Monopole operators, flux, and matter dressing

A flux boundary condition can define a local operator without writing a polynomial of elementary fields. In a Chern–Simons theory, however, flux carries electric gauge weight. The boundary condition must be accompanied by matter that cancels that weight. We will calculate the required dressing, then ask when its R-charge determines a physical dimension.

Read the compact many-brane gauge data, the definition and vacuum charges of a monopole, and three-dimensional chiral shortening. We retain three-dimensional compact $U(N)_k\times U(N)_{-k}$, equal finite $N$, integer $k>0$, ordinary fundamental traces, no extra common-center quotient, and no mass or FI term.

The full generic theory has twelve real Poincaré supercharges; our $\mathcal N=2$ description displays four. At levels one and two the full theory has sixteen, whose enhancement mechanism is a separate topic. Its chiral fields are $A_1,A_2$ in $(N,\overline N)$ and $B_1,B_2$ in $(\overline N,N)$, all with the chosen superconformal R-charge $1/2$. The vector sectors are Chern–Simons sectors, not propagating Yang–Mills multiplets. Constructing a charged monopole alone does not establish the extra currents.

Specify the flux and its remaining gauge symmetry

On a small outward-oriented sphere around an insertion, choose a Cartan representative

$$\frac1{2\pi}\int_{S^2} f =H=\operatorname{diag}(m_1,\ldots,m_N),\qquad \frac1{2\pi}\int_{S^2}\widehat f =\widehat H=\operatorname{diag}(\widehat m_1,\ldots,\widehat m_N). \tag{AB09.1}$$

This notation refers to a gauge choice for the monopole boundary condition. More invariantly, the equatorial transition functions are $\exp(i\varphi H)$ and $\exp(i\varphi\widehat H)$. A cocharacter is such a homomorphism from a circle into the gauge group. For the stated $U(N)$ groups its entries are integers, modulo independent permutations at the two nodes. There is no tracelessness condition.

For example, the northern and southern first-node potentials are $a_N=H(1-\cos\theta)d\varphi/2$ and $a_S=-H(1+\cos\theta)d\varphi/2$. Their difference is $H\,d\varphi$; the transition closes because $e^{2\pi iH}=I_N$. This generalizes the Abelian definition directly.

The flux leaves invariant the subgroup commuting with $H$. If an integer occurs $r$ times, that block retains a full $U(r)$, not merely $r$ independent circles. A repeated flux therefore requires a non-Abelian singlet test. This is the residual gauge group in Bashkirov–Kapustin, §2.5.

Flux alone does not choose a unique state. A supersymmetric radial regulator completes the boundary condition with vector-scalar profiles and selects a vacuum before matter excitations are added. In the site’s chiral convention the local scalar profile is $\sigma=-H/(2\rho)$, as in the monopole prerequisite, with the corresponding hatted profile. The bare monopole below means this regulated vacuum before gauge projection. It is not a solution of the original auxiliary equations with all matter arbitrarily set to zero, and it need not be a physical local operator.

Derive the Chern–Simons gauge weight

The inherited Euclidean connection action is

$$S_{{\rm CS},E}=\frac{ik}{4\pi} \int\left[\omega_3(a)-\omega_3(\widehat a)\right], \qquad \omega_3(a)=\operatorname{Tr}\left(a\,da-\frac{2i}{3}a^3\right), \tag{AB09.2}$$

with weight $e^{-S_E}$. Let $h_i=\oint a_i$ and $\widehat h_j=\oint\widehat a_j$ be Cartan holonomies around an auxiliary circle in $S^1\times S^2$. The globally defined spin Chern–Simons functional obeys

$$\delta S_{{\rm CS},E} =ik\sum_i m_i\,\delta h_i -ik\sum_j\widehat m_j\,\delta\widehat h_j. \tag{AB09.3}$$

The coefficient follows from $\delta S_E=ik\int\operatorname{Tr}(\delta a\wedge f)/(2\pi)$. Using only one local $a\,da$ density on a magnetic bundle would miss patch terms and can lose a factor two.

