From auxiliary fields to a sextic scalar potential

The quartic superpotential produced positive squares of cubic F-term equations. Are those the whole scalar potential? No: the scalar partners of the two Chern–Simons gauge fields also couple to matter. Their algebraic constraints turn those couplings into additional sextic terms.

We will derive the complete classical scalar potential for the same flat three-dimensional compact $U(N)_k\times U(N)_{-k}$ theory, with equal $N\geq1$, positive integer $k$, canonical matter kinetic terms and no masses or FI parameters. Each gauge factor has rank $N$. Keep the ordinary fundamental traces and the compact flux data of the defining action, with no further quotient of the two centers. The generic full theory has twelve real Lorentzian Poincaré supercharges, while the $\mathcal N=2$ formulation used here displays four. Deriving the scalar potential is not by itself a proof of the full extended supersymmetry.

Start with the terms that contain auxiliaries

The scalar matrices $A_i:V_R\to V_L$ and $B_i:V_L\to V_R$, $i=1,2$, retain their two opposite color directions. Define the matrix norm $\|X\|^2=\operatorname{Tr}_{\rm domain}(X^\dagger X)$. It is the sum of absolute squares of entries. The trace is ordinary, without $1/N$.

To discuss a positive potential, select the physical bosonic slice: the antichiral scalar partners are Hermitian conjugates of the chiral scalars, and the vector scalars $\sigma,\widehat\sigma$ are Hermitian. This is a statement about the resulting classical energy. The Euclidean off-shell fields and auxiliary contours must first be treated independently.

Write the two Hermitian matter-density matrices as

$$\mu_L=\sum_i(A_iA_i^\dagger-B_i^\dagger B_i), \qquad \mu_R=\sum_i(B_iB_i^\dagger-A_i^\dagger A_i), \qquad t=\frac{2\pi}{k}. \tag{AB05.1}$$

They act on $V_L$ and $V_R$ respectively. In particular, $\mu_R$ is not obtained by moving every factor of $\mu_L$ through its neighbors. Both matrices have engineering dimension one.

The scalar partners of the gauge connections act on the same sides as their gauge generators. Define

$$X_i=\sigma A_i-A_i\widehat\sigma,\qquad Y_i=\widehat\sigma B_i-B_i\sigma. \tag{AB05.2}$$

The canonical gauge-coupled chiral action supplies $V_\sigma=\sum_i(\|X_i\|^2+\|Y_i\|^2)$. For a single Abelian charge this is the familiar positive term $q^2\sigma^2|\phi|^2$; the two-sided action in (AB05.2) is its bifundamental form. It is not a commutator with one common matrix unless the two vector scalars have actually been identified.

The remaining nondifferentiated bosonic terms needed for elimination are

$$\begin{aligned} \mathcal L_{\rm aux}={}&V_\sigma +\operatorname{Tr}_L\!\left[D\left(\mu_L-\frac{\sigma}{t}\right)\right] +\operatorname{Tr}_R\!\left[\widehat D \left(\mu_R+\frac{\widehat\sigma}{t}\right)\right]\\ &-\widetilde F_I F^I+W_I F^I +\widetilde W^{\,I}\widetilde F_I. \end{aligned} \tag{AB05.3}$$

Here $I$ runs over all scalar entries of the four chirals, and $W_I$ is the entry derivative of $W=2t\,\operatorname{Tr}_L(A_1B_1A_2B_2-A_1B_2A_2B_1)$. The terms linear in $D,\widehat D$ follow from the opposite Chern–Simons levels and the signed matter charges in (AB05.1). The final line uses the Euclidean auxiliary convention established in the superpotential lesson. A tilde is independent until the physical scalar slice is selected.

There is no $D^2/g_{\rm YM}^2$ term in the defining Chern–Simons theory. Adding a Yang–Mills regulator is possible, but that is a different intermediate action whose removal must be specified. We are not adding it here.

