Build the quartic superpotential

Which gauge-invariant quartic interaction belongs to ABJM, and what equations does it impose on its scalar matrices? The quiver lesson found the permitted ordered traces. We will now fix their relative sign and coefficient, then apply the component superpotential construction. Those are the two prerequisites.

Keep the theory of the ABJM introduction: flat three-dimensional compact $U(N)_k\times U(N)_{-k}$, with $N\geq1$, positive integer $k$, canonical matter kinetic terms, and no mass or FI deformation. Each gauge factor has rank $N$. There is no further common-center quotient; the integer diagonal fluxes of each compact $U(N)$ factor remain part of the definition. A local polynomial calculation does not modify those global data.

The full theory generically has $\mathcal N=6$, or twelve real Lorentzian Poincaré supercharges. Our chiral-superfield description displays an $\mathcal N=2$ subalgebra, or four real supercharges. The additional enhancement at levels one and two is described in the introduction; it is not a consequence of the polynomial calculation below. The vector sectors have Chern–Simons couplings, without defining Yang–Mills kinetic terms.

First checks: gauge invariance and R-charge

Let $V_L$ and $V_R$ be the two $N$-dimensional color spaces. The chiral matrices have directions

$$A_i:V_R\longrightarrow V_L,\qquad B_j:V_L\longrightarrow V_R,\qquad i,j=1,2.$$

The same letters will denote their scalar components when we evaluate a polynomial. In particular, $B_j$ is an independent chiral field, not $A_j^\dagger$. Write $\operatorname{Tr}_L$ or $\operatorname{Tr}_R$ for the ordinary fundamental trace at the indicated node. Set

$$\boxed{ W=h\,\operatorname{Tr}_L (A_1B_1A_2B_2-A_1B_2A_2B_1), \qquad h=\frac{4\pi}{k}. } \tag{AB04.1}$$

These are the field phases and normalization of AB v4, (2.5). There is no factor $1/N$ inside the trace.

Under a finite gauge transformation, $A_i\mapsto U A_i\widehat U^{-1}$ and $B_j\mapsto\widehat U B_jU^{-1}$. For example, the first word transforms into $U(A_1B_1A_2B_2)U^{-1}$: the intermediate inverse matrices cancel in their actual order. Its trace is invariant. The second word passes the same test. Neither test permits commuting a $B$ past an $A$.

For the chosen manifest R-symmetry, every chiral scalar has R-charge $1/2$. Each word therefore has charge two, canceled by the charge $-2$ of the chiral measure $d^2\theta$. Canonical engineering dimensions give $[A_i]=[B_j]=1/2$, so $[W]=2$ and $[h]=0$. Thus $\int d^3x\,d^2\theta\,W$ is dimensionless.

These checks are necessary but do not fix the interaction: any relative coefficient between the two traces would pass them. They also do not prove exact conformality or enhanced supersymmetry. We next recover the specific coefficient from the source’s supersymmetric construction.

Recover the coefficient from algebraic fields

An auxiliary chiral superfield is a whole chiral multiplet introduced without a kinetic term. It is distinct from the highest component $F$ of a propagating chiral multiplet. In AB, (2.4), two such adjoint variables, $\Phi_L$ and $\Phi_R$, provide a compact way to specify the coupling:

$$\begin{aligned} W_{\rm aux}={}&\frac{k}{8\pi} \left[\operatorname{Tr}_R(\Phi_R^2) -\operatorname{Tr}_L(\Phi_L^2)\right]\\ &+\sum_{i=1}^2\left[ \operatorname{Tr}_R(B_i\Phi_L A_i) +\operatorname{Tr}_L(A_i\Phi_R B_i)\right]. \end{aligned} \tag{AB04.2}$$

Here $\Phi_L$ acts on $V_L$, $\Phi_R$ on $V_R$, and both have engineering dimension one. Their quadratic coefficients are opposite because the two Chern–Simons levels are opposite. They are algebraic variables in this construction, not additional propagating fields in the theory.

