Why levels one and two are special

An extra supersymmetry requires an extra conserved supercurrent. A monopole with an appealing charge is not yet such a current. The useful route is to find a protected scalar operator, identify its whole superconformal multiplet, and show that the multiplet contains the missing currents.

We use monopole flux and matter dressing in the standard compact three-dimensional $U(N)_k\times U(N)_{-k}$ theory, $N\geq1$, $k>0$, with integer fluxes and no additional common-center quotient. The gauge rank is $N$ per factor, $2N$ in total. The generic theory has twelve real Poincaré supercharges, while its $\mathcal N=2$ description displays four. At $k=1,2$ the full theory has sixteen. We work at the conformal origin with no mass or FI deformation; this is a statement about the quantum theory, not enhancement inferred from a particular vacuum metric.

Fix the charge units before looking for a current

Write a matched flux as $m=\widehat m$, and let $J=\sum_i m_i$ be the integer topological charge used in the preceding lesson. Its bare R-charge vanishes. The inherited Euclidean action is $+ik\,\mathrm{CS}(a)/(4\pi)-ik\,\mathrm{CS}(\widehat a)/(4\pi)$, with weight $e^{-S_E}$. A bare insertion has gauge weights $(km_i,-km_i)$, so positive flux requires B dressing; negative flux requires A dressing.

It is convenient for the enhanced algebra to define a rescaled topological label

$$T=\frac{kJ}{2}. \tag{AB10.1}$$

This does not change the allowed flux lattice. In particular, $T$ can be half-integral at level one. We have not declared a new compact circle with all charges integral. Bashkirov–Kapustin rescale their topological generator by $-k_{\rm BK}/2$; the preceding action dictionary $k_{\rm BK}=-k$, $n_{\rm BK}=m$ gives (AB10.1). Their oscillator bars also encode a choice of creation operators and BPS charge, and should not be copied as literal names of our chiral fields. We derive our dressings using the stated Gauss law. See BK §2.1 and §2.7.

The four complex scalars $Y=(A_1,A_2,B_1^\dagger,B_2^\dagger)$ transform as the $\mathbf4$ of $SU(4)_R\simeq\operatorname{Spin}(6)_R$. The selected $\mathcal N=2$ R-generator acts on them as $\operatorname{diag}(1,1,-1,-1)/2$. Thus

$$\begin{aligned} \mathbf4&=(\mathbf2,\mathbf1)_{1/2}\oplus(\mathbf1,\mathbf2)_{-1/2},\\ \mathbf{10}=\operatorname{Sym}^2\mathbf4 &=(\mathbf3,\mathbf1)_1\oplus(\mathbf2,\mathbf2)_0 \oplus(\mathbf1,\mathbf3)_{-1}. \end{aligned} \tag{AB10.2}$$

Here the two groups are $SU(2)_A\times SU(2)_B$, and these subscripts denote R-charge, not $T$. The decomposition follows by taking two symmetric A indices, one index from each doublet, or two symmetric conjugate-B indices. Its dimensions are $3+4+3=10$. The conjugate representation has the opposite R-charges. We seek a chiral A triplet in $\mathbf{10}_{T=-1}$ and a chiral B triplet in $\overline{\mathbf{10}}_{T=+1}$, both of dimension one.

Worked construction: two scalars cancel the monopole weight

Use radial quantization on a unit two-sphere. Bashkirov–Kapustin introduce a supersymmetric regulator with parameter $t$ that makes the spectrum weakly coupled at large $t$ while retaining the relevant short states. The regulated energy $E$ is not automatically a scaling dimension away from the conformal theory. In a matched flux sector the bare energy is zero, and a scalar component seeing zero relative flux has a spin-zero mode of energy and R-charge $1/2$.

For $k=2$, take

$$m=\widehat m=(1,0,\ldots,0).$$

The distinguished color line has bare weights $(2,-2)$. The operators represented in this flux gauge by

$$\mathcal O_{\dot a\dot b}^{(+)} =\mathcal T_{(1,0,\ldots),(1,0,\ldots)} (B_{\dot a})_{11}(B_{\dot b})_{11}, \qquad \dot a,\dot b=1,2, \tag{AB10.3}$$

cancel both weights. Their effective magnetic flux is $1-1=0$, so no angular excitation raises the energy. The remaining $U(N-1)$ groups act trivially on these components. The notation denotes the residual-gauge projection of a disorder sector; choosing the label 1 is a representative, not a physical preferred color.

The two bosons commute, giving three symmetric flavor components. Their quantum numbers are

$$(E,j,R,T)=(1,0,1,1),\qquad SU(2)_A\times SU(2)_B:\ (\mathbf1,\mathbf3). \tag{AB10.4}$$

The negative-flux A construction gives $(\mathbf3,\mathbf1)$ with $T=-1$. These two chiral constructions are not Hermitian conjugates; conjugation produces the corresponding antichiral states. The relevant chiral classes are nonzero: they are nonzero protected states in the regulated construction, and their scalar dressings are compatible with the quartic F equations.

