— Even-denominator fractional quantum Hall (FQH) states, most notably the half-filled second Landau level at \(\nu = 5/2\) [1], have attracted considerable attention because of their potential non-Abelian nature and applications for topological quantum computing. However, identifying the topological order of the state is a challenging problem [2]–[22], since multiple Abelian and non-Abelian phases [2], [23]–[28] can exist at half-filling with identical electrical conductivities.

Growing interest has focused on daughter states that emerge from the even-denominator states [29]–[32] at nearby fillings. Since the filling of the daughters depends on the topological order of their parent state [31], [33], [34], identifying these daughter states experimentally provides evidence about the parent state topological order. Experiments in wide GaAs quantum wells [35], [36] and graphene-based systems [37]–[42] have observed the daughter states corresponding to Pfaffian [23] and anti-Pfaffian [26] parent states.

Notably, the daughter states appear at the same filling as conventional Jain states. [36] demonstrated that by adjusting carrier density or tilting the magnetic field, a wide GaAs well at \(\nu=8/17\) and \(\nu=7/13\) can be driven between two different states that are believed to be a Jain state and a daughter state. The electrical Hall conductivity \(\sigma_{\mathrm{xy}}\) is identical in the two phases. However, for the Jain states at filling \(\nu=\frac{m}{pm\pm 1}\), the gaps decrease as the order \(m\) increases. In contrast, the observed daughter state energy gap (determined from the temperature dependence of the \(R_{\mathrm{xx}}\) minimum) is larger than the gap of neighboring Jain states at lower order. Moreover, an abrupt change of slope of \(R_{\mathrm{xx}}\) minimum as a function of the tuning parameter is observed. This poses a fundamental question: What is the nature of a quantum phase transition between these two topologically distinct phases?

In this Letter, we formulate a quantum critical theory describing the critical point between these two phases—the Jain and daughter states—as a critical point of \(U(1)^4\) Chern–Simons gauge theory with four Dirac cones obtained via parton construction. Specifically, as we show using the composite fermion picture, the Jain state can be viewed as a stack of a neutral \(E_8\) state (to be elaborated on below) and the daughter state; the process of stacking the \(E_8\) state changes \(\kappa_{\mathrm{xy}}\) by 8 but leaves \(\sigma_{\mathrm{xy}}\) unchanged. Consequently, the transition is equivalent to the transition between the \(E_8\) and the trivial bosonic insulator. We demonstrate, as visualized in 1, that direct continuous transition between the two requires fine-tuning; generically a cascade of intermediate fractionalized (i.e., topologically ordered) phases occurs that changes \(\kappa_{\mathrm{xy}}\) by minimal increments of \(2\), consistent with previously proposed scenarios involving minimal allowed changes in topological invariants [18], [43], [44]. We note that numerical treatment of these phases is a challenging problem since the existing methods [45]–[48] are not well-fitted for high filling (in the composite fermion case) or large denominators (in the fractional case).

Our results not only provide a framework for understanding the transition between the daughter and Jain states, but also predict a broader hierarchy of yet unobserved FQH states near half filling. Moreover, electronic systems near half filling provide an experimental platform for studying the transition between the \(E_8\) and trivial phases.

Quantum phase transitions take place at zero temperature [49], [50]. In the context of FQH, these transitions are topological, meaning there is no local order parameter that changes across the transition. While plateau transitions between states at different fillings have been extensively explored both theoretically [51]–[59] and experimentally [60]–[64], examples at constant filling remain limited and present a more complex landscape. The most prominent example is filling \(\nu=2/3\) [65]–[69], which, in addition to first-order transition [66], [70]–[72] between spin-polarized and spin-unpolarized states, is also predicted to host non-Abelian states [72]–[74]. More recently, experimental evidence for the transition between Abelian and non-Abelian states at \(\nu=1/2\) [75] was also found. Additional candidate transitions were theoretically proposed, described by anyon condensation [76]–[78] and effective Chern–Simons field theories [79]–[84].

We now review the understanding of the Jain and daughter phases within composite fermion theory. At even-denominator filling, the effective magnetic field seen by composite fermions vanishes, and they can form a paired superconducting state. Each of the possible even-denominator states is characterized by the integer-valued chiral central charge \(\mathcal{C}\) equal to the (signed) number of chiral neutral Majorana edge modes. Since each of these modes contributes \(\kappa_{\mathrm{xy}} = \frac{1}{2}\), the different possible states are distinguished by their \(\kappa_{\mathrm{xy}}\).

Upon changing the filling away from half-integer, composite fermions feel a non-zero effective field. The simplest outcome is loss of the composite fermion superconducting order and the formation of an integer quantum Hall (IQH) state of composite fermions at filling \(\nu_{\text{CF}}\), i.e., the Jain state at filling \(\nu^{-1}=2+\nu_{\text{CF}}^{-1}\), where \(2\) is the number of attached flux quanta. An alternative, which may be favored when the magnetic field experienced by the Cooper pairs is small, is the melting of the vortex lattice without the breaking of the composite-fermion Cooper pairs [85]–[89] and the subsequent formation of a bosonic IQH (BIQH) state [80], [81], [90], [91] of the Cooper pairs. These are the daughter states. The charge-conserving BIQH phases have no topological order, carry even filling \(\nu\), and have thermal Hall conductance \(\kappa_{\mathrm{xy}} = 0\). All other known bosonic states without topological order are constructed from the charge-conserving BIQH and the invertible [92]–[99] \(E_8\) state (the state is invertible if there exists a complimentary state such that stacking the two result in topologically trivial state) that has \(\kappa_{\mathrm{xy}} = 8\) (in units of \(\kappa_0 = \frac{\pi^2T}{3h}\)). As such, the bosonic state without topological order has \(\kappa_{\mathrm{xy}} \equiv 0 \mod 8\).

The daughter states considered in this Letter are Abelian, and their filling depends on the value of \(\mathcal{C}\). The simplest (quasiparticle) daughter state appears at filling \(\nu=\frac{8}{17}\) emerging from non-Abelian Pfaffian (\(\mathcal{C}=1\)) and Abelian \(K=8\) (\(\mathcal{C}=0\)) parent states; this daughter state is \(\nu=2\) BIQH state of composite fermion Cooper pairs. The Jain state at the same filling is a \(\nu=8\) IQH state of composite fermions (since \(\nu = \frac{1}{2+\frac{1}{8}}\)). As Cooper pairs have charge 2, the electrical Hall conductance \(\sigma_{\mathrm{xy}}\) of these two states is the same.

Our analysis focuses on the transition at \(\nu=\frac{8}{17}\). Other cases, in which the boson form \(\nu=2m\) states (where \(m\) is the level of daughter state hierarchy) with additional fermionic modes whose number depends on \(\mathcal{C}\), can be analyzed in a similar manner.

— The Abelian states we are interested in can be described within the \(K\)-matrix formalism [28], [100]–[102]. We use \(K\)-matrices built of four building blocks: non-chiral bosonic \(\sigma_x= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) and fermionic \(\sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\) matrices, which carry zero \(\kappa_{xy}\), and the maximally chiral \(8\times 8\) matrices: the identity matrix \(I_8\) (fermionic) and the \(K\)-matrix of the \(E_8\) state \(K_{E_8}\) (bosonic). A \(K\)-matrix with determinant 1 and zero signature defines a topologically trivial state [103]. In the absence of \(U(1)\) symmetry, both \(\sigma_x\) and \(\sigma_z\) describe topologically trivial phases. If \(U(1)\) symmetry is present, an additional requirement of \(\sigma_{\mathrm{xy}} = 0\) appears; \(\sigma_z\) with equal charge of the two modes also describes a topologically trivial phase. Therefore, phases described by \(K\) and \(K \oplus \sigma_z\) are topologically equivalent, representing the addition of a gappable pair of fermionic modes [104].

The fermionic \(\nu=8\) state is described by \(K = I_8\) with charge vector entries \(t_i=1\). The bosonic \(\nu_{\text{pair}}=2\) state is described by \(K = \sigma_x\) with charge vector entries \(t_i=2\). To show the relation between the two, we start from an augmented \(K\)-matrix for the fermionic state: a \(12 \times 12\) matrix \(K_{\text{f}} = I_8 \oplus \sigma_z \oplus \sigma_z\) with charge vector \(t_i=1\).

A key insight comes from applying a \(\mathrm{SL}(12, \mathbb{Z})\) transformation \(W\) [105] to the \(K_{\text{f}}\) such that \[\begin{align} K'_{\text{f}} = W^\top K_{\text{f}} W &= \sigma_x \oplus K_{E_8} \oplus \sigma_z, \label{eq:maptoe8} \end{align}\tag{1}\] where the transformed charge vector \(t'=W^\top t\) has entries \(t'_i=2\) for the \(\sigma_x\) block, \(t'_i=0\) for the \(K_{E_8}\) block, and \(t'_i=1\) for the \(\sigma_z\) block. The transformation is merely a change of basis; it does not change the state. The last \(\sigma_z\) block represents a trivial phase both before and after the transformation, in which single fermion excitations are gapped. Beyond this block, the fermionic \(\nu=8\) state (\(I_8\)) is topologically equivalent to the bosonic \(\nu_{\text{pair}}=2\) state (\(\sigma_x\)) plus a decoupled neutral \(E_8\) state. Hence, the two phases differ by their neutral sector: a chiral \(E_8\) state for the fermionic \(\nu=8\) state and a trivial state for the bosonic \(\nu_{\text{pair}}=2\). As a result, in the transition between the two, the electrical charge gap can remain open.

To describe the critical point between the \(E_8\) state and a topologically trivial state, we employ a parton construction [81], [106], [107]. We express both theories using a single parton construction; they differ in the sign of some parton Chern numbers. This representation allows us to describe the transition as a mass sign change of the partonic Dirac fermions coupled to the gauge fields. We then perform a renormalization group (RG) analysis of the critical point to demonstrate the emergence of intermediate phases.

We begin with a general description of this approach. Parton theory describes a physical state in terms of constituent fermions \(f_j\) (partons), each represented by a bosonic field \(\beta_j\), such that the parton currents are \(j = \frac{1}{2\pi} \curl{\beta}\). When the fermions are gapped, their low-energy dynamics are governed by the topological Chern–Simons action along with the emergent gauge fields \(a_k\), which impose constraints on the parton density: \[\begin{align} \mathcal{L} &= \sum_{j=1}^{N_{\text{p}}} \mathcal{K}_{jj} \beta_j \partial \beta_j + \mathcal{L}_{\text{constraint}}, \label{eq:parton95lagrangian} \end{align}\tag{2}\] where \(\beta \partial \beta\) is a shorthand for \(\epsilon^{\mu\nu\rho} \beta_\mu \partial_\nu \beta_\rho\) and \(\mathcal{L}_{\text{constraint}}\) is the constraint Lagrangian. In the first term, \(\mathcal{K}_{jj}=\pm1\) (we use \(K\) for the \(K\)-matrix for the physical degrees of freedom and calligraphic \(\mathcal{K}\) otherwise), reflecting the partons being in an IQH state. The constraint forces parton densities to be identified: \[\begin{align} \mathcal{L}_{\text{constraint}} &= \sum_{j=1}^{N_{\text{g}}} a_j \partial \sum_{k=1}^{N_{\text{p}}} Q_{jk} \beta_k, \label{eq:constraint} \end{align}\tag{3}\] where \(a_k\) is the set of \(N_{\text{g}}\) gauge fields and \(Q_{jk}\) the charge of the parton \(f_j\) with respect to the gauge field \(a_k\). Since the constraints are linear in \(\beta_k\), we can find a set of gauge-invariant fields for which the constraints are identically satisfied. This process yields a set of physical fields, given by the product of partons, whose Lagrangian, in turn, can represent various Chern–Simons theories depending on \(Q\). For example, three partons constrained to have equal densities generate \(\nu=1/3\) FQH Lagrangian.

