound statesof these excitations are understood to be implicitly dis-played.Again, X will be an Abelian boson that carries halfcharge, and N will be a boson that is a Kramers doubletunder T (cid:48) . Because condensing X makes the surface backto the superfluid coexisting with eT (cid:48) i mZ , the superfluidphase can again be viewed as a condensate of X . Thishas two important consequences.First, the mutual braiding in (J19) still holds, and { e, m } will have trivial braiding with { X, N, X − } . Then N is trivial, and (J20) still holds. However, the mutualbraiding between v and { e, m } is undetermined at thispoint. We only know, because θ E bT ,m = 1, we shouldhave θ v,m = ±
1. Second, no condensate in the superfluid can be multi-plied to the right hand sides of the fusion rules in (J24)without violating charge conservation, which means that(J24) is also the right fusion rules for e and m in thetopologically ordered phase.Now let us determine the fusion rule of two v ’s.Just from charge conservation and vorticity conservation,there seem to be many possible fusion products of two v ’s: E bT , E bT N, E bT m, E bT mN,E bT eX − , E bT eX − N, E bT (cid:15)X − , E bT (cid:15)X − N (J26)However, based on some topological arguments, in thefollowing we can rule out all of them but E bT m and E bT mN .To see this, first notice that for each Abelian anyon, ithas a unique braiding phase factor with all fusion prod-ucts of two v ’s. Because of this, some of the above can-not simultaneously be the fusion products. For example,because E bT and E bT m have different braiding phase fac-tors with e , they cannot both be the fusion products oftwo v ’s. Using this, we see the fusion products can beone of the four possible pairs { E bT , E bT N } , { E bT m, E bT mN } , { E bT eX − , E bT eX − N } , { E bT (cid:15)X − , E bT (cid:15)X − N } (J27)However, because N is a boson with θ v,N = −
1, withineach pair at most one of them can be the fusion productof two v ’s, otherwise the anti-particle of v will have op-posite topological spin of v , and (J14) is violated. Thismeans v has to be Abelian again, and the entire topo-logical order is Abelian. Together with θ E bT ,v = 1, thisalso implies E bT N cannot be a fusion product of two v ’s.Furthermore, because θ X,v = − i , the last four excitationscannot be fusion products of two v ’s, otherwise v wouldhave an antiparticle that has a different topological spinfrom itself.In fact, E bT cannot be a fusion product of two v ’s,either. This is because θ E bT ,e = −
1, which means θ v,e = ± i and θ v,e = −
1, if v × v = E bT . However, e is local,so θ v,e = 1. This contradiction implies that E bT cannotbe the fusion product of two v ’s.So we are finally left with two possibilities: v × v = E bT m (J28)or v × v = E bT mN (J29)In the first possibility, θ v,m = 1, while θ v,m = − topological sectors of e and m are transformed as T (cid:48) : e → eM bT (cid:48) , m → m (J30)under T (cid:48) , and Z : e → e, m → m (J31)4or Z : e → eM bT (cid:48) , m → m (J32)under Z . Notice that the above expressions only implythe action on the topological sectors, and in this casethere are two possibilities for the Z transformation.In the symmetric topologically ordered phase, the righthand sides of these transformation rules can be multipliedby something condensed in the superfluid phase. Also,according to the values of T (cid:48) of e and m and the factthat the π rotation of U (1) is locked with acting T (cid:48) twice, e carries half charge and m carries no charge under U (1)in the topologically ordered phase. In order for T (cid:48) tomaintain the U (1) charge, and for Z to flip the U (1)charge, the unique choice for the transformation rules of e and m in the topologically ordered phase are e → eN, m → m (J33)under T (cid:48) , and e → eM † bT (cid:48) N, m → m (J34)or e → eM † bT (cid:48) , m → m (J35)under Z , corresponding to the two possible Z transfor-mations in the superfluid phase, respectively.To further determine the symmetry assignments, it iscrucial to determine the symmetry actions on v . Let usconsider the Z action on v first, and we will begin withthe case where v × v = E bT m and e → eM † bT (cid:48) N and m → m under Z . Notice in this case θ v,m = 1. As the Z flips both the charge and the vorticity, the options forthe Z partner of v are: v − , v − N, v − m, v − mN,v − eX − , v − eX − N, v − (cid:15)X − , v − (cid:15)X − N (J36)All of them except v − and v − m can be ruled out, be-cause in those cases the Z action cannot keep the topo-logical spin of v invariant. If the Z partner of v is v − ,then under the Z transformation θ e,v becomes θ eNM † bT (cid:48) ,v − = θ eNM † bT (cid:48) ,v = θ e,v θ N,v = θ e,v θ e,v θ N,v θ N,v = θ e,v θ N,v = − θ e,v (J37)which is disallowed. In the above we have used (J15) and(J16). The above discussion implies that the Z partnerof v can only be v − m = vE bT .Just from that T flips the charge but keeps the vortic-ity, the options for the T partner of v are: v, vN, vm, vmN,veX − , veX − N, v(cid:15)X − , v(cid:15)X − N (J38)The last four can be ruled out due to some topologicalreasons. For example, suppose v becomes veX − under T . In order for T to conjugate the topological spin of v , using the (J19) and θ e,v = ±
1, the topological spinof v must be θ v = exp (cid:0) i (cid:0) ± π + nπ (cid:1)(cid:1) , with n an integer.Then the bound state of two v ’s must be a fermion, con-tradicting to the fusion rule v × v = E bT m . This means veX − cannot be the T partner of v . Similar argumentsshow that veX − N , v(cid:15)X − and v(cid:15)X − N cannot be the T partner of v , either.If the T partner of v is vm , then v is either a bosonor a fermion, and θ vvm = 1. But θ vvm should be lockedto the Kramers parity of vvm = E bT , which is −
1. Thiscontradiction implies that vm cannot be the T partnerin this case. The time reversal partner of v can also notbe vN m , because in this case under time reversal θ v,e becomes θ vNm,eM † bT (cid:48) = − θ v,e (cid:54) = θ ∗ v,e (J39)which is disallowed. This is actually another reason why vm cannot be the time reversal partner of v , because vm cannot conjugate θ v,e , either. If the time reversal partnerof v is v , then v has a well defined Kramers parity, butits Z partner, v − m = vE bT , has an opposite Kramersparity. This is disallowed, otherwise Z and time reversalcannot commute for the spin liquid. To see this, suppose v transforms under time reversal as v i → T ij v j (J40)and under Z as v i → C ij ˜ v j (J41)Its Z partner, ˜ v , transforms under time reversal as˜ v i → ˜ T ij ˜ v j (J42)Because v ∗ i M ij v j is a local operator that has no charge orvorticity, its Kramers parity should be 1 and Z shouldcommute with time reversal on this operator, for anymatrix M . This requires T ∗ T = ±
1, ˜ T ∗ ˜ T = ±
1, and
