# Symmetry Protected Topological Order in Open Quantum Systems

Caroline de Groot  Max-Planck-Institut für Quantenoptik, Hans-Kopfermann-
Straße 1, 85748 Garching, Germany  Munich Center for Quantum Science and
Technology, Schellingstraße 4, 80799 München, Germany
<EMAIL_ADDRESS>Alex Turzillo  Max-Planck-Institut für
Quantenoptik, Hans-Kopfermann-Straße 1, 85748 Garching, Germany  Munich Center
for Quantum Science and Technology, Schellingstraße 4, 80799 München, Germany
<EMAIL_ADDRESS>Norbert Schuch  University of Vienna, Faculty of
Mathematics, Oskar-Morgenstern-Platz 1, 1090 Wien, Austria  University of
Vienna, Faculty of Physics, Boltzmanngasse 5, 1090 Wien, Austria  Max-Planck-
Institut für Quantenoptik, Hans-Kopfermann-Straße 1, 85748 Garching, Germany
Munich Center for Quantum Science and Technology, Schellingstraße 4, 80799
München, Germany<EMAIL_ADDRESS>
###### Abstract

We systematically investigate the robustness of symmetry protected topological
(SPT) order in open quantum systems by studying the evolution of string order
parameters and other probes under noisy channels. We find that one-dimensional
SPT order is robust against noisy couplings to the environment that satisfy a
strong symmetry condition, while it is destabilized by noise that satisfies
only a weak symmetry condition, which generalizes the notion of symmetry for
closed systems. We also discuss “transmutation” of SPT phases into other SPT
phases of equal or lesser complexity, under noisy channels that satisfy
twisted versions of the strong symmetry condition.

###### Contents

  1. 1 Introduction
  2. 2 SPT states and their invariants
    1. 2.1 SPTO of matrix product states
    2. 2.2 String order
    3. 2.3 Coherent SPT mixtures
  3. 3 A strong symmetry condition on channels
    1. 3.1 Weak and strong symmetry conditions
    2. 3.2 Charge conservation
    3. 3.3 Symmetric purifications
    4. 3.4 Symmetry conditions on Lindbladians
  4. 4 Strongly symmetric uncorrelated noise
    1. 4.1 Preservation of string order by strongly symmetric channels
    2. 4.2 Numerical study of string order under channels
    3. 4.3 Twisted sector charges
    4. 4.4 Protected edge modes
    5. 4.5 Irrep probabilities and SPT complexity
  5. 5 Causal channels
    1. 5.1 Tensor network representations of causal channels
    2. 5.2 Local realization of the symmetry conditions
    3. 5.3 String operators
    4. 5.4 Preservation of string order by strongly symmetric local Lindbladians
    5. 5.5 Protected edge modes
  6. 6 Transmutation of SPT phases by twisted symmetric channels
    1. 6.1 The action of endomorphisms on SPT phases
    2. 6.2 Patterns of zeros under endomorphisms
    3. 6.3 Twisted symmetric channels
    4. 6.4 Transmutation of SPT phases
  7. 7 Conclusion and outlook
  8. 8 Acknowledgements

##  1 Introduction

The realization that the interplay of symmetries and entanglement can give
rise to novel physics beyond the Landau paradigm has led to an expanding zoo
of topologically ordered phases of matter. A particularly prominent role among
those phases, in particular in one dimension (1D), is played by symmetry
protected topological (SPT) phases, which have their root in Haldane’s
original work elucidating the gapped nature of the spin-1 Heisenberg chain and
its topological origin, nowadays known as the Haldane phase [1, 2, 3]. SPT
phases do not possess intrinsic topological order as their ground states lack
long-range entanglement, yet they nevertheless exhibit topological phenomena
such as non-local order parameters.

SPT phases consist of systems with a unique ground state and a gap, yet which
are distinct from the trivial (mean-field) gapped phase, as witnessed by a
number of characteristic fingerprints: most prominently, string order [4, 5,
6], specific degeneracies in the entanglement spectrum [7], and fractionalized
edge excitations [3]. A key step toward the comprehensive understanding of SPT
phases was made by using Matrix Product State (MPS) representations [8, 9] of
their ground states. This step was based on the fact that MPS faithfully
approximate ground states of gapped systems [10, 11, 12] and that they allow
one to realize global symmetries locally on tensors that carry physical and
entanglement degrees of freedom [9, 13]. Namely, it was understood that in
nontrivial SPT phases, the physical symmetry that protects the phase acts on
entanglement as a projective, rather than linear, representation. This insight
was key in several ways. First, it provided a unified explanation for the
aforementioned fingerprints of SPT phases in terms of this projective action.
Second, it allowed one to obtain a comprehensive classification of SPT phases,
based on the classification of projective representations by group cohomology.
And finally, it connected the characterization of SPT phases based on
fingerprints like string order and edge modes to the characterization based on
the equivalence relation by which two systems are in the same phase if they
can be connected by a path of gapped, symmetric Hamiltonians [14, 15, 16].
Altogether, the representation of SPT states by MPS clarified and unified the
various definitions for SPT order (SPTO) for the ground states of gapped
Hamiltonian systems and allowed for their complete classification [17, 9].

The situation becomes much less clear when moving from pure ground states to
mixed states, which are the states we expect to appear in realistic physical
systems. Several questions arise. _First_ , which states should we consider?
Depending on the scenario, relevant states might be thermal states of
Hamiltonians [18, 19, 20], equilibrium states of dissipative evolutions
(steady states of Lindbladians) [21, 22, 23, 24, 25], or states – for
instance, originally pure states with SPT order – which have been subjected to
noise, which could be either Markovian or discrete-time non-Markovian noise.
_Second_ , what is the correct generalization of the symmetry condition? For
Lindbladian noise, at least two different symmetry conditions have been
considered [26, 27, 28, 29]; they differ in how the symmetry is imposed on the
joint system-bath interaction, and for discrete noise, even further symmetry
conditions are conceivable. _Third_ , which notion should one use for SPTO?
The various fingerprints of SPTO could give divergent results, or might even
be ill-defined, on mixed states. String order parameters can be defined for
any state, but it is a priori not clear whether the patterns they exhibit are
meaningful. For other fingerprints, such as entanglement spectra or edge
modes, it is even unclear how to define them for mixed states. _Fourth_ , are
any of these fingerprints, which are defined on individual systems, compatible
with the notion of SPT phase based on equivalence relations, analogous to
paths of symmetric gapped Hamiltonians? All in all, in the quest to understand
SPTO in the presence of noise, any approach must address these questions.

In this paper, we systematically investigate the robustness of SPTO under
various types of symmetric noise. To this end, we characterize SPTO through
string order parameters. These are constructed by placing local order
parameters (labeled by irreducible representations) at the endpoints of
strings of symmetry operators (labeled by group elements). In gapped phases,
any string order parameter either decays exponentially to zero as the
separation of the endpoints is increased, or converges to a constant whose
value depends on the specific order parameter chosen, and which is generically
nonzero. For ground states, the resulting pattern of zeros and nonzeros, as a
function of the irrep and group element labels of the string, is a fingerprint
of the SPT phase. In many cases, including all abelian symmetry groups, the
pattern is in one-to-one correspondence with the SPT phases protected by the
symmetry [6]. We say that a mixed state has some SPT order if it exhibits the
same pattern of zeros and nonzeros as pure states with the same SPTO; if a
symmetric mixed state (such as a mixture of different pure SPT phases)
exhibits a pattern which cannot appear in pure symmetric states, it is said to
have no SPTO at all (not even trivial SPTO). This definition has several
advantages: it coincides with the pure state definition in the limit of pure
states, and, being an expectation value of an operator with tensor product
structure, it is both simple to compute and to measure.

We study the robustness of SPTO, as witnessed by string order, for systems
subject to evolution by general symmetric and locality-preserving noise. We
consider both discrete-time evolutions, that is, quantum channels, as well as
continuous noise described by Lindbladians. The Lindbladian evolution forms a
special case of quantum channels, where the semigroup structure constrains the
possibilities for the action of the symmetry on the channels at finite times.
Locality-preservation encompasses both noise obtained from local Lindbladians
and locality-preserving evolutions which are not locally generated but which
appear, for example, at the boundaries of two-dimensional systems, in driven
systems, and via coupling to non-Markovian baths. We introduce two different
notions of symmetry of quantum channels – strong symmetry and weak symmetry.
Weakly symmetric channels are invariant under the symmetry action, which seems
the natural definition of a symmetric channel, and corresponds to symmetric
system-bath interactions where the symmetry acts simultaneously on the system
and the bath. Strongly symmetric channels, on the other hand, have the
property that each Kraus operator individually commutes with the symmetry up
to a constant phase factor, and correspond to symmetric system-bath
interactions where the symmetry only acts on the system. Specializing these
concepts to Lindbladian channels, where the semigroup structure imposes
additional restrictions, we recover the notions of strong and weak symmetry
studied by Albert and others [26, 27, 28, 29].

Our main result is that SPTO is robust against locality-preserving noise that
satisfies the stronger of these two symmetry conditions. To be precise, we
prove that the (local) strong symmetry condition on locality-preserving noise
is _sufficient_ to preserve SPTO (Lemma 2) and, conversely, that strong
symmetry is _necessary_ for channels generated in finite time by strictly
local Lindbladian evolution (Theorem 1), which we conjecture to hold for all
local Lindbladian evolutions (section 5.4). This result might appear
surprising in light of the work of Coser and Pérez-García, who show that
symmetric local Lindbladian noise, applied for a short amount of time,
destroys SPTO [30]. As we demonstrate, this is due to the fact that their
noise is only weakly symmetric, and not strongly symmetric. We thus find that
SPTO is robust to noise that satisfies a sufficiently strong yet natural
symmetry condition; namely that the system-bath coupling is invariant under
the symmetries acting on the system alone, as opposed to jointly on the system
and bath.

Finally, we extend our analysis to the scenario where the noise does not
commute with the symmetry but rather acts by exchanging symmetries, either
permuting them or identifying their group actions. We term channels with this
property _twisted strongly symmetric_. In the case of permutation of
symmetries, we demonstrate that the noise acts by permuting SPT phases with
the same complexity, whereas, in the other case, the noise reduces complexity
(Theorem 2). Symmetric channels with a nontrivial twist cannot be generated by
continuous evolution by a symmetric Lindbladian in finite time; thus, they are
only relevant to scenarios with discrete noise and to infinite time evolution.

The paper is structured as follows. In section 2, we review states with SPT
order, and how their SPT phase is characterized through string order
parameters. In section 3, we introduce the weak and strong symmetry conditions
on arbitrary quantum channels, discuss interpretations of the conditions in
terms of conservation laws, purifications, and couplings between the system
and the environment, and finally investigate the form of the symmetry
conditions when applied to Lindbladians. In section 4, we focus on
uncorrelated noise, for which there is a decomposition of the channel as a
tensor product over sites of the lattice. Several SPT order parameters –
string operators, twisted sector charges, edge modes, and irrep probabilities
– are studied analytically and numerically, and it is proven that strong
symmetry is necessary and sufficient for a noisy evolution to preserve SPTO.
In section 5, we discuss the extension of this result from uncorrelated noise
to causal (that is, locality-perserving) noise, which includes the case of
fast, local Lindbladians. In section 6, we broaden our investigation to causal
channels which act by exchanging symmetry actions. Channels satisfying
_twisted symmetry conditions_ are shown to transform between SPT phases, and
we state a necessary and sufficient condition for a channel to preserve a
given SPTO.

##  2 SPT states and their invariants

Let us begin by reviewing the invariants of pure SPT states, including their
manifestation in tensor networks and in patterns of zeros of string operators.
After this review, we will discuss how these invariants appear in a special
class of mixed states we dub _coherent SPT mixtures_.

###  2.1 SPTO of matrix product states

Symmetry protected topological phases of gapped, local Hamiltonians in one
dimension, for a symmetry group $G$, are classified by an invariant
$[\omega]$, a class in the second group cohomology group $H^{2}(G,U(1))$ [14,
15, 16]. A useful method for determining the SPT invariant of a given
Hamiltonian is to represent its ground state as a tensor network and study the
symmetries of its tensor [9]. Let us review this procedure.

Our analysis of one-dimensional SPT states makes use of matrix product states
(MPS), which efficiently approximate states obeying an entanglement entropy
area law, and as such are applicable to ground states of gapped, local
Hamiltonians [10, 31]. A translation-invariant MPS is defined by a single
rank-three tensor $A$ as

$\ket{\Psi[A]}=\sum_{i_{1},\dots,i_{N}}\Tr(A^{i_{1}}\dots
A^{i_{N}})\ket{i_{1}\dots i_{N}},$ (2.1)

where $A^{i}$ are matrices such that $A=\sum_{i}A^{i}\otimes\ket{i}$. As a
tensor diagram, the MPS is written as

$\begin{split}\includegraphics[height=22.76411pt]{MPS1.PNG}\quad\raisebox{5.0561pt}{.}\end{split}$
(2.2)

An MPS tensor is said to be _injective_ if its transfer matrix
$T=\sum_{i}A^{i}\otimes\overline{A^{i}}$ has a nondegenerate eigenspace of
highest weight. For a refresher on MPS technology, we refer readers to more
comprehensive introductory literature [32, 33, 34] as well as to a recent
review [9].

Injective MPS satisfy a fundamental theorem [9], which implies the following.
If a state is invariant under a global symmetry $U_{g}^{\otimes
N}\ket{\Psi[A]}=\ket{\Psi[A]}$, the action of the onsite symmetry $U_{g}$ on
its local tensor $A$ results in an action on the virtual level

$\sum_{j}(U_{g})_{ij}A^{j}=e^{i\phi_{g}}V_{g}A^{i}V_{g}^{\dagger}~{},$ (2.3)

where $V_{g}$ is a projective representation of the symmetry [35], satisfying

$V_{g}V_{h}=\omega(g,h)\,V_{gh}~{},$ (2.4)

for some values $\omega(g,h)$. As a tensor diagram, Eq. (2.3) is written as

$\begin{split}\includegraphics[height=27.82024pt]{MPS2.PNG}\quad\raisebox{12.1424pt}{.}\end{split}$
(2.5)

For a given state, there is always a tensor in a canonical form, where $V_{g}$
is unitary and $\omega(g,h)$ is a phase [9]. The collection of phases
$\omega(g,h)$ constitutes a group cocycle and is defined up to a group
coboundary, meaning it determines a cohomology class $[\omega]$ [16, 15]. It
turns out that $[\omega]$ is invariant along smooth paths of gapped, local,
symmetric Hamiltonians, which is to say it is an SPT phase invariant;
moreover, it is a complete invariant [16]. Physically, the virtual space of
the MPS tensor may be interpreted as the space of edge modes; the fact that
there is a minimal bond dimension on which $V_{g}$ can realize the invariant
$[\omega]$ means that some of the edge modes are protected by the symmetry.

Essential to this definition of the SPT invariant $[\omega]$ is that the state
is well-approximated by an MPS of bond dimension constant in the system size,
a property which comes from the state being a ground state of a gapped, local
Hamiltonian. Generic states in one dimension are not well-approximated by MPS,
so for them a projective action $V_{g}$ – and therefore an invariant
$[\omega]$ – cannot be defined this way.

###  2.2 String order

String order provides an alternate definition of the SPT invariant that does
not rely on an MPS representation of the state. The string order parameter is
a set of expectation values of string operators that can be defined on any
state. On certain well-behaved states, such as the ground states of gapped
Hamiltonians, it yields a well-defined pattern of zeros that is related to the
SPT invariant $[\omega]$.

For the remainder of the paper, we take $G$ to be a finite abelian group. This
assumption ensures that string order is a complete invariant of SPT phases, in
that it uniquely determines $[\omega]$.

Assume $G$ is a finite abelian group, and let $U_{g}$ denote the action of $G$
on an individual site. The string operator is defined as

$s(g,O^{l}_{\alpha},O^{r}_{\alpha})=\mathds{1}\otimes O^{l}_{\alpha}\otimes
U_{g}^{\otimes j}\otimes O^{r}_{\alpha}\otimes\mathds{1}~{},$ (2.6)

for some length $j$, where the end operators $O^{l,r}$ live in _opposite_
irreps of the adjoint action

$U_{h}^{\dagger}O^{l}_{\alpha}U_{h}=\chi_{\alpha}(h)O^{l}_{\alpha}~{},\qquad
U_{h}^{\dagger}O^{r}_{\alpha}U_{h}=\chi_{\alpha}^{*}(h)O^{r}_{\alpha}~{}.$
(2.7)

On some states, the string order parameter obeys a selection rule [6]. This
rule says that, for each $g\in G$, there is a unique character $\alpha_{g}$
such that the expectation value $\langle
s(g,O_{\alpha}^{l},O_{\alpha}^{r})\rangle$ vanishes, for all end operators,
except for $\alpha=\alpha_{g}$. The values forced to vanish by the selection
rule form what is called the _pattern of zeros_ of the state. An SPT state can
be defined as a state for which this selection rule holds; that is, a state
with a well-defined pattern of zeros. The SPT invariant of an SPT state is
extracted by defining the ratios $\omega/\omega$ in terms of the unique
character $\alpha_{g}$ for each $g$ as follows:

$\frac{\omega(h,g)}{\omega(g,h)}=\chi_{\alpha_{g}}(h)~{}.$ (2.8)

The invariant $[\omega]$ may then be recovered from these ratios, as we argue
in section 6.2. We remark that string order is also defined when $G$ is
nonabelian; however, it may not determine $[\omega]$ uniquely [6].

