Generalized Symmetries and Information Diagnostics

Generalized symmetries constrain extended operators through topological defects. Information diagnostics can test those constraints, but the definition depends on whether the defect is invertible and group-like or belongs to a noninvertible fusion category. A charged replica insertion is not automatically a density matrix, and a nonzero twisted quantity is not by itself a phase classifier. The support, fusion data, regulator, and positive operational state must all be specified.

Required background. Symmetry-constrained operations supplies covariance and operation classes.

Helpful background. Charge-resolved entanglement supplies the ordinary-symmetry charged-moment template.

Chapter map. The overview gives the task-to-resource map, the comparison table, and the three gates for an operational claim.

Invertible and higher-form defects

In $d$ spacetime dimensions, a $p$-form symmetry is generated by topological operators supported on closed codimension-$(p+1)$ manifolds, while its charged operators are $p$-dimensional. Ordinary 0-form symmetry already has a codimension-one topological defect; generalized symmetry does not mean replacing a “pointlike symmetry action” by an extended one. Gaiotto et al. 2015, § 1, pp. 1–4, and § 2 establish these support, linking, and selection-rule conventions.

For an invertible group-like defect with a well-defined regional action, one may form the charged moment

$$Z_n(g;A) :=\operatorname{Tr}\!\left(\rho_A^nU_A(g)\right).$$

In the replica path integral, $U_A(g)$ becomes a topological insertion linked with the branch locus. For a higher-form symmetry, whether it factorizes into regional pieces depends on the topology of the region and the support of the symmetry generator. Gatto Lamas, Gliozzi, and Hughes 2026, § II and § III, Eqs. (1)–(3) make that dependence explicit for discrete 1-form symmetry and entanglement asymmetry.

Fourier projection is legitimate only when the defects form an invertible group and the measure and regional action exist. Even then, $\rho_A^nU_A(g)$ need not be positive or normalized. Its trace is a generating function, not a density operator.

Noninvertible fusion-category structure

A noninvertible topological defect is labelled by an object $a$ with fusion

$$D_aD_b =\sum_cN_{ab}^{\ c}D_c.$$

If a product contains several simple objects, there is generally no inverse defect, unitary regional operator $U_A(a)$, or Haar/Fourier formula over the labels. In the framework used here, the fusion algebra acts on untwisted states on a circle, the tube algebra is needed when twisted sectors are included, and a strip algebra describes an interval after its boundary conditions have been chosen.

Benini et al. 2025, Eqs. (2.1)–(2.2) define entanglement asymmetry relative to a symmetry projection. Their Eq. (2.11) and Appendix A.2 establish that the algebraic symmetrizer used there is completely positive, trace-preserving, and idempotent, while §§ 3–4 distinguish the fusion, tube, and strip algebras relevant to noninvertible defects and boundaries. When such a conditional expectation $\mathcal E_{\mathfrak A}$ is available, one may define

$$\Delta S_{\mathfrak A}(\rho_A) :=D\!\left(\rho_A\middle\|\mathcal E_{\mathfrak A}(\rho_A)\right) =S\!\left(\mathcal E_{\mathfrak A}(\rho_A)\right)-S(\rho_A).$$

This is not a universal formula for every generalized symmetry. One must first construct the subsystem Hilbert space, its boundary condition, the represented $C^*$-algebra, and the corresponding symmetrizer. Without those inputs, writing a sum over defect labels does not define a channel.

Positive states versus twisted replica objects

Ordinary mutual information is defined for a normalized positive state:

$$I(A:B)_\rho :=D\!\left(\rho_{AB}\middle\|\rho_A\otimes\rho_B\right) =S(\rho_A)+S(\rho_B)-S(\rho_{AB}).$$

A twisted replica object can be negative, complex, or unnormalized. It therefore has no quantity $I_g(A:B)$ unless an explicit completely positive instrument produces a normalized defect-conditioned state $\rho_{AB}^{(g)}$. If no such state is constructed, compare charged moments, normalized replica ratios, or selection rules—not “defect mutual information.” This distinction prevents a formal insertion from acquiring an operational meaning it does not possess.