At $k=1$, a single line needs flux two to give the same two-scalar construction: replace $(1,0,\ldots)$ in (AB10.3) by $(2,0,\ldots)$. There is a second possibility when $N\geq2$:

$$m=\widehat m=(1,1,0,\ldots,0),\qquad \mathcal O_{\dot a\dot b}^{\prime(+)} =\mathcal T_{m,m}\,\epsilon^{pq}\epsilon^{\widehat r\widehat s} (B_{\dot a})_{\widehat r p}(B_{\dot b})_{\widehat s q}. \tag{AB10.5}$$

All four color indices range over the two-dimensional equal-flux blocks, and $\epsilon^{12}=+1$. The two epsilon tensors make a singlet of both residual $SU(2)$ groups. The dressing transforms as $\det U_R/\det U_L$, cancelling the bare determinant character. Interchanging the flavor labels exchanges both epsilon contractions, giving two minus signs: the flavor tensor is again symmetric. For equal flavors it is twice the determinant of that B block. A product of two selected diagonal entries would fail the continuous residual gauge test.

Why does this construction stop at $k=2$? A nonzero bare energy is a positive integer, and the dressing needed by a charged bare monopole raises it further. An energy-one state must therefore have zero bare offset, requiring matched flux multisets. Two lowest scalar modes can make the required R-charge-one scalar; a single fermion is not a scalar, and two fermions already cost at least energy two. Positive $T=1$ means $kJ=2$; the two B excitations distribute two units of gauge weight. The possible positive flux partitions are one entry with $km=2$, or two equal entries with $km=1$. Integral flux gives precisely

Level	Positive matched flux	Minimum rank
$2$	$(1,0,\ldots)$	$1$
$1$	$(2,0,\ldots)$	$1$
$1$	$(1,1,0,\ldots)$	$2$

At $k>2$ no such charged sector exists. This does not say that all dimension-one operators disappear: the ordinary neutral stress-multiplet scalars remain. The construction and its protection are BK (13)–(18), §§2.8–2.10.

The step that makes these states physical symmetry generators

There are two additional inputs beyond the charge calculation.

First, these scalar states survive removal of the regulator. A short state could disappear only by acquiring a partner and joining a longer multiplet. For these $E=R=1$ scalars, the dangerous spinor partners have $(E,R)=(1/2,0)$ or $(3/2,2)$, with the same flavor and topological labels. BK’s low-energy mode analysis excludes them in the regulated spectrum, and the conformal unitarity bounds exclude the corresponding partners at the conformal endpoint. For example, a lowest matter spinor has $(E,R)=(1,1/2)$; adding an energy-$1/2$ scalar cannot supply the missing R-charge $3/2$. The full protection argument, including vector modes, is the sourced input in §2.10 and Appendix B. We use this result; a bare R-charge formula alone would not prove it.

Second, the protected scalars sit in short multiplets of the full $\mathcal N=6$ superconformal algebra. Restoring $SU(4)_R$ completes their flavor triplets into $\mathbf{10}_{-1}$ and $\overline{\mathbf{10}}_{+1}$ of dimension one. Acting with two supercharges, symmetrizing their spacetime spinor indices and antisymmetrizing their $\mathbf6$ indices, produces a spin-one conformal-primary component in

$$\mathbf6_{-1}\oplus\mathbf6_{+1},\qquad \Delta=1+2\left(\frac12\right)=2. \tag{AB10.6}$$

This is a projection of an ordered $QQ$ descendant, not the anticommutator that generates translations. The $\mathbf6$ projection and its nonzero norm are consequences of the short representation. BK check the nonzero norm using a free $\mathcal N=8$ realization, since the superconformal algebra fixes descendant norms once the primary is normalized. We are invoking that representation argument, not asserting that any operator obtained by writing $QQ\mathcal O$ must survive.

A spin-one primary in a unitary three-dimensional CFT has $\Delta\geq2$. Saturation makes its divergence a null state, hence it is a conserved current. This gives the required extra currents at the conformal endpoint. BK §2.11 explains why merely counting spin-one states in the regulated Fock space would be insufficient: those states can have available pairing partners.

