Many branes and global gauge-group data

The one-brane calculation gave four complex coordinates with a common discrete phase. With several branes, those coordinates begin as matrices. We must explain both why diagonal entries describe a vacuum sector and why the remaining permutation moves an entire four-component tuple. Dividing each matrix’s eigenvalues separately would give the wrong space.

Keep flat three-dimensional compact $U(N)_k\times U(N)_{-k}$, equal $N\geq1$, integer $k>0$, ordinary fundamental traces, canonical matter normalization, and no mass or FI terms. The two factors each have rank $N$. There is no further quotient of their common center. The generic full theory has $\mathcal N=6$, meaning twelve real Poincaré supercharges; the chosen $\mathcal N=2$ notation displays four. A larger gauge rank does not multiply this number. Exceptional-level supersymmetry enhancement is a separate question.

The only matrix input is the two-sided bifundamental transformation law. We first derive a controlled classical vacuum branch. Its quotient geometry will not, by itself, prove an all-orders effective metric or classify every global topological sector of the quantum theory.

A matrix vacuum needs more than F-flatness

The independent chiral scalars are $A_i:V_R\to V_L$ and $B_i:V_L\to V_R$, $i=1,2$. On the physical bosonic slice, define

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

The already derived auxiliary equations give Hermitian matrices $\sigma=t\mu_L$ and $\widehat\sigma=-t\mu_R$. Positivity of the complete scalar potential requires the F gradients to vanish and also

$$\sigma A_i=A_i\widehat\sigma,\qquad \widehat\sigma B_i=B_i\sigma. \tag{AB08.1}$$

These are conditions that the matter maps intertwine the two vector scalars. They are not $\mu_L=\mu_R=0$. In particular, the trace parts have not been removed. The positive-square form and signs follow from AB Appendix A.1, (A.1)–(A.4).

Consider the open sector in which $A_1$ is invertible and $\sigma$ has $N$ distinct eigenvalues. Invertibility and the first equation imply

$$\widehat\sigma=A_1^{-1}\sigma A_1. \tag{AB08.2}$$

Thus the two Hermitian matrices have the same eigenvalues. Diagonalize each by its own unitary gauge transformation, then order the right eigenvalues to match the left: $\sigma=\widehat\sigma=\operatorname{diag}(s_1,\ldots,s_N)$. Equation (AB08.1) becomes, entry by entry,

$$(s_a-s_b)(A_i)_{ab}=0,\qquad (s_a-s_b)(B_i)_{ab}=0. \tag{AB08.3}$$

Since $s_a\ne s_b$ for $a\ne b$, all four matrices are diagonal in these paired bases. We have derived this conclusion for a specified sector, rather than assumed that four arbitrary complex matrices can be diagonalized together.

Conversely, every diagonal choice solves all vacuum equations. On diagonal entries the cubic F gradients are differences of products of complex numbers, so they vanish. Moreover

$$s_a=t\left(|a_{1a}|^2+|a_{2a}|^2-|b_{1a}|^2-|b_{2a}|^2\right)$$

is the same on both nodes. There is no constraint setting it to zero. The diagonal family extends beyond the invertible, distinct-$s_a$ patch, even though the proof using (AB08.3) then stops working. Equality of two $s_a$ is not enough to establish either equality of the four-component tuples or a singularity of their quotient.

For a concrete transverse check, take $N=2$, $A_1=\operatorname{diag}(v,w)$ with real positive $v\ne w$, $B_1=B_2=0$, and perturb $A_2=\varepsilon E_{12}$. Here $E_{12}$ has a single unit entry. The F gradients vanish because they contain a $B$, while

$$\mu_L=\operatorname{diag}(v^2+|\varepsilon|^2,w^2),\qquad \mu_R=-\operatorname{diag}(v^2,w^2+|\varepsilon|^2).$$

The two nonzero cubic expressions in the vector potential are therefore

$$\mu_LA_1+A_1\mu_R =|\varepsilon|^2\operatorname{diag}(v,-w),\qquad \mu_LA_2+A_2\mu_R=(v^2-w^2)\varepsilon E_{12}.$$

Taking their squared norms, with the common factor $t^2$, gives

$$V=t^2\left[ |\varepsilon|^2(v^2-w^2)^2 +|\varepsilon|^4(v^2+w^2) \right]. \tag{AB08.4}$$

Thus this canonically normalized transverse scalar has $m_\perp^2=t^2(v^2-w^2)^2$. It is massive on the chosen patch. Its mass has dimension one because $[v]=[w]=1/2$ and $t$ is dimensionless. A zero quadratic coefficient at $v=w$ would not make the whole perturbation flat: the quartic term remains. Nor does this one fluctuation compute the full gauge and fermion spectrum.