Eliminate the vector fields in the right order

Vary every matrix entry of $D$ and $\widehat D$. Equation (AB05.3) gives

$$\boxed{\sigma=t\mu_L,\qquad \widehat\sigma=-t\mu_R.} \tag{AB05.4}$$

These equations determine the vector scalars. They do not impose $\mu_L=\mu_R=0$. In a pure Chern–Simons multiplet the scalars have no independent kinetic terms, so substituting their algebraic values is appropriate.

For example, the equation multiplying $D$ is $\mu_L-\sigma/t=0$, not $\mu_L=0$. Dropping the $\sigma D$ partner of the Chern–Simons term would change that equation and the theory. Varying $\sigma,\widehat\sigma$ in turn determines $D,\widehat D$; it does not introduce another condition on the matter scalars. Exercise 1 checks this statement explicitly.

On (AB05.4), both terms linear in $D$ vanish. The remaining vector contribution is

$$\boxed{ V_{\rm CS}=t^2\sum_i\left( \|\mu_L A_i+A_i\mu_R\|^2 +\|\mu_R B_i+B_i\mu_L\|^2\right). } \tag{AB05.5}$$

The plus signs inside these squares come from substituting the opposite sign of $\widehat\sigma$. For the $B$ term, $Y_i=-t(\mu_R B_i+B_i\mu_L)$, whose overall sign drops out of the norm.

The squares retain all noncommuting cross terms. For instance,

$$\begin{aligned} \|\sigma A-A\widehat\sigma\|^2 ={}&\operatorname{Tr}_R(A^\dagger\sigma^2 A) -2\operatorname{Tr}_R(A^\dagger\sigma A\widehat\sigma)\\ &+\operatorname{Tr}_R(A^\dagger A\widehat\sigma^2). \end{aligned} \tag{AB05.6}$$

The middle terms combine by trace cyclicity; no matrices were commuted. Substituting (AB05.4) makes this $t^2\operatorname{Tr}_R[A^\dagger\mu_L^2A+ 2A^\dagger\mu_LA\mu_R+A^\dagger A\mu_R^2]$. This agrees directly with the bosonic constraints and ordered potential in AB v4, (A.3)–(A.4).

For this comparison, $\sigma$ and $\widehat\sigma$ have exactly the signs of the two scalars in (A.3). The defining-action lesson separately translated the general equation following the paper’s (2.1). That earlier translation is not a rule to reverse the signs in (A.3). The displayed action (AB05.3) fixes the signs used here.

Add the chiral contribution

Variation of the final line of (AB05.3) yields $\widetilde F_I=W_I$ and $F^I=\widetilde W^{\,I}$. On the physical scalar slice, substitution gives $V_F=\sum_I|W_I|^2$. The negative Euclidean auxiliary square was not a positive function to minimize on real auxiliary variables; this is algebraic elimination with the inherited continuation.

For a complete ordered formula, strip the coefficient $2t$ from the four cyclic gradients and define

$$\begin{aligned} f_{A_1}&=B_1A_2B_2-B_2A_2B_1,& f_{A_2}&=B_2A_1B_1-B_1A_1B_2,\\ f_{B_1}&=A_2B_2A_1-A_1B_2A_2,& f_{B_2}&=A_1B_1A_2-A_2B_1A_1. \end{aligned} \tag{AB05.7}$$

The cyclic gradients have the reversed color directions explained in the prerequisite. The transpose between them and entry derivatives does not alter their summed norm. The result is

$$\boxed{ \begin{aligned} V={}&V_{\rm CS}+V_F\\ ={}&t^2\sum_i\bigl( \|\mu_L A_i+A_i\mu_R\|^2 +\|\mu_R B_i+B_i\mu_L\|^2\bigr)\\ &+4t^2\sum_i\bigl(\|f_{A_i}\|^2+\|f_{B_i}\|^2\bigr). \end{aligned}} \tag{AB05.8}$$

This is the complete classical scalar potential at zero mass and FI parameter, including both auxiliary sectors. Its chiral piece is AB, (A.1)–(A.2), while its vector piece is (A.3)–(A.4). The relative factor four comes from the superpotential coefficient $4\pi/k=2t$, not from a flavor multiplicity.

