-
We start with the 4-form ansatz
4 $ \begin{array}{*{20}{l}} G_4 = f_1\, G_4^{(0)} + R^4\, {\rm d}f_2 \wedge J \wedge e_7 + R^4\, {f}_3\, J^2 \end{array} $
(1) for 11D SUGRA over
$ AdS_4 \times S^7/Z_k $ when the internal space is considered as a$ U(1) $ bundle on$ CP^3 $ , where$G_4^{(0)}={\rm d}{\mathcal{A}}_3^{(0)} = N \mathcal{E}_4$ holds for the ABJM [6] background with$ N=(3/8) R^3 $ units of flux quanta on the internal space,$ R=2 R_{\rm{AdS}} $ is the$ AdS $ curvature radius,$ \mathcal{E}_4 $ is the unit-volume form on$ AdS_4 $ , J is the Kähler form on$ CP^3 $ ,$ e_7 $ is the seventh vielbein5 (of the internal space), and$ f_i $ 's with$ i=1,2,3 $ are scalar functions in bulk coordinates.Taking the ansatz (1) from the Bianchi identity and Euclidean 11D equation
$ {\rm d}G_4=0, \qquad {\rm d} \ast_{11} G_4- \frac{\rm i}{2}\, G_4 \wedge G_4=0, $
(2) we get
$ f_3= f_2 \pm \frac{C_2}{R}, \qquad \bar{f}_1 = {\rm i}\, \frac{3}{16}\, R^5\, f_3^2\, \pm\, {\rm i}\, \frac{3}{8} C_3\, R^3, $
(3) where
$ {C}_i $ 's are real constants and$ N\, f_1 \equiv \bar{f}_1 $ . Note that the plus and minus signs on the last term of the RHS equation indicate considering the Wick-rotated (WR) and skew-whiffed (SW) backgrounds, respectively; the ABJM background is realized with$ C_3=1 $ . In addition, from Eq. (2), using the relationship presented in (3), we get$ \Box_4 f_3 - \frac{1}{R_{\rm{AdS}}^2} \left(1 \pm 3\, {C}_3 \right) {f}_3 - 6\, {f}_3^3=0, $
(4) 6 where$\ast_4 {\rm d} \left(\ast_4 {\rm d}{f}_3 \right)=\Box_4$ is the$ EAdS_4 $ Laplacian. Accordingly, we use the following conventions$ \ast_4 {\bf{1}} =\frac{R^4}{16} \mathcal{E}_4, \quad \ast_7 {\bf{1}} =\frac{R^7}{3!} J^3 \wedge e_7=R^7\, \mathcal{E}_7, \quad \ast_7 \left(J \wedge e_7 \right)= \frac{R}{2} J^2. $
(5) Next, from (3) and (4), we write
7 $ \Box_4 f_2 - m^2\, f_2 \mp \delta\, f_2^2 - \lambda\, f_2^3 \mp F=0, $
(6) 8 where$ \begin{aligned}[b]& m^2=\frac{4}{R^2} \left(1 \pm 3\, C_3 + \frac{9}{2}\, C_2^2 \right), \quad \delta = \frac{18}{R}\, C_2, \\& \lambda=6, \quad F = \frac{4}{R^3}\left(C_2 \pm 3\, C_2\, C_3 + \frac{3}{2}\, C_2^3 \right). \end{aligned} $
(7) To make Eq. (6) homogeneous (that is
$ F=0 $ ), for cases other than$ C_2=0 $ , we have to set$ \delta^2= 27\, m^2 \ \ \ \mathrm{OR} \ \ \ C_2^2 = \frac{\mp 6\, C_3 -2}{3} $
(8) Hence,
$ \begin{array}{*{20}{l}} m^2 R_{\rm{AdS}}^2= -2 (1 \pm 3\, C_3) \equiv -2\, \bar{m}^2 R_{\rm{AdS}}^2, \end{array} $
(9) where
$ \bar{m}^2 $ is indeed the squared mass of$ f_3 $ in (4). To have physically permissible (non-imaginary) masses in this case, we have only to consider the SW version with$ C_3 \geq 1/3 $ . As a result, the SW ABJM background is realized with$ C_3=1 $ ; then,$ m^2 R_{\rm{AdS}}^2= +4 $ ($ C_2 =2/\sqrt{3} $ ).In addition, as noticed in [5],
$ \pm \left(C_2/2 \right) =\pm \sqrt{{-\bar{m}^2}/{\lambda}} $ are in fact homogenous vacua; thus, the (pseudo)scalar is Higgs-like, and the LHS relation in (3) imposes spontaneous symmetry breaking, where f acts as fluctuation around the homogeneous vacua. -
To take backreaction, we should first compute the EM tensors of the corresponding Einstein's equations, the details of which are presented in Appendix A. In fact, because we are looking for instantons that, as topological objects, should not backreact on the background geometry, we solve the main bulk equations using the equations presented in Appendix A, resulting from setting the EM tensors to zero.
Accordingly, we observe that Eq. (A7) is solved with (6) by taking
$ \begin{array}{*{20}{l}} \Box_4\, f_{2}=0, \end{array} $
(10) which means taking the backreaction of the external
$ AdS_4 $ space on the background geometry yields the massless$ m^2 R_{\rm{AdS}}^2=0 $ bulk (pseudo)scalar replying to the boundary exactly marginal operators.9 In the same manner, noting that the Eq. (A9) is the same as the main one (6), from solving the Eqs. (A8) as well as (A7) and (A8) with (6), that is taking the backreaction of the internal (indeed
$ CP^3 $ ) and entire 11D space, we get$ \Box_4 f_2 - \frac{2}{R^2}\, f_2 = \pm\, \frac{2\, C_2}{R^3}, $
(11) $ \Box_4 f_2 - \frac{8}{9\, R^2}\, f_2 = \pm\, \frac{8\, C_2}{9\,R^3}, $
(12) respectively, with
$ m^2 R_{\rm{AdS}}^2=1/2, 2/9 $ corresponding to the marginally irrelevant$ \Delta_{\pm} = {3}/{2} \pm {\sqrt{11}}/{2}, \, {3}/{2} \pm {\sqrt{(89/9)}}/{2} $ boundary operators, of which we encountered the former recently in [4]. -
One may solve Eqs. (11) and (12) using the usual mathematical methods, such as separation in variable. However, a well-known closed solution for the equations- leaving out the inhomogeneous terms that do not contribute to the dynamics
10 - reads [18,19]$ \begin{aligned}[b] f_0(u,\vec{u})=\;& \bar{C}_{\Delta_+} \left(\frac{u}{u^2+(\vec{u}-\vec{u}_0)^2} \right)^{\Delta_+}, \\ \bar{C}_{\Delta_+}=&\frac{\Gamma(\Delta_+)}{\pi^{3/2}\, \Gamma(\nu)}. \end{aligned} $
(13) where
$ \Delta_+ $ ($ \Delta_- $ ) is the larger (smaller) root of$ m^2= \Delta (\Delta-3) $ in$ AdS_4 $ , with$ \Delta_{\pm}=3/2 \pm \nu $ ,$ \sqrt{9 + 4\, m^2}=2 \nu $ , and we use the$ EAdS_4 $ metric$ {\rm d}s^2_{\rm EAdS_4} = \frac{R^2}{4\, u^2} \left({\rm d}u^2 + {\rm d}x^2 + {\rm d}y^2 + {\rm d}z^2 \right), $
(14) noting
$ \vec{u}=(x,y,z) $ in upper-half Poincar$ \acute{e} $ coordinates; therefore,$ \Box_4 f = \frac{4\, u^2}{R^2} \left(\partial_i \partial_i + \partial_u \partial_u - \frac{2}{u} \partial_u \right) f. $