An anomaly can obstruct gauging, defect factorization, or a local symmetric regulator, but these consequences must be imported from the anomaly data. An unusual constant in a twisted entropy is not by itself an anomaly proof.

Executed correlated-sector toy benchmark

To isolate the diagnostic arithmetic from any microscopic geometry, consider an explicit finite toy state motivated by correlated $\mathbb Z_2$ one-form sectors. It is not derived from a particular lattice bipartition. Let $q_A,q_B\in\{0,1\}$ be abstract binary sector labels. Imposing $q_A=q_B$ gives the normalized positive state

$$\rho_{AB}^{\rm corr} =\frac34|00\rangle\langle00| +\frac14|11\rangle\langle11|.$$

A group-like $\mathbb Z_2$ insertion on $A$ is represented by

$$U_A(g)=(-1)^{q_A}.$$

For $p=3/4$, its one-region charged moments are

$$Z_n(e;A)=p^n+(1-p)^n, \qquad Z_n(g;A)=p^n-(1-p)^n.$$

At $n=2$,

$$Z_2(e;A)=\frac58, \qquad Z_2(g;A)=\frac12, \qquad \frac{Z_2(g;A)}{Z_2(e;A)}=\frac45.$$

The identity insertion is normalized correctly. In a microscopic topological realization, invariance under homotopic deformations would be a separate geometric check; this finite model encodes only the assumption that an allowed deformation leaves $q_A$ unchanged. The normalized positive state also has

$$I(A:B)_{\rm corr} =H_2\!\left(\frac34\right) =0.811278\ \text{bits},$$

and perfect joint parity

$$\left\langle(-1)^{q_A+q_B}\right\rangle_{\rm corr}=1.$$

This calculation establishes only the arithmetic of the declared toy sector model. It does not reproduce the geometry or sector weights of Gatto Lamas, Gliozzi, and Hughes 2026, Figs. 1–2 and §§ II–III, who study a single cylindrical region on a torus crossed by a noncontractible symmetry loop.

A uniform $\mathbb Z_5$ control exposes a complementary limitation. For

$$\rho_{AB}^{(5)} =\frac15\sum_{q=0}^{4}|q,q\rangle\langle q,q|,$$

the ordinary mutual information is $I(A:B)=\log_2 5=2.321928$ bits, but at replica index $n=2$ every nontrivial group-like defect has

$$Z_2(k;A) =\frac1{25}\sum_{q=0}^{4}e^{2\pi i kq/5} =0, \qquad k=1,2,3,4.$$

Thus a vanishing nontrivial twisted moment does not imply an absence of sector correlation; identity normalization and an ordinary positive-state diagnostic are indispensable controls.

Executed broken-constraint adversary

A process that violates the joint-sector constraint can decorrelate the two labels. Model that failure by

$$\rho_{AB}(\epsilon) =(1-\epsilon)\rho_{AB}^{\rm corr} +\epsilon\,\rho_A\otimes\rho_B.$$

This control preserves each one-region marginal. Consequently, the one-region charged moments above remain unchanged and would give a false positive if used alone. At $\epsilon=1/2$, the joint probabilities are

$$(P_{00},P_{01},P_{10},P_{11}) =\frac1{32}(21,3,3,5),$$

while

$$\left\langle(-1)^{q_A+q_B}\right\rangle =\frac58, \qquad I(A:B)=0.164996\ \text{bits}.$$

At $\epsilon=1$, the sector labels are independent, the joint parity falls to $1/4$, and $I(A:B)=0$. The strongest claim that survives is therefore conditional: a one-region charged moment diagnoses the chosen insertion response, while a geometry-dependent one-form conclusion requires a microscopic construction, a joint correlation, and deformation and breaking controls.

