Matrix-valued matter, representations, and traces

A field with two color indices can look like an ordinary matrix while carrying very different gauge charges. An adjoint transforms by conjugation. A bifundamental transforms with one gauge group on the left and another on the right. Taking a trace produces a gauge invariant only after those indices form a closed contraction.

This lesson requires ordinary gauge theory and matrix multiplication, with no supersymmetry prerequisite. The color algebra applies in both three and four spacetime dimensions and is independent of spacetime signature. We use classical bosonic matter fields for the main calculations; a final sign rule explains what changes for Grassmann-odd entries.

A representation tells us how indices transform

Let $V=\mathbb C^N$ be a color space. A fundamental field $q^i$ is a column, and a gauge transformation is a unitary matrix $U(x)$:

$$q'^i=U^i{}_j q^j,\qquad q'=Uq.$$

An anti-fundamental is a row $\widetilde q_i$ with $\widetilde q'=\widetilde q\,U^{-1}$. It can be a separate field; the notation does not require $\widetilde q=q^\dagger$. The contraction $\widetilde q_iq^i$ is invariant because the two group matrices cancel.

A representation specifies this linear action of the group on the components. A color index is unrelated to a spinor or spacetime index. For example, $\psi_\alpha{}^i{}_j$ has a spinor index $\alpha$ and two color indices; a color trace sums $i,j$, not $\alpha$. The index form of matter transformations is also the starting point of Bilal, equations (5.19)–(5.20).

An adjoint-valued field can be written

$$X=X^a t^a,\qquad X'=UXU^{-1}.$$

For $SU(N)$, choose traceless Hermitian fundamental generators with

$$[t^a,t^b]=if^{abc}t^c,\qquad \operatorname{Tr}_{N}(t^at^b)=\frac12\delta^{ab}.$$

Here $\operatorname{Tr}_{N}$ is the ordinary trace of an $N$-by-$N$ matrix. The components $X^a$ may be complex, so $X$ need not be Hermitian. There are $N^2-1$ complex components for a complex $SU(N)$ adjoint, or $N^2-1$ real components if a Hermiticity condition is imposed.

The adjoint representation has dimension $N^2-1$ even though we represent its elements by $N$-by-$N$ matrices. Its action is conjugation on that space of traceless matrices; it is not the fundamental action on one column.

Worked example: an adjoint of SU(2)

Use the Pauli matrices in their usual order,

$$t^1=\frac12\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad t^2=\frac12\begin{pmatrix}0&-i\\i&0\end{pmatrix},\quad t^3=\frac12\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$

Thus

$$X=\frac12 \begin{pmatrix} X^3&X^1-iX^2\\ X^1+iX^2&-X^3 \end{pmatrix},\qquad U_\theta=e^{i\theta t^3} =\begin{pmatrix}e^{i\theta/2}&0\\0&e^{-i\theta/2}\end{pmatrix}.$$

Multiplying on both sides gives

$$U_\theta XU_\theta^{-1} =\frac12 \begin{pmatrix} X^3&e^{i\theta}(X^1-iX^2)\\ e^{-i\theta}(X^1+iX^2)&-X^3 \end{pmatrix}.$$

Reading off the coefficients,

$$X'^1=\cos\theta\,X^1+\sin\theta\,X^2,\qquad X'^2=-\sin\theta\,X^1+\cos\theta\,X^2,\qquad X'^3=X^3.$$

The sign is fixed by our choice $U=e^{+i\theta t^3}$. As an independent small-angle check,

$$\delta X=i\theta[t^3,X],\qquad X=v t^1\ \Longrightarrow\ \delta X=-\theta v t^2.$$

At $\theta=\pi/2$, the same field becomes $X'=-vt^2$. Its components have changed, but its color contraction has not:

$$\operatorname{Tr}_2(X^\dagger X) =\frac12\sum_{a=1}^3|X^a|^2 =\operatorname{Tr}_2(X'^\dagger X').$$

For complex components, $\operatorname{Tr}_2(X^2)$ is also invariant but need not be real or positive. Gauge invariance and positivity are different properties.

Two meanings of normalization

Our generator normalization is a choice of basis for the Lie algebra. A normalized trace is a separate definition:

$$\operatorname{tr}_{N}M \equiv\frac1N\operatorname{Tr}_{N}M,\qquad \operatorname{tr}_{N}I_N=1.$$

For example,

$$\operatorname{Tr}_2(t^at^b)=\frac12\delta^{ab}, \qquad \operatorname{tr}_2(t^at^b)=\frac14\delta^{ab}.$$

Replacing $\operatorname{Tr}_{N}$ by $\operatorname{tr}_{N}$ inside an action changes its coefficient unless it is compensated: $c\,\operatorname{Tr}_{N}K=Nc\,\operatorname{tr}_{N}K$. For complex adjoint components with canonical kinetic term $\sum_a|\partial X^a|^2$, our basis gives the matrix expression $2\operatorname{Tr}_{N}(\partial X^\dagger\partial X)$.