(15) Further, because the bulk solutions including the backreaction correspond to variants of marginal operators, for simplicity, we consider the instanton solution for (10) and compute its correction with respect to the background action. To this end, as the background geometry does not change, we use the right parts of the bosonic action of 11D SUGRA in Euclidean space as
$ S_{11}^E = -\frac{1}{4 \kappa_{11}^2} \int \left({G}_4 \wedge \ast_{11} {G}_4 - \frac{\rm i}{3}\, {\mathcal{A}}_3 \wedge {G}_4 \wedge {G}_4 \right), $
(16) where
$ \kappa_{11}^2 = 9 \pi\, \mathcal{G}_{11}=\dfrac{1}{4\pi} (2\pi l_p)^9 $ , and$ \kappa_{11} $ ,$ \mathcal{G}_{11} $ and$ l_p $ are the 11D gravitational constant, Newton's constant, and Plank length, respectively.Next, from the ansatz expressed in (1), using (5), we write its 11D dual 7-form as
$ G_7 = \frac{8}{3} R^3\, \bar{f}_1\, J^3 \wedge e_7 - \frac{R^5}{2} \ast_4 {\rm d}f_2 \wedge J^2 + \frac{R^7}{8} f_3\, \mathcal{E}_4 \wedge J \wedge e_7; $
(17) and
$ \begin{aligned}[b]& {G}_4 = {\rm d}{\mathcal{A}}_3, \quad {\mathcal{A}}_3 = \tilde{\mathcal{A}}_3^{(0)} + R^4\, \left(f_3\, J \wedge e_7 \right), \\& \tilde{G}_4^{(0)} = {\rm d}\tilde{\mathcal{A}}_3^{(0)} = \bar{f}_1\, \mathcal{E}_4. \end{aligned} $
(18) By placing the latter relations in (16), using (3) (noting that, for
$ C_2=0 $ ,$ f_2=f_3 $ ), we get$ \begin{aligned}[b] \tilde{S}_{11}^E =& - \frac{R^{9}}{16\, \kappa_{11}^2} \int \Bigg[-\frac{3}{2}C_3^2\, \mathcal{E}_4 + {\rm d}f_3 \wedge \ast_4 {\rm d}f_3 + \frac{R^2}{2} f_3^2\, \mathcal{E}_4 \\&+ \frac{3}{8} R^4 f_3^4\, \mathcal{E}_4 +{\rm i} \frac{4}{3 R} {\rm d}\left(f_3^2\, \mathcal{A}_3^{(0)} \right) \Bigg] \wedge J^3 \wedge e_7, \end{aligned} $
(19) where the first term on the RHS is the contribution of the ABJM background realized with
$ C_3=1 $ , as one may see from the second term in the RHS relation expressed in (3); the last (surface) term, as a total derivative, does not contribute to the equations, and hence, we discard it.Then, to compute the action expressed in (19) based on the solution presented in (13) with
$ \Delta_+ =3 $ , we use$ \begin{aligned}[b] \mathcal{E}_4 =\;& -\frac{{\rm d}u}{u^4} \wedge {\rm d}x \wedge {\rm d}y \wedge {\rm d}z, \\ \mathrm{vol}_7 = \;&\frac{R^7}{3!} \int J^3 \wedge e_7= \frac{\pi^4\, R^7}{3\, k}, \\ \kappa_{11}^2 =\;& \frac{16}{3} \left(\frac{\pi^{10}\, R^9}{3\, k^3} \right)^{1/2}, \end{aligned} $
(20) and the 3D spherical coordinates, setting
$ |\vec{u}-\vec{u}_0|=r $ . As a result, the finite contribution of the action, after integrating on the external space coordinates, in the unit 7D internal volume reads$ \bar{S}_{11}^{\rm corr.} = - \hat{c}\, \sqrt{\frac{k^3}{R}} \frac{1}{\epsilon^6} \left(1 + \check{c}\, \frac{R^2}{\epsilon^{6}} \right), $
(21) where
$ \hat{c}\simeq 0.000016 $ and$ \check{c}\simeq 0.0033 $ ; because of singularities, we have included$ \epsilon>0 $ as a cutoff parameter to evade the infinity of integrals with respect to (wrt) u; see [20]. Meanwhile, we note that, for finite k and R, the contribution expressed in (21) is small. -
The Higgs-like (pseudo)scalar equation of (6), with (8) and (15), reads
11 $ \begin{aligned}[b]& \left[\partial_i \partial_i + \partial_u \partial_u - \frac{2}{u} \partial_u - \frac{m^2}{u^2} \right] f(u,\vec{u})\\&\quad + \frac{1}{u^2} \left[3 \sqrt{3}\, m\, f(u,\vec{u})^2 - 6\, f(u,\vec{u})^3 \right]=0. \end{aligned} $
(22) For its linear part, using the spherical coordinates with
$ r=|\vec{u}| $ , discarding the angular parts, and separating variables, i.e.,$ f_0(u,r)=f(r) g(u) $ , we can write$ \begin{aligned}[b] & \left[\frac{{\rm d}^2}{{\rm d} r^2} + \frac{2}{r} \frac{\rm d}{{\rm d} r} - k^2 \right] f(r)=0, \\& \left[\frac{{\rm d}^2}{{\rm d} u^2} - \frac{2}{u} \frac{\rm d}{{\rm d} u} - \frac{m^2}{u^2}+ k^2 \right] g(u)=0; \end{aligned} $
(23) with combinations of Hyperbolic and Bessel (or with
$ k=i \kappa $ , Trigonometric and Modified Bessel) functions as solutions for$ f(r) $ and$ g(u) $ , respectively.12 Then, one may use the leading order (LO) solutions to obtain the higher-order solutions of the full NPDE. The resulting solutions always reproduce the right behavior of (pseudo)scalars near the boundary as
$ \begin{array}{*{20}{l}} f(u\rightarrow 0, r) \approx \alpha(r)\, u^{\Delta_-} + \beta(r)\, u^{\Delta_+}. \end{array} $
(25) In contrast, one may employ an ansatz as follows:
$ \begin{array}{*{20}{l}} f(u,r)=F(\xi), \quad \xi=u^{1/2}\ f(r), \end{array} $
(26) which turns (22) into the following NODE
$ \left[ \frac{{\rm d}^2}{{\rm d} \xi^{2}}-\frac{5}{\xi}\, \frac{\rm d}{{\rm d}\xi} -\frac{4\, m^2}{{\xi}^{2}} \right] F(\xi)-\frac{4}{{\xi}^{2}}\, \mathcal{F}(F(\xi)) =0, $
(27) where we define
$ \begin{array}{*{20}{l}} \mathcal{F}(F(\xi)) \equiv - 3 \sqrt{3}\, m\, F(\xi)^{2} + 6\, F(\xi)^{3}. \end{array} $
(28) As a result, the appropriate part of a perturbative solution for (27), up to the first or next-to-leading order (NLO), reads
13 $ f^{(1)}(u,r) = \sum\limits_{{l}=-}^+ \, C_{{l}}\, \left( u\, f(r)^2 \right)^{\Delta_{{l}}}, $
(30) where
$ C_{{l}} $ 's are real constants.Similarly, with
$ \xi=r/u $ (the so-called self-similar reduction method via the scale-invariance of variables; see, for instance, [22]), Eq. (22) turns into$ \left[ \left({\xi}^{2}+1 \right) {\frac{{\rm d }^{2}}{{\rm d}{\xi}^{2}}} + \frac{(2+ 4\, \xi^2)}{\xi}\,\frac{\rm d}{{\rm d} \xi}- m^2 \right] {F}(\xi) - \mathcal{F}(F(\xi)) =0. $
(31) A solution for the linear part of the latter equation is in terms of Legendre functions, and from this, one may build perturbative series solutions for higher-orders; for such a solution, see [4]. Alternatively, we can use