T C = e iφ C ∗ ˜ T , with φ a phase. Taking all these together,we get T ∗ T = C ˜ T ∗ ˜ T C − = ˜ T ∗ ˜ T (J43)That is, v and ˜ v should have the same Kramers parity.Notice to claim that for v ∗ i M ij v j the Kramers parity is 1and Z commutes with T , it is important that this oper-ator is not only local, but also carries no charge or vor-ticity, otherwise by a gauge transformation its Kramersparity and the commutation relation can be changed.So the time reversal partner can only be vN . Noticein this case v must be a semion or anti-semion, because θ v,N = − v . Then θ vvN = 1, which means the Kramersparity of vvN = E bT mN is 1, so N has to be a Kramersdoublet.The above discussion implies that if v × v = E bT m and the Z action on e and m is given by (J34), the Z partner of v can only be vE bT , and the T partner of v can5only be vN . Using similar arguments, one can actuallycheck that if v × v = E bT mN and the Z action on e and m is given by (J34), the Z partner of v can still onlybe vE bT , and the T partner of v can only be vN . Inboth cases, the entire symmetry assignments are largelydetermined, as shown in Table XIX.This surface state is actually problematic. To see this,notice under T the topological sector of eX − v is in-variant, so eX − v has a well defined Kramers parity.However, under the Z this topological sector becomes eX − vE bT , which carries an opposite Kramers parity.As discussed above, this is disallowed, otherwise T and Z cannot commute for the spin liquid. X N v E bT e mU (1) T X − N vN E bT eM † bT (cid:48) mZ X − N vE bT E bT eNM † bT (cid:48) m TABLE XIX. Symmetry assignments of the surface topologi-cal order if the Z action on e and m is given by (J34), for boththe case with v × v = E bT m and the case with v × v = E bT mN .The first row lists all nontrivial excitations, from which thesymmetry assignments on all their bound states can be in-ferred. The second row lists the charges of these excitationsunder U (1). The third row lists the time reversal partners ofthese excitations. And the fourth row lists the Z partners ofthese excitations. The above discussion implies that if the Z action on e and m is given by (J34), the surface state of this SPTis problematic. Now we are only left with the case wherethe Z action on e and m is given by (J35). In thiscase, one can use similar method to constrain the restof the symmetry assignments, and the resulting symme-try assignment is shown in Table XX. There is also aproblem of this topological order: the Z partner of v is v − , so it has a well defined value for charge-conjugationsquared. However, its T partner has an opposite value ofcharge-conjugation squared, because the values of charge-conjugation squared for N and m are 1 and −
1, respec-tively. This contradicts the fact that Z and T shouldcommute for the spin liquid. X N v E bT e mU (1) T X − N vm/vmN E bT eM † bT (cid:48) N mZ X − N v − E bT eM † bT (cid:48) m TABLE XX. Symmetry assignments of the surface topologi-cal order if the Z action on e and m is given by (J35). Thefirst row lists all nontrivial excitations, from which the sym-metry assignments on all their bound states can be inferred.The second row lists the charges of these excitations under U (1). The third row lists the time reversal partners of theseexcitations. And the fourth row lists the Z partners of theseexcitations. To see this, suppose T acts as v i → T ij ˜ v j (J44) and Z acts as v i → C ij v ∗ j , ˜ v i → ˜ C ij ˜ v ∗ j (J45)where T , C and ˜ C are three unitary matrices (in fact,being invertible is enough for the following argument).For an arbitrary matrix M , because v ∗ i M ij v j is a neu-tral local operator that carries no vorticity , Z and timereversal should commute on it. Demanding the resultsof acting time reversal first and Z later and the resultsof acting Z first and time reversal later to be the samegives ( T ˜ C ) † M ∗ ( T ˜ C ) = ( C ∗ T ∗ ) † M ∗ C ∗ T ∗ (J46)For this equation to be true for arbitrary M , we musthave T ˜ C = e iφ C ∗ T ∗ (J47)with φ a phase. Or equivalently,˜ C = e iφ T − C ∗ T ∗ (J48)Taking the complex conjugation on both sides yields˜ C ∗ = e − iφ ( T − ) ∗ CT . So˜ C ˜ C ∗ = T − C ∗ T ∗ ( T − ) ∗ CT = T − C ∗ CT (J49)Notice because v ∗ i M ij v j is local and neutral, Z squaresto 1 for it because the system is made of M bZ . Combinedwith the discussion in Appendix A, the above equationimplies that ˜ C ˜ C ∗ = C ∗ C . That is, v and ˜ v should havethe same charge-conjugation squared.Taking all these arguments together, the secondstatement is established: in the scenario of eT (cid:48) i mZ reduced bulk SPT, there is no SPT made of M bT (cid:48) thatcan become ( E bT M bT (cid:48) ) − . eT (cid:48) i mZ reduced bulk SPT for ( E f M bT (cid:48) ) − Finally, we turn to the last statement. Using similararguments as before, the surface topological order can bewritten as { , X, N, X − , v, E f , v − , e, m } × { , M bT (cid:48) } (J50)with the similar symbols representing parallel excitationsas before.As before, in this topological order, the fusion rulesfor e and m are still given by (J24). The symmetry as-signments for e and m are such that e carries half chargeunder U (1), while m carries no charge, and the othersymmetry assignments for e and m are given by (J33)and (J34), or (J33) and (J35).In this case, most of the topological data will be thesame as the case with eT (cid:48) i mZ reduced bulk SPT for( E bT M bT (cid:48) ) − . The only difference is in the fusion productof two v ’s. Modifying the arguments before while keepingin mind that now E f is a fermion, we find two possiblefusion rules for two v ’s v × v = E f m (J51)6or v × v = E f N m (J52)In the first possibility, θ v,m = −
1, while θ v,m = 1 in thesecond possibility. In both cases the right hand side ofthe fusion rules are fermions, so the topological spin of v must be θ v = exp (cid:0) i (cid:0) π + nπ (cid:1)(cid:1) , with n an integer.In this case, in order for T to keep the vorticity of v and conjugate the topological spin of v , the T partner of v can be one of the following: veX − , veX − N, v(cid:15)X − , v(cid:15)X − N (J53)Which one of these can conjugate the topological spin of v depends on the values of θ v = exp (cid:0) i (cid:0) π + nπ (cid:1)(cid:1) , θ e,v = ± θ (cid:15),v = ±
1. However, no matter which one ofthe above four excitations is the T partner of v , θ v,m becomes − θ v,m (cid:54) = θ ∗ v,m . This means there is no consistentsymmetry assignment for this topological order.This establishes the third statement: in the scenarioof eT (cid:48) i mZ reduced bulk SPT, there is no SPT made of M bT (cid:48) that can become ( E f M bT (cid:48) ) − .Taking all the arguments above together, we have es-tablished that ( E bT M (cid:48) bT ) − is anomalous with Z ×T sym-metry. Notice unlike ( E bZ M bZ ) − , which is anomalouseven if there is only the Z symmetry, here both the T and Z symmetries are responsible for the anomaly.
