Define and charge the simplest monopole operators

A local operator need not be a polynomial in the elementary fields. We can also insert an operator by requiring fields to have a prescribed singularity at its position. In three dimensions, an especially useful example inserts quantized magnetic flux. Its charge under the magnetic symmetry follows directly from that definition. Its charges under matter symmetries require a quantum calculation.

Start with the SQED branches and compact dual photon and radial quantization. We retain three-dimensional $\mathcal N=2$ SQED, with four real Lorentzian supercharges, compact gauge group $U(1)$, and two independent chiral multiplets $Q,\widetilde Q$ of gauge charges $+1,-1$. They are not conjugate multiplets. The gauge Chern–Simons level, superpotential, masses, and FI parameter vanish. The paired regulator preserves this $k=0$ theory.

The common axial charge of both matter multiplets is $+1$. Write their scalar trial R-charge as $r$; their fermions then have R-charge $r-1$. Our axial and reference R-generators are even under the symmetry exchanging $Q,\widetilde Q$ while sending $a,\sigma$ to $-a,-\sigma$. This fixes their possible mixing with the odd topological generator. We add no mixed gauge–background counterterm that changes that choice.

Insert flux at a point

Excise a small ball of radius $\rho$ about the insertion. On its outward-oriented boundary impose

$$\frac{1}{2\pi}\int_{S^2_\rho}f=m,\qquad m\in\mathbb Z,\qquad f=da.$$

The integer follows from the compact gauge group with unit electric charges. For example, potentials regular on northern and southern patches are

$$a_N=\frac m2(1-\cos\theta)d\varphi,\qquad a_S=-\frac m2(1+\cos\theta)d\varphi.$$

They give $f=(m/2)\sin\theta\,d\theta\wedge d\varphi$ and differ by $m\,d\varphi$. With $D^{(q)}=d-iqa$, a charge-$q$ field has transition factor $e^{iqm\varphi}$. Single-valuedness for $q=1$ requires $m\in\mathbb Z$.

This is a local disorder insertion: the path integral integrates over fields satisfying the boundary condition. Its short-distance normalization must be renormalized. It is not a smooth finite-energy magnetic soliton in four-dimensional spacetime, whose particle trajectory is a worldline. Here a point is surrounded by a two-sphere in three-dimensional Euclidean spacetime. No smooth monopole core is being assumed.

Normalize the topological current by

$$j_J=\frac{*f}{2\pi},\qquad j_J^\mu=\frac{1}{4\pi}\epsilon^{\mu\nu\rho}f_{\nu\rho}.$$

Away from insertions its conservation is the Bianchi identity. The integrated flux says that an insertion in sector $m$ has $J=m$. Equivalently, the dual photon has $\gamma\sim\gamma+2\pi$, and a local factor $e^{im\gamma}$ transforms with charge $m$ under $\gamma\mapsto\gamma+\alpha$. This identifies a magnetic charge, not a complete quantum operator or its other charges.

Flux alone does not specify a unique operator. One may choose different states in that flux sector or attach matter excitations. We seek the simplest scalar chiral monopole $V_m$, with a supersymmetric boundary condition and no extra matter excitations. “Chiral” means that the same family of supercharges that annihilates a chiral scalar also annihilates this insertion.

Complete the boundary condition supersymmetrically

Set matter, fermions, and the vector auxiliary $D$ to zero in the boundary background. Define $b_\mu=\tfrac12\epsilon_{\mu\nu\rho}f^{\nu\rho}$. The already established flat vector transformation contains

$$\delta_{\widetilde\zeta}\widetilde\lambda =i\gamma^\mu\widetilde\zeta(\partial_\mu\sigma-b_\mu).$$

Thus preserving this family for arbitrary constant $\widetilde\zeta$ requires $\nabla\sigma=\mathbf b$. Since $\mathbf b=m\widehat{\boldsymbol\rho}/(2\rho^2)$, choose

$$\sigma=-\frac{m}{2\rho}.$$

The opposite sign preserves the other family and defines the anti-chiral boundary condition. Both positive and negative $m$ are allowed in the chiral choice above.

For comparison with the spectral calculation of Borokhov, Kapustin, and Wu, define $\chi=-\sigma$. Their chiral background obeys $f=-*d\chi$ and $\chi=m/(2\rho)$. This is an explicit scalar-sign translation, not a change of the earlier vector rules. Under the radial map to a cylinder with sphere radius $L$, the corresponding background is $\chi_{\rm cyl}=m/(2L)$. See their supersymmetric construction, section 2.2.

The sphere used for quantization surrounds the insertion; it is not a new physical compactification of the original theory. At a conformal fixed point the insertion prepares a state on this sphere. We will use its charged matter modes to calculate symmetry charges, without treating a free fluctuation spectrum as the exact interacting energy spectrum.