We formulate the parton construction in \(K\)-matrix formalism. Consider an electronic state defined by \(K\)-matrix \(K_{\text{e}}\) involving \(N_{\text{e}}\) fields \(\psi_i\) that are constructed from \(N_{\text{p}}\) partons \(f_j\) via \(\psi_i = \prod_j f_j^{P_{ij}}\). The matrix \(P\) is an \(N_{\text{e}}\times N_{\text{p}}\) full-rank matrix, and its null space defines the coupling of the partons to the gauge fields \(a_k\): \(Q\) is an \(N_{\text{p}} \times N_{\text{g}}\) matrix satisfying \(PQ=0\). Substituting these into 2 and solving the constraints results in the Chern–Simons Lagrangian with \(K\)-matrix \(K_{\text{e}} = P\mathcal{K} P^\top\). We restrict \(P\) to have values in the set \(\require{physics} \qty{-1,0,1}\), meaning each parton appears at most once in the definition of \(\psi_i\).

Generally, there are multiple possible partonic constructions (possibly with different \(N_{\text{p}}\)) for each quantum Hall state. In what follows, we point out which of our conclusions depend on the particular construction choice and which are universal.

We now use a specific construction for the \(E_8\) and trivial states using \(N_{\text{p}}=12\) partons. The trivial state is realized with \(\mathcal{K}_1 = I_6 \oplus (-I_6)\) and the \(E_8\) state with \(\mathcal{K}_2 = I_{10} \oplus (-I_2)\), using the \(8 \times 12\) projection matrix \(P\) and \(12 \times 4\) gauge coupling matrix \(Q\). These matrices (and an alternative partonic construction) are given in the supplementary materials [105]. Changing the sign of a single element of \(\mathcal{K}\) amounts to a rank-1 update of \(K_e\), which can change the sign of at most one eigenvalue of \(K_e\) [108]; as a result, \(\kappa_{\mathrm{xy}}\) is changed either by 2 or by 0. The total change in \(\kappa_{\mathrm{xy}}\) between \(E_8\) and the trivial phase is 8, so it has to change by 2 with every such sign change.

The phase transition between the two states closes the gap and thus requires a description that includes the partons. In particular, it involves flipping the sign of the mass for \(N_{\text{f}}=4\) Dirac fermions. We analyze this transition by including those partons in the Lagrangian, while integrating out all the others. The Lagrangian describing the low-energy physics near the critical point is \[\begin{align} \mathcal{L} = \mathcal{L} = \sum_{i=1}^{N_{\text{f}}} \sum_{j=1}^{N_{\text{g}}} &\bar{f}_{i}\gamma_\mu (\partial^\mu - {R}_{ij} a^\mu_j +m)f_{i}+ \nonumber\\+ &\frac{1}{4\pi} \sum_{i,j=1}^{N_{\text{g}}}\mathcal{K}^{\text{eff}}_{ij} a_i \partial a_j + \mathcal{L}_{\text{Maxwell}}, \label{eq:Ltwophases} \end{align}\tag{4}\] where \(\mathcal{K}^{\text{eff}}_{ij} = \sum_k Q_{ki} \mathcal{K}_{kk} Q_{kj}\) with \(k\) running over partons we integrate out, \(R\) is a \(N_{\text{f}} \times N_{\text{g}}\) matrix of rows of \(Q^\top\) corresponding to the remaining partons, and \(\mathcal{L}_{\text{Maxwell}}\) includes non-universal higher-order terms. Far from the transition, the Lagrangians defined in 2 4 are equivalent after integration out of the fields \(\beta\) in 2 and the Dirac fermions in 4 . The phase in question is determined by the signs of \(\mathcal{K}_{jj}\) in 2 , or, equivalently, by the sign of \(m\) in 4 .

In our case, the matrices \(\mathcal{K}^{\text{eff}}\) and \(R\) are given by [105]: \[\begin{align} \mathcal{K}^{\text{eff}} &= \begin{pmatrix} 0 & 0 & 1 & 1 \\ 0 & -1 & 2 & 0 \\ 1 & 2 & 0 & -1 \\ 1 & 0 & -1 & -1 \\ \end{pmatrix} \tag{5}\\ R &= \begin{pmatrix} 0 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. \tag{6} \end{align}\]

A direct transition from the \(E_8\) to the trivial state requires the \(N_{\text{f}}=4\) Dirac fermion masses to vanish simultaneously. Yet, the simultaneous vanishing of four masses should not happen generically. It may be imposed by a symmetry, either on a microscopic level or an emergent one. The system in question, however, lacks a suitable microscopic symmetry. Another possibility is the emergence of a larger gauge group [83] due to the symmetry between the partons, i.e., the linear dependence between the columns of \(P\). However, since twice the rank of the \(P(\mathcal{K}_1-\mathcal{K}_2)P^\top\) provides an upper bound on the change in \(\kappa_{\mathrm{xy}}\) between two phases, in a transition between \(E_8\) and trivial states there are at least four linearly independent partons.

Alternatively, all four masses can vanish simultaneously if, at low energies, the mass difference operators are irrelevant in the RG sense, resulting in an emergent symmetry. The relevance of the operator is determined by its scaling during the RG flow. In general, the coefficient \(h\) of the operator \(\mathcal{M}\) flows as \[\begin{align} \dv{h}{\ell} &= (d+1-\Delta_{\mathcal{M}})h, \label{eq:rg95scaling} \end{align}\tag{7}\] where \(d+1\) is spacetime dimension, \(\ell\) is the RG time and \(\Delta_{\mathcal{M}}\) is the scaling dimension of \(\mathcal{M}\): \(h\) decreases if \(\mathcal{M}\) is irrelevant, i.e., \(d-\Delta_{\mathcal{M}}<0\) [109], [110].

Thus, we want to check whether near the critical point of 4 there are relevant operators beyond the one that controls the phase transition. If any such mass-like operator \(\mathcal{M} = \sum_{ij} \bar{\psi}_i m_{ij} \psi_j\) is relevant, there is no direct transition between the two phases. In a non-interacting 2+1D fermionic theory, the bilinear \(\mathcal{M}\) has scaling dimension \(\Delta^{(0)}_\mathcal{M} = 2\) (twice the fermion’s), hence it is relevant.

One way to tackle the interactions between the Dirac fermions mediated by the gauge fields \(a\) is to use a large-\(N\) expansion [82], [111]–[114] by rewriting the Lagrangian (4 ) as \[\begin{align} \mathcal{L} = \sum_{k=1}^{N}\Big[ \sum_{i=1}^{N_{\text{f}}} \sum_{j=1}^{N_{\text{g}}} &\bar{f}_{i,k}\gamma_\mu (\partial^\mu - {R}_{ij} a^\mu_j +m)f_{i,k}+ \nonumber\\+ &\frac{1}{4\pi} \sum_{i,j=1}^{N_{\text{g}}}\mathcal{K}^{\text{eff}}_{ij} a_i \partial a_j ], \label{eq:LtwophasesNf} \end{align}\tag{8}\] In this Lagrangian, there are still \(N_{\text{g}}\) gauge fields, but \(N_{\text{f}} \cdot N\) fermionic species, and \(N\) species with the same \(k\) are coupled to the gauge fields in the same way. Moreover, the Chern–Simons coupling between the gauge fields is also linear in \(N\). The original Lagrangian corresponds to \(N=1\), and the observables can be calculated using perturbation theory in \(1/N\).

While applying the large-\(N\) limit for \(N=1\) is not rigorously justified, it yields corrections consistent with other methods, such as \(\epsilon\)-expansion for QED [115], even for \(N=1\). Note that \(\epsilon\)-expansion can not be applied in our case due to the presence of the Chern–Simons term.

Due to the mixing between the scalar mass terms \(\mathcal{M}^\alpha(x)=\frac{1}{\sqrt{N}}\sum_{k=1}^N \bar{f}_{\alpha,k} f_{\alpha,k}\), the operators with well-defined scaling at \(\order{1/N}\) are certain linear combinations of \(\mathcal{M}^\alpha(x)\) [113], [116]–[118]. To find the corrections to the tree-level scaling dimension \(\Delta^{(0)}=2\) we calculate the leading-order correction to the correlator \[\require{physics} \begin{align} \expval{\mathcal{M}^\alpha(x)(x)\mathcal{M}^\beta(0)} = \frac{\delta_{\alpha\beta} - \frac{1}{N}M_{\alpha\beta}\log(\abs{x}^2 \Lambda^2) + \order{\frac{1}{N^2}}}{\abs{x}^{2\Delta^{(0)}}}. \label{eq:corr95log} \end{align}\tag{9}\] The operators with well-defined scaling are given by the eigenvectors \(\mu_{\alpha \beta}\) of \(M_{\alpha \beta}\) via \(\tilde{\mathcal{M}}^\alpha =\mu_{\alpha \beta} \mathcal{M}^\beta\). The scaling dimension can then be extracted using 9 as an expansion of \(\require{physics} \expval{\tilde{\mathcal{M}}^\alpha (x)\tilde{\mathcal{M}}^\alpha(0)} =\abs{x}^{-2\Delta_\alpha}\) around \(\Delta_{\alpha}=\Delta^{(0)}\). While off-diagonal mass terms can, in principle, appear in the action, such terms are gauge-invariant only if two fermions \(f_i\) and \(f_j\) have the same charge under all gauge fields.

To evaluate 9 , we first calculate the photon propagator by integrating out the Dirac fermions at the critical point of 8 , resulting in a momentum-space Lagrangian \[\require{physics} \begin{align} \mathcal{L}_{\text{crit}}(p) = a_{i \mu}(-p)\Big[ &\frac{\abs{p}}{16} G^{ij} \delta^{\mu\nu} + \frac{i}{4 \pi} \epsilon^{\mu\nu\rho}p_\rho \qty(\mathcal{K}^{\text{eff}})^{ij} + \nonumber\\&\qquad + \frac{\abs{p}}{16} G^{ij} \frac{3-2\zeta}{2(\zeta-1)} \frac{p^\mu p^\nu}{p^2} \Big] a_{j \nu}(p). \end{align}\] where \(G = \frac{N}{2}\bar{R}^\top\bar{R}\), with \(\bar{R}\) being a \(N_{\text{f}} N \times N_{\text{g}}\) matrix acquired from \(R\) by stacking it \(N\) times, \(\mathcal{K}^{\text{eff}}\) is redefined to incorporate \(N\) factor and \(\zeta\) is an arbitrary real gauge-fixing parameter. The photon propagator \(\require{physics} (D_{\mu\nu})_{ij} = \expval{a_{i\mu}(p) a_{j\nu}(0) }\) is then an \(N_{\text{g}} \times N_{\text{g}}\) matrix. When multiplied by the charge matrices, it is given by \[\require{physics} \begin{align} \tilde{D}_{\mu\nu}(p) &= \bar{R} D_{\mu\nu}(p) \bar{R} ^\top= \frac{\tilde{A}}{\abs{p}}\qty( \delta_{\mu\nu} -\frac{p_\mu p_\nu}{\abs{p}^2})+\tilde{C} \frac{p^\rho}{p^2}\epsilon_{\rho\mu\nu}\label{eq:final95propagator} \\ V &= \qty[\frac{\pi}{64} G {\mathcal{K}^{\text{eff}}}^{-1}G + \frac{1}{4\pi}\mathcal{K}^{\text{eff}}]^{-1} \\ \tilde{A}&=\frac{\pi}{4}\bar{R} {\mathcal{K}^{\text{eff}}}^{-1}G V \bar{R} ^{\top} \\ \tilde{C}&= -\bar{R} V \bar{R} ^{\top} \end{align}\tag{10}\] where we use Feynman gauge (\(\zeta=1\), enforcing the condition \(a_\mu p^\mu=0\)) and the matrix multiplication is over gauge field indices \(i\), \(j\). The elements of \(\tilde{A}\) and \(\tilde{C}\) are all \(\order{1/N}\).