3. Anomaly of E bTT (cid:48) M b − In this subsection we show the anomaly of E bT T (cid:48) M b − ,by showing that no SPT made of M b − will become E bT T (cid:48) M b − after gauging. As a reminder, in this case Z does not act as a charge-conjugation, and it anticom-mutes with T on M b − . As before, we will first condensedouble charge on the surface to get a surface superfluid,whose minimal trivial vortex is the 4 π vortex. We willthen proliferate these 4 π vortices to restore the full sym-metry and get a surface topological order. Again, thereare two scenarios for the surface superfluid: it can eitherbe a simple superfluid, or this superfluid has to coex-ist with another anomalous topological order. We willdiscuss these cases in turn. Simple superfluid
We begin with the case of a simple superfluid. Usingsimilar argument as before, we see the symmetric surfacetopological order obtained by condensing 4 π vortices canbe written as { , X, N, X − , v, E bT T (cid:48) , v − } × { , M b − } (J54)The symbols stand for parallel meanings as before, whilenow the 2 π vortex is E bT T (cid:48) , a Kramers doublet underboth T and T (cid:48) .Similar arguments as before show that the braidingand fusion are similar to the surface state obtained from the simple superfluid for the corresponding SPTof ( E bT M bT (cid:48) ) − , and the symmetry assignment is shownin Table XXI. In particular, in order for E bT T (cid:48) = v × v to be a Kramers doublet under both T and T (cid:48) , both ofthe T and T (cid:48) partners of v should be vN . Notice Xv will become XvM † b − under T , and it is invariant under Z . Below we show this is disallowed. X N v E bTT (cid:48) U (1) T X − N vN E bTT (cid:48) Z X N v E bTT (cid:48) T (cid:48) X − N vN E bTT (cid:48)
TABLE XXI. Symmetry assignments of the surface topologi-cal order from the simple superfluid surface of the correspond-ing SPT of E bTT (cid:48) M b − . The first row lists all nontrivial ex-citations, from which the symmetry assignments on all theirbound states can be inferred. The second row lists the chargesof these excitations under U (1). The third row lists the T partners of these excitations. The fourth row lists the Z partners of these excitations. The fifth row lists the T (cid:48) part-ners of these excitations. For notational simplicity, denote Xv and XvM † b − by x and y , respectively. Suppose the T action is T : x i → T ij y j , y i → ˜ T ij x j (J55)and the Z action is Z : x i → Z ij x j , y i → ˜ Z ij y j (J56)with T , ˜ T , Z and ˜ Z four invertible matrices.Because, for any matrix M , x ∗ i M ij x j is local and neu-tral, Z and T should commute on this operator. Thisgives a condition T ˜ Z = e iφ Z ∗ T (J57)Because, for any matrix M , x ∗ i M ij y j is local and carriescharge-1, Z and T should anti-commute on this opera-tor, because the system is made of M b − . This, togetherwith the above condition, yields˜ T Z = e iφ ˜ Z ∗ ˜ T (J58)with e i ( φ − φ ) = − Z = e iφ T ∗ ˜ Z ∗ T ∗− = e iφ ˜ T − ˜ Z ∗ ˜ T (J59)Using that e i ( φ − φ ) = −
1, the above equation yields˜ Z ∗ = − ˜ T T ∗ ˜ Z ∗ T ∗− ˜ T − (J60)Now notice y ∗ i M ij y j should have Kramers parity 1 forany matrix M , this implies that ˜ T ∗ T = e iφ , or ˜ T T ∗ = e − iφ , where φ is another phase. Plugging this into theabove equation yields ˜ Z ∗ = − ˜ Z ∗ (J61)so ˜ Z = 0. This is disallowed.The above argument shows that in the scenario of asimple superfluid, there is no SPT made of M b − that canbecome E bT T (cid:48) M b − after gauging.7 Superfluid coexisting with another anomalous topologicalorder
Next we turn to the more complicated case wherethe superfluid has to coexist with another anomaloustopological order. Again, this happens when the bulkremains as a nontrivial SPT when the bulk symmetry isalso reduced. For this purpose, let us first clarify whatthe reduced symmetry is. The reduced symmetry grouphas a Z T subgroup, with an anti-unitary generator t thatsatisfies t = 1. It also has a Z subgroup, with a unitarygenerator g that satisfies g = 1. However, gt + tg = 0.Denote t (cid:48) = gt , then t (cid:48) = gtgt = −
1, so t (cid:48) generates a Z T (cid:48) symmetry. Notice t (cid:48) gets inverted when conjugatedwith both t and g , that is, gt (cid:48) g = ggtg = tg = − gt = − t (cid:48) and tt (cid:48) t = tgtt = tg = − gt = − t (cid:48) . So we will denotethis group by D T ≡ Z T (cid:48) (cid:111) Z = Z T (cid:48) (cid:111) Z T . Now it isalso easy to get the full symmetry, which also has U (1)charge conservation. In the gauge where t and g arefixed to be identity, because each charge-1 boson has Z and T anti-commuting, the π rotation of the U (1)is locked with t (cid:48) , and the full symmetry group can bewritten as ( U (1) × D T ) /Z .