Now let us consider a state represented as an MPS and show that the definition
of the SPT invariant $[\omega]$ by the projective action on edge modes agrees
with its definition by string order parameters [6]. Injectivity may be
achieved by blocking, which does not change the form of the string operator as
long as its length is assumed to be large compared to the size of the blocks.
On such a state, the string operator evaluates to

$\langle
s(g,O^{l}_{\alpha},O^{r}_{\alpha})\rangle=E_{l}(g,O_{\alpha}^{l})E_{r}(g,O_{\alpha}^{r})~{},$
(2.9)

where

$E_{l,r}(g,O_{\alpha}^{l,r})=\Tr[N_{l,r}^{g}O_{\alpha}^{l,r}]$ (2.10)

are defined as

$\begin{split}\includegraphics[height=45.52824pt]{stringop1.PNG}\quad\raisebox{20.2322pt}{.}\end{split}$
(2.11)

Here, $\rho_{R}$ is the unique (by injectivity) right fixed point of the MPS
transfer matrix, and we have used the canonical form where the left fixed
point is the identity. The evaluation (2.9) can be seen with the following
diagrammatic argument (due to Ref. [6]):

$\begin{split}\includegraphics[height=45.52824pt]{sop2.PNG}\quad\raisebox{20.2322pt}{.}\end{split}$
(2.12)

Evaluating the above

$\begin{split}\includegraphics[height=45.52824pt]{stringop3b.PNG}\quad\raisebox{20.2322pt}{.}\end{split}$
(2.13)

The operators $N_{l,r}^{g}$ transform as

$U_{h}^{\dagger}N_{l}^{g}U_{h}=\frac{\omega(g,h)}{\omega(h,g)}N_{l}^{g}~{},\qquad
U_{h}^{\dagger}N_{r}^{g}U_{h}=\left(\frac{\omega(g,h)}{\omega(h,g)}\right)^{*}N_{r}^{g}~{},$
(2.14)

The operators $N_{l,r}^{g\dagger}$ and $O_{\alpha}^{l,r}$ are orthogonal (and
so $E_{l,r}$ is zero) unless they transform the same way; that is, unless the
selection rule (2.8) is satisfied. This completes the argument.

If the operators $N_{l,r}^{g\dagger}$ and $O_{\alpha}^{l,r}$ transform the
same way, then they are not orthogonal _generically_ (and so $E_{l,r}$ is
nonzero generically). This is because $N_{l,r}^{g}$ picks out a single
direction in the multiplicity space for $\alpha$, so the subspace orthogonal
to this direction is codimension one in the full space of end operators
$O_{\alpha}^{l,r}$. In other words, for a generic choice of end operators, the
only expectation values $\langle s(g,O_{\alpha}^{l},O_{\alpha}^{r})\rangle$
that vanish are those that belong to the pattern of zeros determined by the
SPT invariant. The nonzero values of the string order parameter depend on the
choice of end operators, but the pattern of zeros does not.

###  2.3 Coherent SPT mixtures

Now we turn to SPT invariants of open systems. In the formalism of Lindbladian
evolution [30], phases of open systems are defined in terms of states, as
opposed to with some open systems analog of gapped paths of Hamiltonians. We
do not attempt to answer the questions of which mixed states generalize the
ground states of gapped, local Hamiltonians and what is their phase
classification under an appropriate equivalence relation. Instead we seek to
motivate the strong symmetry condition on Lindbladian evolution by focusing on
a special class of mixed states for which we can define an invariant:

Definition: A coherent SPT mixture is a mixed state with a well-defined
pattern of zeros. (2.15)

Such states have a well-defined SPT invariant $[\omega]$ that can be extracted
from the pattern of zeros, as discussed above. Mixed states that are ensembles

$\rho=\sum_{i}p_{i}|\psi_{i}^{\omega}\rangle\langle\psi_{i}^{\omega}|$ (2.16)

of SPT pure states $|\psi_{i}^{\omega}\rangle$ all in the same SPT phase
$[\omega]$ are examples of coherent SPT mixtures. We leave open the
possibility that there exist exotic coherent SPT mixtures that are not covered
by this example.111Components of the ensemble that have a different SPTO or no
SPTO at all could cancel exactly the expectation values of the string
operators, yielding a well-defined pattern of zeros.

The main claim of the paper is that translation-invariant pure SPT states (and
more generally, states of the form (2.16) where each component
$|\psi_{i}^{\omega}\rangle$ is translation-invariant) are transformed into
coherent SPT mixtures (2.15) with the same SPT invariant by a Lindbladian
evolution if and only if the evolution is strongly symmetric. We prove this
claim for uncorrelated noise in section 4 and show its ‘if’ direction (while
conjecturing its ‘only if’ direction) for fast, local Lindbladians in section
5. In particular, this result means that SPTO of pure states is robust in open
systems described by strongly symmetric Lindbladians.

##  3 A strong symmetry condition on channels

We begin by introducing the _weak_ (3.2) and _strong_ (3.8) _symmetry
conditions_ and discussing their various formulations. The latter is motivated
by showing that weak symmetry is insufficient to preserve SPTO. The argument
that strong symmetry is necessary and sufficient to preserve SPTO is reserved
for section 4, section 5.

The symmetry conditions are first formulated and studied for arbitrary quantum
channels. Then in section 3.4, for the particularly important case of
Lindbladian evolution $\mathcal{E}_{t}=e^{t\mathcal{L}}$, the conditions are
reformulated in terms of $\mathcal{L}$; the Lindbladian formulations have been
discussed previously [26, 27, 28, 29]. The definitions and results in this
section apply to general systems on finite-dimensional Hilbert spaces, not
just spin chains; $U_{g}$ denotes the action of the symmetry on the full
system, not on a single site of a spin chain.

###  3.1 Weak and strong symmetry conditions

A channel $\mathcal{E}$ is said to satisfy the _weak symmetry (WS) condition_
if it commutes, as a superoperator, with the symmetry-implementing channels

$\mathcal{U}_{g}(\rho)=U_{g}\rho\,U_{g}^{\dagger}~{};$ (3.1)

that is, if

$\boxed{\qquad\mathcal{U}_{g}\circ\mathcal{E}\circ\mathcal{U}_{g}^{\dagger}=\mathcal{E}~{},\quad\forall\,g~{}.\qquad(\text{weak
symmetry condition})\qquad}$ (3.2)

The channel $\mathcal{E}$ can be expressed in terms of a Kraus representation,
$\mathcal{E}(\rho)=\sum K_{i}\rho K_{i}^{\dagger}$, where we can interpret the
$K_{i}$ as representing different trajectories. In terms of a Kraus
representation of $\mathcal{E}$, the weak symmetry condition reads

$\sum_{i}(U_{g}K_{i}U_{g}^{\dagger})\,\rho\,\,(U_{g}K_{i}U_{g}^{\dagger})^{\dagger}=\sum_{i}K_{i}\rho\,K_{i}^{\dagger}~{},\quad\forall\,g~{}.$
(3.3)

Since $K_{i}$ and $U_{g}K_{i}U_{g}^{\dagger}$ define Kraus representations of
the same channel, they are related by a unitary $x^{g}$ [36]:

$U_{g}K_{i}U_{g}^{\dagger}=\sum_{j}x_{ji}^{g}K_{j}~{},\quad\forall\,i,g~{}.$
(3.4)

Since $U_{g}$ forms a representation of $G$, so does $x^{g}$:

$\sum_{k}x_{ki}^{gh}K_{k}=U_{gh}K_{i}U_{gh}^{\dagger}=U_{g}U_{h}K_{i}U_{h}^{\dagger}U_{g}^{\dagger}=\sum_{jk}x_{ji}^{h}x_{kj}^{g}K_{k}\
,\quad\forall\,i,g~{}.$ (3.5)

In a basis of Kraus operators $K_{i}^{g}$ that diagonalizes $x^{g}$ as a
collection of phases $\theta_{i}(g)$, (3.4) amounts to

$U_{g}K_{i}^{g}U_{g}^{\dagger}=e^{i\theta_{i}(g)}K_{i}^{g}~{},\quad\forall\,i,g~{}.$
(3.6)

The existence of a basis $K_{i}^{g}$, for each $g$, such that this relation
holds is equivalent to the WS condition.

Observe that the phases $\theta_{i}$ in the WS condition (3.6) may differ
across the trajectories (labeled by $i$). It will be demonstrated in section 4
that this interference between the trajectories is the source of the
destruction of SPTO, as eliminating it by setting the phases equal is
sufficient to ensure that a channel preserves SPTO. Let us now take the phases
to be equal: $\theta_{i}(g)=\theta(g)$ for all $i,g$. Under this restriction,
the condition (3.6) is independent of the basis of Kraus operators because
$K_{i}^{g}{}^{\prime}=\sum_{j}v_{ij}K_{j}^{g}$ (for any unitary $v_{ij}$)
satisfies

$U_{g}K_{i}^{g}{}^{\prime}U_{g}^{\dagger}=\sum_{j}v_{ij}U_{g}K_{j}^{g}U_{g}^{\dagger}=e^{i\theta(g)}\sum_{j}v_{ij}K_{j}^{g}=e^{i\theta(g)}K_{i}^{g}{}^{\prime}~{}.$
(3.7)

Basis-independence means we can also drop the group label on the Kraus
operators: $K_{i}^{g}=K_{i}$ for all $i,g$. We arrive at what we call the
_strong symmetry (SS) condition_ :

$\boxed{\qquad
U_{g}K_{i}U_{g}^{\dagger}=e^{i\theta(g)}K_{i}~{},\quad\forall\,i,g~{}.\qquad(\text{strong
symmetry condition})\qquad}$ (3.8)

Note that there is no distinction between the WS and SS conditions for a
reversible (unitary) channel, as such a channel is realized by a single Kraus
operator, and so it has only a single phase $\theta(g)$.

By Schur’s lemma, the SS condition (3.8) may be restated as the condition that
each $K_{i}$ is block-diagonal in the irrep basis:
$K_{i}=\oplus_{\alpha}K_{i}^{\alpha}$, where $K_{i}^{\alpha}$ acts on the
multiplicity space of the (isomorphism class of the) irrep $\alpha$. The
completeness relation $\sum_{i}K_{i}^{\dagger}K_{i}=\mathds{1}$ is equivalent
to a completeness relation on each block, so the channel decomposes as
$\mathcal{E}=\oplus_{\alpha}\mathcal{E}_{\alpha}$, where
$\mathcal{E}_{\alpha}$ is the channel with Kraus operators $K_{i}^{\alpha}$.
This decomposition is a stronger constraint than the decomposition of WS
channels, which, when viewed as matrices on the space of operators, have a
block-diagonal form $\mathcal{E}=\sum_{\alpha}\Phi_{\alpha}$ in the irrep
basis of the action $U_{g}\otimes U_{g}^{\dagger}$; note that the operators
$\Phi_{\alpha}$ are different from $\mathcal{E}_{\alpha}$ and are not
themselves channels.

###  3.2 Charge conservation

The strong symmetry condition may be alternatively characterized as

$\boxed{\qquad\mathcal{E}^{\dagger}(U_{g})=e^{i\theta(g)}U_{g}~{},\quad\forall\,g~{}.\qquad(\text{strong
symmetry condition})\qquad}$ (3.9)

where the dual channel $\mathcal{E}^{\dagger}$ is the channel with Kraus
operators $K_{i}^{\dagger}$, $\mathcal{E}^{\dagger}(X)=\sum
K_{i}^{\dagger}XK_{i}$. This alternative statement may be interpreted as
conservation of symmetry charge. The charge of a state under a symmetry $g$ is
the expectation value $\langle U_{g}\rangle_{\rho}=\Tr[\rho\,U_{g}]$ of the
operator $U_{g}$ on the state, and the strong symmetry condition means this
expectation value is the same (up to a phase) for $\rho$ and
$\mathcal{E}(\rho)$:

$\langle
U_{g}\rangle_{\mathcal{E}(\rho)}=\langle\mathcal{E}^{\dagger}(U_{g})\rangle_{\rho}\stackrel{{\scriptstyle
SS}}{{=}}e^{i\theta(g)}\langle U_{g}\rangle_{\rho}~{}.$ (3.10)

The connection between strong symmetry and conservation laws has been noted
previously [26].

The equivalence of the two statements may be seen as follows. If a channel
satisfies Eq. (3.8), then

$\mathcal{E}^{\dagger}(U_{g})=\sum_{i}K_{i}^{\dagger}U_{g}K_{i}=e^{i\theta(g)}\sum_{i}K_{i}^{\dagger}K_{i}U_{g}=e^{i\theta(g)}U_{g}~{},$
(3.11)

which is Eq. (3.9). For the converse, we need a lemma: if
$\mathcal{E}^{\dagger}(X)=Y$ and
$\mathcal{E}^{\dagger}(X^{\dagger}X)=Y^{\dagger}Y$, then $XK_{i}=K_{i}Y$. If
this is true, the statement that Eq. (3.9) implies Eq. (3.8) follows from
taking $X=U_{g}$ and $Y=e^{i\theta(g)}U_{g}$. The lemma is proved by borrowing
the argument for Theorem 6.13 of Ref. [37]:

$\sum_{i}(XK_{i}-K_{i}Y)^{\dagger}(XK_{i}-K_{i}Y)=\mathcal{E}^{\dagger}(X^{\dagger}X)-\mathcal{E}^{\dagger}(X^{\dagger})Y-Y^{\dagger}\mathcal{E}^{\dagger}(X)+Y^{\dagger}\mathcal{E}^{\dagger}(\mathds{1})Y=0~{}.$
(3.12)

Then, since the left hand side is a sum of positive terms, each of them must
individually vanish: $XK_{i}=K_{i}Y$.

###  3.3 Symmetric purifications

The symmetry conditions can be restated in terms of purifications:

Claim: A channel is weakly symmetric if it has a purification to a unitary
that commutes, up to a phase, with some diagonal symmetry $U_{g}\otimes
U_{g}^{A}$, for which the action $U_{g}^{A}$ on the ancillary space leaves the
ancilla state invariant. (3.13) Claim: A channel is strongly symmetric if and
only if it has a purification to a unitary that commutes, up to the phase
$e^{i\theta(g)}$, with the symmetry $U_{g}\otimes\mathds{1}^{A}$. (3.14)

We do not prove a converse to the first claim, though we expect it or a
similar statement to hold. The second claim means that one may take
$U_{g}^{A}=\mathds{1}^{A}$ precisely when the channel is strongly symmetric.
These statements will come in handy in section 5, when we discuss causal
channels in terms of their purifications to matrix product unitaries. The
statements also have interpretations in terms of couplings between the system
and environment, which we discuss here. First let us review the basics of
purifications and justify the claims.

