Chern–Simons gauge fields with non-Abelian matter

In electrodynamics, a charged field sees one connection. An ABJM matter matrix carries two independent color indices, so transporting it requires two connections. Their signs are fixed by the representation, while the signs of the two Chern–Simons levels are an additional property of the theory. Keeping those two statements separate prevents a common error: reversing a matter charge does not automatically reverse a gauge-field action.

Use the ABJM definition as preparation. We retain flat three-dimensional space, compact $U(N)_k\times U(N)_{-k}$ with $k>0$, no further quotient, and no masses or FI terms. The full generic theory has twelve real Poincaré supercharges; its manifest $\mathcal N=2$ presentation has four. This lesson constructs its gauge sector and the color dependence of matter couplings. The quartic superpotential and elimination of auxiliaries are separate calculations.

Transport a matrix, not two unrelated columns

Let $V_L$ and $V_R$ be the two $N$-dimensional color spaces. A scalar $X$ of either $A$ multiplet is a map $V_R\to V_L$. Thus

$$X'=U X\widehat U^{-1}, \qquad \mathcal D_\mu X =\partial_\mu X-ia_\mu X+iX\widehat a_\mu. \tag{AB02.1}$$

The connections $a_\mu,\widehat a_\mu$ are Hermitian matrices. Their engineering dimension is one; $[X]=1/2$. Hats label the second group, not conjugation. We use the ordinary fundamental trace $\operatorname{Tr}_N$, with Hermitian generators normalized by $\operatorname{Tr}_N(t^at^b)=\delta^{ab}/2$.

To derive the derivative, impose $\mathcal D'_\mu X'=U(\mathcal D_\mu X)\widehat U^{-1}$. The terms differentiating $U$ and $\widehat U^{-1}$ cancel precisely when

$$a'_\mu=Ua_\mu U^{-1}-i(\partial_\mu U)U^{-1}, \qquad \widehat a'_\mu =\widehat U\widehat a_\mu\widehat U^{-1} -i(\partial_\mu\widehat U)\widehat U^{-1}. \tag{AB02.2}$$

For example, $\partial_\mu\widehat U^{-1} =-\widehat U^{-1}(\partial_\mu\widehat U)\widehat U^{-1}$. This minus sign is canceled by the right connection in (AB02.1). Using a minus sign for both connection terms would fail even for two Abelian gauge groups.

The $B$ multiplets have the reverse direction, $Y:V_L\to V_R$:

$$\mathcal D_\mu Y =\partial_\mu Y-i\widehat a_\mu Y+iYa_\mu, \qquad Y'=\widehat UYU^{-1}. \tag{AB02.3}$$

The derivative on each matter fermion has the same color action. Its spinor index is separate from these matrix indices. The representation assignment is that of AB, section 2.2; its two opposite sets of diagonal weights appear explicitly in KWY, (4.2).

Curvature checks both signs

Define

$$f_{\mu\nu} =\partial_\mu a_\nu-\partial_\nu a_\mu-i[a_\mu,a_\nu], \qquad f=da-i\,a\wedge a, \tag{AB02.4}$$

and the hatted counterpart. Expanding the two derivatives, the terms containing one derivative of $X$ cancel. The left and right matrix actions commute with each other. The result is

$$[\mathcal D_\mu,\mathcal D_\nu]X =-if_{\mu\nu}X+iX\widehat f_{\mu\nu}. \tag{AB02.5}$$

The right action has the opposite curvature sign because it is the dual representation. In particular, a constant non-Abelian connection need not be flat: its matrix commutator can be nonzero.

For an explicit $U(2)\times U(2)$ example, take $a_1=u\tau_1/2$, $a_2=v\tau_2/2$, $a_3=0$, and all hatted connections zero. Here $\tau_a$ are the ordinary Pauli matrices and $u,v$ are real constants of dimension one. Since $[\tau_1,\tau_2]=2i\tau_3$,

$$f_{12}=\frac{uv}{2}\tau_3, \qquad [\mathcal D_1,\mathcal D_2]X =-\frac{iuv}{2}\tau_3X.$$

Computing the commutator directly from $\mathcal D_1X=-iu\tau_1X/2$ and $\mathcal D_2X=-iv\tau_2X/2$ gives the same answer, including its sign.