Each matched diagonal entry retains a compact pair

Package the diagonal scalar entries as

$$Y^I=\operatorname{diag}(y_1^I,\ldots,y_N^I),\qquad y_a=(a_{1a},a_{2a},b_{1a}^*,b_{2a}^*)\in\mathbb C^4. \tag{AB08.5}$$

The conjugated $B$ entries have the same left-to-right color transformation as the $A$ entries. These are scalar coordinates, not four chiral superfields of the manifest subalgebra.

Write the diagonal connections as $a=\operatorname{diag}(a_1,\ldots,a_N)$ and $\widehat a=\operatorname{diag}(\widehat a_1,\ldots,\widehat a_N)$. The ordinary trace makes the restricted Chern–Simons action a sum of $N$ terms with levels $k,-k$, not $Nk,-Nk$:

$$S_{\rm top}=\frac{ik}{4\pi} \sum_{a=1}^N\int(a_a\wedge da_a-\widehat a_a\wedge d\widehat a_a). \tag{AB08.6}$$

For the stated global $U(N)$ groups, diagonal flux representatives have $n_a,\widehat n_a\in\mathbb Z$, modulo permutations. The $a$th component of a fundamental Wilson probe has charge one under the $a$th Cartan circle. There is no condition $\sum_a n_a=0$. A local labeling of the eigenvalues fixes these Cartan labels; a different labeling permutes the entire flux vectors too.

For each pair use the integer basis $b_a=a_a-\widehat a_a$, $c_a=\widehat a_a$. The one-brane first-order calculation applies separately:

$$S_{\rm first}=\sum_a\left[ \frac{ik}{4\pi}\int b_a\wedge db_a +\frac{ik}{2\pi}\int b_a\wedge f_a -\frac{i}{2\pi}\int\tau_a\,df_a\right], \qquad \tau_a\sim\tau_a+2\pi. \tag{AB08.7}$$

Here $f_a$ replaces $dc_a$, with unit flux $2\pi$. Its equation gives $kb_a+d\tau_a=0$. Under the relative gauge angle $\alpha_a$, $y_a\mapsto e^{i\alpha_a}y_a$ and $\tau_a\mapsto\tau_a-k\alpha_a$. On a local lift,

$$X_a=e^{i\tau_a/k}y_a,\qquad X_a\sim\omega X_a,\qquad \omega=e^{2\pi i/k}. \tag{AB08.8}$$

The phases are independent for the $N$ pairs. The continuous gauge angle removes the dual angle rather than removing one more scalar coordinate: $8+1-1=8$ real coordinates per pair. The residual common-node circles act trivially on these coordinates but still participate in the topological gauge theory.

Permute whole tuples, then take the quotient

A simultaneous permutation matrix $P$ on both nodes sends $Y^I\mapsto P Y^I P^{-1}$ for every $I$. It reorders the $N$ tuples. An independent permutation on only one node generally takes a generic paired diagonal configuration out of diagonal form. After restoring that form, it does not supply a second independent permutation of the tuples.

The finite group therefore acts by

$$X_a^I\longmapsto \omega^{n_a}X_{\pi(a)}^I, \qquad n_a\in\mathbb Z_k,\quad \pi\in S_N. \tag{AB08.9}$$

$S_N$ is the group of permutations of $N$ labels. Permutations also reorder the phase assignments, so the group is $(\mathbb Z_k)^N\rtimes S_N$, a semidirect product, with $k^N N!$ elements. For example, exchanging labels 1 and 2 conjugates the phase rotation of tuple 1 into that of tuple 2.

The diagonal vacuum branch is

$$\boxed{ \mathcal M_{\rm diag} =\frac{(\mathbb C^4)^N}{(\mathbb Z_k)^N\rtimes S_N} =\operatorname{Sym}^N(\mathbb C^4/\mathbb Z_k). } \tag{AB08.10}$$

The notation $\operatorname{Sym}^N$ means an unordered collection of $N$ points, allowing coincidences. Its complex dimension is $4N$, or real dimension $8N$. A finite identification changes neither dimension. This is the standard many-brane quotient reported in AB section 2.3, the paragraph after (2.9). Our explicit argument proves the generic diagonal sector and its closure; it does not infer a classification of arbitrary non-diagonal degenerate solutions merely from a generic mass check.

The canonical kinetic term restricts to $\sum_a\sum_I|D y_a^I|^2$. Equation (AB08.8) turns it locally into

$$ds^2=\sum_{a,I}dX_a^I\,d\overline X_a^I. \tag{AB08.11}$$

Both discrete phases and permutations are isometries. Hence this is a locally flat classical quotient metric away from fixed points. It is not a claim that every higher-derivative interaction vanishes, or a proof of quantum nonrenormalization. The original paper explicitly begins its moduli-space discussion by separating its classical analysis from the expectation concerning quantum corrections.

