Global gauge groups and generalized symmetries

Two theories can have the same local gauge fields and different exact observables. The distinction appears before adding any matter: which Wilson probes are genuine, and which bundles are included in the path integral?

We will hold the Lie algebra $\mathfrak{su}(2)$ fixed and change the global group from $SU(2)$ to a specified $SO(3)=SU(2)/\mathbb Z_2$ quotient. In the example below the partition function on a three-torus changes from five to three. Merely deleting the disallowed Wilson lines does not reproduce the quotient theory, even though it accidentally gives the same torus count.

Prepare with connections, flux and holonomy and the Seifert partition-function calculation. This dossier is an established reproduction with controlled training variations. It introduces the extra language of topological lines as it is needed; a course in topological quantum field theory is not assumed.

Specify the two theories

Begin with pure three-dimensional N=2 $SU(2)$ Chern–Simons theory at positive supersymmetric level $K=6$. There is one vector multiplet, no charged matter and no FI parameter. The gauge rank is one; the full flat-space supersymmetry has four real supercharges. For a product $\Sigma_g\times S^1$, use the A-twist and periodic circle spin prescription of the preceding dossier. The torus case admits constant supersymmetry parameters; we use the same opposite-R pair throughout the comparison.

Normalize Hermitian generators by $\operatorname{Tr}_2(t_at_b)=\delta_{ab}/2$, where $\operatorname{Tr}_2$ is the ordinary fundamental trace. The gaugino shifts the infrared bosonic level to

$$k=K-h^\vee=6-2=4. \tag{D05.1}$$

This is the source convention of Q25, equation (2.40). $K$ and $k$ are not interchangeable. The infrared topological theory has no propagating local degrees of freedom; its lines and finite state spaces retain information about the original gauge theory.

The second theory is obtained by gauging the bosonic center symmetry specified below. Its group is $SO(3)$, with the level measured in the same $SU(2)$ trace normalization and with the inherited regulator/contact prescription. We add no extra weighting of discrete sectors. This defines the quotient more precisely than a bare label “$SO(3)$ level four,” whose level unit can differ between references. In the supersymmetric description it is the corresponding quotient of the $K=6$ theory, including the gaugino contribution.

Our principal observables are $Z_{\Sigma_g\times S^1}$, with the periodic supersymmetric trace understood. In this bosonic infrared sector they count ground states. These product observables avoid an additional comparison between sphere framing conventions. Nothing here changes the absolute phase prescription of the preceding Seifert calculation.

See the group in a transition function

Set $t_3=\operatorname{diag}(1,-1)/2$. A Cartan-valued monopole on a test two-sphere can be described by

$$a_N=\frac m2(1-\cos\theta)t_3\,d\phi,\qquad a_S=-\frac m2(1+\cos\theta)t_3\,d\phi, \qquad \int_{S^2}f=2\pi m t_3. \tag{D05.2}$$

With $D=d-ia$, their overlap is $g_{NS}=e^{im\phi t_3}$. At $\phi=2\pi$ this is $(-1)^m\mathbf1$. It is a closed transition in $SU(2)$ only for even $m$. It is closed in $SO(3)$ for every integer $m$, because the two central elements are identified there.

The same calculation tests the probes. A spin-$j$ representation has weights $-j,-j+1,\ldots,j$. For integer $j$, the central element acts as $+1$; for half-integer $j$, it acts as $-1$. Thus the fundamental spin-$1/2$ Wilson loop is a genuine probe of the $SU(2)$ theory even without fundamental dynamical matter. It is not a genuine Wilson loop of the quotient. “Genuine” means it can be defined without an additional attached surface.

Odd $m$ labels an $SO(3)$ bundle that cannot be lifted to an $SU(2)$ bundle. Its obstruction is denoted $w_2=1$ on this sphere; $w_2$ is a binary cohomology class recording the failure to choose consistent lifts of the transitions. This is a bundle test, not a claim to have found a finite-energy monopole particle.

An equally explicit flat example uses a spatial torus. The lifts $U=i\sigma_1$ and $V=i\sigma_2$ obey $UV=-VU$. Their images in $SO(3)$ are commuting rotations by $\pi$ about orthogonal axes, so they define a flat quotient connection. Their lifted commutator is $-\mathbf1$, detecting $w_2=1$. Such a sector is absent from an $SU(2)$ bundle sum. Changing the group therefore changes more than the labels on an unchanged list of states.

