Monodromy Defects from Hyperbolic Space
Simone Giombi, Elizabeth Helfenberger, Ziming Ji, Himanshu Khanchandani
MMonodromy Defects from Hyperbolic Space
Simone Giombi, Elizabeth Helfenberger, Ziming Ji, and HimanshuKhanchandani
Department of Physics, Princeton University, Princeton, NJ 08544, USA
Abstract
We study monodromy defects in O ( N ) symmetric scalar field theories in d dimensions.After a Weyl transformation, a monodromy defect may be described by placing the theoryon S × H d − , where H d − is the hyperbolic space, and imposing on the fundamental fields atwisted periodicity condition along S . In this description, the codimension two defect lies atthe boundary of H d − . We first study the general monodromy defect in the free field theory,and then develop the large N expansion of the defect in the interacting theory, focusingfor simplicity on the case of N complex fields with a one-parameter monodromy condition.We also use the (cid:15) -expansion in d = 4 − (cid:15) , providing a check on the large N approach.When the defect has spherical geometry, its expectation value is a meaningful quantity, andit may be obtained by computing the free energy of the twisted theory on S × H d − . Itwas conjectured that the logarithm of the defect expectation value, suitably multiplied bya dimension dependent sine factor, should decrease under a defect RG flow. We check thisconjecture in our examples, both in the free and interacting case, by considering a defectRG flow that corresponds to imposing alternate boundary conditions on one of the low-lyingKaluza-Klein modes on H d − . We also show that, adapting standard techniques from theAdS/CFT literature, the S × H d − setup is well suited to the calculation of the defectCFT data, and we discuss various examples, including one-point functions of bulk operators,scaling dimensions of defect operators, and four-point functions of operator insertions on thedefect.February 24, 2021 a r X i v : . [ h e p - t h ] F e b ontents S × H d − . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.1.1 One-point functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2 Twisted free energy on S d . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 H d − and a defect RG flow 214 Monodromy defect at large N H d − . . . . . . . . . . . . . . . . . . . . . 324.2 Conformal weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 d = 4 − (cid:15) S d with twisted boundary conditions 54B Bulk Integrals 55 B.1 For bulk two-point function . . . . . . . . . . . . . . . . . . . . . . . . . . . 55B.2 For coefficient of displacement . . . . . . . . . . . . . . . . . . . . . . . . . . 58B.3 For bulk OPE coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
Conformal defects are extended objects in a conformal field theory, which preserve a subgroupof the original conformal symmetry. A codimension q flat or spherical conformal defectin a d -dimensional conformal field theory preserves a SO ( q ) × SO ( d − q + 1 ,
1) subgroupof the original SO ( d + 1 ,
1) symmetry group (see [1] for an introduction to the subject). SO ( d − q + 1 ,
1) represents the conformal group on the defect, and SO ( q ) corresponds to1otations around the defect. For the correlation functions of operators on the defect, SO ( q )acts as a global symmetry. Even though only a subgroup of the full conformal symmetry ispreserved, the correlation functions are highly constrained. There has been a lot of recentactivity in analyzing the consequences of these constraints [1–9].In this paper we study monodromy defects, working with scalar field theories as ourmain example. A monodromy defect is a codimension two defect, so q = 2. To describe thiscodimension two defect in flat space, we can parametrize coordinates as x = ( r, θ, (cid:126)y ) where (cid:126)y represents d − r, θ are polar coordinates perpendicularto the defect. The defect is located at r = 0 in these coordinates. A monodromy defect in aCFT with a global symmetry group G may be defined by requiring that the fields satisfy φ ( r, θ + 2 π, (cid:126)y ) = φ g ( r, θ, (cid:126)y ) (1.1)where g is a non-identity element of G . For a free scalar field and 3 d Ising model, G = Z .Such a monodromy defect in the 3d Ising model (where it is a line defect) was introducedin [12, 13], and further studied in [8, 14]. In this paper, we consider more generally conformalfield theories consisting of N scalars that preserve an O ( N ) symmetry, so G = O ( N ). Wediscuss both free and interacting O ( N ) model, using the large N and (cid:15) expansions . Alreadyin the free theory, the structure is richer compared to the case of a single free scalar. Someresults for the monodromy defect in critical O ( N ) model in d = 4 − (cid:15) were also obtainedin [15].A codimension two defect may also be studied by mapping the problem to a hyperboliccylinder, S × H d − (for previous examples of conformal mapping to H m × S n spaces todescribe conformal defects, see e.g. [2, 11, 16, 17]). A flat defect in flat space can be relatedby a Weyl transformation to S × H d − as follows ds R d = r (cid:18) ( d(cid:126)y ) + dr r + dθ (cid:19) = r ds S × H d − (1.2)with (cid:126)y and r being the Poincar´e coordinates on H d − and θ being the coordinate on S .The monodromy defect is then simply described by imposing twisted periodicity conditions(1.1) along S in the path integral for the theory on S × H d − (for instance, in the Z case,this just means taking the scalar to be antiperiodic on S ). The defect is located at r = 0,which is the boundary of the hyperbolic space. The hyperbolic cylinder setup that we usehere is similar to that used in [18–20] to study Renyi entropies for a spherical entangling Conformal defects in free scalar theories were recently discussed also in [10, 11]. Starting with the theory on S × H d − , one can then perform a Kaluza-Kleinreduction on S to obtain a tower of massive fields on H d − with the defect theory on itsboundary. Standard techniques from the AdS/CFT literature may then be used to obtainresults for the defect CFT data. For example, the scaling dimensions of the defect operatorscan be related to the masses on H d − . Other examples of defect CFT data may also beextracted conveniently from the hyperbolic space setup, and we will discuss a few explicitsuch calculations below. Similar ideas have been used in the literature for boundaries inconformal field theory in [21–25] and for higher codimension defects in [11]. Field theory ona hyperbolic cylinder was also studied in [26, 27].A monodromy defect with spherical geometry may be described in the same way bysimply using, instead of the Poincar´e metric in (1.2), the hyperbolic ball metric for H d − ,so that the boundary is the sphere S d − . For a spherical defect, it is natural to define itsexpectation value (cid:104)D(cid:105) . In the hyperbolic space setup, this expectation value can be obtainedin terms of the free energy of the twisted theory on S × H d − as − log (cid:104)D(cid:105) = F twisted − F untwisted (1.3)where F twisted and F untwisted are the free energies on S × H d − in the presence and absenceof a monodromy defect respectively. The subtraction of the untwisted theory free energycorresponds to normalizing (cid:104)D(cid:105) by the partition function of the theory without defect.For a range of masses, massive fields on H d − have two allowed boundary conditions [28],which in our setup are interpreted as different defect CFTs, which sit at the endpoints of aRG flow localized on the defect. It was conjectured in [29] (generalizing a similar proposal [30]in the absence of defects) that for a codimension q defect, the quantity defined by˜ D ≡ sin (cid:18) π ( d − q )2 (cid:19) log (cid:104)D(cid:105) (1.4)decreases under RG flows localized on the defect. Note that, when the dimension of thedefect d − q is even, the expectation value has a logarithmic divergence related to one ofthe conformal anomaly coefficients of the defect. This divergence appears as a pole whenworking in dimensional regularization. The sine factor in (1.4) cancels the pole and ˜ D is afinite quantity proportional to the anomaly coefficient. On the other hand, for odd d − q , In that case, rather than a twisted periodicity condition, one lets the inverse temperature, i.e. the lengthof S , be 2 πq to describe the q th Renyi entropy. This setup can also be thought in terms of defect CFT, asdiscussed in [17]. D decreases under defect RG flows in the examples we study, bycalculating the defect expectation value on S × H d − . We verify that whenever the defectRG flow is such that scaling dimensions obey unitarity bounds, ˜ D decreases under the flow,but this does not hold true for non-unitary flows.We now provide a summary of the rest of this paper. We start in section 2 by studyingmonodromy defects in the free O ( N ) model. We discuss the most general monodromydefect, and show that it is essentially sufficient to study the monodromy defect in a singlefree complex scalar field theory, and the results for the most general monodromy defect infree O ( N ) model follow from the results for this case. We start by describing the monodromydefect in flat space, and then explain how to map the problem to the hyperbolic cylinder S × H d − by a Weyl transformation. We also discuss how an alternative Weyl transformationmaps the problem to the d -dimensional sphere (with twisted periodicity along one angle). Wecalculate various defect CFT observables, including the dimensions of the defect operatorsand one-point functions of bulk operators with spin. We define and calculate the expectationvalue of a spherical defect using both the S × H d − and S d setups, and show that theyagree. In section 3, we then study a defect RG flow in the free theory. The two defect CFTsconnected by the RG flow correspond to the two possible boundary conditions for one of thelow-lying KK modes on H d − in the S × H d − setup. A similar defect RG flow in the caseof non-monodromy defects in a free scalar theory was discussed in [11].In section 4, we study the monodromy defect in the interacting theory of N scalars witha φ O ( N ) invariant interaction. In the interacting case, we study the simple situation of 2 N scalars with N pairs mixing into each other as they go around the defect. This is equivalent tostudying a monodromy defect for N interacting complex scalars with the same monodromyfor all of them, and it preserves a U ( N ) symmetry. The case of Z monodromy on all scalars,generalizing the Ising case of [13], can be obtained as a special case. We develop the large N description of the monodromy defect, and calculate the scaling dimensions of defect operatorsto leading order at large N . We also study the defect expectation value and the expectationvalue of the bulk stress-tensor, which is proportional to the “conformal weight” [16, 17] ofthe defect, to leading order at large N . We then study the same defect in the Wilson-Fisher (cid:15) expansion in d = 4 − (cid:15) , and provide some checks of the large N analysis. Wealso calculate correlation functions of the defect operators by computing Witten diagramsin H d − to leading order in (cid:15) . We then conclude in section 6 and mention some futuredirections. Appendices contain some technical details and some useful integrals.4 Monodromy defect in free field theory
Consider an O ( N ) symmetric theory of N free scalars in flat space. The most generalmonodromy defect that we can define imposes that the scalars satisfy [15] φ I ( r, θ + 2 π, (cid:126)y ) = G IJ φ J ( r, θ, (cid:126)y ) , G IJ ∈ O ( N ) . (2.1)The most general O ( N ) matrix G IJ , can always, by a change of basis, be brought to thefollowing form G IJ ( ϑ ) = R ( ϑ ) . . . 0 R ( ϑ k ) ±
10 . . . ± , R ( ϑ ) = (cid:34) cos ϑ − sin ϑ sin ϑ cos ϑ (cid:35) . (2.2)So there are k pair of scalars that mix into each other and the rest either remain unchangedor pick up a minus sign as they go around the defect. We can then combine each pair intoa complex combination Φ = φ + iφ and the monodromy can be represented asΦ( r, θ + 2 π, (cid:126)y ) = e iϑ Φ( r, θ, (cid:126)y ) , ϑ ∼ ϑ + 2 π. (2.3)Hence ϑ = 0 describes the trivial defect while ϑ = π describes the special case when the twofields change a sign as they go around the defect. So in the rest of this section, we will considera single complex scalar with the monodromy defined in (2.3). It has a U (1) ∼ SO (2) internalsymmetry which is enhanced to O (2) for ϑ = 0 and π (Φ → ¯Φ, which is a part of O (2)but not SO (2), is also a symmetry for these values of ϑ ). One may combine these complexscalars with different ϑ ’s to obtain results for free O ( N ) model with a general monodromydefect (for each minus sign in the monodromy matrix (2.2), one can simply set ϑ = π inthe result for a complex scalar below, and include an extra factor of 1 / v = ϑ/ π and use either v or ϑ , whichever is convenient. There is a periodicity in v whichimplies that everything should be invariant under v → v + 1, but for many calculations, we Not to be confused with the group of rotations around the defect, which is a spacetime symmetry in thebulk and is also SO (2). v to be 0 ≤ v <
1. We will write expressions specializing to this rangeof v , so they may not look periodic in v .In a conformal field theory with a defect, in addition to the usual short distance OPE inthe bulk, a bulk operator can also be expanded in terms of operators living on the defect.For the complex scalar with monodromy given by (2.3), it takes the following form [1, 13]Φ( r, θ, (cid:126)y ) = (cid:88) O C Φ O e is O θ r ∆ Φ − ∆ O B ∆ O ( r, (cid:126)∂ y ) O ( (cid:126)y ) , s O ∈ Z + v. (2.4)As we mentioned in the introduction, SO (2) symmetry of rotations around the defect actsas a global symmetry on the defect. s O is the charge of the operator O under this globalsymmetry and we will call it transverse spin or just spin. There is also a longitudinal spin l ,which is the charge under rotations along the defect, but we will only consider l = 0 defectoperators in this paper. The remaining conformal invariance fixes the bulk-defect two pointfunction (cid:104) Φ( x ) ¯ O ( (cid:126)y ) (cid:105) = C Φ O C ∆ O e iθs O r ∆ Φ − ∆ O ( r + ( (cid:126)y ) ) ∆ O (2.5)where ¯ O is a defect operator that has spin − s O and dimension ∆ O . Consistency of (2.4) and(2.5) fixes the form of the function B s ( r, (cid:126)∂ y ) B ∆ O ( r, (cid:126)∂ y ) = ∞ (cid:88) m =0 ( − m r m ( (cid:126)∂ y ) m m !2 m (cid:0) ∆ O + 2 − d (cid:1) m . (2.6)This is similar to what was done for BCFT in [31]. In general, there could be several defectoperators of a given spin. But since Φ is a free field, it satisfies the bulk equation of motion ∇ Φ = 0, which implies (cid:18) ∂ ∂r + 1 r ∂∂r + 1 r ∂ ∂θ + ∂ ∂(cid:126)y (cid:19) (cid:104) Φ( x ) ¯ O ( (cid:126)y ) (cid:105) = 0= ⇒ ∆ O = ∆ Φ ± | s O | = d − ± | s O | . (2.7)The unitarity bound for the CFT on the defect requires the dimensions of the defect operatorto satisfy ∆ O ≥ max (cid:18) d − , (cid:19) . (2.8)This is always satisfied for the positive sign above (as long as d >