$ F(\xi)= {\rm e}^{\int\, G(\xi)\, {\rm d}\xi}, \quad \frac{1}{F(\xi)}\frac{{\rm d}F(\xi)}{{\rm d}\xi}=G(\xi), $
(32) which turns Eq. (31) into the following first-order Riccati equation:
$ \left({\xi}^{2}+1 \right) \left[ \frac{{\rm d}G(\xi)}{{\rm d}\xi} + G(\xi)^2 \right]+\frac{1}{{\xi}} (2+ 4\, \xi^2)\, G(\xi) - m^2 =0. $
(33) For massive modes, a common series solution for the latter equation, keeping the normalizable term appropriate for the corresponding boundary analyses of AdS
$ _4 $ / CFT$ _3 $ , reads$ f_0(u,r) = \tilde{C}_{\Delta_+} \left(\frac{u}{r} \right)^{\Delta_+}, $
(34) from which one may build higher-order solutions. For example, for the mode
$ m^2=18 $ that we consider, a series expansion around$ u=0 $ , up to NLO, reads$ f^{(1)}(u,r) = \left[ \check{C}_{\Delta_-}\, \ln\left(\frac{r}{u}\right) \right] \left(\frac{u}{r}\right)^{\Delta_-=-3} + \hat{C}_{\Delta_+} \left(\frac{u}{r}\right)^{\Delta_+=6} $
(35) with real constants
$ \check{C}_{\Delta_-} $ and$ \hat{C}_{\Delta_+} $ – when doing boundary analyses, we return to this solution as well. -
Here, we employ the ADM formulation, as shown in Appendix 8, to build series solutions appropriate for near the boundary analyses of the specific Higgs-like mode
$ m^2=18 $ . This mode could be realized with$ C_3=17/3 $ in the WR version of (4) (equally for (6) in addition to$ C_2=0 $ ) and with$ C_3=10/3 $ and$ C_2=\sqrt{6} $ in the SW version of (6) for$ F=0 $ . As a result, the series solutions of these equations about$ u=0 $ , with the initial or near the boundary data from (B1) with$ \Delta_+=6 $ and the Adomian polynomials (B5) with$ \delta=0 $ and$ \delta= 9\, \sqrt{6} $ , respectively, up to NNNLO, wrt (29), read$ \begin{aligned}[b] f^{(3)}(u,r)=& -5 f(r) \Big[1+20\,\ln(u) + 400\, \ln(u)^2\\&-8000\, \ln(u)^3 \Big] {u}^{6} + O({u}^{8}), \end{aligned} $
(36) $ \begin{aligned}[b] f^{(3)}(u,r)=& -20\,f(r)\, {u}^{6} + \frac{17965}{43904} \Bigg(\frac{{\rm d}^2 f(r)}{{\rm d} r^2}\\& + \frac{2}{r} \frac{{\rm d} f(r)}{{\rm d} r} \Bigg) {u}^{8} + O({u}^{10}). \end{aligned} $
(37) Meanwhile, from near the boundary behavior of the closed solution of (13),
$ f_0(u\rightarrow0,r) \approx \bar{C}_{\Delta_+}\, \left( \frac{u}{r^{2}} \right)^{\Delta_+}, $
(38) we consider
$ f(r)=\bar{C}_6/{r^{12}} $ to rewrite the series solutions clearly.Moreover, we can use (B8) with (B9) and near the boundary behavior of the closed solution of (B7), yielding
$ g_0(u\rightarrow 0,r) = \frac{2}{\sqrt{3}} \frac{{b}_0}{(a_0^2-b_0^2+ r^2)} \left[1 - \frac{2\, a_0}{(a_0^2-b_0^2 + r^2)}\, u \right], $
(39) as the initial data, which might also be obtained from the LHS relation in (B1), in the ADM, to obtain approximate solutions. As a result, we arrive at a series solution about
$ u=0 $ , up to the first iteration of ADM or NLO of the expansion, as$ f^{(1)}(u,r) = \sum\limits_{\Delta_+=1}^6\, \frac{\mathcal{H}_{\Delta_+}(r, a_0, b_0, m)\, u^{\Delta_+}}{(a_0^2-b_0^2 + r^2)^{\Delta_+}}, $
(40) where
$ \mathcal{H}_{\Delta_+}(r, a_0, b_0, m) $ is a polynomial of its arguments; in particular, for the term corresponding to the bulk mode$ m^2=18 $ , it becomes$ \mathcal{H}_{6} = - \frac{64}{\sqrt{3}}\, a_0^3\, b_0^3. $
(41) -
First, we remind that the truncation here is consistent, considering that our ansatz (1) is H-singlet, given that
$ e_7 $ , J and the (pseudo)scalars in resulting equations respect the same symmetry. Second, the setups here are as if we add$ \ell $ probe (anti)M-branes to the (WR)SW M2-branes background and so, the resultant theory is for anti-M2-branes with the quiver gauge group of$S U(N+\ell)_k \times S U(N)_{-k}$ . Indeed the (anti)M-branes wrap around mixed internal and external directions, breaking all SUSY's and parity.14 and that to realize the latter we focus on$ U(1) \subset U(\ell) $ part of the gauge group (in the large k limit) and keep G as a spectator- a so-called novel Higgs mechanism; see for instance [25].15 Third, the bulk settings break the inversion (and so, the special conformal transformation$ K_\mu $ ) symmetry and scale-invariance (denoted by the dilation operator D) because of the mass and nonlinear terms in the bulk action and translational-invariance (denoted by the translation operator$ P_\mu $ ) due to the non-constant solutions.16 As a result, the conformal symmetry$S O(4,1)$ (as the isometry of$ EAdS_4 $ ) breaks into$S O(4)$ , which in turn includes six generators consisting of three Lorentz transformations (denoted by the operator$ L_{\mu \nu} $ ) and$ R_\mu \approx \left(K_\mu + a^2 P_\mu \right) $ 17 corresponding to rotations on$ S^3 $ , where a is the scale parameter. The four generators of the broken symmetries (translations and scale transformations)- and therefore the four free parameters a (or$ b_0 $ ) and$ \vec{u}_0 $ - move$S O(4)$ - symmetric ($S O(3, 1)$ in Lorentzian signature) bubble around the 4D bulk.In contrast, the mass spectrum of 11D SUGRA over