Define the two holomorphic matrices $S_L=\sum_i A_iB_i$ and $S_R=\sum_i B_iA_i$. Trace cyclicity rewrites the linear terms as $\operatorname{Tr}_L(\Phi_L S_L)+\operatorname{Tr}_R(\Phi_R S_R)$. Varying each adjoint entry gives

$$-\frac{k}{4\pi}\Phi_L+S_L=0,\qquad \frac{k}{4\pi}\Phi_R+S_R=0, \qquad \Phi_L=hS_L,\quad \Phi_R=-hS_R. \tag{AB04.3}$$

Substitution, including the quadratic terms, yields

$$W_{\rm eff}=\frac h2 \left[\operatorname{Tr}_L(S_L^2) -\operatorname{Tr}_R(S_R^2)\right]. \tag{AB04.4}$$

The terms with the same flavor twice cancel, since $\operatorname{Tr}_L(A_iB_iA_iB_i) =\operatorname{Tr}_R(B_iA_iB_iA_i)$. The two mixed terms in the first square both give $\operatorname{Tr}_L(A_1B_1A_2B_2)$. Those in the second square both give $\operatorname{Tr}_L(A_1B_2A_2B_1)$. The factor two cancels the $1/2$ in (AB04.4), proving (AB04.1). This elimination is holomorphic algebra; it does not assume a convergent real Euclidean integration contour for an auxiliary field.

The relative minus sign also displays a flavor symmetry. With $\epsilon^{12}=+1$,

$$W=\frac h2\,\epsilon^{ij}\epsilon^{mn} \operatorname{Tr}_L(A_iB_mA_jB_n). \tag{AB04.5}$$

Expanding the four nonzero epsilon terms gives twice the trace difference in (AB04.1). Independent $SU(2)$ rotations of the two $A$ flavors and the two $B$ flavors preserve their respective epsilon tensors, hence preserve $W$. This is AB, (2.6). The paper’s argument for the larger R-symmetry also uses transformations of the rest of the action; invariance of $W$ alone is not that argument.

Differentiate the trace without losing its order

For an ordinary complex variable, the coefficient of $\delta\phi$ is $W'$. For a matrix, specify the pairing that defines its derivative. Define the cyclic gradients $G_{A_i}$ and $G_{B_j}$ by

$$\delta W= \sum_i\operatorname{Tr}_L(\delta A_i\,G_{A_i}) +\sum_j\operatorname{Tr}_R(\delta B_j\,G_{B_j}). \tag{AB04.6}$$

The word “cyclic” means that we move the varied factor to the beginning of its trace while retaining the order of the remaining factors. The gradient has the reverse direction to the field: $G_A:V_L\to V_R$ and $G_B:V_R\to V_L$. In components,

$$\frac{\partial W}{\partial(A_i)^a{}_{\widehat b}} =(G_{A_i})^{\widehat b}{}_a. \tag{AB04.7}$$

Thus an array of entry derivatives is the transpose of the displayed cyclic-gradient matrix. No Hermitian conjugation is involved in holomorphic differentiation.

For example, varying $A_2$ gives

$$\begin{aligned} \delta_{A_2}W &=h\,\operatorname{Tr}_L (A_1B_1\,\delta A_2 B_2-A_1B_2\,\delta A_2 B_1)\\ &=h\,\operatorname{Tr}_L \left[\delta A_2(B_2A_1B_1-B_1A_1B_2)\right]. \end{aligned}$$

The first residual factor is $B_2$, because it lay immediately to the right of $\delta A_2$ before the cyclic rotation. Applying this operation to every field gives

$$\begin{aligned} G_{A_1}&=h(B_1A_2B_2-B_2A_2B_1),\\ G_{A_2}&=h(B_2A_1B_1-B_1A_1B_2),\\ G_{B_1}&=h(A_2B_2A_1-A_1B_2A_2),\\ G_{B_2}&=h(A_1B_1A_2-A_2B_1A_1). \end{aligned} \tag{AB04.8}$$

These are ordered cubic equations, not scalar equations obtained by ignoring matrix multiplication. Their gauge transformations are $G_A\mapsto\widehat U G_AU^{-1}$ and $G_B\mapsto U G_B\widehat U^{-1}$, so (AB04.6) is invariant. A second check is the homogeneity identity $\sum_i\operatorname{Tr}_L(A_iG_{A_i})+ \sum_j\operatorname{Tr}_R(B_jG_{B_j})=4W$, obtained by varying all four fields by the same infinitesimal scale.