2) and we defer thediscussion of negative sign until next section. So for every spin, there is a single operator on6he defect with dimension ∆ s = d/ − | s | . Hence, the bulk-defect OPE of the fundamentalfields may be written as a sum over spinsΦ( r, θ, (cid:126)y ) = (cid:88) s ∈ Z + v C Φ s e isθ r ∆ Φ − ∆ s B s ( r, (cid:126)∂ y )Ψ s ( (cid:126)y )¯Φ( r, θ, (cid:126)y ) = (cid:88) s ∈ Z + v ( C Φ s ) ∗ e − isθ r ∆ Φ − ∆ s B s ( r, (cid:126)∂ y ) ¯Ψ s ( (cid:126)y ) . (2.9)In terms of original real scalar fields, Ψ s = ψ s + iψ s while ¯Ψ s = ¯ ψ s − i ¯ ψ s = ψ − s − iψ − s where ψ is appear in the bulk-defect OPE of the real scalars φ i .In the presence of a defect, the two-point function of bulk scalars is fixed up to a functionof cross-ratios [1, 8] (cid:104) Φ( x ) ¯Φ( x ) (cid:105) = F ( θ , ξ )( r r ) d − , θ = θ − θ , ξ = ( (cid:126)y − (cid:126)y ) + ( r − r ) r r . (2.10)Corresponding to the two OPE limits (i.e. the bulk OPE and the bulk-defect OPE), thefunction F can be expanded into bulk and defect channel conformal blocks F ( θ , ξ ) = (cid:88) O C ¯Φ;Φ; O C O g ∆ O ,J O ( θ , ξ ) = (cid:88) O | C Φ O | f ∆ O ,s O ( θ , ξ ) (2.11)where g and f are the bulk channel and defect channel conformal blocks respectively. Thesum on the left runs over the bulk operators that get a non-zero one-point function, andthe coefficient is the product of the usual bulk OPE coefficient times the one-point functioncoefficient of the bulk operator. The sum on the right runs over the defect operators thatappear in the bulk-defect OPE of Φ.As we determined above, the operators appearing in the defect channel have spin s anddimension ∆ s = d/ − | s | . The defect channel blocks are known in general [1, 8]. For thecase of a codimension two defect, they simplify and the resulting expression for the two-pointfunction can be written as a sum over defect operators (cid:104) Φ( x ) ¯Φ( x ) (cid:105) = G ¯ΦΦ ( x , x ) = (cid:88) s ∈ Z + v Γ (∆ s ) e isθ F (cid:16) ∆ s , ∆ s + − d ; 2∆ s + 3 − d ; − ξ (cid:17) r r ) d − π d/ Γ(∆ s + 2 − d )(4 ξ ) ∆ s . (2.12)7he sum can be explicitly performed in d = 4 to get (cid:104) Φ( x ) ¯Φ( x ) (cid:105) = (cid:88) s ∈ Z + v e isθ π r r (cid:112) ξ (1 + ξ )( √ ξ + √ ξ ) s − = ( ξ (1 + ξ )) − / π r r e iθ v (cid:0) √ ξ + √ ξ + 1 (cid:1) v − e iθ (cid:16) ξ + 2 (cid:112) ξ (1 + ξ ) + 1 (cid:17) + e iθ v (cid:0) √ ξ + √ ξ + 1 (cid:1) − v − e iθ + 2 ξ + 2 (cid:112) ξ (1 + ξ ) + 1 . (2.13)Note that we are using a normalization, such that in the bulk OPE limit, when x → x ,the correlator goes like (cid:104) Φ( x ) ¯Φ( x ) (cid:105) ∼ Γ (cid:0) d − (cid:1) π d/ | x − x | d − . (2.14)We normalize defect operators such that C Φ s = 1 in the free theory. The two-point functionof the defect operators is then given by (cid:104) Ψ s ( (cid:126)y ) ¯Ψ s ( (cid:126)y ) (cid:105) = δ s ,s C ∆ s ( (cid:126)y ) ∆ s , C ∆ s = Γ (∆ s )2 π d/ Γ (cid:0) ∆ s + 2 − d (cid:1) . (2.15)In the bulk channel conformal block decomposition, the operators that appear are thebulk scalar ¯ΦΦ and the conserved currents of all spins, which can be schematically writtenas ¯Φ( ∂ µ ) J Φ. To extract the bulk expansion coefficients, one may use the inversion formulaof [8]. Here, we restrict to calculating the one-point function of the first few operators of lowspin. The one-point function of the operator ¯ΦΦ can be extracted from the short distancelimit of the correlator (2.12) (cid:104) ¯ΦΦ( x ) (cid:105) = C ¯ΦΦ1 r d − , C ¯ΦΦ1 = ( d − (cid:0) d − v (cid:1) Γ (cid:0) d − v (cid:1) sin ( vπ ) Γ (cid:0) d (cid:1) π d +1 Γ( d )(2 − d ) . (2.16)The conserved currents, which are spinning operators, also get one-point functions. The spinone current, which corresponds to the global U (1) symmetry of the theory is given by J µ = i (cid:0) Φ ∂ µ ¯Φ − ¯Φ ∂ µ Φ (cid:1) . (2.17)The one-point function of a parity odd spin one operator in the presence of a defect is fixed8y conformal symmetry [1] (cid:104) J i (cid:105) = C J (cid:15) ij n j r d , (cid:104) J a (cid:105) = 0 , n i = x i /r. (2.18)We parametrize the coordinates as x = ( x i , x a ) = with i, j now being Cartesian transversecoordinates, a, b being directions along the defect and (cid:15) ij is the antisymmetric tensor intransverse directions. We can calculate this one point function by calculating derivativesof (2.12), and then taking the short distance limit. Since it is fixed up to a constant, it isenough to do the calculation just for one component. We do it for the θ component, (cid:104) J θ (cid:105) = − C J r d − = i (cid:104) Φ ∂ θ ¯Φ − ¯Φ ∂ θ Φ (cid:105) = 2 − d π − ( d +1)2 Γ (cid:0) − d (cid:1) r d − ∞ (cid:88) k = −∞ ( k + v )Γ (cid:0) d − | k + v | (cid:1) Γ (cid:0) − d + | k + v | (cid:1) = ( d − v − d − C ¯ΦΦ1 r d − . (2.19)At v = 1 /
2, we expect the internal U (1) symmetry to be enhanced to full O (2) symmetrywhich includes ¯Φ → Φ, under which J µ → − J µ . So we expect the correlators containing oddpowers of J µ to vanish at v = 1 /
2, and indeed the one-point function vanishes at v = 1 / (cid:104) T ij (cid:105) R d = h π ( d − δ ij − dn i n j r d , (cid:104) T ab (cid:105) R d = − h π δ ab r d , (cid:104) T ai (cid:105) R d = 0 . (2.20)In analogy with the scaling dimensions of local operators, h is referred to as the conformalweight of the defect [16, 17, 32]. It can be determined by doing explicit calculation of anycomponent of the stress tensor and we choose T θθ . The canonical stress energy tensor for afree complex scalar in flat space is T µν = ∂ µ ¯Φ ∂ ν Φ − g µν ∂ ¯Φ · ∂ Φ − ( d − d − (cid:0) ∂ µ ∂ ν − g µν ∂ (cid:1) | Φ | . (2.21)This gives (cid:104) T θθ (cid:105) = (cid:104) ∂ θ ¯Φ ∂ θ Φ (cid:105) − r d −
1) ( (cid:104) ∂ r ¯Φ ∂ r Φ (cid:105) + (cid:104) (cid:126)∂ y ¯Φ (cid:126)∂ y Φ (cid:105) ) − d − d − (cid:104) ¯Φ ∂ θ Φ (cid:105) − r ( d − d − (cid:104) ¯Φ ∂ r Φ (cid:105) . (2.22)9sing the two-point function in (2.12) and taking appropriate derivatives, we get (cid:104) T θθ (cid:105) = Γ (cid:0) − d (cid:1) (1 − v ) v (cid:0) csc π (cid:0) d − v (cid:1) − csc π (cid:0) d + v (cid:1)(cid:1) d (4 π ) d − Γ (cid:0) − d − v (cid:1) Γ (cid:0) − d + v (cid:1) r d − = − ( d − v (1 − v ) d C ¯ΦΦ1 r d − . (2.23)Comparing with (2.20), it is easy to see that (cid:104) T θθ (cid:105) = ( d − h π r d − = ⇒ h = − π Γ (cid:0) − d (cid:1) (1 − v ) v (cid:0) csc π (cid:0) d − v (cid:1) − csc π (cid:0) d + v (cid:1)(cid:1) d (4 π ) d − Γ (cid:0) − d − v (cid:1) Γ (cid:0) − d + v (cid:1) . (2.24)We checked numerically that this conformal weight h is always positive for d >
2. This isconsistent with the conjecture proposed in [4] which says that h ≥ . We can follow this logic and calculate the one-point function of any higher spin current.We just do it for one more case here, namely the spin 3 symmetric current. The current isgiven by (explicit expression in d = 4 can be found in, for example, [33]) J µνρ = 6 i (cid:18) ¯Φ ∂ µ ∂ ν ∂ ρ Φ − d + 2) d − ∂ ( µ ¯Φ ∂ ν ∂ ρ ) Φ + 6 d − g ( µν ∂ γ ¯Φ ∂ γ ∂ ρ ) Φ (cid:19) + c.c. (2.25)where () in the subscript means that the indices are symmetrized. Its one-point function isalso fixed by conformal symmetry up to a number, so we only look at one of its componentswith all indices equal to θ . We act with these derivatives on (2.12) and expand them in thebulk limit ξ → (cid:104) J θθθ (cid:105) = 6(1 − v ) (3 d − d (2( v − v + 5) + 4( v − v + 8) d − C ¯ΦΦ1 r d − . (2.26) Displacement operator
The presence of a defect breaks the translational symmetry perpendicular to the defect. Thisleads to the presence of a displacement operator in the spectrum of the defect theory whichmay be defined as the divergence of stress tensor ∂ µ T µi = D i ( (cid:126)y ) δ ( x i ) (2.27)where i represents directions perpendicular to the defect and the delta function is localizedon the defect, at x i = 0. This equation fixes the scaling dimension of the displacement equal In [4], stress tensor one-point function was written in terms of a T which is related to h by h = − πa T /d ,so they conjectured that a T ≤ d −
1, and its SO (2) spin equal to 1. In the free theory, this requires the displacement tobe proportional to the operator Ψ v ¯Ψ − v because this is the only operator with the requiredspin and conformal dimension. The proportionality constant is also fixed by (2.27). Inthis subsection, we find this constant and calculate the normalization of the displacementoperator which is a piece of the defect CFT data.It is convenient to work with complex coordinates in directions transverse to the defect z = re − iθ and call the two components of displacement D and ¯ D defined by ∂ µ T µz = 2 D ( (cid:126)y ) δ ( z, ¯ z ) , ∂ µ T µ ¯ z = 2 ¯ D ( (cid:126)y ) δ ( z, ¯ z ) (2.28)since 2 δ ( z, ¯ z ) = δ ( x ). In terms of these coordinates, the bulk defect OPE takes the formΦ( z, ¯ z, (cid:126)y ) = (cid:88) s ∈ Z + v z | s |− s ¯ z | s | + s B s ( | z | , (cid:126)∂ y )Ψ s ( (cid:126)y ) . (2.29)Using these definitions, it is easy to see that only the ∂ z T zz term in the divergence of thestress tensor contains a delta function T zz = 4 T ¯ z ¯ z (cid:51) v (1 − v )¯ z Ψ v ¯Ψ − v ( (cid:126)y ) = ⇒ ∂ z T zz (cid:51) πv (1 − v )Ψ v ¯Ψ − v ( (cid:126)y ) δ ( z, ¯ z )= ⇒ D ( (cid:126)y ) = 4 πv (1 − v )Ψ v ¯Ψ − v ( (cid:126)y ) . (2.30)We can use this to directly calculate coefficient of the displacement operator (cid:104) D ( (cid:126)y ) ¯ D ( (cid:126)y ) (cid:105) = C D ( (cid:126)y ) d − = 16 π v (1 − v ) C ∆ v C ∆ v − ( (cid:126)y ) d − = ⇒ C D = 4Γ (cid:0) d − v (cid:1) Γ (cid:0) d − v (cid:1) π d − Γ( − v )Γ( v − . (2.31)The displacement operator satisfies a Ward identity involving correlators of the displace-ment with the bulk operators [1]. It implies that if the displacement appears in the bulk-defect OPE of a bulk operator O , then the corresponding bulk-defect OPE coefficient C O D must satisfy [1] ∆ O C O = − (cid:16) π (cid:17) d − √ π Γ (cid:0) d − (cid:1) C O D C D (2.32)where C O is the coefficient of the one-point function of O . The displacement operator doesappear in the bulk defect OPE of the operator ¯ΦΦ. It can be seen from the defect channeldecomposition of the two-point function of ¯ΦΦ, which can be obtained by Wick contraction11nd contains the following term (cid:104) ¯ΦΦ( x ) ¯ΦΦ( x ) (cid:105) (cid:51) e i ( θ − θ ) Γ (∆ v ) Γ (∆ v − )4 v +∆ v − π d Γ (cid:0) ∆ v + 2 − d (cid:1) Γ (cid:0) ∆ v − + 2 − d (cid:1) ξ ∆ v +∆ v − (cid:51) e i ( θ − θ ) Γ (cid:0) d − v (cid:1) Γ (cid:0) d − v (cid:1) d π d Γ(1 + v )Γ(2 − v ) ξ d − . (2.33)Comparing it to the form we expect, (cid:104) ¯ΦΦ( x ) ¯ΦΦ( x ) (cid:105) (cid:51) ( C ¯ΦΦ D ) C D e i ( θ − θ ) (4 ξ ) d − = ⇒ ( C ¯ΦΦ D ) C D = Γ (cid:0) d − v (cid:1) Γ (cid:0) d − v (cid:1) π d Γ(1 + v )Γ(2 − v ) . (2.34)Using results in (2.16), (2.31) and (2.34), it is is easy to check that the Ward identity (2.32)is satisfied. S × H d − As explained in the introduction, the monodromy defect may also be studied on a hyperboliccylinder by a Weyl transformation as in (1.2). The operators also get rescaled under thisWeyl transformation. The scalars, for instance, transform as O S × H d − = r ∆ O O R d .In order to describe a spherical defect, one may use the hyperbolic ball coordinates on H d − , obtained from the Poincar´e coordinates by the following coordinate transformation r = 1cosh η − Ω sinh η , y a = Ω a +1 sinh η cosh η − Ω sinh η (2.35)where (Ω , . . . , Ω d − ) are the coordinates on a d − | Ω a | = 1 and0 ≤ η < ∞ . The metric in these coordinates takes the following simple form ds S × H d − = dθ + dη + sinh η ds S d − . (2.36)Note that the defect is compact and is located at the boundary of hyperbolic ball, η → ∞ ,which is a d − S d − .The complex scalar on S × H d − is described by the action S = 12 (cid:90) d d x (cid:112) g ( x ) (cid:18) g µν ∂ µ Φ ∂ ν ¯Φ + (cid:18) ( d − d − R + m (cid:19) | Φ | (cid:19) = 12 (cid:90) d d x (cid:112) g ( x ) (cid:18) g µν ∂ µ Φ ∂ ν ¯Φ − (cid:18) ( d − − m (cid:19) | Φ | (cid:19) (2.37)12ith the field Φ obeying twisted boundary conditions along S , Φ( r, (cid:126)y, θ + 2 π ) = e iϑ Φ( r, (cid:126)y, θ ).We will be interested in the conformally coupled case with m = 0. An equivalent descriptionof the system can be written in terms of untwisted field Ψ defined by Φ( x ) = e ivθ Ψ( x ). Ψhas the usual periodic boundary conditions Ψ( r, (cid:126)y, θ + 2 π ) = Ψ( r, (cid:126)y, θ ). The action in termsof Ψ can be written as S = 12 (cid:90) d d x (cid:112) g ( x ) (cid:32)(cid:12)(cid:12)(cid:12)(cid:12) ( ∂ θ + iv ) Ψ (cid:12)(cid:12)(cid:12)(cid:12) + g αβ ∂ α Ψ ∂ β ¯Ψ − ( d − | Ψ | (cid:33) (2.38)where α, β are the coordinates on H d − . This shows that having a monodromy defect isequivalent to having a constant background gauge field in the θ direction. Taking derivativeswith v is equivalent to inserting the θ component of the U (1) current − δ log Zδv = i (cid:90) d d x (cid:112) g ( x ) (cid:104) (cid:0) Φ ∂ θ ¯Φ − ¯Φ ∂ θ Φ (cid:1) (cid:105) = 12 (cid:90) d d x (cid:112) g ( x ) (cid:104) J θ (cid:105) (2.39)where Z is the partition function in presence of the defect.We then perform a Kaluza-Klein (KK) reduction on S to get a tower of massive scalarfields on H d − . The bulk field can be expanded into KK modes as Φ( r, (cid:126)y, θ ) = (cid:80) e isθ Φ s ( r, (cid:126)y )where s ∈ Z + v and modes Φ s ( r, (cid:126)y ) have mass s − ( d − /
4. Since the defect is locatedon the boundary of H d − , we can use the standard AdS/CFT dictionary to calculate thedimensions of the defect operators of spin s induced by Φ∆ s (∆ s − d + 2) = s − ( d − / ⇒ ∆ s = d − ± | s | . (2.40)As before, we leave the discussion of − sign until the next section. The two-point functionon S × H d − can then be written as a sum over KK modes with the two-point function ofeach KK mode being just the usual bulk-bulk propagator on H d − . This gives (cid:104) Φ( (cid:126)y , r , θ ) ¯Φ( (cid:126)y , r , θ ) (cid:105) = G ¯ΦΦ ( x , x ) = (cid:88) s ∈ Z + v e isθ π G bb ∆ s = (cid:88) s ∈ Z + v
