Commuting matrices and the Coulomb branch

A theory with a vanishing beta function can have vacua containing a mass scale. The scale can be a scalar expectation value rather than a running coupling. We will solve for those vacua and identify precisely where their low-energy Abelian description fails.

Continue from couplings and conformality. Use the same four-dimensional, flat Lorentzian $SU(N)$ theory, $N\geq2$, with sixteen real Poincaré supercharges, no masses and fixed $g>0$. An $\mathcal N=1$ presentation makes four supercharges manifest. The signature is $(+---)$, and the ordinary fundamental trace satisfies $\operatorname{Tr}_N(t^at^b)=\delta^{ab}/2$. We analyze constant vacuum representatives on uncompactified space, with no field strength, fermion background or prescribed holonomy. The explicit spectrum and metric below are classical calculations.

Solve the zero-energy conditions

Recall the derived scalar action:

$$\begin{aligned} \mathcal L_{\mathrm{scalar}} &=\sum_{I=1}^6\operatorname{Tr}_N (\mathscr D_\mu X^I\mathscr D^\mu X^I)-U,\\ U&=g^2\sum_{I<J}\operatorname{Tr}_N \bigl([X^I,X^J]^\dagger[X^I,X^J]\bigr),\\ \mathscr D_\mu X^I&=\partial_\mu X^I-ig[A_\mu,X^I]. \end{aligned} \tag{N05.1}$$

Each $X^I$ is Hermitian and traceless, with mass dimension one. For any matrix $C$, $\operatorname{Tr}(C^\dagger C)=\sum_{a,b}|C_{ab}|^2$. Thus every term in $U$ is nonnegative, and

$$U=0\quad\Longleftrightarrow\quad [X^I,X^J]=0\quad\text{for every }I,J. \tag{N05.2}$$

This uses the physical Hermitian slice. Treating six unrelated complex matrices as if the same positivity argument applied would change the problem.

The F and D auxiliaries vanish on these configurations. All sixteen Poincaré supercharges are preserved: the constant commuting conditions are invariant under the scalar R rotations, and applying the manifest supersymmetry test to any R-rotated configuration again gives zero. The rotated manifest supercharges span the full set. R rotations can nevertheless move us to a different vacuum; preserving supersymmetry does not require preserving the full R symmetry.

The matrices are gauge-dependent representatives. Physical vacuum data are their common conjugacy classes, or gauge-invariant functions such as traces of products. This is the adjoint transformation law used in Minahan, (4.1)–(4.4), not an assertion that a gauge-variant expectation value is an observable.

Diagonalize all six matrices together

Diagonalize $X^1$ by a unitary matrix. If $v$ lies in its eigenvalue-$\lambda$ subspace, commutativity gives $X^1(X^Jv)=X^JX^1v=\lambda X^Jv$. Every other $X^J$ therefore preserves that subspace. Diagonalize $X^2$ inside each degenerate block, then continue with the remaining Hermitian matrices. This constructs a common orthonormal eigenbasis.

The diagonalizing matrix can be chosen in $SU(N)$: multiplying a unitary matrix by a scalar phase fixes its determinant without changing its conjugation action. We obtain

$$X^I_0=\operatorname{diag}(x_1^I,\ldots,x_N^I),\qquad \mathbf x_a=(x_a^1,\ldots,x_a^6)\in\mathbb R^6,\qquad \sum_{a=1}^N\mathbf x_a=0. \tag{N05.3}$$

A color label $a$ now identifies a joint eigenvalue, one six-vector. There is no physical ordering of these labels. Any permutation is implemented by an $SU(N)$ conjugation: correct a permutation matrix’s determinant by a diagonal phase, which commutes with the diagonal scalars. The residual identification is one common permutation group $S_N$, the Weyl group:

$$\boxed{\mathcal M_{\mathrm{cl},SU(N)} =\left\{(\mathbf x_1,\ldots,\mathbf x_N)\in(\mathbb R^6)^N: \sum_a\mathbf x_a=0\right\}/S_N.} \tag{N05.4}$$

Its generic real dimension is $6(N-1)$. Dividing by a finite group makes identifications; it does not subtract continuous dimensions. We must not sort the eigenvalues of each $X^I$ independently, because that loses which entries belong to the same joint eigenvector.

Nor do we divide by the global $SO(6)$ scalar rotations. They relate physically distinct vacuum orientations. For example, $\operatorname{Tr}(X^IX^J)=\sum_a x_a^Ix_a^J$ is gauge invariant and retains information about those orientations.