Let $\mathcal{E}$ be a channel on a system with Hilbert space ${\mathcal{H}}$.
A purification of $\mathcal{E}$ is a unitary $W$ on a space
${\mathcal{H}}\otimes A$ – the original space appended with an ancillary space
– such that, for some ancilla state $|a\rangle\in A$,

$\Tr_{A}(W(\rho\otimes|a\rangle\langle a|)W^{\dagger})=\mathcal{E}(\rho)~{}.$
(3.15)

A purification $W$ always exists. Given a set of Kraus operators $K_{i}$,
indexed in a set $\mathcal{I}$, form the ancillary space $A$ spanned by an
orthonormal basis $|e_{i}\rangle$, $i\in\mathcal{I}$ and the operator
$V:{\mathcal{H}}\rightarrow{\mathcal{H}}\otimes A$ that acts as

$V:|\psi\rangle\mapsto\sum_{i}K_{i}|\psi\rangle\otimes|e_{i}\rangle~{}.$
(3.16)

The operator $V$ is called a Stinespring dilation of $\mathcal{E}$ and is an
isometry since

$\langle\phi|V^{\dagger}V|\psi\rangle=\sum_{ij}\langle
e_{i}|\langle\phi|K_{j}^{\dagger}K_{i}|\psi\rangle|e_{j}\rangle=\sum_{i}\langle\phi|K_{i}^{\dagger}K_{i}|\psi\rangle=\langle\phi|\psi\rangle~{}.$
(3.17)

Then use the fact that any isometry $V$ on
$\mathcal{H}\cong\mathcal{H}\otimes\lvert a\rangle$, $\lvert a\rangle\in A$,
can always be extended to a unitary $W$ on ${\mathcal{H}}\otimes A$.
Conversely, if we expand the expression (3.15) in an orthonormal basis
$|e_{i}\rangle$ of $A$ to obtain

$\sum_{i}\langle e_{i}|W|a\rangle\rho\langle
a|W^{\dagger}|e_{i}\rangle=\mathcal{E}(\rho)~{},$ (3.18)

we see that $K_{i}:=\langle e_{i}|W|a\rangle$ are candidates for a set of
Kraus operators for $\mathcal{E}$. To see that they are actually Kraus
operators, check completeness:

$\sum_{i}K_{i}^{\dagger}K_{i}=\sum_{i}\langle
a|W^{\dagger}|e_{i}\rangle\langle e_{i}|W|a\rangle=\langle
a|(\mathds{1}\otimes\mathds{1}^{A})|a\rangle=\mathds{1}~{}.$ (3.19)

The purification can be expressed diagrammatically as a tensor

$\begin{split}\includegraphics[height=73.3562pt]{purification1.PNG}\quad\raisebox{32.88408pt}{.}\end{split}$
(3.20)

The claim (3.13) about weak symmetry (3.2) may be expressed as

$\begin{split}\includegraphics[height=53.11629pt]{purification_ws.PNG}\raisebox{22.76411pt}{,}\end{split}$
(3.21)

while the claim (3.14) about strong symmetry (3.8) may be expressed as

$\begin{split}\includegraphics[height=53.11629pt]{purification_ss.PNG}\raisebox{22.76411pt}{.}\end{split}$
(3.22)

Let us now prove the claim (3.13). Suppose $W$ is symmetric with the symmetry
$U_{g}\otimes U_{g}^{A}$ and the state $\lvert a\rangle$ is invariant:
$U_{g}^{A}|a\rangle=|a\rangle$. Then the $K_{i}=\langle e_{i}|W|a\rangle$
satisfy weak symmetry (3.4):

$\begin{split}\includegraphics[height=45.52824pt]{purification3.png}\raisebox{20.2322pt}{,}\end{split}$
(3.23)

where $x^{g}_{ji}$ are the matrix elements $x^{g}_{ji}=e^{i\theta(g)}\langle
e_{i}|(U_{g}^{A})^{\dagger}|e_{j}\rangle=e^{i\theta(g)}(U_{g}^{A})_{ji}^{*}$.

One direction of the claim (3.14) follows from a similar argument. Suppose $W$
is symmetric with $U_{g}\otimes\mathds{1}^{A}$. Then the $K_{i}=\langle
e_{i}|W|a\rangle$ satisfy the strong symmetry condition (3.8):

$\begin{split}\includegraphics[height=45.52824pt]{purification4.PNG}\raisebox{20.2322pt}{.}\end{split}$
(3.24)

Conversely, suppose the Kraus operators satisfy the SS condition and construct
a symmetric $W$ as follows. Without loss of generality, take $|a\rangle$ to be
$|e_{1}\rangle$. A unitary extension of the Stinespring dilation $V$ is a
square matrix $W$ consisting of blocks $K_{i}^{j}$, where the blocks of the
first column are the Kraus operators $K_{i}^{1}:=K_{i}$, and we choose the
remaining blocks so that $W$ is unitary:

$\sum_{i}K_{i}^{j\dagger}K_{i}^{k}=\delta^{jk}\mathds{1}_{\mathcal{H}}~{}.$
(3.25)

The remaining blocks may be chosen to be symmetric (so that $W$ is symmetric)
as follows. Build linear independent columns by adding signs like
$K_{i}^{j}:=(-1)^{\delta(i<j)}K_{i}$, then make them orthogonal by applying
the Gram-Schimdt process, and finally normalize. The result is manifestly
symmetric.

The claims (3.13) and (3.14) may be interpreted in terms of the coupling
between the system and the environment (the ancillary space). Suppose $W$
represents unitary evolution by a Hamiltonian:

$W=e^{-itH/\hbar}~{},\qquad H=\sum_{i}H_{i}^{S}\otimes H_{i}^{E}~{},$ (3.26)

where the $H_{i}^{S}$ and $H_{i}^{E}$ are each assumed to be linearly
independent. If the unitary evolution $W$ satisfies the formulation of the
weak symmetry condition in the claim (3.13) at all times $t$, then
$U_{g}\otimes U_{g}^{A}$ is a symmetry of $H$. If the unitary evolution $W$
satisfies the strong symmetry condition (3.14) at all times $t$, then

$0=(U_{g}\otimes\mathds{1}^{A})H-e^{i\theta(g)}H(U_{g}\otimes\mathds{1}^{A})=\sum_{i}(U_{g}H_{i}^{S}-e^{i\theta(g)}H_{i}^{S}U_{g})\otimes
H_{i}^{E}~{},\,\forall\,g~{},$ (3.27)

which, since the $H_{i}^{E}$ are linearly independent, implies
$U_{g}H_{i}^{S}=e^{i\theta(g)}H_{i}^{S}U_{g}~{},\,\forall\,i,g$. By
conjugating both sides of the equation, we find that
$e^{i\theta(g)}=e^{-i\theta(g)}$, so $\theta(g)=0,\pi$. Since $W$ is a
continuous function of $t$, the phase $\theta(g)$ must vary continuously from
zero at $t=0$ to its values at nonzero times (this can be formalized by
including the constant order in the expansion of (3.26) in the symmetry
condition (3.27)); this means it must be zero at all times. Therefore, for
channels arising from a continuous coupling of system to environment,

$U_{g}H_{i}^{S}=H_{i}^{S}U_{g}~{},\,\forall\,i,g~{},$ (3.28)

which is to say that the system alone, rather than merely its composite with
the environment, is symmetric.

###  3.4 Symmetry conditions on Lindbladians

Let us now discuss semigroups of channels generated by continuous time
evolution by a Lindbladian

$\mathcal{L}(\rho)=-\frac{i}{\hbar}[H^{S},\rho]+\sum_{i=1}^{\ell}\left(L_{i}\rho
L_{i}^{\dagger}-\frac{1}{2}L_{i}^{\dagger}L_{i}\rho-\frac{1}{2}\rho
L_{i}^{\dagger}L_{i}\right)~{}.$ (3.29)

Here, the $L_{i}$ are jump operators and $H^{S}$ is the Hamiltonian of the
system. If a semigroup consists of channels satisfying the weak or strong
symmetry condition, the Lindbladian generating it satisfies, respectively,

$\boxed{\qquad\mathcal{U}_{g}\circ\mathcal{L}\circ\mathcal{U}_{g}^{\dagger}=\mathcal{L}~{},\quad\forall\,g~{}.\qquad\text{(WS
condition on $\mathcal{L}$)}\qquad}$ (3.30) $\boxed{\qquad
U_{g}L_{i}=L_{i}U_{g}~{},\qquad
U_{g}H^{S}=H^{S}U_{g}~{},\qquad\forall\,i,g~{}.\qquad\text{(SS condition on
$\mathcal{L}$)}\qquad}$ (3.31)

In particular, this implies that semigroups of strongly symmetric channels
generated by Lindbladian evolution necessarily have $\theta(g)=0,\,\forall\,g$
at all times.

To see the weak symmetry condition (3.30), observe that the channels
$\mathcal{E}_{t}=e^{t\mathcal{L}}$ commute with $\mathcal{U}_{g}$ at all times
$t$ if and only if $\mathcal{L}$ commutes with $\mathcal{U}_{g}$. To see the
strong symmetry condition (3.31), observe that

$\displaystyle\begin{split}\mathcal{E}_{\delta t}(\rho)&=\rho+\delta
t\,\mathcal{L}(\rho)\\\ &=\rho+\delta
t\,\left[\left(-\frac{i}{\hbar}H^{S}-\frac{1}{2}\sum_{i=1}^{\ell}L^{\dagger}_{i}L_{i}\right)\rho+\rho\left(\frac{i}{\hbar}H^{S}-\frac{1}{2}\sum_{i=1}^{\ell}L^{\dagger}_{i}L_{i}\right)\right]+\sum_{i=1}^{\ell}\delta
t\,L_{i}\rho L_{i}^{\dagger}~{},\end{split}$ (3.32)

at small times $\delta t$, and thus

$\mathcal{E}_{\delta t}(\rho)=\sum_{i=0}^{\ell}K_{i}(\delta
t)\,\rho\,K_{i}(\delta t)^{\dagger}$ (3.33)

with Kraus operators

$K_{0}(\delta t)=\mathds{1}+\delta
t\,\left(-\frac{i}{\hbar}H^{S}-\frac{1}{2}\sum_{i=1}^{\ell}L^{\dagger}_{i}L_{i}\right)~{},\qquad
K_{i>0}(\delta t)=\sqrt{\delta t}\,L_{i}\ .$ (3.34)

This relationship between Kraus operators and jump operators lets us translate
our strong symmetry condition (3.8) on channels into the strong symmetry
condition (3.31) on the Lindbladians that generate them: First, from the
commutation relation $U_{g}K_{i>0}=e^{i\theta(g)}K_{i>0}U_{g}$, we infer that
$U_{g}L_{i}=e^{i\theta(g)}L_{i}U_{g}$. It follows that $X:=\mathds{1}-\delta
t\,\tfrac{1}{2}\sum L_{i}^{\dagger}L_{i}$ commutes with $U_{g}$. Second, from
the commutation relation $U_{g}K_{0}=e^{i\theta(g)}K_{0}U_{g}$, we get
$U_{g}(-(i/\hbar)H^{S}\delta
t+X)U_{g}^{\dagger}=e^{i\theta(g)}(-(i/\hbar)H^{S}\delta t+X)$. Taking the
Hermitian part and using that $X$ commutes with $U_{g}$, we obtain
$X=U_{g}XU_{g}^{\dagger}=\cos\theta(g)\,X+\sin\theta(g)\,(\delta
t/\hbar)\,H^{S}$, or

$(1-\cos\theta(g))X=\sin\theta(g)\,(\delta t/\hbar)\,H^{S}\ .$ (3.35)

As this must hold for all $g\in G$ simultaneously, the proportionality factor
$(1-\cos\theta(g))/\sin\theta(g)$ must not depend on $g$, which is only
possible if $\theta(g)$ is constant. As $\theta(g)$ is a representation, this
implies $\theta(g)\equiv 0$ for all $g\in G$ (and in particular, the above
equation then imposes no constraints on $X$ and $H^{S}$).222There is an
alternative topological argument for $\theta(g)=0$. Since the $K_{i}(t)$ are
continuous functions in $t$, the phase $\theta(g)$ in their commutation
relations must also be continuous in $t$. But $e^{i\theta}$ is a one-
dimensional representation of $G$, and there are only discretely many of these
(even if $G$ is continuous), so $\theta(g)$ must be a constant function in
$t$. Then, since $\theta(g)=0$ for all $g$ at $t=0$ (because
$W(t=0)=\mathds{1}$), the condition $\theta(g)=0$ must also be true at all
times $t$. It thus follows that both $L_{i}$ and $H^{S}$ must commute with
$U_{g}$. We conclude that a Lindbladian generates a family of SS channels if
and only if it satisfies the condition (3.31).

We also obtain a characterization of strong symmetry for Lindbladians by
applying the charge conservation condition (3.9) to $\mathcal{E}_{\delta
t}=e^{\delta t\mathcal{L}}$ (using that we now know that $\theta(g)\equiv 0$
for Lindbladian channels):

$U_{g}=\mathcal{E}_{\delta t}^{\dagger}(U_{g})=U_{g}+\delta
t\,\mathcal{L}^{\dagger}(U_{g})\ ,$ (3.36)

and thus

$\boxed{\qquad\mathcal{L}^{\dagger}(U_{g})=0~{}.\qquad\text{(SS condition on
$\mathcal{L}$)}\qquad}$ (3.37)

The Lindblad master equation (3.29) can also be recovered from the Hamiltonian
that couples the system to the environment, under the Born (weak coupling,
large environment) and Markov (memoryless environment) approximations. We
refer readers to Chapter 6.2.1 of Ref. [38] for a detailed analysis of this
procedure. The jump operators appear in the coupling Hamiltonian as

$H=H^{S}\otimes\mathds{1}^{E}+\mathds{1}^{S}\otimes
H^{E}+\sum_{i>0}L_{i}\otimes B_{i}~{}.$ (3.38)

In this picture, the strong symmetry condition (3.31) on Lindbladians is
equivalent to our previous result about the strong symmetry condition on
purifications (3.28).

####  3.4.1 Destruction of SPTO by weakly symmetric coupling

Let us now demonstrate that having merely weakly symmetric noise is
insufficient to preserve SPTO. Specifically, the fast, local Lindbladian
evolution defined by Coser and Pérez-García [30], which they showed to destroy
SPTO, is weakly symmetric. However, as we will also see, it lacks the strong
symmetry condition, leaving open the possibility that the latter preserves
SPTO.

The local Lindbladian of Ref. [30] is given as a sum over sites on a spin
chain

$\mathcal{L}=\sum_{s}\mathcal{L}_{s}~{},$ (3.39)

where

$\mathcal{L}_{s}={\mathcal{T}}_{s}-\mathds{1}_{s}~{},\qquad{\mathcal{T}}_{s}(\rho)=\Tr_{s}[\rho]|\phi\rangle_{s}\langle\phi|$
(3.40)

for some single-site state $|\phi\rangle$. It is shown to drive any one-
dimensional SPT state toward the product state $|\phi\rangle^{\otimes L}$,
approximating it well in short time. This result suggests that no SPTO is
robust in open systems.

Any time $\mathcal{L}$ is given as a sum of single site terms (3.39), the SS
condition (3.37) reads

$0=\mathcal{L}^{\dagger}(U_{g}^{\otimes
L})=\sum_{s}\mathcal{L}_{s}^{\dagger}(U_{g}^{\otimes
L})=\sum_{s}\mathcal{L}_{s}^{\dagger}(U_{g}^{(s)}\otimes\mathds{1}^{(L\backslash
s)})\otimes U_{g}^{(L\backslash s)}~{},$ (3.41)

where $U_{g}$ now denotes the action of the symmetry on a single site. This
condition is equivalent to each of the single site terms satisfying the local
condition
$\mathcal{L}_{s}^{\dagger}(U_{g}^{(s)}\otimes\mathds{1}^{(L\backslash s)})=0$.
The channel (3.40) has
$\mathcal{L}_{s}^{\dagger}(X)=\langle\phi|X|\phi\rangle_{s}\otimes\mathds{1}^{(L\backslash
s)}-X$, so it fails this condition for any $g\neq 1$ and therefore is not SS.
On the other hand, $\mathcal{L}$ commutes with $\mathcal{U}_{g}$ and so is WS,
as long as $|\phi\rangle$ is taken to be symmetric.

##  4 Strongly symmetric uncorrelated noise

We begin by considering uncorrelated noise – channels that decompose into
onsite operations as $\mathcal{E}=\otimes_{s}\mathcal{E}_{s}$. If
$\\{K_{i_{s}}^{s}\\}$ is a Kraus representation of $\mathcal{E}_{s}$, a Kraus
representation of $\mathcal{E}$ is by operators

$K_{i}=\otimes_{s}K_{i_{s}}^{s}~{}.$ (4.1)

The full channel satisfies the WS or SS condition if and only if all of the
single site channels do. Uncorrelated noise is the simplest class of channels,
and include the example of section 3.4.1, so they are a natural place to
start.

The following subsections consider several typical probes of SPTO, in
particular, string order, twisted sector charges, edge modes, and irrep
probabilities. We demonstrate that these probes are preserved by strongly
symmetric uncorrelated noise, indicating that SPTO is preserved. We show in
Theorem 1 that, for semigroups of noise generated by Lindbladians, the strong
symmetry condition on the Lindbladians is both necessary and sufficient for
string order to be preserved at all finite times. We present analytical
arguments as well as numerical investigations of example states and channels.