Recognize the center symmetry by linking lines

A topological line can be moved without changing a correlator as long as it does not cross another insertion. Bringing two parallel lines together defines their fusion. In a non-Abelian theory this can be a sum of distinct line types. A line is invertible if fusing it has a unique inverse.

A one-form symmetry acts on line operators, in contrast to an ordinary symmetry acting on local operators. In three dimensions its symmetry operators are themselves topological lines. Linking a symmetry line once with another line measures the latter’s charge. This geometrical operation replaces the familiar action of a charge operator on a local field.

The exact infrared $SU(2)_k$ input is a finite set of lines $j=0,\tfrac12,\ldots,k/2$. The spin and normalized linked-loop matrix are

$$h_j=\frac{j(j+1)}{k+2}\pmod1,\qquad S_{jl}=\sqrt{\frac2{k+2}} \sin\frac{\pi(2j+1)(2l+1)}{k+2}. \tag{D05.3}$$

Here $e^{2\pi ih_j}$ is the phase from one full framing twist of a line, not its spacetime particle spin in a four-dimensional theory. $S$ is the amplitude for exchanging the two cycles in torus quantization, equivalently the normalized Hopf-link amplitude. These are the exact CS data supplied by Q25 (3.112)–(3.113). We use this input rather than claiming to derive non-Abelian CS quantization from the classical action.

One can check the sine matrix’s normalization directly. Start with Fourier modes on $\mathbb Z_{2(k+2)}$ and form the antisymmetric pairs $|r\rangle-|-r\rangle$, $r=1,\ldots,k+1$. Their Fourier matrix elements are differences of two exponentials, giving D05.3 after the common phase is fixed. Discrete sine orthogonality then gives $S^\dagger S=1$.

The center generator is $J=k/2$, with $J\times j=k/2-j$ and $J\times J=0$. It has unit quantum dimension, where $d_j=S_{0j}/S_{00}$ is the value of an unknotted line relative to the empty diagram. At $k=4$,

$$h_J=\frac k4=1,\qquad \frac{S_{Jj}}{S_{0j}}=(-1)^{2j}. \tag{D05.4}$$

The second relation follows by replacing $\sin\alpha$ by $\sin[(2j+1)\pi-\alpha]$. It reproduces the central charge of each representation directly from a linked-loop experiment. The first relation says $J$ is a boson. Thus it can be summed over without introducing spin-dependent line-twist phases. This is the bosonic gauging condition used here, not a claim that every center quotient is allowed with the same data.

For reference, the complete parent data are small:

$j$	$0$	$1/2$	$1$	$3/2$	$2$
$h_j$ modulo one	$0$	$1/8$	$1/3$	$5/8$	$0$
$d_j$	$1$	$\sqrt3$	$2$	$\sqrt3$	$1$
Linked-center phase	$+$	$-$	$+$	$-$	$+$

Gauge the symmetry, including the missing sector

Gauging means summing the symmetry-line networks, with their gauge redundancy divided out. It has three effects in this example.

First, a surviving genuine line must have trivial linked-center phase: only $j=0,1,2$ survive. Second, fusing with a summed $J$ line becomes an identification. Thus $0$ and $2$ are one orbit, giving the new vacuum line. The line $j=1$ is fixed by $J$.

Third, a fixed orbit must be resolved. There are two ways to specify the action of the stabilizing $\mathbb Z_2$ on that line, its characters $+1$ and $-1$. Equivalently, there is an untwisted and a twisted sector, and their two Fourier combinations carry these characters. Both must be included. Denote the resulting lines by $x$ and $y$:

$$\{0,2\}\longrightarrow\mathbf1,\qquad \{1\}\longrightarrow x,y, \qquad \dim\mathcal H_{T^2}:5\longrightarrow3. \tag{D05.5}$$

These are Q25’s neutral-orbit and stabilizer rules, (3.26)–(3.28), applied explicitly. Restriction alone gives three old labels; restriction followed by identification gives two orbits. Neither procedure is the complete gauging. The final answer is three for a different reason: one vacuum orbit plus two resolved fixed sectors.