2Γ (∆ s ) e isθ F (cid:16) ∆ s , ∆ s + − d ; 2∆ s + 3 − d ; − ξ (cid:17) π d/ Γ(∆ s + 2 − d )(4 ξ ) ∆ s . (2.41)This is related by a Weyl transformation to the two-point function in flat space (2.12).13 quantity of interest is the free energy on the hyperbolic space, since this is related tothe expectation value of the spherical monodromy defect. In the free theory, it is given bythe following determinant F twisted ( ϑ ) = tr log (cid:18) −∇ + m − ( d − (cid:19) . (2.42)The eigenfunctions of the Laplacian on S × H d − may be written as Φ H d − ( y i , r ) e isθ withΦ H d − being the eigenfunction on the d − λ + ( d − / s with a degeneracy given by [34, 35] D ( λ ) dλ = (Vol( H d − ))(4 π ) d − Γ( d − ) | Γ( i √ λ + d − ) | | Γ( i √ λ ) | dλ √ λ . (2.43)Using this, we can compute the twisted free energy on the hyperbolic space as F twisted ( ϑ ) = (cid:90) ∞ dλD ( λ ) (cid:88) s ∈ Z + v log (cid:0) λ + m + s (cid:1) = Vol( H d − )(4 π ) d − Γ( d − ) (cid:90) ∞−∞ dν | Γ( iν + d − ) | | Γ( iν ) | (cid:88) s ∈ Z + v log (cid:0) ν + s + m (cid:1) . (2.44)This can be used to calculate the expectation value of the defect, and it is natural to normalizeit by the partition function of the untwisted theory. In the conformally coupled case, it gives − log (cid:104)D(cid:105) = F twisted − F untwisted = Vol( H d − )(4 π ) d − Γ( d − ) (cid:90) ∞−∞ dν | Γ( iν + d − ) | | Γ( iν ) | (cid:32) (cid:88) n ∈ Z + v − (cid:88) n ∈ Z (cid:33) log (cid:0) ν + n (cid:1) = Vol( H d − )(4 π ) d − Γ( d − ) (cid:90) ∞−∞ dν | Γ( iν + d − ) | | Γ( iν ) | log (cid:18)
12 csch ( πν ) (cosh(2 πν ) − cos(2 πv )) (cid:19) . (2.45)To derive the above formula, we had to use the sum [18] (cid:88) k ∈ Z log (cid:0) ( k + α ) + a (cid:1) = log (2 cosh(2 πa ) − πα )) . (2.46)When d is even, the analytic form is easy to obtain by doing the integral over ν first in the14 d ˜ Figure 1: A plot of ˜ D for a single free complex scalar between dimensions 2 < d < v = 1 / (cid:104)D(cid:105) = (cid:32) (cid:88) n ∈ Z + v − (cid:88) n ∈ Z (cid:33) ∂∂α (cid:34) Vol( H d − )(4 π ) d − Γ( d − ) (cid:90) ∞−∞ dν | Γ( iν + d − ) | | Γ( iν ) | ν + n ) α (cid:35) α → . (2.47)It gives − log (cid:104)D(cid:105) d =2 = Vol( H ) (cid:18) ζ ( − , v ) + ζ ( − , − v ) + 16 (cid:19) = v (1 − v )Vol( H ) − log (cid:104)D(cid:105) d =4 = Vol( H ) (cid:18) − ζ ( − , v ) − ζ ( − , − v ) + 1360 π (cid:19) = v (1 − v ) π Vol( H ) − log (cid:104)D(cid:105) d =6 = Vol( H ) (cid:18) ζ ( − , v ) + ζ ( − , − v )60 π − ζ ( − , v ) + ζ ( − , − v )36 π + 11680 π (cid:19) = − v (1 − v ) ( − − v (1 − v ))180 π Vol( H ) . (2.48)Note that the factors of the hyperbolic space volume here are logarithmically divergent[20,36], see eq. (2.50). The quantity ˜ D defined in (1.4) is however finite and it is proportionalto the quantities multiplying the volume factors above. Indeed using the above result (2.45),˜ D can be seen to be a smooth and finite function of d . We plot it for 2 < d < Z monodromy, v = 1 /
2, in figure 1. For future reference, let us also list some15xplicit values of ˜ D in various d for v = 1 / D| d =3 = log 24 − ζ (3)8 π , ˜ D| d =4 = π , ˜ D| d =5 = log 264 − ζ (3)192 π − ζ (5)128 π , ˜ D| d =6 = 13 π . (2.49)To do the above calculation and obtain the plot, we had to use the regularized volume ofthe hyperbolic space [20, 36] Vol( H d − ) = π d − Γ (cid:18) − d (cid:19) . (2.50)For odd values of d and generic v , it is more convenient to use the sphere geometry to obtainanalytic results, as we show in (2.64) for d = 3. Using the round sphere setup below, we willalso obtain an expression for log (cid:104)D(cid:105) valid in continuous d . One-point functions of bulk operators can also be readily obtained in the hyperbolic space.For the scalar ¯ΦΦ, it is a constant given by C ¯ΦΦ1 = 1 π Vol( H d − ) ∂F twisted ( ϑ ) ∂m (cid:12)(cid:12)(cid:12)(cid:12) m =0 (2.51)The mass derivative of the free energy can be calculated as follows ∂F twisted ( ϑ ) ∂m = ∞ (cid:88) k = −∞ Vol( H d − )(4 π ) d − Γ (cid:0) d − (cid:1) (cid:90) ∞−∞ dν | Γ( iν + d − ) | | Γ( iν ) | (cid:18) ν + ( k + v ) + m (cid:19) = Vol( H d − )Γ (cid:0) − d (cid:1) (4 π ) d − ∞ (cid:88) k = −∞ Γ (cid:18) d − (cid:113) m + ( k + v ) (cid:19) Γ (cid:18) − d + (cid:113) m + ( k + v ) (cid:19) = Vol( H d − )Γ (cid:0) − d (cid:1) (4 π ) d − (cid:88) s Γ (∆ s )Γ (3 − d + ∆ s ) . (2.52)To perform the integral, we had to close the contour in the ν plane and sum over residues [22].The arc at infinity can only be dropped for d <
3, but the final result can be analyticallycontinued in dimensional regularization. One of the Gamma function introduces poles at ν = i ( d/ − κ ) for integer κ , which all lie in the upper half plane for d > m →
0, we get ∂F twisted ( ϑ ) ∂m (cid:12)(cid:12)(cid:12)(cid:12) m =0 = Vol( H d − )Γ (cid:0) − d (cid:1) (4 π ) d − ∞ (cid:88) k = −∞ Γ (cid:0) d − | k + v | (cid:1) Γ (cid:0) − d + | k + v | (cid:1) = ( d − H d − )Γ (cid:0) d − v (cid:1) Γ (cid:0) d − v (cid:1) sin ( vπ )(4 π ) d − (2 − d )Γ (cid:0) d +12 (cid:1) = ⇒ C ¯ΦΦ1 = ( d − (cid:0) d − v (cid:1) Γ (cid:0) d − v (cid:1) sin ( vπ ) Γ (cid:0) d (cid:1) π d +1 Γ( d )(2 − d ) . (2.53)This of course agrees with the flat space result (2.16). For the spin one U (1) current, usingits general form in (2.18), it is easy to see that the one-point function of its θ component is aconstant on hyperbolic cylinder. It may be calculated as in (2.39), by taking the derivativeof free energy with v (cid:104) J θ (cid:105) = 1 π Vol( H d − ) ∂F twisted ( ϑ ) ∂v = ∞ (cid:88) k = −∞ − d π − ( d +1)2 Γ (cid:0) d − (cid:1) (cid:90) ∞−∞ dν | Γ( iν + d − ) | | Γ( iν ) | (cid:18) k + vν + ( k + v ) (cid:19) = 2 − d π − ( d +1)2 Γ (cid:18) − d (cid:19) ∞ (cid:88) k = −∞ ( k + v )Γ (cid:0) d − | k + v | (cid:1) Γ (cid:0) − d + | k + v | (cid:1) = ( d − v − d − C ¯ΦΦ1 , (2.54)which again agrees with the flat space result in (2.19).Similarly, for the stress tensor, the general form (2.20) tells us that T θθ should have aconstant one-point function on the hyperbolic cylinder. There is a simpler way to calculateit on the hyperbolic cylinder [17, 37]. We start by keeping the length of S to be a variable β instead of fixing it to 2 π . This is equivalent to rescaling the metric component g θθ by( β/ π ) . So if we compute the free energy for arbitrary β and then take a derivative withrespect to β , this is the same as inserting T θθ in the path integral T µν = 2 √ g δSδg µν = ⇒ (cid:104) T θθ (cid:105) S × H d − = − H d − ) ∂F twisted ( ϑ, β ) ∂β (cid:12)(cid:12)(cid:12)(cid:12) β =2 π . (2.55)In practice, we can calculate the free energy for a general β by rescaling n by 2 π/βF twisted ( ϑ, β ) = Vol( H d − )(4 π ) d − Γ( d − ) (cid:90) ∞−∞ dν | Γ( iν + d − ) | | Γ( iν ) | (cid:88) n ∈ Z + v log (cid:18) ν + 4 π n β (cid:19) . (2.56)17e can use this to calculate the stress-tensor one-point function (cid:104) T θθ (cid:105) S × H d − = ∞ (cid:88) k = −∞ π (4 π ) d − Γ (cid:0) d − (cid:1) (cid:90) ∞−∞ dν | Γ( iν + d − ) | | Γ( iν ) | ( k + v ) ν + ( k + v ) = Γ (cid:0) − d (cid:1) π (4 π ) d − ∞ (cid:88) k = −∞ Γ (cid:0) d − | k + v | (cid:1) ( k + v ) Γ (cid:0) − d + | k + v | (cid:1) = Γ (cid:0) − d (cid:1) (1 − v ) v (cid:0) csc π (cid:0) d − v (cid:1) − csc π (cid:0) d + v (cid:1)(cid:1) d (4 π ) d − Γ (cid:0) − d − v (cid:1) Γ (cid:0) − d + v (cid:1) (2.57)which is consistent with what we got above by a direct calculation in flat space in (2.23).This hyperbolic space technique will be useful below, when we try to calculate the conformalweight in the interacting theory. S d Another useful way to study a codimension two defect is to map the problem to a d -dimensional sphere S d which is related by a Weyl transformation to flat space. Indeedstarting from the hyperbolic cylinder described by (2.36), we perform a coordinate transfor-mation sinh η = tan τ to get ds S × H d − = 1cos τ (cos τ dθ + dτ + sin τ ds S d − ) = 1cos τ ds S d . (2.58)The sphere is spanned by 0 ≤ θ < π , 0 ≤ τ < π/ S d − coordinates. Thedefect is located at τ = π/
2, and the monodromy action along θ is the same as in thehyperbolic cylinder, Φ( τ, θ + 2 π ) = e iϑ Φ( τ, θ ), where we suppressed the S d − coordinates.These coordinates are related to the usual S d coordinates by the transformation sin τ =sin θ sin θ , cos τ sin θ = cos θ which gives us the usual S d metric ds S d = dθ + sin θ ( dθ + sin θ ds S d − ) (2.59)where 0 ≤ θ , < π .The twisted free energy of a complex scalar on d − dimensional unit sphere is F twisted ( ϑ ) = tr log (cid:18) −∇ + d ( d − (cid:19) (2.60)where the trace is taken over the eigenfunctions obeying twisted boundary condition. We18alculate their eigenvalues and degeneracy in appendix A. Using the result in (A.6), thetwisted free energy in d = 3 is given by F twisted ( ϑ ) = (cid:32) (cid:88) n = k + v + (cid:88) n = k +1 − v (cid:33) d n log (cid:18) n ( n + d −
1) + d ( d − (cid:19) = ∞ (cid:88) k =0 ( k + 1)( k + 2)2 log (cid:18) ( k + v ) ( k + v + 2) + 34 (cid:19) + v → − v = ∞ (cid:88) k =1 k (cid:20) log (cid:18) k + v − (cid:19) (cid:18) k − v + 12 (cid:19)(cid:21) . (2.61)The sum can be regulated using zeta-function regularization F twisted ( ϑ ) = − dds (cid:34) ∞ (cid:88) k =1 k (cid:0) k + v − (cid:1) s + ∞ (cid:88) k =1 k (cid:0) k − v + (cid:1) s (cid:35) = − ζ (cid:48) (cid:18) − , v + 12 (cid:19) + 2 (cid:18) v − (cid:19) ζ (cid:48) (cid:18) − , v + 12 (cid:19) − (cid:18) v − (cid:19) ζ (cid:48) (cid:18) , v + 12 (cid:19) + v → − v. (2.62)Using identities for derivatives of Hurwitz zeta function (see for e.g. [18]), we may write theresult as F twisted ( ϑ ) = ∞ (cid:88) n =1 ( − n (cid:18) cos nϑ π n + ( ϑ − π )2 π sin nϑn − ( ϑ − π ) π cos nϑn (cid:19) = ( π − ϑ ) log (2 cos ϑ/ − i ( π − ϑ ) (cid:0) Li (cid:0) − e − iϑ (cid:1) − Li (cid:0) − e iϑ (cid:1)(cid:1) + Li (cid:0) − e − iϑ (cid:1) + Li (cid:0) − e iϑ (cid:1) π (2.63)where Li s ( z ) is the Polylog. This gives us the expectation value of the defect in 3 dimensions − log (cid:104)D(cid:105) d =3 = ( π − ϑ ) log (2 cos ϑ/ − i ( π − ϑ ) (cid:0) Li (cid:0) − e − iϑ (cid:1) − Li (cid:0) − e iϑ (cid:1)(cid:1) π + Li (cid:0) − e − iϑ (cid:1) + Li (cid:0) − e iϑ (cid:1) π − log 24 + 3 ζ (3)8 π (2.64)where we used the result for the untwisted free energy of a complex scalar in three dimensionsfrom [30]. It can be numerically checked to agree with the result obtained from hyperbolic19pace calculation in (2.45). For general d , we get F twisted ( ϑ ) = ∞ (cid:88) k =0 Γ( k + d )Γ( k + 1)Γ( d ) log (cid:18) ( k + v ) ( k + v + d −
1) + d ( d − (cid:19) + ( v → − v )= ∞ (cid:88) k =0 Γ( k + d )Γ( k + 1)Γ( d ) log (cid:18) k + v + d (cid:19) (cid:18) k + v + d − (cid:19) + ( v → − v )= ∞ (cid:88) k =0 Γ( k + d )Γ( k + 1)Γ( d ) log Γ ( k + v + d − δ )Γ ( k + v + δ ) + ( v → − v ) , δ = d − δ . To do the sum, it is more convenient to treat δ as a new variable,and perform the sum after taking a derivative with respect to δ [30, 36] ∂F twisted ( ϑ, δ ) ∂δ = − ∞ (cid:88) k =0 Γ( k + d )Γ( k + 1)Γ( d ) ( ψ ( k + v + d − δ ) + ψ ( k + v + δ )) + ( v → − v )= Γ( − d ) (cid:18) Γ ( v + d − δ )Γ ( v − δ ) + Γ ( v + δ )Γ ( v + δ − d ) (cid:19) + ( v → − v ) (2.66)which implies that F twisted ( ϑ ) = Γ( − d ) (cid:90) d/ − d/ dδ (cid:18) Γ ( v + d − δ )Γ ( v − δ ) + Γ ( v + δ )Γ ( v + δ − d ) + ( v → − v ) (cid:19) = − Γ( − d ) (cid:90) dw (cid:32) Γ (cid:0) v + d + w (cid:1) Γ (cid:0) v − d + w (cid:1) + Γ (cid:0) v + d − w (cid:1) Γ (cid:0) v − d − w (cid:1) + ( v → − v ) (cid:33) . (2.67)We can use this to calculate the defect expectation value − log (cid:104)D(cid:105) = F twisted − F untwisted = − (cid:90) dw (cid:20) Γ( − d ) (cid:32) Γ (cid:0) v + d + w (cid:1) Γ (cid:0) v − d + w (cid:1) + Γ (cid:0) v + d − w (cid:1) Γ (cid:0) v − d − w (cid:1) + ( v → − v ) (cid:33) − (cid:0) d (cid:1) Γ (cid:0) − d (cid:1) π Γ(1 + d ) w sin( πw )Γ (cid:18) d w (cid:19) Γ (cid:18) d − w (cid:19) (cid:21) (2.68)In d = 3, it agrees with the result in (2.64). It can also be checked numerically, to agreewith the result obtained on hyperbolic cylinder in (2.45).Similar to what we did on the hyperbolic cylinder, we can also calculate the one-point20unction of ¯ΦΦ on the sphere by introducing a mass term and calculating the derivative ofthe free energy with mass ∂F twisted ( ϑ ) ∂m = ∞ (cid:88) k =0 Γ( k + d )Γ( k + 1)Γ( d ) (cid:34) (cid:0) k + v + d − (cid:1) + m − + ( v → − v ) (cid:35) = Γ (cid:16) d +1 − v + (cid:113) − m (cid:17) Γ (cid:16) d − v − (cid:113) − m (cid:17) sin π (cid:16)(cid:113) − m + − v (cid:17) d ) sin (cid:0) πd (cid:1) (cid:113) − m + ( v → − v ) . (2.69)The Weyl factor involved in going from hyperbolic cylinder to S d is cos τ , so the one-pointfunction on S d is C ¯ΦΦ1 / (cos τ ) d − . Hence, ∂F∂m (cid:12)(cid:12)(cid:12)(cid:12) m =0 = C ¯ΦΦ1 π Vol( S d − ) (cid:90) π/ dτ (sin τ ) d − (cos τ ) − d +3 = ⇒ C ¯ΦΦ1 = Γ (cid:0) d + 1 − v (cid:1) Γ (cid:0) d − v (cid:1) sin ( πv ) Γ (cid:0) d (cid:1) π d +1 Γ( d )(2 − d ) + ( v → − v )= ( d − (cid:0) d − v (cid:1) Γ (cid:0) d − v (cid:1) sin ( πv ) Γ (cid:0) d (cid:1) π d +1 Γ( d )(2 − d ) (2.70)in agreement with the flat space (2.16) and hyperbolic space results (2.53). H d − and a defectRG flow As we discussed in the previous section in (2.7) and (2.40), dimensions of the defect operatorsare related to their spin by the relation∆ s (∆ s − d + 2) = s − ( d − / ⇒ ∆ ± s = d − ± | s | . (3.1)So far, we considered the + sign above for all values of s ∈ Z + v . But the − sign may alsobe consistent with the defect unitarity bound (2.8) as long as | s | < s = v or − v . In the free theory, the two cases are allowed by unitarity in the following In most of this paper, we use + boundary condition for all the modes, so we avoid using the superscript+. We use superscript − whenever we impose a − boundary condition. v ∆ − v allowed : v < d − , ∆ −− v allowed : v > − d . (3.2)Note that there is some range of dimensions between 3 and 4 where both s = v and − v modes are allowed to have a ∆ − boundary condition. In d = 3 , s = v mode is allowedto have a ∆ − boundary condition only for v < /
2, while s = − v mode is allowed tohave a ∆ − boundary condition only for v > /
2. In the hyperbolic space setup, we canthink of the two possible values of ∆ as the two possible boundary conditions for the bulkmassive scalar after we have performed KK reduction on S . In the defect theory, it definesa defect RG flow where we start with a ∆ − v operator in the UV and flow with a relevantdeformation ¯Ψ v Ψ v to the IR where the original operator is replaced by an operator withdimension ∆ + v . This is similar to the study of double-trace flows in AdS/CFT [38]. We candefine a similar flow with ¯Ψ − v Ψ − v operator under which the dimension of s = − v operator changes from ∆ −− v to ∆ + − v . Both of these flows preserve the reality of the actionand the U (1) ∼ SO (2) internal symmetry of the original action. As we mentioned in theprevious section, at v = 1 /