###  4.1 Preservation of string order by strongly symmetric channels

The first indicator of SPTO we consider is the string order parameter, which
was introduced in section 2.2. A channel $\mathcal{E}$ preserves string order
if the evolved state $\mathcal{E}(\rho)$ has the same pattern of zeros as the
initial state $\rho$. In the Heisenberg picture, this means that the
collection of values
$\langle\mathcal{E}^{\dagger}(s(g,O_{\alpha}^{l},O_{\alpha}^{r}))\rangle$ has
the same pattern of zeros as $\langle
s(g,O_{\alpha}^{l},O_{\alpha}^{r})\rangle$, where $\mathcal{E}^{\dagger}$ is
the dual channel to $\mathcal{E}$. We save for section 6.4 the question of
precisely which channels preserve the string order of a state in a given SPT
phase. For now, we show

Lemma 1: _A channel of uncorrelated noise maps string operators to other
string operators of the same type $(g,\alpha)$ if and only if the channel is
strongly symmetric._

Note that the evolved string operators are not guaranteed to be
nonvanishing.333It is possible that a strongly symmetric channel annihilates
some string operators by mapping their end operators to zero. This phenomenon
is unrelated to the symmetry selection rules for string order, discussed in
section 2.2, which happens on the level of expectation values. For channels
that are generic in the sense of Eq. (6.24), it does not occur for generic end
operators.

To see the lemma, consider evolving the string operator (2.6) by the
uncorrelated noise. It becomes

$\mathcal{E}^{\dagger}(s(g,O_{\alpha}^{l},O_{\alpha}^{r}))=\mathds{1}\otimes\mathcal{E}_{l}^{\dagger}(O_{\alpha}^{l})\otimes\left(\bigotimes\mathcal{E}_{s}^{\dagger}(U_{g})\right)\otimes\mathcal{E}_{r}^{\dagger}(O_{\alpha}^{r})\otimes\mathds{1}~{}.$
(4.2)

If $\mathcal{E}_{s}$ is SS, each of the terms in the bulk of the string
becomes $\mathcal{E}_{s}^{\dagger}(U_{g})=e^{i\theta_{s}(g)}U_{g}$. Since an
SS channel is in particular WS (3.2), it maps the end operators to other end
operators with the same charge:
$\tilde{O}_{\alpha}^{l,r}:=\mathcal{E}_{s}^{\dagger}(O_{\alpha}^{l,r})$ has
$U_{g}^{\dagger}\tilde{O}_{\alpha}^{l}U_{g}=\chi_{\alpha}(g)\tilde{O}_{\alpha}^{l}$
and similarly for $\tilde{O}_{\alpha}^{r}$. Then in total, we have

$\mathcal{E}^{\dagger}(s(g,O_{\alpha}^{l},O_{\alpha}^{r}))=e^{i\sum_{s}\theta_{s}(g)}s(g,\tilde{O}_{\alpha}^{l},\tilde{O}_{\alpha}^{r})~{}.$
(4.3)

Conversely, asking that
$\mathcal{E}^{\dagger}(s(g,O_{\alpha}^{l},O_{\alpha}^{r}))$ is a string
operator of type $(g,\alpha)$ for all $(g,\alpha)$ requires,444The choice
$O_{\alpha}^{l,r}=\mathds{1}$ always results in nonvanishing
$\tilde{O}_{\alpha}^{l,r}$, so the bulk $\mathcal{E}_{s}^{\dagger}(U_{g})$ can
be compared to $U_{g}$. in particular, that $\mathcal{E}_{s}^{\dagger}(U_{g})$
is proportional to $U_{g}$, which is the strong symmetry condition.

####  4.1.1 A necessary and sufficient condition on Lindbladians

Our discussion has so far focused on the string operators. We now turn toward
analyzing their expectation values, which encode the invariant of SPT states.
In the following, a coherent SPT mixture is a mixed state with a well-defined
pattern of zeros (and thus a well-defined invariant $[\omega]$), in the sense
of Definition (2.15). A “coherent SPT phase” is a class consisting of all
coherent SPT mixtures with a given invariant. Preserving a phase means that
every mixture of translation-invariant pure SPT states in the phase (states of
the form Eq. (2.16)) is mapped to a coherent SPT mixture in the same
phase.555We expect that strongly symmetric Lindbladian evolution takes every
coherent SPT mixture to a coherent SPT mixture in the same phase; however, we
restrict ourselves to initial states that are mixtures of translation-
invariant pure SPT states in order to make use of tensor network methods in
the proof of the theorem.

Our main result is stated as a theorem:

Theorem 1: _Fix any coherent SPT phase. A semigroup of channels of
uncorrelated noise generically preserves the phase at all finite times if and
only if the semigroup is generated by a strongly symmetric Lindbladian._

In other words, the notion of coherent SPT phase defined by patterns of zeros
coincides with the notion of phase defined by strongly symmetric Lindbladian
evolution.

To make a claim as strong as Theorem 1, it is necessary to work on the level
of phases rather than the states (equivalently, systems) that compose them. A
phase protected by a symmetry $G$ consists of systems together with the data
of embeddings of $G$ into the systems’ full groups of symmetries. For example,
the AKLT system may be regarded as lying in a
$G={\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2}$ SPT phase if one specifies this
group’s embedding into the system’s larger $SO(3)$ intrinsic symmetry group.
Also, there is no restriction on the physical degrees of freedom, or the
symmetry action on these degrees of freedom, that a system in a phase may
have. For example, the ${\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2}$ SPT phase to
which the AKLT system belongs also contains systems that are not built of
spin-$1$ degrees of freedom. Whether two systems lie in the same phase has not
to do with their intrinsic symmetry groups or degrees of freedom but on the
values taken by their order parameters. This means that MPS representations of
states within the phase will have tensors of various physical dimensions and
symmetries. The theorem concerns which symmetry condition channels must
satisfy so that they preserve the string order of _every system in the SPT
phase_. To do this, we consider generic states, whose MPS tensors have exactly
the symmetry $G$, are injective, and have physical Hilbert space no larger
than the tensor’s image. We emphasize that this result does not preclude the
possibility of nongeneric systems within the phase for which a condition
weaker than strong symmetry is sufficient.

With these remarks behind us, we are ready to prove the theorem, starting with
the ‘if’ direction. If the semigroup is generated by a strongly symmetric
Lindbladian, each channel $\mathcal{E}_{t}$ is itself strongly symmetric. Then
by Lemma 1, $\mathcal{E}_{t}$ preserves the type of string operators.
Moreover, finite time Lindbladian evolution defines channels that are
invertible as linear maps666Note that invertibility of the channel as a linear
map is a weaker condition than reversibility of the channel, as the inverse
linear map is not required to be a channel. since
$\det(e^{t\mathcal{L}})=e^{\Tr(t\mathcal{L})}\neq 0$ for finite $t$, so these
channels do not annihilate any string operators. Thus the pattern of zeros of
the expectation values of string operators is preserved, as long as one uses
end operators that are generic in the sense of section 2.2 – namely, that
$O_{\alpha}^{l,r}$ is orthogonal neither to $N_{l,r}^{g\dagger}$ nor to
$\mathcal{E}(N_{l,r}^{g\dagger})$. The assumption that the end operators are
generic can be dropped if one is interested in generic evolution times. This
is because the expectation value of a string operator is analytic as a
function of time, and so it either vanishes at all times – as occurs where the
initial pattern of zeros has a zero – or is zero only at isolated points in
time. The assumption of finite time cannot be dropped, as an infinite time
evolution may annihilate some string operators and therefore alter the pattern
of zeros. An evolution for which this phenomenon occurs is the fully dephasing
channel introduced below.

Next we turn to the ‘only if’ direction. We show it for initial states that
are translation-invariant pure SPT states. Then it also holds for their
mixtures. If a channel $\mathcal{E}$ preserves the string order of an SPT
state, there is, for every symmetry $g$, a string operator with $g$ in the
bulk whose expectation value is nonvanishing. This expectation value may be
written as $\langle L|T_{\mathcal{E}_{s}^{\dagger}(U_{g})}^{j}|R\rangle$,
where $T_{\mathcal{E}_{s}^{\dagger}(U_{g})}$ is the transfer matrix

$\begin{split}\includegraphics[height=48.06017pt]{transferop.PNG}\raisebox{20.2322pt}{.}\end{split}$
(4.4)

This operator must have $\lambda=1$ as its maximum eigenvalue in order for the
expectation value to be nonvanishing in the thermodynamic limit.777At finite
string lengths, the difference between strongly symmetric and non-strongly
symmetric Lindbladians is reflected in the length-dependence of the order
parameters: only for the latter is there decay as a function of length. By
Lemma 2 of Ref. [34], for injective MPS, in order for $T_{X}$ to have
$\lambda=1$, $X$ must be a symmetry of the MPS, which by assumption is some
element $h_{g}\in G$. Thus
$\mathcal{E}_{s}^{\dagger}(U_{g})=e^{i\theta(g)}U_{h_{g}}$. Taking $X=U_{g}$
and $Y=e^{i\theta(g)}U_{h_{g}}$ in the argument at the end of section 3.2, we
see that $U_{g}K_{i}=e^{i\theta(g)}K_{i}U_{h_{g}}$ for all $i$. The map
$\sigma:g\mapsto h_{g}$ is an endomorphism888An endomorphism is a map from the
group to itself that is compatible with the group structure. of $G$ since $U$
is faithful and, up to phases $\theta$,

$U_{\sigma(g)}U_{\sigma(h)}=\sum_{i}K_{i}^{\dagger}K_{i}U_{\sigma(g)}U_{\sigma(h)}\sim\sum_{i}K_{i}^{\dagger}U_{g}U_{h}K_{i}=\sum_{i}K_{i}^{\dagger}U_{gh}K_{i}\sim\sum_{i}K_{i}^{\dagger}K_{i}U_{\sigma(gh)}=U_{\sigma(gh)}~{}.$
(4.5)

If $\mathcal{E}$ is connected to the trivial channel by a semigroup of
channels $\mathcal{E}_{t}$ satisfying the above at all times, the endomorphism
$\sigma$ must be connected to the identity endomorphism $\sigma=1$ by a
continuous path $\sigma_{t}$. Since we have assumed $G$ is abelian, the only
identity-connected endormophism is $\sigma=1$ itself. Thus, we have
$\mathcal{E}_{s}^{\dagger}(U_{g})=U_{g}$, which is the strong symmetry
condition (3.9) on the site $s$. This holds for all $s$, so the full channel
$\mathcal{E}$ is strongly symmetric. It holds at all finite times, so by the
discussion in section 3.4, the Lindbladian that generates it is strongly
symmetric. This completes the proof of Theorem 1.

It is interesting to note that, by Theorem 1 and Lemma 1, any semigroup that
preserves an SPT phase, even the trivial SPT phase, also preserves the type of
all string operators. This reflects how special SPT states, even trivial SPT
states, are among mixed states, most of which lack valid patterns of zeros.

The proof of Theorem 1 involved only the identity-connected endomorphisms of
the symmetry group. In section 6 we lift the restriction that the channel
belongs to a semigroup of SPT-preserving channels; in this broader setting,
more general group endomorphisms play an important role.

Let us comment on the generalization to a nonabelian symmetry group $G$.
Recall that in this case, string order is not guaranteed to capture the full
SPT invariant $\omega$; nevertheless, we can ask about which channels preserve
string order. If $G$ is a finite group, the only identity-connected
endomorphism is again $\sigma=1$ itself. On the other hand, if $G$ is a
semisimple Lie group, the only identity-connected endomorphisms are _inner
automorphisms_ [39]. An automorphism $\sigma$ is said to be inner if there
exists an element $h\in G$ such that

$\sigma(g)=\text{conj}_{h}(g):=h^{-1}gh~{},\quad\forall g\in G~{}.$ (4.6)

The element $h$ is defined up to elements of the center. An example of a
channel with automorphism $\sigma=\text{conj}_{h}$ is the symmetry-
implementing channel $\mathcal{U}_{h}$ (3.1), which has a single Kraus
operator $K=U_{h}$ satisfying

$U_{\sigma(g)}:=K^{\dagger}U_{g}K=U_{h}^{-1}U_{g}U_{h}=U_{h^{-1}gh}~{}.$ (4.7)

Any channel $\mathcal{E}$ with inner automorphism $\sigma=\text{conj}_{h}$ may
be expressed as the composition of the symmetry-implementing channel
$\mathcal{U}_{h}$ and a strongly symmetric channel $\mathcal{E}_{SS}$:

$\mathcal{E}=\mathcal{E}_{SS}\circ\mathcal{U}_{h}~{}.$ (4.8)

To see this, let $K_{i}$ denote the Kraus operators for $\mathcal{E}$ and
define Kraus operators for $\mathcal{E}_{SS}$ as

$K_{i}^{\prime}=U_{h}^{-1}K_{i}~{}.$ (4.9)

One can easily verify that the latter define a strongly symmetric channel
$\mathcal{E}_{SS}$. The semigroup of channels of the form (4.8) is generated
by the sum of of a strongly symmetric Lindbladian $\mathcal{L}_{SS}$ and a
generator ${\mathcal{Q}}$ of the continuous symmetry
$\mathcal{U}_{h}=e^{\mathcal{Q}}$. We obtained this condition by asking that a
string operator expectation value not vanish, which is weaker than asking that
string order is preserved; in fact, string order may be modified by such
channels. The $\mathcal{U}_{h}$ factor changes the bulks of string operators
from $U_{g}$ to $\mathcal{U}_{h}^{\dagger}(U_{g})=U_{h^{-1}gh}$ and the ends
from $O_{\alpha}$ to $\mathcal{U}_{h}^{\dagger}(O_{\alpha})$, which transform
as $(\sigma^{-1})^{*}\alpha$. It may happen, however, that string order is
preserved despite the permutation of string operators; this phenomenon is
discussed in section 6.

###  4.2 Numerical study of string order under channels

In this subsection, we simulate the behavior of string order under the action
of two example channels – the depolarising channel and the dephasing channel.
We find, in agreement with the analytical results, that evolutions with strong
symmetry preserve string order, while weakly symmetric channels do not. We
also use the example of the fully dephasing channel, which destroys SPTO
despite being strongly symmetric, to demonstrate the importance of the finite
time assumption in the theorem.

The depolarising channel is a severely noisy channel, as it contracts the
Bloch sphere to the origin, driving states towards the maximally mixed state.
This channel is written as

$\mathcal{E}(\rho)=\left(1-\lambda\right)\,\rho+\lambda\,(\Tr{\rho})\,\tfrac{1}{d}\mathbbm{1}~{},$
(4.10)

with the decay rate parametrized by $\lambda\in[0,1]$ [40]. The value
$\lambda=1$ gives the fully depolarising channel. The channel satisfies the
weak symmetry condition (3.2) for any symmetry.

The Kraus decomposition for this channel can be written as a twirling
operation since the normalised $d$-identity can be decomposed as an average
over the generators of the Lie algebra of $SO(d)$ [36]. In $d=2$, these
operators are Paulis, while in higher $d$ they are the Heisenberg-Weyl
matrices: the shift operator $X\ket{j}=\ket{j+1\mod d}$ and the phase operator
$Z\ket{j}=e^{i2\pi j/d}\ket{j}$, which have commutation relation
$Z^{m}X^{n}=e^{i2\pi nm/d}X^{n}Z^{m}$. The Kraus decomposition for spin-$1$
($d=3$) systems such as the AKLT state is given by

$\mathcal{E}(\rho)=(1-\lambda)\,\rho+\frac{\lambda}{9}\sum_{i}N_{i}\rho
N_{i}^{\dagger}~{},\qquad
N_{i}=\\{\mathbbm{1},Z,Z^{2},X,ZX,Z^{2}X,X^{2},ZX^{2},Z^{2}X^{2}\\}~{}.$
(4.11)

A $G={\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2}$ symmetry acts on the spin-$1$
system as $U_{g}=e^{i\pi S_{j}}$, where $S_{j}$ are the spin-$1$ operators.
Since the Kraus operators $K_{i}\sim N_{i}$ only commute with the symmetry
action up to different phases, the channel does not satisfy the strong
symmetry condition (3.8); it is only weakly symmetric.

The second channel we discuss is given by

$\mathcal{E}(\rho)=(1-\lambda)\rho+\frac{\lambda}{4}\sum_{j}N_{j}\rho
N_{j}^{\dagger}~{},\qquad N_{j}=\\{\mathbbm{1},e^{i\pi S_{x}},e^{i\pi
S_{y}},e^{i\pi S_{z}}\\}~{},$ (4.12)

where $S_{j}$ are the spin-$1$ operators. Since the Kraus operators $K_{j}\sim
N_{j}$ commute with the symmetry action, the channel is strongly symmetric
(3.8). The channel is the fully dephasing channel when $\lambda=1$.

####  4.2.1 String order after a single time-step

Let us consider the AKLT state, which belongs to the Haldane SPT phase. This
state has nontrivial $SO(3)$ SPT order, but can be protected by just the
subgroup ${\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2}$ [6, 41]. The AKLT state has
an exact tensor network representation given by the Pauli operators in the
basis $\\{\ket{+},\ket{0},\ket{-}\\}$.