Other references choose another quadratic norm. For example, Bilal (8.4) writes a unit generator norm. If the ordinary matrix trace is held fixed, a concrete conversion is

$$T^a=\sqrt2\,t^a,\qquad \operatorname{Tr}_{N}(T^aT^b)=\delta^{ab},\qquad X=X^at^a=\frac{X^a}{\sqrt2}T^a.$$

The matrix field stays the same. The structure constants also change: for these $SU(2)$ matrices, $[T^a,T^b]=i\sqrt2\,\epsilon^{abc}T^c$. Alternatively one can keep the basis fixed and rescale the trace convention. Specify which operation is being performed before comparing component coefficients.

A square matrix can contain a singlet

An unconstrained complex $N$-by-$N$ field $M$ transforming by conjugation splits as

$$M=X+\frac{\operatorname{Tr}_{N}M}{N}I_N,\qquad \operatorname{Tr}_{N}X=0.$$

The first part transforms in the $SU(N)$ adjoint. The identity part is invariant. Thus $N^2=(N^2-1)+1$ counts an adjoint plus a singlet, not an extra component of a traceless adjoint.

The two parts are orthogonal in the matrix norm:

$$\operatorname{Tr}_{N}(M^\dagger M) =\operatorname{Tr}_{N}(X^\dagger X) +\frac{|\operatorname{Tr}_{N}M|^2}{N}.$$

Both cross terms vanish because $X$ is traceless. For $M=\begin{pmatrix}1&i\\0&3\end{pmatrix}$, the singlet part is $2I_2$ and $X=\begin{pmatrix}-1&i\\0&1\end{pmatrix}$. Their squared norms are $8$ and $3$, adding to $11$.

A $U(N)$ adjoint includes this central component; an $SU(N)$ adjoint excludes it. Removing the trace therefore changes field content. In a theory with only commutator interactions the identity component commutes with everything, but that algebraic observation alone does not identify the full global gauge theories or their allowed line operators.

A bifundamental connects two color spaces

Take gauge group $U(M)\times U(N)$, with separate color spaces $V_L=\mathbb C^M$ and $V_R=\mathbb C^N$. Let $U$ act on $V_L$ and $V$ on $V_R$. A field

$$Y:V_R\longrightarrow V_L,\qquad Y^i{}_{\hat a},\qquad Y'=UYV^{-1}$$

is an $M$-by-$N$ matrix. The hatted index belongs to the right group. It transforms as a fundamental of the left group and an anti-fundamental of the right, denoted $(\mathbf M,\overline{\mathbf N})$. The rectangular case is a representation example; no specific interacting two-node theory is being assumed.

To see both charges explicitly, set $M=N=2$ and use

$$U=e^{i\theta t^3},\qquad V=e^{i\varphi t^3},\qquad Y=\begin{pmatrix}a&b\\c&d\end{pmatrix}.$$

The transformation is

$$Y'= \begin{pmatrix} e^{i(\theta-\varphi)/2}a&e^{i(\theta+\varphi)/2}b\\ e^{-i(\theta+\varphi)/2}c&e^{-i(\theta-\varphi)/2}d \end{pmatrix}.$$

Each entry carries a left weight and the negative of a right weight. In general $\operatorname{Tr}_2Y'$ differs from $\operatorname{Tr}_2Y$: the left and right spaces are independently transformed even when they have the same dimension.

A norm closes both indices correctly:

$$\begin{aligned} \operatorname{Tr}_{M}(YY^\dagger) &=Y^i{}_{\hat a}(Y^\dagger)^{\hat a}{}_i =\sum_{i,\hat a}|Y^i{}_{\hat a}|^2\\ &=\operatorname{Tr}_{N}(Y^\dagger Y). \end{aligned}$$

Indeed, $YY^\dagger\mapsto UYY^\dagger U^{-1}$ and $Y^\dagger Y\mapsto VY^\dagger YV^{-1}$. The ordinary traces agree although they are taken over different spaces. Their normalized traces are generally different when $M\ne N$.