2, the theory has an enhanced O (2) symmetry, and to preservethe full O (2) symmetry, we have to add a sum of these two deformations. We now study howsome of the defect CFT observables change under such a flow. We only do the calculationfor the ¯Ψ v Ψ v flow so only s = v operator has a ∆ − boundary condition. The generalizationto the case when either just s = − v or both s = v and s = − v operators have a∆ − boundary condition is straightforward. Another interesting scenario is when the defectspectrum contains both ∆ + and ∆ − operators for the same s . This defines a more non-trivialdefect CFT with extra interactions localized on the defect. Some remarks on this possibilitywere made in [10]. We will not study this scenario here.First, let us calculate how the defect expectation value (2.45) changes under this flow.The change is equal to the change in twisted free energy. As we said above, this defect RGflow is analogous to a double trace flow in the d − F d − − ∆ − F ∆ = − (cid:16) π ( d − (cid:17) Γ( d − (cid:90) ∆+1 − d du u sin πu Γ (cid:18) d − u (cid:19) Γ (cid:18) d − − u (cid:19) (3.3)where the operator driving the flow has dimension ∆ in the UV and d − − ∆ in the IR.22n our case, we consider the flow by operator ¯Ψ v Ψ v , so the change in the defect expectationvalue is given bylog (cid:104)D(cid:105) − − log (cid:104)D(cid:105) + = 1sin (cid:16) π ( d − (cid:17) Γ( d − (cid:90) v du u sin πu Γ (cid:18) d − u (cid:19) Γ (cid:18) d − − u (cid:19) . (3.4)Here and elsewhere the superscript − corresponds to the case when s = v operator has a∆ − boundary condition while all the other operators have a ∆ + boundary condition. Aswe mentioned in the introduction, the quantity ˜ D defined in (1.4) should decrease under adefect RG flow. So we expect that˜ D − − ˜ D + = 1Γ( d − (cid:90) v du u sin πu Γ (cid:18) d − u (cid:19) Γ (cid:18) d − − u (cid:19) (3.5)should always be positive. It can be checked numerically for various values of v and d thatthis is always positive whenever the flow is allowed by the unitarity (3.2). But this is nottrue when the flow is non-unitary, so for instance, in d = 3, (3.5) is not positive for v > / s = v pieceand is given by G − ¯ΦΦ − G +¯ΦΦ = e iv ( θ − θ ) π d/
2Γ (∆ − v ) F (cid:16) ∆ − v , ∆ − v + − d ; 2∆ − v + 3 − d ; − ξ (cid:17) ∆ − v Γ(∆ − v + 2 − d ) ξ ∆ − v − ∆ − v → ∆ + v = e iv ( θ − θ ) − d Γ (cid:0) − d (cid:1) (cid:0) csc (cid:0) π ( d + 2 v ) (cid:1) − csc (cid:0) π ( d − v ) (cid:1)(cid:1) π d − Γ (cid:0) − d − v + 2 (cid:1) Γ (cid:0) − d + v + 2 (cid:1) (1 + O ( ξ )) (3.6)where we expanded the result in small ξ . The constant ξ independent piece tells us thechange in the one-point function coefficient of ¯ΦΦ( C ¯ΦΦ1 ) − − ( C ¯ΦΦ1 ) + = Γ (cid:0) − d (cid:1) (cid:0) csc (cid:0) π ( d + 2 v ) (cid:1) − csc (cid:0) π ( d − v ) (cid:1)(cid:1) d − π d − Γ (cid:0) − d − v + 2 (cid:1) Γ (cid:0) − d + v + 2 (cid:1) = ⇒ ( C ¯ΦΦ1 ) − = Γ (cid:0) d − v − (cid:1) Γ (cid:0) d + v (cid:1) d − π d − ( d − (cid:0) d − (cid:1) Γ( v )Γ(1 − v ) (3.7)where we used the result from (2.16).Using the change in the bulk two-point function, it is possible to also calculate the bulk23ne-point function of other operators. Let’s just mention the result for the stress tensor. Itsone-point function is specified by the conformal weight (2.20) and the change in conformalweight under the flow is as follows: h − − h + = − v Γ (cid:0) − d (cid:1) (cid:0) csc (cid:0) π ( d + 2 v ) (cid:1) − csc (cid:0) π ( d − v ) (cid:1)(cid:1) d − π d − Γ (cid:0) − d − v + 2 (cid:1) Γ (cid:0) − d + v + 2 (cid:1) h − = − v ( v + 1)Γ (cid:0) − d (cid:1) (cid:0) csc (cid:0) π ( d − v ) (cid:1) − csc (cid:0) π (cid:0) d + v (cid:1)(cid:1)(cid:1) d − π d − d Γ (cid:0) − d − v + 1 (cid:1) Γ (cid:0) − d + v + 2 (cid:1) . (3.8)As long as the flow is unitary, the conformal weight h − can be checked to be positive,consistent with the conjecture made in [4].Finally, let’s discuss the displacement operator in the theory with ∆ − v boundary condition.Based on the fact that it must have dimension d − v Ψ v . To more directly calculate it similar to what we did in for the ∆ + case in section2, note that the bulk defect OPE containsΦ (cid:51) z − v Ψ v + ¯ z v Ψ v . (3.9)Then, using the definition of stress -tensor in the free theory, it is easy to see that T zz = 4 T ¯ z ¯ z (cid:51) − v (1 + v )¯ z ¯Ψ v Ψ v ( (cid:126)y ) = ⇒ ∂ z T zz = − πv (1 + v ) ¯Ψ v Ψ v ( (cid:126)y ) δ ( z, ¯ z )= ⇒ D ( (cid:126)y ) = − πv (1 + v ) ¯Ψ v Ψ v ( (cid:126)y ) . (3.10)This definition gives the following two-point function coefficient C − D = 4 v (1 + v )Γ (cid:0) d − − v (cid:1) Γ (cid:0) d + v (cid:1) π d − Γ(1 − v )Γ( v ) . (3.11)We can also check that these coefficients satisfy the ward identity (2.32) by noting that thedisplacement contribution to the two-point function of ¯ΦΦ now takes the form (cid:104) ¯ΦΦ( x ) ¯ΦΦ( x ) (cid:105) (cid:51) e i ( θ − θ ) Γ (∆ − v ) Γ (∆ v )4 − v +∆ v π d Γ (cid:0) ∆ − v + 2 − d (cid:1) Γ (cid:0) ∆ + 2 − d (cid:1) ξ d − . (3.12)This gives a relation between bulk defect OPE coefficient for ¯ΦΦ with displacement and thedisplacement two-point function coefficient( C ¯ΦΦ − D ) C − D = Γ (cid:0) d − − v (cid:1) Γ (cid:0) d + v (cid:1) π d Γ(1 − v )Γ(2 + v ) . (3.13)24he coefficients (3.7), (3.11) and (3.13) satisfy the Ward identity (2.32). N In this section, we study the monodromy defect in the interacting O (2 N ) model (it will soonbe clear why we choose 2 N instead of N here) at large N . The S × H d − setup provides aconvenient way to study the problem. The action may be written as S = (cid:90) d d x (cid:112) g ( x ) (cid:18) g µν ∂ µ φ A ∂ ν φ A − ( d − φ A φ A + λ φ A φ A ) (cid:19) (4.1)where A now goes from 1 to 2 N . We again consider monodromy defect defined as in (2.1).We want to do a large N analysis, and to accomplish that, we want to preserve a largesymmetry group. The simplest such case is when we fix the matrix G AB in (2.2) to consistof N identical 2 × ϑ i = ϑ . This is the only case we consider in thispaper. Then, as before, it is convenient to package these 2 N real scalars into N complexscalars as Φ I = φ I − + iφ I , where I goes from 1 to N and all N complex scalars havethe same monodromy as in (2.3). The original theory has O (2 N ) symmetry, and the defectbreaks it down to U ( N ). However for ϑ = 0 and π which correspond to a trivial defect and Z monodromy defect respectively, the symmetry is enhanced and the defect preserves full O (2 N ) symmetry. The action in terms of complex variables is S = (cid:90) d d x (cid:112) g ( x ) (cid:18) g µν ∂ µ ¯Φ I ∂ ν Φ I − ( d − I Φ I + λ I Φ I ) (cid:19) . (4.2)At large N , we can use the well-known Hubbard-Stratonovich transformation to write thisin terms of auxiliary field σ ( x ) S = (cid:90) d d x (cid:112) g ( x ) (cid:18) g µν ∂ µ ¯Φ I ∂ ν Φ I − ( d − I Φ I + 12 σ ¯Φ I Φ I (cid:19) . (4.3)We dropped a σ / λ term above, which can be consistently done in the critical limit (seefor example [39], for a review). We can then integrate out the fields Φ I since the action isquadratic in Φ I to get Z = exp[ − F twisted ] = (cid:90) [ dσ ] exp (cid:20) − N tr log (cid:18) −∇ + σ − ( d − (cid:19)(cid:21) (4.4) We assume that the mass terms have been tuned away so that the bulk is always critical.
25t large N , we can use a saddle point approximation to do the integral over σ and look fora saddle with a constant value for the field σ ( x ). This constant is the one-point function of σ ( x ) which is a constant on the hyperbolic cylinder . So at leading order at large N , thefield σ ( x ) only contributes through its one-point function, and acts as a mass term for Φ I .Similar to the case of free theory (2.44), the free energy in the interacting theory at leadingorder at large N may then be written as F twisted ( ϑ ) = N Vol( H d − )(4 π ) d − Γ( d − ) (cid:90) ∞−∞ dν | Γ( iν + d − ) | | Γ( iν ) | (cid:88) n ∈ Z + v log (cid:0) ν + n + σ (cid:1) . (4.5)The value of σ at the large N fixed point, σ ∗ can be obtained by solving the saddle pointequation which says that the following derivative should vanish ∂F twisted ( ϑ ) ∂σ (cid:12)(cid:12)(cid:12)(cid:12) σ = σ ∗ = N Vol( H d − )(4 π ) d − Γ( d − ) (cid:90) ∞−∞ dν | Γ( iν + d − ) | | Γ( iν ) | (cid:88) n ∈ Z + v ν + n + σ ∗ = N Vol( H d − )Γ (cid:0) − d (cid:1) (4 π ) d − ∞ (cid:88) k = −∞ Γ (cid:18) d − (cid:113) σ ∗ + ( k + v ) (cid:19) Γ (cid:18) − d + (cid:113) σ ∗ + ( k + v ) (cid:19) = N Vol( H d − )Γ (cid:0) − d (cid:1) (4 π ) d − (cid:88) s ∈ Z + v Γ (∆ s )Γ (3 − d + ∆ s ) (4.6)where the integral over ν is similar to the one in (2.52) and can be performed with similartechniques. In the last line, we used the usual AdS/CFT dictionary to write the expressionin terms of the dimensions of defect operators∆ s (∆ s − d + 2) = s + σ ∗ − ( d − ⇒ ∆ ± s = d − ± √ σ ∗ + s . (4.7)Note that we used ∆ + s solution to write the above expression of ∂F/∂σ in (4.6). This isbecause the spectral representation of the free energy is only valid for ∆ s > d/ −
1. However,written in terms of ∆ s , the expression in (4.6) can be analytically continued and also usedfor the case when we impose ∆ − boundary condition on one or more of the operators. We In the flat space, this one-point function is (cid:104) σ ( x ) (cid:105) = σ ∗ r with σ ∗ being the constant one-point functionon the hyperbolic cylinder σ ∗ ≥ max (cid:0) − v , − (1 − v ) (cid:1) . (4.8)Another equivalent way to derive this large N saddle point equation is to look at thetwo-point function of Φ in the bulk OPE limit. As we discussed above (2.41), the two-pointfunction on the hyperbolic cylinder is given by the sum over bulk-bulk propagators (cid:104) Φ I ( x ) ¯Φ J ( x ) (cid:105) = (cid:88) s ∈ Z + v δ I J
Γ (∆ s ) e isθ F (cid:16) ∆ s , ∆ s + − d ; 2∆ s + 3 − d ; − ξ (cid:17) π d/ Γ(∆ s + 2 − d )(4 ξ ) ∆ s (4.9)and here ∆ s is given by (4.7). In the bulk OPE limit, the two point function behaves as (cid:104) Φ I ( x ) ¯Φ J ( x ) (cid:105) = δ I J π (4 π ) d − (cid:88) s ∈ Z + v (cid:34) Γ (cid:0) d − (cid:1) ξ d − (1 + O ( ξ )) + Γ (cid:0) − d (cid:1) Γ (∆ s )Γ (3 − d + ∆ s ) (1 + O ( ξ )) (cid:35) . (4.10)The constant ξ independent piece in the second term represents the presence of operatorΦ I ¯Φ I of dimension d − N critical U ( N ) model,this operator is replaced by the operator σ of dimension 2. This should still be true in thepresence of the defect, and demanding that this term vanishes is equivalent to the saddlepoint equation written in (4.6).When we impose ∆ + boundary condition on all the operators, σ ∗ can be determined bysolving the following equation from (4.6)Vol( H d − )Γ (cid:0) − d (cid:1) (4 π ) d − ∞ (cid:88) k = −∞ Γ (cid:18) d − (cid:113) σ ∗ + ( k + v ) (cid:19) Γ (cid:18) − d + (cid:113) σ ∗ + ( k + v ) (cid:19) = 0 . (4.11)It is hard to perform this sum analytically as a function of d . To proceed, we separate outthe sum into a divergent piece at large k and a finite piece. The divergent piece of the sumcan be performed by dimensional regularization and analytically continued in d . For thefinite piece, it is harder to perform the sum as a function of d . However in d = 4 − (cid:15) , wecan first do a series expansion in (cid:15) and then it can be performed up to first two orders in (cid:15) .Hence, the saddle point equation can be solved order by order in (cid:15) and it gives σ ∗ = v ( v − (cid:15) + 3 v ( v −
1) + 12 (cid:15) + O ( (cid:15) ) . (4.12)27otice that the order (cid:15) term does not vanish at v = 0 and 1. But the defect becomes trivialat v = 0 and 1, so all one-point functions, including the one-point function of σ should vanishat these values of v . This problem arises because we are doing an expansion in (cid:15) , and weexpect this problem to be resolved by higher order terms in (cid:15) . Indeed when we calculate σ ∗ numerically in d = 3 . v = 0 and 1. One possibilityis that the higher order terms in (cid:15) are singular at v = 0 and 1. For example, consider the (cid:15) expansion of the following simple function (cid:15)v √ v + (cid:15) = (cid:15)v − (cid:15) (cid:15) v + O ( (cid:15) ) . (4.13)The function vanishes at v = 0 for any fixed (cid:15) , but when we expand at small (cid:15) , the limit v → σ ∗ into (4.7), we get the dimension of defect operators at large N andleading orders in (cid:15) ∆ + s = 1 − (cid:15) | s | + v ( v − | s | (cid:15) + 3 v ( v −
1) + 14 | s | (cid:15) − v ( v − | s | (cid:15) + O ( (cid:15) ) . (4.14)This is consistent with what we find in the (cid:15) expansion in (5.15). We can also calculate thetwisted free energy at the large N fixed point F twisted = F twisted (cid:12)(cid:12)(cid:12)(cid:12) σ =0 + (cid:90) σ ∗ dσ ∂F twisted ∂σ = F twisted (cid:12)(cid:12)(cid:12)(cid:12) σ =0 + N Vol( H )4 π (cid:90) σ ∗ dσ (cid:16) − σ(cid:15) + v ( v − (cid:17) = N F freetwisted + N Vol( H ) v ( v − π (cid:15) (4.15)where we used the fact that F twisted ( σ = 0) is the same as twisted free energy in the freetheory. This is also consistent with the (cid:15) expansion result we obtain below in (5.3). Thecomparison of the large N and the epsilon expansion calculations of free energy and defectdimensions only involves the O ( (cid:15) ) piece in σ ∗ . In order to check (cid:15) piece, we look at thebulk OPE coefficient times the one-point function coefficient C ¯Φ;Φ σ σ ∗ . Since the operator σ replaces ¯ΦΦ in the large N theory, this should match C ¯Φ;Φ ¯ΦΦ C ¯ΦΦ1 calculated in (cid:15) expansion.28 .2 0.4 0.6 0.8 1.0 v - - - - - σ * Numerical Result ϵ expansion result with ϵ = Figure 2: Saddle point value of σ ∗ in d = 3 . + boundary conditionat leading order in large N . Dashed line represents smooth interpolation of the numericalresult. For comparison, we also plot the analytic result in d = 4 − (cid:15) at (cid:15) = 0 . O (2 N ) model is [40], C ¯Φ;Φ σ = (cid:115) N C σ Γ( d − − d )Γ ( d − C Φ (4.16)where C Φ and C σ are the normalization of the bulk two-point functions, and in our conven-tions, they are given by (see for e.g [39]) C σ = 2 d +1 Γ( d − ) sin( πd ) N π Γ( d − , C Φ = Γ (cid:0) d − (cid:1) π d/ . (4.17)Combining it with the result for σ ∗ in (4.12), this gives the following result C ¯Φ;Φ σ σ ∗ = ( v − v π + ( v − v ( γ + 3 + log π ) + 18 π (cid:15) + O ( (cid:15) ) (4.18)which is consistent with the (cid:15) expansion result in (5.28).Away from d = 4, the finite piece of the sum in (4.11) can be performed numerically, fora given d, σ ∗ and v . We start with d = 3 .
