Global gauge groups and line operators

Two theories can have the same gauge bosons, scalars, fermions and local interaction vertices, yet disagree about whether a fundamental Wilson line exists. Electric–magnetic duality must compare that information too.

Continue from electric–magnetic duality. Our interacting example has gauge algebra $\mathfrak{su}(2)$ in four dimensions, rank one, with the full N=4 adjoint field content and sixteen real Poincaré supercharges; four are manifest in an N=1 description. We will compare global groups $SU(2)$ and $SO(3)=SU(2)/\{\pm1\}$. There is no dynamical fundamental matter. Work on spin spacetimes with a specified spin structure. The curved example below checks gauge topology and the action, not preserved curved-space supersymmetry.

The line spectra and their duality action follow Aharony, Seiberg and Tachikawa, §§1.1–1.2 and 2.4. The small lattice calculation will be derived explicitly. It is a consistency test of the global duality statement, not a proof of non-Abelian S-duality.

What the local fields cannot distinguish

The center of a group consists of elements commuting with every group element. The center of $SU(2)$ is $\{1,-1\}$. Conjugation by $-1$ leaves every adjoint field unchanged, so the same local adjoint action can be written with either global group.

A Wilson line asks a different question. It traces the parallel transport around a contour in a chosen representation. The spin-$j$ representation of $SU(2)$ assigns the central element the value $(-1)^{2j}$. It descends to an $SO(3)$ representation only when $j$ is an integer. Thus a fundamental, $j=1/2$, Wilson line is available for $SU(2)$ but is not an ordinary Wilson line of $SO(3)$. Choosing a lift of an $SO(3)$ holonomy to $SU(2)$ leaves an undetermined sign in that fundamental trace.

A magnetic, or ‘t Hooft, line instead imposes a prescribed magnetic singularity. A small two-sphere linking the line measures its flux. A dyonic line carries both electric and magnetic data. These are external probes, not assertions that a dynamical particle of the same charge exists. A genuine line can be specified without choosing an extra two-dimensional surface ending on it. We classify their allowed charge sectors; additional scalar couplings, spin, and BPS decorations of a particular line are separate data.

Derive the charge units from a transition function

Take the Hermitian Cartan generator $H=\operatorname{diag}(1,-1)/2$, with $\operatorname{Tr}H^2=1/2$ in the ordinary fundamental trace. Locally write the absorbed connection as $\mathcal A=aH$, so the covariant derivative is $\partial-i\mathcal A$. A weight-$e$ electric component couples as $\exp[i(e/2)\oint a]$. Here $e$ is an integer weight label, not the gauge coupling: fundamental weights are $e=\pm1$, while the charged adjoint fields have $e=\pm2$. An integer lattice is simply the set of all integer sums of specified charge generators.

On the linking sphere, choose regular north and south potentials

$$a_N=\frac m2(1-\cos\vartheta)\,d\varphi,\qquad a_S=-\frac m2(1+\cos\vartheta)\,d\varphi. \tag{N18.1}$$

Their overlap has $a_N-a_S=m\,d\varphi$, and $\int_{S^2}da=2\pi m$. The corresponding transition function is

$$g_{NS}(\varphi)=e^{i m\varphi H},\qquad g_{NS}(2\pi)=(-1)^m1. \tag{N18.2}$$

For $SU(2)$ it must close to $1$, requiring even $m$. For $SO(3)$ the two signs are identified, so every integer $m$ closes. In both cases $m$ is measured in the same magnetic weight unit. This calculation derives the difference in magnetic flux sectors rather than guessing it from the Lie algebra.

Moving a weight-$e$ electric probe around the transition gives the phase $e^{i\pi e m}$. For two dyons $\gamma=(e,m)$ and $\gamma'=(e',m')$, the two electric–magnetic contributions enter with opposite orientation. Their Dirac pairing and associated linking phase are

$$\langle\gamma,\gamma'\rangle =\frac{em'-me'}2,\qquad \exp\!\left(2\pi i\langle\gamma,\gamma'\rangle\right) =(-1)^{em'-me'}. \tag{N18.3}$$

Mutual locality requires this phase to be one. It requires an integer pairing, not a pairing equal to zero. Otherwise a supposedly auxiliary Dirac surface would be detectable in the correlation of the two lines. This is source (1.4) with center $\mathbb Z_2$.

The preceding lesson’s unit-Maxwell labels ranged over $\mathbb Z^2$ with pairing $nm'-mn'$. They are not identified component by component with the present $e,m$: the Cartan normalization here gives the factor $1/2$, and a global form supplies parity restrictions. Copying just the Maxwell coupling or just its integral labels would miss this difference.