The corrections to the scaling dimensions depend on the parton construction and are given by the eigenvalues of the matrix \[\require{physics} \begin{align} M_{\alpha\beta} = \frac{-16\tilde{A}_{\alpha\beta}\delta_{\alpha\beta} + 3N\qty(\tilde{A}^\top\tilde{A} - \tilde{C}^\top\tilde{C})_{\alpha\beta} }{12 \pi^2}. \label{eq:finalm} \end{align}\tag{11}\] For the parton construction corresponding to 5 6 , we find two pairs of operators \(\tilde{\mathcal{M}}^\alpha\) with equal scaling dimensions \(\Delta_1=\Delta_2\) and \(\Delta_3=\Delta_4\), where the correction to the \(\Delta^{(0)}_\mathcal{M}\) is negative. As a result, the operators become even more relevant at \(\order{1/N}\) and the fixed point is unstable to mass-like perturbations, meaning that direct transition requires fine-tuning and intermediate phases appear generically. The values of \(\Delta_k\) are given in supplementary material [105].

These results, as expected, are independent of choice of \(Q\) in 3 : if \(Q' = Q \cdot O\) for some (invertible) matrix \(O\), then \(\mathcal{K}^{\text{eff}}\mapsto O^\top K^{\text{eff}}O\), \(G \mapsto O^\top G O\), and \(R\mapsto R O\) so that 10 is unaltered.

We conclude that generically there is no direct transition between the \(E_8\) and the trivial bosonic states. The generic path in parameter space between these two phases has to pass through three different intermediate phases with \(\kappa_{\mathrm{xy}} = \{2,4,6\}\). Since integer bosonic phases have \(\kappa_{\mathrm{xy}} \equiv 0 \mod 8\), any intermediate states would necessarily be fractional [90], [119]. It should be noted that different choices of parton construction predict different topological orders in the intermediate phases. Even for a specific parton construction, the multiple possible assignments of mass signs lead to multiple phases with the same value of \(\kappa_{\mathrm{xy}}\). Moreover, from the Cauchy–Binet formula [120], changing the sign of \(\mathcal{K}_{jj}\) in 2 cannot change the parity of the determinant of \(K_{\text{e}}\), meaning that all the intermediate phases have \(K\)-matrices with odd determinants [105]. In addition, the smallest odd determinant allowed for the bosonic model with \(\kappa=4\) is \(\pm 5\) [105], [121].

For the transition between the daughter state and the Jain state of a given filling \(\nu=\frac{p}{q}\), we expect the appearance of the intermediate phases with the ground state degeneracy equal to \(r\cdot q\), where \(r\) is the degeneracy of the fractionalized neutral sector and is odd. In that sense, the construction we used can be considered “minimal” [122], [123], generating a single-parameter path between the \(E_8\) and trivial phases passing through the \(\kappa_{\mathrm{xy}}=6\) and \(\kappa_{\mathrm{xy}}=2\) phases with \(r=3\), and the \(\kappa_{\mathrm{xy}}=4\) phase with \(r=5\).

1 Details of the basis transformation and parton construction↩︎

Here, we give the full definition of the matrices used for our construction. First, to map between the \(I_8 \oplus \sigma_z \oplus \sigma_z\) fermionic state and \(\sigma_x \oplus E_8 \oplus \sigma_z\) state, we use the \(\mathrm{SL}(\mathbb{Z})\) transformation that maps between \(K_{\text{f}}\) and \(K'_{\text{f}}\) as defined in 1 : \[\begin{align} W &= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 & 2 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 & 1 \\ 0 & 0 & -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & -1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 & 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 1 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 1 & 1 & 0 & -1 \\ 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 1 & 2 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 3 \\ \end{pmatrix} \end{align}\]

To construct the physical state, we use 12 total partons, defined by matrices \(P\) and \(Q\): \[\begin{align} P &= \begin{pmatrix} 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 & -1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & -1 & -1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 0 \\ \end{pmatrix} \tag{12}\\ Q^\top&= \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 & 1 & 0 & 0 & -1 & 0 & -1 & -2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 \\ \end{pmatrix}. \tag{13} \end{align}\]

An alternative construction is \[\begin{align} P&= \begin{pmatrix} 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & -1 & 0 \\ 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & -1 & -1 & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & -1 &1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 1 & -1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix} \tag{14}\\ Q^\top&= \begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & -2 \\ 1 & 2 & 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 1 & -1 \\ 1 & 2 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & 0 \\ \end{pmatrix} \tag{15} \end{align}\]

2 Properties of intermediate phases↩︎

2.1 Rank-1 updates↩︎

A sign change of the single parton \(f_i\) Chern number maps the physical \(K\)-matrix as \[\begin{align} K_{\text{e}} = P^\top\mathcal{K} P \mapsto P^\top(\mathcal{K} + E) P \end{align}\] where \(E\) is a matrix with a single non-zero element \(E_{ii} = \pm 2\). We can rewrite the new \(K_{\text{e}}\) as \(P^\top\mathcal{K}P + P^\top E P\). Since \(E\) is rank-1 matrix and \(P\) is full rank, \(P^\top E P\) is also rank-1. We now use a variation of the Cauchy interlacing theorem [108]: for a symmetric \(n\times n\) matrix \(D\) with eigenvalues \(\lambda_i\), the eigenvalues \(\mu_i\) of \(D+\sigma \vb{u}\vb{u}^\top\) for \(\sigma>0\) satisfy the following inequality \[\begin{align} \lambda_1 \leqslant \mu_1 \leqslant \lambda_2 \leqslant \dots \leqslant \lambda_n \leqslant \mu_n. \end{align}\] If \(\sigma<0\) an equivalent inequality is obtained by changing the roles of \(D\) and \(D+\sigma \vb{u}\vb{u}^\top\).

Now consider a concrete case of \(K_{E_8}\), which has eight positive eigenvalues. After rank-1 update arising from the one parton Chern number changing sign, only the smallest eigenvalue can become negative; as such, the signature of the new \(K\)-matrix differs at most by \(2\). After repeating the process four times, we arrive at the matrix with signature \(0\); the change is required to be exactly \(2\) at each mass change.

The same argument shows that there cannot be linear dependence between the columns of \(P\) corresponding to partons that change sign. We can write the trivial phase \(K\)-matrix as \(K_{E_8} - P^\top\tilde{E} P\), where \(\tilde{E}_{ii} = 2\) for all partons that change sign; and since the difference between the signature of \(K_{E_8}\) and trivial one is \(8\), the rank of \(P^\top\tilde{E} P\) is at least 4, meaning all four columns of \(P\) corresponding to partons that change sign are linearly independent.

2.2 \(K\)-matrix determinant parity↩︎

Let \(A\) and \(B\) be two matrices of size \(m\times n\) and \(n \times m\) correspondingly. Denote by \(\binom{[n]}{m}\) the set of all subsets of \(\require{physics} \qty{1,\dots,n}\) of size \(m\). Now for \(S \in \binom{[n]}{m}\) we denote by \(A_{[m], S}\) a matrix with \(m\) columns of \(A\) at indices from \(S\) and by \(B_{S, [m]}\) a matrix with \(m\) rows of \(B\) at indices from \(S\). The Cauchy–Binet formula [120] then gives the determinant of \(AB\) as \[\begin{align} \det(AB) = \sum_{S\in \binom{[n]}{m}} \det(A_{[m],S})\det(B_{S,[m]}). \label{appeq:cauchybinet} \end{align}\tag{16}\]

We apply it to \(K_e\) by choosing \(A=P^\top\) and \(B=\mathcal{K}P\), with \(m=N_{\text{e}}\) and \(n=N_{\text{p}}\). Since \(\mathcal{K}\) is a diagonal matrix, when we change the sign of some diagonal elements of \(\mathcal{K}\), the signs of the corresponding rows of \(B\) are changed. As a result, \(\det(B_{S,[m]})\) is multiplied by \(-1\) if there is an odd number of rows that changed sign in \(S\) and unchanged otherwise. This means that the determinant of the new \(K_e\) is the same sum (16 ), but with some elements having a different sign, which means the parity of the determinant is unchanged.

2.3 Minimal odd determinant at \(\kappa_{\mathrm{xy}}=4\)↩︎

To show that the minimal odd \(K\)-matrix determinant at \(\kappa_{\mathrm{xy}}=4\) is 5, we will show that the bosonic phase with ground state degeneracy 3 can not have \(\kappa_{\mathrm{xy}}=4\). We denote the absolute value of the determinant as \(\abs{K}\) in what follows.

We use the decomposition into prime anyon models [121]. Since the determinant of the \(K\)-matrix is multiplicative, the prime decomposition of the state with \(\abs{K}=3\) has to have one component with \(\abs{K}=3\) and some number of components with \(\abs{K}=1\). The latter cannot change \(\kappa_{\mathrm{xy}} \mod 8\) and thus can be ignored.

The only prime anyonic models with \(\abs{K}=3\) are \(A_3\) or \(B_3\), with \(K\)-matrices being Cartan matrix of \(E_6\) (\(\kappa_{\mathrm{xy}}=6\)) and \(K_{B_3}=\begin{pmatrix} 2&1\\1&2 \end{pmatrix}\) (\(\kappa_{\mathrm{xy}}=2\)), correspondingly. As a result, \(\abs{K}\neq 3\) for \(\kappa_{\mathrm{xy}}=4\).

3 RG analysis of \(U(1)^k\) Chern–Simons–QED\(_{3}\)↩︎

In this section, we perform an RG analysis of a \(U(1)^k\) Chern–Simons–QED\(_3\) with large-\(N\) expansion. Specifically, we compute the scaling dimension of the mass term for a generic Chern–Simons coupling and fermionic charges, and then specialize the results to the theory obtained in the main text. To calculate the scaling dimension, we are interested in the logarithmic divergence of the leading-order (in \(1/N\)) correction to the mass term.