9, so that we can compare it with the prediction in d = 4 − (cid:15) . We evaluate the sum for a range of values of σ and v , and then find the root ofthe equation on the real σ axis for different values of v . We then interpolate in v and plotthe value of σ ∗ in figure 2. We also compare the result with the result in d = 4 − (cid:15) in (4.12)at (cid:15) = 0 . d = 3 to solve the saddle point equation in (4.11)numerically. We plot the solution in figure 3. Once we know σ ∗ , we can calculate thedimensions of defect operators using (4.7). We also plot the dimensions of three low-lying29 .0 0.2 0.4 0.6 0.8 1.0 - - - v σ * v Δ s Δ v Δ v - Δ v + Figure 3: Saddle point value of σ ∗ and the dimensions of defect operators in three dimensionalcritical theory at large N . The solid lines are smooth interpolation of the numerical results.Note that the defect is one-dimensional, therefore the unitarity bound just requires the defectdimensions to be positive.defect operators in figure 3. For v = 1 /
2, corresponding to Z monodromy on all scalars, weget σ ∗ = − . , ∆ / = 0 . , ∆ / = 1 .
943 (4.19)to leading order at large N . We are doing a large N analysis, but it is interesting to comparethe result with the Monte Carlo results for monodromy defect in d = 3 Ising model in [12].They found ∆ / = 0 .
918 and ∆ / = 1 . σ ∗ , we can also calculate the expectation value of the defect with spherical geom-etry. It is defined in the same way as the free theory (2.45), as the negative of the differencebetween the twisted and the untwisted free energy. However, recall that in the untwistedtheory, the one-point functions vanish, so σ ∗ = 0. This implies that the corrections due tothe interactions for the untwisted free energy start at order 1 /N . So at leading order, wecan just use the untwisted free energy of the free theory and the interacting corrections tothe defect expectation value are due to the corrections in the twisted free energy − log (cid:104)D(cid:105) = − N log (cid:104)D(cid:105) free + (cid:90) σ ∗ dσ ∂F twisted ∂σ . (4.20)The first term above is the free theory result we have from (2.45). We can numericallyintegrate (4.6) using the numerical results for σ ∗ to evaluate the second term. We plot theresult in figure 4 for the case of d = 3, corresponding to a circular defect. We used thestandard regularized volume of H , Vol( H ) = − π , that can be obtained (2.50). We alsoplot the quantity ˜ D defined in (1.4) as a function of d between 3 < d < v in figure 5. It shows that this quantity is a smooth function of d in this range.30 .2 0.4 0.6 0.8 v 〈〉 N Figure 4: Defect expectation value in d = 3 critical theory at leading order in large N . Solidline interpolates the numerical results. d ˜ N v = = = Figure 5: A plot of ˜ D in the large N critical theory between 3 < d <
4. Solid and dashedlines interpolate numerical results. 31 .1 Alternate boundary condition on H d − When we impose ∆ − boundary condition for s = v operator, we instead get the followingsaddle point equation from (4.6)Vol( H d − )Γ (cid:0) − d (cid:1) (4 π ) d − Γ (cid:0) d − − √ σ ∗ + v (cid:1) Γ (cid:0) − d − √ σ ∗ + v (cid:1) + (cid:88) k (cid:54) =0 Γ (cid:18) d − (cid:113) σ ∗ + ( k + v ) (cid:19) Γ (cid:18) − d + (cid:113) σ ∗ + ( k + v ) (cid:19) = 0 . (4.21)Note that since we are working in 3 < d < − v > ⇒ σ ∗ ≤ (cid:18) d − (cid:19) − v . (4.22)So we have to look for a solution to the saddle point equation (4.21) satisfying both (4.8)and (4.22).In d = 4 − (cid:15) the saddle point equation in (4.21) can be solved perturbatively in (cid:15) similarto what we did in the ∆ + case. The result is σ ∗ = v ( v + 1) (cid:15) + 3 v ( v + 1) + 12 (cid:15) + O ( (cid:15) ) . (4.23)As in the ∆ + case, the order (cid:15) term does not vanish at v = 0 and 1, and the explanationgiven below (4.12) still applies. This σ ∗ gives the dimensions of the defect operators at large N ∆ − s = 1 − (cid:15) | s | + v ( v + 1)2 | s | (cid:15) + O ( (cid:15) ) , s (cid:54) = v ∆ − v = 1 − (cid:15) − v − ( v + 1)2 (cid:15) + O ( (cid:15) ) . (4.24)These results are consistent with the results in (5.17) and (5.19) found using (cid:15) expansion.Note that the notation is such that − sign in the superscript above means that we areworking in a theory where s = v mode has ∆ − boundary condition, but all the other modesstill have ∆ + boundary condition. As in ∆ + case in (4.18), to check the order (cid:15) term in σ ∗ ,we look at the product of bulk OPE coefficient and the one-point function C ¯Φ;Φ σ σ ∗ = v ( v + 1)4 π + v ( v + 1)( γ + 3 + log π ) + 18 π (cid:15) + O ( (cid:15) ) . (4.25)32 .1 0.2 0.3 0.4 0.5 0.6 0.7 v σ * Numerical Result ϵ expansion result with ϵ = Figure 6: Saddle point value of σ ∗ in d = 3 . − boundary conditionand for comparison, the analytic result in d = 4 − (cid:15) at (cid:15) = 0 . (cid:15) expansion result (5.33).Far from d = 4, we have to perform the sum numerically as we did before. In d = 3 .
9, itis possible to numerically find a saddle point subject to constraints (4.8) and (4.22) between0 < v < .
71. We plot the result in figure 6. The match with the analytic exression in d = 4 − (cid:15) is not as good as ∆ + boundary condition, but this may just indicate that O ( (cid:15) )terms are important.Using the value of σ ∗ , we can also calculate the expectation value of the defect (4.20). Aquantity of interest is the difference in the logarithm of defect expectation value between the∆ − and ∆ + boundary condition. It should be positive according to the conjectured defectF-theorem (1.4) between 3 < d < . We plot it in figure 7 in d = 3 . v where the ∆ − saddle exists.As we go down to d = 3, there is no solution to the saddle point equation (4.21) con-sistent with the constraints (4.8) and (4.22). So an interacting unitary fixed point in threedimensions does not exist for ∆ − boundary condition. We can also calculate the conformal weight of the defect in the interacting theory. It isrelated to T θθ (2.24), which in the hyperbolic cylinder approach is given by (2.55). So weneed to know the dependence of the twisted free energy on β . Similar to the case of free The sin factor in front in the definition of ˜ D in (1.4) is positive between 3 < d <
4, so the difference of˜ D and log (cid:104)D(cid:105) should both be positive. .1 0.2 0.3 0.4 0.5 0.6 0.7 v 〈〉 - - Log 〈〉 + N Figure 7: The difference in the defect expectation value between the theories with ∆ − and∆ + boundary condition in d = 3 . S to β , the expression for the free energy changes to F twisted ( ϑ, β ) = N Vol( H d − )(4 π ) d − Γ( d − ) (cid:90) ∞−∞ dν | Γ( iν + d − ) | | Γ( iν ) | (cid:88) n ∈ Z + v log (cid:18) ν + 4 π n β + σ (cid:19) . (4.26)If we impose ∆ + boundary condition on all the operators, then the large N saddle pointequation is ∂F twisted ∂σ (cid:12)(cid:12)(cid:12)(cid:12) σ = σ ∗ = N Vol( H d − )Γ (cid:0) − d (cid:1) (4 π ) d − ∞ (cid:88) k = −∞ Γ (cid:16) d − (cid:113) π β ( k + v ) + σ (cid:17) Γ (cid:16) − d + (cid:113) π β ( k + v ) + σ (cid:17) = 0 (4.27)Proceeding in the same way as β = 2 π , we first expand the sum in general d in large k andisolate the pieces that diverge as k → ∞ . The divergent piece of the sum can be performedin dimensional regularization and analytically continued in d . And for the finite piece, wehave to either expand in (cid:15) or turn to numerical methods. In d = 4 − (cid:15) , the solution to thesaddle point equation to leading order in (cid:15) is σ ∗ = (cid:18) π (6( v − v + 1)3 β − (cid:19) (cid:15) + O ( (cid:15) ) (4.28)34orrection to the twisted free energy, to leading order in (cid:15) and N is F twisted ( ϑ, β ) = F twisted ( ϑ, β ) (cid:12)(cid:12)(cid:12)(cid:12) σ =0 + N (cid:90) σ ∗ dσ ∂F twisted ∂σ = F twisted ( ϑ, β ) (cid:12)(cid:12)(cid:12)(cid:12) σ =0 + Vol( H ) N ( β − π (6( v − v + 1)) π β (cid:15) + O ( (cid:15) ) (4.29)Using (2.55) and (2.24), we get the conformal weight in terms of the free energy h = − H d − ) 2 π ( d − ∂F twisted ∂β (cid:12)(cid:12)(cid:12)(cid:12) β =2 π . (4.30)This gives the conformal weight in the interacting theory to leading order in (cid:15)h = − N π Γ (cid:0) − d (cid:1) (1 − v ) v (cid:0) csc π (cid:0) d − v (cid:1) − csc π (cid:0) d + v (cid:1)(cid:1) d (4 π ) d − Γ (cid:0) − d − v (cid:1) Γ (cid:0) − d + v (cid:1) + N (1 − v ) v (9(1 − v ) v − π (cid:15) = N v (1 − v ) π + (cid:15)N (1 − v ) v π (cid:18) − H − v − − ψ ( v −
1) + 37 + 6 log( π ) − v (1 − v ) (cid:19) (4.31)where ψ is the Polygamma function and H n is the n th harmonic number. We used the freetheory result for conformal weight (2.24) in d = 4 − (cid:15) . This agrees with the result fromepsilon expansion calculation in the large N limit (5.8).Away from d = 4, we can still work numerically. For a given d , we now have threevariables in the sum (4.27), namely β, v and σ . We are interested in calculating a derivativewith β at β = 2 π . So we choose three values of β near 2 π as β = { π − . , π, π + 0 . } and then calculate the sum in (4.27) over a range of values of σ and v . We do an interpolationin σ and find the root for several values of v and all three values of β . So we have an analogueof figure 3 but for three different values of β . We then use this saddle point solution for σ ∗ to calculate the integral for free energy in (4.29). We finally calculate the conformal weightusing (4.30) where for the derivative, we use the numerical analogue h = − H d − ) 2 π ( d − F twisted (2 π + 0 . − F twisted (2 π − . . . (4.32)We plot the result in d = 3 in figure 8. It is positive in accordance with the conjecture madein [4]. 35 .2 0.4 0.6 0.8 1.0 v0.0050.0100.0150.020 hN Figure 8: Numerical result and a smooth interpolation for conformal weight in d = 3. d = 4 − (cid:15) In this section, we study monodromy defect in the critical O (2 N ) model described in (4.2)in a perturbation theory near 4 dimensions. In d = 4 − (cid:15) , there is a fixed point with thefixed point value of the coupling constant given by λ ∗ = 8 π N + 8 (cid:15). (5.1)We will compute defect CFT observables at this fixed point. To start with, let’s calculatethe twisted free energy on the hyperbolic space. To leading order in λ , it is given by F twisted = N F freetwisted + λ (cid:90) d d x √ g x (cid:104) ( ¯Φ I Φ I ( x )) (cid:105) = N F freetwisted + λN ( N + 1)4 2 π Vol( H )( C ¯ΦΦ1 ) (5.2)where F freetwisted is the free energy of a single free complex scalar in the presence of the twistdefect. Working to first order in λ , we only need C ¯ΦΦ1 in the free theory (2.16). Using thatin d = 4 and plugging in the fixed point value of the coupling gives F twisted = N F freetwisted + (cid:15)N ( N + 1)Vol( H ) (1 − v ) v π ( N + 4) . (5.3)This is consistent with the large N result (4.15). Using this, we can compute expectationvalue of the defect defined in (2.45), by subtracting the free energy of the untwisted energyfrom the above expression. Note that in the theory without the twist defect, leading cor-36ection to the free energy is of order λ , because the one-point functions vanish. So at thisorder − log (cid:104)D(cid:105) = F twisted − F untwisted = − N log (cid:104)D(cid:105) free + (cid:15)N ( N + 1)Vol( H ) (1 − v ) v π ( N + 4)= N Vol( H − (cid:15) ) (1 − v ) v π + Vol( H ) (cid:15)N ( N + 1) (1 − v ) v π ( N + 4) − Vol( H ) (cid:15)N π × (cid:90) ∞ dνν (cid:0) H − iν + ψ ( iν + 1) − − log π (cid:1) log (cid:18) csch ( πν )2 (cosh(2 πν ) − cos(2 πv )) (cid:19) (5.4)where we expanded the free theory result in (2.45) in d = 4 − (cid:15) .We can proceed in the same way to calculate the conformal weight of the defect, definedin (2.55). It can be calculated from the free energy on H d − × S as in (4.30) by first keepingthe length of S , β , to be arbitrary and then taking a derivative with respect to β at β = 2 π .Generalizing what we did above to arbitrary β gives F twisted = N F free twisted + λN ( N + 1)4 β Vol( H )( C ¯ΦΦ1 ( β )) (5.5)Since λ is order (cid:15) , we only need to do the calculation of one-point function of ¯ΦΦ as afunction of β in d = 4. As in (2.51), we can calculate this one-point function coefficient bytaking a mass derivative of the free energy C ¯ΦΦ1 ( β ) = 1 N β Vol( H d − ) ∂F twisted ∂m (cid:12)(cid:12)(cid:12)(cid:12) m =0 = 2Γ (cid:0) − d (cid:1) β (4 π ) d − ∞ (cid:88) k = −∞ Γ (cid:16) d − πβ | k + v | (cid:17) Γ (cid:16) − d + πβ | k + v | (cid:17) (5.6)To do the sum we first expand the sum in general d in large k and isolate the pieces thatdiverge as k → ∞ . We compute the sum for the divergent pieces keeping d arbitrary,and thenanalytically continue to d = 4. The finite piece starts contributing at order (cid:15) in d = 4 − (cid:15) ,so we do not need to add it here. The end result is C ¯ΦΦ1 ( β ) = 6( v − v + 16 β − π . (5.7)37e can then use (4.30), to find the the conformal weight in d = 4 − (cid:15)h = − N π Γ (cid:0) − d (cid:1) (1 − v ) v (cid:0) csc π (cid:0) d − v (cid:1) − csc π (cid:0) d + v (cid:1)(cid:1) d (4 π ) d − Γ (cid:0) − d − v (cid:1) Γ (cid:0) − d + v (cid:1) + N ( N + 1)(1 − v ) v (9(1 − v ) v − π ( N + 4) (cid:15) = N v (1 − v ) π + (cid:15)N (1 − v ) v π (cid:18) − H − v − − ψ ( v −