There is a useful operator check. In the five-state parent basis let $U$ multiply $|j\rangle$ by $(-1)^{2j}$ and let $V$ send it to $|2-j\rangle$. They commute and square to one. Projection in the old space gives

$$\operatorname{Tr}\frac{(1+U)(1+V)}4 =\frac{5+1+1+1}{4}=2. \tag{D05.6}$$

The extra state comes from the nontrivial spatial $w_2$ sector, exemplified by the anticommuting lifts above. An ordinary invariant-subspace calculation cannot create that sector.

Reproduce the full flux sum and change the genus

The background for a $\mathbb Z_2$ one-form symmetry is a flat discrete two-form $B$. Its gauge-equivalence classes on $T^3$ are the eight elements of $H^2(T^3,\mathbb Z_2)$, one binary flux through each independent two-torus. No additional phase is assigned to those sectors in our chosen gauging.

The finite gauge normalization is important. On a cellular description write $C^r$ for the finite group of degree-$r$ cochains and $Z^r$ for its closed cochains. One-form gauge transformations of $B$ themselves have zero-form redundancy. Their counting gives

$$\frac{|C^0|}{|C^1|}\sum_{B\in Z^2}Z[B] =\frac{|H^0|}{|H^1|}\sum_{[B]\in H^2}Z[B] =\frac14\sum_{[B]\in H^2(T^3,\mathbb Z_2)}Z[B]. \tag{D05.7}$$

For the middle equality use $|Z^1|=|C^0||H^1|/|H^0|$ after grouping exact two-cochains. On the connected three-torus, $|H^0|=2$ and $|H^1|=8$. This is not an average with weight $1/8$.

The zero-background answer is five. A nonzero background is dual to an inserted $J$ loop. For a representative running in the time direction, the dimension of the punctured torus is $\operatorname{Tr}N_J=1$: the fusion permutation $j\mapsto2-j$ has the single fixed label $j=1$. Every nonzero binary background is related to this representative by a change of torus cycles. Since $J$ has trivial self-braiding and twist, this introduces no anomaly phase. Consequently

$$Z_{T^3}[SO(3)]=\frac{5+7\cdot1}{4}=3. \tag{D05.8}$$

This is the same result as the resolved-orbit calculation, now obtained by summing all bundles rather than discarding charged states.

For a controlled variation, keep both theories and all prescriptions fixed, and change the base from genus one to genus two. The exact CS sewing formula is

$$Z_{\Sigma_g\times S^1}=\sum_j S_{0j}^{\,2-2g} =\sum_j\mathcal H_j^{\,g-1},\qquad \mathcal H_j=S_{0j}^{-2}.$$

Each extra handle multiplies a given channel by $\mathcal H_j$; at genus one each line contributes one state. The parent handle factors are $(12,4,3,4,12)$. The quotient rule is $S'_{0,(\omega,\chi)}=|\omega|S_{0j}$, or $\mathcal H'=\mathcal H_j/|\omega|^2$. The length-two vacuum orbit and the two resolved length-one sectors therefore all have $\mathcal H'=3$. It follows that

$$Z_{\Sigma_2\times S^1}[SU(2)]=35,\qquad Z_{\Sigma_2\times S^1}[SO(3)]=9. \tag{D05.9}$$

On a sphere both give one, as required for the vacuum state space. On general genus the quotient gives $3^g$. The change is not a constant factor multiplying every parent partition function.

For an independent finite calculation, take $X$ to denote the fundamental fusion operator. The five parent lines are represented by the polynomials $1,X,X^2-1,X^3-2X,X^4-3X^2+1$, with relation $X^5-4X^3+3X=0$. Multiplying and reducing these polynomials constructs the fusion multiplicities $N_{ab}^{c}$. Cutting a genus-two surface into two three-holed spheres counts $\sum_{a,b,c}(N_{ab}^{c})^2=35$. The quotient has three unit-dimension lines with fusion $x^2=y$, $xy=1$, $y^2=x$; the same sewing counts nine. This check uses integer fusion spaces, not another evaluation of the handle-factor sum.

What can be varied next?

Literature checked 20 October 2026. AST explains why local gauge data do not specify genuine probes and bundles; its four-dimensional $SO(3)$ discrete-theta labels are not a classification of this three-dimensional example. Q25 gives the pure-CS quotient construction used here. Its general Seifert comparisons retain counterterm data and explicitly identify some broader identities as checked or conjectured; our product-manifold calculation does not prove them all.

The recent Cheng–Seiberg study develops one-form symmetries and transitions in bosonic topological theories. Its section 2.2 gives a useful next diagnostic: a single puncturing line charged under the invertible one-form symmetry has no states on a closed surface. A neutral line can have states at sufficiently large genus. This is an established constraint, not a new conjecture suggested by our numbers.