From the gradients to a vacuum test

Collect the scalar entries of the four matrices into coordinates $\phi^I$. The index $I$ includes flavor and both color labels. This converts the matrix problem into the multi-field construction of the superpotential lesson, without changing matrix order. Its chiral interaction is

$$\mathcal L_{\rm hol} =W_I F^I-\frac12W_{IJ}\psi^I\psi^J, \qquad W_I=\frac{\partial W}{\partial\phi^I},\quad W_{IJ}=\frac{\partial^2W}{\partial\phi^I\partial\phi^J}. \tag{AB04.9}$$

The second derivative fixes the Yukawa couplings as well as the linearized F equations. The spinor contraction is the one in the prerequisite; the entries $\phi^I$ themselves commute. No Grassmann sign was needed when rotating the purely scalar words above.

In the inherited Euclidean convention the auxiliary terms are $-\widetilde F_I F^I+W_I F^I+\widetilde W^{\,I}\widetilde F_I$. Their algebraic equations give $\widetilde F_I=W_I$ and $F^I=\widetilde W^{\,I}$. Antichiral variables remain independent in Euclidean signature. On the bosonic physical slice, $\widetilde W^{\,I}=(W_I)^*$, so the resulting positive potential contribution is

$$\boxed{ V_F=\sum_i\operatorname{Tr}_L(G_{A_i}^\dagger G_{A_i}) +\sum_j\operatorname{Tr}_R(G_{B_j}^\dagger G_{B_j})\geq0. } \tag{AB04.10}$$

The sum of squared entries is insensitive to the transpose in (AB04.7), but the ordered gradient itself is not. Each $G$ has dimension $3/2$; $V_F$ has dimension three and is degree six in scalar fields. Vanishing of every $G$ is called F-flatness.

F-flatness is necessary for a zero-energy supersymmetric vacuum. To find the full vacuum space one must also include the vector-sector constraints and divide by the appropriate gauge transformations. Equation (AB04.10) is only the chiral auxiliary contribution, not the complete scalar potential.

Worked example: zero interaction value, nonzero force

For $N=2$, choose complex numbers $v,b,c$ and

$$A_1=v\,1_2,\qquad A_2=0,\qquad B_1=b\,\sigma_1,\qquad B_2=c\,\sigma_2,$$

where $\sigma_1=\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$, $\sigma_2=\left(\begin{smallmatrix}0&-i\\i&0\end{smallmatrix}\right)$, and $\sigma_3=\operatorname{diag}(1,-1)$. These Pauli matrices are color matrices in this example, not spacetime gamma matrices.

Every term in $W$ contains $A_2$, so $W=0$. Three gradients also vanish. However,

$$G_{A_2}=h\,vbc(\sigma_2\sigma_1-\sigma_1\sigma_2) =-2ih\,vbc\,\sigma_3, \qquad V_F=8|h\,vbc|^2. \tag{AB04.11}$$

Here $[\sigma_1,\sigma_2]=2i\sigma_3$ and $\operatorname{Tr}_2(\sigma_3^2)=2$ fix both the sign and the factor eight. For nonzero $v,b,c$, this configuration fails the vacuum test. A function can vanish at a point while its derivatives do not.

Conversely, if all four matrices are diagonal in the chosen color bases, each diagonal entry of (AB04.8) cancels between commuting numbers. This gives F-flat configurations. At $N=1$, $W$ and all four gradients vanish identically, as noted in AB, section 2.3. Neither statement removes the compact gauge fields, their levels, or the residual global identifications of a vacuum.

Exercises

1. A relative coefficient that the charge test misses

Replace (AB04.1) by $W_\rho=h\operatorname{Tr}_L(A_1B_1A_2B_2-\rho A_1B_2A_2B_1)$, where $\rho$ is a fixed complex number. Gauge invariance and the R-charge test still pass. Use the infinitesimal flavor operation $\delta A_1=\alpha A_2$, $\delta A_2=0$, $\delta B_j=0$ to determine which $\rho$ preserves the full $SU(2)$ acting on the $A$ doublet. This operation is one complex linear combination of the real $SU(2)$ generators, which may be tested on a holomorphic polynomial. Explain what this test does and does not fix.

Hint 1

Vary the two appearances of $A_1$. After replacing them by $A_2$, the resulting traces are equal by cyclic rotation.

Hint 2

The trace need not vanish. You can set $A_2=B_1=B_2=1_N$ to test whether its coefficient must vanish as a polynomial identity.