1) + 37 + 6 log( π ) − v (1 − v ) (cid:19) − (cid:15)N (1 − v ) v (9(1 − v ) v − π ( N + 4) . (5.8)This is also consistent with the large N expansion results (4.31). In this subsection, we calculate the two-point function of the defect fields Ψ Is induced by thebulk field Φ I on the defect. In the free theory, it is just given by the N field generalizationof (2.15) (cid:104) ¯Ψ I s ( (cid:126)y )Ψ Js ( (cid:126)y ) (cid:105) = δ I J δ s ,s C ∆ s ( (cid:126)y ) ∆ s , C ∆ s = Γ (∆ s )2 π d/ Γ (cid:0) ∆ s + 2 − d (cid:1) . (5.9)In the interacting theory, it gets corrected by the bulk one-loop Witten diagram (cid:104) ¯Ψ I s ( (cid:126)y )Ψ Js ( (cid:126)y ) (cid:105) = Ψ Js ( (cid:126)y )¯Ψ I s ( (cid:126)y ) x . (5.10)The integral involved is (cid:104) ¯Ψ I s ( (cid:126)y )Ψ Js ( (cid:126)y ) (cid:105) = − πλ ( N + 1) δ I J δ s ,s C ¯ΦΦ1 (cid:90) H d − ( C ∆ s K ∆ s )( C ∆ s K ∆ s ) (5.11)where we already performed the bulk integral over S . The bulk-boundary propagator isnormalized as K ∆ si ( (cid:126)y i , (cid:126)y, r ) = (cid:18) rr + ( (cid:126)y i − (cid:126)y ) (cid:19) ∆ si . (5.12)38lugging in d = 4, the integral may be written as (cid:104) ¯Ψ I s ( (cid:126)y )Ψ Js ( (cid:126)y ) (cid:105) = − λ ( N + 1) δ I J δ s ,s ( C ¯ΦΦ1 )4 π (cid:90) r s − η drd (cid:126)y ( r + ( (cid:126)y − (cid:126)y ) ) ∆ s ( r + ( (cid:126)y − (cid:126)y ) ) ∆ s = − λ ( N + 1) δ I J δ s ,s ( C ¯ΦΦ1 )4 π (cid:90) r | s |− η drd (cid:126)y ( r + ( (cid:126)y − (cid:126)y ) ) | s | +1 ( r + ( (cid:126)y − (cid:126)y ) ) | s | +1 (5.13)where we added a regulator η to regulate the r → (cid:126)y can be performed by introducing Feynman parameter α . Theremaining integral is (cid:104) ¯Ψ Is ( (cid:126)y )Ψ Js ( (cid:126)y ) (cid:105) = − λ ( N + 1) δ I J δ s ,s ( C ¯ΦΦ1 )Γ(1 + 2 | s | )4 π Γ(1 + | s | ) × (cid:90) dα (cid:90) dr ( α (1 − α )) | s | r | s |− η ( r + α (1 − α ) (cid:126)y ) | s | = − λ ( N + 1) δ I J δ s ,s ( C ¯ΦΦ1 )4 π | s | ( (cid:126)y ) | s | (cid:18) η + log (cid:0) (cid:126)y (cid:1) − | s | (cid:19) = (cid:15) ( N + 1) v (1 − v ) δ I J δ s ,s π ( N + 4) | s | ( (cid:126)y ) | s | (cid:18) η + log (cid:0) (cid:126)y (cid:1) − | s | (cid:19) (5.14)up to terms that vanish as η →
0. The anomalous dimension of the defect operators can beread off from the coefficient of the log, which gives the corrected dimension∆ s = 1 + | s | − (cid:15) v ( v − N + 1)2( N + 4) | s | (cid:15). (5.15)This result is consistent with the results from large N expansion (4.14). Let’s also do thiscalculation when the KK mode with s = v has a ∆ − v boundary condition. First let us lookat the correction to the two-point function of defect operators Ψ s with s (cid:54) = v . Then theintegrals above remain the same, but the one-point function coefficient of ¯ΦΦ changes. Sothe correction to the two-point function is (cid:104) ¯Ψ Is ( (cid:126)y )Ψ Js ( (cid:126)y ) (cid:105) − = − λ ( N + 1) δ I J δ s ,s ( C ¯ΦΦ1 ) − π | s | ( (cid:126)y ) | s | (cid:18) κ + log (cid:0) (cid:126)y (cid:1) − | s | (cid:19) = − (cid:15) ( N + 1) v (1 + v ) δ I J δ s ,s π ( N + 4) | s | ( (cid:126)y ) | s | (cid:18) κ + log (cid:0) (cid:126)y (cid:1) − | s | (cid:19) (5.16)39here we used the value of the one-point function of ¯ΦΦ in the free theory with ∆ − boundarycondition (3.7). This gives us the dimension of the s (cid:54) = v operators to leading order in (cid:15) ∆ − s = 1 + | s | − (cid:15) v (1 + v )( N + 1)2( N + 4) | s | (cid:15). (5.17)For the s = v operator, the leading correction to the two-point function is given by (cid:104) ¯Ψ Iv ( (cid:126)y )Ψ Jv ( (cid:126)y ) (cid:105) − = − λ ( N + 1) δ I J ( C ¯ΦΦ1 ) − π (cid:90) r − v − κ drd (cid:126)y ( r + ( (cid:126)y − (cid:126)y ) ) − v ( r + ( (cid:126)y − (cid:126)y ) ) − v = λ ( N + 1) δ I J ( C ¯ΦΦ1 ) − π v ( (cid:126)y ) − v (cid:18) κ + log (cid:0) (cid:126)y (cid:1) + 1 v (cid:19) = (cid:15) ( N + 1)(1 + v ) δ I J π ( N + 4)( (cid:126)y ) − v (cid:18) κ + log (cid:0) (cid:126)y (cid:1) + 1 v (cid:19) . (5.18)This gives the corrected dimension of the s = v operator∆ − v = 1 − v − (cid:15) − (1 + v )( N + 1)2( N + 4) (cid:15). (5.19)These results are also consistent with the large N expansion calculation (4.24). We now calculate the correction to the bulk two-point function of Φ. In the bulk OPE limit,when the two fields are close to each other, we expect the following behavior (cid:104) ¯Φ I ( x )Φ J ( x ) (cid:105) = Γ (cid:0) d − (cid:1) π d | x | d − + C ¯Φ;Φ ; ¯ΦΦ ( C ¯ΦΦ1 ) + ... (5.20)up to terms subleading in x = x − x . We will calculate the bulk one-point function C ¯ΦΦ1 to leading order in the perturbation theory in λ . In the free theory, the two-point functionis given by (2.12). The leading correction in the interacting theory is given by a tadpolediagram in H d − × S (cid:104) ¯Φ I ( x )Φ J ( x ) (cid:105) = − λ ( N + 1)( C ¯ΦΦ1 ) δ I J (cid:90) d d x √ g x G ¯ΦΦ ( x , x ) G ¯ΦΦ ( x, x ) . (5.21)40o leading order in λ , we only need the bulk propagator in d = 4, G ¯ΦΦ ( x , x ) = e isθ (cid:88) s ∈ Z + v ( ξ (1 + ξ )) − π ( √ ξ + √ ξ ) s − . (5.22)Then following [13], we place the two operators at the same position on S , ( θ = θ ) andat the same distance from the defect ( r = r = r (cid:48) ). We place them at a separation of y (cid:48) along the defect and parameterize their position as (cid:126)y = ( y (cid:48) / ,
0) and (cid:126)y = ( − y (cid:48) / , S and are left with the following integral on H d − (cid:104) ¯Φ I ( x )Φ J ( x ) (cid:105) = − λ ( N + 1)( C ¯ΦΦ1 ) δ I J π (cid:88) s (cid:90) drdydzr ( r (cid:48) ) (4 r (cid:48) r ) s − d + d − e + e − (( d + + d − )( e + + e − )) s − (5.23)where the integral runs over the H d − coordinated 0 ≤ r < ∞ and −∞ < y, z < ∞ . Asin [13], we defined d ± = (cid:115)(cid:18) y − y (cid:48) (cid:19) + z + ( r ± r (cid:48) ) , e ± = (cid:115)(cid:18) y + y (cid:48) (cid:19) + z + ( r ± r (cid:48) ) . (5.24)We can then perform the sum over spins to get (cid:104) ¯Φ I ( x )Φ J ( x ) (cid:105) = − λ ( N + 1)( C ¯ΦΦ1 ) δ I J π × (cid:90) drdydzr ( r (cid:48) ) (4 r (cid:48) r ) − v ( d + + d − ) v ( e + + e − ) v d + d − e + e − (( d + + d − ) ( e + + e − ) − (4 r (cid:48) r ) ) + ( v → − v ) . (5.25)We are interested in extracting the one-point function of the operator ¯ΦΦ from this two-pointfunction. For that purpose, we define µ = y (cid:48) /r (cid:48) and then look at the integral in the limit µ →
0. Such a analysis was done in [13] for v = 1 / v . We do this in appendix B.1, and here just report the result (B.10) (cid:104) ¯Φ I ( x )Φ J ( x ) (cid:105) = − λ ( N + 1)( C ¯ΦΦ1 ) δ I J π (cid:18) − µ − (cid:0) H − v + H v (cid:1) + 1 v (1 − v ) (cid:19) (5.26)where H n is the n -th harmonic number. The constant, µ independent, piece above con-tributes to the one-point function coefficient of ¯ΦΦ. Combining this with the free theory41esult (2.16) in d = 4 − (cid:15) gives the result C ¯Φ;Φ ; ¯ΦΦ ( C ¯ΦΦ1 ) = ( v − v π + ( v − v ( γ + 3 + log π ) + 18 π (cid:15) + 3 (cid:15) v ( v − π ( N + 4) (cid:18) − (cid:0) H − v + H v (cid:1) + 1 v (1 − v ) (cid:19) (5.27)At large N , this gives C ¯Φ;Φ ; ¯ΦΦ ( C ¯ΦΦ1 ) = ( v − v π + ( v − v ( γ + 3 + log π ) + 18 π (cid:15) (5.28)consistent with (4.18). For general N , but for the Z twist defect with v = 1 /
2, this gives C ¯Φ;Φ ; ¯ΦΦ ( C ¯ΦΦ1 ) = − π + (cid:15) π (1 − γ − log π ) − (cid:15) π ( N + 4) log 2 . (5.29)This is the result for O (2 N ) model, which should give the Ising result for N = 1 /
2, and itmatches with the result in [13] for that case up to a difference in normalization. Using resultfor C ¯Φ;Φ ; ¯ΦΦ from (B.31), we get C ¯ΦΦ1 = N ( v − v π + N (( v − v ( γ + 4 + log π ) + 1)8 π (cid:15) + 3 (cid:15)N v ( v − π ( N + 4) (cid:18) − (cid:0) H − v + H v (cid:1) + 1 v (1 − v ) − (cid:19) . (5.30)There is an extra factor of N in the free part compared to (2.16) because we have N complexfields .Let’s also discuss how the calculation changes when we use ∆ − boundary condition forthe s = v mode. We now have to use the ( C ¯ΦΦ1 ) − from (3.7) for the free theory one-pointfunction. And when we perform the sum in (5.23), we need to use ∆ − for the s = v mode.So the correction due to interactions, to the bulk two-point function is given by Note that we work in a normalization such that C ¯Φ;Φ ; ¯ΦΦ = 1 in the free theory of a single complexscalar. We hope that using the same symbol, C ¯ΦΦ1 , for the one-point function of ¯ΦΦ in both single field and N field case does not lead to confusion. ¯Φ I ( x )Φ J ( x ) (cid:105) − = − λ ( N + 1)( C ¯ΦΦ1 ) − δ I J π (cid:18) − µ − (cid:0) H − v + H v (cid:1) + 1 v (1 − v ) (cid:19) − λ ( N + 1)( C ¯ΦΦ1 ) − δ I J π (cid:90) drdydzr ( r (cid:48) ) d + d − e + e − (cid:18) (4 r (cid:48) r ) − v (( d + + d − )( e + + e − )) − v − ( v → − v ) (cid:19) . (5.31)So we now have an additional integral to perform which we also perform in appendix B.1 withthe result in (B.14). Overall, the result for the interacting correction to the bulk two-pointis (cid:104) ¯Φ I ( x )Φ J ( x ) (cid:105) − = − λ ( N + 1)( C ¯ΦΦ1 ) − δ I J π (cid:18) − µ − (cid:0) H − v + H v (cid:1) + 2 v − v (1 − v ) (cid:19) . (5.32)The µ independent piece is the contribution of the bulk interaction to the product of thebulk OPE coefficient and the one-point function of ¯ΦΦ. Adding it to the free theory resultin d = 4 − (cid:15) (3.7) gives C ¯Φ;Φ ; ¯ΦΦ ( C ¯ΦΦ1 ) − = ( v + 1) v π + ( v + 1) v ( γ + 3 + log π ) + 18 π (cid:15) + 3 (cid:15) v ( v + 1)8 π ( N + 4) (cid:18) − (cid:0) H − v + H v (cid:1) + 2 v − v (1 − v ) (cid:19) . (5.33)In the large N limit, the leading order in N piece agrees with the result from the large N calculation (4.25). In this subsection, we calculate the coefficient of two-point function of the displacementoperator to leading order in (cid:15) . To do that, we use the fact that the displacement appears inthe bulk-defect OPE of ¯ΦΦ as we saw in the free theory (2.33). So we need to now look atcorrections to the bulk two-point function of ¯ΦΦ. In the free theory, it is given by (cid:104) ¯Φ I Φ I ( x ) ¯Φ J Φ J ( x ) (cid:105) = N ( G ¯ΦΦ ( x , x )) . (5.34)43t gets corrected in the interacting theory, and the leading correction is given by (cid:104) ¯Φ I Φ I ( x ) ¯Φ J Φ J ( x ) (cid:105) = − λN ( N + 1)2 (cid:90) d d x √ g x G ( x , x ) G ( x, x )= − (cid:88) s ,s ,s s λN ( N + 1)2(8 π ) (cid:90) d d x √ g x e i ( s − s )( θ − θ )+ i ( s − s )( θ − θ ) × ξ − x x (1 + ξ x x ) − ξ − x x (1 + ξ x x ) − ( (cid:112) ξ x x + (cid:112) ξ x x ) | s | + | s | ) ( (cid:112) ξ x x + (cid:112) ξ x x ) | s | + | s | ) . (5.35)The integral over θ gives a delta function δ ( s − s + s − s ). The remaining sum over spinsthen captures the contribution of various defect operators appearing in the bulk defect OPEof ¯ΦΦ. Let us then isolate the contribution of the displacement operator, which appears for s = s = v and s = s = − v (cid:104) ¯Φ I Φ I ( x ) ¯Φ J Φ J ( x ) (cid:105) (cid:51) − λN ( N + 1)2(8 π ) (cid:90) d d − x √ g x e iθ ξ − x x (1 + ξ x x ) − ξ − x x (1 + ξ x x ) − ( (cid:112) ξ x x + (cid:112) ξ x x ) ( (cid:112) ξ x x + (cid:112) ξ x x ) (cid:51) − λN ( N + 1) π (cid:90) drdydz ( r (cid:48) ) r d d − e e − ( d + + d − ) ( e + + e − ) . (5.36)To go from the first line to the second line, we fixed the position of operators x , x as writtenfor the bulk two-point function of Φ above (5.23). d ± , e ± are defined in the same way asbefore (5.24). But now, we are interested in the contribution of the defect operator, so wedefine κ = r (cid:48) /y (cid:48) = 1 / (cid:112) ξ x x and look at the integral in the limit κ →
0. We do this in theappendix B.2, with the result being (B.22) (cid:104) ¯Φ I Φ I ( x ) ¯Φ J Φ J ( x ) (cid:105) (cid:51) λN ( N + 1) κ π (cid:18) log κ + 14 (cid:19) (cid:51) (cid:15)N ( N + 1)8 π ( N + 4)(4 ξ ) (cid:18) − log(4 ξ ) + 12 (cid:19) . (5.37)We should combine this result with the result in (2.33) in d = 4 − (cid:15) . Note that the result inthe first line of (2.33) is written in terms of ∆ s which also gets corrected by the interactions445.15), so we need to use the corrected ∆ s . This gives, to leading order in (cid:15) (cid:104) ¯Φ I Φ I ( x ) ¯Φ J Φ J ( x ) (cid:105) (cid:51) N π (4 ξ ) − (cid:15) + (cid:15)N ( N + 1)8 π ( N + 4)(4 ξ ) log(4 ξ ) − (cid:15)N π (4 ξ ) ( ψ (1 + v ) + ψ (2 − v ) − π ) (5.38)where ψ ( z ) is the polygamma function. Adding (5.38) and (5.37), the log piece cancels,which implies that, as expected, the displacement has protected dimension d −
1. It givesthe following result for the product of bulk-defect OPE coefficient and the displacementtwo-point function( C ¯ΦΦ D ) C D = N π − (cid:15)N π (cid:18) ψ (1 + v ) + ψ (2 − v ) − − π (cid:19) − (cid:15)N π ( N + 4) . (5.39)Then, we can use the ward identity (2.32)∆ ¯ΦΦ C ¯ΦΦ1 = − (cid:16) π (cid:17) d − √ π Γ (cid:0) d − (cid:1) C ¯ΦΦ D C D (5.40)and the result for C ¯ΦΦ1 in (5.30) to get C D = 4 v (1 − v ) Nπ (cid:20) (cid:15) (cid:18) γ + 74 + log π − v (1 − v ) + ( H − v + H v )2 (cid:19) + 3 (cid:15)N + 4 (cid:18) − (cid:0) H − v + H v (cid:1) + 1 v (1 − v ) − (cid:19) (cid:21) . (5.41)Note that we had to use the bulk result for O (2 N ) model [41]∆ ¯ΦΦ = 2 − (cid:15)N + 4 . (5.42) In this subsection we calculate the four-point function of the defect operators Ψ s in the d − s ( (cid:126)∂ ) m ( (cid:126)∂ ) l ¯Ψ s or of form Ψ s ( (cid:126)∂ ) m ( (cid:126)∂ ) l Ψ s .Each of these have free theory dimensions ∆ s + ∆ s + 2 m + l and longitudinal spin l . Forthe former, the transverse SO (2) spin is s − s and hence is an integer, while for the latter,it is s + s and hence, is fractional, of the form Z + 2 v . We will study both these cases45elow. Integer spin operators on the defect
Let’s first look at the integer spin operators on the defect by considering the following four-point function in the 12 →
34 channel (cid:104) Ψ Is ( (cid:126)y ) ¯Ψ J s ( (cid:126)y )Ψ Ks ( (cid:126)y ) ¯Ψ L s ( (cid:126)y ) (cid:105) = δ I J δ K L G sing + (cid:18) δ I L δ J K − δ I J δ K L N (cid:19) G adj (5.43)where we have decomposed the correlator into singlet and adjoint representations of U ( N ).The operators that appear in this channel are of form Ψ Is ( (cid:126)∂ ) m ( (cid:126)∂ ) l ¯Ψ J s . Let’s restrict tothe longitudinal spin l being zero, because to the order we will consider, we will only be ableto see anomalous dimensions for l = 0 operators. There will be mixing among operatorswhich have the same dimension in the free theory and same SO (2) spin | s | + | s | + d − m = | s | + | s | + d − n, s − s = s − s . (5.44)To keep the analysis simple, we will also restrict to the case when { s , s } > { s , s } < SO (2) spins implies m = n , and hence there will only be mixingbetween operators having equal number of derivatives. So we are only considering com-posite operators of the form form O s,mα = Ψ α + v ( (cid:126)∂ ) m ¯Ψ − s + α + v for positive integer s and α = 0 , , . . . s −