Worked example: two triples of phases and one exchange

For $N=2,k=3$, let $X_1=(1,0,0,0)$ and $X_2=(0,2,0,0)$ in fixed scalar units. There are $3^2\cdot2=18$ distinct representatives in this orbit. The configuration $(X'_1,X'_2)=(\omega X_2,\omega^2X_1)$ represents the same vacuum. The common phase of each tuple rotates all four components together.

Now try to exchange only the second components. The resulting pair is $(1,2,0,0)$ and $(0,0,0,0)$, which is not equivalent. The invariant

$$\mathcal I =\operatorname{Tr}_L(Y^1Y^{1\dagger}Y^2Y^{2\dagger}) =\sum_a|X_a^1|^2|X_a^2|^2 \tag{AB08.12}$$

is zero for the original pair and four for the altered one. This is a fully contracted color invariant, not just a function of separately unordered component lists.

When two nonzero points in $\mathbb C^4/\mathbb Z_k$ coincide, choose nearby lifts with the same phase. Set $X_{\rm c}=(X_1+X_2)/\sqrt2$ and $X_{\rm r}=(X_1-X_2)/\sqrt2$. Exchange fixes $X_{\rm c}$ and reverses $X_{\rm r}$. Locally the collision therefore has the geometry $\mathbb C^4\times(\mathbb C^4/\mathbb Z_2)$. At an individual origin there is also a $\mathbb Z_k$ stabilizer for $k>1$. Thus permutation collisions remain singular even at $k=1$, when the individual one-brane space is smooth. The generic Abelian description should not be used to integrate out modes all the way through a collision.

The Lie algebra does not fix the global answer

Locally, a $U(N)$ connection can be split into a traceless part and a central part $a_0 I_N$. Its central CS coefficient is $Nk$, since $\operatorname{Tr}I_N=N$. This does not mean that $a_0$ is an independent unit-flux $U(1)$ connection alongside an independent $SU(N)$ connection. Globally,

$$U(N)=\frac{SU(N)\times U(1)}{\mathbb Z_N}. \tag{AB08.13}$$

For example a valid $U(2)$ Cartan flux $(1,0)$ has central flux $(1+0)/2=1/2$ and traceless flux $(1/2,-1/2)$. Around the transition circle, each proposed separate factor ends at minus the identity, while their product closes. Removing that common identification would exclude an allowed $U(2)$ bundle. Replacing the groups by $SU(N)$ instead imposes traceless flux and determinant constraints. The warning in AB section 2.5 is precisely that the local splitting does not preserve the global flux problem.

Here is a concrete alternative, derived directly rather than asserted to be the standard theory. At rank one replace the gauge group by $(U(1)\times U(1))/\mathbb Z_p^{\rm diag}$, where the divided subgroup is generated by $(e^{2\pi i/p},e^{2\pi i/p})$. The bifundamental still exists. A Wilson probe of charges $(q_L,q_R)$ is allowed only if $q_L+q_R\in p\mathbb Z$; in particular an individual unit fundamental probe is lost when $p>1$.

The flux lattice becomes

$$(n_L,n_R)\in\mathbb Z^2+\frac1p(1,1)\mathbb Z. \tag{AB08.14}$$

Use unit-flux connections $b=a-\widehat a$ and $c'=p\widehat a$. Relative to this new lattice, the unchanged local level form is

$$K'= \begin{pmatrix}k&k/p\\ k/p&0\end{pmatrix}. \tag{AB08.15}$$

For this spin-CS theory the integral bilinear condition requires $p$ to divide $k$. Odd diagonal entries are allowed with the spin structure; a fractional mixed entry is not. We are not adding another discrete counterterm to repair a different choice.

Dualizing $c'$ with a $2\pi$-periodic angle now gives $(k/p)b+d\tau'=0$. The residual phase group is $\mathbb Z_{k/p}$, and the rank-one scalar target is $\mathbb C^4/\mathbb Z_{k/p}$. For $k=p=2$ it is $\mathbb C^4$, instead of the standard theory’s $\mathbb C^4/\mathbb Z_2$. The matter representation and local equations looked similar; the allowed probes, fluxes, and global theory changed. This calculation is a controlled comparison, not a classification of all global forms at arbitrary rank.

Exercises

1. Does a repeated auxiliary eigenvalue locate a collision?

For $N=2$, take $A_1=\operatorname{diag}(1,0)$, $A_2=\operatorname{diag}(0,1)$ and $B_i=0$. Calculate both vector scalars and the potential. Are the two points equal in $\mathbb C^4/\mathbb Z_k$? Explain what fails in the diagonalization proof, and what does not follow from that failure.