Every expression inside a norm is cubic in scalar entries, so $V$ is sextic. With scalar dimension $1/2$, each square has dimension three, as required in three dimensions. Rescaling all scalars by a real $\lambda$ multiplies $V$ by $\lambda^6$; changing $k$ at fixed fields multiplies its coefficient by $k^{-2}$. Neither engineering check independently proves a quantum scaling dimension.

The norm form makes nonnegativity manifest on the stated slice. Each matrix also transforms with its appropriate pair of unitary gauge matrices, so the norms are gauge invariant. Thus zero potential requires all four F equations and both sets of vector equations inside (AB05.8) to vanish. Gauge equivalence and compact global identifications must still be imposed when constructing a vacuum space.

Worked example: F-flatness is insufficient

Let $N=2$, set $B_1=B_2=0$, and choose $A_1=vE_{11}$, $A_2=wE_{12}$, where $E_{ab}$ has one unit entry in row $a$, column $b$. Each F equation vanishes because it contains a $B$. However,

$$\mu_L=(|v|^2+|w|^2)E_{11},\qquad \mu_R=-|v|^2E_{11}-|w|^2E_{22}.$$

Using $E_{ab}E_{cd}=\delta_{bc}E_{ad}$ gives

$$\begin{aligned} \mu_LA_1+A_1\mu_R&=v|w|^2E_{11},\\ \mu_LA_2+A_2\mu_R&=w|v|^2E_{12},\\ V&=t^2|v|^2|w|^2(|v|^2+|w|^2). \end{aligned} \tag{AB05.9}$$

For nonzero $v,w$, the full potential is positive even though $V_F=0$. The two matrices send different right-node basis vectors into the same left-node direction. Their density matrices therefore do not intertwine in the way required by (AB05.5).

If instead all four scalar matrices are diagonal in the chosen color bases, each diagonal entry has $\mu_R=-\mu_L$, and the cubic F equations contain only commuting numbers. Every term in (AB05.8) then vanishes. This constructs a zero-potential family; it neither proves that all vacua have been classified nor determines their residual quotient.

Exercises

1. What happens to the remaining vector auxiliaries?

Before substituting (AB05.4), vary $V_\sigma$ with respect to the Hermitian matrices $\sigma,\widehat\sigma$. Define $N_L=\sum_i(A_iA_i^\dagger+B_i^\dagger B_i)$ and $N_R=\sum_i(A_i^\dagger A_i+B_iB_i^\dagger)$. Find the equations determining $D,\widehat D$ from (AB05.3). Explain why they do not demand that $V_\sigma$ be stationary with respect to a freely propagating Coulomb scalar.

Hint 1

Use (AB05.6) and move each varied matrix to the beginning of its own trace. For a fixed matrix $N$, the variation of $\operatorname{Tr}(\sigma^2N)$ is $\operatorname{Tr}[\delta\sigma(\sigma N+N\sigma)]$.

Hint 2

The terms linear in $D,\widehat D$ also vary. Their coefficients are $-D/t$ and $+\widehat D/t$.

Solution

With $\{X,Y\}=XY+YX$, the two trace derivatives are

$$\begin{aligned} K_L&=\{\sigma,N_L\} -2\sum_i(A_i\widehat\sigma A_i^\dagger+B_i^\dagger\widehat\sigma B_i),\\ K_R&=\{\widehat\sigma,N_R\} -2\sum_i(A_i^\dagger\sigma A_i+B_i\sigma B_i^\dagger). \end{aligned}$$

Thus the equations are $K_L-D/t=0$ and $K_R+\widehat D/t=0$, giving $D=tK_L$ and $\widehat D=-tK_R$. For Hermitian vector scalars, both $K$ matrices are Hermitian and have dimension two, consistent with $[D]=2$.