We must also fix the sign of the matter character. Define the gauge generator by $[Q,\phi]=q\phi$, so the field transforms as $e^{i\alpha Q}\phi e^{-i\alpha Q}=e^{iq\alpha}\phi$. For $D_\tau=\partial_\tau-iqa_\tau$, remove a constant connection by $\chi=e^{-iqh\tau/\beta}\phi$. A periodic field becomes $\chi(\beta)=e^{-iqh}\chi(0)$. The resulting trace has insertion $e^{-ihQ}$. Equivalently, a charged wavefunction $\psi_q(re^{i\varphi})\propto e^{iq\varphi}$ acquires $e^{-iqh}$ when its endpoint is rotated by $-h$. This distinguishes a field-transformation weight from a convention chosen for a trace fugacity.

Combining this character with (AB09.3), the Cartan projection contains

$$\int\frac{dh_i}{2\pi}\, e^{-ih_i(km_i+Q_{L,i})},\qquad \int\frac{d\widehat h_j}{2\pi}\, e^{-i\widehat h_j(-k\widehat m_j+Q_{R,j})}. \tag{AB09.4}$$

Thus the bare insertion has field-transformation weights

$$\boxed{ Q_{L,i}(T_{m,\widehat m})=k m_i,\qquad Q_{R,j}(T_{m,\widehat m})=-k\widehat m_j. } \tag{AB09.5}$$

Matter must cancel these weights. For $A_{ij}$ the two weights are $(+e_i,-\widehat e_j)$; for $B_{ji}$ they are $(-e_i,+\widehat e_j)$, where $e_i$ selects one Cartan component. In this convention positive matched flux is dressed by B fields. Equal negative flux is dressed by A fields.

Summing both nodes’ constraints gives a necessary condition for a local operator dressed only with the theory’s fields:

$$\sum_i m_i=\sum_j\widehat m_j. \tag{AB09.6}$$

Every matter excitation has zero charge under the common central transformation, and adjoint excitations do too. An unequal total flux cannot be repaired by adding more bifundamentals. Equality of these totals is not the full residual-group singlet condition. We normalize the two topological charges as $J_L=\sum_i m_i$ and $J_R=\sum_j\widehat m_j$; on these physical sectors their common value is $J$. This is a flux normalization, not BK’s separately rescaled topological generator.

Check the dual angle and the source signs

At rank one the compact first-order action contains $-i\int\tau\,df_c/(2\pi)$, with $c=\widehat a$, $\tau\sim\tau+2\pi$ and $\tau\mapsto\tau-k\alpha$ for the relative gauge angle. To enforce positive flux $df_c=2\pi m\,\delta_x$, the Fourier delta is

$$\int D\tau\, \exp\left[\frac{i}{2\pi}\int\tau\,df_c-im\tau(x)\right]. \tag{AB09.7}$$

The insertion factor is therefore $e^{-im\tau}$, with gauge weight $+km$, agreeing with (AB09.5). Inserting $e^{+im\tau}$ into this same first-order integral creates the opposite flux. This identifies the phase and its orientation; a complete renormalized supersymmetric monopole also includes its boundary condition and vacuum response.

Source formulas must be compared with their actions. With fixed orientation and $t_L=-i\tau_E$, a real Lorentzian coefficient $k_L\,\mathrm{CS}/(4\pi)$ continues to $-ik_L\,\mathrm{CS}/(4\pi)$. Our choice (AB09.2) therefore corresponds to $k_L=-k$. The adjacent gauge-sector dictionary is