The matter norm fixes the source currents

Let $\widetilde X:V_L\to V_R$ be the independent antichiral scalar. Its derivative is $\mathcal D_\mu\widetilde X =\partial_\mu\widetilde X-i\widehat a_\mu\widetilde X+i\widetilde Xa_\mu$. The scalar kinetic term is

$$\mathcal L_{\rm kin,X} =\operatorname{Tr}_{R} [(\mathcal D_\mu\widetilde X)(\mathcal D^\mu X)]. \tag{AB02.6}$$

The subscript identifies which color space is being traced; both spaces have dimension $N$. On the usual bosonic slice $\widetilde X=X^\dagger$, this is the positive sum of absolute squares. In Euclidean algebra the tilded variable remains independent until a contour is specified.

The same representation determines the vector-scalar and auxiliary couplings. Define the linear operators

$$\Sigma X=\sigma X-X\widehat\sigma, \qquad \mathscr D X=DX-X\widehat D.$$

The inherited chiral action contains $\operatorname{Tr}_{R}(\widetilde X\Sigma^2X+\widetilde X\mathscr D X)$. For example,

$$\Sigma^2X =\sigma^2X-2\sigma X\widehat\sigma+X\widehat\sigma^2. \tag{AB02.7}$$

This expands the charge-squared scalar term from the Abelian action without treating a bifundamental as an adjoint of either node. Reverse $L,R$ for $Y$. Fermionic interactions use these representations too, with the additional Grassmann signs when odd coefficients are reordered.

We can also derive the scalar contribution to each gauge equation directly. Define currents by $\delta\mathcal L_{\rm kin,X} =\operatorname{Tr}_L(J_L^\mu\delta a_\mu) +\operatorname{Tr}_R(J_R^\mu\delta\widehat a_\mu)$. Varying (AB02.6), then closing each trace in its proper color space, gives

$$\begin{aligned} J_L^\mu &=i[(\mathcal D^\mu X)\widetilde X -X(\mathcal D^\mu\widetilde X)],\\ J_R^\mu &=i[(\mathcal D^\mu\widetilde X)X -\widetilde X(\mathcal D^\mu X)]. \end{aligned} \tag{AB02.8}$$

Each is adjoint-valued at its own node. Their traces add to zero, although the matrices cannot generally be equated. The diagonal central gauge transformation acts trivially on $X$; this trace cancellation is its local current check. The full theory adds the other scalars and the fermion currents.

Why the Chern–Simons action needs a cubic term

For a single Abelian connection, $a\wedge da$ suffices. For matrices it misses the commutator in (AB02.4). Consider a matrix-valued differential form, where multiplication means matrix multiplication and exterior multiplication. Cyclicity of its trace includes the exterior-degree sign.

The required three-form is

$$\omega_3(a) =\operatorname{Tr}_N \left(a\wedge da-\frac{2i}{3}a\wedge a\wedge a\right). \tag{AB02.9}$$

Its coefficient can be found without guessing. Vary the first term: $\delta\operatorname{Tr}(a\,da) =2\operatorname{Tr}(\delta a\,da) -d\operatorname{Tr}(a\,\delta a)$. The three variations of $\operatorname{Tr}(a^3)$ are equal after graded cyclicity, giving $3\operatorname{Tr}(\delta a\,a^2)$. Therefore

$$\boxed{\delta\omega_3(a) =2\operatorname{Tr}_N(\delta a\wedge f) -d\operatorname{Tr}_N(a\wedge\delta a).} \tag{AB02.10}$$

The factor $-2i/3$ is precisely what assembles $f=da-ia^2$. Another check, now in four dimensions, is $d\omega_3=\operatorname{Tr}_N(f\wedge f)$.

For $U=e^{i\alpha}$, the infinitesimal connection change is $\delta_\alpha a=d\alpha-i[a,\alpha]$. The Bianchi identity gives $\operatorname{Tr}[(\delta_\alpha a)\wedge f]=d\operatorname{Tr}(\alpha f)$. Thus the action changes by a boundary term. We use closed spaces or variations and gauge transformations with the required boundary behavior. On a physical boundary, this calculation leaves an actual boundary term to handle.

The expression also requires its compact global definition. Integer levels and the ordinary trace are those fixed in the flat Chern–Simons lesson and the ABJM definition. A local three-form is not a replacement for the spin Chern–Simons functional on a nontrivial bundle.

Opposite levels multiply complete multiplets

The Euclidean connection action is

$$S_{{\rm conn},E} =\frac{ik}{4\pi}\int [\omega_3(a)-\omega_3(\widehat a)]. \tag{AB02.11}$$

The relative minus sign is a defining choice of ABJM, not a consequence of bifundamental covariance. It is specified in AB, section 2.2 and in the invariant trace pairing of KWY, (4.3). The latter uses $k/(4\pi)$ times the difference of ordinary fundamental traces.