3D bosonic SPTs with D T symmetry Now let us discuss 3D bosonic SPTs with D T symme-try. By group cohomology, these SPTs are classified by Z . There should be another root state, ef mf , whichis beyond group cohomology. So the full classification ofthese SPTs is expected to be Z . Among the other rootstates, eT mT should be one of them. The U (1) symmetrycan be added to ef mf and eT mT with a trivial action, soreduced bulk SPTs with such anomalies can be easily can-celled, and we will ignore these two states from now on.We propose two other root states: ( eT T (cid:48) i mT (cid:48) i ) ZT (cid:48) (cid:15)T T (cid:48) and ( eT (cid:48) i mT (cid:48) i ) ZT (cid:48) .The root state ( eT T (cid:48) i mT (cid:48) i ) ZT (cid:48) (cid:15)T T (cid:48) is actually the de-scendant of the corresponding SPT of E fT T (cid:48) M b − (viewedfrom the perspective of M b − ), when the U (1) symme-try is broken to its Z subgroup generated by its π ro-tation. The surface state of this corresponding SPT of E fT T (cid:48) M b − can be a Z topological order, with symme-tries assigned as in Table XXII. These symmetry assign-ments can be derived from the corresponding SPT of E fT T (cid:48) M b − viewed from the perspective of E fT T (cid:48) , whichis described in Appendix J 1. Notice the fusion rules are e × e = m × m = M b − .When the U (1) symmetry is reduced to its Z subgroupgenerated by its π rotation, the symmetry becomes D T .The resulting state is still anomalous, which can be seenby checking its Z T (cid:48) domain wall. Consider breaking the Z T (cid:48) symmetry in two different ways on the two sides of a2D domain wall, while keeping a Z subgroup generatedby t (cid:48) intact across the entire system. Then this domainwall has a Z × Z symmetry, and by relating it to thecorresponding SPT of E fT T (cid:48) M b − , we see this time re-versal domain wall is a Levin-Gu state (protected by the e m (cid:15) = em † U (1)
12 12 T e † m † (cid:15)T − − Z m e (cid:15)Z T (cid:48) m † e † (cid:15)T (cid:48) ± i ± i − Z surface topo-logical order of the bosonic SPT made of M b − that will be-come E fTT (cid:48) M b − after gauging. The first row lists all nontriv-ial excitations, from which the symmetry assignments on alltheir bound states can be inferred. The second row lists thecharges of these excitations under U (1). The third row liststhe time reversal partners of these excitations. The fourthrow lists the values of T of these excitations, with emptyentries representing that T is not well-defined. The fifth rowlists the Z partners of these excitations. The sixth row liststhe values of Z of these excitations. The seventh row liststhe T (cid:48) partners of these excitations. And the last row liststhe values of T (cid:48) of these excitations, with ± i standing forthat T (cid:48) = − Z generated by t (cid:48) ). The existence of this decorated do-main wall shows the descendant state is still a nontrivialSPT with the D T symmetry.Next we turn to ( eT (cid:48) i mT (cid:48) i ) ZT (cid:48) , whose symmetry assign-ments are shown in Table XXIII. Or more precisely, thesymmetry assignments are T : e → ie, m → mZ : e → m, m → e (J62)Notice the fusion rules are e × e = m × m = M b − . e m (cid:15) = em † T e m (cid:15)T Z m e (cid:15)Z T (cid:48) m e (cid:15)T (cid:48) ± i ± i Z surface topo-logical order ( eT (cid:48) i mT (cid:48) i Z ) TT (cid:48) (cid:15)T Z . The first row lists all non-trivial excitations, from which the symmetry assignments onall their bound states can be inferred. The second row liststhe T partners of these excitations. The third row lists thevalues of T of these excitations. The fourth row lists the Z partners of these excitations. The fifth row lists the values of Z of these excitations. The sixth row lists the T (cid:48) partnersof these excitations. And the last row lists the values of T (cid:48) of these excitations, with ± i standing for that T (cid:48) = − To show that this state is anomalous, consider break-ing the D T symmetry to Z × Z T , where the first Z isgenerated by t (cid:48) , and the Z T is generated by t . Noticein this case the Z generated by g is broken. This canbe done, for example, by considering that T acts on the8local boson as M → M , M → M (J63)and Z acts on the local boson as M → M , M → − M (J64)Giving M M a nonzero expectation value will break the D T to Z × Z T in the above way.With this reduced symmetry, this state becomes eZmZ , where the symbol Z stands for half charge underthe Z (generated by t (cid:48) ). This reduced state is anoma-lous. To see this, define ˜ t = t (cid:48) · t , which generates anotheranti-unitary symmetry, ˜ T . The state eZmZ can then berelabelled as e ˜ T m ˜ T , so it is anomalous. Enhancing thesymmetry back to D T will not add Kramers doublet bo-son under ˜ T to the system, so it will not remove thisanomaly. This also shows that ( eT (cid:48) i mT (cid:48) i ) ZT (cid:48) is distinctfrom the previous ( eT T (cid:48) i mT (cid:48) i ) ZT (cid:48) (cid:15)T T (cid:48) , because when thesymmetry is reduced to Z × T in the above way, thelatter becomes eZT mZ , which is a distinct anomalousstate.For later purpose, let us now consider the U (1) chargeof e and m when the 4 π vortices are condensed and thefull symmetry is restored. Again, because the π rotationof the U (1) is locked with t (cid:48) , e and m should carry halfcharge in the topological order.The above discussion implies that, in the scenariowhere the surface