3.1 RG analysis of the critical point↩︎

We first remind the reader how the logarithmic divergence of the two-point correlator is related to the scaling dimension of the operator. For some operator \(\mathcal{O}\) define \[\require{physics} \begin{align} \mathcal{M}(x) &= \expval{\mathcal{O}(x) \bar{\mathcal{O}}(0)} = \mathcal{M}^{(0)}(x) + \frac{1}{N} \mathcal{M}^{(1)}(x) + \dots \end{align}\] Expanding conformal correlator around \(\Delta=\Delta_0\), we get \[\require{physics} \begin{align} \mathcal{M}(x) = \frac{M_0}{\abs{x}^{2\Delta}} = \frac{M_0}{\abs{x}^{2\qty(\Delta^{(0)} + \frac{1}{N}\Delta^{(1)})}} = \frac{M_0}{\abs{x}^{2 \Delta^{(0)}}}\qty(1-\frac{1}{N}\Delta^{(1)} \log(\abs{x}^2\Lambda^2) + \order{\frac{1}{N^2}}), \label{appeq:correlator} \end{align}\tag{17}\] where \(\Lambda\) is the momentum cutoff. By calculating the logarithmic divergence of the correlator, we can extract the scaling dimension correction \[\begin{align} \Delta^{(1)} &= \frac{M_1}{M_0}\\ \Delta &= \Delta^{(0)} + \Delta^{(1)} \cdot \frac{1}{N} +\order{\frac{1}{N^2}}. \end{align}\]

The logarithmic divergence (17 ) is the same both in momentum and position space. Starting from \[\require{physics} \begin{align} \int \dd[d]{p} e^{ipx} \abs{x}^{-\alpha} = C(\alpha) \abs{p}^{-(d-\alpha)} \end{align}\] for \(0<\alpha<d\), and differentiating this equation with respect to \(\alpha\), we get \[\require{physics} \begin{align} \int \dd[d]{p} e^{ipx} \abs{x}^{-\alpha} \log(\abs{x}) = \abs{p}^{-(d-\alpha)} \log(1/\abs{p}) + \order{\abs{p}^{-(d-\alpha)} }. \end{align}\]

3.2 Useful identities↩︎

Gamma matrix algebra in 3D: \[\require{physics} \begin{align} \gamma^\mu\gamma^\eta\gamma^\nu &= \delta^{\mu\eta}\gamma^\nu - \delta^{\mu\nu}\gamma^\eta + \delta^{\eta\nu}\gamma^\mu + i\epsilon^{\mu\eta\nu} \tag{18}\\ \gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma &= \delta^{\mu\nu}\delta^{\rho\sigma} - \delta^{\mu\rho}\delta^{\nu\sigma} + \delta^{\mu\sigma}\delta^{\nu\rho} + i\qty[\delta^{\mu\nu}\epsilon^{\rho\sigma\lambda}+\delta^{\rho\sigma}\epsilon^{\mu\nu\lambda}-\delta^{\nu\lambda}\epsilon^{\rho\sigma\mu}+\delta^{\mu\lambda}\epsilon^{\rho\sigma\nu}]\gamma_\lambda. \tag{19} \end{align}\] We will use the fact that \(\delta^{\mu\nu}\delta^{\mu\nu} = d\) in \(d\) dimensions. As we are working in Eucledian space, \(\gamma^{\mu\nu} = \delta^{\mu\nu}\gamma_{\mu\nu}\).

Radial integrals \[\require{physics} \begin{align} \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{1}{\abs{k}^3} &= \int_0^\Lambda \frac{\dd{k}}{2\pi^2} \frac{1}{k} = \frac{1}{4\pi^2}\log(\Lambda^2 x^2)\\ \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{k_\mu k_\nu}{\abs{k}^5} &= \frac{\delta_{\mu\nu}}{3}\int \frac{\dd[3]{k}}{(2\pi)^3} \frac{k^2}{\abs{k}^5} = \frac{\delta_{\mu\nu}}{12\pi^2}\log(\Lambda^2 x^2), \end{align}\] where in the second integral we used the reflection symmetry to establish that the integral vanishes if \(\mu\neq \nu\) and the rotational symmetry to replace \(k^\mu k^\nu\) by \(\frac{\delta_{\mu\nu}}{3}k^2\).

Momentum integrals \[\require{physics} \begin{align} \int \frac{\dd[3]{q}}{(2\pi)^3} \frac{1}{q^2(q-k)^2} &= \frac{1}{8\abs{k}} \tag{20}\\ \int \frac{\dd[3]{q}}{(2\pi)^3} \frac{q^\mu}{q^2(q-k)^2} &= \frac{k^\mu}{16\abs{k}} \tag{21} \end{align}\] (the second can be obtained by contracting it with \(k_\mu\) and observing that it equals half the first one when completing a square to \((q-k)^2\)).

For \(\abs{k}\gg \abs{p}\) we can expand \[\require{physics} \begin{align} \frac{1}{(p-k)^2} = \frac{1}{k^2} \cdot \frac{1}{1 - 2\frac{p\vdot k}{k^2} + \qty(\frac{p}{k})^2} = \frac{1}{k^2}\qty[1 + 2\frac{p\vdot k}{k^2} - \qty(\frac{p}{k})^2 + 4\qty(\frac{p\vdot k}{k^2})^2 + \order{\qty(\frac{p}{k})^3}] \label{eq:frac95expansion} \end{align}\tag{22}\]

Fermion propagator and vertex for the Lagrangian (8 ) \[\require{physics} \begin{align} \mathcal{L} &= \sum_{k=1}^{N}\qty[\sum_{i=1}^4\bar{f}_{i,k}\gamma_\mu (\partial_\mu - {R}_{ij} a_j +m)f_{i,k}+ \frac{1}{4\pi} \mathcal{K}^{\text{eff}}_{ij} a^i \partial a^j] \end{align}\] are given by \[\begin{align} G^0(p)=\begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/mpyiqrsv.png}\tag{23}\end{figure} &= \frac{\gamma^\mu p_\mu}{p^2}\\ G^0(x_1,x_2)=\begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/xdqorkny.png}\tag{24}\end{figure} &= \frac{\gamma^\mu (x_1-x_2)_\mu}{2\pi \abs{x_1-x_2}^3}\\ \begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/yoijsdcr.png}\tag{25}\end{figure} &= R_{ij} \gamma_\mu. \end{align}\]

3.3 Photon propagator↩︎

There are two methods for calculating the photon propagator. We can either calculate the bare photon propagator and use it to evaluate geometric series that include polarization bubbles, or we can integrate out the fermions first and then calculate the photon propagator in the effective theory.

We start from a simpler calculation involving one gauge field, and then generalize it to the case of multiple \(U(1)\) gauge fields.

3.3.1 QED\(_3\) photon propagator↩︎

As a warm-up, we calculate the propagator for the QED\(_3\) [113]. The Lagrangian density for photons is \[\begin{align} \mathcal{L} &= A^\mu O^{\mathrm{QED}}_{\mu\nu} A^\nu\\ O^{\mathrm{QED}}_{\mu\nu} &= q^2\delta_{\mu\nu} - \frac{N}{16(\zeta-1)\abs{q}}q_{\mu}q_{\nu}, \end{align}\] where \(\zeta\) is a gauge-fixing parameter. The bare photon propagator \(D^{\mathrm{QED},0}_{\rho\nu}\) is given by \[\begin{align} O^{\mathrm{QED}}_{\mu\rho} D^{\mathrm{QED},0}_{\rho\nu} &= \delta_{\mu\nu} \end{align}\]

Since the fermion loop contributions are not suppressed for the photon, we need to sum the geometric series. The polarization bubble (fermion loop) of massless Dirac fermions is given by \[\require{physics} \begin{align} \Pi^{\mu \nu}(q) &= \int \frac{\dd[d]{p}}{(2\pi)^d} \Tr[i\gamma^\mu \frac{ p_\alpha \gamma^\alpha}{p^2} i \gamma^\nu \frac{ (p+q)_\beta \gamma^\beta}{(p+q)^2}]= \frac{N \abs{q}}{16}\qty(\delta^{\mu\nu} - \frac{q^\mu q^\nu}{q^2}). \end{align}\] The full photonic propagator given by the geometric series sum \[\require{physics} \begin{align} D^{\mathrm{QED}}_{\mu\nu}(p) &= D_{\mu\nu}^{\mathrm{QED},0}(p) - D_{\mu\alpha}^{\mathrm{QED},0}(p) \Pi^{\alpha \beta}(p) D_{\beta\nu}^{\mathrm{QED},0}(p) + \dots = \qty\Big(\qty[D_{\mu\nu}^{\mathrm{QED},0}(p)]^{-1}+\Pi^{\mu \nu}(p) )^{-1} = \qty\Big(O^{\mathrm{QED}}_{\mu\nu}(p) +\Pi^{\mu \nu}(p) )^{-1}. \label{appeq:geomser} \end{align}\tag{26}\]

The same expression can be obtained by integrating out the fermions. In this case, an additional term of the form \(\Pi^{\mu \nu}(p)\) appears in the effective Lagrangian for photons, and the photon propagator again acquires the form of 26 . We guess an ansatz of the form \[\begin{align} D^{\mathrm{QED}}_{\rho\nu} &= a(q) \delta_{\rho\nu} + b(q) q_\rho q_\nu, \end{align}\] which gives us \[\require{physics} \begin{align} &\qty(O^{\mathrm{QED}}_{\mu\rho}(p) +\Pi^{\mu \rho}(p) )^{-1} D^{\mathrm{QED}}_{\rho\nu} = \qty[q^2\delta_{\mu\rho} - \frac{N}{16(\zeta-1)\abs{q}}q_{\mu}q_{\rho} + \frac{N \abs{q}}{16}\qty(\delta^{\mu\rho} - \frac{q^\mu q^\rho}{q^2})]\qty[\alpha(q) \delta_{\rho\nu} + \beta(q) q_\rho q_\nu] = \nonumber \\&= q^2\delta_{\mu\nu} \alpha(q) - \frac{N}{16(\zeta-1)\abs{q}}q_{\mu}q_{\nu} \alpha(q) + \frac{N \abs{q}}{16}\qty(\delta^{\mu\nu} - \frac{q^\mu q^\nu}{q^2})\alpha(q) + q^2 \beta(q) q_\mu q_\nu - \frac{N\abs{q}}{16(\zeta-1)}q_{\mu} q_\nu \beta(q) = \nonumber \\&= \qty(q^2 + \frac{N \abs{q}}{16})\delta^{\mu\nu}\alpha(q) +\qty[- \frac{N}{16(\zeta-1)\abs{q}} \alpha(q) -\frac{N}{16 \abs{q}} \alpha(q) + q^2 \beta(q) - \frac{N\abs{q}}{16(\zeta-1)} \beta(q) ] q_\mu q_\nu = \nonumber \\&= \qty(q^2 + \frac{N \abs{q}}{16}) \delta^{\mu\nu}\alpha(q) +\qty[ - \frac{N}{16\abs{q}} \frac{\zeta}{\zeta-1} \alpha(q) + \qty(q^2 - \frac{N\abs{q}}{16(\zeta-1)}) \beta(q) ] q_\mu q_\nu, \end{align}\] meaning \(\require{physics} a(q) = \qty(q^2 + \frac{N \abs{q}}{16})^{-1} = \frac{16}{N\abs{q}} + \order{1/N^2}\) and \[\require{physics} \begin{align} \qty(q^2 - \frac{N\abs{q}}{16(\zeta-1)}) b(q) &= \frac{\zeta}{\zeta-1}\frac{1}{q^2} \\ b(q) &= -\frac{16\zeta}{N\abs{q}^3} + \order{1/(q^2 N^2)}, \end{align}\] and resulting in a photon propagator \[\require{physics} \begin{align} D^{\mathrm{QED}}_{\rho\nu} &= \frac{16}{N\abs{q}} \qty(\delta_{\rho\nu} -\zeta \frac{q_\rho q_\nu}{q^2}). \end{align}\]

3.3.2 Chern–Simons–QED\(_3\)↩︎

We now turn to the Chern–Simons–QED\(_3\) with a single \(U(1)\) gauge field. The action is \[\require{physics} \begin{align} S_E[a, \psi] = \int \dd[3]{x} \qty[ \sum_{i=1}^{N} \bar{\psi}_{i}\gamma_\mu (\partial_\mu - a)\psi_{i} + \frac{k_{\text{eff}}}{4\pi} \epsilon_{\mu\nu\rho} a_\mu \partial_\nu a_\rho ] \label{appeq:cs95action} \end{align}\tag{27}\] Here the charge \(e=1\). Note that since the Maxwell term is less relevant than the Chern–Simons term, it can be neglected.