A bounded continuation is therefore to keep $k=4$ and add one specified line puncture. For the parent’s $j=1$ line, $N_{1j}^{j}$ is nonzero at $j=1/2,1,3/2$, so the punctured-torus dimension is three. For $j=1/2$ it is zero. In the quotient, each nontrivial line $x$ or $y$ has zero one-puncture dimension: translation by a nontrivial element of its three-element fusion group has no fixed point. These tests distinguish a projected old Wilson label from a resolved genuine line.

After reproducing that calculation at another genus, a matter deformation would require a new specification of representations, surviving one-form symmetry, anomaly and background counterterms. Fundamental matter already obstructs the quotient used here. The pure-CS formulas do not provide a general matter-coupled answer. Disagreement should first trigger checks of the level shift, fixed-sector resolution and finite gauge normalization; agreement establishes the stated reproduction, not novelty.

Exercises

1. Test the magnetic patch with an actual probe

In D05.2 take $m=1$. Test the overlap on weights $\pm1/2$ and on weights $-1,0,1$. Decide which global group permits the bundle. Repeat the group test at $m=2$, and explain why omitting fundamental dynamical matter does not by itself turn an $SU(2)$ theory into an $SO(3)$ theory.

Hint 1

Evaluate $e^{2\pi imw}$ for each weight $w$.

Hint 2

The group fixes which external Wilson probes are allowed, not only the representations of fields appearing in the Lagrangian.

Solution

At $m=1$, the two fundamental weights give $e^{\pm i\pi}=-1$, whereas the three adjoint weights all give one. The transition ends at the nontrivial center element in $SU(2)$, so it is not an $SU(2)$ transition around the overlap circle. Its image closes in $SO(3)$ and has $w_2=1$. At $m=2$ the fundamental phases also equal one, and the transition is allowed for both groups, with trivial lift obstruction. An $SU(2)$ theory without fundamentals still permits a fundamental external Wilson loop and still sums only over its own bundles. Changing those choices is an additional operation on the theory.

2. Diagnose two incomplete quotient calculations

One program retains only the three center-neutral parent labels and reports three states. Another projects with D05.6 and reports two. Identify the error in each. Compute the genus-two result that the first program would produce if it also retained the old handle factors, and compare it with the full quotient.

Hint 1

Neutrality, orbit identification and fixed-sector resolution are different operations.

Hint 2

The neutral old handle factors are $12,3,12$. The length-two orbit changes its handle factor by a factor of four.

Solution

The first program gets the torus count right accidentally. It still treats $j=0$ and $j=2$ as separate lines and fails to resolve $j=1$. Its genus-two answer is $12+3+12=27$, not nine. The second program correctly identifies the invariant states in the old, liftable sector, but omits the extra twisted sector; hence two instead of three. Full gauging gives one vacuum orbit with handle factor $12/2^2=3$, and two resolved fixed states with factor three each. Their genus-two sum is nine. This changed-genus test detects an error that a single torus number conceals.

3. Change the level before using the same recipe

Consider first supersymmetric $K=8$, then $K=10$, with the same gauge algebra and no matter. For each case compute the infrared level and the center-line twist. Does the bosonic quotient prescription above apply? In the allowed case count the quotient torus states by neutral orbits and their stabilizers.

Hint 1

Use $k=K-2$ and $h_J=k/4$. A twist of minus one requires spin information absent from the bosonic prescription.

Hint 2

At $k=8$, the neutral labels are the integer spins from zero to four. The center sends $j$ to $4-j$.

Solution

At $K=8$, $k=6$ and $h_J=3/2$, so the twist is $-1$. The same bosonic condensation cannot be used. A spin-dependent quotient may be defined with the appropriate refinement; the calculation above does not specify it. This is not a statement that all $SO(3)$ theories at that level are meaningless.

At $K=10$, $k=8$ and $h_J=2$, so the twist is one. The neutral orbits are $\{0,4\}$, $\{1,3\}$ and $\{2\}$. The first two have trivial stabilizer and contribute one state each. The fixed orbit has a $\mathbb Z_2$ stabilizer and contributes two. Therefore the quotient torus dimension is four, compared with nine for its parent. Using $K$ in place of $k$ would misidentify both the line set and this count.