superfluid has to coexist with anotheranomalous topological order, in order to show there isno SPT made of M b − that can become E bT T (cid:48) M b − aftergauging, it is sufficient to show:1. In the scenario of a simple superfluid surface, thereis no SPT made of M b − that can become E f M b − .2. In the scenario of ( eT (cid:48) i mT (cid:48) i ) ZT (cid:48) reduced bulk SPT,there is no SPT made of M b − that can become E bT T (cid:48) M b − .3. In the scenario of ( eT (cid:48) i mT (cid:48) i ) ZT (cid:48) reduced bulk SPT,there is no SPT made of M b − that can become E f M b − .Below we will show these three statements in turn. Simple superfluid surface for E f M b − We start with the first statement. Using similar argu-ments as before, the surface topological order in this casecan be written as { , X, N, X − , v, E f , v − } × { , M b − } (J65)The symbols stand for parallel meanings as in the previ-ous cases, and now the 2 π vortex, E f , is a fermion andis a Kramers singlet under both T and T (cid:48) .In this case, most of the topological data will be thesame as in the case with simple superfluid surface for E bT T (cid:48) M b − . The only difference is in the fusion rule of v . The similar arguments as before implies that the fusionrule of two v ’s in this case is v × v = E f N (J66)otherwise the antiparticle of v will have an oppositetopological spin as itself. Because the right hand sideis a fermion, the topological spin of v will be θ v =exp (cid:0) i (cid:0) π + nπ (cid:1)(cid:1) , with n an integer.Now let us consider the T partner of v . As T shouldkeep the vorticity and flip the charge, there are twopossible T partners of v : v and vN . Because θ v =exp (cid:0) i (cid:0) π + nπ (cid:1)(cid:1) and θ v,N = −
1, neither of these optionswill conjugate θ v under T . So this is inconsistent.This establishes the first statement: in the scenarioof a simple superfluid surface, there is no SPT made of M b − that can become E f M b − .( eT (cid:48) i mT (cid:48) i ) ZT (cid:48) reduced bulk SPT for E bT T (cid:48) M b − Next we turn to the second statement. In this case,similar arguments as before show the surface topologicalorder can be written as { , X, N, X − , v, E bT T (cid:48) , v − , e, m } × { , M b − } (J67)with the symbols standing for parallel meanings as be-fore.Recall that both e and m will carry half charge under U (1). By charge conservation and vorticity conservation,the possible fusion products of two v ’s are E bT T (cid:48) , E bT T (cid:48) N, E bT T (cid:48) (cid:15), E bT T (cid:48) (cid:15)NE bT T (cid:48) eX − , E bT T (cid:48) eX − N,E bT T (cid:48) mX − , E bT T (cid:48) mX − N (J68)Similar arguments as before imply that there are twopossible fusion rules for v : v × v = E bT T (cid:48) (cid:15) (J69)or v × v = E bT T (cid:48) (cid:15)N (J70)In both cases, the bound state of two v ’s is fermionic, sothe topological spin of v should be θ v = exp (cid:0) i (cid:0) π + nπ (cid:1)(cid:1) ,with n an integer. In order for T to conjugate θ v , the T partner of v should be one of veX − , veX − N, vmX − , vmX − N (J71)From the symmetry actions on e and m in the super-fluid phase, by multiplying something condensed in thesuperfluid to be consistent with charge conservation, weget the symmetry actions on e and m in the topologicallyordered phase: T : e → eM † b − N, m → mM † b − NZ : e → m, m → e (J72)9Combining this with that θ v,N = −
1, we see no matterwhich one of the four excitations is the T partner of v , θ (cid:15),v cannot be conjugated by T , which is disallowed.This establishes the second statement: in the scenarioof ( eT (cid:48) i mT (cid:48) i ) ZT (cid:48) reduced bulk SPT, there is no SPTmade of M b − that can become E bT T (cid:48) M b − .( eT (cid:48) i mT (cid:48) i ) ZT (cid:48) reduced bulk SPT for E f M b − Finally we come to the last statement. Similar argu-ments as before imply the topological order can be writ-ten as { , X, N, X − , v, E f , v − , e, m } × { , M b − } (J73)with symbols standing for parallel meanings as before.Notice the 2 π vortex, E f , is a fermion, and it is a Kramerssinglet under both T and T (cid:48) .Most of the topological data can be easily determinedusing similar arguments as before: θ X,v = − i, θ X,E bTT (cid:48) = − , θ v,e = ± , θ v,m = ± ,X × X = N M b − , N × N = 1 , E bT T (cid:48) × E bT T (cid:48) = 1(J74)There are two possible fusion rules for v : v × v = E f (cid:15) (J75)or v × v = E f (cid:15)N (J76)In both cases, the bound state of two v ’s is a boson, so v is a boson, fermion, semion or anti-semion.Knowing that T should flip the charge and keep thevorticity, the T partner of v can be one of v, vN, v(cid:15), v(cid:15)N,veX − , veX − N, vmX − , vmX − N (J77)Because v is a boson, fermion, semion, or anti-semion,and θ X,v = − i , θ v,e = ± θ v,m = ±