Count the angular zero modes

Let a fermion have gauge charge $q$, and set $n=qm$. Work on the unit sphere for this angular calculation; physical inverse lengths are restored by $1/L$. Use Pauli matrices $\tau_1,\tau_2,\tau_3$ for the two-dimensional tangent Clifford algebra. In the polar orthonormal frame, the Hermitian angular Dirac operator is

$$\mathscr D_n =-i\tau_1\left(\partial_\theta+\frac12\cot\theta\right) -\frac{i\tau_2}{\sin\theta} \left[\partial_\varphi-\frac{in}{2}(1-\cos\theta)\right].$$

The $\tfrac12\cot\theta$ is the spin connection. It also makes the derivative formally Hermitian with the measure $\sin\theta\,d\theta\,d\varphi$. Define angular chirality $\Gamma=\tau_3$; then $\{\mathscr D_n,\Gamma\}=0$.

We can count the zero modes directly. For $n>0$, try an upper-component spinor $e^{ik\varphi}(u(\theta),0)^T$. The zero-mode equation becomes

$$\frac{u'}{u} =-\frac12\cot\theta +\frac{k-\frac n2(1-\cos\theta)}{\sin\theta}.$$

The polar spin frame makes the components antiperiodic under one turn, so $k=j+\tfrac12$ with integer $j$. Integration gives

$$u_j(\theta) =C_j\sin^j\frac\theta2\, \cos^{n-1-j}\frac\theta2.$$

Regularity in smooth frames at the two poles requires $j\geq0$ and $n-1-j\geq0$. Hence there are precisely $n$ such modes, $j=0,\ldots,n-1$. For the lower component the two powers would instead be $-j-1$ and $j-n$; they cannot both be nonnegative. There are no lower-component zero modes for $n>0$.

For $n<0$, the calculation exchanges the components and reverses the azimuthal phase. It gives $|n|$ lower-component modes. Therefore

$$\dim\ker\mathscr D_n=|n|,\qquad \Gamma\big|_{\ker\mathscr D_n}=\operatorname{sgn}(n).$$

At $n=0$ neither component has an allowed zero mode. Every nonzero eigenvalue has an opposite partner, because $\Gamma$ maps an eigenstate of eigenvalue $\lambda$ to one of $-\lambda$. The unpaired modes are the essential information. The angular-momentum spectrum gives the same count: its lowest spin is $(|n|-1)/2$ and its multiplicity is $|n|$. This agrees with the monopole-harmonic analysis in BKW, section 4.1 and Appendix.

For unit flux and either matter charge there is one angular zero mode. Its angular dependence in the polar frame is only $e^{\pm i\varphi/2}$, with normalized constant amplitude $1/\sqrt{4\pi}$. The frame dependence does not make it a spin-half rotational multiplet: the combined spatial and gauge rotation acts with total spin zero.

The supersymmetric background lifts these modes

Angular zero modes are not zero-energy states of the full supersymmetric radial problem. In a Hermitian one-particle convention its fermion operator is unitarily equivalent to

$$L H_n=\mathscr D_n+\frac n2\Gamma.$$

The second term is the coupling to $q\chi_{\rm cyl}$. On an angular zero mode it gives

$$E_0=\frac{n\,\operatorname{sgn}(n)}{2L} =\frac{|n|}{2L}>0.$$

For a nonzero angular pair, choose a basis in which $\mathscr D_n=\lambda\tau_1$ and $\Gamma=\tau_3$. Then

$$L H_n= \begin{pmatrix}n/2&\lambda\\ \lambda&-n/2\end{pmatrix}, \qquad E_\pm=\pm\frac{\sqrt{\lambda^2+n^2/4}}L.$$

The paired levels remain symmetric, while all $|n|$ unpaired levels have positive energy. Reversing the scalar boundary condition reverses the unpaired energies. This explains why the supersymmetric matter vacuum is selected, rather than a freely chosen filling of degenerate zero-energy levels.

These are fluctuation operators in the prescribed background. Their use below is a vacuum-charge calculation. In particular, summing these free-mode energies at one flavor would not determine an exact interacting scaling dimension.

Derive the vacuum charges by symmetric ordering

For an unpaired fermionic oscillator with $\{c,c^\dagger\}=1$, suppose $c^\dagger$ adds charge $F$ under a chosen symmetry. Symmetric ordering assigns the generator

$$\widehat F_{\rm mode} =\frac F2(c^\dagger c-cc^\dagger) =F\left(c^\dagger c-\frac12\right).$$

The empty and occupied states then carry $-F/2,+F/2$. Assigning zero charge to the empty state would be a different prescription. We use symmetric ordering and subtract the zero-flux vacuum charge, consistently with the paired regulator and the symmetry convention at the start.