Check the enlarged algebra and the number of supercharges

The existing R-currents give $\mathbf{15}_0$, and the topological current gives $\mathbf1_0$. Adding (AB10.6) yields

$$\mathbf{28}=\mathbf{15}_0\oplus\mathbf1_0 \oplus\mathbf6_{+1}\oplus\mathbf6_{-1}. \tag{AB10.7}$$

This is more than the arithmetic $15+1+6+6=28$. To see the Lie-algebra structure, let $M_{ij}=E_{ij}-E_{ji}$ be real antisymmetric matrices on $\mathbb R^8$, where $E_{ij}$ has a single nonzero entry. For $a,b=1,\ldots,6$, the matrices $M_{ab}$ generate $\mathfrak{so}(6)$. Put $J_0=M_{78}$ and $E_a^\pm=M_{a7}\pm iM_{a8}$. Direct matrix multiplication gives

$$\begin{aligned} [J_0,E_a^\pm]&=\pm iE_a^\pm,\\ [E_a^+,E_b^-]&=-2M_{ab}+2i\delta_{ab}J_0,\\ [E_a^+,E_b^+]&=[E_a^-,E_b^-]=0. \end{aligned} \tag{AB10.8}$$

The Hermitian charge generator is $-iJ_0$, so the charged generators have labels $T=\pm1$. Opposite-charge commutators recover the original rotations and the topological generator. These 28 independent matrices span $\mathfrak{so}(8)$. In the physical argument, the short-multiplet structure and current algebra supply this enhancement; the matrices check its representation content and normalization. They do not manufacture currents that have not been shown to exist.

The enhanced R-symmetry is conventionally called $\operatorname{Spin}(8)_R$; its Lie algebra is what we have established here, without deciding every faithful global quotient on line operators. The supercharges organize as

$$\mathbf8_Q=\mathbf6_0\oplus\mathbf1_{+1}\oplus\mathbf1_{-1}. \tag{AB10.9}$$

Each real three-dimensional Majorana supercharge has two real components. The original six give twelve; the two additional real directions give four more, for sixteen Poincaré supercharges. There are also sixteen conformal supercharges at the fixed point. This does not change the gauge rank. The scalar eight is a different $\operatorname{Spin}(8)$ representation: it branches as $\mathbf4_{-1/2}\oplus\overline{\mathbf4}_{+1/2}$, not as (AB10.9).

A free calculation checks nonvanishing, not the whole interacting theorem

Take four canonical free complex scalars $X^I$ with $\langle X^I(x)\overline X_J(0)\rangle=\delta^I_JG(x)$, $G(x)=1/(4\pi|x|)$. They realize the scalar eight just described. The symmetric bilinears $X^IX^J$ have dimension one and transform as $\mathbf{10}_{-1}$. A bosonic piece of the charged R-current is

$$j_\mu^{IJ}=X^I\partial_\mu X^J-X^J\partial_\mu X^I.$$

Its divergence is $X^I\partial^2X^J-X^J\partial^2X^I=0$ away from insertions. For $I=1,J=2$, Wick contraction gives

$$\begin{aligned} \langle j_\mu^{12}(x)\overline j_\nu^{12}(0)\rangle &=2\left[-G\,\partial_\mu\partial_\nu G +(\partial_\mu G)(\partial_\nu G)\right]\\ &=\frac{1}{8\pi^2|x|^4} \left(\delta_{\mu\nu}-2\frac{x_\mu x_\nu}{|x|^2}\right). \end{aligned} \tag{AB10.10}$$

The coefficient is positive and nonzero. The full supersymmetric R-current also contains a fermion bilinear; its action on the scalar fields already forbids it from being the zero operator. This explicit free check supports the representation-theory step. It does not calculate the current normalization of an interacting ABJM sector or prove protection by a free-field Wick contraction.

Rank one, the free sector, and the global group

At $k=1$, flux $(1,0,\ldots)$ needs just one B scalar, and the opposite flux needs one A scalar. They give four complex scalar primaries of dimension $1/2$, including their flavor components. A scalar saturating the three-dimensional bound $\Delta\geq1/2$ obeys a free field equation. Their supersymmetric completion therefore supplies a decoupled free $\mathcal N=8$ sector for every $N$. For $N>1$ there is also an interacting sector. The two independent dimension-one constructions above are consistent with separate stress multiplets; they should not be counted as a single theory with “$\mathcal N=16$” supersymmetry. The factorwise algebras act on separate sectors, while the common spacetime supersymmetry is the usual $\mathcal N=8$ algebra. These are the free-sector statements of BK §2.9.

For $N=1,k=1$, the local matter theory is free. For $N=1,k=2$, it is the discrete gauging $\mathbb C^4/\mathbb Z_2$, with $X\mapsto-X$. Linear scalar operators are projected out, while the quadratic scalars and R-currents survive. Thus it has $\mathcal N=8$ without gauge-invariant free scalar primaries of dimension $1/2$. Calling it four ungauged free chirals would erase this projection and the accompanying global gauge data. The rank-one identification is explicit in AB §2.3.

Finally, the integral flux lattice is essential. Changing the compact group changes which disorder operators exist. The preceding lesson’s rank-one common-center quotient permits fractional matched flux and can change the minimum dressing degree. Its local CS polynomial is not enough to import the present standard-group spectrum. Likewise, a moduli-space orbifold suggests a symmetry but does not by itself establish the non-Abelian current multiplets. Here those multiplets were supplied by the protected-state argument.