The $|G|^{2}=16$ string operators $s_{ij}$ for the symmetry
$G={\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2}$ are built out of end operators
$O_{\alpha}^{l,r}=S_{i}$ and bulk operators $U_{g}=e^{i\pi S_{j}}$, where
$S_{i}$ are the spin-$1$ matrices with $j=\\{e,x,y,z\\}$. On the AKLT state,
the diagonal string operators take nonzero values $s_{zz},s_{xx},s_{yy}=-4/9$
and $s_{\mathbbm{1},\mathbbm{1}}=\mathbbm{1}$ independent of string length in
the limit of infinite system size, while the off-diagonal string operators
elements $s_{ij},i\neq j$ have vanishing expectation values in the limit of
long string length.

Recall that the string order parameter may be manipulated into the form (2.13)

$\langle
s(g,O_{\alpha}^{l,r})\rangle=\Tr(\rho_{l}T_{O_{\alpha}^{l}}(T_{U_{g}})^{N-2}T_{O_{\alpha}^{r}}\rho_{r})~{},$
(4.13)

where $T$ is the transfer matrix $T=\sum_{i}A^{i}\bigotimes\overline{A}^{i}$
for the MPS tensor $A$, and $\rho_{l,r}$ are its fixed points. Now consider
evolving the string operator under a channel. The operator becomes

$\langle\mathcal{E}^{\dagger}(s(g,O_{\alpha}^{l},O_{\alpha}^{r}))\rangle=\Tr(\rho_{l}T_{\mathcal{E}^{\dagger}(O_{\alpha}^{l})}T_{\mathcal{E}^{\dagger}(U_{g})}^{N-2}T_{\mathcal{E}^{\dagger}(O_{\alpha}^{r})}\rho_{r})~{}.$
(4.14)

For the SS channel (4.12), the evolved string order pattern for the AKLT state
as a function of the decay rate $\lambda$ is

$\langle\mathcal{E}^{\dagger}(s(e^{i\pi
S_{j}},S_{i}))\rangle={\small\begin{pmatrix}1&-\frac{4}{9}(1-\lambda)^{2}(-\frac{1}{3})^{N}&-\frac{4}{9}(1-\lambda)^{2}(-\frac{1}{3})^{N}&-\frac{4}{9}(1-\lambda)^{2}(-\frac{1}{3})^{N}\\\
(-\frac{1}{3})^{N}&-\frac{4}{9}(1-\lambda)^{2}&-\frac{4}{9}(1-\lambda)^{2}(-\frac{1}{3})^{N}&-\frac{4}{9}(1-\lambda)^{2}(-\frac{1}{3})^{N}\\\
(-\frac{1}{3})^{N}&-\frac{4}{9}(1-\lambda)^{2}(-\frac{1}{3})^{N}&-\frac{4}{9}(1-\lambda)^{2}&-\frac{4}{9}(1-\lambda)^{2}(-\frac{1}{3})^{N}\\\
(-\frac{1}{3})^{N}\
&-\frac{4}{9}(1-\lambda)^{2}(-\frac{1}{3})^{N}&-\frac{4}{9}(1-\lambda)^{2}(-\frac{1}{3})^{N}&-\frac{4}{9}(1-\lambda)^{2}\end{pmatrix}}~{}.$
(4.15)

In the limit of large string length $N\rightarrow\infty$, the diagonal entries
are nonzero while the off-diagonal entries are zero, which indicates that the
string order of the AKLT state was preserved, as is expected for SS channels.
Note that if we were to repeatedly evolve the state by $\mathcal{E}$ a large
number of times, the entire string order set would decay exponentially to
zero, so the preservation of the pattern is only visible at finite times. This
exponential decay in time is visible in Fig. 1.

In contrast, consider the WS depolarising channel (4.11). The string order for
the evolved state is

$\langle\mathcal{E}^{\dagger}(s(e^{i\pi
S_{j}},S_{i}))\rangle={\small\begin{pmatrix}1&-\frac{4}{9}(1-\lambda)^{2}(-\frac{1}{3})^{N}&-\frac{4}{9}(1-\lambda)^{2}(-\frac{1}{3})^{N}&-\frac{4}{9}(1-\lambda)^{2}(-\frac{1}{3})^{N}\\\
(-\frac{1}{3})^{N}&-\frac{4(1-\lambda)^{2}}{9}(1-\frac{8\lambda}{9})^{N}&-\frac{4(1-\lambda)}{9}(-\frac{1}{3}+\frac{4\lambda}{9})^{N}&-\frac{4(1-\lambda)}{9}(-\frac{1}{3}+\frac{4\lambda}{9})^{N}\\\
(-\frac{1}{3})^{N}&-\frac{4(1-\lambda)}{9}(-\frac{1}{3}+\frac{4\lambda}{9})^{N}&-\frac{4(1-\lambda)^{2}}{9}(1-\frac{8\lambda}{9})^{N}&-\frac{4(1-\lambda)}{9}(-\frac{1}{3}+\frac{4\lambda}{9})^{N}\\\
(-\frac{1}{3})^{N}\
&-\frac{4(1-\lambda)}{9}(-\frac{1}{3}+\frac{4\lambda}{9})^{N}&-\frac{4(1-\lambda)}{9}(-\frac{1}{3}+\frac{4\lambda}{9})^{N}&-\frac{4(1-\lambda)^{2}}{9}(1-\frac{8\lambda}{9})^{N}\end{pmatrix}}~{}.$
(4.16)

For any $\lambda>0$, the string order goes to zero instantaneously (with a
single time-step, application of $\mathcal{E}$) in the limit
$N\rightarrow\infty$. This is because, in contrast to the SS channel, this WS
channel receives an exponential contribution to its decay from the bulk of the
string.

Figure 1: Evolution of the absolute value of the $s_{zz}$ component of the
${\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2}$ string order parameter for the AKLT
state (initial $|s_{zz}|>0$) under the dephasing channel (4.12) and the
depolarising channel (4.11).

####  4.2.2 Master equation simulation of string order

We simulate the time-evolved master equation on pure SPT states by acting with
the channel at each time-step

$\mathcal{E}_{t}=\mathcal{E}_{T}\circ\cdots\circ\mathcal{E}_{t1}\circ\mathcal{E}_{t_{0}}.$
(4.17)

Figure 1 depicts the results of simulating the evolution of the $s_{zz}$
component of the string order parameter under the following channels: the
dephasing channel (4.12) with $\lambda=0.5$, the dephasing channel with
$\lambda=1$ (fully dephasing), and the depolarising channel (4.11). The first
two of these channels are SS, while the third is only WS. The first exhibits
slow decay, while the others exhibit instantaneous decay.

The reason that the fully dephasing channel fails to preserve SPT order
despite being SS can be traced to how it annihilates the end operators $S_{i}$
with nontrivial labels and thus sets the corresponding rows of the pattern of
zeros to zero. This phenomenon cannot occur for finite time evolutions, as
these channels are invertible as linear maps, so Theorem 1 is safe. The fully
dephasing channel, however, is only realized as an _infinite_ time evolution,
so SS is insufficient to protect SPTO. In section 6.4, we discuss a
genericness condition on SS channels, failed by the fully dephasing channel,
that says whether or not they preserve SPTO.

###  4.3 Twisted sector charges

The SPTO of a state may alternatively be detected in the charges of its
twisted sector states [42, 43]. Consider a state on a closed chain,
represented as an MPS

$\langle i_{1}\cdots i_{L}|\psi\rangle=\Tr[\,A^{i_{1}}\cdots A^{i_{L}}\,]~{},$
(4.18)

with the symmetry represented projectively on the virtual space by operators
$V_{g}$. The states

$\langle i_{1}\cdots i_{L}|\psi_{h}\rangle=\Tr[\,V_{h}\,A^{i_{1}}\cdots
A^{i_{L}}\,]$ (4.19)

are the twisted sector states. In closed systems, they appear as ground states
of the Hamiltonians obtained by twisting the original Hamiltonian by the
insertion of a symmetry flux through the closed chain.

The charge of a twisted sector state is obtained by acting on the state with
the charge operator $U_{g}^{\otimes L}$:

$\begin{split}\includegraphics[height=43.00404pt]{twisted_sectors1.PNG}\quad\raisebox{10.11995pt}{.}\end{split}$
(4.20)

For simplicity, assume $G$ is abelian. The charges of the twisted sector
states are given by

$\langle i_{1}\cdots i_{L}|U_{g}^{\otimes
L}|\psi_{h}\rangle=\Tr[\,V_{g}^{-1}V_{h}V_{g}\,A^{i_{1}}\cdots
A^{i_{L}}\,]=\frac{\omega(h,g)}{\omega(g,h)}\Tr[\,V_{h}\,A^{i_{1}}\cdots
A^{i_{L}}\,]~{}.$ (4.21)

As we saw in the previous subsection, this ratio $\omega/\omega$ determines
the cohomology class $[\omega]$. This means that the collection of twisted
sector charges completely characterizes the SPTO.

The twisted sector charges of an initial state and the state reached by
evolving by a channel are the expectation values of $U_{g}^{\otimes L}$ and
$\mathcal{E}^{\dagger}(U_{g}^{\otimes L})$, respectively, on the initial
twisted sector states $|\phi_{h}\rangle$. Since for a strongly symmetry
channel (3.9) these are equal, such a channel preserves the SPTO.

###  4.4 Protected edge modes

SPT phases are also characterized by their topologically protected edge modes.
Consider a pure state $|\psi^{\omega}\rangle$ in an SPT phase characterized by
a cocycle $\omega$. The SPT invariant is encoded in the projective action of
the symmetry on the edge introduced by cutting the system:

$(\mathds{1}^{l}\otimes
U_{g}^{r})|\psi^{\omega}\rangle=\sum_{a}|\psi^{\omega}_{l,a}\rangle\otimes
U_{g}^{r}|\psi^{\omega}_{r,a}\rangle=\sum_{a,b}(V_{g})_{ab}|\psi^{\omega}_{l,a}\rangle\otimes|\psi^{\omega}_{r,b}\rangle~{},\qquad
V_{g}V_{h}=\omega(g,h)V_{gh}~{}.$ (4.22)

After evolving through a channel, $|\psi^{\omega}\rangle$ becomes a mixture of
the (unnormalized) states $K_{i}|\psi^{\omega}\rangle$. We claim that, if the
channel is strongly symmetric and onsite, each of these states has SPT
invariant $\omega$. To see this, write the Kraus operators as
$K_{i}^{l}\otimes K_{i}^{r}$, so that
$K_{i}|\psi^{\omega}_{i}\rangle=\sum_{a}K_{i}^{l}|\psi^{\omega}_{l,a}\rangle\otimes
K_{i}^{r}|\psi^{\omega}_{r,a}\rangle$. Then, since $K_{i}^{r}$ commutes with
$U_{g}^{r}$, the projective representation $V_{g}$ on this cut is the same as
for $|\psi^{\omega}\rangle$; in particular, $\omega$ is the same.

###  4.5 Irrep probabilities and SPT complexity

In this section, we investigate the behavior of irrep probabilities – another
probe of SPTO – under channels, strongly symmetric and not. Irrep
probabilities measure the weight of the state in each symmetry sector. They
are given by Fourier transforms of the string operators with identity end
operators

$p_{\alpha}=\frac{1}{|G|}\sum_{g}\chi_{\alpha}(g)\langle
s(U_{g},\mathds{1},\mathds{1})\rangle~{},$ (4.23)

Using orthogonality of characters, one can confirm that the irrep
probabilities sum to $1$:

$\sum_{\alpha}p_{\alpha}=\frac{1}{|G|}\sum_{g}\langle
s(U_{g},\mathds{1},\mathds{1})\rangle\sum_{\alpha}\chi_{\alpha}(g)=\frac{1}{|G|}\sum_{g}\langle
s(U_{g},\mathds{1},\mathds{1})\rangle\,|G|\,\delta_{g,1}=\langle
s(\mathds{1},\mathds{1},\mathds{1})\rangle=1~{},$ (4.24)

Irrep probabilities partially distinguish SPT phases by capturing the _SPT
complexity_

$D_{\omega}=\sqrt{|G|/|K_{\omega}|}~{},$ (4.25)

where

$K_{\omega}=\\{\,g\,:\,\omega(g,h)=\omega(h,g)~{}\,\forall\,h\,\\}$ (4.26)

is subgroup of $G$ called the _projective center_. A phase with
$K_{\omega}=\\{1\\}$ has maximum complexity and is said to be _maximally
noncommutative_ (MNC) [44]. An example of an MNC phase is the Haldane phase of
systems with ${\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2}$ symmetry. MPS
representations of states in this phase have virtual symmetries related to the
Pauli operators, which do not commute, meaning that $K_{\omega}=\\{1\\}$. On
the other end of the spectrum is the trivial phase $\omega=1$, with complexity
$D_{1}=1$. The complexity of a generic MPS was shown numerically to appear in
the degeneracy $D_{\omega}^{2}$ of the irrep probabilities [45]. In an MNC
phase, the probabilities are all $p_{\alpha}=1/|G|$ with degeneracy $|G|$; in
the trivial phase, they are generically distinct. The value $D_{\omega}$ also
appears as the degree of the projective representation $V_{g}$ (c.f. [46],
theorem VI.6.39), which means $D_{\omega}^{2}$ is the number of topologically
protected edge modes. For a given symmetry $G$, there may be multiple phases
with the same complexity and these cannot be distinguished from each other by
their irrep probabilities; nevertheless, because they can distinguish phases
with different complexities, irrep probabilities are a useful tool.

The values of the irrep probabilities are preserved by strongly symmetric
channels. To see this, observe that such channels preserve the string
operators $s(U_{g},\mathds{1},\mathds{1})$, and therefore their Fourier
transforms, exactly. The exact preservation of irrep probabilities stands in
contrast with general string order parameters, whose end operators
$\mathcal{O}_{\alpha}\neq\mathds{1}$ may cause decay toward zero in infinite
time.

In contrast with strongly symmetric channels, non-SS Lindbladian channels map
all irrep probabilities to the maximally degenerate values $p_{\alpha}=1/|G|$,
for any phase. This is because, as was established in section 4.1.1, such
channels annihilate all of the string order parameters accept for those with
$g=1$. The result is

$p_{\alpha}=\frac{1}{|G|}\sum_{g}\chi_{\alpha}(g)\,\delta_{g,1}=\frac{1}{|G|}~{}.$
(4.27)

This means that, outside of states which already have maximally degenerate
irrep probabilities (for example, states in MNC phases), the effect of a
Lindbladian channel on irrep probabilities is a diagnostic for whether or not
a channel is strongly symmetric. In section 6.4.2, we discuss non-Lindbladian
channels, some of which preserve irrep probabilities and SPT complexity
despite not being SS.

##  5 Causal channels

Let us extend our analysis from uncorrelated noise to causal channels. A
channel is said to be _causal_ if there is a range $r$ such that it maps
operators supported on a compact region $A$ to operators supported on the
region of sites within distance $r$ of $A$ [47].999The terms “causal” and
“locality-preserving” have different meanings in Ref. [47], and we are
interested in the former. Channels that are not causal can create long-range
correlations, and so are expected to destroy topological order and SPTO, no
matter the symmetry condition imposed on them. For this reason, we will not
consider non-causal channels here.

A subset of causal channels, labeled “dQC”101010The “d” stands for “dilation,”
as in the Stinespring dilation of the channel. Our “sd” stands for “symmetric
dilation.” in Ref. [47], have purifications that are causal. In addition, one
can consider their convex combinations, which are also causal. It has been
suggested that these convex combinations might constitute all causal channels,
but this remains an open question [47]. Another open question is whether every
channel in dQC that has a symmetric purification (and so by Claim (3.13) is
weakly symmetric) has a purification that is _both_ causal and symmetric with
respect to an on-site symmetry. Rather than attempting to answer this
question, we let “sdQC” denote the channels with such a purification. We
restrict our focus to channels in dQC (in section 5.1) and sdQC (in section
5.2 and section 5.3) and leave open the possibility that channels outside of
these classes exhibit different behaviors. As we remark in section 5.4, our
expectation is that only channels in these classes are relevant to local
Lindbladian evolution.

Low depth circuits of local unitary gates are a special class of unitary
causal channels. Causal unitaries have a topological index111111This index may
be computed locally from the tensor of the MPU discussed below [48, 49]. that
takes values $\log(p/q)$ for natural numbers $p,q$, capturing the flow of
information to the left and to the right [50], and low depth circuits of local
unitary gates are precisely the causal unitaries for which this topological
index vanishes. From the perspective of phase classification, circuits are the
only causal unitaries one should care about, as they approximate fast local
unitary evolution. Nevertheless, it is most convenient for us to work with
causal unitaries in general – forgetting whether or not they are circuits –
because causal unitaries have convenient tensor network representations (which
we will describe shortly). Similarly, a special class of causal channels is
given by low depth circuits of local channels.121212By this, we mean that the
channel consists of a small number of layers of disjoint channels supported on
small intervals. It may be the case that these are the causal channels which
approximate fast local Lindbladian evolution, just as unitary circuits do for
unitary evolution, and that they are in dQC and are characterized by a
vanishing topological index of their purification. We do not attempt to prove
this conjecture. Regardless, our analysis considers causal channels in
general, even if most of them are unrelated to Lindbladian evolution.