This whole six-scalar family is called the $\mathcal N=4$ Coulomb branch: generic vacua retain Abelian gauge fields. A fixed $\mathcal N=2$ decomposition singles out only two real scalars as its vector scalar; the other four lie in the adjoint hyper. Its narrower vector-only Coulomb slice and its mixed-branch terminology should not be confused with the full family here. Tachikawa, §2.2, explains that distinction for an $\mathcal N=2$ presentation.

A degenerate matrix need not mean enhancement

Consider the concrete $SU(3)$ vacuum, with real parameters $v,w$,

$$X^1_0=v \begin{pmatrix}1&0&0\\0&1&0\\0&0&-2\end{pmatrix}, \qquad X^2_0=w \begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix}, \qquad X^{3,\ldots,6}_0=0. \tag{N05.5}$$

The first matrix has a two-dimensional degenerate block, inside which the second acts. They commute. An explicit common diagonalizer is

$$S=\begin{pmatrix} 1/\sqrt2&1/\sqrt2&0\\ 1/\sqrt2&-1/\sqrt2&0\\ 0&0&-1 \end{pmatrix},\qquad \det S=1.$$

Conjugating by $S$ leaves $X^1_0$ unchanged and sends $X^2_0$ to $\operatorname{diag}(w,-w,0)$. The joint vectors, with four suppressed zero components, are

$$\mathbf x_1=(v,w),\qquad \mathbf x_2=(v,-w),\qquad \mathbf x_3=(-2v,0). \tag{N05.6}$$

For $w\ne0$ all three are distinct, even when $v=0$. The unbroken group is then the determinant-one diagonal torus $U(1)^2$. The degeneracy of $X^1_0$ alone did not restore a non-Abelian group.

More generally, a matrix element $G_{ab}$ of an unbroken gauge transformation can be nonzero only if $\mathbf x_a=\mathbf x_b$: commuting with every diagonal $X^I_0$ gives $(x_a^I-x_b^I)G_{ab}=0$ for every $I$. If the distinct joint vectors occur with multiplicities $n_1,\ldots,n_k$, the exact centralizer in our chosen gauge group is

$$G_{\mathrm{unbroken}} =S\bigl(U(n_1)\times\cdots\times U(n_k)\bigr). \tag{N05.7}$$

Here $S$ means that the product of block determinants is one. Its Lie algebra is $\bigoplus_j\mathfrak{su}(n_j)\oplus\mathfrak u(1)^{k-1}$; that algebraic decomposition need not be a direct-product identity for the global group. In (N05.6), $w=0$, $v\ne0$ gives $S(U(2)\times U(1))$, while $v=w=0$ restores $SU(3)$.

Read the separation as a physical mass

For one color pair $a<b$, define $\mathbf d=\mathbf x_a-\mathbf x_b$. Let $E_{ab}$ have a single unit entry, and use the two Hermitian generators $T_1=(E_{ab}+E_{ba})/2$ and $T_2=(E_{ab}-E_{ba})/(2i)$, each with trace norm $1/2$. The identity $[X^I_0,E_{ab}]=d^I E_{ab}$ turns the scalar kinetic term into

$$\mathcal L_{\mathrm{vector\ mass}} =\frac{g^2|\mathbf d|^2}{2} \bigl(A_\mu^1A^{1\mu}+A_\mu^2A^{2\mu}\bigr), \qquad M_{ab}=g|\mathbf x_a-\mathbf x_b|. \tag{N05.8}$$

The canonical gauge kinetic term is $-\sum_c(F_{\mu\nu}^c)^2/4$, so comparison with a Proca coefficient $M^2/2$ fixes the mass without an extra $\sqrt2$. Both sides have dimension one. On a single-complex-scalar slice, the source normalization $\Phi_{\mathrm T}=g(X^1+iX^2)$ makes this exactly the eigenvalue-difference mass in Tachikawa, (2.2.8)–(2.2.12).

The scalar spectrum checks the coefficient independently. Write an off-diagonal fluctuation as $Y^I=y_1^IT_1+y_2^IT_2$. The quadratic part of (N05.1) is

$$\begin{aligned} U^{(2)}_{ab} &=\frac{g^2}{2}\sum_{c=1}^2\sum_{I<J} (d^Iy_c^J-d^Jy_c^I)^2\\ &=\frac12\sum_{c=1}^2\mathbf y_c^T \,g^2\bigl(|\mathbf d|^2\mathbf1_6-\mathbf d\mathbf d^T\bigr) \mathbf y_c . \end{aligned} \tag{N05.9}$$

The second line follows by expanding the squares. For $\mathbf d\ne0$, five scalar directions in each real generator have mass $M_{ab}$; the direction along $\mathbf d$ has zero potential. It is a gauge-orbit direction, since an infinitesimal conjugation gives $Y^I=i[\Lambda,X^I_0]$. Each of the two massive vector fields eats one such Goldstone direction. The pair has six massive-vector polarizations and ten physical scalar modes, sixteen bosonic on-shell degrees, as required by the original two gauge-generator multiplets. The fermionic partners have the matching spectrum by the unbroken supersymmetry.