After integrating out the fermions, which adds a polarization bubble term to the action and switching to momentum space, we get an effective Lagrangian \[\require{physics} \begin{align} \mathcal{L}[a(q)] &= \mathcal{L}_{\text{fermion}} + \mathcal{L}_{\text{CS}} + \mathcal{L}_{\text{gauge fixing}}\\ \mathcal{L}_{\text{fermion}} &= \frac{1}{2} a_\mu(-q) \qty[ \frac{N}{16}\abs{q}\qty(\delta^{\mu\nu} - \frac{q^\mu q^\nu}{q^2}) ] a_\nu(q)\\ \mathcal{L}_{\text{CS}} &= \frac{k_{\text{eff}}}{4\pi} \epsilon_{\rho\mu\nu} q_\rho a_\mu(q) a_\nu(-q) \label{appeq:cslagrangian}\\ \mathcal{L}_{\text{gauge fixing}} &= \frac{N}{32(\zeta-1)}a_\mu(-q) \frac{q_\mu q_\nu}{\abs{q}}a_\nu(q). \end{align}\tag{28}\] We write the quadratic part of the action in the form \[\begin{align} \mathcal{L} &= \frac{1}{2} a_\mu O_{\mu\nu} a_\nu\\ O_{\mu\nu} &= \alpha \abs{q} \delta_{\mu\nu} + \beta \epsilon_{\mu\nu\rho} q_\rho + \gamma q_\mu q_\nu\\ \alpha &= \frac{N}{16}\\ \beta &= \frac{k_{\text{eff}}}{4\pi}\\ \gamma &= \frac{N}{\xi \abs{q}} \end{align}\] with \(\xi = \frac{32(\zeta-1)}{3-2\zeta}\).

We introduce projections on transverse and longitudinal directions \[\begin{align} P^T_{\mu\nu} &= \delta_{\mu\nu} - \frac{q_\mu q_\nu}{q^2}\\ P^L_{\mu\nu} &= \frac{q_\mu q_\nu}{q^2} \end{align}\] that satisfy \(P_T+P_L = \delta_{\mu\nu}\), \(P_T^2 = P_T\), \(P_L^2=P_L\), and \(P_L P_T = 0\). In addition, we introduce an antisymmetric transversal tensor \[\begin{align} S_{\mu\nu} &= \epsilon_{\mu\nu\rho} \frac{q_\rho}{\abs{q}} \end{align}\] satisfying \[\require{physics} \begin{align} S^2 &= S_{\mu\sigma} S_{\sigma\nu} = -\qty(\delta_{\mu\nu}\delta_{\rho\alpha} - \delta_{\mu\alpha}\delta_{\nu\rho})\frac{q_\rho q_\alpha}{q^2} = \frac{q_\mu q_\nu}{q^2} - \delta_{\mu\nu} = -P^T\\ S P_T &= \epsilon_{\mu\sigma\rho} \frac{q_\sigma}{\abs{q}} \qty( \delta_{\mu\nu} - \frac{q_\mu q_\nu}{q^2}) = \epsilon_{\mu\nu\rho} \frac{q_\sigma}{\abs{q}} = S. \end{align}\]

We rewrite \(O\) as \[\begin{align} O_{\mu\nu} &= \alpha \abs{q} \delta_{\mu\nu} + \beta \epsilon_{\mu\nu\rho} q_\rho + \gamma q_\mu q_\nu = \alpha \abs{q} P^T + \beta \abs{q} S + \alpha \abs{q}P^L + \gamma q^2 P^L \end{align}\] and we write the propagator as \[\begin{align} D_{\mu\nu} &= a(q) P^T + b(q) P^L + c(q) S. \label{appeq:prop95generic} \end{align}\tag{29}\] Note that \(a\) and \(b\) have a similar meaning as in the previous section, but different numerical values.

Multiplying the two, we get \[\require{physics} \begin{align} O_{\mu\eta} D_{\eta\nu} &= \qty(\alpha \abs{q} P^T + \beta \abs{q} S + \alpha \abs{q}P^L + \gamma q^2 P^L)\qty(a(q) P^T + b(q) P^L + c(q) S) = \nonumber \\&= \qty(\alpha \abs{q} P^T + \beta \abs{q} S)a(q) P^T + \qty(\alpha \abs{q}P^L + \gamma q^2 P^L) b(q) P^L + \qty(\alpha \abs{q} P^T + \beta \abs{q} S)c(q) S = \nonumber\\&= \qty(\alpha \abs{q} a(q) - \beta \abs{q}c(q)) P^T + \qty(\alpha \abs{q} + \gamma q^2 ) b(q) P^L + \qty(\alpha \abs{q} c(q) + \beta \abs{q}a(q)) S \end{align}\] which gives us three equations \[\require{physics} \begin{align} \alpha \abs{q} a(q) - \beta \abs{q}c(q) &= 1\\ \qty(\alpha \abs{q} + \gamma q^2 ) b(q) &= 1\\ \alpha \abs{q} c(q) + \beta \abs{q}a(q) &= 0 \end{align}\] with solution \[\begin{align} b(q) &= \frac{1}{\alpha \abs{q} + \gamma q^2}\\ a(q) &= - \frac{\alpha}{\beta} c(q) \tag{30}\\ c(q) &= -\frac{\beta}{ \abs{q} (\alpha^2 + \beta^2 )}, \tag{31} \end{align}\] i.e., \[\require{physics} \begin{align} D_{\mu\nu} &= -c(q) \qty(\frac{\alpha}{\beta} P^T - S) + b(q) P^L. \end{align}\] The form of the 30 31 can be understood in a way analogous to 26 . However, for action (27 ), the transversal part of the bare propagator is acquired by inverting 28 and as such is antisymmetric, i.e., \(D^{0}_{\mu\nu} \propto S\), while \(\Pi^{\mu\nu}\) is symmetric; \(D_{\mu\nu}\) has both symmetric and antisymmetric components. That means that the symmetric part of \(D_{\mu\nu}\) has to contain an even number of \(D^{0}_{\mu\alpha}\Pi^{\alpha\nu}\) factors, and the antisymmetric part has to contain an odd number of such factors. This amounts to summing geometric series with the term squared relatively to one in 26 ; as a result, the denominator in 31 contains the \(\alpha^2\) and \(\beta^2\) terms.

We define \(\lambda = \frac{4k_{\text{eff}} }{\pi N}\), i.e., \(k_{\text{eff}} = \frac{\pi N \lambda }{4}\) \[\begin{align} \beta &= \frac{N \lambda }{16} = \alpha \lambda \end{align}\] Substituting these, we get \[\begin{align} b(q) &= \frac{16\xi}{N (\xi + 16 ) \abs{q}}\\ c(q) &= -\frac{\lambda }{ \abs{q} \alpha (1 + \lambda ^2 )} \end{align}\] to acquire the propagator \[\begin{align} D_{\mu\nu} &= \frac{16 }{ \abs{q} N (1 + \lambda ^2 )} ( P^T - \lambda S) + \frac{16\xi}{N (\xi + 16 ) \abs{q}} P^L. \end{align}\] A good choice of \(\xi\) is such that \(\frac{q_\mu q_\nu}{q^2}\) cancels, i.e., \(\xi=\frac{16}{\lambda^2}\): \[\require{physics} \begin{align} D_{\mu\nu} &= \frac{16 }{ \abs{q} N (1 + \lambda ^2 )} \qty( \delta_{\mu\nu} - \lambda \frac{q_\rho}{\abs{q}} \epsilon_{\rho\mu\nu}) \end{align}\]

3.3.3 Propagator of \(U(1)^k\) QED–Chern–Simons↩︎

For more general \(U(1)^k\) case we have Lagrangian given by \[\require{physics} \begin{align} \mathcal{L}_{\text{crit}} &= a_{i \mu}(-q)\qty[ \frac{\abs{q}}{16} G_{ij} \delta^{\mu\nu} + \frac{i}{4 \pi} \epsilon^{\mu\nu\rho}q_\rho K^{\text{eff}}_{ij} + \frac{\abs{q}}{16} G_{ij} \frac{3-2\zeta}{2(\zeta-1)} \frac{q^\mu q^\mu}{q^2} ] a_{j \nu}(q). \end{align}\] with \(G = \bar{R}^\top\bar{R}\) and \(\bar{R}\) is a \(N_{\text{f}} N \times N_{\text{g}}\) matrix acquired from the charge matrix \(R\) by repeating it \(N\) times. For the example in the main text, defined by 12 13 , after integration out we get 5 6 : \[\begin{align} \mathcal{K}^{\text{eff}} &= \begin{pmatrix} 0 & 0 & 1 & 1 \\ 0 & -1 & 2 & 0 \\ 1 & 2 & 0 & -1 \\ 1 & 0 & -1 & -1 \\ \end{pmatrix} \tag{32} \\ R &= \begin{pmatrix} 0 & -1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \tag{33}. \end{align}\] For convenience we redefine \(K^{\text{eff}} = N \mathcal{K}^{\text{eff}}\).

We define the matrices \[\begin{align} \alpha &= \frac{G}{16}\\ \beta &= \frac{K^{\text{eff}}}{4\pi}\\ \gamma &= \frac{G}{\xi \abs{q}} \end{align}\] and repeat the same process with a generic (29 ), where \(a\), \(b\), \(c\) are now matrices too. Specifically, we get the matrix equations \[\require{physics} \begin{align} \alpha \abs{q} a(q) - \beta \abs{q}c(q) &= 1\\ \qty(\alpha \abs{q} + \gamma q^2 ) b(q) &= 1\\ \alpha \abs{q} c(q) + \beta \abs{q}a(q) &= 0 \end{align}\] are now matrix equations with a solution which gives us \[\require{physics} \begin{align} a(q) &= - \beta^{-1} \alpha c(q) \\ b(q) &= \qty(\alpha \abs{q} + \gamma q^2 )^{-1}\\ c(q) &= -\frac{1}{ \abs{q} }\qty[ \alpha \beta^{-1} \alpha + \beta]^{-1} \end{align}\] or equivalently \[\require{physics} \begin{align} a(q) &= \frac{ 1}{\abs{q}}\qty[\alpha + \beta \alpha^{-1}\beta ]^{-1}\\ b(q) &= \qty(\alpha \abs{q} + \gamma q^2 )^{-1}\\ c(q) &=- \alpha^{-1}\beta a(q). \end{align}\] Note that the expression for \(b\) is well-defined only if \(G\) is invertible; however, with appropriate gauge choice (\(\zeta=1\)), it will not appear in the propagator. The two different expressions for \(a\) and \(c\) are well-defined if either \(K^{\text{eff}}\) or \(G\) is invertible, correspondingly. In the case in the main text, \(K^{\text{eff}}\) is invertible and thus we use the first version. Thus \[\require{physics} \begin{align} D_{\mu\nu} &= \frac{1}{ \abs{q} }\frac{\pi}{4}\qty(K^{\text{eff}})^{-1} G \qty[ \frac{\pi }{64} G\qty(K^{\text{eff}})^{-1} G - \frac{1}{4\pi}K^{\text{eff}}]^{-1} P^T + \frac{16\xi}{(16+\xi)\abs{q}}G^{-1} P^L -\frac{1}{ \abs{q} }\qty[ \frac{\pi }{64} G\qty(K^{\text{eff}})^{-1} G - \frac{1}{4\pi}K^{\text{eff}}]^{-1} S \end{align}\] Substituting \(\frac{16\xi}{(16+\xi)} = 32(\zeta-1)\), this gives us 10 \[\require{physics} \begin{align} \tilde{D}_{\mu\nu}(p) = \bar{R} D_{\mu\nu}(p) \bar{R}^\top&= \frac{\tilde{A}}{\abs{p}} \delta_{\mu\nu} + \tilde{B}\frac{p_\mu p_\nu}{\abs{p}^3}+\tilde{C} \frac{p^\rho}{p^2}\epsilon_{\rho\mu\nu} \\ V &= \qty[\frac{\pi}{64} G {\mathcal{K}^{\text{eff}}}^{-1}G + \frac{1}{4\pi}\mathcal{K}^{\text{eff}}]^{-1} \\ \tilde{A}&=\frac{\pi}{4} \bar{R} {\mathcal{K}^{\text{eff}}}^{-1}G V \bar{R}^{\top} \\ \tilde{B}&= 32(\zeta-1) \bar{R} G^{-1}\bar{R}^{\top} - \tilde{A}\\ \tilde{C}&= -\bar{R} V \bar{R}^{\top}, \end{align}\] where \(\tilde{A}\), \(\tilde{B}\), \(\tilde{C}\) are all \(\order{1/N}\) independent of \(p\), and we can make convenient choice of \(\zeta=1\) in what follows. As a result, for the diagrams we plan to calculate, each bosonic propagator contributes a factor of \(1/N\), and each fermionic loop contributes a factor of \(N\), which allows us to calculate correlators order-by-order in \(1/N\).

