Read the two-node ABJM quiver

A quiver is a directed graph that records representations. Its practical use is to answer a question before doing a long calculation: can these fields be multiplied and contracted to make a gauge invariant? Equal-sized matrices can conceal an invalid contraction. The arrows retain the distinction between their two color spaces.

Start from Chern–Simons gauge fields with non-Abelian matter. We retain three-dimensional ABJM with compact $U(N)_k\times U(N)_{-k}$, positive integer $k$, and no additional quotient. The generic theory has twelve real Poincaré supercharges, while the chiral multiplets below display a three-dimensional $\mathcal N=2$ subalgebra with four. Each gauge factor has rank $N$; the product has rank $2N$. The arrows do not count supercharges.

Four fields before four arrows

Let $V_L$ and $V_R$ be distinct $N$-dimensional complex color spaces. Write $U\in U(N)_L$ and $\widehat U\in U(N)_R$ for their gauge transformations. The hat names the second gauge group. An unhatted color index $a$ belongs to $V_L$; a hatted index $\hat b$ belongs to $V_R$. Both range from $1$ to $N$, but they are different kinds of index.

The four chiral multiplets have the following representations. Here $\mathbf N$ denotes the fundamental representation and $\overline{\mathbf N}$ its complex conjugate.

Chiral multiplet	Representation under $U(N)_L\times U(N)_R$	Map and matrix indices
$A_1$	$(\mathbf N,\overline{\mathbf N})$	$V_R\to V_L$, $(A_1)^a{}_{\hat b}$
$A_2$	$(\mathbf N,\overline{\mathbf N})$	$V_R\to V_L$, $(A_2)^a{}_{\hat b}$
$B_1$	$(\overline{\mathbf N},\mathbf N)$	$V_L\to V_R$, $(B_1)^{\hat a}{}_b$
$B_2$	$(\overline{\mathbf N},\mathbf N)$	$V_L\to V_R$, $(B_2)^{\hat a}{}_b$

Their transformations are explicitly

$$\begin{aligned} A_1'&=UA_1\widehat U^{-1}, &A_2'&=UA_2\widehat U^{-1},\\ B_1'&=\widehat U B_1U^{-1}, &B_2'&=\widehat U B_2U^{-1}. \end{aligned} \tag{AB03.1}$$

Each row describes a whole chiral multiplet: a complex scalar, a two-component complex spinor, and a complex auxiliary field, all in the indicated gauge representation. In the matrix calculations below, $A_i,B_j$ denote the scalar components. The flavor labels $i,j=1,2$ distinguish arrows; they are not color indices. This field content and the opposite levels are specified in the original paper, section 2.2 before (2.4).

We draw an arrow from the domain of its map to its codomain. Thus an $A$ arrow points from the right node to the left node.

Each arrow is one chiral multiplet. Its tail is the input color space and its head is the output color space. The node levels are additional action data; reversing an arrow does not change a Chern–Simons level.

Some quiver conventions reverse every arrow while defining the associated representation accordingly. Equation (AB03.1), rather than a remembered picture, settles the convention here.

A reverse representation is not a conjugacy condition

The phrase “conjugate representation” in the table does not mean $B_i=A_i^\dagger$. For unitary transformations,

$$(A_i^\dagger)'=\widehat U A_i^\dagger U^{-1}. \tag{AB03.2}$$

So $A_i^\dagger$ and $B_j$ transform in the same gauge representation. Nevertheless, $A_i^\dagger$ is the conjugate of a specified field, whereas $B_j$ is an independent chiral field. Conjugating an entire chiral superfield produces an antichiral superfield. The four chiral multiplets therefore contain four independent complex scalar matrices before gauge identifications and vacuum constraints.

On a physical Lorentzian conjugate slice, each scalar has its own Hermitian adjoint. In Euclidean supersymmetric calculations the corresponding antichiral variables can instead be treated independently, with a contour specified when needed. Neither convention identifies a $B$ chiral multiplet with the conjugate of an $A$ multiplet.

The familiar scalar list $(A_1,A_2,B_1^\dagger,B_2^\dagger)$ puts four scalars in the same gauge representation. It is not a list of four chiral superfields of this manifest subalgebra. The original paper explicitly distinguishes superfield and scalar notation in section 2.2, footnote 4. We use the quiver to preserve that distinction.

Close a path and contract its color indices

Consider the matrix product $A_iB_j$. The rightmost matrix acts first:

$$V_L\xrightarrow{\ B_j\ }V_R \xrightarrow{\ A_i\ }V_L. \tag{AB03.3}$$

It is an endomorphism of $V_L$, meaning a map from that space back to itself. Under a gauge transformation the middle factors cancel:

$$A_i'B_j' =UA_i\underbrace{\widehat U^{-1}\widehat U}_{1}B_jU^{-1} =U(A_iB_j)U^{-1}.$$

Its ordinary fundamental trace is invariant. In indices, $\operatorname{Tr}_L(A_iB_j)=(A_i)^a{}_{\hat b}(B_j)^{\hat b}{}_a$, with both repeated color indices summed. There are four such quadratic words, labeled by the two independent flavor choices.