Exercises

1. Change the level and count the candidate sectors

For the standard equal-rank group, find the positive matched flux sectors that can supply the two-B scalar triplet with $T=1$ at $k=1,2,3$, first for $N=1$ and then for $N=2$. Give their dimensions after protection. Explain why the result is not a count of every dimension-one operator.

Hint 1

Each B cancels one positive left-node weight. The total weight to cancel is $kJ=2$.

Hint 2

At rank two, distinguish a partition with both units on one color line from a repeated-flux block. Only the latter needs two epsilon tensors.

Solution

At rank one, level one uses $m=\widehat m=2$ and level two uses $m=\widehat m=1$. Level three would require $m=2/3$, outside the specified lattice. Each allowed sector has the symmetric pair $B_1^2,B_1B_2,B_2^2$, so there are three chiral flavor components with $R=\Delta=1$.

At rank two, level one has $(2,0)$ and $(1,1)$, both matched at the second node. These are distinct flux orbits and each supplies a triplet. The repeated block uses (AB10.5), which is not a product of preferred diagonal entries. Level two has only $(1,0)$; level three has none. Negative sectors supply the corresponding A triplets. The statement concerns charged enhancement multiplets. Neutral operators in the ordinary stress multiplet exist at all these levels, and completing an entire $SU(4)$ multiplet includes components outside the chosen chiral ring.

2. Distinguish a free theory from its discrete gauging

At rank one, consider the local lift $X^I\mapsto\omega X^I$, $\omega=e^{2\pi i/k}$. Determine which of $X^I$, $X^IX^J$, $X^I\overline X_J$ and $j_\mu^{IJ}$ survive the projection. Count the complex symmetric quadratic operators at $k=1,2,3$. Does the absence of linear scalars at $k=2$ remove the enhanced currents?

Hint 1

A derivative does not change the discrete charge. The holomorphic pair and the holomorphic current both acquire $\omega^2$.

Hint 2

Use the projector $k^{-1}\sum_{r=0}^{k-1}\omega^{rd}$ for discrete degree $d$, and count symmetric pairs of four labels.

Solution

The projector is one precisely when $k$ divides $d$. The linear fields have $d=1$, the holomorphic quadratics and charged currents have $d=2$, and $X^I\overline X_J$ has $d=0$. Thus linear fields survive only at $k=1$, while the holomorphic quadratic multiplet survives at $k=1,2$ and is absent at $k=3$.

There are $4(4+1)/2=10$ complex symmetric quadratics at each of $k=1,2$, and none at $k=3$. The mixed bilinears have $16=15+1$ real Hermitian components; the traceless fifteen are the neutral scalars in the stress multiplet. At $k=2$ the charged currents are even and survive despite the absence of linear gauge-invariant scalars. This is consistent with enhancement in a discretely gauged free theory; it does not turn its projected operator algebra into the unprojected one. This projection check is exact for the rank-one model and is not a proof for general $N$.

3. A level-six quotient changes the answer’s premise

Replace the rank-one standard group at $k=6$ by $(U(1)\times U(1))/\mathbb Z_3^{\rm diag}$ with the allowed spin-CS action and no extra discrete counterterm used in the preceding lesson. Its cocharacters are $\mathbb Z^2+\frac13(1,1)\mathbb Z$. Find the smallest positive matched flux and its two-field dressing, evaluate $T=kJ/2$, and identify the local rank-one matter quotient. Does this contradict the standard-group level criterion?

Hint 1

The level form acts on the new lattice; the smallest common flux is no longer one.

Hint 2

The previously derived residual discrete group has order $k/3$. Separate its local matter projection from additional global gauge-sector statements.

Solution

The minimum flux is $(1/3,1/3)$, giving bare weights $(2,-2)$. Two B scalars cancel them, and $T=6(1/3)/2=1$. They have $R=\Delta=1$ in the rank-one chiral sector. The local matter quotient is $\mathbb C^4/\mathbb Z_2$, so its even charged currents survive just as in the preceding exercise. The unquotiented level-six theory instead needs six B fields at minimum positive flux and has $T=3$ for that sector.

There is no contradiction: the global gauge group, hence the local disorder-operator lattice, was changed. The conclusion concerns the specified rank-one local matter sector. It is not an equivalence of every line operator or partition-function normalization, nor an enhancement theorem for arbitrary non-Abelian quotients.

You can now distinguish three statements that are often compressed into one: a flux sector admits a dressing; its scalar short multiplet survives in the conformal theory; and nonzero current descendants enlarge the symmetry algebra. The last two steps explain why the exceptional monopoles imply $\mathcal N=8$, while the first alone does not.

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