research35 min studyLesson 779 in Physics

Noninvertible symmetry in solvable examples

Use the critical Ising topological-line algebra to distinguish a simple noninvertible symmetry defect from a reducible sum or a unitary time-evolution operator.

Before you begin

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The question for today

What you’ll understand

Calculate the Ising duality-line fusion algebra and positive quantum dimension.
Identify a physical kernel using its action on conformal primary sectors.
Distinguish exact continuum fusion from a differently normalized finite-lattice duality operator.

The critical Ising conformal field theory in one spatial and one time dimension has a simple topological line whose square is a sum of two different lines. This duality line is noninvertible even though the underlying quantum theory is unitary. It is neither a time-evolution operator nor merely the reducible sum of ordinary symmetry elements. The example is solvable enough to calculate powers, quantum dimensions, and a kernel exactly, while still illustrating why fusion rules and microscopic duality maps need careful normalization.

A simple line can fuse to more than one outcome

The three simple topological lines of the bosonic diagonal Ising conformal field theory are the identity, the invertible spin-flip line eta, and the duality line D. Eta squares to the identity and fusing it with D returns D. Fusing D with itself gives the identity plus eta. The plus sign represents a direct sum of fusion sectors; it is not an instruction to choose either outcome with an assumed fifty-percent probability.

The duality line is simple because it is not itself a direct sum of smaller defects in this theory. Its positive quantum dimension solves the fusion-character equation with identity and eta both of dimension one, giving square root of two. This contrasts with the reducible identity-plus-eta object, whose dimension is two. The same broad appearance of a sum in an algebra should therefore not erase the distinction between a simple defect and a sum of existing group defects.

\eta^2=I,\qquad \eta D=D\eta=D,\qquad D^2=I\oplus\eta,\qquad d_D^2=2,\quad d_D=\sqrt2

These are exact topological-line data of the D = 2 bosonic diagonal critical Ising CFT in its standard line normalization. The line has codimension one and generalizes zero-form symmetry; it is not a higher-form charged Wilson line in D = 4.↔ Scroll wide equations horizontally.

A kernel is compatible with unitary bulk dynamics

On the untwisted Ising Hilbert space, the spin-flip line has eigenvalue plus one in the identity and energy-primary sectors and minus one in the spin-primary sector. The duality line acts by plus square root of two, minus square root of two, and zero respectively. Its square agrees with identity plus eta in every sector. In particular, the spin-primary sector lies in its kernel, so no inverse on the whole Hilbert space can exist.

A topological defect operator need not preserve the norm of a state when inserted in a correlation function. Applying it is not automatically a deterministic quantum channel or a measurement outcome with probability equal to the resulting squared norm. Ordinary unitary time evolution remains intact. A finite-dimensional restriction to primary states is a useful exact algebra check, but the untwisted Hilbert space contains the full descendant towers, not only three primary states. Spatial defects or twisted boundary conditions define additional, distinct defect Hilbert spaces.

\eta\big|_{(1,\varepsilon,\sigma)}=\operatorname{diag}(1,1,-1),\qquad D\big|_{(1,\varepsilon,\sigma)}=\operatorname{diag}(\sqrt2,-\sqrt2,0)

The notation specifies eigenvalues on the three primary sectors, or on a chosen orthonormal primary-state subspace. It is not the complete CFT state space. The zero eigenvalue proves noninvertibility of this operator; its nonunitarity as an insertion does not violate bulk unitarity.↔ Scroll wide equations horizontally.

Microscopic duality and continuum fusion have different presentations

A Kramers–Wannier transformation exchanges order and disorder descriptions, and on a finite periodic lattice it interacts with boundary conditions, twisted sectors, and translations. A conserved finite-lattice operator can square to a spin-flip projection times a lattice translation, with an additional normalization factor. One cannot replace that relation by the continuum line fusion without taking the appropriate low-energy limit and matching the operator normalization.

The continuum algebra gives exact identities for powers of the topological line and constrains symmetry-preserving deformations. It does not determine the entire nearby phase diagram or every possible lattice realization. Associator data can distinguish categories sharing the same fusion coefficients, and boundary compatibility imposes further constraints. The solvable example is therefore best used as a complete check of a stated algebraic claim, followed by a separate analysis of which physical construction realizes it.

D^3=2D,\qquad D^4=2(I\oplus\eta),\qquad D\sigma=0\quad\text{on the spin-primary state}

Powers use the standard continuum line/operator normalization. The last expression denotes the action on its corresponding primary state, not pointwise multiplication of local fields. A differently normalized lattice duality operator can obey a translation-dependent relation before the continuum limit.↔ Scroll wide equations horizontally.

Work it through

Compute powers and dimension of the Ising duality line

Use the continuum relations D² = I ⊕ eta and eta D = D. Find D³, D⁴, and the positive quantum dimension of D. Explain why an ordinary group inverse cannot be inferred from D².

Multiply once more to get $D^3=(I+\eta)D=2D$ .
Then $D^4=2D^2=2I+2\eta$ . These coefficients count fusion multiplicities.
The dimension equation is $d_D^2=d_I+d_\eta=2$ , giving dD = √2. Since D annihilates the spin-primary sector, it cannot have an inverse on the full Hilbert space despite these exact power identities.

The result

D³ = 2D, D⁴ = 2I ⊕ 2eta, and dD = √2; the kernel rules out an ordinary global inverse.

The living notebook

Explore the model

Interactive lab

Interactive lab · markov

A map can lose information without describing time evolution in this field theory

Reset and set BOTH Transition p · 0 → 1 to 0.2 and Transition q · 1 → 0 to 0.2, then Step once; predict state-one probability 0.2, below its stationary 0.5. Then Reset and set BOTH p to 0.5 and q to 0.5, then Step once; predict 0.5 and stationarity after one transition. The second stochastic matrix has rank one. This exact classical probability analogy does not represent the Ising defect, a quantum channel for the CFT, duality fusion, or quantum dimensions.

P(state 0)P(state 1)Stationary distribution

P(state 1) at step 0: 0%
Stationary P(state 1): 60%
Chain behavior: Monotone convergence

Transition p · 0 → 10.3Transition q · 1 → 00.2

t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forNoninvertibility is an algebraic property whose physical interpretation depends on the map being studied.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The line acts by zero on the sigma sector while acting nontrivially on other sectors. This kernel prevents an inverse without implying any failure of bulk quantum unitarity.

Your turn

Try it yourself

Restrict to orthonormal primary states in the order (1, epsilon, sigma). For the normalized vector v = (1,1,2)/√6, compute Dv and its squared norm using the displayed diagonal action. Why is that norm not automatically a transition probability?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Apply the three eigenvalues separately, and remember that a defect insertion is not stipulated to be a trace-preserving process.

Compare with the solution

The result is $Dv=(\sqrt2,-\sqrt2,0)/\sqrt6$ , so its squared norm is $(2+2)/6=2/3$ . The sigma component is annihilated. This is the norm of an unnormalized vector after an operator insertion, not automatically an outcome probability or deterministic evolution. Interpreting it operationally would require a specified measurement or channel normalization beyond the fusion algebra.

Keep the idea

Take it with you

A simple noninvertible Ising line differs from a reducible sum of group lines.
Its exact sector eigenvalues exhibit a kernel without compromising unitary bulk dynamics.
Continuum fusion, finite-lattice duality, and operational quantum maps require separate normalization and scope checks.

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