###  5.1 Tensor network representations of causal channels

Causal unitary operators, which in this context are the purifications of
channels in dQC, have finite bond dimension tensor network representations
called matrix product unitaries (MPUs) [48, 49]:

$\begin{split}\includegraphics[height=22.76411pt]{MPU1.PNG}\quad\raisebox{10.11995pt}{.}\end{split}$
(5.1)

As in section 3.3, the red legs of the tensor carry ancilla indices.

Here, unlike in the previous section, we simplify the analysis by restricting
to channels that are translation invariant, which means that the tensor
network representations of their purifications consist of the same tensor at
every site. We expect, however, that our results hold without this assumption.

The theory of MPUs says that there exists a length $r\leq\delta^{4}$ (where
$\delta$ is the bond dimension of $W$) such that on blocks of $r$ sites, the
tensor satisfies the following “simpleness relations” [48]:

$\begin{split}\includegraphics[height=48.06017pt]{MPU2.PNG}\raisebox{25.29604pt}{,}\end{split}$
(5.2)

where $\Lambda$ denotes the right fixed point of the MPU transfer matrix.

The fact that the causal unitaries have tensor network representations means
that channels in dQC do as well. The Kraus operator $K_{i}$ is realized as the
matrix product operator obtained by plugging in the ancilla state
$|a\rangle=\otimes_{s}|a_{s}\rangle$ (take all $a_{s}$ to be the same) and the
ancillary space basis vector $|e_{i}\rangle=\otimes_{s}|e_{i_{s}}\rangle$:

$\begin{split}\includegraphics[height=30.35216pt]{MPOdefinition.PNG}\raisebox{12.64415pt}{.}\end{split}$
(5.3)

Then, in terms of the tensors for the Kraus operators, the simpleness
relations become

$\begin{split}\includegraphics[height=48.06017pt]{MPU4.PNG}\raisebox{32.88408pt}{.}\end{split}$
(5.4)

The uncorrelated noise considered in section 4 appears here as the channels
whose MPUs have $\delta=1$:

$\begin{split}\includegraphics[height=55.64821pt]{MPU5.PNG}\raisebox{27.82024pt}{\quad.}\end{split}$
(5.5)

###  5.2 Local realization of the symmetry conditions

It will be demonstrated in the following subsections that a causal channel
preserves SPTO if it satisfies not just the strong symmetry condition, but the
strong symmetry condition _realized locally_ (5.9). The present subsection is
dedicated to describing what is meant by local realization of the symmetry
conditions and to motivating it. We consider channels in the class sdQC we
defined earlier, meaning that the causal purification is symmetric under
acting with the symmetry operator $U_{g}\otimes U_{g}^{A}$ on every site.

As always, the symmetry conditions refer to how the Kraus operators transform
under conjugation by a symmetry. If the Kraus operators do not decompose as
products of uncorrelated terms, we must work out what these global conditions
mean locally, in terms of their tensor network representation. Local
properties of a symmetry action may be studied by cutting the spin chain into
two halves, so that the purified channel decomposes as
$W=\sum_{\mu}W_{l}^{\mu}\otimes W_{r}^{\mu}$, and acting with the symmetry on
the right half:

$(\mathds{1}^{l}\otimes(U_{g}\otimes
U_{g}^{A})^{r})W(\mathds{1}^{l}\otimes(U_{g}\otimes
U_{g}^{A})^{r})^{\dagger}=\sum_{\mu,\nu}(Q_{g})_{\mu\nu}W_{l}^{\mu}\otimes
W_{r}^{\nu}~{}.$ (5.6)

The operators $Q_{g}$ are defined up to redefinition by phases and form a
projective representation: $Q_{g}Q_{h}=\nu(g,h)Q_{gh}$. By folding the MPU
representing the purification $W$ into a normal MPS [48] and applying the
usual arguments [9, 34], it can be shown that the $Q_{g}$ satisfy131313The
version of the Fundamental Theorem of MPS in Theorem IV.4 of Ref. [9] can be
used to show that this condition holds not just on blocks of size $r$ but on
individual sites; however, we will not require this stronger condition in our
arguments.

$\begin{split}\includegraphics[height=50.59209pt]{purificationsym1.PNG}\raisebox{25.29604pt}{.}\end{split}$
(5.7)

In particular, when the channel is SS, we have $U_{g}^{A}=\mathds{1}$, so the
Kraus operators satisfy

$\begin{split}\includegraphics[height=50.59209pt]{purificationsym2.PNG}\raisebox{25.29604pt}{.}\end{split}$
(5.8)

We note that symmetric MPUs have been studied previously [51].

Now we can state the condition of local realization of the symmetry:

$\boxed{\qquad Q_{g}Q_{h}=Q_{gh}~{},\quad\forall\,g,h~{},\quad\text{up to
redefinition by phases.}\qquad(\text{locally realized symmetry})\qquad}$ (5.9)

Under redefinition of $Q$ by phases, the cocycle $\nu$ that captures the
projectivity of $Q$ shifts by a coboundary, so local realization is the
condition that $[\nu]$ is trivial in cohomology. In the case of uncorrelated
noise, the symmetry conditions are automatically realized locally because $Q$
acts on a one-dimensional space. When a channel is WS or SS and its symmetry
is realized locally, we say it is “locally-WS” or “locally-SS”.

To build intuition for local realization, let us see that it is satisfied by
the circuits of symmetric gates that define phase equivalence for states of
closed systems. For a circuit, symmetry of the gates means that $Q$ can be
extracted from a single gate by acting on half of its legs with the symmetry
$g$:

$\begin{split}\includegraphics[height=55.64821pt]{circuit.PNG}\raisebox{30.35216pt}{.}\end{split}$
(5.10)

Since the adjoint action of $U_{g}$ on half of the gate is a linear (non-
projective) representation, $Q$ is linear as well. More generally, consider
the circuits of local channels that were mentioned briefly above. In analogy
to the condition that the unitary gates are symmetric, these local channels
can be made to satisfy the weak or strong symmetry condition. If they satisfy
the WS or SS condition, the argument we just used for unitary circuits
demonstrates that the causal channel as a whole is locally-WS or locally-SS,
respectively.

###  5.3 String operators

Let us generalize Lemma 1 of section 4.1 to causal channels by showing the
following lemma.

Lemma 2: _A translation-invariant channel in sdQC maps string operators to
sums of string operators of the same type $(g,\alpha)$ if and only if the
channel satisfies the local strong symmetry condition._

Doing so requires working with a slight generalization of string operators
where the end operators are supported intervals, rather than sites. It is
assumed that the system size and the length of the string are large compared
to the length of these intervals and the range $r$ of the channel.

After evolution by the channel, the string operator (2.6) becomes

$\begin{split}\includegraphics[height=45.52824pt]{stringorder_channel.PNG}\quad\raisebox{10.11995pt}{.}\end{split}$
(5.11)

If the SS condition is satisfied, the symmetry pulls through at the cost of
operators $Q_{g}$:

$\begin{split}\includegraphics[height=55.64821pt]{stringorder_channel2.PNG}\quad\raisebox{20.2322pt}{,}\end{split}$
(5.12)

which cancel except near the ends of the string. Then the simpleness relations
(5.4) can be applied to obtain

$\begin{split}\includegraphics[height=55.64821pt]{stringorder_channel3.PNG}\quad\raisebox{30.35216pt}{.}\end{split}$
(5.13)

This is a string operator with the symmetry $g$ in the bulk. The new end
operators are the result of acting on the original end operators by the
superoperators (generalizing the $\mathcal{E}_{l,r}^{\dagger}$ of uncorrelated
noise)

$\begin{split}\includegraphics[height=55.64821pt]{stringorder_channel5b.png}\quad\raisebox{20.2322pt}{.}\end{split}$
(5.14)

and are supported on $2r$ more sites than the original end operators. The
superoperators $\mathcal{S}_{l}^{g}$ and $\mathcal{S}_{r}^{g}$ transform in
the representations $h\mapsto\nu(g,h)/\nu(h,g)$ and
$h\mapsto\left(\nu(g,h)/\nu(h,g)\right)^{*}$, respectively:

$\begin{split}\includegraphics[height=131.52861pt]{stringorder_channel6.png}\quad\raisebox{25.29604pt}{.}\end{split}$
(5.15)

This means that, if and only if $\nu$ is trivial, as is the case when the
symmetry condition is locally realized, the new end operators
$\mathcal{S}_{l,r}^{g}(\mathcal{O}_{\alpha}^{l,r})$ transform in the same
representations as the original ones $\mathcal{O}_{\alpha}^{l,r}$. We conclude
that the evolved string operator is of type $(g,\alpha)$ if the channel
satisfies the locally realized SS condition. Note that, in contrast with the
case of uncorrelated noise, WS is not enough to ensure the correct
transformation of the end operators. This is because the WS condition states
only that the charge of operators is conserved _globally_. When correlations
between sites are present, charge can flow between regions of the system, such
as between the two end operators, changing their individual charges.

It remains to show the converse: that, assuming the evolved string operator
(5.11) is a string operator of type $(g,\alpha)$, the channel must have been
locally SS. The string operator with bulk $U_{g}^{\otimes j^{\prime}}$ evolves
into

$\begin{split}\includegraphics[height=60.70432pt]{stringorder_channel7.png}\quad\raisebox{17.708pt}{.}\end{split}$
(5.16)

Compose it with $(U_{g}^{\dagger})^{\otimes j^{\prime}}$ on the string bulk
and take the trace to obtain

$\begin{split}\includegraphics[height=80.94424pt]{stringorder_channel8.PNG}\quad\raisebox{20.2322pt}{,}\end{split}$
(5.17)

neglecting the part of the tensor network outside the support of the evolved
string. Meanwhile, doing the same to some string operator results in

$\begin{split}\includegraphics[height=70.82428pt]{stringorder_channel9.PNG}\quad\raisebox{20.2322pt}{,}\end{split}$
(5.18)

where $j=j^{\prime}-2r^{\prime}$ for $r^{\prime}$ the spread of the end
operators under the channel (which turns out to be $r^{\prime}=r$ due to SS).
Setting (5.17) and (5.18) equal (by our assumption), and defining

$\begin{split}\includegraphics[height=48.06017pt]{transferopMPU.PNG}\quad\raisebox{20.2322pt}{.}\end{split}$
(5.19)

we find that $\langle\rho_{\ell}|{\mathbb{E}}^{j}|\rho_{r}\rangle=vd^{j}$ for
some relative normalization $v\neq 0$, where $\langle\rho_{\ell}|$ and
$|\rho_{r}\rangle$ are the boundary conditions imposed by the end operators in
Eq. (5.17). Expressing ${\mathbb{E}}$ in terms of its distinct nonzero
eigenvalues $\lambda_{k}$, this amounts to the condition $\sum
w_{k}\lambda_{k}^{j}=vd^{j}$ for all $j\geq 1$, which implies that there must
be an eigenvalue $\lambda_{1}=d$.141414 First, note that by moving the
$vd^{j}$ to the other side of the equation, one obtains $\sum
w_{k}\lambda_{k}^{j}-vd^{j}=0$ for all $j\geq 1$, and thus, one is left with
showing the following
Lemma. _Given $K$ distinct $\mu_{k}\neq 0$, then _
$\sum_{k=1}^{K}c_{k}\mu_{k}^{j}=0\quad\forall\,j=J_{0},\dots,K+J_{0}-1\quad\Rightarrow\quad
c_{k}=0\quad\forall\,k=1,\dots,K\,.$ ($\star$) __ Proof. The matrix with
entries $M_{jk}\equiv\big{(}\mu_{k}^{j}\big{)}_{jk}$ is the product of the
diagonal matrix $\mathrm{diag}(\mu_{1}^{J_{0}},\dots,\mu_{K}^{J_{0}})$ with
the Vandermonde matrix $\big{(}\mu_{k}^{j-J_{0}}\big{)}_{jk}$, both of which
are invertible. Thus, the linear system ($\star$ ‣ 14), $M\vec{c}=\vec{0}$,
has the unique solution $c_{k}\equiv 0$.$\square$
Note that this also provides a concise proof of the often-used Lemma in the
MPS literature that $\sum a_{k}^{j}=\sum b_{k}^{j}$ implies that the $a_{k}$
and $b_{k}$ must be pairwise equal.  On the other hand, considering the MPU
$W$ on a periodic ring of length $N$, we have – using Cauchy-Schwarz – that

$\big{|}\sum_{k}\lambda_{k}^{N}\big{|}=\big{|}\mathrm{tr}\,{\mathbb{E}}^{N}\big{|}=\big{|}\Tr[U_{g}^{\otimes
N}W(U_{g}^{\otimes N})^{\dagger}W]\big{|}\leq\sqrt{\big{|}\Tr[(U_{g}^{\otimes
N}W(U_{g}^{\otimes
N})^{\dagger})(\cdots)^{\dagger}]\big{|}\,\big{|}\mathrm{tr}[WW^{\dagger}]\big{|}}=d^{N}\
.$ (5.20)

Thus, $\left|m_{1}d^{N}+\sum_{k>1}\lambda_{k}^{N}\right|\leq d^{N}$ for all
$N$, where $m_{1}\geq 1$ is the multiplicity of $\lambda_{1}=d$ and the
$\lambda_{k>1}\neq d$ are the other eigenvalues. This implies that
${\mathbb{E}}$ has one nondegenerate eigenvalue $\lambda_{1}=d$, and all other
eigenvalues are $0$.151515To see this, write $\lambda_{k}=e^{2\pi
i\xi_{k}}|\lambda_{k}|$, and fix $M=8^{K}$. Dirichlet’s approximation theorem
states that there are integers $p_{k}$ and $1\leq q\leq M$ such that
$|\xi_{k}-p_{k}/q|\leq 1/(qM^{1/K})$. Then, $|2\pi(q\xi_{k}-p_{k})|\leq\pi/4$,
and thus $\mathrm{Re}\big{[}(e^{2\pi i\xi_{k}})^{q}\big{]}>0$. It follows that
$\mathrm{Re}\big{[}\sum_{k>1}\lambda_{k}^{q}\big{]}>0$ and thus
$\left|m_{1}d^{q}+\sum_{k>1}\lambda_{k}^{q}\right|>d^{q}$, unless $m_{1}=1$
and there are no other nonzero eigenvalues $\lambda_{k}\neq 0$. Thus, the
Cauchy-Schwarz inequality (5.20) is saturated, which implies that
$U_{g}^{\otimes N}W(U_{g}^{\otimes N})^{\dagger}=e^{i\phi}W$, which is to say
that the channel with purification $W$ is strongly symmetric. To see that the
SS condition is realized locally, apply the ‘if’ direction to obtain the
evolved end operators. By assumption, they transform in the same irrep
$\alpha$ as the inital end operators; therefore, the local-SS condition
$[\nu]=0$ must hold.

###  5.4 Preservation of string order by strongly symmetric local Lindbladians

Following Ref. [30], we say a Lindbladian is _local_ if it can be written as a
sum of local terms, each supported on an interval of length at most some
constant.

The preservation of string operators by locally-SS channels in sdQC (Lemma 2)
means that we can state the following analog to Theorem 1, where short times
are times that are small compared to the system size.

Conjecture: _Evolution generated by a local Lindbladian preserves SPTO at
short times if and only if the Lindbladian is strongly symmetric._

The conjecture is inspired by a plausible connection between local
Lindbladians and causal channels. Just as local unitary evolution is
approximated by locally-symmetric causal unitaries (in particular, circuits of
symmetric local unitary gates) precisely when the generating Hamiltonian is
symmetric, we expect that

Evolution by a local Lindbladian is approximated by locally-WS/SS channels in
sdQC precisely when the Lindbladian is WS/SS. (5.21)

Let us motivate this statement nonrigorously. Local Lindbladian evolution is
subject to Lieb-Robinson bounds [30], so we expect it to be described by
causal channels (with range $r$ linear in time), up to exponentially small
errors outside of the lightcone. Moreover, we expect such causal channels to
live in dQC since nontrivial convex combinations of channels in dQC (which
plausibly are arbitrary causal channels [47]) seem to introduce unphysically
long-range correlations. We established in section 3.4 that WS/SS of a
Lindbladian implies WS/SS of the channels it generates, but the question
remains whether locality of the symmetric Lindbladian implies that the channel
is in sdQC and that the symmetry of the channel is locally realized. As
mentioned previously, it may be the case that causal channels approximating
local Lindbladian evolution are circuits of local channels, just as causal
unitaries approximating local unitary evolution are circuits of local
unitaries, and that these local ‘gates’ are WS/SS precisely when the
generating Lindbladian is WS/SS. Then an argument like Eq. (5.10) would
translate the symmetry of the gates into locally realized symmetry of the
channel.