For the example in the main text, for \(N=1\), i.e., using 32 33 we get \[\begin{align} \tilde{A} &= \frac{8\pi^2 }{(20 + \pi^2)N} \begin{pmatrix} 1&-1&0&0\\ -1&1&0&0\\ 0&0&1&1\\ 0&0&1&1 \end{pmatrix} \tag{34}\\ \tilde{C} &= \frac{8\pi }{(20 + \pi^2)N} \begin{pmatrix} 2&-2&-4&-4\\ -2&2&4&4\\ -4&4&-2&-2\\ -4&4&-2&-2\\ \end{pmatrix} \tag{35} \end{align}\] For \(N>1\), \(\tilde{A}\) and \(\tilde{C}\) are block-diagonal matrices with \(N\) blocks each equal the above.

3.4 Fermion self-energy↩︎

We start by calculating the self-energy of the fermion. It is given by a diagram \[\require{physics} \begin{align} \Sigma(p)&=\begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/werifjzo.png}\label{wijvftcm}\end{figure} = - \int \frac{\dd[3]{k}}{(2\pi)^3} \gamma^\nu G^0(p-k) \gamma^\mu \tilde{D}_{\mu\nu}(k). \end{align}\tag{36}\] Substituting the propagators and using 18 , we get \[\require{physics} \begin{align} \Sigma(p)&= - \int \frac{\dd[3]{k}}{(2\pi)^3} \gamma^\mu \frac{\gamma^\eta (p-k)_\eta}{(p-k)^2} \gamma^\nu \qty[\frac{\tilde{A}}{\abs{k}} \delta_{\mu\nu} + \tilde{B}\frac{k_\mu k_\nu}{\abs{k}^3}+\tilde{C} \frac{k^\rho}{\abs{k}}\epsilon_{\rho\mu\nu}] =\\&= - \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{(p-k)_\eta}{(p-k)^2} \qty[\delta^{\mu\eta}\gamma^\nu - \delta^{\mu\nu}\gamma^\eta + \delta^{\eta\nu}\gamma^\mu + i\epsilon^{\mu\eta\nu}] \qty[\frac{\tilde{A}}{\abs{k}} \delta_{\mu\nu} + \tilde{B}\frac{k_\mu k_\nu}{\abs{k}^3}+\tilde{C} \frac{k^\rho}{k^2}\epsilon_{\rho\mu\nu}] =\\&= - \int \frac{\dd[3]{k}}{(2\pi)^3} \qty[ \frac{\gamma^\nu (p-k)^\mu}{(p-k)^2} - \delta^{\mu\nu} \frac{\gamma^\eta (p-k)_\eta}{(p-k)^2} + \frac{\gamma^\mu (p-k)^\nu}{(p-k)^2} + i\epsilon^{\mu\eta\nu} \frac{(p-k)_\eta}{(p-k)^2} ] \qty[\frac{\tilde{A}}{\abs{k}} \delta_{\mu\nu} + \tilde{B}\frac{k_\mu k_\nu}{\abs{k}^3}+\tilde{C} \frac{k^\rho}{k^2}\epsilon_{\rho\mu\nu}]=\\&= - \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{1}{(p-k)^2}\qty[ - \tilde{A}\frac{\gamma^\eta (p-k)_\eta}{\abs{k}} +\tilde{B}\qty[ 2\frac{(p-k)^\mu k_\mu \gamma^\nu k_\nu }{ \abs{k}^3} - \frac{\gamma^\eta (p-k)_\eta}{\abs{k}} ] -2 \tilde{C} \frac{(p-k)_\eta k^\eta}{k^2 }] =\\&= - \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{1}{(p-k)^2}\qty[ - (\tilde{A}+\tilde{B})\frac{\gamma^\eta p_\eta}{\abs{k}}+ 2\tilde{B} \frac{p^\mu k_\mu \gamma^\nu k_\nu }{ \abs{k}^3} + (\tilde{A}- \tilde{B})\frac{\gamma^\eta k_\eta}{\abs{k}} -2 i \tilde{C} \frac{p_\eta k^\eta}{k^2 } +2 i \tilde{C} ] \end{align}\] We are interested in logarithmic divergence, i.e., terms proportional to \(k^{-3}\). We thus split the integrand by powers of \(k\) and keep only logarithmic divergent parts: \[\require{physics} \begin{align} \mathcal{I}_1 = - \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{1}{(p-k)^2}\qty[ - (\tilde{A}+\tilde{B})\frac{\gamma^\eta p_\eta}{\abs{k}}+ 2\tilde{B} \frac{p^\mu k_\mu \gamma^\nu k_\nu }{ \abs{k}^3}-2 i \tilde{C} \frac{p_\eta k^\eta}{k^2 } ] &= \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{1}{k^2}\qty[ (\tilde{A}+\tilde{B})\frac{\gamma^\eta p_\eta}{\abs{k}} - 2\tilde{B} \frac{p^\mu k_\mu \gamma^\nu k_\nu }{ \abs{k}^3}]\\ \mathcal{I}_2 = - \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{1}{(p-k)^2}\qty[ (\tilde{A}- \tilde{B})\frac{\gamma^\eta k_\eta}{\abs{k}} +2 i \tilde{C} ] &= - 2 p^\mu \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{ k_\mu}{\abs{k}^5} (\tilde{A}- \tilde{B}) \gamma^\eta k_\eta, \end{align}\] where we drop numerators with \(k\)-odd terms. The first integral gives \[\begin{align} \mathcal{I}_1 = \frac{3\tilde{A}+\tilde{B}}{12\pi^2} \gamma^\mu p_\mu \log(\Lambda^2 / p^2) \end{align}\] and the second gives \[\begin{align} \mathcal{I}_2 = - \frac{2\tilde{A} - 2\tilde{B}}{12\pi^2}\gamma^\mu p_\mu \log(\Lambda^2 / p^2) \end{align}\] giving in total \[\begin{align} \mathcal{I}_1 + \mathcal{I}_2 = \frac{\tilde{A}+3\tilde{B}}{12\pi^2} \gamma^\mu p_\mu \log(\Lambda^2 / p^2). \end{align}\] As a result, the fermion propagator is given by \[\require{physics} \begin{align} \begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/rqhuakwg.png}\label{sydfxhve}\end{figure} &= G^0(p) \qty[1-\frac{\Delta^{(1)}_\psi}{N} \log(\Lambda^2 / p^2) + \order{1/N^2}]\\ \frac{\Delta^{(1)}_\psi}{N} &= \frac{\tilde{A}+3\tilde{B}}{12\pi^2} \end{align}\tag{37}\] In real space, we get \[\require{physics} \begin{align} \begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/nbeyumfl.png}\tag{38}\end{figure} &= G^0(x) \qty[1-\frac{\Delta^{(1)}_\psi}{N} \log(\Lambda^2 x^2) + \order{1/N^2}] \tag{39} \end{align}\]

3.5 Mass correction↩︎

Now we will calculate the leading order correction to the scaling dimension of the mass operator in the form of 17 . For that, we are interested in logarithmic divergences of the correlator \(\require{physics} \expval{\mathcal{M}^\alpha(x)\mathcal{M}^\beta(0)}\) proportional to \(\log(\Lambda^2 x^2)\), where \[\begin{align} \mathcal{M}^\alpha= \sum_{ij} \bar{f}_{i} m^{\alpha}_{ij} f_{\alpha,j} = \frac{1}{\sqrt{N}}\sum_{k=1}^N \bar{f}_{\alpha,k} f_{\alpha,k} \end{align}\] is a mass operator and \(\Lambda\) is the momentum cutoff.

The final values of all the diagrams contributing to the correlator \(\require{physics} \expval{\mathcal{M}^\alpha(x)\mathcal{M}^\beta(0)}\) (calculated below) are \[\require{physics} \begin{align} \begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/kbgwfrhc.png}\tag{40}\end{figure} &= \delta_{\alpha\beta} \Gamma^{(0)}(x) = \frac{2}{4\pi \abs{x}^4} \delta_{\alpha\beta} \tag{41}\\ \begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/pkmxvnej.png}\tag{42}\end{figure} = \begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/hbotekxf.png}\tag{43}\end{figure} &= -\frac{\Delta^{(1)}_\psi}{N} \delta_{\alpha\beta} \Gamma^{(0)}(x) \log(x^2 \Lambda^2 ) \tag{44}\\ \begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/pjuqbwec.png}\tag{45}\end{figure} &= \qty[\frac{3\tilde{A}+\tilde{B}}{2\pi^2 }]_{\alpha\beta} \delta_{\alpha\beta} \Gamma^{(0)}(x) \log(x^2 \Lambda^2 ) \tag{46}\\ \begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/cwbsyevt.png}\tag{47}\end{figure} = \begin{figure}\includegraphics[width=0.8\textwidth]{_pdflatex/bslfnmkx.png}\tag{48}\end{figure} &= -N\qty[\frac{ \tilde{A}^\top\tilde{A}-\tilde{C}^\top\tilde{C}}{4\pi^2}]_{\alpha\beta} \Gamma^{(0)}(x) \log(x^2 \Lambda^2 ) \tag{49} \end{align}\] The dot denotes the insertion of the bilinear.

We summarize the calculation of these diagrams below.

The first diagram (41 ) gives \(\order{1}\) value of the vertex and is given by \[\require{physics} \begin{align} \expval{\mathcal{M}^\alpha(x)\mathcal{M}^\beta(0)} &= \expval{\bar{\psi}_i(x) m^\beta_{ij} \psi_j(x) \bar{\psi}_{i'}(0) m^\beta_{i'j'} \psi_{j'}(0)} = G^0_{ij'}(x)m^\alpha_{ij}m^\beta_{i'j'}G^0_{i'j}(-x) =\\&= \Tr(m^\alpha m^\beta) G^0(x)G^0(-x) =\frac{2}{4\pi \abs{x}^4} \Tr(m^\alpha m^\beta), \end{align}\] where we used explicit indices on fermion propagators for clarity. The trace is given by \(\delta_{\alpha\beta}\) in our case. We denote \[\begin{align} \Gamma^{(0)}(x) =G^0(x)G^0(-x) =\frac{2}{4\pi \abs{x}^4}. \end{align}\]

We now calculate the \(\order{1/N}\) diagrams. The second two diagrams (44 ) are symmetric; they are acquired by replacing one of the fermion propagators in 41 by the self-energy one (39 ).