For (N05.6) the masses are

$$M_{12}=2g|w|,\qquad M_{13}=M_{23}=g\sqrt{9v^2+w^2}. \tag{N05.10}$$

A collision of all six joint coordinates makes the corresponding fields massless. The quotient also has a fixed point under their permutation. These are the same enhancement loci seen from the spectrum and from the vacuum geometry. At a simple two-vector collision, the relative six-vector is identified with its negative, giving a local $\mathbb R^6/\mathbb Z_2$ singularity transverse to the remaining directions.

The origin, the low-energy theory, and the free center

Away from collisions, energies $E\ll\min_{a<b}M_{ab}$ probe $N-1$ massless Abelian $\mathcal N=4$ multiplets. A weak-coupling expansion makes integrating out the off-diagonal fields controlled. It fails near a collision unless those additional light fields are retained. A singular Abelian description does not mean that the original matrix theory has ceased to exist.

Slowly varying diagonal fields give the classical kinetic metric

$$\mathcal L_{\mathrm{moduli}} =\sum_a\partial_\mu\mathbf x_a\cdot\partial^\mu\mathbf x_a, \qquad ds_{\mathrm{cl}}^2=2\sum_a d\mathbf x_a\cdot d\mathbf x_a, \quad \sum_a d\mathbf x_a=0, \tag{N05.11}$$

before the common permutation quotient. The factor two uses the convention $\mathcal L=\frac12G_{AB}\partial q^A\partial q^B$. The commuting-matrix argument establishes the classical vacuum space and this tree-level metric. It is not a proof of quantum nonrenormalization, nor a claim that the effective action has no higher-derivative interactions.

At the origin, all $\mathbf x_a=0$, the full $SU(N)$ symmetry is restored and the conformal symmetry discussed in the prerequisite is unbroken. A nonzero vacuum value introduces physical masses (N05.8) and spontaneously breaks scale invariance while leaving the coupling’s beta function zero. A dilation moves to another point of the branch. R rotations can also move the vacuum; sixteen preserved Poincaré supercharges should not be confused with unbroken superconformal symmetry.

For $U(N)$, drop the zero-sum constraint. The scalar vacuum space becomes $(\mathbb R^6)^N/S_N$. Write $\mathbf x_a=\mathbf x_{\mathrm{cm}}+\mathbf y_a$ with $\sum_a\mathbf y_a=0$. All commutator interactions and masses depend only on the $\mathbf y_a$. The center contributes $N(\partial\mathbf x_{\mathrm{cm}})^2$, or six canonical free scalars $\mathbf X^0=\sqrt{2N}\,\mathbf x_{\mathrm{cm}}$, alongside the free central gauge and fermion fields. The vacuum space therefore factors into the center $\mathbb R^6$ and the traceless quotient. The global gauge theory still has $U(N)=(SU(N)\times U(1))/\mathbb Z_N$; factorization of these local scalar data does not discard its bundle and line-operator identifications.

Exercises

1. Resolve the rank-one quotient

In $SU(2)$, write a commuting representative as $X^I_0=v^I t^3$, where $t^3=\operatorname{diag}(1,-1)/2$. Determine the residual identification, metric and vector mass. Show that the gauge-invariant matrix $Q^{IJ}=\operatorname{Tr}(X^IX^J)$ determines the vacuum, including the origin. Describe its constraints.

Hint 1

The single nontrivial Weyl permutation exchanges the two joint eigenvectors simultaneously.

Hint 2

For nonzero $\mathbf v$, the outer product $\mathbf v\mathbf v^T$ determines its direction up to a common sign. Use polar coordinates $\mathbf v=\rho\mathbf n$.

Solution

The joint vectors are $\pm\mathbf v/2$, so the identification is $\mathbf v\sim-\mathbf v$ and the vacuum space is $\mathbb R^6/\mathbb Z_2$. Equations (N05.8) and (N05.11) give

$$M_W=g|\mathbf v|,\qquad ds^2=d\mathbf v^2=d\rho^2+\rho^2d\Omega_5^2, \qquad \mathbf n\sim-\mathbf n.$$

At fixed nonzero radius the angular space is $\mathbb{RP}^5$, the sphere with antipodal points identified. Only the origin is a fixed point of the sign action. It has the restored $SU(2)$ group, whereas a nonzero point has unbroken $U(1)$.