Supersymmetry completes each node with the partners derived in the flat Chern–Simons lesson. Keeping its Euclidean auxiliary and spinor conventions, the entire vector action is

$$\begin{aligned} S_{{\rm vec},E} ={}&S_{{\rm conn},E}\\ &+\frac{k}{4\pi}\int d^3x\,\operatorname{Tr}_N \left[-2\sigma D+2i\lambda\widetilde\lambda +2\widehat\sigma\widehat D -2i\widehat\lambda\widetilde{\widehat\lambda}\right]. \end{aligned} \tag{AB02.12}$$

Here $\lambda\widetilde\lambda=\lambda^{\mathsf T}C\widetilde\lambda$ contracts spinor indices with $C=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$; the trace contracts color indices. The two gauginos at each node are independent Euclidean variables. Every term has dimension three, and changing a level reverses all its partners. Reversing only the hatted connection term would spoil its relative supersymmetric coefficients.

For source comparison, KWY uses $\partial_\mu+i[A_\mu,\,\cdot\,]$ after (2.2). Setting $A_{\rm KWY}=-a$ maps its connection polynomial with $+2iA^3/3$ to (AB02.9). Our overall Euclidean factor and auxiliary continuation remain those of the flat Chern–Simons lesson; a connection substitution alone is not a dictionary for every source field.

A source equation, not a Maxwell wave equation

Define $\mathcal B^\mu=\epsilon^{\mu\nu\rho}f_{\nu\rho}/2$, with $\epsilon^{123}=+1$. Equations (AB02.10)–(AB02.11) give, including the complete matter currents,

$$\frac{ik}{2\pi}\mathcal B^\mu+J_L^\mu=0, \qquad -\frac{ik}{2\pi}\widehat{\mathcal B}^\mu+J_R^\mu=0. \tag{AB02.13}$$

The connection has first-order Chern–Simons dynamics; matter sources its curvature. There is no $\partial_\nu f^{\nu\mu}/g^2$ term. These are Euclidean stationary equations, which can require complex continued connections; they are not a claim that a nonzero real current balances an imaginary term on a real contour.

With no sources, the equation is locally $f=0$ and supplies no Maxwell photon. With matter, the curvature need not vanish. A negative level is therefore not a negative Maxwell energy: the connection term is a phase in the Euclidean weight, while (AB02.6) retains its positive scalar norm.

Worked limit: the two Abelian connections remain linked

For $N=1$, define

$$b=a-\widehat a,\qquad c=\frac{a+\widehat a}{2}.$$

Then $\mathcal D X=(d-ib)X$ and $\mathcal D Y=(d+ib)Y$. The scalar matter is neutral under $c$, but the connection action is

$$a\,da-\widehat a\,d\widehat a =c\,db+b\,dc =2b\,dc-d(c\,b), \qquad S_{{\rm conn},E}=\frac{ik}{2\pi}\int b\wedge dc \tag{AB02.14}$$

when the boundary term vanishes. This is often called a BF coupling: one one-form multiplies the curvature of another. Varying $c$ gives $db=0$ locally. Varying $b$ relates $dc$ to the matter current. Equivalently, (AB02.8) has $J_R=-J_L$ in the Abelian case, and the two equations (AB02.13) imply $da=d\widehat a$.

The neutral connection thus enforces a constraint; dropping it removes an equation. Also, $db=0$ is a local statement, not permission to discard compact holonomies. The combinations $b,c$ inherit correlated periods from the original compact product, as Exercise 3 checks. The vacuum quotient requires that global information and the matter potential in addition to this local calculation.

The useful checkpoint is now operational: given a matter matrix, place the connections on the correct sides, recover its curvature, and vary the two level terms with their full signs. These ingredients will support the ordered quiver interactions without changing the gauge theory along the way.

Exercises

1. Test a derivative with a common central shift

At a point, let a constant $U(2)\times U(2)$ matter matrix and one component of each connection be

$$X=\begin{pmatrix}1&i\\2&-1\end{pmatrix}, \qquad a_\mu=\operatorname{diag}(2,-1),\qquad \widehat a_\mu=\operatorname{diag}(3,1).$$

Use a common mass unit for the displayed connection entries. Compute $\mathcal D_\mu X$ and $\operatorname{Tr}[(\mathcal D_\mu X)^\dagger\mathcal D_\mu X]$. Does shifting both connections by $qI_2$ change the answer? Diagnose the proposed derivative $\partial_\mu X-ia_\mu X-iX\widehat a_\mu$.