Solution

The variation is

$$\begin{aligned} \delta W_\rho &=\alpha h\left[ \operatorname{Tr}_L(A_2B_1A_2B_2) -\rho\operatorname{Tr}_L(A_2B_2A_2B_1)\right]\\ &=\alpha h(1-\rho)\operatorname{Tr}_L(A_2B_1A_2B_2). \end{aligned}$$

With the matrices in Hint 2 the trace is $N$, so invariance for all field values requires $\rho=1$. At that value, the epsilon form (AB04.5) proves invariance under the entire $SU(2)_A\times SU(2)_B$.

Flavor symmetry fixes the relative coefficient in this two-term ansatz but leaves the overall $h$ free. Its value $4\pi/k$ came from the matched action construction. Neither the charge check nor this flavor check alone establishes the full extended supersymmetry or the quantum fixed point.

2. Turn on the missing matrix

In the worked example, replace $A_2=0$ by $A_2=t\sigma_3$, with complex $t$. Calculate $W$, every cyclic gradient, and $V_F$. Verify that the derivative of $W$ with respect to $t$ equals $\operatorname{Tr}_L(\sigma_3G_{A_2})$. Can varying $t$ restore F-flatness while $v,b,c$ remain nonzero?

Hint 1

Use $\sigma_i\sigma_j=\delta_{ij}1_2+i\epsilon_{ijk}\sigma_k$. The products $\sigma_1\sigma_3\sigma_2$ and $\sigma_2\sigma_3\sigma_1$ have opposite signs.

Hint 2

Each gradient is a complex number times either $1_2$ or one Pauli matrix. Each such matrix has squared norm $\operatorname{Tr}(M^\dagger M)=2$.

Solution

The two triple products are $-i1_2$ and $+i1_2$. Therefore $W=-4ihvbct$. The four gradients are

$$\begin{aligned} G_{A_1}&=-2ihbct\,1_2,& G_{A_2}&=-2ihvbc\,\sigma_3,\\ G_{B_1}&=-2ihvct\,\sigma_1,& G_{B_2}&=-2ihvbt\,\sigma_2. \end{aligned}$$

Their squared norms give

$$V_F=8|h|^2 \left(|bct|^2+|vbc|^2+|vct|^2+|vbt|^2\right).$$

Indeed, $\partial_tW=-4ihvbc= \operatorname{Tr}[\sigma_3(-2ihvbc\,\sigma_3)]$. This is the directional-derivative pairing, with no conjugation. For nonzero $v,b,c$, the second positive term never vanishes, regardless of $t$. Thus no point on this one-parameter family is F-flat. At $t=0$ the result reduces to (AB04.11).

3. A trace normalization is part of the coupling

Suppose a formula is rewritten using the normalized trace $\operatorname{tr}_N=\operatorname{Tr}_N/N$, while the canonical kinetic action and the level $k$ are held fixed. What coefficient $h_{\rm norm}$ reproduces (AB04.1)? If one instead keeps the coefficient $4\pi/k$, how do the entry gradients and $V_F$ change? Does the set of F-flat configurations detect the mistake?

Hint 1

Match the two functions $h_{\rm norm}\operatorname{tr}_N(\cdots)$ and $h\operatorname{Tr}_N(\cdots)$ before taking a derivative.

Hint 2

The kinetic metric used to eliminate the auxiliaries has not changed. Scaling $W$ scales each gradient once and its positive norm twice.

Solution

Equality requires $h_{\rm norm}=Nh=4\pi N/k$. If the coefficient is left equal to $h$, the new superpotential is $W/N$, its entry derivatives and cyclic gradients are $1/N$ times the original ones, and its chiral potential is $V_F/N^2$.

For finite $N\geq1$, multiplication by the nonzero constant $1/N$ leaves the equations $G=0$ unchanged. A list of F-flat configurations therefore cannot detect this interaction-normalization error. The value of the potential away from that locus, or the Yukawa coefficients $W_{IJ}$, does detect it. Changing the kinetic normalization too would require a separate field and coupling translation; it is excluded by the question. At $N=1$, the trace conventions coincide, and this particular superpotential vanishes in either convention.

The quartic interaction now has a fixed coefficient, a typed derivative, and a concrete F-flatness test. Its next use is to combine the chiral contribution with the vector-sector auxiliary equations. That step determines the complete sextic scalar potential; only then can the full symmetry and vacuum analysis use all the interactions together.

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