$ AdS_4 \times S^7/Z_k $ 18 includes three generations of scalars ($ 0_1^+, 0_2^+, 0_3^+ $ ) and two generations of pseudoscalars ($ 0_1^-, 0_2^- $ ). In fact, the massless multiplet ($ n=0 $ ) includes a graviton ($ {\bf{1}} $ ), a gravitino ($ {\bf{8}}_s $ ), 28 spin-1 fields ($ {\bf{28}} $ ), 56 spin-$ \frac{1}{2} $ fields ($ {\bf{56}}_{s} $ ), 35 scalars ($ {\bf{35}}_{v} $ ) of$ 0_1^+ $ arising from the external components ($ \mathcal{A}_{\mu \nu \rho} $ ), and 35 pseudoscalars ($ {\bf{35}}_{c} $ ) of$ 0_1^- $ arising from the internal components ($ \mathcal{A}_{m n p} $ ), without any H-singlet under the branching$ G\rightarrow H $ for scalars ($ {\bf{35}}_v \rightarrow \bar{{\bf{10}}}_{-2} \oplus {\bf{10}}_{2} \oplus {\bf{15}}_{0} $ ) and pseudoscalars ($ {\bf{35}}_c \rightarrow {\bf{10}}_{-2} \oplus \bar{{\bf{10}}}_{2} \oplus {\bf{15}}_{0} $ ).19 In massive or higher KK multiplets ($ n>0 $ ), the massless ($ m^2=0 $ ) pseudoscalar and scalar set in$ \acute{{\bf{840}}}_{s} $ of$ 0_1^- $ with$ n=2 $ and$ {\bf{1386}}_{v} $ of$ 0_1^+ $ with$ n=4 $ of G, again without any H-singlet under the branching. In the same manner, the massive ($ m^2=18 $ ) pseudoscalar sets in$ \acute{{\bf{840}}}_{c} $ of$ 0^-_2 $ with$ n=4 $ and$ {\bf{75075}}_{vc} $ of$ 0^-_1 $ with$ n=6 $ of G, while as scalar, it sets in$ {\bf{30940}}_{v} $ of$ 0^+_1 $ with$ n=10 $ and$ {\bf{23400}}_{v} $ of$ 0^+_3 $ with$ n=6 $ , as well as$ {\bf{1}} $ of$ 0^+_2 $ with$ n=2 $ of G, again without any H-singlet under the branching, except for the last one$ {\bf{1}}(0,0,0,0) \rightarrow {\bf{1}}_{0}[0,0,0]$ .However, because of the triality of G
20 , one can exchange its three inequivalent reps$ {\bf{8}}_v $ ,$ {\bf{8}}_s $ , and$ {\bf{8}}_c $ . In fact, to find the desired singlet modes and realize SUSY breaking in the boundary theory, we swap the three reps21 . Therefore, swapping$ {\bf{8}}_s \leftrightarrow {\bf{8}}_c $ and keeping$ {\bf{8}}_v $ constant, which means exchanging spinors(supercharges) with fermions and keeping scalars unchanged, the massless and massive pseudoscalar reps change accordingly without any H-singlet under the branching of the resulting reps, while the scalar reps do not change. In the same manner, after swapping$ {\bf{8}}_s \leftrightarrow {\bf{8}}_v $ and keeping$ {\bf{8}}_c $ constant, which means exchanging spinors with scalars and keeping fermions unchanged, the resulting reps of both modes as pseudoscalar do not include any H-singlet under the branching. However, we have$ {\bf{1386}}_{s} $ and$ {\bf{30940}}_{s} $ ,$ {\bf{23400}}_{s} $ from the massless and massive scalar modes, respectively, while rep$ {\bf{1}} $ of$ 0^+_2 $ with$ n=2 $ of G remains the same as before, with the latter swapping. For the latter reps, the branching$ G \rightarrow H $ reads$ \begin{aligned}[b] {\bf{30940}}_{s} \rightarrow {\bf{1}}_{0} &\oplus \acute{{\bf{20}}}_{0} \oplus {\bf{105}}_{0} \oplus {\bf{336}}_{0} \oplus \acute{{\bf{825}}}_{0} \oplus \acute{{\bf{1716}}}_{0} \oplus {\bf{3185}}_{0} \oplus ... \, , \\ {\bf{23400}}_{s} \rightarrow {\bf{1}}_{0}& \oplus {\bf{15}}_{0} \oplus 3\, (\acute{{\bf{20}}}_{0}) \oplus {\bf{84}}_{0} \oplus 3\, ({\bf{105}}_{0}) \oplus 2\,({{\bf{175}}}_{0}) \\ & \oplus {{\bf{336}}}_{0} \oplus {{\bf{729}}}_{0} \oplus 2\,({{\bf{735}}}_{0}) \oplus {{\bf{3640}}}_{0} \oplus ...\ , \\[-15pt]\end{aligned} $
(42) where we have only written
$ U(1) $ -neutral reps. The corresponding reps for$ {\bf{1386}}_{s} $ remain the same as the first four terms of the reps above for$ {\bf{30940}}_{s} $ under the branching. As a result, we see that, after exchanging$ {\bf{s}} \leftrightarrow {\bf{v}} $ , the desired H-singlet rep ($ {\bf{1}}_{0} $ ) occurs for both massless and massive (pseudo)scalars we consider here.In contrast, a bulk (pseudo)scalar with near the boundary behaviour of (25) could be quantized with either the Neumann or alternate (
$ \delta\beta =0 $ ) boundary condition for the masses in the range of$ -9/4\leq m^2 \leq -5/4 $ or the Dirichlet or standard ($ \delta\alpha=0 $ ) boundary condition that can in turn be applied to any mass (see for instance [40, 41]), while the regularity (that$ \Delta_+ $ is real) and stability require that the mass is above the Breitenlohner–Freedman (BF) bound$ m^2\geq m_{\rm BF}^2=-9/4 $ [42, 43]. As a result, for the massless and massive modes, only mode β is normalizable; α and β have holographic expositions as source and vacuum expectation value of the one-point function of the operator$ \Delta_+ $ , and vice versa for the operator$ \Delta_- $ . Then, we write the Euclidean AdS/CFT dictionary as$ \begin{aligned}[b] & \langle \mathcal{{O}}_{\Delta_+} \rangle_{\alpha} = - \frac{\delta W[\alpha]}{\delta\alpha} = \beta, \quad \langle \mathcal{{O}}_{\Delta_-} \rangle_{\beta} = - \frac{\delta \tilde{W}[\beta]}{\delta\beta}= \alpha, \\ & \tilde{W}[\beta] = - W[\alpha] - \int {\rm d}^3 \vec{u}\ \alpha(\vec{u})\, \beta(\vec{u}), \end{aligned} $
(43) where
$ W[\alpha] $ ($ \tilde{W}[\beta] $ ) is the generating functional of the connected correlator of the operator$ \mathcal{{O}}_{\Delta_+} $ ($ \mathcal{{O}}_{\Delta_-} $ ) on the usual (dual) boundary CFT$ _3 $ with$ {\Delta_+} $ ($ \Delta_- $ ) quantization. -
The bulk setups with the symmetries discussed in the previous section, including parity breaking, are dual to the boundary CS
$ O(N) $ or$ U(N) $ interacting vector models.22 However, we usually consider elements of ABJM's model with, depending on the case, only one scalar (say$ Y =\varphi= h(r)\,{\bf{I}}_N $ , with$ h(r) $ as the scalar profile) or fermion (say ψ)23 resulting in zero scalar and fermion potentials, and the following deformation:$ \mathcal{L}^{(p)} = \mathcal{L}_{CS}^+ - \mathrm{tr} \left({\rm i} \bar{\psi}\, \gamma^k D_k \psi \right) - \mathrm{tr}\left(D_k Y^{\dagger} D^k Y \right) - \mathcal{W}_\Delta^{(p)}, $
(44) where the CS Lagrangian reads
$ \mathcal{L}_{CS}^+ = \frac{{\rm i} k}{4\pi}\ \varepsilon^{ij k}\ \mathrm{tr} \left(A_i^+ \partial_j A_k^+ + \frac{2i}{3} A_i^+ A_j^+ A_k^+ \right), $
(45) which is attributed to the remaining