1, the last fouroptions can be ruled out, because in those cases T willnot conjugate θ v . Because the T partner of e is eM † b − N ,in order to conjugate θ e,v , the T partner of v cannot be v or vN . That is, the T partner of v is either v(cid:15) or v(cid:15)N .If v × v = E f (cid:15) and the T partner of v is v(cid:15) , then v × v(cid:15) = E f . Because v and v(cid:15) are T partners and E f isa fermion, E f must be a Kramers doublet under T , whichcontradicts the original assumption. The same reasoningrules out the possibility that v × v = E f (cid:15)N and the T partner of v is v(cid:15)N .If v × v = E f (cid:15) and the T partner of v is v(cid:15)N , then v × v(cid:15)N = E f N . This means N is a Kramers doublet under T . On the other hand, the T partner of e is eM † b − N ,so N can be viewed as the bound state of e and its T partner. This implies that N is a Kramers singlet under T , which leads to a contradiction. The same reasoningalso rules out the possibility that v × v = E f (cid:15)N and the T partner of v is v(cid:15) . Putting all these analysis together, we have establishedthe last statement: in the scenario of ( eT (cid:48) i mT (cid:48) i ) ZT (cid:48) re-duced bulk SPT, there is no SPT made of M b − that canbecome E f M b − .Therefore, we have shown that no SPT made of M b − can become E bT T (cid:48) M b − after gauging, which means E bT T (cid:48) M b − is anomalous with Z × T symmetry.As discussed before, we have already shown all statesin Table XIV and Table X are anomalous. K. Z × T symmetric U (1) quantum spin liquidswith θ = π and Z not acting as a chargeconjugation In this appendix, we discuss Z × T symmetric U (1)quantum spin liquids with θ = π and Z not acting as acharge conjugation.As discussed in Sec. VII A, in this case the quantumnumbers of the (cid:0) , (cid:1) dyon determine the phase. Sincethere is no nontrivial projective representation of the Z ×T symmetry on the (cid:0) , (cid:1) dyon, we expect only one state:( E fT T (cid:48) M f ) θ . In this state, the electric charge must bea Kramers doublet under both T and T (cid:48) , and M (the(0 ,
2) dyon in this context) has Z and T commutingwith each other. One may wonder whether it is possibleto have Z and T anti-commuting with each other in thiscase. Below we show this is not possible.Denote the (cid:0) , (cid:1) dyon by D (+) , and its time reversalpartner, the (cid:0) , − (cid:1) dyon, by D ( − ) . Notice M is a boundstate of D (+) and D ( − ) † . Suppose the Z action on D (+) and D ( − ) is Z : D (+) i → Z ij D (+) j , D ( − ) i → ˜ Z ij D ( − ) j (K1)and the T action on D (+) and D ( − ) is T : D (+) i → T ij D ( − ) j , D ( − ) i → ˜ T ij D (+) j (K2)For any matrix M , the operator D (+) † i M ij D (+) j is local,so the actions of Z and T should commute on it. Thisgives the condition( Z ∗ T ) † M ∗ ( Z ∗ T ) = ( T ˜ Z ) † M ∗ ( T ˜ Z ) (K3)In order for this equation to be satisfied by any matrix M , we need to have Z ∗ T = e iφ T ˜ Z (K4)or, equivalently, Z = e − iφ T ∗ ˜ Z ∗ ( T ∗ ) − (K5)Consider the local operator D ( − ) † i M ij D ( − ) j in a similarway, we get the condition˜ Z ∗ ˜ T = e iφ ˜ T Z (K6)or, equivalently, Z = e − iφ ˜ T − ˜ Z ∗ ˜ T (K7)0On the other hand, acting time reversal twice on thelocal operator D (+) ∗ i M ij D (+) j should results in a trivialaction, which implies that T ∗ ˜ T = e iφ (K8)or, equivalently, ˜ T = e iφ ( T ∗ ) − (K9)Combining this equation and (K7) yields Z = e − iφ T ∗ ˜ Z ∗ ( T ∗ ) − (K10)Comparing this equation and (K5) yields e iφ = e iφ (K11)Now consider the (0 ,
2) dyon, which is represented as D (+) i M ij D ( − ) † j . Acting on this operator by Z and T with different orders using (K1) and (K2), and using theconstraints (K5), (K7) and (K11), we see that the Z and T commute on D (+) i M ij D ( − ) † j , which proves the assertionstated at the beginning of this appendix.The above argument can also be applied to O (2) × T symmetric U (1) quantum spin liquids discussed in Sec.L 1. If such a spin liquid has θ = π and the improper Z component not acting as a charge conjugation, then on M the actions of Z and T should commute. L. U (1) quantum spin liquids with some othersymmetries In this appendix we briefly discuss U (1) quantum spinliquids with some other symmetries. The classificationsof these symmetry enriched U (1) quantum spin liquidsare quite complicated, which we leave for future work.In this appendix we only lay out the principle of enumer-ating the putative states and make some comments, butwe will not attempt to finish the procedure of examiningthe anomalies. O (2) × T symmetry First we consider the case where the SO (3) × T sym-metry is broken down to O (2) × T ∼ = ( U (1) (cid:111) Z ) × T .Physically, here U (1) can represent spin rotations aroundone axis, while the Z transformation is a π spin rotationaround another axis perpendicular to this one. This caseis rather complicated, and we do not attempt to completethe anomaly detection and determine the final classifica-tion. Instead, we will just give a way to systematicallylist all putative (possibly anomalous) states and makesome comments.The structure of projective representations of O (2) ×T is rich. On the electric charge, it is classified by Z , andthe three root projective representations physically cor-respond to having half charge under the U (1) subgroup, being a Kramers doublet under T and being a Kramersdoublet under T (cid:48) (the anti-unitary symmetry whose gen-erator is the product of the generators of Z and T ).If θ = 0, then on the magnetic monopole the projectiverepresentations are classified by Z , which physically cor-respond to having half charge under the U (1) subgroupand having the discrete Z anti-commuting with T . Soat θ = 0, if the discrete Z symmetry does not act as acharge conjugation, there are 3 × × = 96 putativestates. We will not write down the long list of all thesestates, since it is straightforward and not particularly il-luminating. Notice some of these are descendant statesof an SO (3) × T symmetric state. Because the anomalyargument for the SO (3) ×T symmetric states should alsoapply to these O (2) ×T symmetric states, their anomaliescan be determined immediately.At θ = π , there are only two putative states,( E fT T (cid:48) M f ) θ and ( E fT T (cid:48) M f ) θ , which are the descen-dants of ( E fT M f ) θ and ( E fT M f ) θ with SO (3) ×T sym-metry, respectively. In the former state the (cid:0) , (cid:1) dyoncarries integer U (1) charge, while in the latter it carrieshalf charge. As discussed in Sec. III, the former stateis anomaly-free while the latter is anomalous even with O (2) symmetry. Here the actions of Z and T commuteon M (the (0 ,