Hint 1

Each entry has connection difference $(a_\mu)_{ii}-(\widehat a_\mu)_{jj}$.

Hint 2

The diagonal central transformation acts trivially on $X$. Its connection must disappear from this derivative.

Solution

The matrix of differences is $\left(\begin{smallmatrix}-1&1\\-4&-2\end{smallmatrix}\right)$. Multiplying each entry of $X$ by $-i$ times its difference gives

$$\mathcal D_\mu X= \begin{pmatrix}i&1\\8i&-2i\end{pmatrix}, \qquad \operatorname{Tr}[(\mathcal D_\mu X)^\dagger\mathcal D_\mu X] =1+1+64+4=70.$$

The common shift cancels in every difference. The proposed derivative instead uses sums, gives $\left(\begin{smallmatrix}-5i&3\\-4i&0\end{smallmatrix}\right)$ and norm $50$ at the displayed point, and acquires $-2iqX$ under the common shift. It assigns an unwanted diagonal central charge. Numerical agreement at some special matrix would not repair this covariance failure.

2. Detect the cubic coefficient without derivatives

For one $U(2)$ node take constant connections $a_1=u\tau_1/2$, $a_2=v\tau_2/2$, $a_3=w\tau_3/2$. Compute the coefficient of $dx^1\wedge dx^2\wedge dx^3$ in $\omega_3(a)$ and vary it with respect to $u$. Compare with the bulk term in (AB02.10). What changes for a hatted node with the same three matrices?

Hint 1

Use $\operatorname{Tr}(\tau_a\tau_b\tau_c)=2i\epsilon_{abc}$. All six exterior permutations contribute.

Hint 2

Although $da=0$, the curvature $f_{23}=-i[a_2,a_3]$ is nonzero.

Solution

The cubic trace has coefficient $\epsilon^{abc}\operatorname{Tr}(a_a a_b a_c)=3iuvw/2$. Consequently $\omega_3$ has coefficient $(-2i/3)(3iuvw/2)=uvw$, and its variation is $vw\,\delta u$.

Independently, $f_{23}=vw\tau_1/2$, so $2\operatorname{Tr}[(\delta u\,\tau_1/2)f_{23}] =vw\,\delta u$. The boundary derivative vanishes for these constant data. Omitting the cubic term would give zero action variation despite nonzero curvature.

For the hatted node the level-free polynomial is identical, but its contribution to $S_E$ is $-ikuvw/(4\pi)$ instead of $+ikuvw/(4\pi)$. Its auxiliary and gaugino partners reverse with it. Two identical connections cancel in the difference of connection actions; they would add for equal levels.

3. Preserve the compact periods in the Abelian change of variables

Let $U=e^{i\alpha}$ and $\widehat U=e^{i\widehat\alpha}$ with each angle of period $2\pi$. Define $\beta=\alpha-\widehat\alpha$ and $\gamma=(\alpha+\widehat\alpha)/2$, the parameters for $b,c$ in (AB02.14). Find their simultaneous lattice of identifications and the fluxes of $b,c$ when the original fluxes are $(1,0)$. Finally, replace the level pair $(k,-k)$ by $(k,k)$ and compare the local connection polynomial.

Hint 1

Shift the original angles by $(2\pi m,2\pi n)$ before changing variables.

Hint 2

The sum and difference of two integers have the same parity. Expand both the sum and the difference of the original Chern–Simons forms.

Solution

The simultaneous identifications are

$$(\beta,\gamma)\sim (\beta+2\pi(m-n),\,\gamma+\pi(m+n)), \qquad m,n\in\mathbb Z.$$

Thus writing the shifts as $(2\pi u,\pi v)$ requires $u\equiv v\pmod2$. A shift of $\beta$ by $2\pi$ accompanies an odd multiple of $\pi$ in $\gamma$; it is not an independent rectangular period lattice. The fluxes, in units of $2\pi$, are $n_b=n_L-n_R=1$ and $n_c=(n_L+n_R)/2=1/2$. This is an allowed original product-group configuration. Treating $c$ as a separately normalized compact $U(1)$ connection would incorrectly reject it.

For equal levels, direct expansion gives

$$a\,da+\widehat a\,d\widehat a =2c\,dc+\frac12b\,db.$$

This has two self-couplings rather than the cross-coupling (AB02.14). Matter still sees the same $b$ charges, but the connection equations have changed. Reversing a representation charge cannot convert this action into the opposite-level ABJM action.

Continue with Read the two-node ABJM quiver to turn the two-sided matter transformations into arrows, ordered matrix products and gauge-invariant operators.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading the original ABJM paper