Close the contraction in the correct order

An independent opposite-orientation field $Z:V_L\to V_R$ transforms as $Z'=VZU^{-1}$. It need not equal $Y^\dagger$. The compositions have distinct destinations:

Product	Acts on	Transformation
$YZ$	$V_L$	$YZ\mapsto U(YZ)U^{-1}$
$ZY$	$V_R$	$ZY\mapsto V(ZY)V^{-1}$
$Y_1Z_1Y_2Z_2$	$V_L$	Conjugation by $U$

For example, the complete contraction

$$\operatorname{Tr}_{M}(Y_1Z_1Y_2Z_2) =(Y_1)^i{}_{\hat a}(Z_1)^{\hat a}{}_j (Y_2)^j{}_{\hat b}(Z_2)^{\hat b}{}_i$$

is invariant: every $V^{-1}V$ and $U^{-1}U$ pair cancels, and the final $U,U^{-1}$ cancel under the trace. Matrix multiplication follows composable maps. Two $Y$ fields cannot in general be composed; when their dimensions happen to match, their product still has uncancelled gauge matrices.

This is the color structure of the alternating products of $A_i$ and $B_j$ in the ABJM paper, equations (2.4)–(2.6): $Y$ has the orientation of $A_i$, and $Z$ that of $B_j$. The $B_j$ are independent matter fields, not a condition $B_j=A_i^\dagger$. The standard equal-rank theory uses $U(N)_k\times U(N)_{-k}$; changing these groups to $SU(N)$ factors changes the theory.

Ordinary bosonic trace cyclicity permits moving a complete factor from one end to the other. It does not permit arbitrary reordering. For rectangular maps,

$$\operatorname{Tr}_{M}(YZ) =\sum_{i,\hat a}Y^i{}_{\hat a}Z^{\hat a}{}_i =\operatorname{Tr}_{N}(ZY).$$

The index sum proves the identity without pretending that $YZ$ and $ZY$ are the same matrix.

Differentiate a field with both connections

Use Hermitian connections with the gauge coupling included in their definition. Then $D_\mu q=\partial_\mu q-iA_\mu q$. Covariance fixes

$$A'_\mu=UA_\mu U^{-1}-i(\partial_\mu U)U^{-1}.$$

For the bifundamental, the derivative is

$$\boxed{ D_\mu Y=\partial_\mu Y-iA^L_\mu Y+iYA^R_\mu. }$$

The right connection has the opposite sign because the right index is anti-fundamental. To verify it, differentiate $UYV^{-1}$:

$$\partial_\mu(UYV^{-1}) =(\partial_\mu U)YV^{-1} +U(\partial_\mu Y)V^{-1} -UYV^{-1}(\partial_\mu V)V^{-1}.$$

The inhomogeneous term in $-iA'^L_\mu Y'$ cancels the first term. The inhomogeneous term in $+iY'A'^R_\mu$ cancels the last. The remainder is precisely $U(D_\mu Y)V^{-1}$. Therefore

$$\operatorname{Tr}_{N}\!\left[(D_\mu Y)^\dagger D^\mu Y\right]$$

is gauge invariant. In Euclidean signature its kinetic quadratic form is nonnegative. For a canonically normalized scalar in $d=3$ or $4$, $[Y]=(d-2)/2$, $[A^{L,R}]=1$ in this connection convention, and this kinetic term has dimension $d$.

Setting the two spaces and transformations equal gives the adjoint derivative

$$D_\mu X=\partial_\mu X-i[A_\mu,X].$$

Minahan’s equations (4.1)–(4.4) use an anti-Hermitian convention: his gauge parameter and connection correspond to $i\alpha$ and $iA_\mu$ here. With that translation, both the commutator derivative and the traced operator products agree.

For center transformations $U=e^{i\alpha}I_M$, $V=e^{i\beta}I_N$, the field has phase $Y'=e^{i(\alpha-\beta)}Y$. The common phase acts trivially on this matter field. It does not follow that the corresponding gauge field can be dropped from a full theory: Chern–Simons couplings and flux conditions can still involve it. The Abelian discussion in ABJM §2.3, (2.7)–(2.9) supplies a concrete example.

A sign when the entries are fermionic

The preceding cyclic identities used bosonic entries. For matrices $P,Q$ whose entries have definite Grassmann parities $|P|,|Q|$, the color trace instead obeys

$$\operatorname{Tr}(PQ) =\sum_{i,j}P^i{}_jQ^j{}_i =(-1)^{|P||Q|}\operatorname{Tr}(QP).$$

This is the sign from interchanging the entries, not a new color-group rule. Two odd matrices acquire a minus sign. Spinor indices still require their own contractions. Thus a color trace alone never licenses exchanging two fermions without checking parity.