Here $Q^{IJ}=v^Iv^J/2$. It is a real symmetric positive-semidefinite matrix of rank at most one, so all its two-by-two minors vanish. Conversely any such nonzero matrix has one positive eigenvalue $\lambda$ and unit eigenvector $\mathbf n$, yielding $\mathbf v=\pm\sqrt{2\lambda}\,\mathbf n$. The two signs are gauge equivalent. The zero matrix gives the origin. Thus these invariant data recover exactly the quotient; replacing the entire matrix $Q$ by its trace would incorrectly identify distinct R-symmetry orientations.

2. Count the fields at a collision

Take an $SU(4)$ vacuum with joint vectors $(\mathbf p,\mathbf p,\mathbf q,-2\mathbf p-\mathbf q)$, where the three displayed distinct values really are different. Find its unbroken group and the nonzero masses, including their color-pair multiplicities. Why does counting every massless scalar in the quadratic action overcount the dimension of the vacuum branch?

Hint 1

There are six unordered color pairs. One is inside the repeated block.

Hint 2

The restored non-Abelian block has scalar interactions even though its quadratic scalar masses vanish. Split the two coincident six-vectors to find the local vacuum coordinates.

Solution

The exact group is $S(U(2)\times U(1)\times U(1))$, with algebra $\mathfrak{su}(2)\oplus\mathfrak u(1)^2$ of dimension five. The five massive unordered pairs have

Pairs	Mass	Multiplicity
$(1,3),(2,3)$	$g\lvert\mathbf p-\mathbf q\rvert$	2
$(1,4),(2,4)$	$g\lvert3\mathbf p+\mathbf q\rvert$	2
$(3,4)$	$2g\lvert\mathbf p+\mathbf q\rvert$	1

They account for ten broken real generators; $15-10=5$ remain. The five unbroken generators have thirty real scalar components with zero quadratic mass. But the three $SU(2)$ generators still have the quartic commutator potential. Arbitrary motion in those eighteen components is not motion through vacua.

Locally write the first two joint vectors as $\mathbf p+\mathbf d/2$ and $\mathbf p-\mathbf d/2$. The center and the two remaining distinct vectors supply twelve independent real coordinates after the zero-sum condition; the relative vector supplies six, with $\mathbf d\sim-\mathbf d$. The local space is a smooth twelve-dimensional factor times $\mathbb R^6/\mathbb Z_2$, of dimension eighteen, equal to $6(4-1)$. Massless fields at a singular vacuum are not all independent flat directions.

3. Test whether separate spectra suffice

Let $v,w$ be nonzero real numbers. Compare the $SU(3)$ vacua

$$\begin{array}{c|cc} &X^1&X^2\\ \hline \mathrm A&\operatorname{diag}(-v,0,v)&\operatorname{diag}(-w,0,w)\\ \mathrm B&\operatorname{diag}(-v,0,v)&\operatorname{diag}(0,w,-w) \end{array}$$

with all other scalars zero. Each separate scalar matrix has the same eigenvalues in the two configurations. Are the vacua gauge equivalent? Compute a mixed invariant and their three vector masses.

Hint 1

Because $X^1$ has distinct eigenvalues, a gauge transformation preserving this ordered matrix is diagonal.

Hint 2

Evaluate $\operatorname{Tr}(X^1X^2)$ and then use distances between joint pairs, not between separately sorted lists.

Solution

The mixed invariant is $2vw$ in A and $-vw$ in B. They differ for the stipulated nonzero parameters, so the vacua are not gauge equivalent. Equivalently, no diagonal transformation preserving $X^1$ changes $X^2$.

In A the joint vectors are $(-v,-w),(0,0),(v,w)$, giving masses $g\sqrt{v^2+w^2}$ twice and $2g\sqrt{v^2+w^2}$ once. In B they are $(-v,0),(0,w),(v,-w)$, giving

$$g\sqrt{v^2+w^2},\qquad g\sqrt{4v^2+w^2},\qquad g\sqrt{v^2+4w^2}.$$

All three are positive, so both vacua have the same generic unbroken torus, while their gauge-invariant expectation values and spectra differ. The common Weyl quotient retains exactly the pairing information that six independent quotients would lose.

The vacuum is now specified by unordered joint eigenvalue vectors, and their separations determine which fields can be omitted at low energy. Gauge-invariant operators probe these vacua. Next, distinguish protected and unprotected operators at the conformal origin, where representation theory constrains their scaling dimensions.