The third diagram (46 ) is given by \[\begin{align} G^0_{i}(y-x)G^0_{i}(0-y)m^\alpha_{ij}G^0_{j}(w-0)G^0_{j}(x-w) m^\beta_{ji} \tilde{D}_{ij}(w-y) \end{align}\] The log divergence occurs when \(y\) and \(w\) are close to \(0\) or \(x\). We calculate one of the cases and multiply the result by 2. In this case we can amputate \(G^0(-x)G^0(x)\) to get \[\begin{align} \Gamma^{(0)}(x) G^0_{i}(-y)m^\alpha_{ij}G^0_{j}(w) m^\beta_{ji} \tilde{D}_{ij}(w-y) \end{align}\] We first ignore the indices \(i\), \(j\), and calculate matrix elements for arbitrary fermion indices. After the Fourier transform, we need to calculate the integral \[\require{physics} \begin{align} \Gamma^{(1)}_c(x) &\equiv \Gamma^{(0)}(x) \int \frac{\dd[3]{k}}{(2\pi)^3} \gamma^{\mu} G^0(k)G^0(-k) \gamma^\sigma \tilde{D}_{\mu\sigma}(k) =\\&= - \Gamma^{(0)}(x)\int \frac{\dd[3]{k}}{(2\pi)^3} \gamma^{\mu} \gamma^\nu \frac{k_\nu}{k^2} \gamma^\rho \frac{k_\rho }{k^2}\gamma^\sigma \qty[\frac{\tilde{A}}{\abs{k}} \delta_{\mu\sigma} + \tilde{B}\frac{k_\mu k_\sigma}{\abs{k}^3}+\tilde{C} \frac{k^\eta}{k^2}\epsilon_{\eta\mu\sigma}] \end{align}\] We substitute 19 : \[\require{physics} \begin{align} \Gamma^{(1)}_c(x) &= -\Gamma^{(0)}(x)\int \frac{\dd[3]{k}}{(2\pi)^3} \frac{k_\nu k_\rho }{k^4}\qty[\delta^{\mu\nu}\delta^{\rho\sigma} - \delta^{\mu\rho}\delta^{\nu\sigma} + \delta^{\mu\sigma}\delta^{\nu\rho} + i\qty[\delta^{\mu\nu}\epsilon^{\rho\sigma\lambda}+\delta^{\rho\sigma}\epsilon^{\mu\nu\lambda}-\delta^{\nu\lambda}\epsilon^{\rho\sigma\mu}+\delta^{\mu\lambda}\epsilon^{\rho\sigma\nu}]\gamma_\lambda] \times \nonumber \\& \quad \quad \times \qty[\frac{\tilde{A}}{\abs{k}} \delta_{\mu\sigma} + \tilde{B}\frac{k_\mu k_\sigma}{\abs{k}^3}+\tilde{C} \frac{k^\eta}{k^2}\epsilon_{\eta\mu\sigma}] =\\&= -\Gamma^{(0)}_m(x) \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{\tilde{A}}{\abs{k}} \frac{k_\nu k_\rho }{k^4} \qty[ 3\delta^{\nu\rho} + i\epsilon^{\rho\nu\lambda} \gamma_\lambda] +\tilde{B} \frac{k_\nu k_\rho }{k^4} \qty[ \frac{1}{\abs{k}}\delta^{\nu\rho} + i\frac{k_\mu }{\abs{k}^3}\qty[k^\nu \epsilon^{\rho\mu\lambda}+k^\rho \epsilon^{\mu\nu\lambda}+k^\lambda \epsilon^{\rho\mu\nu}]\gamma_\lambda]+ \nonumber \\&\quad \quad +\tilde{C} \frac{k_\nu k_\rho }{k^4} \frac{k^\eta}{k^2}\qty[\epsilon^{\nu\rho}_{\eta}- \epsilon^{\rho\nu}_{\eta}+ i\qty[ 2\delta^{\rho\nu}\delta^{\lambda}_{\eta}-\delta^{\lambda\nu}\delta^{\rho}_{\eta}]\gamma_\lambda] =\\&= -\Gamma^{(0)}_m(x) \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{3\tilde{A}+\tilde{B}}{\abs{k}^3} \end{align}\] which gives \[\require{physics} \begin{align} \Gamma^{(1)}_c(x) &= (3\tilde{A}+\tilde{B})\Gamma^{(0)}(x) \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{1}{\abs{k}^3} = \frac{3\tilde{A}+\tilde{B}}{4\pi^2} \Gamma^{(0)}(x) \log(\Lambda^2 x^2). \end{align}\] We now reinstate the fermionic indices. We are interested in singlet mass term \(\mathcal{M}_i(x)=\frac{1}{\sqrt{N}}\sum_{k=1}^N \bar{f}_{i,k} f_{i,k}\). The product \(m^\alpha_{ij} m^{\beta}_{ji}\) is zero if \(\alpha\neq \beta\). Thus, the diagram is given by the diagonal part of \(\Gamma^{(1)}_{c}\).

The sum of the first three diagrams is thus \[\require{physics} \begin{align} \qty[ -2\frac{\tilde{A} + 3\tilde{B}}{12 \pi^2} + \frac{3\tilde{A} + \tilde{B}}{2 \pi^2} ] \delta_{\alpha \beta} \log(\Lambda^2 / p^2) = \frac{4\tilde{A}}{3 \pi^2} \delta_{\alpha \beta}\log(\Lambda^2 / p^2). \end{align}\] Note that the contribution of the \(\tilde{B}\) term cancels out as expected since it contains a gauge-dependent parameter.