Source convention	Flux and level used in its integrated constraint	Result in site field weights
AB (2.10), with $A_{\rm AB}=-a$	$n_{\rm AB}=-m$, $k_{\rm AB}=-k$	$k_{\rm AB}n_{\rm AB}+Q_L=km+Q_L$
BK §2.7, with the same minus-cubic connection polynomial as (AB09.2)	$n_{\rm BK}=m$, $k_{\rm BK}=-k$	$-k_{\rm BK}n_{\rm BK}+Q_L=km+Q_L$

The hatted constraints give $-k\widehat m+Q_R$ in both cases. The first row includes the connection sign already used in the one-brane lesson; that substitution alone would leave the Lorentzian-to-Euclidean level sign untranslated. This is a gauge-sector comparison, not an identification of every source spinor, oscillator bar, or R-generator. See AB (2.10) and BK §2.7.

Worked example: one flux unit at level three

Let $N=1$, $k=3$ and $(m,\widehat m)=(1,1)$. The bare weights are $(3,-3)$. Each $B_i$ has $(-1,+1)$, so the four lowest scalar dressings are

$$\mathcal O_p=T_{1,1}\,B_1^pB_2^{3-p}, \qquad p=0,1,2,3. \tag{AB09.8}$$

All matter sees relative flux $m-\widehat m=0$. Its lowest scalar mode on the radial sphere has angular momentum zero; three such modes can make a scalar chiral operator. At rank one the quartic superpotential vanishes, so it does not set these monomials to zero. They are degree-three polynomials in the appropriate local free chiral coordinates, invariant under the residual discrete identification. They form the symmetric third power of the two B flavors.

For negative matched flux, replace the three B fields by three A fields. These are independent chiral constructions, not Hermitian conjugates: conjugation also changes chirality. A product of A and B fields with the wrong net relative charge cannot repair the Gauss law merely because it has the correct total degree.

There is a useful finite counting check. Give every scalar a formal degree weight $x$, with $0<x<1$. In this matched Abelian sector the zero-angular-momentum dressing count is

$$\begin{aligned} P_d(x) &=\int_0^{2\pi}\frac{dh}{2\pi} \frac{e^{-idh}}{(1-xe^{-ih})^2(1-xe^{ih})^2}\\ &=\sum_{p\geq0}(p+1)(p+d+1)x^{2p+d}\\ &=\frac{x^d[(d+1)+(1-d)x^2]}{(1-x^2)^3}, \qquad d=km\geq0. \end{aligned} \tag{AB09.9}$$

The phase projects $N_B-N_A=d$. At $d=3$ the leading coefficient is four, as in (AB09.8). This is a count of specified scalar dressing polynomials, not the full superconformal index or a count of all monopole states.

Calculate the bare R-charge

The regulated vacuum has an R-charge offset even before dressing. Recall the symmetric-ordering result: a fermion of R-charge $r_f$ and flux weight $w$ contributes $-\tfrac12 r_f|w|$. The absolute value counts angular unpaired modes; it is not a signed flux. Opposite weights cancel the analogous gauge-charge offsets in the present matter content and adjoint root pairs. We retain the supersymmetric level prescription defining the original integer $k$.

The four matter fermions have $r_f=-1/2$. For every $(i,j)$, two A and two B fermions see opposite versions of $m_i-\widehat m_j$. Their combined contribution is $|m_i-\widehat m_j|$. The first-node gaugino has R-charge $+1$ and roots $m_i-m_j$, $i\ne j$. Its two opposite roots contribute $-|m_i-m_j|$ for each unordered pair; the second node gives the same expression with hats. Hence

$$\boxed{ R_{\rm bare}(m,\widehat m) =\sum_{i,j}|m_i-\widehat m_j| -\sum_{i<j}|m_i-m_j| -\sum_{i<j}|\widehat m_i-\widehat m_j|. } \tag{AB09.10}$$

This matches the regulated bare-vacuum result BK (12), §2.6. Its vector contribution is a regulated vacuum contribution, not a count of physical Yang–Mills photons added to ABJM.