2) dyon in this context), and it is shownin Appendix K that, in this case, it is impossible to havethe actions of Z and T anti-commuting on M .In the case where the discrete Z symmetry acts asa charge conjugation, the possible fractional quantumnumbers on the electric charge have a structure of Z :having half charge under the U (1) subgroup, being aKramers doublet under T , and having charge conjuga-tion squaring to −
1. If θ = 0, the possible fractionalquantum numbers on the magnetic monopole also have astructure of Z : being a Kramers doublet under T (cid:48) , andhaving charge conjugation squaring to −
1. So there are3 × × = 96 such putative states. At θ = π , there aretwo putative states: ( E fT M fT (cid:48) ) θ − and ( E fT M fT (cid:48) ) θ − Z .In the former, the (cid:0) , (cid:1) dyon carries a linear represen-tation of the symmetry, while this dyon has charge con-jugation squaring to − Z × T symmetric cousin with a further U (1) symmetry.The latter is anomalous, because even if the symmetry isbroken to Z × T it is still anomalous.We finish this subsection by briefly commenting ona few models that realize O (2) × T symmetric U (1)quantum spin liquids. Ref. 15–17 studied a couple ofdifferent lattice models that realize three dimensional U (1) quantum spin liquid phases with O (2) × T symme-try, and the particular phase realized in these works is( E b M b ) − . Ref. 70 constructed two models of O (2) × T symmetric U (1) quantum spin liquids, where one of themhas a bosonic monopole and the other has a fermionicmonopole. These two states are ( E b M b ) − and ( E b M f ) − ,respectively.1 Z × Z × T symmetry Now consider the case where the SO (3) × T symmetryis broken down to Z × Z × T , where these two Z ’scan represent π spin rotations around two perpendicu-lar axes. It is known that the projective representations(on the electric charge) of Z × Z × T symmetry areclassified by Z , where two of the four Z ’s are descen-dants of the projective representations of SO (3) ×T , andthe other two Z ’s come from the nontrivial interplay be-tween Z × Z and time reversal. More precisely, eachof the two Z ’s together with time reversal can form anew anti-unitary symmetry, and the other two root non-trivial projective representations can be viewed as havingKramers doublets under such new anti-unitary symme-tries. If θ = 0, on the magnetic monopole the projectiverepresentations are classified by Z , and the nontrivialroot projective representations physically correspond tothe two Z ’s and T anti-commuting.If θ = π , the phase is determined by the (cid:0) , ± (cid:1) dyons, which has one nontrivial projective representa-tion that corresponds to that the two Z symmetriesanti-commute. The state with the two Z symmetriescommuting is a descendant of the SO (3) × T symmetric( E fT M f ) θ , so it is still anomaly-free. The state with thetwo Z symmetries anti-commuting is a descendant of the SO (3) × T symmetric ( E fT M f ) θ , which we conjectureis still anomalous with Z × Z × T symmetry.The descendants of the 15 non-anomalous spin liq-uid states with SO (3) × T still remain non-anomalousand distinct, and we conjecture all the anomalous statesremain anomalous even if the symmetry is broken to Z × Z × T . This is of course just a partial classifi-cation, because, on the one hand, states labelled by theother projective representations of Z × Z × T should betaken into account, and on the other hand, states where Z × Z can act as charge conjugation should also beconsidered. This is not attempted in this paper. Z × Z symmetry Parallel considerations as in Sec. VIII B can be appliedto the case with Z × Z symmetry. Here the projectiverepresentations are only classified by Z , and the non-trivial projective representation is the descendant of theprojective representation of SO (3). Therefore, the de-scendants of E b M b and E b M b will remain distinct andnon-anomalous when the symmetry is broken from SO (3)to Z × Z . We conjecture that the descendant of E b M b remains anomalous. Again, to have a complete classifica-tion, states where Z × Z permutes fractional excitationsshould be taken into account. We do not attempt it here. [1] M. Levin and A. Stern, Phys. Rev. B , 115131 (2012).[2] T. Neupert, L. Santos, S. Ryu, C. Chamon, andC. Mudry, Phys. Rev. B , 165107 (2011), 1106.3989.[3] L. Santos, T. Neupert, S. Ryu, C. Chamon, andC. Mudry, Phys. Rev. B , 165138 (2011), 1108.2440.[4] A. M. Essin and M. Hermele, Phys. Rev. B , 104406(2013), 1212.0593.[5] A. Mesaros and Y. Ran, Phys. Rev. B , 155115 (2013),1212.0835.[6] L.-Y. Hung and Y. Wan, Phys. Rev. B , 195103(2013), 1302.2951.[7] Y.-M. Lu and A. Vishwanath, Phys. Rev. B , 155121(2016), 1302.2634.[8] C. Wang and T. Senthil, Phys. Rev. B , 235122 (2013).[9] M. Barkeshli, P. Bonderson, M. Cheng, and Z. Wang,(2014), 1410.4540.[10] N. Tarantino, N. H. Lindner, and L. Fidkowski, NewJournal of Physics, Volume , Issue3,035006 (2015),1506.06754.