$ U(1) $ part of the original quiver gauge group discussed in Section 424 .$D_k \Phi = \partial_k \Phi + {\rm i} A_k\, \Phi - {\rm i} \Phi\, \hat{A}_k$ ,$F_{ij}=\partial_i A_j - \partial_j A_i + {\rm i} \left[A_i, A_j \right]$ , and$ \mathcal{W}_\Delta^{(p)} $ , whose integral is W, as depicted in (43), which stands for (with p marking) deformations we make with various H-singlet operators. -
For the bulk solutions in subsections II.A and II.B, arising from taking the backreaction, which correspond to (exactly and irrelevant
25 ) marginal operators, except for the$ \Delta_+=3 $ operators of$ \mathcal{O}_{3}^{(a)} =\mathrm{tr}(\varphi \bar{\varphi})^3 $ ,$ \mathcal{O}_{3}^{(b)} =\mathrm{tr}(\varphi \bar{\varphi}) \mathrm{tr}(\psi \bar{\psi}) $ ,$ \mathcal{O}_{3}^{(c)}=\mathrm{tr}({A} \wedge {F}) $ , and$ \mathcal{O}_3^{(d)}=\mathrm{tr}(\varphi \bar{\varphi})\, \varepsilon^{ij}\, F_{ij}^+ $ already considered in [2−5, 16, 17, 45], we include two new ones:$ \begin{array}{*{20}{l}} \mathcal{O}_{3}^{(e)}= \mathrm{tr}(\varphi \bar{\varphi})^2\, \varepsilon^{k ij} \varepsilon_{ij}\, A_k^+, \quad \mathcal{O}_{3}^{(f)}= \mathrm{tr}(\psi \bar{\psi})\, \varepsilon^{k ij} \varepsilon_{ij}\, A_k^+. \end{array} $
(46) Next, we consider the following deformation:
$ \begin{array}{*{20}{l}} \mathcal{W}_3^{(abf)}=\lambda_6\, \mathcal{O}_{3}^{(a)} + \hat{\lambda}_6\, \mathcal{O}_{3}^{(b)} + \check{\lambda}_6\, \mathcal{O}_{3}^{(f)}, \end{array} $
(47) where the λ's are the coupling constants. We set
$ \alpha = 1 $ for now. Then, if we take both CS terms$ \mathcal{L}_{CS}+ \hat{\mathcal{L}}_{CS} $ instead of$ \mathcal{L}_{CS}^+ $ in (44), after some mathematical manipulations on the resultant scalar$ \bar{\varphi}=\varphi^{\dagger} $ , fermion$ \bar{\psi} $ , and gauge$ A_k^+ $ field equations, we get$ \begin{array}{*{20}{l}} \partial_k \partial^k \varphi -3\, \lambda_6\, \varphi^5 =0, \end{array} $
(48) $ {\rm i}\, \bar{\psi} \gamma^k \partial_k \psi + 2\, \bar{\psi} \gamma^k \psi\, A_k^+ + \frac{{\rm i} k}{4\pi}\, \varepsilon^{ijk} F_{ij}^+\, A_k^+ =0, $
(49) where
$ Y=Y^{\dagger} $ and$ A_i^-=0 $ are also set. Subsequently, a closed solution for (48) reads$ h = \left( \frac{1}{g_6} \right)^{1/4} \left( \frac{a}{a^2 + (\vec{u}-\vec{u}_0)^2} \right)^{1/2}, $
(50) where
$ g_6 \equiv -\lambda_6 $ . Accordingly, by employing the ansatz$ \begin{array}{*{20}{l}} A_k^+ = \varepsilon_{k ij}\, \varepsilon^{ij} A^+(r), \end{array} $
(51) where
$ A^+(r) $ is a scalar function on the boundary, a solution for (49) reads$ A^+ = \frac{3}{4} \left( \frac{a}{a^2 + (\vec{u}-\vec{u}_0)^2} \right), $
(52) $ \psi= \tilde{a} \left( \frac{{a} + {\rm i} (\vec{u} - \vec{u}_0). \vec{\gamma} }{\left[ {a}^2 + (\vec{u} - \vec{u}_0)^2 \right]^{\varsigma=3/2}} \right) \chi, $
(53) where
$ \vec{\gamma}=(\sigma_2, \sigma_1, \sigma_3) $ are the Euclidean gamma matrices, and χ with$ \chi^{\dagger} \chi=1 $ is a constant dimensionless spinor26 . Finally, from computing the corresponding boundary action$ \begin{aligned}[b] {S}_{(3)}^{\rm modi.} =& - \int ( \left(\partial_i \varphi \right)^2 -2\, \mathrm{tr}(\bar{\psi} \gamma^3 \psi) A_3^+ + \lambda_6\, \varphi^6 \\&+ \hat{\lambda}_6\, \varphi^2\, \mathrm{tr}(\psi \bar{\psi}) + 12\, \check{\lambda}_6\, A^+\, \mathrm{tr}(\psi \bar{\psi}) ) \end{aligned} $
(54) based on the solutions (50), (52), and (53) and setting the couplings equal to 1 and
$ \tilde{a}= a = a^{\dagger} $ for simplicity, we get$ {S}_{(3)}^{\rm modi.} = - 36 \int_0^\infty \frac{\pi\, a^3\, r^2}{\left( {a}^2+r^2 \right)^3}\, {\rm d}r = - \frac{9}{4}{\pi^2}, $
(55) 27 which is finite, indicating an instanton with size$ {a} \geq 0 $ at the origin ($ \vec{u}_0=0 $ ) of a three-sphere with radius r at infinity ($ S_\infty^3 $ ).Consequently, as a basic test of the correspondence, wrt (43), we have
$ \langle \mathcal{O}_{3}^{(a,b,f)} \rangle_{{\alpha}} = a_1 \left( \frac{{a}}{{a}^2+(\vec{u} - \vec{u}_0)^2} \right)^3, $
(56) with
$ a_1, a_2,... $ being the boundary constants, as compatible with near the boundary behavior of (34) with$ \Delta_+ =3 $ wrt (25), in the limit of$ {a}\rightarrow 0, r\rightarrow \infty $ . Meanwhile, compared with the bulk closed solution of (13), this boundary solution may be considered as an instanton sitting at the conformal point of$ u =a $ .28 Note also that, with
$ \hat{\lambda}_6=\check{\lambda}_6=0 $ in (47) and including a mass-deformation term ($ m_b^2\, \mathrm{tr}(\varphi \bar{\varphi}) $ ),29 we have, in general, the RB model (look also at [4]), whose$ \bar{\varphi} $ equation reads$ \begin{array}{*{20}{l}} \left( \partial_i \partial^i -m_b^2 \right) h + 3\, g_6\, h^5 = 0. \end{array} $
(57) Solutions for its free massive equation are available in terms of (modified) Bessel functions; an explicit expression is stated as
$ h_c(r) \cong \frac{a_2}{\sqrt{m_b}} \frac{{\rm e}^{-m_b\, r}}{r}, $
(58) which satisfies the condition
$ h_c(r\rightarrow \infty)\rightarrow 0 $ , resulting in a finite action. Solutions for the interaction equation could be obtained in the context of constrained instantons; see [21, 54−56 ]. In fact, considering (58) as the initial data, one may employ perturbative methods and get solutions with a simple structure, such as$ h \sim 1/r $ . Thus, we have the single-operator correspondence$ \langle \mathcal{O}_{3}^{(a)} \rangle_{\alpha} \sim {1}/{r^6} $ with the typical near the boundary solution of (34) for$ \Delta_+ =3 $ .In particular, if we use only the operator