Now construct a longer word:

$$\begin{aligned} \mathcal O_1 &=\operatorname{Tr}_L(A_1B_1A_2B_2)\\ &=(A_1)^a{}_{\hat b} (B_1)^{\hat b}{}_c (A_2)^c{}_{\hat d} (B_2)^{\hat d}{}_a. \end{aligned} \tag{AB03.4}$$

The walk starts at $V_L$ and follows $B_2,A_2,B_1,A_1$ in that order. Every output index feeds an input of the same color type, including the final contraction back to the starting node. Equivalently,

$$\begin{aligned} A_1'B_1'A_2'B_2' &=(UA_1\widehat U^{-1})(\widehat U B_1U^{-1}) (UA_2\widehat U^{-1})(\widehat U B_2U^{-1})\\ &=U(A_1B_1A_2B_2)U^{-1}. \end{aligned} \tag{AB03.5}$$

Taking the trace removes the remaining change of basis. This gives a direct gauge-invariance proof, without imposing equations of motion.

Why insist on alternating arrows? The numerical product $A_1A_2$ exists when both matrices happen to be $N$ by $N$. As a product of maps, however, it attempts to feed the $V_L$ output of $A_2$ into the $V_R$ input of $A_1$. No identification between those independently transforming spaces has been supplied. Equal dimensions do not supply one.

There is also a quick necessary test. Under the central transformations $U=e^{i\alpha}1_N$ and $\widehat U=e^{i\beta}1_N$,

$$A_i\longmapsto e^{i(\alpha-\beta)}A_i, \qquad B_j\longmapsto e^{-i(\alpha-\beta)}B_j. \tag{AB03.6}$$

A polynomial built only from these scalar fields must have equal numbers of $A$ and $B$ factors to be invariant under this relative center. That charge test is necessary, but it does not perform the non-Abelian index contractions. Exercise 1 exhibits the difference.

Move the starting point, not the order

Trace cyclicity lets us start the same closed walk at another point:

$$\operatorname{Tr}_L(A_1B_1A_2B_2) =\operatorname{Tr}_R(B_2A_1B_1A_2). \tag{AB03.7}$$

The second trace belongs to $V_R$, because its rightmost $A_2$ starts there. One can prove the equality by relabeling the four summed indices in (AB03.4). No commutativity assumption enters.

Cyclicity does not allow an arbitrary permutation. Define a different closed word $\mathcal O_2=\operatorname{Tr}_L(A_1B_2A_2B_1)$. To see the distinction concretely, take $N=2$ and let $E_{ab}$ denote the matrix with entry one in row $a$, column $b$, and zero elsewhere. Choose scalar values

$$A_1=E_{11},\qquad A_2=E_{22}, \qquad B_1=E_{12},\qquad B_2=E_{21}. \tag{AB03.8}$$

These are allowed field values at a point; no vacuum condition is being asserted. The multiplication rule $E_{ab}E_{cd}=\delta_{bc}E_{ad}$ gives

$$A_1B_1A_2B_2=E_{11}, \qquad A_1B_2=0, \qquad \mathcal O_1=1,\quad \mathcal O_2=0. \tag{AB03.9}$$

For $N=1$, each field is an ordinary commuting complex number, and the same two words coincide. Thus the Abelian check is useful but cannot verify non-Abelian ordering.

These two ordered words occur in the original superpotential, (2.5). Their coefficients and the resulting auxiliary equations require the action; the quiver alone only establishes that both words are permitted gauge contractions. With the canonical engineering dimension $[A_i]=[B_j]=1/2$, a quadratic scalar word has engineering dimension one and a quartic word has engineering dimension two. This counting does not replace a calculation of quantum operator dimensions.

The same distinction matters for the chiral ring. We have constructed gauge-invariant holomorphic scalar polynomials before imposing F-term relations. The original paper’s section 2.4 imposes those relations when discussing chiral operators. Our word list is not a count of independent chiral-ring classes, and it does not include operators defined by magnetic flux insertions.

Exercises

1. A neutral expression that is not invariant

Explain why $\operatorname{Tr}(A_1A_2B_1B_2)$ passes the relative-center charge test but fails the color-space test. Demonstrate the failure for $N=2$ with

$$A_1=E_{11},\quad A_2=E_{22},\quad B_1=B_2=1_2, \qquad U=1_2,\quad \widehat U=\frac1{\sqrt2} \begin{pmatrix}1&1\\-1&1\end{pmatrix}.$$

Compare it with the properly contracted $\operatorname{Tr}_L(A_1B_1A_2B_2)$.

Hint 1

Count the two central charges, then separately ask where the output of the rightmost $A$ would go.

Hint 2

Set $R=\widehat U$. The transformed malformed product reduces numerically to $E_{11}R^{-1}E_{22}R$. The correctly alternating product has adjacent $R^{-1}R$ factors.