Taking the statement (5.21) for granted and neglecting the issue of
approximation, the ‘if’ direction of the conjecture follows from Lemma 2. The
strongly symmetric local Lindbladian generates locally-SS channels in sdQC,
which by Lemma 2 preserve the types of string operators and therefore their
patterns of zeros with generic end operators. The short time of the evolution
is a crucial assumption, as it was necessary in Lemma 2 that the ranges of the
causal channels were small compared to the string length; otherwise, the bulks
of the strings were swallowed up by the end operators. The ‘only if’ direction
of the conjecture might require analyzing the transfer matrix of the
expectation value of the evolved string operator, as in section 4.1.1.

###  5.5 Protected edge modes

As in section 4.4, consider a pure state $|\psi^{\omega}\rangle$ in an SPT
phase characterized by the invariant $\omega$. Under the channel, it evolves
into a mixture of states $K_{i}|\psi^{\omega}\rangle$. The SPT invariant will
be obtained by cutting the system into left and right halves and acting on the
right half by the symmetry. Across the cut, the state and Kraus operators
decompose as
$|\psi^{\omega}\rangle=\sum_{a}|\psi_{l,a}^{\omega}\rangle\otimes|\psi_{r,a}^{\omega}\rangle$
and $K_{i}=\sum_{\mu}K_{i,l}^{\mu}\otimes K_{i,r}^{\mu}$. The SS condition
means that Eq. (5.6) holds. Therefore, the states in the mixture transform as

$(\mathds{1}^{l}\otimes
U_{g}^{r})K_{i}|\psi^{\omega}\rangle=\sum_{a,b,\mu,\nu}(V_{g})_{ab}(Q_{g})_{\mu\nu}K_{i,l}^{\mu}|\psi_{l,a}^{\omega}\rangle\otimes
K_{i,r}^{\nu}|\psi_{l,b}^{\omega}\rangle~{},$ (5.22)

so their SPT invariants are captured by the projectivity class $[\omega\nu]$
of $V\otimes Q$. If the SS condition is realized locally, $[\nu]$ is trivial,
so the SPT invariant $[\omega\nu]=[\omega]$ is unchanged.

This argument has a simple diagrammatic representation when the state
$|\psi^{\omega}\rangle$ is an MPS. In this case, the MPO tensor for $K_{i}$ is
contracted with the MPS tensor for $|\psi^{\omega}\rangle$ to obtain an MPS
tensor for $K_{i}|\psi^{\omega}\rangle$. The virtual space of the new MPS
tensor has symmetry action $V\otimes Q$ on blocks of size $r$.

$\begin{split}\includegraphics[width=281.85034pt]{edgemodes1a.PNG}\quad\raisebox{20.2322pt}{.}\end{split}$
(5.23)

Crucial to the preservation of protected edge modes is the fastness
assumption, which means that the range $r$ of the MPO is small compared to the
system size; without it, there is no invariant to be extracted locally.

##  6 Transmutation of SPT phases by twisted symmetric channels

This section discusses versions of the symmetry conditions that are twisted by
an endomorphism $\sigma:G\rightarrow G$ of the symmetry group. The twisted
conditions are stated in Eqs. (6.16), (6.17), and (6.18). For $\sigma$ that
are not identity-connected (every nontrivial $\sigma$ when $G$ is finite and
all but those of the form (4.6) otherwise [39]), channels twisted by $\sigma$
are not generated by continuous symmetric Lindbladian evolution in finite
time. These channels therefore describe infinite time evolution (for example
in section 6.4.3) and discrete noise.

First we discuss an action of group endomorphisms $\sigma$ on the SPT
invariant $[\omega]$. We then argue in Theorem 2 that $\sigma$-twisted SS
channels have the effect of changing the SPTO according to this action. In
particular, when an endomorphism $\sigma$ does not change $[\omega]$, channels
satisfying the $\sigma$-twisted SS condition preserve the phase with invariant
$[\omega]$. This allows us to answer a question we had previously deferred –
of the _necessary_ condition for a channel to preserve a given SPTO. The
answer is that the channel must be a mixture of $\sigma$-twisted SS channels
for $\sigma$ that fix the SPTO. We also discuss the general situation where a
channel does not preserve the phase but rather transmutes it into one of equal
or lesser complexity, defined in Eq. (4.25).

###  6.1 The action of endomorphisms on SPT phases

An endomorphism $\sigma:G\rightarrow G$ acts on the cocycle $\omega$ as a
pullback. Concretely,

$\sigma:\omega\mapsto\sigma^{*}\omega~{},\qquad(\sigma^{*}\omega)(g,h)=\omega(\sigma(g),\sigma(h))~{}.$
(6.1)

The action of endomorphisms has the following property:

An automorphism preserves the complexity of phases. (6.2)

This is because, if $\sigma$ is an automorphism, the transformed projective
center

$K_{\sigma^{*}\omega}=\\{\,g\,:\,\omega(\sigma(g),\sigma(h))=\omega(\sigma(h),\sigma(g))~{}\forall\,h\,\\}~{}.$
(6.3)

equals $\sigma^{*}K_{\omega}$, since $\sigma(h)$ runs over the whole $G$, and
this in turn is isomorphic to $K_{\omega}$ by $\sigma^{*}$. The converse to
Claim (6.2) is false because noninvertible endomorphisms may also preserve
complexity. As a counterexample, take any $G,\sigma$ with $\omega=1$. Less
trivially, take $G=H_{1}\times H_{2}$ and $\omega=P^{*}\omega_{1}$, where $P$
projects onto $H_{1}$ and $\omega_{1}$ is any cocycle on $H_{1}$. The
endomorphism $\sigma=P$ is not an automorphism, yet it fixes $\omega$ and its
complexity. Despite the lack of a full converse, one can make the following
weaker claim:

An endomorphism maps MNC phases, and only MNC phases, to MNC phases if and
only if it is an automorphism. (6.4)

In other words, an endomorphism preserves the distinction between MNC and non-
MNC phases precisely when it is an automorphism. The ‘if’ direction follows
from Claim (6.2). To see the ‘only if’ direction, note that the kernel of
$\sigma$ is contained in $K_{\sigma^{*}\omega}$, so the MNC condition
$K_{\sigma^{*}\omega}=\\{1\\}$ implies that $\ker\sigma=\\{1\\}$. The
properties (6.2) and (6.4) appear in Figure 2 as constraints on the arrows
between nodes. The special case of
$G={\mathbb{Z}}_{12}\times{\mathbb{Z}}_{12}$ is explored in complete detail in
Figure 3.

Figure 2:  Phases $\omega$ appear as nodes and the action of endomorphisms
$\sigma$ as arrows between nodes. From left to right, the columns depict the
identity endormorphism, automorphisms, noninvertible endomorphisms, and the
constant endomorphism. In anticipation of the implementation of these
endomorphism actions by twisted symmetric channels (c.f. section 6.4), the
columns are labeled by the corresponding symmetry conditions.

Let us examine in detail one of the most studied settings for investigations
of one-dimensional SPT phases – that of symmetry group
$G={\mathbb{Z}}_{n}\times{\mathbb{Z}}_{n}$, where phases are classified by
$H^{2}(G;U(1))={\mathbb{Z}}_{n}$.

Elements of ${\mathbb{Z}}_{n}\times{\mathbb{Z}}_{n}$ are “vectors” $(w,x)$
with $w,x\in{\mathbb{Z}}_{n}$. Endomorphisms of
${\mathbb{Z}}_{n}\times{\mathbb{Z}}_{n}$ are matrices with entries in
${\mathbb{Z}}_{n}$ that act on these vectors by matrix multiplication:

$\displaystyle\begin{split}&\,{\rm
End}({\mathbb{Z}}_{n}\times{\mathbb{Z}}_{n})=M_{2}({\mathbb{Z}}_{n})=\left\\{\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\,:\,a,b,c,d\in{\mathbb{Z}}_{n}\right\\}~{},\\\
&g=\left(\begin{array}[]{c}w\\\
x\end{array}\right)~{},\qquad\sigma(g)=\left(\begin{array}[]{cc}a&b\\\
c&d\end{array}\right)\left(\begin{array}[]{c}w\\\
x\end{array}\right)=\left(\begin{array}[]{c}aw+bx\\\
cw+dx\end{array}\right)~{}.\end{split}$ (6.5)

Automorphisms are those with invertible matrix, i.e. where the determinant
$ad-bc$ is relatively prime to $n$.

Now let’s discuss cocycles. The $n$ classes of
$H^{2}({\mathbb{Z}}_{n}\times{\mathbb{Z}}_{n};U(1))={\mathbb{Z}}_{n}$ are
represented by cocycles

$\omega_{k}[(w,x),(y,z)]=\exp(\tfrac{2\pi i}{n}k\,xy)~{}.$ (6.6)

Note that $\exp(\tfrac{2\pi i}{n}(-k)\,wz)$ is cohomologous to $\omega_{k}$ by
the coboundary of $\phi[w,x]=\exp(\tfrac{2\pi i}{n}\,wx)$ and that
$\exp(\tfrac{2\pi i}{n}\,wy)$ and $\exp(\tfrac{2\pi i}{n}\,xz)$ are
trivialized by $\phi[w,x]=\exp(\tfrac{2\pi i}{n}\,w)$ and
$\phi[w,x]=\exp(\tfrac{2\pi i}{n}\,x)$, respectively.

An endomorphism $\sigma$ on ${\mathbb{Z}}_{n}\times{\mathbb{Z}}_{n}$ induces
an endomorphism $\sigma^{*}$ on $H^{2}={\mathbb{Z}}_{n}$ as follows:

$\displaystyle\begin{split}(\sigma^{*}\omega_{k})[(w,x),(y,z)]&=\omega_{k}[\sigma(w,x),\sigma(y,z)]\\\
&=\omega_{k}[(aw+bx,cw+dx),(ay+bz,cy+dz)]\\\ &=\exp(\tfrac{2\pi
i}{n}k\,(cw+dx)(ay+bz))\\\ &=\exp(\tfrac{2\pi
i}{n}k\,(ac\,wy+bc\,wz+ad\,xy+bd\,xz))\\\ &\sim\exp(\tfrac{2\pi i}{n}k\,(ad-
bc)\,xy)\\\ &=\omega_{k(ad-bc)}[(w,x),(y,z)]~{}.\end{split}$ (6.7)

The penultimate line holds up to coboundaries. We conclude that the action of
$\sigma^{*}$ on the group of SPT phases is multiplication of the SPT index $k$
by the determinant $(ad-bc)$ of $\sigma$.

Endomorphisms of ${\mathbb{Z}}_{n}$ are given by multiplication by an element
of ${\mathbb{Z}}_{n}$, while automorphisms are those where the multiplication
is by a generator of ${\mathbb{Z}}_{n}$, i.e. by a number relatively prime to
$n$. This means that $\sigma^{*}$ is an automorphism of the group of SPT
phases precisely when $\sigma$ is an automorphism of $G$.

For example, the automorphism $\sigma(w,x)=(x,w)$ that exchanges the two
factors has the effect of inverting SPT phases since it has determinant $-1$.
(For $n=2$, inversion is the identity, so the two phases – trivial and Haldane
– are fixed by the exchange automorphism.) On the other hand, the endomorphism
$\sigma(w,x)=(w,e)$ that collapses the second factor to the identity has
determinant $0$, so it destroys all SPT phases.

Let us compute the projective center $K_{\omega_{k}}$, the set of elements
$(w,x)$ such that

$\exp(\tfrac{2\pi
i}{n}k\,xy)=\omega_{k}[(w,x),(y,z)]=\omega_{k}[(y,z),(w,x)]=\exp(\tfrac{2\pi
i}{n}k\,wz)~{},\quad\forall\,(y,z)~{},$ (6.8)

i.e. such that $k\,xy\equiv k\,wz\mod n~{},\,\forall\,y,z$. Taking $z=0$ while
varying $y$ and vice versa, we find

$K_{\omega_{k}}=\\{\,(w,x)\in{\mathbb{Z}}_{n}\times{\mathbb{Z}}_{n}\,:\,k\cdot(w,x)\equiv
0\\}~{}.$ (6.9)

In particular, when $k$ is coprime to $n$, the projective center is trivial,
so the cocycle $\omega_{k}$ is MNC. The invariant $(\det\sigma)\,k$ of the
transformed phase is coprime to $n$ precisely when $k$ and $(\det\sigma)$ and
both coprime to $n$; that is, when the original phase is MNC and $\sigma$ is
an automorphism, in agreement with Claim 6.4.

Figure 3:  The $12$ phases of the group
$G={\mathbb{Z}}_{12}\times{\mathbb{Z}}_{12}$ are depicted as nodes. Arrows
represent the actions of two endomorphisms: on the left, an automorphism with
$\det\sigma=5$; on the right, a noninvertible endomorphism with
$\det\sigma=3$. Observe that the automorphism preserves complexity, as
required by Claim (6.2), by preserving phases $3,6,9,0$ and exchanging the
remaining phases with others of equal complexity. On the other hand, the
noninvertible endomorphism reduces the SPT complexity of all MNC phases
($1,5,7,11$), as required by Claim (6.4) but nevertheless preserves the SPT
complexity of phases $3,6,9$, even fixing phase $6$.

###  6.2 Patterns of zeros under endomorphisms

Let $G$ be abelian. The cocycle $\omega$ defines a “pattern of zeros”
$\zeta_{\omega}:G\rightarrow G^{*}$ given by

$\zeta_{\omega}:g\mapsto\chi_{g}^{\omega}(\cdot)=\frac{\omega(\cdot,g)}{\omega(g,\cdot)}~{}.$
(6.10)

The image is indeed linear characters (one-dimensional representations) since

$\chi_{g}^{\omega}(h)\chi_{g}^{\omega}(k)=\frac{\omega(h,g)}{\omega(g,h)}\frac{\omega(k,g)}{\omega(g,k)}=\frac{\omega(k,g)\omega(h,gk)}{\omega(g,h)\omega(hg,k)}\overset{\text{abelian}}{=}\frac{\omega(k,g)\omega(h,kg)}{\omega(g,h)\omega(gh,k)}=\frac{\omega(hk,g)}{\omega(g,hk)}=\chi_{g}^{\omega}(hk)~{}.$
(6.11)

The kernel of $\zeta_{\omega}$ is the projective center $K_{\omega}$ (4.26).

The pattern of zeros $\zeta_{\omega}$ determines the cohomology class
$[\omega]$ of the cocycle $\omega$. To see this, note that the map
$\omega\mapsto\zeta_{\omega}$ is a group homomorphism:
$\zeta_{\omega_{1}}\zeta_{\omega_{2}}=\zeta_{\omega_{1}\omega_{2}}$;
therefore, is suffices to check that its kernel consists of coboundaries.
Suppose $\omega\mapsto 1$, i.e. $\omega(g,h)=\omega(h,g)$ for all $g,h$. Then
any projective representation with class $\omega$ satisfies
$V_{g}V_{h}=V_{h}V_{g}$. By Schur’s lemma, $V_{g}$ must be proportional to the
identity by a scalar $\lambda(g)$. Then $\omega$ is the coboundary of
$\lambda$ since

$\lambda(g)\lambda(h)\mathds{1}=V_{g}V_{h}=\omega(g,h)V_{gh}=\lambda(gh)\omega(g,h)\mathds{1}~{}.$
(6.12)

It is convenient to represent a pattern of zeros $\zeta_{\omega}$ as a two-
dimensional array with columns indexed by group elements $g$ and rows indexed
by linear characters $\alpha$. The entry $(g,\alpha)$ in this array is $\star$
if $\zeta_{\omega}(g)=\chi_{\alpha}$ and zero otherwise. Since
$\zeta_{\omega}(g)$ is a particular linear character, there is exactly one
$\star$ per column. The row indexed by $\alpha$ has either $|K_{\omega}|$ or
zero $\star$’s depending on whether $\chi_{\alpha}$ is in the image of
$\zeta_{\omega}$. The rank of the array is $|G|/|K_{\omega}|$. For example,
the two phases of symmetry $G={\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2}$ have
patterns of zeros

$\zeta_{\text{trivial}}=\left(\begin{array}[]{cccc}\star&\star&\star&\star\\\
0&0&0&0\\\ 0&0&0&0\\\
0&0&0&0\end{array}\right)~{},\quad\zeta_{\text{Haldane}}=\left(\begin{array}[]{cccc}\star&0&0&0\\\
0&\star&0&0\\\ 0&0&\star&0\\\ 0&0&0&\star\end{array}\right)~{},$ (6.13)

with columns indexed by $g=(0,0),(0,1),(1,0),(1,1)$ and rows by $\alpha$ with
$\tfrac{1}{\pi i}\log\chi_{\alpha}(w,x)=0,w,x,w+x$. Now consider the action of
an endomorphism $\sigma$. It acts on $\omega$ as Eq. (6.1) and on
$\zeta_{\omega}$ as

$\sigma\cdot\zeta_{\omega}=\zeta_{\sigma^{*}\omega}:g\mapsto\chi_{g}^{\sigma^{*}\omega}=\sigma^{*}\chi_{\sigma(g)}^{\omega}~{}.$
(6.14)

This rule tells us how the array for $\omega$ transforms into the array for
$\sigma^{*}\omega$:

For each group element $g$, look up the unique row $\beta$ of the old pattern
$\zeta_{\omega}$ such that the entry $(\sigma(g),\beta)$ is $\star$. Then
compose $\beta$ with $\sigma$ to obtain the row $\alpha$ of the new pattern
$\sigma\cdot\zeta_{\omega}$ such that the entry $(g,\alpha)$ is $\star$.
(6.15)

Using this rule for transforming patterns of zeros, one can check the examples
of endomorphisms introduced above. The exchange automorphism swaps the middle
two rows and swaps the middle two columns, fixing both the trivial and Haldane
patterns. On the other hand, the endomorphism that collapses the second factor
copies the first the third columns, which have $g$ in the image of $\sigma$,
and moves their $\star$ entries up according to $\sigma$; the result is that
both patterns are mapped to the trivial one.