$ \mathcal{O}_{3}^{(e)} $ to deform the action of (44), discarding its fermion kinetic term, the equations for$ \bar{\varphi} $ and$ A_k^+ $ read$ \partial_i \partial^i \varphi -2 \varphi\, \mathrm{tr}(\varphi \bar{\varphi})\, \varepsilon^{k ij} \varepsilon_{ij}\, A_k^+=0, $
(59) $ \frac{{\rm i} k}{4\pi} \varepsilon^{k ij} F^+_{ij} - \mathrm{tr}(\varphi \bar{\varphi})\, \varepsilon^{k ij} \varepsilon_{ij} + {\rm i} \left[\varphi \left(\partial^k \bar{\varphi} \right)- \left(\partial^k \varphi \right) \bar{\varphi}\right]=0, $
(60) respectively. Next, with
$ \varphi=\bar{\varphi} $ , from the last two equations, we can write$ \partial_i \partial^i h(r)=0 \Rightarrow h(r)= {a}_3 + \frac{{a}_4}{r}, $
(61) while for the gauge field, we may use the ansatz (51) with
$ A^+(r) \sim 1/r^2 $ . Thus, with$ {a}_3=0 $ , the basic correspondence$ \langle \mathcal{O}_{3}^{(e)} \rangle_{\alpha}\sim {1}/{r^6} $ is realized, with near the boundary solution of (34) with$ \Delta_+ =3 $ wrt (25). In contrast, if$ \varphi \neq \bar{\varphi} $ , which is allowed due to being in Euclidean space, and explicitly with$ \varphi = h(r)\, I_N, \quad \varphi^{\dagger} = {a}_5\, I_N, $
(62) from Eqs. (59) and (60), we get
$ \begin{array}{*{20}{l}} \partial_k \left(\varepsilon^{k ij} F^+_{ij} \right)=0, \quad A_k^+ = {\rm i}\, \partial_k \ln h. \end{array} $
(63) The latter solution is reminiscent of the duality (
$ {A}_k=\eta_{k j}\, \partial^j h/h $ ) between the instanton solution of the pure$S U(2)$ from Yang-Mills theory [57],$ {A}_k \approx \eta_{k j}\, \frac{(x-x_0)^j}{a^2+(x-x_0)^2} \Rightarrow {F}_{i j} \approx \eta_{i j} \left(\frac{a}{a^2+(\vec{u}-\vec{u}_0)^2}\right)^2, $
(64) with
$ \eta_{i j} $ as 't Hooft symbols [58], and the$S O(4)$ -invariant solution of the so-called$ \varphi^4 $ model, expressed as30 $ \nabla^2 h + \lambda_4\, h^3=0 \Rightarrow h = \sqrt{\frac{8}{\lambda_4}} \left( \frac{a}{a^2 + (\vec{u}-\vec{u}_0)^2} \right). $
(65) As a result, with the latter solutions, we have the same correspondence as (56) for
$ \mathcal{O}_{3}^{(e)} $ . -
For the Higgs-like (pseudo)scalar
$ m^2=18 $ , except for the$ \Delta_+=6 $ operators introduced in [2], including$ \mathcal{O}_{6}^{(b)}= (\mathcal{O}_{3}^{(b)})^2 $ (a double-trace deformation), we consider a few new ones. The operator$ \mathcal{O}_{6}^{(a)}=\mathrm{tr}(\psi \bar{\psi})^3 $ , first considered in [4], is interesting in that it can also be taken in the CF model. In fact, if we consider the deformation$ \begin{array}{*{20}{l}} \mathcal{W}_6^{(a)}=m_f\, \mathcal{O}_{2}^{(a)} + \tilde{g}_6\, \alpha\, \mathcal{O}_{6}^{(a)}, \end{array} $
(66) where
$ \mathcal{O}_{2}^{(a)}=\mathrm{tr}(\psi \bar{\psi}) $ , by excluding the scalar kinetic term in (44), the$ \bar{\psi} $ equation reads$ \begin{array}{*{20}{l}} {\rm i}\, \gamma^k \partial_k \psi + m_f\, \psi +3\, \tilde{g}_6\, \alpha\, \psi\, \mathrm{tr}(\psi \bar{\psi})^2 =0. \end{array} $
(67) The solution of (53) with
$ \tilde{a}=a/(\tilde{g}_6)^{1/4} $ is also valid for the latter equation provided that$ \begin{array}{*{20}{l}} \alpha = \mathrm{tr}(\psi \bar{\psi})^{-3/2} \end{array} $
(68) for the massless case (
$ m_f = 0 $ ) and$ m_f \rightarrow \tilde{\alpha}(\vec{u})= \mathrm{tr}(\psi\bar{\psi})^{1/2} $ for the massive case. Thus, the deformation might in fact be considered as a triple-trace one. As a result,$ \langle \mathcal{O}_{6}^{(a)} \rangle_{{\alpha}} = \left( \frac{\tilde{a}}{{a}^2+(\vec{u} - \vec{u}_0)^2} \right)^6, $
(69) which, wrt (25), corresponds to (40) for
$ \Delta_+=6 $ , considering (41),$ a_0^2-b_0^2=a^2 $ , and$ - a_0\, b_0 \sim \tilde{a}^2 $ . Moreover, we can obtain an explicit profile for$ f(r) $ in (30) from this solution. In fact, according to the above discussions, with$ \Delta_\pm =6, -3 $ and the correspondence rules of (43), we get$ f(r)= \left[ \frac{\tilde{a}}{a^2 + r^2} \right]^{1/2}, $
(70) where
$ C_- =\sqrt{C_+} $ , with$ C_+=1 $ for simplicity.It is also interesting to check the BF duality (or 3D Bosonization)- see for instance [66, 67]- from our setups attributed to RB and CF models
31 at the level of the solutions. Indeed, under the BF duality, the coupling of$ \tilde{g}_6\, \mathrm{tr}(\psi \bar{\psi})^3\sim \tilde{g}_6\, \sigma_f^3 $ , where$ \sigma_f $ is the so-called Hubbard-Stratonovich field, is mapped into the coupling of$ {g}_6\, \mathrm{tr}(\varphi \bar{\varphi})^3 $ ($ \mathcal{W}_3^{(a)} $ of (47)); see [68, 69].32 In this regard, from the solutions of the boson model (50) and fermion model (53), we have$ \mathrm{tr}({\psi} \bar{{\psi}}) = \left(\frac{1}{\tilde{g}_6} \right) \left( \frac{a}{{a}^2+(\vec{u} - \vec{u}_0)^2} \right)^2= \mathrm{tr}(\varphi \bar{\varphi})^2, $
(71) where
$ \tilde{g}_6 \leftrightarrow g_6 $ , thus realizing the BF duality with$ \psi \leftrightarrow \varphi^2 $ or$ \psi \leftrightarrow \varphi $ when including α in the fermion model.In this way, we now examine two new operators
$ \begin{array}{*{20}{l}} \mathcal{O}_{6}^{(c)}= \mathrm{tr}(\psi \bar{\psi})^2\, \varepsilon^{ij}\, F_{ij}^+, \quad \mathcal{O}_{6}^{(d)}= \mathrm{tr}(\psi \bar{\psi})\, \mathrm{tr}(F_{ij}^+\, F^{+\, ij}); \end{array} $
(72) with the associated deformations
$ \begin{array}{*{20}{l}} \mathcal{W}_6^{(q)}=\alpha\, \mathcal{O}_{6}^{(q)}, \end{array} $
(73) where
$ q=c,d,...,h $ from now on. Next, discarding the scalar kinetic term of (44), with$ \mathcal{W}_6^{(c)} $ , the fermion$ \bar{\psi} $ and gauge$ A_i^+ $ equations read$ \begin{array}{*{20}{l}} {\rm i}\, \gamma^k \partial_k \psi + 2\, \alpha\, \psi\, \mathrm{tr}(\psi \bar{\psi}) \varepsilon^{ij} F_{ij}^+ =0, \end{array} $
(74) $ \frac{{\rm i} k}{4\pi} \varepsilon^{k ij} F^+_{ij} + 2\, \bar{\psi}\, \gamma^k\, \psi= 0, $