Reduce the problem to four parity classes

Electric weights differing by an adjoint root differ by two. The magnetic root lattice likewise has even labels. To isolate the discrete information, reduce both labels modulo two:

$$(z_e,z_m)=(e\bmod2,m\bmod2) \in\mathbb Z_2\times\mathbb Z_2. \tag{N18.4}$$

The symbol $\mathbb Z_2$ means addition with $1+1=0$. These parity classes do not identify every detailed line operator. They retain exactly the part relevant to the linking phase. The Weyl reflection, the residual gauge interchange of the two Cartan eigenvalues, identifies $(e,m)$ with $(-e,-m)$ simultaneously; it does not flip the two signs independently.

Allowed classes must contain $(0,0)$ and be closed under addition and reversal. Require also a maximal mutually local choice: no further charge class can be added while preserving locality. This completeness condition is part of the standard standalone gauge-theory specification in AST §1.3 and footnote 5. Locality alone would also permit the smaller, incomplete choice $\{(0,0)\}$.

There are three nonzero parity vectors, $v_e=(1,0)$, $v_m=(0,1)$, and $v_d=(1,1)$. Each has zero pairing with itself modulo two. Any two distinct nonzero vectors have pairing one modulo two. Hence a maximal choice contains zero and exactly one of them:

Theory at a fixed theta convention	Allowed parity classes	Full weight lattice	Nontrivial representative
$SU(2)$	$(0,0),(1,0)$	$e\in\mathbb Z,\ m\in2\mathbb Z$	Fundamental Wilson $(1,0)$
$SO(3)_+$	$(0,0),(0,1)$	$e\in2\mathbb Z,\ m\in\mathbb Z$	Magnetic $(0,1)$
$SO(3)_-$	$(0,0),(1,1)$	$e,m\in\mathbb Z,\ e\equiv m\pmod2$	Dyonic $(1,1)$

For example, $(1,1)$ and $(3,-1)$ belong to $SO(3)_-$ and have pairing $-2$, so they are mutually local despite neither being purely electric. But adding $(1,0)$ would give a half-integer pairing with $(1,1)$. The two $SO(3)$ choices have the same ordinary Wilson representations and different magnetic or dyonic completions.

This is a classification under the stated adjoint-matter, spin-theory assumptions. It is not a classification of every theory obtained by coupling extra topological sectors or changing the dynamical matter.

Follow S and T through the three theories

Keep the non-Abelian coupling $\tau=\theta/(2\pi)+4\pi i/g^2$ with $g>0$. The specified exchange is

$$S:\quad \tau\mapsto-\frac1\tau,\qquad (e,m)\mapsto(m,-e). \tag{N18.5}$$

It preserves (N18.3). Applying it to the allowed sets, rather than to one preferred charge, gives $SU(2)\leftrightarrow SO(3)_+$ and $SO(3)_-\mapsto SO(3)_-$. This is AST (2.10)–(2.12).

There are two comparisons associated with a $2\pi$ theta shift. The source’s active Witten-effect transport is

$$T_{\rm W}:\quad (e,m)\mapsto(e+m,m). \tag{N18.6}$$

The passive endpoint relabeling used in the Maxwell energy calculation, when $\tau'=\tau+1$, instead uses $T_{\rm P}:(e,m)\mapsto(e-m,m)$. For instance, a minimal magnetic label becomes $(1,1)$ under the former and $(-1,1)$ under the latter. These are different full labels, even though their difference is an electric root. They define the same parity class, so both comparisons induce the same permutation of the three spectra:

Map of the coupling	Image of $SU(2)$	Image of $SO(3)_+$	Image of $SO(3)_-$
$S:\tau\mapsto-1/\tau$	$SO(3)_+$	$SU(2)$	$SO(3)_-$
$T:\tau\mapsto\tau+1$	$SU(2)$	$SO(3)_-$	$SO(3)_+$

The source’s plus sign and the passive minus sign have not been equated. Their agreement in this table follows specifically from reduction modulo two.

At $\tau=i$, $S$ leaves the numerical coupling unchanged but still exchanges $SU(2)$ and $SO(3)_+$. Equal coupling values therefore do not make these the same theory. Conversely, the $SO(3)_-$ charge set is preserved, while its particular line $(1,1)$ maps to $(1,-1)$; set preservation need not fix every operator.