The positive-energy unpaired modes are empty in the chiral background vacuum. Opposite nonzero energy levels cancel in the vacuum offset. More explicitly, with a positive regulator $\varepsilon$, one fermion contributes

$$F_{\rm vac}(\varepsilon) =-\frac F2\sum_E \operatorname{sgn}(E)e^{-\varepsilon L|E|} =-\frac F2\,|qm|e^{-\varepsilon |qm|/2}.$$

Taking $\varepsilon\to0^+$ gives $-|qm|F/2$. This is spectral asymmetry: a charge induced by the imbalance of positive and negative fermion levels. The scalar spectrum is paired and has no corresponding vacuum-charge offset. The Abelian gaugino is neutral, so it contributes no flux-dependent term. Thus for the undressed chiral vacuum,

$$\boxed{F(V_m)=-\frac12\sum_{\text{matter fermions }i} |q_i m|F_i.}$$

The spectral argument is the charge calculation in BKW2, section 2.2, equation (2). Its section 4.1 explains the index argument against perturbative corrections and states the limitations of its large-flavor derivation. We do not use that approximation for an exact energy.

There is a separate check applicable to this finite-flavor theory: the supersymmetric index formula contains the vacuum flavor charge $q_{0F}=-\sum_i|q_i s|F_i$, where its magnetic label $s$ is half-integral. Our flux is $m=2s$, so it gives the boxed coefficient exactly. No index calculation is required here; this is a translation check on the vacuum factor, not a derivation of the full trace. See IDX3, equation (2.4) and its definitions.

Worked example: the meson and the two elementary monopoles

For the two charged matter fermions,

Fermion	Gauge charge	Axial charge	Trial R-charge
$\psi_Q$	$+1$	$+1$	$r-1$
$\psi_{\widetilde Q}$	$-1$	$+1$	$r-1$

Each has $|m|$ angular zero modes. Applying the charge formula gives

$$\begin{aligned} Q_{\rm gauge}(V_m) &=-\frac{|m|}{2}(1-1)=0,\\ A(V_m)&=-\frac{|m|}{2}(1+1)=-|m|,\\ R_{\rm trial}(V_m) &=-\frac{|m|}{2}\bigl[(r-1)+(r-1)\bigr] =|m|(1-r). \end{aligned}$$

At $k=0$ the vanishing gauge charge lets the bare state satisfy the integrated Gauss law without charged dressing. The magnetic charge is instead fixed by the insertion: $J(V_m)=m$.

Define $V_+=V_{m=1}$ and $V_-=V_{m=-1}$. The meson is the scalar bottom component of $Q\widetilde Q$, namely $M=\phi_+\phi_-$. Its charges add directly:

Operator	Gauge charge	Topological charge $J$	Axial charge $A$	Trial R-charge
$M$	$0$	$0$	$2$	$2r$
$V_+$	$0$	$+1$	$-1$	$1-r$
$V_-$	$0$	$-1$	$-1$	$1-r$

The absolute value in the fermion count explains the equal last two columns for opposite fluxes. Using $m$ instead of $|m|$ would give the wrong axial charge for $V_-$.

Two source conventions provide useful checks. AHISS (3.1) uses $R_0(Q)=R_0(\widetilde Q)=0$; it gives $R_0(V_\pm)=1$. Our $R_{\rm trial}=R_0+rA$ then gives $1-r$. Willett–Yaakov’s charge table instead starts at matter R-charge $1/2$; setting $N_c=N_f=1$ gives $R_{\rm UV}(V_\pm)=1/2$. Mixing by $(r-1/2)A$ gives the same result.

What the charges do and do not determine

Both $V_+$ and $V_-$ are chiral. They are exchanged by the charge-exchange symmetry, but they are not Hermitian conjugates. For example, $V_+^\dagger$ has axial charge $+1$ and is anti-chiral, whereas $V_-$ has axial charge $-1$. Equal and opposite magnetic charges alone do not identify two operators.

Nor does the semiclassical phase $e^{\pm i\gamma}$ imply $V_+V_-=1$. Quantum monopoles include the matter-vacuum response we just calculated: the product has axial charge $-2$, while the identity has zero. Determining products in the chiral ring requires dynamics beyond this charge calculation.

Finally, $r$ is a choice of trial R-generator. At a unitary three-dimensional superconformal fixed point, a scalar chiral primary obeys $\Delta=R_{\rm SC}$ for the R-generator in the superconformal algebra. Hence the charge expressions would give $\Delta(M)=2r_*$ and $\Delta(V_\pm)=1-r_*$ once that generator is identified. They do not determine $r_*$, and arbitrary trial charges are not measured scaling dimensions. The distinction is also explicit in IDX3, section 2.1, footnote 1.