(75) respectively, reminding that the second term on the LHS of (75) exists when we include both CS terms (for
$ A_i $ and$ \hat{A}_i $ ) in (44) and that$ F_{ij}^-=0, A_i^-=0 $ is set. With only the CS term of (45), the ansatz$ \begin{array}{*{20}{l}} A_\mu^+ = \omega_{\mu \nu}\, x^\nu A(r), \quad \omega_{\mu \nu}= \left\{ \begin{split} & 1 \ \ \ : \ \nu>\mu, \\ & 0 \ \ \ : \ \nu=\mu,\ \ \mu,\nu \neq i, j , \end{split} \right. \end{array} $
(76) for the
$ U(1) $ gauge field, with$ \mu,\nu $ for the boundary indices as well and$ A(r) $ as another boundary scalar function, we obtained the following desired solution (see also [5]):$ A(r)= \frac{{a}_6 + 4\, a_7\, r}{4\, r^4} \Rightarrow \varepsilon^{ij} F_{ij}^+ \equiv F^+ = \frac{a_6}{r^4}. $
(77) In this case, a solution for ψ is taken from (53) with
$ a=0 $ and$\tilde{a} = \dfrac{\rm i}{2} \sqrt[3]{\dfrac{4}{5}}$ . As a result,$ \langle \mathcal{O}_{6}^{(c)} \rangle_{\alpha} = \frac{a_6\, \tilde{a}^4}{r^{12}} \cong f(r), $
(78) with
$ f(r) $ in (37). Accordingly, one can also adjust$ \bar{C}_6 =a_6\, \tilde{a}^4 $ of (38) wrt (25).However, combining (74) and (75), we get
$ \gamma^k\, \partial_k \psi + \frac{16 \pi}{k}\, \alpha\, \mathrm{tr}(\psi \bar{\psi})^2\, \gamma^3\, \psi =0, $
(79) where taking the third component of the gamma matrices is for compatibility with the solution we take for ψ, which in turn reads from (53) with
$\tilde{a} =a \sqrt[1/2]{\dfrac{-3 {\rm i} k}{16 \pi}}$ . Thus, from (74), we have$ F_{ij}^+ = a_8\, \varepsilon_{ij} \left( \frac{a}{{a}^2+(\vec{u} - \vec{u}_0)^2} \right)^2, $
(80) where
$a_8=(3 \pi {\rm i}/k)^{1/2}$ , reminding that$ F^+(r\rightarrow \infty) \rightarrow 0 $ .33 As a result,$ \langle \mathcal{O}_{6}^{(c)} \rangle_{{\alpha}} = \frac{3}{2} \frac{a^2\, \tilde{a}^2}{\left[a^2+(\vec{u} - \vec{u}_0)^2\right]^6}, $
(81) which, for
$ \Delta_+=6 $ , can be made to correspond to (40) with (41) and to (13) with an instanton at the conformal point of$ u=a $ .Moreover, to confirm the instanton nature of the Euclidean solutions, we compute the value of the corresponding action as follows:
$ S_{(6c)}= \int \mathcal{W}_6^{(c)}\, {\rm d}^3\vec{u} \quad \Rightarrow S_{(6c)}^{\rm modi.}= \frac{3 \pi^2}{8} \frac{a}{\tilde{a}}, $
(82) where we have used the result of the integral presented in (55) and the same interpretation.
Similarly, for the deformation
$ \mathcal{W}_6^{(d)} $ of (73), discarding the scalar kinetic term of (44) and taking both CS terms, the fermion$ \bar{\psi} $ and gauge$ A_i^+ $ equations read$ {\rm i}\, \gamma^k \partial_k \psi + \alpha\, \psi\, \mathrm{tr}(F_{ij}^+\, F^{+\, ij}) =0, $
(83) $ \frac{{\rm i} k}{4\pi} \varepsilon^{k ij} F^+_{ij} + 2\, \bar{\psi}\, \gamma^k\, \psi +4\, \alpha\, \mathrm{tr}(\psi \bar{\psi})\, \partial_j F^{+\, jk}= 0. $
(84) Then, using (68), solutions for the fermion and gauge fields are read from (53) with
$\tilde{a} =a \sqrt[1/2]{\dfrac{-9 {\rm i} k}{8 \pi}}$ and from (80) with$ a_8=1 $ . However, if we set$ \alpha=1 $ in the equations, a solution for ψ is read from (53) with$ \varsigma=0 $ instead of$ 3/2 $ along with the gauge solution (80) with$ a_8=1/a $ to obtain$ \tilde{a} $ the same as before. As a result, we have$ \langle \mathcal{O}_{6}^{(d)} \rangle_{{\alpha}} = \frac{a^2\, \tilde{a}^2}{\left[a^2+(\vec{u} - \vec{u}_0)^2\right]^3}, $
(85) which corresponds to the bulk near the boundary solution of (35) with
$ \check{C}_{-3}=0 $ and$ \hat{C}_{6}=a^2\, \tilde{a}^2 $ , in the limit of$ {a}\rightarrow 0, r\rightarrow \infty $ (see footnote 5 on page 6).Another operator we consider is
$ \mathcal{O}_{6}^{(e)}= \mathrm{tr}(\varphi \bar{\varphi})^4\, \varepsilon^{ij}\, F_{ij}^+; $
(86) we deform the action of (44), discarding its fermion term, with (73) having
$ q=e $ . As a result, the scalar$ \bar{\varphi} $ equation reads$ \begin{array}{*{20}{l}} \partial_i \partial^i \varphi - 4\, \alpha\, \varphi\, \mathrm{tr}(\varphi \bar{\varphi})^3\, \varepsilon^{ij}\, F_{ij}^+=0, \end{array} $
(87) where the gauge
$ A_i^+ $ equation is the same as (60) apart from omitting the middle term on the LHS. Next, with$ \varphi = \bar{\varphi} $ , we can obtain for the gauge part a similar solution to (80) with$ a_8=1 $ . Then, taking$ \alpha \sim \mathrm{tr}(\varphi \bar{\varphi})^{-3} $ and$ F^+ \sim h^4 $ , we obtain a similar solution to (50) for h with$ g_6=1 $ ; thus, the same correspondence as (69) with$ a=\tilde{a} $ for$ \mathcal{O}_{6}^{(e)} $ is confirmed.However, when
$ \varphi \neq \bar{\varphi} $ we take (62). Subsequently, from (87) and (60) without the middle term on the LHS of the latter, we can write$ \partial_k \left(\varepsilon^{k ij} F^+_{ij} \right) - \frac{16 \pi}{k}\, \alpha\, \mathrm{tr}(\varphi \bar{\varphi})^4\, \varepsilon^{ij} F^+_{ij}=0. $
(88) Then, using the ansatz (76),
$F^+= -2 \left(6 A(r) + 2 r \acute{A}(r) \right)$ , with$ a_5 =\sqrt[1/2]{\dfrac{-3 \sqrt{3} k}{16 \pi}} $ and$ \alpha=1 $ , we get$ \frac{{\rm d}^2 A(r)}{{\rm d}r^2} + \left( \tilde{h}(r) + \frac{4}{r} \right) \frac{{\rm i}A(r)}{{\rm i}r} + \frac{3}{r}\, \tilde{h}(r)\, A(r)=0, $
(89) for which we can write the following solution:
$ \begin{aligned}[b] h^4(r) \equiv \;& \tilde{h}(r) = \frac{n}{r} \Rightarrow A(r) = \frac{{a}_7}{r^3} + \frac{{a}_9}{r^n} \Rightarrow F^+ \\=&\frac{4\, a_9}{r^n} (n-3), \end{aligned} $
(90) with n being a real number. As a result, we have
$ \langle \mathcal{O}_{6}^{(e)} \rangle_{{\alpha}} = \frac{4\,n\, a_5^4\, a_9}{r^{n+1}} (n-3), $
(91) which, with