One can identify the subgroup preserving the $SU(2)$ choice. For $G=\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ with integer entries and $ad-bc=1$, use the passive convention

$$\tau'=\frac{a\tau+b}{c\tau+d},\qquad \binom{e'}{m'}= \begin{pmatrix}a&-b\\-c&d\end{pmatrix}\binom em. \tag{N18.7}$$

For every even $m$, the result $m'=-ce+dm$ is even exactly when $c$ is even. The determinant condition then also ensures that the inverse preserves the lattice. This subgroup is called $\Gamma_0(2)$. It preserves the global line choice, not necessarily the numerical value of $\tau$.

See the theta distinction in a real bundle

Why does $T$ exchange the two $SO(3)$ choices? Their path integrals can distinguish bundles which cannot be lifted to $SU(2)$. The obstruction is a binary two-cycle flux, traditionally denoted $w_2$. It is a property of the gauge bundle, distinct from the spacetime obstruction to a spin structure.

For an explicit example, take the spin manifold $X=S^2\times S^2$, oriented by the product of the standard sphere orientations. Put magnetic integers $p,q$ through its two spheres. On each of the four product patches use the potentials (N18.1) with $m=p$ or $q$, and set $\mathcal A=H(a_1+a_2)$. The commuting transition functions define an $SO(3)$ bundle for all integers $p,q$. This bundle lifts to $SU(2)$ precisely when both are even.

Let $\alpha,\beta$ be closed two-forms with unit integrals on the first and second sphere. Then $\int_X\alpha\wedge\beta=1$ and $\alpha\wedge\alpha=\beta\wedge\beta=0$. The curvature is $\mathcal F=2\pi H(p\alpha+q\beta)$. With the fixed instanton normalization,

$$k=\frac1{8\pi^2}\int_X\operatorname{Tr}(\mathcal F\wedge\mathcal F) =\frac{(2\pi)^2}{8\pi^2} \underbrace{\operatorname{Tr}H^2}_{1/2}\,2pq =\frac{pq}{2}. \tag{N18.8}$$

The factor two comes from the two cross terms of the commuting two-forms. For $p=q=1$, $k=1/2$. The weight $e^{i\theta k}$ changes sign when $\theta$ increases by $2\pi$ and returns after $4\pi$. No singular gauge field or non-spin spacetime was needed. This reducible Cartan family does not exhaust all bundles; in particular, its liftable examples do not construct every integer-instanton $SU(2)$ bundle.

The general spin-manifold statement in AST (6.1)–(6.5) is $k=\mathcal P(w_2)/4\pmod1$, where the integrated Pontryagin square $\mathcal P(w_2)$ is a value modulo four. In an integrally lifted example such as ours, compute it by squaring an integral lift of $w_2$ and reducing modulo four. Here that lift is $p\alpha+q\beta$, giving $\mathcal P(w_2)=2pq\bmod4$. On a closed spin four-manifold it is even.

At a fixed $\theta$ modulo $2\pi$, write the two choices of weight as

$$\exp(i\theta k)\, (-1)^{\nu\,\mathcal P(w_2)/2}, \qquad \nu=0,1. \tag{N18.9}$$

Thus $\theta\mapsto\theta+2\pi$ exchanges $\nu=0$ and $\nu=1$. Equivalently one can describe a single $SO(3)$ family with theta range $4\pi$. For this group the binary choice and the extended theta range are alternative descriptions, not two independent extra parameters.

To identify $\nu$ with the line table, surround a magnetic line by $S^2$ and test a large gauge transformation in the remaining directions. In AST (6.16)–(6.21), closing those directions gives a second two-cycle, with binary flux $b$. The same cross-term calculation gives $\mathcal P(w_2)/2=z_m b$. Its discrete phase is $(-1)^{\nu z_m b}$, whereas an electric class transforms by $(-1)^{z_e b}$. Equality for both $b=0,1$ requires $z_e=\nu z_m$. Hence $\nu=0$ gives $SO(3)_+$ and $\nu=1$ gives $SO(3)_-$.

Spin is essential to this statement. The source’s footnote 4 allows quarter-instanton numbers and a different periodicity on non-spin manifolds; extending the adjoint-fermion theory there needs additional structure. We have not made that extension. Nor does the topological sign alone calculate a complete supersymmetric partition function: zero modes and the observable can make individual contributions vanish.

Exercises

1. Test a proposed line spectrum

A proposed theory includes genuine lines $\gamma=(3,2)$ and $\delta=(2,1)$. Test mutual locality, identify a maximal spectrum containing each separately, and decide whether attaching adjoint electric charge can repair the pair. Contrast this with including $(1,1)$ and $(3,-1)$. Explain why keeping only the neutral parity class would not produce a fourth maximal choice.

Hint 1

Reduce the charges modulo two, but also compute the full pairing to see its physical normalization.

Hint 2

An adjoint electric attachment changes $e$ by an even integer. Test whether that can change the linking phase.