Hint 1

The density matrices sum the two squared amplitudes. The quotient rotates the whole tuple by one phase; it does not interchange its flavor components.

Hint 2

Compare $(1,0,0,0)$ and $(0,1,0,0)$. A repeated $s_a$ removes information from (AB08.3).

Solution

One finds $\mu_L=I_2$, $\mu_R=-I_2$ and $\sigma=\widehat\sigma=tI_2$. Every intertwining equation and F gradient vanishes, so $V=0$. The two tuples are $(1,0,0,0)$ and $(0,1,0,0)$, which no common phase relates. They are distinct nonzero points of the one-brane quotient.

$A_1$ is not invertible, and the vector scalars have repeated eigenvalues, so both sufficient hypotheses of the chosen diagonalization proof fail. The configuration is nonetheless already diagonal and lies on a regular point of the finite quotient: neither a nontrivial phase nor a phase-assisted exchange fixes this ordered pair. Failure of that proof is not a new vacuum constraint or evidence of a quotient singularity. It also does not compute a physical mass spectrum.

2. Count representatives at a collision

For $N=2,k=3$, find the orbit sizes of an ordered pair of generic distinct nonzero orbifold points, two equal nonzero points, one nonzero point and the origin, and two origins. Derive the local relative-coordinate identification at a nonzero collision and its real codimension.

Hint 1

The group has eighteen elements. Its stabilizer consists of transformations that leave the particular ordered pair fixed; orbit size is group size divided by stabilizer size.

Hint 2

At a nonzero vector, only the identity phase fixes it. At the origin every phase does. Exchange fixes a pair of equal vectors.

Solution

The stabilizer orders are respectively $1,2,3,18$, giving orbit sizes $18,9,6,1$. For coincident nonzero orbifold points first choose equal lifts; a different choice conjugates the stabilizer and leaves its order unchanged. For one origin and one nonzero vector, exchange does not fix the pair, but all three phases on the origin do.

Using the orthonormal coordinates $X_{\rm c},X_{\rm r}$ preserves the flat kinetic metric. Exchange acts as $(X_{\rm c},X_{\rm r})\mapsto(X_{\rm c},-X_{\rm r})$. The fixed set is $X_{\rm r}=0$, four complex equations or real codimension eight inside the sixteen-real-dimensional two-brane branch. The local singularity is therefore not a one-dimensional reflection boundary; it is a quotient of eight real relative directions. At two origins, the independent phase stabilizers combine with exchange, so using only the exchange stabilizer would miss identifications.

3. Change the flux lattice consistently

Start with the alternate rank-one quotient group above. For $k=6,p=3$, find the level matrix, an allowed new fractional original flux, a forbidden old Wilson probe, and the scalar target. Would the same unchanged level form define this spin-CS theory at $k=6,p=4$?

Hint 1

Use the lattice basis $(1,0)$ and $(1/p,1/p)$ in the original two flux coordinates. Its pairings under $\operatorname{diag}(k,-k)$ are the entries of (AB08.15).

Hint 2

Integrality is a condition on pairings of every two allowed fluxes. The self-pairing of $(1/p,1/p)$ alone is zero and is not a sufficient check.

Solution

For $k=6,p=3$, $K'=\left(\begin{smallmatrix}6&2\\2&0\end{smallmatrix}\right)$. The flux $(1/3,1/3)$ is allowed in the new group. A charge-$(1,0)$ Wilson probe is forbidden because the divided center acts on it by $e^{2\pi i/3}$. The bifundamental charge $(1,-1)$ remains allowed. Dualization gives $2b+d\tau'=0$, hence the residual group is $\mathbb Z_2$ and the scalar target is $\mathbb C^4/\mathbb Z_2$. Keeping the old $\mathbb Z_6$ answer would have kept the wrong flux lattice.

For $p=4$ the pairing of $(1,0)$ with $(1/4,1/4)$ equals $6/4=3/2$. The proposed mixed level is not integral, so the unchanged compact spin-CS action is not defined on all of those bundles. A spin structure permits odd integral self-levels; it does not turn this mixed half-integer into an integer. Additional changes would define a different problem and have not been supplied.

The many-brane answer now includes paired matrix coordinates, compact fluxes and a finite group acting on complete tuples. Next, derive monopole charges and their matter dressing: a local flux insertion acquires electric gauge weight from the Chern–Simons term, and its dressing must satisfy the full remaining gauge symmetry. A coordinate on this vacuum quotient is not yet a derivation of that local operator.

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