Together with the equations from varying $D,\widehat D$, these determine all four algebraic vector variables. Setting $K_L=K_R=0$ independently would discard their couplings to $D,\widehat D$ and add a spurious matter restriction. The substitution of (AB05.4) already accounts for the vector-sector contribution to the matter potential.

2. Change one Chern–Simons level

At $N=1$, let $u=|A_1|^2+|A_2|^2-|B_1|^2-|B_2|^2$. First show that the potential vanishes for arbitrary scalar values in the opposite-level theory. Then change the second level from $-k$ to $+k$, leaving the matter kinetic normalization fixed. What vector potential results? The quartic superpotential still vanishes at this rank. This changed theory is an $\mathcal N=2$ Chern–Simons–matter theory; it is not the fixed ABJM theory.

Hint 1

Initially $\mu_L=u$, $\mu_R=-u$ and $\sigma=\widehat\sigma=tu$. Every scalar matrix is now an ordinary complex number.

Hint 2

The changed second level changes its linear auxiliary term to $\widehat D(\mu_R-\widehat\sigma/t)$. Recompute $\widehat\sigma$ before using (AB05.2).

Solution

For opposite levels, equality of the two vector scalars makes every $X_i,Y_i$ zero. The four F expressions vanish for commuting numbers, so $V=0$ without imposing $u=0$.

For equal positive levels, $\sigma=tu$ and $\widehat\sigma=t\mu_R=-tu$. Hence $X_i=2tu A_i$ and $Y_i=-2tu B_i$, giving

$$V_{\rm equal\ levels} =4t^2u^2\sum_i(|A_i|^2+|B_i|^2).$$

The sign change has altered the scalar interaction. It is not a harmless redefinition of the hatted field while holding its action fixed. In the original theory, zero local potential does not eliminate the compact gauge sector or prove that the physical vacuum space is an unrestricted copy of four complex coordinates. Those questions also involve gauge and flux data.

3. A zero quadratic term need not be a flat direction

Take real $u,v>0$ and

$$A_1=\operatorname{diag}(u,v),\qquad A_2=\varepsilon E_{12},\qquad B_1=B_2=0,$$

with complex $\varepsilon$. Find the exact potential on this family and the coefficient of $|\varepsilon|^2$. What happens when $u=v\ne0$? Retain the response of $\sigma,\widehat\sigma$ to the changed matter fields.

Hint 1

The density matrices are $\mu_L=\operatorname{diag}(u^2+|\varepsilon|^2,v^2)$ and $\mu_R=-\operatorname{diag}(u^2,v^2+|\varepsilon|^2)$.

Hint 2

Although $A_1$ was part of the original diagonal background, its cubic vector equation is no longer zero. Include its norm as well as that of $A_2$.

Solution

All F terms remain zero. The two nonzero cubic vector matrices are

$$\begin{aligned} \mu_LA_1+A_1\mu_R &=|\varepsilon|^2\operatorname{diag}(u,-v),\\ \mu_LA_2+A_2\mu_R &=(u^2-v^2)\varepsilon E_{12}. \end{aligned}$$

Their norms yield the exact expression

$$V=t^2\left[ (u^2-v^2)^2|\varepsilon|^2 +(u^2+v^2)|\varepsilon|^4\right].$$

For the canonically normalized complex scalar entry, the quadratic potential coefficient is $m_\varepsilon^2=t^2(u^2-v^2)^2$, with dimension two. This identifies a scalar quadratic coefficient; it is not a complete diagonalization of the coupled Chern–Simons gauge and matter dynamics.

At $u=v\ne0$, the quadratic term vanishes but the quartic term is $2t^2u^2|\varepsilon|^4>0$ for $\varepsilon\ne0$. Thus a zero eigenvalue of a quadratic approximation does not establish an exact flat direction. Freezing the vector scalars at their original values would incorrectly miss this response.

The complete potential is now a sum of ordered positive squares with its coefficient fixed. Continue with Where does N=6 supersymmetry come from? to reorganize the fields and combine the vector and superpotential contributions under the full R-symmetry.

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