The last two diagrams (49 ) are \[\begin{align} G^0_i(w-x) G^0_i(y-w) G^0_i(x-y) G^0_j(0-z) G^0_j(u-0) G^0_j(z-u) m^\alpha_{ii} \tilde{D}_{ij}(z-w) m^\beta_{jj} \tilde{D}_{ji}(y-u) \end{align}\] Taking again \(u,w,y,z\) close to zero, factoring out \(\Gamma^{(0)}\) we need to calculate the matrix \[\require{physics} \begin{align} \Gamma^{(1)}_2(x) = \Gamma^{(0)}(x) \iint \dd{y}\dd{w}\dd{z}\dd{u}G^0_i(y-w) G^0_j(-z) G^0_j(u) G^0_j(z-u) \tilde{D}_{ij}(z-w) \tilde{D}_{ji}(y-u) \end{align}\] and after Fourier transform \[\require{physics} \begin{align} \Gamma^{(1)}_2(x) = \Gamma^{(0)}(x) \iint \dd{k}\dd{q} G^0_i(-k) G^0_j(q) G^0_j(-q) G^0_j(q-k) \tilde{D}_{ij}(k) \tilde{D}_{ji}(k) \end{align}\] Dropping fermionic indices for convenience, we write \[\require{physics} \begin{align} \Gamma^{(1)}_2(x) &= \Gamma^{(0)}(x) \iint \frac{\dd[3]{k}}{(2\pi)^3} \frac{\dd[3]{q}}{(2\pi)^3} \gamma^\mu G^0(-k) \tilde{D}_{\mu\rho}(k) \gamma^\nu \tilde{D}_{\nu\sigma}(k) \gamma^\rho G^0(q-k) \gamma^\sigma G^0(q) G^0(-q). \end{align}\] We define \[\require{physics} \begin{align} \Gamma^{(1)}_2(x) &= \Gamma^{(0)}(x) \iint \frac{\dd[3]{k}}{(2\pi)^3} \gamma^\mu G^0(-k) \tilde{D}^\top_{\mu\rho}(k) \gamma^\nu \tilde{D}_{\nu\sigma}(k) \Pi^{\rho\sigma}(k) \label{appeq:gamma2oneint} \end{align}\tag{50}\] where \(\Pi^{\rho\sigma}(k)\), calculated below, is given by \[\require{physics} \begin{align} \Pi^{\rho\sigma}(k) &= \int \frac{\dd[3]{q}}{(2\pi)^3} \Tr[G^0(q) \gamma^\rho G^0(q-k) \gamma^\sigma G^0(-q)] = \int \frac{\dd[3]{q}}{(2\pi)^3} \Tr[\frac{ \gamma^\beta q_\beta \gamma^\rho \gamma^\alpha (q-k)_\alpha \gamma^\sigma \gamma^\delta q_\delta}{q^4 (q-k)^2}] = \nonumber\\&= i\epsilon^{\rho \sigma \alpha } \frac{k_\alpha}{8\abs{k}} \end{align}\] To calculate \[\require{physics} \begin{align} \int \frac{\dd[3]{q}}{(2\pi)^3} \Tr[\frac{ \gamma^\beta \gamma^\rho \gamma^\alpha \gamma^\sigma \gamma^\delta q_\beta(q-k)_\alpha q_\delta}{q^4 (q-k)^2}] \end{align}\] we use \[\begin{align} \gamma^\beta \gamma^\rho = \delta^{\beta\rho} + i \epsilon^{\beta\rho\kappa} \gamma_\kappa \end{align}\] and write \[\require{physics} \begin{align} \Tr[\gamma^\beta \gamma^\rho \gamma^\alpha \gamma^\sigma \gamma^\delta] = \delta^{\beta\rho} \Tr[ \gamma^\alpha \gamma^\sigma \gamma^\delta]+ i \epsilon^{\beta\rho\kappa} \Tr[\gamma_\kappa \gamma^\alpha \gamma^\sigma \gamma^\delta] \end{align}\] Using \[\require{physics} \begin{align} \Tr[\gamma^\mu \gamma^\nu \gamma^\rho] &= 2i \epsilon^{\mu\nu\rho} \end{align}\] we get \[\require{physics} \begin{align} \Tr[\gamma^\beta \gamma^\rho \gamma^\alpha \gamma^\sigma \gamma^\delta] = 2i\delta^{\beta\rho}\epsilon^{\alpha\sigma\delta}+ i \epsilon^{\beta\rho\kappa} \Tr[\gamma_\kappa \gamma^\alpha \gamma^\sigma \gamma^\delta] \end{align}\] Repeating the two-trace exercise \[\require{physics} \begin{align} \Tr[\gamma_\kappa \gamma^\alpha \gamma^\sigma \gamma^\delta] = \delta_{\kappa}^\alpha \Tr[\gamma^\sigma \gamma^\delta] + i{\epsilon_{\kappa}^{\alpha}}_{\eta} \Tr[\gamma^\eta \gamma^\sigma \gamma^\delta] \end{align}\] Substituting also \[\require{physics} \begin{align} \Tr[\gamma^\sigma \gamma^\delta] &= \delta^{\sigma \delta} \end{align}\] we get \[\require{physics} \begin{align} \Tr[\gamma_\kappa \gamma^\alpha \gamma^\sigma \gamma^\delta] &= 2\delta_{\kappa}^\alpha \delta^{ \sigma \delta} -2{\epsilon_{\kappa}^{\alpha}}_{\eta} \epsilon^{ \sigma \delta\eta} = 2\delta_{\kappa}^\alpha \delta^{ \sigma \delta} -2\delta_{\kappa}^{\sigma }\delta^{\alpha \delta} + 2\delta_{\kappa}^{ \delta }\delta^{\alpha \sigma} \end{align}\] contracting we get \[\require{physics} \begin{align} \epsilon^{\beta\rho\kappa} \Tr[\gamma_\kappa \gamma^\alpha \gamma^\sigma \gamma^\delta] &= 2 \epsilon^{\beta\rho\alpha} \delta^{ \sigma \delta} -2 \epsilon^{\beta\rho\sigma} \delta^{\alpha \delta} + 2 \epsilon^{\beta\rho\delta}\delta^{\alpha \sigma} \end{align}\] which gives us \[\require{physics} \begin{align} \Tr[\gamma^\beta \gamma^\rho \gamma^\alpha \gamma^\sigma \gamma^\delta] = 2i \qty[ \delta^{\beta\rho}\epsilon^{\alpha\sigma\delta} + \epsilon^{\beta\rho\alpha} \delta^{ \sigma \delta} - \epsilon^{\beta\rho\sigma} \delta^{\alpha \delta} + \epsilon^{\beta\rho\delta}\delta^{\alpha \sigma}] \end{align}\] Finally \[\require{physics} \begin{align} 2i &\int \frac{\dd[3]{q}}{(2\pi)^3} \frac{ q_\beta (q-k)_\alpha q_\delta}{q^4 (q-k)^2} \qty[ \delta^{\beta\rho}\epsilon^{\alpha\sigma\delta} + \epsilon^{\beta\rho\alpha} \delta^{ \sigma \delta} - \epsilon^{\beta\rho\sigma} \delta^{\alpha \delta} + \epsilon^{\beta\rho\delta}\delta^{\alpha \sigma}] = \nonumber \\=-2i &\int \frac{\dd[3]{q}}{(2\pi)^3} \qty[ \epsilon^{\alpha\sigma\beta} \frac{ q^\rho k_\alpha q_\beta}{q^4 (q-k)^2} + \epsilon^{\beta\rho\alpha} \frac{ q_\beta k_\alpha q^\sigma}{q^4 (q-k)^2} + \epsilon^{\beta\rho\sigma} \frac{ q_\beta }{q^2 (q-k)^2} - \epsilon^{\beta\rho\sigma} \frac{ q_\beta k^\alpha q_\alpha}{q^4 (q-k)^2} ] \end{align}\] Using rotational invariance, we get \[\require{physics} \begin{align} \Pi^{\rho\sigma} &=2i \epsilon^{\alpha\sigma\rho} \int \frac{\dd[3]{q}}{(2\pi)^3} \frac{ k_\alpha -q_\alpha }{q^2 (q-k)^2} \end{align}\] We now use the integrals 20 21 to get \[\begin{align} \Pi^{\rho\sigma}(k) &= i \epsilon^{\alpha\sigma\rho} \frac{k^\alpha}{8\abs{k}} \label{appeq:pi95result} \end{align}\tag{51}\] and thus, substituting \(\Pi^{\rho\sigma}(k)\) into 50 \[\require{physics} \begin{align} \Gamma^{(1)}_2(x) &= \Gamma^{(0)}(x) \iint \frac{\dd[3]{k}}{(2\pi)^3} \gamma^\mu G^0(-k) \tilde{D}^\top_{\mu\rho}(k) \gamma^\nu \tilde{D}_{\nu\sigma}(k) \Pi^{\rho\sigma} =\\&= -i \Gamma^{(0)}(x) \int \frac{\dd[3]{k}}{(2\pi)^3} \gamma^\mu \frac{\gamma^\lambda k_\lambda}{k^2} \qty[\frac{\tilde{A}}{\abs{k}} \delta_{\mu\rho} + \tilde{B}\frac{k_\mu k_\rho}{\abs{k}^3}+\tilde{C} \frac{k^\alpha}{k^2}\epsilon_{\alpha\mu\rho}]^\top\gamma^\nu \qty[ \frac{\tilde{A}}{\abs{k}} \delta_{\nu\sigma} + \tilde{B}\frac{k_\nu k_\sigma}{\abs{k}^3}-\tilde{C} \frac{k^\beta}{k^2}\epsilon_{\beta\nu\sigma} ] \epsilon^{\rho \sigma \eta } \frac{k_\eta}{8\abs{k}} =\\&= -i \Gamma^{(0)}(x) \int \frac{\dd[3]{k}}{(2\pi)^3} \gamma^\mu \frac{\gamma^\lambda k_\lambda}{k^2} \qty[\frac{\tilde{A}}{\abs{k}} \delta_{\mu\rho} + \tilde{B}\frac{k_\mu k_\rho}{\abs{k}^3}+\tilde{C} \frac{k^\alpha}{k^2}\epsilon_{\alpha\mu\rho}]^\top\gamma^\nu \qty[ -\frac{\tilde{A}}{\abs{k}}\epsilon_{\nu}^{\rho \eta } \frac{k_\eta}{8\abs{k}}+\tilde{C} \qty(\frac{k^\rho}{k^2}\frac{k_\nu}{8\abs{k}}-\frac{1}{8\abs{k}}\delta_{\nu}^{\rho }) ] =\\&= -i \Gamma^{(0)}(x) \int \frac{\dd[3]{k}}{(2\pi)^3} \gamma^\mu \frac{\gamma^\lambda k_\lambda}{k^2} \qty{ -\qty[\frac{\tilde{A}^\top\tilde{A} k_\eta}{8\abs{k}^3 } \epsilon_{\nu\mu}^{ \eta } -\frac{\tilde{A}^\top\tilde{C}}{8k^2}\qty(\frac{k_\mu k_\nu}{k^2} - \delta_{\mu\nu}) ] +\tilde{C}^\top\frac{1}{8\abs{k}} \qty[\frac{\tilde{A}}{\abs{k}} \qty(\frac{k_\mu k_\nu}{k^2}-\delta_{\mu\nu})-\tilde{C} \frac{k^\alpha}{k^2}\epsilon_{\nu\alpha\mu}] }\gamma^\nu =\\&= i \Gamma^{(0)}(x) \int \frac{\dd[3]{k}}{(2\pi)^3} \gamma^\mu \frac{\gamma^\lambda k_\lambda}{k^2} \qty[\frac{(\tilde{A}^\top\tilde{A}-\tilde{C}^\top\tilde{C}) k^\alpha}{8\abs{k}^3 } \epsilon_{\nu\mu \alpha } \gamma^\nu + \frac{\tilde{C}^\top\tilde{A} - \tilde{A}^\top\tilde{C}}{8k^2}\qty(\frac{k_\mu k_\nu}{k^2}-\delta_{\mu\nu})] \end{align}\] The the second term is \(k\)-odd, so we can drop it. We now substitute 18 : \[\require{physics} \begin{align} \gamma^\mu \gamma^\lambda \gamma^\nu \epsilon_{\nu\mu \alpha } = \qty[\delta^{\mu\lambda}\gamma^\nu + \delta^{\lambda\nu}\gamma^\mu ]\epsilon_{\nu\mu \alpha }+ 2i\delta^{\lambda}_{\alpha } \end{align}\] giving us (the first two terms cancel) \[\require{physics} \begin{align} \Gamma^{(1)}_2(x) &= - \Gamma^{(0)}(x) \frac{\tilde{A}^\top\tilde{A}-\tilde{C}^\top\tilde{C}}{4} \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{ k_\alpha}{k^2} \frac{ k^\alpha}{\abs{k}^3 } = - \Gamma^{(0)}(x) \frac{\tilde{A}^\top\tilde{A}-\tilde{C}^\top\tilde{C}}{4} \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{ 1}{\abs{k}^3} \end{align}\] resulting in \[\require{physics} \begin{align} \Gamma^{(1)}_2(x) &= - \Gamma^{(0)}(x) \frac{\tilde{A}^\top\tilde{A}-\tilde{C}^\top\tilde{C}}{4} \int \frac{\dd[3]{k}}{(2\pi)^3} \frac{ 1}{ \abs{k}^3} = - \Gamma^{(0)}(x) \frac{\tilde{A}^\top\tilde{A}-\tilde{C}^\top\tilde{C}}{16\pi^2} \log(\Lambda^2 / p^2) \end{align}\] For \(N>1\), we can use 34 35 to calculate the value and then multiply it by the number of species, \(N\).

Finally, the second diagram in 49 is identical to the first one, but with \(q+k\) instead of \(q-k\) in the fermion propagator. As a result, in 51 we get \(i \epsilon^{\alpha\sigma\rho} \frac{3 k^\alpha}{8\abs{k}}\) for a total multiplicative factor of \(4\).

In total, we get \[\require{physics} \begin{align} \Gamma^{(1)}(x) = \qty[\frac{4\tilde{A}_{\alpha\beta}}{3 \pi^2}\delta_{\alpha\beta} - N \qty[\frac{ \tilde{A}^\top\tilde{A}-\tilde{C}^\top\tilde{C}}{4\pi^2}]_{\alpha\beta}] \log(\Lambda^2 x^2),\label{appeq:final95log95mass} \end{align}\tag{52}\] as claimed in 11 .

3.6 Scaling dimensions↩︎

To calculate the scaling dimensions, we need the eigenvalues of the matrix (52 ). We also include the calculations of the adjoint mass term, which we do not use in our paper, to verify our results against the literature [82], [113].

3.6.1 QED\(_3\)↩︎

For QED\(_3\), the \(\Gamma^{(1)}(x)\) is a scalar with \(a = \frac{16}{N}\) and \(c=0\) and thus we get \[\begin{align} \Delta_{\text{scalar}}^{(1)} = -\frac{16a - 3a^2 N + 3c^2 N}{12 \pi^2} = \frac{128}{3N \pi^2}\\ \Delta_{\text{adj}}^{(1)} = -\frac{16a }{12 \pi^2} = -\frac{64}{3N \pi^2} \end{align}\]

3.6.2 Chern–Simons–QED\(_3\)↩︎

For Chern–Simons–QED\(_3\), we substitute the scalars \(a=\frac{16}{N(1+\lambda^2)}\) and \(c= - a \lambda\)

\[\begin{align} \Delta_{\text{scalar}}^{(1)} = -\frac{16a - 3a^2 N + 3c^2 N}{12 \pi^2} &= -\frac{64 }{3 N(1+\lambda^2)\pi^2}+\frac{ 64 (1-\lambda^2)}{ N(1+\lambda^2)^2 \pi^2}\\ \Delta_{\text{adj}}^{(1)} &= -\frac{64 }{3 N(1+\lambda^2)\pi^2} \end{align}\]

3.6.3 \(U(1)^k\) Chern–Simons–QED\(_3\)↩︎

In this case, \(\Delta\) depends on flavor, so to get the scaling dimensions, we calculate the eigenvalues of the matrix in 52 . For the example used in the main text, defined by 12 13 , we substitute the values from 34 35 and acquire the eigenvalues \[\require{physics} \begin{align} \Delta^{(1)}_{1,2} &= -\frac{32}{3 \qty(20 +\pi^2 )}\\ \Delta^{(1)}_{3,4} &= -\frac{160 \qty(28 - \pi ^2 )}{3 \qty(20 +\pi ^2 )^2}. \end{align}\] For \(N=1\) we get \(\Delta^{(1)}_{1,2} \approx -1.083\) and \(\Delta^{(1)}_{3,4} \approx -0.357\).

For the second example, given by 14 15 , the eigenvalues can be calculated only numerically, and are equal \(\require{physics} \qty{-1.35558, -1.20776, -1.11313, -0.421461}\).

References↩︎