Executed noninvertible algebra control

The Ising fusion category has simple objects $\{1,\psi,\sigma\}$ with

$$\psi\times\psi=1, \qquad \psi\times\sigma=\sigma, \qquad \sigma\times\sigma=1+\psi.$$

In the ordered basis $(1,\psi,\sigma)$, left fusion by $\sigma$ is

$$N_\sigma =\begin{pmatrix} 0&0&1\\ 0&0&1\\ 1&1&0 \end{pmatrix}, \qquad N_\sigma^2=I+N_\psi, \qquad \det N_\sigma=0.$$

Thus $\sigma$ has no group inverse. Moreover, the naive “twirl”

$$\widetilde{\mathcal G}(\rho) :=\frac13\sum_{a\in\{1,\psi,\sigma\}} N_a\rho N_a^\dagger$$

is not trace preserving because

$$\frac13\sum_aN_a^\dagger N_a =I+\frac13N_\psi \ne I.$$

This exact failure is why the represented fusion/tube/strip $C^*$-algebra and its genuine symmetrizer are required. Replacing the symmetrizer by a uniform sum over defect labels is not a harmless approximation.

Common pitfalls

Using group notation for a noninvertible object. Fusion labels need not have inverses or unitary regional representatives. Build the appropriate algebra first.

Computing mutual information from a twisted trace. Mutual information requires a positive normalized state. A charged moment can be negative or complex.

Treating a one-region response as topological order. The adversary leaves all one-region charged moments fixed while destroying most joint sector correlation. Test topology, joint support, and explicit breaking separately.

Exercises

1. Correlated-sector moments and information

Derive $Z_2(e)$, $Z_2(g)$, and $I(A:B)$ for $\rho_{AB}^{\rm corr}$.

Solution

The marginal is $\rho_A=\operatorname{diag}(3/4,1/4)$, so

$$Z_2(e)=\operatorname{Tr}\rho_A^2 =\frac9{16}+\frac1{16}=\frac58,$$

and insertion of $U_A(g)=\operatorname{diag}(1,-1)$ gives

$$Z_2(g)=\frac9{16}-\frac1{16}=\frac12.$$

The joint state and each marginal have the same two nonzero probabilities. Hence

$$I(A:B)=H_2(p)+H_2(p)-H_2(p)=H_2(3/4)=0.811278\ \text{bits}.$$

2. Why a twisted operator is not a state

Show directly that $\rho_A^2U_A(g)$ in the benchmark is neither positive nor normalized.

Solution

The operator is

$$\rho_A^2U_A(g) =\operatorname{diag}\!\left(\frac9{16},-\frac1{16}\right).$$

It has a negative eigenvalue, so it is not positive. Its trace is $1/2$, not $1$. The scalar charged moment $Z_2(g)=1/2$ is therefore a generating-function value, not the trace of a density state to which ordinary entropy or mutual information may be applied.

3. Failure of the naive noninvertible twirl

Verify $N_\sigma^2=I+N_\psi$ and show that the uniform Kraus sum over $1,\psi,\sigma$ is not trace preserving.

Solution

The fusion matrix for $\psi$ is

$$N_\psi =\begin{pmatrix} 0&1&0\\ 1&0&0\\ 0&0&1 \end{pmatrix}.$$

Direct multiplication gives

$$N_\sigma^2 =\begin{pmatrix} 1&1&0\\ 1&1&0\\ 0&0&2 \end{pmatrix} =I+N_\psi.$$

Since $N_\psi^\dagger N_\psi=I$ and $N_\sigma^\dagger N_\sigma=I+N_\psi$,

$$\frac13\sum_aN_a^\dagger N_a =\frac13(3I+N_\psi) \ne I.$$

The proposed Kraus operators therefore fail the trace-preservation condition. A normalized $C^*$-algebra symmetrizer must replace the group-style average.

References

• Benini, Francesco, Pasquale Calabrese, Michele Fossati, Amartya Harsh Singh, and Marco Venuti. “Entanglement Asymmetry for Higher and Noninvertible Symmetries.” arXiv preprint arXiv:2509.16311, submitted 19 September 2025; version accessed 26 August 2026. Abstract. Open PDF.
• Gaiotto, Davide, Anton Kapustin, Nathan Seiberg, and Brian Willett. “Generalized Global Symmetries.” Journal of High Energy Physics 2015, no. 2 (2015): 172. DOI. Open PDF.
• Gatto Lamas, Amanda, Jacopo Gliozzi, and Taylor L. Hughes. “Higher-Form Entanglement Asymmetry and Topological Order.” Physical Review B 114 (2026): 105119. DOI. Open PDF.