1. These are all degenerate, and we expect this degeneracy to be lifted byinteractions. So we expect the following form for their two-point function (cid:104) O s,mα ( (cid:126)y ) ¯ O s,mβ ( (cid:126)y ) (cid:105) = 1( (cid:126)y ) s + d − m (cid:0) δ αβ − (cid:15) ∆ s,mαβ log( (cid:126)y ) (cid:1) . (5.45)Note that we have to properly normalize the operators such that their two-point functionis unit normalized in the free theory. ∆ s,mαβ is the matrix of anomalous dimensions, and itseigenvalues give the anomalous dimensions of all the spin s composite operators with 2 m derivatives. To extract this anomalous dimension from the four-point function, we note thatin general, such a four-point function can be decomposed into conformal partial waves, or46quivalently the conformal blocks (cid:104) Ψ Is ( (cid:126)y ) ¯Ψ J s ( (cid:126)y )Ψ Ks ( (cid:126)y ) ¯Ψ L s ( (cid:126)y ) (cid:105) = (cid:88) O C Ψ s Ψ s O C O Ψ s Ψ s W ∆ O ,l ( (cid:126)y i ) W ∆ ,l ( (cid:126)y i ) = (cid:18) (cid:126)y (cid:126)y (cid:19) ∆122 (cid:18) (cid:126)y (cid:126)y (cid:19) ∆342 G ∆ ,l ( u, v )( (cid:126)y ) ∆ s s ( (cid:126)y ) ∆ s s (5.46)where ∆ = ∆ s − ∆ s and the conformal block depends on the cross ratios u = (cid:126)y (cid:126)y (cid:126)y (cid:126)y , v = (cid:126)y (cid:126)y (cid:126)y (cid:126)y . (5.47)The particular correlator we are considering then, in perturbation theory in (cid:15) will contain (cid:104) Ψ Iα + v ( (cid:126)y ) ¯Ψ J α − s + v ( (cid:126)y )Ψ K − s + β + v ( (cid:126)y ) ¯Ψ L β + v ( (cid:126)y ) (cid:105) (cid:51) (cid:88) m C Ψ s ¯Ψ s ¯ O s,mα C O s,mβ Ψ s ¯Ψ s (cid:16) δ αβ + (cid:15) s,mαβ ∂ m (cid:17) W d − s +2 m, ( (cid:126)y i ) . (5.48)With these generalities in mind, we can go on and calculate the four-point function. Attree level, this correlator is just given by the disconnected pieces (cid:104) Ψ Is ( (cid:126)y ) ¯Ψ J s ( (cid:126)y )Ψ Ks ( (cid:126)y ) ¯Ψ L s ( (cid:126)y ) (cid:105) = C ∆ s C ∆ s (cid:20) δ I J δ K L δ s ,s δ s ,s ( (cid:126)y ) ∆ s ( (cid:126)y ) ∆ s + δ I L δ J K δ s ,s δ s ,s ( (cid:126)y ) ∆ s ( (cid:126)y ) ∆ s (cid:21) . (5.49)The first term above just represents the contribution of the identity operator and the secondterm represents the contribution of composite operators of the type we mentioned. Todecompose it into the conformal blocks, we can use the following result from [42] δ s ,s δ s ,s ( (cid:126)y ) ∆ s ( (cid:126)y ) ∆ s = δ s ,s δ s ,s (cid:88) n,l c (12)0 ( n, l ) W ∆=∆ s +∆ s +2 n + l,l ( (cid:126)y i ) . (5.50)The squared OPE coefficients are given by c (12)0 ( n, l ) = (cid:0) ∆ s + 2 − d (cid:1) n (cid:0) ∆ s + 2 − d (cid:1) n (∆ s ) l + n (∆ s ) l + n (∆ s + ∆ s + 2 n + 2 l − − l l ! n ! (cid:0) l + d − (cid:1) n (∆ s + ∆ s + n − d + 3) n (cid:0) ∆ s + ∆ s + n + l − d + 1 (cid:1) n . (5.51)Note that the individual dimensions appearing in the subscript of partial wave ∆ s i includecorrections from (5.15). Also, the delta function constraint δ s ,s δ s ,s is equivalent to δ αβ so the disconnected piece only contributes to the diagonal part of the anomalous dimension Note that all the irreducible representations of U ( N ) will individually have this decomposition. U ( N ) singlet sector for example( C Ψ s ¯Ψ s ¯ O s,mα ) = c (12)0 ( m, C ∆ s C ∆ s N = (∆ s ) m (∆ s ) m π N ( m !) ( s + m + 1) m (5.52)The leading correction to it comes from the contact Witten diagram (cid:104) Ψ Is ( (cid:126)y ) ¯Ψ J s ( (cid:126)y )Ψ Ks ( (cid:126)y ) ¯Ψ L s ( (cid:126)y ) (cid:105) = Ψ Ks ( (cid:126)y )¯Ψ J s ( (cid:126)y )Ψ Is ( (cid:126)y ) ¯Ψ L s ( (cid:126)y ) x = − πλ ( δ I J δ K L + δ I L δ J K ) δ ( s + s − s − s ) × (cid:90) (cid:89) i =1 C ∆ si K ∆ si ( (cid:126)y i , (cid:126)y, r ) (5.53)where the remaining integral is only over H d − . The above integral can be evaluated interms of the well known D -functions [43–45] (cid:104) Ψ Is ( (cid:126)y ) ¯Ψ J s ( (cid:126)y )Ψ Ks ( (cid:126)y ) ¯Ψ L s ( (cid:126)y ) (cid:105) = − πλ ( δ I J δ K L + δ I L δ J K ) δ ( s + s − s − s ) × (cid:32) (cid:89) i =1 C ∆ si (cid:33) D ∆ s , ∆ s , ∆ s , ∆ s ( (cid:126)y i ) (5.54)The D -function has the following conformal block decomposition [46, 47] D ∆ s , ∆ s , ∆ s , ∆ s ( (cid:126)y i ) = (cid:88) m P (12)1 ( m, W ∆ m , ( (cid:126)y i ) + (cid:88) n P (34)1 ( n, W ∆ n , ( (cid:126)y i ) . (5.55)The dimensions of the operators appearing are ∆ m = ∆ s +∆ s +2 m and ∆ n = ∆ s +∆ s +2 n and the squared OPE coefficients are given by P (12)1 ( m,
0) =( − m π d − (∆ s ) m (∆ s ) m Γ (cid:16) ∆ s +∆ s +∆ m +2 − d (cid:17) Γ (cid:16) ∆ s +∆ s − ∆ m (cid:17) Γ (cid:0) ∆ +∆ m (cid:1) Γ (cid:0) ∆ +∆ m (cid:1) m !(∆ s + ∆ s + m + 1 − d ) m Γ (∆ s ) Γ (∆ s ) Γ (∆ m ) (5.56)and similarly for P (34)1 ( n, s +∆ s − ∆ s − s = 2 k , an even integer. This is equivalent to the condition that | s | + | s |−| s |−| s | = 2 k .Note that we also have the s + s − s − s = 0 constraint coming from the delta functionin the contact diagram. For the case we are considering when s , s > s , s <
0, thedelta function constraint implies that ∆ s + ∆ s = ∆ s + ∆ s . In that particular case, theabove decomposition for the D -function becomes D ∆ s , ∆ s , ∆ s , ∆ s ( (cid:126)y i ) = (cid:88) m π (cid:0)(cid:81) i =1 (∆ s i ) m (cid:1) (∆ s + ∆ s + 2 m − − m m !) (∆ s + ∆ s + 2 m − (cid:18) s + ∆ s + 2 m −
1+ 2 H m − H m +∆ s +∆ s − + 4 H m +∆ s +∆ s − − (cid:88) i =1 H m +∆ si − − ∂ m (cid:19) W ∆ s +∆ s +2 m, ( (cid:126)y i ) . (5.57)Notice that as we said before, to this order, the contact term only gives anomalous dimensionsto l = 0 operators.As we mentioned briefly above, in addition to the contact interaction, there is also adisconnected diagram that contributes to the anomalous dimension of O s,mα . This just cor-responds to the anomalous dimensions of the two individual Ψ s i which make up O s,mα . Thisdisconnected piece just contributes to the matrix for α = β . Combining the result from thecorrections to the individual dimensions (5.15), and the contact diagram (5.57), we get thefollowing anomalous dimension matrix in the singlet U ( N ) sector∆ s,mαβ = ( N +1)( N +4)( s +2 m +1) if α (cid:54) = β ( N +1)( N +4)( s +2 m +1) + v ( v − N +1) s N +4)( α + v )( s − α − v ) if α = β. (5.58)To get the dimensions of all the spin s operators of this kind then, we have to find theeigenvalues of the above matrix and add that to the bare dimension 2 − (cid:15) + s + 2 m . Inparticular, the displacement operator is non degenerate and corresponds to | s | = v and | s | = 1 − v . So for the displacement, s = 1 , m = 0 and α = β = 0 and it is easy to see thatit does not get anomalous dimension to this order. In general, we expect the displacement tostay protected to all orders in perturbation theory. Also, notice that the contribution of thecontact interaction is the same order in N as the contribution of the disconnected diagram.This is because we are talking about the singlet sector of U ( N ). In order to calculate thiscorrection in the large N expansion, we have to know the correlator of σ operator. We donot do it here. 49 ractional spin operators on the defect In order to analyze the operators of the form Ψ s ( (cid:126)∂ ) m ∂ l Ψ s , we have to look at the abovecorrelator in 13 →
24 channel. These operators have SO (2) spin s = s + s ∈ Z + 2 v .Alternatively, we find it more convenient to consider the following four-point function, andlook at it in 12 →
34 channel (cid:104) Ψ Is ( (cid:126)y )Ψ Js ( (cid:126)y ) ¯Ψ K s ( (cid:126)y ) ¯Ψ L s ( (cid:126)y ) (cid:105) = (cid:18) δ I K δ J L + δ I L δ J K (cid:19) G sym. + (cid:18) δ I K δ J L − δ I L δ J K (cid:19) G ant (5.59)and now we are looking at the operators in the tensor product of two fundamentals of U ( N ), so the four-point function decomposes into the symmetric and anti-symmetric rep-resentations. Let’s again restrict to the l = 0 case, and also consider the case when { s , s , s , s } >
0. This is the case when mixing happens only among the operatorswith the same number of derivatives. So we are going to consider operators of the form O s,mα = Ψ α + v ( (cid:126)∂ ) m Ψ s − α − v where s = k + 2 v for some positive integer k and α = 0 , . . . (cid:98) k (cid:99) .We again use the normalization such that the two-point function of the composite operatorsis given by (5.45).At tree level, as before, the correlator is given by the disconnected pieces, which can bewritten as (cid:104) Ψ Is ( (cid:126)y )Ψ Js ( (cid:126)y ) ¯Ψ K s ( (cid:126)y ) ¯Ψ L s ( (cid:126)y ) (cid:105) = C ∆ s C ∆ s (cid:20) δ I K δ J L δ s ,s δ s ,s ( (cid:126)y ) ∆ s ( (cid:126)y ) ∆ s + δ I L δ J K δ s ,s δ s ,s ( (cid:126)y ) ∆ s ( (cid:126)y ) ∆ s (cid:21) = C ∆ s C ∆ s (cid:88) n,l (cid:0) ( − l δ I K δ J L δ s ,s δ s ,s + δ I L δ J K δ s ,s δ s ,s (cid:1) c (12)0 ( n, l ) W ∆=∆ s +∆ s +2 n + l,l ( (cid:126)y i )(5.60)The difference from the previous case is that when we consider operators in the symmetricsector, for s = s or equivalently α = s/ − v , both the terms above will contribute andhence the OPE coefficient squared will be twice as much. So, in the symmetric sector, theOPE coefficients are( C Ψ s Ψ s ¯ O s,mα ) = c (12)0 ( m, C ∆ s C ∆ s = (∆ s ) m (∆ s ) m π ( m !) ( s + m + 1) m ( C Ψ s Ψ s ¯ O s,mα ) = 2 c (12)0 ( m, C s = (∆ s ) m π ( m !) ( s + m + 1) m . (5.61)50he first order correction coming from the interaction is also similar to before (cid:104) Ψ Is ( (cid:126)y )Ψ Js ( (cid:126)y ) ¯Ψ K s ( (cid:126)y ) ¯Ψ L s ( (cid:126)y ) (cid:105) = − πλ ( δ I K δ J L + δ I L δ J K ) δ ( s + s − s − s ) × (cid:32) (cid:89) i =1 C ∆ si (cid:33) D ∆ s , ∆ s , ∆ s , ∆ s ( (cid:126)y i ) (5.62)Again, for the case when all { s , s , s , s } >
0, above contact interaction is only non-zerowhen ∆ s + ∆ s = ∆ s + ∆ s and the decomposition in (5.57) can be used. Using that, andadding the result of the disconnected pieces from (5.15), we can see that in the symmetricsector, we have the following anomalous dimension matrix∆ s,mαβ = N +4)( s +2 m +1) if α (cid:54) = β and neither α nor β is s − v √ N +4)( s +2 m +1) if α (cid:54) = β and either α or β is s − v N +4)( s +2 m +1) + v ( v − N +1) s N +4)( α + v )( s − α − v ) if α = β (cid:54) = s − v N +4)( s +2 m +1) + v ( v − N +1) s ( N +4) if α = β = s − v. (5.63) Z monodromy defect For completeness, we also specialize this four-point function calculation to the case of Z twist defect where v = 1 /
2. We consider N real scalars and the monodromy is defined suchthat they pick up a minus sign as they go around the defect. We have the full unbroken O ( N ) invariance. We only need to talk about one set of operators on the defect since thereality condition on scalars implies ¯Ψ s = Ψ − s . These defect operators have the two-pointfunction given by (cid:104) Ψ Is ( (cid:126)y )Ψ J − s ( (cid:126)y ) (cid:105) = δ IJ C ∆ s ( (cid:126)y ) ∆ s , C ∆ s = Γ (∆ s )4 π d/ Γ (cid:0) ∆ s + 2 − d (cid:1) , ∆ s = 1 + | s | − (cid:15) v ( v − N + 2)2( N + 8) | s | (cid:15). (5.64)where s is an half-integer now. We then consider the following four-point function in the12 →
34 channel (cid:104) Ψ Is ( (cid:126)y )Ψ Js ( (cid:126)y )Ψ Ks ( (cid:126)y )Ψ Ls ( (cid:126)y ) (cid:105) = δ IJ δ KL G S + (cid:18) δ IK δ JL + δ IL δ JK − δ IJ δ KL N (cid:19) G T + δ IK δ JL − δ IL δ JK G A (5.65)51nd we have now decomposed it into singlet, traceless symmetric and anti-symmetric rep-resentations of O ( N ). We restrict to the case with s , s > s , s < O s,mα = Ψ α + ( (cid:126)∂ ) m ∂ l Ψ s − α − for α = 0 , . . . (cid:98) s − (cid:99) . These operatorsare normalized to have unit two-point function as before.As in the previous case, we can calculate the correlator as (cid:104) Ψ Is ( (cid:126)y )Ψ Js ( (cid:126)y )Ψ Ks ( (cid:126)y )Ψ Ls ( (cid:126)y ) (cid:105) = C ∆ s C ∆ s (cid:20) δ IK δ JL δ s , − s δ s , − s ( (cid:126)y ) ∆ s ( (cid:126)y ) ∆ s + δ IL δ JK δ s ,s δ s ,s ( (cid:126)y ) ∆ s ( (cid:126)y ) ∆ s (cid:21) = (cid:0) δ IK δ JL δ s , − s δ s , − s ( − l + δ IL δ JK δ s , − s δ s , − s (cid:1) × C ∆ s C ∆ s (cid:88) n,l c (12)0 ( n, l ) W ∆=∆ s +∆ s +2 n + l,l ( (cid:126)y i ) (5.66)and the first order correction (cid:104) Ψ Is ( (cid:126)y )Ψ Js ( (cid:126)y )Ψ Ks ( (cid:126)y )Ψ Ls ( (cid:126)y ) (cid:105) = − πλ ( δ IJ δ KL + δ IK δ JL + δ IL δ JK ) δ ( s + s + s + s ) × (cid:32) (cid:89) i =1 C ∆ si (cid:33) D ∆ s , ∆ s , ∆ s , ∆ s ( (cid:126)y i ) . (5.67)We can extract the anomalous dimension from the above results similar to the previous cases,and in the O ( N ) singlet sector, we find∆ s,mαβ = N +2)( N +8)( s +2 m +1) if α (cid:54) = β and neither α nor β is s − √ N +2)( N +8)( s +2 m +1) if α (cid:54) = β and either α or β is s − N +2)( N +8)( s +2 m +1) − ( N +2) s N +8)(2 α +1)(2 s − α − if α = β (cid:54) = s − N +2)( N +8)( s +2 m +1) − ( N +2)2 s ( N +8) if α = β = s − . (5.68)For N = 1, they give the results for the twist defect in Ising model, which were also discussedin [13]. The authors in [13] did not discuss the case in the second line of the above equation. In this paper, we studied monodromy defects in free and critical O ( N ) model. We usedconformal symmetry to map the problem to S × H d − , and saw that the hyperbolic space52s well suited to many of the calculations. In particular, we computed the expectation valueof a spherical monodromy defect (in 3d CFT, a circular defect) from the free energy on thehyperbolic cylinder, and also from an alternative approach based on mapping to S d . We thenstudied a defect RG flow related to choosing alternate boundary conditions on H d − , andverified the conjectured “defect C-theorem” for this RG flow. We studied the monodromydefect in the critical O ( N ) model both in an large N expansion keeping d arbitrary and in d = 4 − (cid:15) dimensions with (cid:15) expansion techniques keeping N arbitrary. We performed severalconsistency checks among these two approaches.As a next step, it would be interesting to do a systematic study of the bulk data (thebulk one-point functions for instance). In [8], a defect CFT inversion formula was used tolearn more details about the bulk data for twist defect in the Ising model. It would beinteresting to extend the techniques to the more general monodromy defect we considered.It would also be interesting to get more analytic control over the large N results. We reliedon numerics to solve the saddle point equation (4.11), but there might be a way to dothis analytically. Understanding the large N solution better may also be a useful first steptowards understanding the holographic description of the system. Moreover, as we saw insection 4, an interacting unitary fixed point does not exist in d = 3 when we choose ∆ − boundary condition for s = v mode. However, there does not seem to be any issues indescribing it close to d = 4. It would be useful to understand why this happens, and ananalytic understanding of the large N solution may help us understand it better. It wouldbe also interesting to go beyond the leading order at large N , which would entail includingthe σ fluctuations around the saddle point and finding the corresponding propagator. Suchanalysis in the BCFT case was done in [31] (see also [25]), and similar techniques may applyin this case. Acknowledgments