Solution

Two $A$ factors and two $B$ factors give zero relative-center charge. But $A_1A_2$ would contract an unhatted output with a hatted input, which is not an invariant tensor of the product group.

Before the transformation the numerical expression is $\operatorname{Tr}(E_{11}E_{22})=0$. Equation (AB03.1) gives

$$\begin{aligned} \operatorname{Tr}(A_1'A_2'B_1'B_2') &=\operatorname{Tr}(E_{11}R^{-1}E_{22}R)\\ &=\operatorname{Tr}\left[ E_{11}\frac12 \begin{pmatrix}1&-1\\-1&1\end{pmatrix} \right]=\frac12. \end{aligned}$$

The proposed quantity changes under a unitary gauge transformation. In contrast,

$$\operatorname{Tr}_L(A_1'B_1'A_2'B_2') =\operatorname{Tr}(E_{11}R^{-1}R E_{22}R^{-1}R)=0.$$

This failure is genuinely non-Abelian: a central phase alone would not detect it.

2. Count cyclic words, then change the rank

There are sixteen flavor assignments for $\mathcal O_{ij;k\ell}=\operatorname{Tr}_L(A_iB_jA_kB_\ell)$. How many classes remain when we identify words only by trace cyclicity? At $N=1$, how many distinct commuting monomials remain? Identify the extra equality that accounts for the difference. Do not impose superpotential equations.

Hint 1

For a word written with an $A$ first, a two-place cyclic rotation exchanges the two pairs $(i,j)$ and $(k,\ell)$. Count the words fixed by that exchange.

Hint 2

At rank one, choose a degree-two monomial in $A_1,A_2$ and independently a degree-two monomial in $B_1,B_2$.

Solution

There are four choices for each flavor pair. Four of the sixteen words have $(i,j)=(k,\ell)$ and are fixed by pair exchange. The other twelve form six two-element classes. Hence there are

$$4+\frac{16-4}{2}=10$$

cyclic classes. Odd cyclic rotations give the same walks written with a $B$ first and the trace based at the other node; they add no further identification between these ten representatives.

At $N=1$, the $A$ monomials are $A_1^2,A_1A_2,A_2^2$, and the $B$ monomials are $B_1^2,B_1B_2,B_2^2$. Their products give nine distinct monomials. The extra equality is

$$A_1B_1A_2B_2=A_1B_2A_2B_1.$$

These two words belong to different cyclic classes but both become $A_1A_2B_1B_2$ for commuting numbers. The example (AB03.8) proves that this equality is not valid for general matrices. We counted words with specified kinematic identities, not independent interacting chiral operators.

3. An epsilon contraction tests the gauge group

A determinant can contract $N$ copies of an $A$ matrix using epsilon tensors. Derive the transformation of $\det A_1$. Is it invariant for our product of $U(N)$ groups? Compare with $\det(A_1B_1)$.

For $N=2$, derive

$$\det M=\frac12\left[(\operatorname{Tr}M)^2-\operatorname{Tr}(M^2)\right], \qquad M=A_1B_1,$$

and verify it for $A_1=\operatorname{diag}(2,3)$ and $B_1=\left(\begin{smallmatrix}1&1\\0&4\end{smallmatrix}\right)$.

Hint 1

Take determinants of the full transformation law. An epsilon tensor is invariant under $SU(N)$ but acquires a determinant factor under $U(N)$.

Hint 2

For the second part, trace the two-by-two characteristic identity $M^2-(\operatorname{Tr}M)M+(\det M)1_2=0$.

Solution

Multiplicativity gives

$$\det A_1'=\frac{\det U}{\det\widehat U}\det A_1.$$

Under the centers this is the phase $e^{iN(\alpha-\beta)}$. Thus $\det A_1$ alone is not a gauge invariant of the specified $U(N)\times U(N)$ theory. Changing both factors to $SU(N)$ would remove these determinant phases, but that changes the gauge-group data. We have not made that change.

The opposite phase of $\det B_1$ cancels in $\det(A_1B_1)=\det A_1\det B_1$. This is invariant, also directly because $M'=UMU^{-1}$. Taking the trace of the stated characteristic identity yields $\operatorname{Tr}(M^2)-(\operatorname{Tr}M)^2+2\det M=0$. In the example,

$$M=\begin{pmatrix}2&2\\0&12\end{pmatrix}, \qquad \operatorname{Tr}M=14,\quad \operatorname{Tr}(M^2)=148,\quad \det M=24.$$

Indeed $(14^2-148)/2=24$. A product of two quadratic traces and a single quartic trace can therefore encode the same determinant invariant at this rank. This is a matrix-size identity, not a dynamical F-term relation.

You can now read a proposed interaction as a closed, ordered color walk and explain which equalities follow from indices alone. Continue to build the quartic superpotential, fixing its relative sign and coefficient before deriving the ordered F-term equations. The arrows supply the representation data needed for that calculation.