$ n=5 $ wrt (25), corresponds to the normalizable part of the bulk solution (35), after adjusting the constants of both sides. In the same manner, with$ n=11 $ , the expression can be made to correspond to$ f(r) $ in (37).Among other similar operators, if we use any of the following three operators
$ \begin{aligned}[b] & \mathcal{O}_{6}^{(f)} = \mathrm{tr}(\varphi \bar{\varphi})\, \mathrm{tr}(\psi \bar{\psi})^2\, \varepsilon^{ijk}\, \varepsilon_{ij}\, A_k^+, \\& \mathcal{O}_{6}^{(g)} = \mathrm{tr}(\varphi \bar{\varphi})^2\, \mathrm{tr}(\psi \bar{\psi})\, \varepsilon^{ij}\, F_{ij}^+, \\ & \mathcal{O}_{6}^{(h)} = \mathrm{tr}(\varphi \bar{\varphi})\, \mathrm{tr}(\psi \bar{\psi})\, \varepsilon^{ijk}\, F_{ij}^+\, A_k^+, \end{aligned} $
(92) to deform (44) wrt (73), the fermion solution may correspond to (53) with
$ a=0 $ , scalar solution may correspond to (61), gauge solution may correspond to$ F^+(r) $ in (77), and$ A^+(r) \sim 1/r^2 $ (for$ \mathcal{O}_{6}^{(f)} $ ) according to the ansatz of (51). Consequently, as a primary test of the correspondence,$ \langle \mathcal{O}_{6}^{(f,g,h)} \rangle_{{\alpha}} \sim 1/r^{12} $ matches with the bulk solution (37) and (38) for$ \Delta_+=6 $ as before. -
We use the Einstein's equations
$ \mathcal{R}_{MN} - \frac{1}{2} g_{MN} \mathcal{R} = 8 \pi\, \mathcal{G}_{11} T_{MN}^{{G}_4}, \tag{A1} $
where
$ T_{MN}^{{G}_4} = \frac{1}{4!} \left[4\, {G}_{MPQR}\, {G}_N^{PQR} - \frac{1}{2} g_{MN}\, {G}_{PQRS}\, {G}^{PQRS} \right], \tag{A2} $
and the capital
$ M, N,... $ , small$ m, n,... $ , and Greek$ \mu, \nu,.... $ indices are for the entire 11D, 6D internal$ CP^3 $ , and 4D external$ AdS_4 $ spaces, respectively.Next, using the conventions and performing computations similar to those expressed in Appendix B of [3], we get
$ {G}_{PQRS}\, {G}^{PQRS} = 96 \left[ \frac{8}{3 R^8} \bar{f}_1^2 + \frac{R^2}{32} (\partial_\mu f_2)(\partial^\mu f_2) + \frac{1}{8} {f}_3^2\right], \tag{A3} $
$ {G}_{\mu PQR}\, {G}_\nu^{PQR} = \frac{64}{R^8} \bar{f}_1^2\, g_{\mu \nu} + \frac{3 R^2}{4} (\partial_\mu f_2)(\partial_\nu f_2), \tag{A4} $
$ {G}_{m PQR}\, {G}_n^{PQR} = \left[2 {f}_3^2 + \frac{R^2}{4} (\partial_\mu f_2)(\partial^\mu f_2) \right] g_{m n} , \tag{A5} $
$ {G}_{7 PQR}\, {G}_7^{PQR} = \frac{3 R^2}{4} (\partial_\mu f_2)(\partial^\mu f_2)\ g_{7 7}, \tag{A6} $
with a
$ 4! $ factor for all terms.Then, by plugging (A3) with (A4), (A5), and (A6) back into (A2); using (3) with the conventions of (7); taking the traces; and using the Euler-Lagrange equation, we finally get
$ \Box_4 f_2 + 4\, m^2\, f_2 \pm\, {4}\, \delta\, f_2^2 + 4\, \lambda\, f_2^3 \pm 4\, F=0, \tag{A7} $
$ \Box_4 f_2 + \left( 3\, m^2 - \frac{8}{R^2} \right) f_2 \pm\, 3\, \delta\, f_2^2 + 3\, \lambda\, f_2^3 \pm 3\, F=\pm\, \frac{8\, C_2}{R^3}, \tag{A8} $
$ \Box_4 f_2 - m^2\, f_2 \mp \delta\, f_2^2 - \lambda\, f_2^3 \mp F=0, \tag{A9} $
for the external
$ AdS_4 $ space, internal$ CP^3 $ , and eleventh$ S^1/Z_k $ components, respectively, noting that (A9) is the same as the main expression in Eq. (6). -
The Adomian decomposition method or the inverse operator method [13] is a mathematical method specially used to solve NPDEs; see [79]. Because we are looking for solutions near the boundary (
$ u=0 $ ), we use$ f_0(0,r)= f(0,r) - u\, f_u (0,r), \quad f(0,r)= {f}(r)\, u^{\Delta_+} \tag{B1} $
as the initial data for the second order NPDE (22), corresponding to the Dirichlet boundary condition. Accordingly, from the main equation, we can write
36 $ \Box_4 f_0 - m^2\, f_0 =0, \tag{B2} $
with the closed solution of (13), and
$ \Box_4 f_{n+1} - m^2\, f_{n+1} = \sum\limits_{n=0}^\infty A_n, \tag{B3}$
where the nonlinear terms are written as the sum of Adomian polynomials
$ A_n $ 's,$ A_n=\frac{1}{n!} \frac{d^n}{{\rm d}\lambda^n} \left[\mathcal{F}\left(\sum\limits_{n=0}^n \lambda^n\, f_n \right) \right]_{\lambda=0}, \quad n=0,1,2,... \ , \tag{B4} $
with
$ \mathcal{F}(f) $ for the nonlinear terms of (28). As a result,$ \begin{aligned}[b] & \ A_0=6 f_0^3 - \delta\, f_0^2, \quad A_1= 18\, f_0^2\, f_1- 2\, \delta\, f_0^2\, f_1, \\ & A_2=18\, f_0^2\, f_2 + 18\,f_0\, f_1^2 - 2\, \delta\, f_0\, f_2- \delta\, f_1^2 , ... \ , \end{aligned}\tag{B5} $
where
$ \delta= 3\sqrt{3}\, m $ . Accordingly, a series solution up to the nth order of the iteration processes may be written according to (29).In contrast, with
$ f=(u/R_{\rm AdS})\, g $ and$R_{\rm AdS}=1$ from (22), we can write$ \left(\partial_i \partial_i + \partial_u \partial_u \right) g_0 - 6\, g_0^3 =0, \tag{B6} $
with the exact solution of
$ g_0(u,\vec{u}) ={\frac{2}{\sqrt{3}}}\, \left(\frac{b_0}{-b_0^2 + (u+a_0)^2 + (\vec{u}-\vec{u}_0)^2} \right), \tag{B7} $
which is indeed for the so-called conformally coupled (pseudo)scalar
$ m^2=-2 $ in the SW version of (4) with$ C_3=1 $ - with$ a_0, b_0 $ being physically meaningful constants and$ |\vec{u}-\vec{u}_0| \equiv r $ when using the spherical coordinates- and$ \left(\partial_i \partial_i + \partial_u \partial_u \right)\, g_{n+1}- 6\, g_{n+1}^3 =\sum\limits_{n=0}^\infty A_n, \tag{B8} $
with the Adomian polynomials
$ \begin{aligned}[b]& A_0 = \frac{(2+m^2)}{u^2}\, g_0 - \frac{3 \sqrt{3}\, m}{u}\, g_0^2, \\& A_1 = \frac{(2+m^2)}{u^2}\, g_1 - \frac{6 \sqrt{3}\, m}{u}\, g_0\, g_1, .... \ , \end{aligned}\tag{B9} $
to obtain series solutions near the boundary
$ u=0 $ .
Higgs-like (pseudo)scalars in AdS4, marginal and irrelevant deformations in CFT3, and instantons on S3
- Received Date: 2023-11-30
- Available Online: 2024-04-15
Abstract: Employing a 4-form ansatz of 11-dimensional supergravity over a non-dynamical