[11] T. Lan, L. Kong, and X.-G. Wen, Phys. Rev. B ,235140 (2017), 1602.05946.[12] C. Xu, Phys. Rev. B , 205137 (2013), 1307.8131.[13] X. Chen and M. Hermele, Phys. Rev. B , 195120(2016), 1602.00187.[14] S.-Q. Ning, Z.-X. Liu, and P. Ye, Phys. Rev. B ,245120 (2016), 1609.00985.[15] O. I. Motrunich and T. Senthil, Phys. Rev. Lett. ,277004 (2002). [16] M. Hermele, M. P. A. Fisher, and L. Balents, Phys. Rev.B , 064404 (2004), cond-mat/0305401.[17] O. I. Motrunich and T. Senthil, Phys. Rev. B , 125102(2005).[18] M. A. Levin and X.-G. Wen, Phys. Rev. B , 045110(2005).[19] L. Savary and L. Balents, Phys. Rev. Lett. , 037202(2012), 1110.2185.[20] L. Savary and L. Balents, Phys. Rev. Lett. , 087203(2017), 1604.04630.[21] J. Rochner, L. Balents, and K. P. Schmidt, Phys. Rev.B , 201111 (2016), 1609.00484.[22] I. Affleck and J. B. Marston, Phys. Rev. B , 3774(1988).[23] P. A. Lee and N. Nagaosa, Physical Review B , 5621(1992).[24] X.-G. Wen and P. A. Lee, Phys. Rev. Lett., , 503(1996), cond-mat/9506065.[25] M. Hermele, T. Senthil, M. P. A. Fisher, P. A. Lee, N. Na-gaosa, and X.-G. Wen, Phys. Rev. B , 214437 (2004).[26] S.-S. Lee and P. A. Lee, Physical review letters ,036403 (2005).[27] S.-S. Lee, Phys. Rev. B , 085129 (2008), 0804.3800.[28] T. Senthil, Physical Review B , 045109 (2008).[29] L. Zou and T. Senthil, Phys. Rev. B , 115113 (2016).[30] C. Wang and T. Senthil, Phys. Rev. X , 011034 (2016).[31] M. A. Metlitski and A. Vishwanath, Phys. Rev. B ,245151 (2016). [32] M. A. Metlitski, ArXiv e-prints (2015), arXiv:1510.05663[hep-th].[33] C. Xu and Y.-Z. You, Phys. Rev. B , 220416 (2015),cond-mat/0010440.[34] C. Wang and T. Senthil, Phys. Rev. X , 041031 (2015).[35] D. F. Mross, J. Alicea, and O. I. Motrunich, Phys. Rev.Lett. , 016802 (2016).[36] D. T. Son, Phys. Rev. X , 031027 (2015).[37] C. Wang and T. Senthil, Phys. Rev. B , 085110 (2016).[38] S. D. Geraedts, M. P. Zaletel, R. S. K. Mong, M. A.Metlitski, A. Vishwanath, and O. I. Motrunich, Science , 197 (2016).[39] C. Wang and T. Senthil, Phys. Rev. B , 245107 (2016).[40] C. Wang, N. R. Cooper, B. I. Halperin, and A. Stern,Phys. Rev. X , 031029 (2017), 1701.00007.[41] P.-S. Hsin and N. Seiberg, Journal of High EnergyPhysics , 95 (2016).[42] C. Wang, A. Nahum, M. A. Metlitski, C. Xu, andT. Senthil, Phys. Rev. X , 031051 (2017), 1703.02426.[43] N. Seiberg, T. Senthil, C. Wang, and E. Witten, Annalsof Physics , 395 (2016), 1606.01989.[44] A. Karch and D. Tong, Physical Review X , 031043(2016), arXiv:1606.01893 [hep-th].[45] J. Murugan and H. Nastase, Journal of High EnergyPhysics , 159 (2017).[46] S. Kachru, M. Mulligan, G. Torroba, and H. Wang, Phys.Rev. D , 085009 (2016), arXiv:1608.05077 [hep-th].[47] A. Vishwanath and T. Senthil, Phys. Rev. X , 011016(2013).[48] M. A. Metlitski, C. L. Kane, and M. P. A. Fisher, Phys.Rev. B , 035131 (2013).[49] C. Wang, A. C. Potter, and T. Senthil, Science , 629(2014).[50] S. Kravec, J. McGreevy, and B. Swingle, Phys. Rev. D , 085024 (2015).[51] R. Thorngren, Journal of High Energy Physics , 152(2015).[52] T. T. Wu and C. N. Yang, Phys. Rev. D , 3845 (1975).[53] A. S. Goldhaber, Physical Review Letters , 1122(1976).[54] X. Chen, Z.-C. Gu, Z.-X. Liu, and X.-G. Wen, Phys.Rev. B , 155114 (2013).[55] T. Senthil and M. P. A. Fisher, Physical Review B ,7850 (2000).[56] J. Maciejko, X.-L. Qi, A. Karch, and S.-C. Zhang, Phys.Rev. Lett. , 246809 (2010).[57] B. Swingle, M. Barkeshli, J. McGreevy, and T. Senthil,Phys. Rev. B , 195139 (2011), arXiv:1005.1076 [cond-mat.str-el].[58] A. P. Schnyder, S. Ryu, and A. W. W. Ludwig, Phys.Rev. Lett. , 196804 (2009). [59] Y.-M. Lu and A. Vishwanath, Phys. Rev. B , 125119(2012).[60] T. Senthil and M. Levin, Phys. Rev. Lett. , 046801(2013).[61] Y. Ran, A. Vishwanath, and D.-H. Lee, Phys. Rev. Lett. , 086801 (2008).[62] X.-L. Qi and S.-C. Zhang, Phys. Rev. Lett. , 086802(2008).[63] X.-G. Wen, Quantum field theory of many-body systems (Oxford University Press, 2004) Chap. 7.[64] X.-G. Wen, Phys. Rev., B , 165113 (2002), cond-mat/0107071.[65] L. Fu, C. L. Kane, and E. J. Mele, Phys. Rev. Lett. ,106803 (2007), cond-mat/0607699.[66] S. M. Kravec and J. McGreevy, Phys. Rev. Lett. ,161603 (2013), 1306.3992.[67] X.-G. Wen, Phys. Rev. B , 205101 (2015), 1410.8477.[68] Z. Bi and C. Xu, Phys. Rev. B , 184404 (2015),1501.02271.[69] M. Cheng, privite communication (2017).[70] M. Levin and X.-G. Wen, Phys. Rev. B , 035122(2006).[71] Y. D. Li and G. Chen, Phys. Rev. B , 041106 (2017),1607.02287.[72] G. Chen, Phys. Rev. B , 085136 (2017).[73] G. Chen, Phys. Rev. B , 195127 (2017).[74] L. Zou, Phys. Rev. B , 045130 (2017), 1711.03090.[75] C.-M. Jian and X.-L. Qi, Physical Review X , 041043(2014), arXiv:1405.6688 [cond-mat.str-el].[76] Z. Bi, K. Slagle, and C. Xu, (2015), 1504.04373.[77] C. Wang and T. Senthil, Phys. Rev. B , 195124 (2014).[78] V. Borokhov, A. Kapustin, and X. Wu, JHEP , 049(2002), hep-th/0206054.[79] T. Senthil, J. B. Marston, and M. P. A. Fisher, PhysicalReview B , 4245 (1999).[80] C. L. Kane and E. J. Mele, Phys. Rev. Lett. , 226801(2005).[81] A. C. Potter, C. Wang, M. A. Metlitski, and A. Vish-wanath, (2016), 1609.08618.[82] E. Witten, Physics Letters B , 283 (1979).[83] C. Wang, C.-H. Lin, and Z.-C. Gu, Phys. Rev. B ,195147 (2017), 1610.08478.[84] M. Cheng, N. Tantivasadakarn, and C. Wang, Phys. Rev.X , 011054 (2018).[85] M. Levin and Z.-C. Gu, Phys. Rev. B , 115109 (2012),1202.3120.[86] C. Xu and T. Senthil, Phys. Rev. B87,