Exercises

1. Reverse and double the flux

Take the same chiral boundary condition with $m=-2$. Determine the angular zero-mode count and chirality for each matter fermion, then all four charges of the bare monopole. Explain why replacing $|m|$ by $m$ in the charge formula fails.

Hint 1

The angular Dirac operator sees $n=qm$, not just the magnetic flux.

Hint 2

The mass term is $(n/2)\Gamma$. Evaluate it on a zero mode before deciding which levels are occupied.

Solution

For $\psi_Q$, $n=-2$: there are two lower-component zero modes with $\Gamma=-1$. For $\psi_{\widetilde Q}$, $n=+2$: there are two upper-component zero modes with $\Gamma=+1$. Both sets have $E_0=1/L$, so they are empty in the bare chiral vacuum.

Each set contributes $-F_i$, giving

$$Q_{\rm gauge}=-(+1)-(-1)=0,\qquad A=-1-1=-2, \qquad R_{\rm trial}=-2(r-1)=2(1-r),\qquad J=-2.$$

The normal-order offset counts oscillators, whose number cannot be negative. Reversing flux changes angular chirality and the sign of the scalar mass simultaneously, leaving their product positive. Using signed $m$ would incorrectly turn the empty-state offset into the occupied-state offset.

2. Gauge neutrality does not select the bare chiral state

At $m=1$, retain only the two unpaired fermionic oscillators. Let $c_+^\dagger,c_-^\dagger$ add the charges of the two rows in the fermion table. Calculate gauge, axial, and trial R-charges for the empty state, each singly occupied state, and the doubly occupied state. Which satisfy the integrated Gauss law? Does that make all of them bare chiral monopoles?

Hint 1

Start at $(Q_{\rm gauge},A,R)=(0,-1,1-r)$ and add a fermion’s charges for each occupation.

Hint 2

Each occupation also raises the energy of this fluctuation problem by $1/(2L)$. The angular zero mode has already been lifted.

Solution

The four occupation states have

State	$Q_{\rm gauge}$	$A$	$R_{\rm trial}$	Added fluctuation energy
$\lvert0\rangle$	$0$	$-1$	$1-r$	$0$
$c_+^\dagger\lvert0\rangle$	$+1$	$0$	$0$	$1/(2L)$
$c_-^\dagger\lvert0\rangle$	$-1$	$0$	$0$	$1/(2L)$
$c_+^\dagger c_-^\dagger\lvert0\rangle$	$0$	$+1$	$r-1$	$1/L$

All remain in sector $J=1$. The empty and doubly occupied states pass the integrated gauge-charge constraint; the singly occupied states need additional charged excitations to do so. Passing this necessary constraint alone does not specify the full gauge-invariant interacting state or make it chiral. The bare chiral construction selects the empty supersymmetric matter vacuum. The doubly occupied configuration is an excitation, not a second choice of degenerate vacuum and not $V_-$, whose flux is negative.

3. Change the trial generator without changing the matter charges

Allow a trial R-generator $R'=R_0+rA+tJ$ with real $t$. Find the charges of $M,V_\pm$, and explain what fixes $t=0$ when the trial generator respects the charge-exchange symmetry. Compare $V_-$ with $V_+^\dagger$ and decide whether their magnetic charges alone can establish an operator identity.

Hint 1

The elementary matter fields have $J=0$. The monopoles do not.

Hint 2

Charge exchange preserves the chiral family and exchanges $V_+,V_-$. Hermitian conjugation instead reverses every Abelian charge and changes chirality.

Solution

The elementary scalar R-charges still equal $r$, but

$$R'(M)=2r,\qquad R'(V_+)=1-r+t,\qquad R'(V_-)=1-r-t.$$

Thus elementary matter assignments alone would leave a further mixing ambiguity. Since charge exchange sends $J\mapsto-J$ and leaves $R_0,A$ invariant, a generator invariant under that symmetry has $t=0$.

In contrast, $V_+^\dagger$ has $(J,A,R')=(-1,+1,-1+r-t)$ and is anti-chiral. The chiral $V_-$ has $(-1,-1,1-r-t)$. Their axial charges already forbid identifying them, regardless of $r,t$. An accidental equality of one trial R-charge would not remove this obstruction. None of this determines the actual superconformal value of $r$.

You can now derive the charges of the three simplest gauge-invariant chiral coordinates. The next question is dynamical: what infrared theory describes them together, and which products survive in its chiral ring? Build the SQED–XYZ dictionary by deriving the cubic model’s three branches and matching their operators and charges.