We can prove a nontrivial property of this expression without diagonalizing a fluctuation Hamiltonian. For real $t$ define the integer step functions

$$M(t)=\#\{i:m_i>t\},\qquad \widehat M(t)=\#\{j:\widehat m_j>t\}.$$

The identity $|u-v|=\int_{\mathbb R}(\mathbf1_{u>t}-\mathbf1_{v>t})^2dt$ counts the length between two real numbers. At fixed $t$ the cross sum in (AB09.10) gives $N(M+\widehat M)-2M\widehat M$. The two within-node sums give $NM-M^2$ and $N\widehat M-\widehat M^2$. Subtracting them proves

$$\boxed{ R_{\rm bare}=\int_{\mathbb R}[M(t)-\widehat M(t)]^2\,dt\geq0. } \tag{AB09.11}$$

Both counts are $N$ far to the left and zero far to the right, so the integral is finite. It vanishes precisely when their jumps, including multiplicities, agree: the two flux multisets must be equal. Equality of total flux is weaker.

For example, $(m_1,m_2)=(2,0)$ and $(\widehat m_1,\widehat m_2)=(1,1)$ both have total two. On $(0,1)$ the difference of counts is $-1$; on $(1,2)$ it is $+1$. Thus $R_{\rm bare}=2$, whereas matched multisets give zero.

Apply shortening only after constructing a physical operator

In (AB09.8), the matched rank-one vacuum has $R_{\rm bare}=0$ and the three B scalars add $3/2$. The nonzero scalar chiral operators therefore have

$$R(\mathcal O_p)=\Delta(\mathcal O_p)=\frac32. \tag{AB09.12}$$

Here the final equality uses the actual superconformal R-generator and chiral shortening. The elementary charged fields and the unprojected bare vacuum are not gauge-invariant local operators to which one can separately apply the scalar unitarity bound. In particular, a bare value zero does not imply an extra dimension-zero physical scalar.

For other fluxes, Gauss-law neutrality is only one test. One must also construct the full residual-group singlet, account for the monopole angular modes, and establish a nonzero scalar chiral class before setting its dimension equal to its R-charge. A nonzero effective magnetic flux gives scalar harmonics with minimum angular momentum $|m_i-\widehat m_j|/2$; arbitrary charged dressing is not automatically a scalar.

The local definition here is the flux boundary condition followed by gauge projection. It does not rely on making an arbitrary open Wilson line unobservable in a non-Abelian theory with matter. The original paper explicitly corrects that assertion in AB v4, footnote 6.

Exercises

1. Keep the same total flux but change its distribution

At $N=2$, compare $m=(3,-1)$ with $\widehat m=(1,1)$. Check the common-center Gauss-law condition, calculate $R_{\rm bare}$ in two ways, and decide whether these checks alone produce a scalar chiral primary. Compare with $\widehat m=(-1,3)$.

Hint 1

The common-center test uses the sum of entries. The normal-order formula uses their pairwise differences.

Hint 2

The count difference is nonzero on two intervals, each of length two. Permuting entries does not change a multiset.

Solution

Both totals are two, so (AB09.6) passes. The cross sum is $8$, the first within-node sum is $4$, and the second is zero. Thus $R_{\rm bare}=4$. Equivalently, $M-\widehat M=-1$ on $(-1,1)$ and $+1$ on $(1,3)$, giving $2+2=4$ from (AB09.11).

The right node retains $U(2)$, while the left retains $U(1)\times U(1)$. All bifundamental entries have flux magnitude two and hence minimum scalar angular momentum one. A choice of matter occupations must still satisfy every Cartan constraint, combine into a right-node singlet, and pass the scalar/chiral tests. The two computed quantities do not establish such an operator or its dimension.

For $\widehat m=(-1,3)$, the multisets agree after a permutation, so $R_{\rm bare}=0$. Entries connecting equal fluxes now have zero effective flux. This removes that angular obstruction for suitable entries; it still does not remove their CS gauge weights.