We thank Xinan Zhou and Nathan Benjamin for discussions on related topics. This researchwas supported in part by the US NSF under Grants No. PHY-1914860.53
Spectrum of Laplacian on S d with twisted boundaryconditions In this appendix, we determine the spectrum of Laplacian on S d with twisted boundaryconditions. In the coordinate system specified in (2.58), we may write the eigenfunctionsof Laplacian as e im θ θ f ( τ ) κ S d − , where κ S d − is the eigenfunction of Laplacian on S d − withthe usual periodic boundary condition. The required monodromy in θ direction implies m θ ∈ Z + v . The eigenvalues and their degeneracies for the Laplacian with periodic boundarycondition on S d − , are well known − m φ ( m φ + d − , d m φ = (2 m φ + d − m φ + d − m φ !Γ( d − , m φ = 0 , , , ... (A.1)Hence, the eigenfunction equation for f ( τ ) is f (cid:48)(cid:48) ( τ ) + (( d −
2) cot τ − tan τ ) f (cid:48) ( τ ) − (cid:18) m θ cos τ + m φ ( m φ + d − τ (cid:19) f ( τ ) = λf ( τ ) . (A.2)The solution regular at τ = π/ f ( τ ) = (cos τ ) | m θ | (sin τ ) − d − m φ F (cid:18) − d + 2 | m θ | − m φ − (cid:112) ( d − − λ , − d + 2 | m θ | − m φ + (cid:112) ( d − − λ , | m θ | + 1 , cos ( τ ) (cid:19) . (A.3)We also want the solution to be regular at τ = 0 in all dimensions. The eigenvalue λ can beparametrized as λ = − n ( n + d − , (A.4)and regularity of the solution at τ = 0 demands n = | m θ | + m φ + 2 a, a ∈ N (A.5)where N represents the set of non-negative integers (includes 0). As mentioned previously,the definition of defect requires m θ ∈ Z + v which implies that n = k + v or n = k + 1 − v with k being a non-negative integer. Using the degeneracies of m φ in (A.1), it is easy to54ount the degeneracies of nn ∈ N , d n = (2 n + d − n + d − n !Γ( d ) n = k + v, k ∈ N , d n = Γ( k + d ) k !Γ( d ) . (A.6)When summing over the eigenvalues, we need to sum over both n = k + v and n = k + 1 − v .When v = 1 /
2, they both coincide. This effectively means that for v = 1 /
2, the degeneracyof a given n is twice of what we wrote above. B Bulk Integrals
B.1 For bulk two-point function
In order to calculate the one-point function of ¯ΦΦ to leading order in (cid:15) in subsection 5.2, weneed to do the following integral J = (cid:90) drdydzr ( r (cid:48) ) (4 r (cid:48) r ) − v ( d + + d − ) v ( e + + e − ) v d + d − e + e − (( d + + d − ) ( e + + e − ) − (4 r (cid:48) r ) ) + ( v → − v ) (B.1)where the integral runs over 0 ≤ r < ∞ , −∞ < y < ∞ , −∞ < z < ∞ and d ± = (cid:115)(cid:18) y − y (cid:48) (cid:19) + z + ( r ± r (cid:48) ) , e ± = (cid:115)(cid:18) y + y (cid:48) (cid:19) + z + ( r ± r (cid:48) ) . (B.2)This integral was analyzed in the limit µ → v = 1 /
2. The same method goesthrough for arbitrary values of v , with some changes. We will do that here. First, let’s makethe integral over dimensionless variables by making a substitution y = r (cid:48) a, z = r (cid:48) b, r = r (cid:48) c ,and then extend the integration over all of R J = 2 − v (cid:90) R dadbdc ( c ) − v ( ˜ d + + ˜ d − ) v (˜ e + + ˜ e − ) v ˜ d + ˜ d − ˜ e + ˜ e − (cid:16) ( ˜ d + + ˜ d − ) (˜ e + + ˜ e − ) − (4 c ) (cid:17) + ( v → − v )˜ d ± = (cid:114)(cid:16) a − µ (cid:17) + b + ( c ± , ˜ e ± = (cid:114)(cid:16) a + µ (cid:17) + b + ( c ± . (B.3)When µ →
0, this integral has logarithm divergences around ( a, b, c ) = (0 , , ± α log µ + β up to terms that vanish as µ →
0. We areinterested in calculating α and β . To do that, we introduce an auxiliary parameter N and55ivide the integration region into two parts and write the integral as J = J ( µ, N )+ J ( µ, N ). J ( µ, N ) is the integral over spheres of radius µ N around the two points ( a, b, c ) = (0 , , ± J ( µ, N ) is the integral over rest of R . The integrals J , ( µ, N ) simplify in the limit N → ∞ , µ
N → µ →
0. Let’s first look at J ( µ, N ) and focus on the sphere around (0 , , a = µx, b = µy, c =1 + µz gives˜ d + = ˜ e + = 2 + µz + O ( µ )˜ d − = µ (cid:115)(cid:18) x − (cid:19) + y + z + O ( µ ) , ˜ e − = µ (cid:115)(cid:18) x + 12 (cid:19) + y + z + O ( µ ) . (B.4)Similar things can also be said about the sphere around (0 , , − µ → J ( µ, N ) = 12 (cid:90) x + y + z ≤N dxdydz f + f − ( f + + f − ) ; f ± = (cid:115)(cid:18) x ± (cid:19) + y + z = 2 π (cid:90) N dx x + − | x − | + (cid:113) N + − x − (cid:113) N + + x x = π log 2 N + O (1 / N ) . (B.5)Now, let’s shift our attention to J ( µ, N ) which runs over the domain D = { ( a, b, c ) ∈ R | a + b + ( c ± ≥ ( µ N ) } . In the limit N → ∞ , we can take ˜ d ± = ˜ e ± = (cid:112) a + b + ( c ± and hence 4 c = ˜ d − ˜ d − . The integral then simplifies to J ( µ, N ) = 2 − v (cid:90) D dadbdc ( c ) − v ( ˜ d + + ˜ d − ) v ( ˜ d + ˜ d − ) (cid:16) ( ˜ d + + ˜ d − ) − (4 c ) (cid:17) + ( v → − v )= 12 (cid:90) D dadbdc
1( ˜ d + ˜ d − ) (cid:32) ˜ d + + ˜ d − | ˜ d + − ˜ d − | (cid:33) v − + ( v → − v ) . (B.6)We then scale the variables by two followed by an inversion to change the variables as a (cid:48)(cid:48) = 2 aa + b + ( c − , b (cid:48)(cid:48) = 2 ba + b + ( c − , c (cid:48)(cid:48) −
12 = 2( c − a + b + ( c − = ⇒ ˜ d + = 2 (cid:118)(cid:117)(cid:117)(cid:116) a (cid:48)(cid:48) + b (cid:48)(cid:48) + (cid:0) c (cid:48)(cid:48) + (cid:1) a (cid:48)(cid:48) + b (cid:48)(cid:48) + (cid:0) c (cid:48)(cid:48) − (cid:1) , ˜ d − = 2 (cid:113) a (cid:48)(cid:48) + b (cid:48)(cid:48) + (cid:0) c (cid:48)(cid:48) − (cid:1) (B.7)56nd the integral simplifies to J ( µ, N ) = 116 (cid:90) D (cid:48)(cid:48) da (cid:48)(cid:48) db (cid:48)(cid:48) dc (cid:48)(cid:48) r (cid:32)(cid:18) r + + 1 | r + − | (cid:19) v − + (cid:18) r + + 1 | r + − | (cid:19) − v (cid:33) ,r + = (cid:115) a (cid:48)(cid:48) + b (cid:48)(cid:48) + (cid:18) c (cid:48)(cid:48) + 12 (cid:19) . (B.8)When we scale by two, the integration domain gets mapped to everywhere outside thespheres centered at (0 , , /
2) and (0 , , − /
2) with radius µ N /
2. Then, when we invertabout (0 , , / µ N , it gets mapped to everywhere between the sphereof radius µ N / , , − /
2) and the sphere of radius 2 /µ N centered at (0 , , / D (cid:48)(cid:48) = { ( a (cid:48)(cid:48) , b (cid:48)(cid:48) , c (cid:48)(cid:48) ) ∈ R | a (cid:48)(cid:48) + b (cid:48)(cid:48) + ( c (cid:48)(cid:48) ± / ≷ ( µ N / ± } . Again, to leading order in µ N , we can shift the outer sphere of radius 2 /µ N so that it is also centered at (0 , , − / J ( µ, N ) = π (cid:90) µ N µ N dr + r + (cid:32)(cid:18) r + + 1 | r + − | (cid:19) v − + (cid:18) r + + 1 | r + − | (cid:19) − v (cid:33) = π (cid:18) − − (cid:0) H − v + H v (cid:1) + 1 v (1 − v ) − µ N (cid:19) (B.9)So N cancels and the full integral is J = π (cid:18) − µ − (cid:0) H − v + H v (cid:1) + 1 v (1 − v ) (cid:19) . (B.10)For the same calculation with ∆ − boundary condition, we need an additional integral J − = (cid:90) drdydzr ( r (cid:48) ) d + d − e + e − (( d + + d − )( e + + e − )) v (4 r (cid:48) r ) v − ( v → − v ) . (B.11)Doing the same subsitutions as before, we get J − = 2 − − v (cid:90) R dadbdc ( c ) − − v ( ˜ d + + ˜ d − ) v (˜ e + + ˜ e − ) v ˜ d + ˜ d − ˜ e + ˜ e − − ( v → − v ) . (B.12)This integral in fact does not have diverge as µ →
0, so we can just plug in µ = 0, sincewe are only interested in µ independent piece for this calculation. For µ = 0, we have57 d ± = ˜ e ± = (cid:112) a + b + ( c ± and 4 c = ˜ d − ˜ d − which gives a simpler integral J − = 2 (cid:90) R dadbdc ( ˜ d + + ˜ d − ) v − ( ˜ d + ˜ d − ) | ˜ d + − ˜ d − | v +1 − ( v → − v ) . (B.13)We then do a change of variables as in (B.7) to get J − = 14 (cid:90) R da (cid:48)(cid:48) db (cid:48)(cid:48) dc (cid:48)(cid:48) ( r + + 1) v − r | r + − | v +1 − ( v → − v ) , r + = (cid:115) a (cid:48)(cid:48) + b (cid:48)(cid:48) + (cid:18) c (cid:48)(cid:48) + 12 (cid:19) = π (cid:90) ∞ dr + ( r + + 1) v − | r + − | v +1 − ( v → − v )= − πv . (B.14) B.2 For coefficient of displacement
For extracting the coefficient of the displacement in subsection 5.3, we have to perform thefollowing integral I = (cid:90) drdydz ( r (cid:48) ) r d d − e e − ( d + + d − ) ( e + + e − ) (B.15)in the limit κ = r (cid:48) /y (cid:48) →
0. As before, we rescale the coordinates as y = ay (cid:48) , z = by (cid:48) , r = cy (cid:48) to rewrite the integral as I = κ I, I = 12 (cid:90) R dadbdc ( c ) / ˜ d ˜ d − ˜ e ˜ e − ( ˜ d + + ˜ d − ) (˜ e + + ˜ e − ) ˜ d ± = (cid:115)(cid:18) a − (cid:19) + b + ( c ± κ ) , ˜ e ± = (cid:115)(cid:18) a + 12 (cid:19) + b + ( c ± κ ) . (B.16)As before, this integral has a logarithmic divergence as κ → a, b, c ) = ( ± / , , I up to terms that vanish as κ →
0. As before, wedivide the integration region into two parts I = I ( κ, N ) + I ( κ, N ) with I ( κ, N ) being theintegral within two spheres of radius κ N around ( ± / , , N → ∞ , N κ → / , , a = 1 / κx, b = κy, c = κz , and notethat ˜ e ± = 1 + O ( κ ) I ( κ, N ) = 14 (cid:90) x + y + z ≤N dxdydz ( z ) / f f − ( f + + f − ) , f ± = (cid:112) x + y + ( z ± (B.17)58he integral over x, y can be directly performed, and the rest can be performed after weexpand in large N . Up to terms that vanish as N →
0, we get I ( κ, N ) = − π (cid:90) N dz (cid:18) z log (cid:18) | z − | ( z + 1)2 z (cid:19) + z log 2 + z N (cid:19) = π
16 log
N − π . (B.18)Let’s now look at the other part of the integral I ( κ, N ) = 12 (cid:90) D dadbdc ( c ) / ˜ d ˜ d − ˜ e ˜ e − ( ˜ d + + ˜ d − ) (˜ e + + ˜ e − ) (B.19)where the domain of the integral is D = { ( a, b, c ) ∈ R | ( a ± / + b + c ≥ ( κ N ) } . Atlarge N , we can use ˜ d + = ˜ d − , ˜ e + = ˜ e − , so the integral becomes I ( κ, N ) = 132 (cid:90) D dadbdc ( c ) / r r − , r ± = (cid:115)(cid:18) a ± (cid:19) + b + c . (B.20)Next, we do an inversion centered at (1 / , ,
0) which maps the integration domain to D (cid:48) which is the region between a sphere of radius κ N centered at ( − / , ,
0) and a sphereof radius 1 /κ N centered at (1 / , ,
0) up to terms of order κ N . Under this inversion, theintegral simplifies to I ( κ, N ) = 132 (cid:90) D (cid:48) dadbdc ( c ) / r . (B.21)We can then shift the outer sphere so that it is also centered at ( − / , ,
0) which onlychanges the result at subleading order in κ N . Then the integral can be evaluated to be I ( κ, N ) = − π
16 log κ N = ⇒ I = − πκ (cid:18) log κ + 14 (cid:19) . (B.22) B.3 For bulk OPE coefficient
In this appendix, we show how to calculate the bulk OPE coefficient C ¯Φ;Φ ; ¯ΦΦ . It can beextracted from the bulk four-point function of Φ in the absence of the defect in flat space.As we saw in section 5.4, such a four-point function can be decomposed into singlet andadjoint representations of U ( N ) (cid:104) ¯Φ I ( x )Φ J ( x ) ¯Φ K ( x )Φ L ( x ) (cid:105) = δ I J δ K L G sing + (cid:18) δ I L δ J K − δ I J δ K L N (cid:19) G adj . (B.23)59e expect the operator ¯ΦΦ to be present in the singlet sector. For simplicity, we place thefour operators on a line with | x | = | x | = r , and | x | = | x | = r/µ and look at it in thelimit µ << G sing = 1 r Φ (cid:104) C + ( C ¯Φ;Φ ; ¯ΦΦ ) C ¯ΦΦ µ ∆ ¯ΦΦ (1 + O ( µ )) (cid:105) . (B.24)In the following, we fix the normalization of the operator such that C ¯ΦΦ = N C . In the freetheory, we have ∆ Φ = d/ − , ∆ ¯ΦΦ = d −
2. The result for ∆ Φ does not get corrected tofirst order in interaction. In the free theory, the correlator is just (cid:104) ¯Φ I ( x )Φ J ( x ) ¯Φ K ( x )Φ L ( x ) (cid:105) = δ I J δ K L C r Φ + µ Φ δ I L δ J K C r Φ (1 − µ ) Φ . (B.25)In the interacting theory, it gets corrected to (cid:104) ¯Φ I ( x )Φ J ( x ) ¯Φ K ( x )Φ L ( x ) (cid:105) = − λ δ I J δ K L + δ I L δ J K ) (cid:90) d x C x x x x = − λπ δ I J δ K L + δ I L δ J K ) µ r C ¯ D , , , ( u, v ) (B.26)where we used the fact that this conformal integral can be performed in terms of well known¯ D functions (see for instance [48]). Conformal cross-ratios u, v can be defined by u = x x x x = µ , v = x x x x = (1 − µ ) . (B.27)This ¯ D function can be expanded into a power series around u = 0 , v = 1 as [45, 49]¯ D , , , ( u, v ) = ∞ (cid:88) m,n =0 (( m + n )!) n !(2 m + n + 1)! u m (1 − v ) n (2 ψ (2 + 2 m + n ) − ψ (1 + m + n ) − log u )(B.28)The singlet sector correlator, to leading order in interaction, and to leading order in µ isthus given by G sing = C r d − (cid:18) µ d − N (cid:19) − λπ ( N + 1)2 N µ r C ¯ D , , , ( u, v )= C r d − (cid:18) µ d − N (cid:19) − π ( N + 1) (cid:15)N ( N + 4) µ r C ( − µ + 2) . (B.29)60sing our convention that C Φ = Γ (cid:0) d − (cid:1) π d/ (B.30)we get ∆ ¯ΦΦ = 2 − N + 4 (cid:15)C ¯Φ;Φ ; ¯ΦΦ = 1 N (cid:18) − N + 12( N + 4) (cid:15) (cid:19) . (B.31) References [1] M. Bill`o, V. Gon¸calves, E. Lauria, and M. Meineri, “Defects in conformal fieldtheory,”
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