Practice

1. Cyclic order is not arbitrary order

For $X=t^1$, $Y=t^2$, calculate $\operatorname{Tr}_2(XYXY)$ and $\operatorname{Tr}_2(X^2Y^2)$, then their normalized traces. Explain why their difference is consistent with trace cyclicity.

Hint 1

Use $(t^a)^2=I_2/4$ and $t^1t^2=it^3/2$.

Hint 2

The cyclic rotations of the word $XYXY$ alternate the two letters; none is $XXYY$.

Complete solution

The products are

$$XYXY=-\frac1{16}I_2,\qquad X^2Y^2=\frac1{16}I_2.$$

Their ordinary traces are $-1/8$ and $+1/8$. Dividing each by the color-space dimension gives normalized traces $-1/16$ and $+1/16$. Cyclicity equates $\operatorname{Tr}(XYXY)$ with $\operatorname{Tr}(YXYX)$, but it never exchanges only the middle two factors. This difference is a direct check that commuting scalar coefficients do not make their color matrices commute.

2. Trace over the right space

For the $U(2)\times U(3)$ bifundamental

$$Y=\begin{pmatrix}1&0&2i\\0&3&0\end{pmatrix},$$

calculate $YY^\dagger$, $Y^\dagger Y$, both ordinary traces, and both normalized traces. Which are gauge invariant? Is $\operatorname{Tr}(Y)$ defined?

Hint 1

The two products are respectively two-by-two and three-by-three.

Hint 2

Their ordinary traces both equal the sum of the squared absolute values of all entries of $Y$.

Complete solution

Direct multiplication gives

$$YY^\dagger=\begin{pmatrix}5&0\\0&9\end{pmatrix},\qquad Y^\dagger Y= \begin{pmatrix}1&0&2i\\0&9&0\\-2i&0&4\end{pmatrix}.$$

Both ordinary traces are $14$. The normalized traces are

$$\operatorname{tr}_{2}(YY^\dagger)=7,\qquad \operatorname{tr}_{3}(Y^\dagger Y)=\frac{14}{3}.$$

Each of these four scalars is invariant under the two gauge groups. Gauge invariance does not force the two normalization choices to agree. The matrix $Y$ itself is rectangular, so it is not an endomorphism of either space and has no ordinary matrix trace.

3. A determinant tests the gauge group

Let $Y$ be a square bifundamental of $U(2)\times U(2)$ and $Z$ an independent opposite-orientation field. Under $U=e^{i\alpha}I_2$, $V=e^{i\beta}I_2$, find the transformations of $\operatorname{Tr}Y$, $\det Y$, and $\operatorname{Tr}(YZ)$. What changes for $\det Y$ if the groups are instead $SU(2)\times SU(2)$?

Hint 1

$Y$ and $Z$ carry opposite relative phases.

Hint 2

For a general transformation, use $\det(UYV^{-1})=(\det U)(\det Y)/(\det V)$.

Complete solution

For the center transformations,

$$\operatorname{Tr}Y\mapsto e^{i(\alpha-\beta)}\operatorname{Tr}Y,\qquad \det Y\mapsto e^{2i(\alpha-\beta)}\det Y.$$

Neither is generally invariant under $U(2)\times U(2)$. In contrast $YZ$ transforms by conjugation, so $\operatorname{Tr}(YZ)$ is invariant under the entire group, not just its center.

For $SU(2)\times SU(2)$, both group determinants equal one. Thus $\det Y$ is invariant under every gauge transformation of that different group. The trace of a single bifundamental still has no corresponding general invariance. Changing the group has changed the available invariant contractions; it is not just a new notation for the same theory. The calculation does not determine any additional global quotient or flux prescription.

Carry the indices into the next theory

Before taking a trace, identify its vector space, follow the orientation of every map, and state whether the trace is normalized. Before comparing an action coefficient, also fix the generator basis and the location of the gauge coupling.

These rules prepare the independent four-dimensional action lessons and the three-dimensional ABJM lessons. Their supersymmetry and interactions will be specified there; the matrix manipulations here require neither a supersymmetric multiplet nor a large-color approximation.

Continue with matrix integrals, roots, weights, and eigenvalues to diagonalize a Hermitian matrix inside an integral. The first checkpoint is to recover the same absolute Gaussian normalization before and after diagonalization.

Use these representations in What is ABJM, and what can we calculate? to specify the opposite-level gauge factors, four matter matrices and their auxiliary constraints.