2. Dress a repeated flux without replacing U(2) by its Cartan

Let $N=2$, $k=3$ and $m=\widehat m=(1,1)$. Using only the scalar matrix $B_1$, find a polynomial that cancels the bare character under the full $U(2)_L\times U(2)_R$. Why is $(B_{1,11}B_{1,22})^3$ inadequate? Determine the R-charge of your dressing.

Hint 1

The bare character is $(\det U_L)^3(\det U_R)^{-3}$. The matrix transforms as $B_1\mapsto U_RB_1U_L^\dagger$.

Hint 2

The determinant contracts both color indices antisymmetrically. Every entry has zero effective flux in this example.

Solution

Use

$$\mathcal O=T_{(1,1),(1,1)}(\det B_1)^3,\qquad \det B_1=B_{1,11}B_{1,22}-B_{1,12}B_{1,21}.$$

Since $\det B_1$ transforms by $(\det U_R)(\det U_L)^{-1}$, its third power cancels the complete bare character. It is also invariant under the determinant-one parts of both groups.

For a direct failure test, start with $B_1=\operatorname{diag}(u,v)$ and take $U_L=\left(\begin{smallmatrix}\cos\gamma&\sin\gamma\\-\sin\gamma&\cos\gamma\end{smallmatrix}\right)$, $U_R=I$. The proposed diagonal product becomes $uv\cos^2\gamma$, while the determinant remains $uv$. Its third power therefore fails the residual $SU(2)$ test although it had the correct diagonal weights. Summing only over permutations would not fix invariance under continuous rotations.

The matched multisets give $R_{\rm bare}=0$. The determinant cubed contains six B scalars and has R-charge three. It uses zero-flux scalar modes and is nonzero on the diagonal vacuum branch with invertible $B_1$; the A fields may vanish, so the quartic F equations do not remove it. Its scalar chiral primary has $\Delta=3$. The determinant supplies the singlet tensor; no choice of one preferred diagonal component is a substitute for it.

3. Change the compact group before choosing the minimum flux

At rank one take $k=6$, but replace the standard group by $(U(1)\times U(1))/\mathbb Z_3^{\rm diag}$, with the same local level form and no added discrete counterterm. The preceding global-data lesson established that this spin-CS choice is allowed. Find the smallest positive matched flux, its bare gauge weights, and a lowest-degree B dressing. Compare with the standard unquotiented group.

Hint 1

The new cocharacter lattice is $\mathbb Z^2+\tfrac13(1,1)\mathbb Z$. Do not impose the old integer lattice afterward.

Hint 2

The level multiplies the flux. A valid character of the quotient has the sum of its two charges divisible by three.

Solution

The new minimum is $(m,\widehat m)=(1/3,1/3)$, with bare weights $(2,-2)$. A product of two B fields cancels them; there are three symmetric flavor monomials. Both the bare character and the B representation descend to the quotient because their two charges sum to zero.

At matched rank-one flux the effective matter flux and $R_{\rm bare}$ vanish. The dressing has R-charge one, giving dimension one for its nonzero scalar chiral primary. This agrees with the degree-two identification expected from the residual $\mathbb Z_{6/3}=\mathbb Z_2$ in the local vacuum quotient.

For the unquotiented group the smallest positive matched flux is $(1,1)$, its weights are $(6,-6)$, and the minimum B dressing has degree six and R-charge three. The local CS polynomial did not change, but the minimum permitted disorder insertion did. The quotient result must not be imported into the standard ABJM spectrum.

The useful output is a sequence of tests: specify an allowed flux, impose the full Gauss law, calculate its vacuum charges, and only then identify a physical short multiplet. Why levels one and two are special identifies the monopole multiplets that supply additional conserved currents and explains the protection and nonzero-descendant inputs needed for that conclusion.

The original ABJM paper guide locates this calculation in the source argument and gives a changed example to check while reading.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading the original ABJM paper