Solution

The first pair has $\langle\gamma,\delta\rangle=(3\cdot1-2\cdot2)/2=-1/2$, so its linking phase is $-1$. The classes are $(1,0)$ and $(0,1)$. The first belongs to the $SU(2)$ spectrum and the second to $SO(3)_+$, but they cannot both be genuine in one of the stated theories. Adding $2r$ to an electric label changes the pairing by an integer; it does not remove the sign.

For the second pair the pairing is $(1\cdot(-1)-1\cdot3)/2=-2$. Both have class $(1,1)$, and the $SO(3)_-$ spectrum contains them. Locality does not demand a zero pairing.

The set containing only $(0,0)$ is closed and mutually local but not maximal. Any one of the three nonzero classes can be added. Adding two different nonzero classes fails locality. These alternatives exhaust the four-element parity space and produce the three rows of the table.

2. Transform the entire theory, not just its coupling

Start with the $SU(2)$ spectrum at $\tau=i$. Apply (N18.7) with $G=\left(\begin{smallmatrix}1&0\\1&1\end{smallmatrix}\right)$. Find the new coupling, the images of the generators $(1,0),(0,2)$, and the resulting maximal spectrum. Check their pairing. Does this matrix belong to the subgroup preserving the original global choice?

Hint 1

The two generators produce every original weight by integer addition. Transforming them determines the full image lattice.

Hint 2

An image has $e'=e$ and $m'=m-e$. Test the parity of $e'+m'$.

Solution

The new coupling is $\tau'=i/(1+i)=(1+i)/2$. The charge generators become $(1,-1)$ and $(0,2)$. Their full integer span consists exactly of pairs with $e'+m'$ even: given such a pair, set the original $e=e'$ and $m=m'+e'$, which is even. The image is therefore $SO(3)_-$.

The original generator pairing is $(1\cdot2-0)/2=1$; after transformation it is also $(1\cdot2-(-1)\cdot0)/2=1$. Nevertheless, $c=1$ is odd, so the matrix is not in $\Gamma_0(2)$. Preserving the pairing is weaker than preserving one chosen maximal lattice. The transformation relates global theories as well as their couplings; it is not a renormalization-group flow.

3. Change the product fluxes and radii

In the $S^2\times S^2$ construction, let the radii be $R_1,R_2>0$ and the fluxes be arbitrary integers $p,q$. Compute the positive gauge action $S_{\rm kin}=g^{-2}\int\operatorname{Tr}(\mathcal F\wedge*\mathcal F)$. Compare it with $8\pi^2|k|/g^2$ and find the equality condition. Which flux pairs acquire a minus sign under a $2\pi$ theta shift? Does a half-integer instanton number mean this spacetime has no spin structure?

Hint 1

In orthonormal frames on the two spheres, the two curvature coefficients are $p/(2R_1^2)$ and $q/(2R_2^2)$. The product volume is $16\pi^2R_1^2R_2^2$.

Hint 2

Use $\operatorname{Tr}H^2=1/2$ and complete the square in $pR_2/R_1$ and $qR_1/R_2$.

Solution

The two area forms are orthogonal, so

$$S_{\rm kin}=\frac{2\pi^2}{g^2} \left(p^2\frac{R_2^2}{R_1^2} +q^2\frac{R_1^2}{R_2^2}\right) \geq\frac{4\pi^2}{g^2}|pq| =\frac{8\pi^2}{g^2}|k|.$$

Equality requires $|p|/R_1^2=|q|/R_2^2$. For nonzero fluxes of the same sign the curvature is self-dual; for opposite signs it is anti-self-dual. If just one flux vanishes, equality requires the other to vanish too. For $p=q=1,R_1=R_2$, the action is $4\pi^2/g^2$, consistent with $k=1/2$.

The theta factor is multiplied by $e^{2\pi i k}=(-1)^{pq}$, so exactly the odd-odd pairs change sign. For example $(1,2)$ does not change sign, although its bundle still fails to lift because one flux is odd. Non-liftability and half-integral instanton number are related but not equivalent conditions.

The spacetime $S^2\times S^2$ is spin. The obstruction here belongs to the gauge bundle, not the tangent bundle. Self-duality minimizes this gauge action in its topological sector; we have not solved any curved-space Killing-spinor equation or claimed a preserved supercharge on this background.

The useful duality datum is now a coupling together with a complete line spectrum. A fixed Lie algebra or a local perturbative correlator does not specify both. The supersymmetric Wilson-loop lesson adds scalar couplings and a preserved-charge condition to an allowed electric line; those additions refine an operator without replacing the global charge-lattice test.