The MNC property has a meaning in terms of patterns of zeros: the only column
with a $\star$ in the $\alpha=1$ row is the $g=1$ column. Claim 6.4 can be
shown in this language. Consider the ‘if’ direction. We wish to find the
entries $(g,1)$ of the new pattern that are $\star$. If $\sigma$ is an
automorphism, these entries are the entries
$(\sigma(g),(\sigma^{-1})^{*}1)=(\sigma(g),1)$ of the old pattern. Precisely
when the old pattern is MNC, the only of these entries with $\star$ is the one
with $\sigma(g)=1$ and so, since $\sigma$ is an automorphism, $g=1$ is the
only solution and the new pattern is MNC. Consider the ‘only if’ direction.
The entry $(\sigma(g),1)$ of the old pattern is the entry
$(g,\sigma^{*}1)=(g,1)$ of the new pattern. When the new pattern is MNC, it
has a $\star$ in this row only for $h=1$. Precisely when the old pattern is
MNC, it does only for $\sigma(g)=1$, which means the new pattern does for all
$h\in\ker\sigma$; therefore, precisely in this case do we have
$\ker\sigma=\\{1\\}$, which is to say that $\sigma$ is an automorphism.

###  6.3 Twisted symmetric channels

Having understood the action of group endomorphsisms on phases, we turn to
studying the channels that implement it. Here, we introduce twisted symmetry
conditions and discuss the structure of Kraus operators of twisted symmetric
channels. Later we will argue that the $\sigma$-SS condition implements the
action of $\sigma$.

The $\sigma$-twisted weak symmetry ($\sigma$-WS) condition is

$\boxed{\qquad\mathcal{U}_{g}\circ\mathcal{E}\circ\mathcal{U}_{\sigma(g)}^{\dagger}=\mathcal{E}~{},\quad\forall\,g~{}.\qquad\text{($\sigma$-twisted
weak symmetry condition)}\qquad}$ (6.16)

By setting the phases $\theta_{i}$ equal as before, we obtain the
$\sigma$-twisted strong symmetry ($\sigma$-SS) condition:

$\boxed{\qquad
U_{g}K_{i}U_{\sigma(g)}^{\dagger}=e^{i\theta(g)}K_{i}~{},\quad\forall\,i,g~{}.\qquad\text{($\sigma$-twisted
strong symmetry condition)}\qquad}$ (6.17)

Using the argument from before but with $X=U_{g}$,
$Y=e^{i\theta(g)}U_{\sigma(g)}$, we obtain the alternative statement

$\boxed{\qquad\mathcal{E}^{\dagger}(U_{g})=e^{i\theta(g)}U_{\sigma(g)}~{},\quad\forall\,g~{}.\qquad\text{($\sigma$-twisted
strong symmetry condition)}\qquad}$ (6.18)

The untwisted SS condition means that the channel decomposes as a sum of
channels on irrep blocks, acting only within multiplicity spaces. A similar
statement holds for $\sigma$-SS channels: each Kraus operator has a block
decomposition $K_{i}=\oplus_{\alpha\beta}K_{i}^{\alpha\beta}$ such that the
component $K_{i}^{\alpha\beta}$, which is a map from the multiplicity space of
$\beta$ to that of $\alpha$, vanishes unless $\alpha=\sigma^{*}\beta$. This is
because the $\sigma$-SS condition says that $K_{i}$ maps to a space where $g$
acts as $\sigma(g)$ did before mapping. This means the completeness condition
on $\mathcal{E}$ implies

$\sum_{i}(K_{i}^{\sigma^{*}\alpha,\alpha})^{\dagger}K_{i}^{\sigma^{*}\alpha,\beta}=\sum_{i}(K_{i}^{\dagger}K_{i})^{\alpha\beta}=\mathds{1}\delta^{\alpha\beta}~{},$
(6.19)

which in particular enforces a completeness condition on the channels

$\mathcal{E}_{\alpha}(\rho)=\sum_{i}(K_{i}^{\sigma^{*}\alpha,\alpha})^{\dagger}\rho\,K_{i}^{\sigma^{*}\alpha,\alpha}~{}.$
(6.20)

When $\sigma$ is not an automorphism, each term $K_{i}^{\dagger}K_{i}$ may
have off-diagonal components: $K_{i}$ maps $\alpha$ to $\sigma^{*}\alpha$,
which $K_{i}^{\dagger}$ maps back to any $\beta$ in the preimage. Eq. (6.19)
implies these must cancel in the sum.

Extending the untwisted class sdQC of causal channels, one can define
$\sigma$-sdQC as the set of channels with a purification that is both causal
and $\sigma$-twisted symmetric under an on-site symmetry. The folded MPS of
the MPU representing the purification has a symmetry $U_{g}\otimes
U_{\sigma(g)}^{\dagger}\otimes U_{g}^{A}\otimes(U_{\sigma(g)}^{A})^{\dagger}$,
which defines a projective representation $Q$. Local realization of the
symmetry is again the condition that $Q$ is linear.

While strong symmetry twisted by an automorphism is possible in reversible
channels, strong symmetry twisted by a noninvertible endomorphism is
not.161616A stronger statement also holds: noninvertible twists are impossible
not just in reversible channels but in all channels that are invertible as
linear maps. To see this, note that $\mathcal{E}^{\dagger}$ annihilates
$U_{g}-\mathds{1}$ for $g\in\ker\sigma$ and that $\mathcal{E}$ is invertible
iff $\mathcal{E}^{\dagger}$ is. To see this, suppose $g$ belongs to the kernel
of $\sigma$. Then the single Kraus operator $K$ of the reversible channel
satisfies $K^{\dagger}U_{g}K=\mathds{1}$, but this implies $U_{g}=\mathds{1}$,
so $g=1$ by faithfulness. In light of Theorem 2 (below), this means that
reduction of SPT complexity – as opposed to change of phase at a fixed SPT
complexity level – is a phenomenon unique to irreversible channels. On the
other hand, any automorphism can be realized by a reversible channel: let
$U_{g}$ contain one copy of each irrep and let $K$ be the permutation matrix
that implements the induced action $\sigma^{*}$ on irreps.

The impossibility of noninvertible twists for reversible channels is reflected
in purifications. If $\sigma$ is an automorphism, the construction in section
3.3 yields purifications $W$ of $\sigma$-SS channels that satisfy

$(U_{g}\otimes\mathds{1}^{A})W=W(U_{\sigma(g)}\otimes\mathds{1}^{A})~{}.$
(6.21)

However, if $\sigma$ is not invertible, the irrep block structure of the Kraus
operators means that some rows of $W$ constructed this way must be zero,
meaning it is not unitary and so not a valid purification.

Endomorphisms compose contravariantly under the composition of channels. If
$\mathcal{E}$ is a $\sigma$-WS channel and $\mathcal{E}^{\prime}$ a
$\sigma^{\prime}$-WS channel, their composition
$\mathcal{E}\circ\mathcal{E}^{\prime}$ has WS twisted by
$\sigma^{\prime}\circ\sigma$, as can be seen by

$\mathcal{U}_{g}\circ\mathcal{E}\circ\mathcal{E}^{\prime}\circ\mathcal{U}_{(\sigma^{\prime}\circ\sigma)(g)}^{\dagger}=\mathcal{U}_{g}\circ\mathcal{E}\circ\mathcal{U}_{\sigma(g)}^{\dagger}\circ\mathcal{E}^{\prime}=\mathcal{E}\circ\mathcal{E}^{\prime}~{}.$
(6.22)

If both $\mathcal{E}$ and $\mathcal{E}^{\prime}$ have $\sigma$\- and
$\sigma^{\prime}$\- twisted-SS, respectively, the composition
$\mathcal{E}\circ\mathcal{E}^{\prime}$ has $(\sigma^{\prime}\circ\sigma)$-SS
since

$(\mathcal{E}\circ\mathcal{E}^{\prime})^{\dagger}(U_{g})=(\mathcal{E}^{\prime\dagger}\circ\mathcal{E}^{\dagger})(U_{g})=\mathcal{E}^{\prime\dagger}(U_{\sigma(g)})=U_{(\sigma^{\prime}\circ\sigma)(g)}~{}.$
(6.23)

We also note that convex combinations and tensor products of $\sigma$-WS/SS
channels are $\sigma$-WS/SS.

###  6.4 Transmutation of SPT phases

This section is dedicated to showing that certain $\sigma$-twisted strongly
symmetric channels have the effect of transmuting one coherent SPT phase into
another, according to the action of $\sigma$ on the SPT invariants.

It is not the case that all $\sigma$-SS channels perform transmutation of SPT
phases, but most of them do. We state our result for $\sigma$-SS channels
satisfying a genericness condition

$\Phi_{\alpha}\neq 0~{},\quad\forall\,\alpha\in{\rm im}\,\sigma~{},$ (6.24)

where $\Phi_{\alpha}$ is the $\alpha$-labeled component of $\mathcal{E}$ in
the irrep block decomposition of $\mathcal{E}$ discussed in section 3.1. This
condition excludes, for example, the fully dephasing channel of section 4.2,
which destroys SPT order despite having strong symmetry. It includes all
channels generated by Lindbladians in finite time. In the following theorem,
“coherent SPT phase” means a class of coherent SPT states with a given pattern
of zeros (which, by Theorem 1 and the conjecture of section 5.4, is a phase
defined by strongly symmetric Lindbladians). Mapping an SPT phase to another
refers to mapping every state in one phase to a state in the other.

Theorem 2: _Generic locally $\sigma$-twisted strongly symmetric channels map
the coherent SPT phase with invariant $\omega$ to the phase with invariant
$\omega^{\prime}=\sigma^{*}\omega$.

For uncorrelated noise, the converse holds:
if a channel maps the coherent SPT phase with invariant $\omega$ to the phase
with invariant $\omega^{\prime}$, it is $\sigma$-twisted strongly symmetric
for some $\sigma$ with $\sigma^{*}\omega=\omega^{\prime}$. _

As we will discuss later, the converse statement is false for causal
channels.171717We also remark about the converse statement that the
genericness condition (6.24) is sufficient but not necessary: a weaker
$\omega^{\prime}$-dependent genericness condition on only the subset of the
$\alpha$ that appear in the pattern of $\omega^{\prime}$ is enough.

A consequence of Theorem 2 is that, among channels of uncorrelated noise,
twisted strongly symmetric channels are precisely those that map within the
space of SPT states. Focusing on the MNC phases discussed in section 6.2, we
can also conclude that, among channels of uncorrelated noise, _automorphism-
twisted_ strongly symmetric channels are precisely those that map within the
space of MNC SPT states.

Theorem 2 tells us when a channel preserves a given SPTO:

Corollary: _The SPTO with invariant $\omega$ is preserved by generic locally
$\sigma$-twisted strongly symmetric channels with $\sigma$ that fix $\omega$.
Among channels of uncorrelated noise, $\sigma$-twisted strongly symmetric
channels with such $\sigma$ are the only channels that preserve this SPTO._

To prove the theorem, we need the following lemma, which generalizes Lemma 1
(for uncorrelated noise, not necessarily symmetric) and Lemma 2 (for channels
in $\sigma$-sdQC, which in particular are $\sigma$-WS) by adding a twist
$\sigma$. Let $s(g,\alpha_{l},\alpha_{r})$ denote a string operator with end
operators transforming in $\alpha_{l}$ and $\alpha_{r}^{*}$,
respectively.181818We previously considered only string operators with
$\alpha_{l}=\alpha_{r}$ since these are the ones with nonvanishing patterns of
zeros.

Lemma 3: _Consider either a channel of uncorrelated noise or a translation-
invariant causal channel in $\sigma$-sdQC.

The channel satisfies the $\sigma$-twisted (local) strong symmetry condition
if and only if it maps each string operator $s(g,\alpha,\alpha)$ to a sum of
string operators $s(\sigma(g),\beta_{l},\beta_{r})$, where
$\sigma^{*}\beta_{l,r}=\alpha$ (if no $\beta_{l,r}$ exists, the sum is
empty)._

Let us now prove the lemma. The label on the string bulk is changed from $g$
to $\sigma(g)$ if and only if $\mathcal{E}$ is $\sigma$-SS. In the case of
uncorrelated noise, this is because each $\mathcal{E}_{s}$ is $\sigma$-SS and
so $\mathcal{E}_{s}^{\dagger}(U_{g})=e^{i\theta(g)}U_{\sigma(g)}$. For a
channel in $\sigma$-sdQC, the argument is essentially that of section 5.3. The
label on the ends of the string are changed from $\alpha$ to a sum of irreps
$\beta_{l,r}$ satisfying $\sigma^{*}\beta_{l,r}=\alpha$. This is because the
superoperators $S_{l,r}^{g}$ (5.14) (which are simply
$\mathcal{E}_{l,r}^{\dagger}$ for uncorrelated noise) are invariant under
acting on the inner legs with $U_{h}$ and on the outer legs with $U_{k}$ for
$\sigma(k)=h$, as can be seen by an argument like Eq. (5.15). This means that
they map the representation space $\alpha$ on the inner legs to its preimage
under $\sigma^{*}$ on the outer legs. We have shown that the string operator
evolves into an operator
$s(\sigma(g),\mathcal{S}_{l}^{g}(\mathcal{O}^{l}_{\alpha}),\mathcal{S}_{r}^{g}(\mathcal{O}^{r}_{\alpha}))$
and that $\mathcal{S}_{l,r}^{g}(\mathcal{O}^{l,r}_{\alpha})$ is a sum of end
operators that transform with $\beta_{l,r}$ such that
$\sigma^{*}\beta_{l,r}=\alpha$, proving the lemma. For uncorrelated noise, an
alternative way of understanding the change in representation labeling the end
operators is with the block decomposition of the Kraus operators. Since
$\mathcal{E}_{l,r}$ are $\sigma$-SS, their Kraus operators vanish outside of
the irrep blocks $K_{i}^{\lambda\tau}$ with $\tau$ in the preimage of
$\lambda$ under $\sigma^{*}$. Meanwhile $\mathcal{O}_{\alpha}^{l,r}$ have
nonvanishing blocks for irreps $\lambda^{\prime},\lambda$ such that
$\lambda^{\prime}\otimes\lambda^{*}=\alpha$. Putting these together, the
nonvanishing blocks of each term
$K_{i}^{\dagger}\mathcal{O}_{\alpha}^{l,r}K_{i}$ in the evolved end operator
occur at irreps $\tau^{\prime},\tau$ in the preimage of
$\lambda^{\prime},\lambda$ with $\lambda^{\prime}\otimes\lambda^{*}=\alpha$.
Each of these blocks has $\tau^{\prime}\otimes\tau^{*}$ in the preimage of
$\alpha$, so we conclude that
$\mathcal{E}_{l,r}^{\dagger}(\mathcal{O}_{\alpha}^{l,r})$ lives in the sum of
irreps $\beta$ with $\sigma^{*}\beta=\alpha$.

With Lemma 3 in hand, let us turn toward proving Theorem 2 by first
reformulating it in terms of patterns of zeros. The pattern $\zeta_{\omega}$
of a state is understood as the collection of pairs $(g,\alpha)$ such that
$\langle s(g,\alpha,\alpha)\rangle$ is generically nonvanishing on the state.
By the rule (6.15), the pattern of zeros $\sigma\cdot\zeta_{\omega}$ consists
of pairs $(g,\alpha)$ such that $\alpha=\sigma^{*}\beta$ for the (unique)
$\beta$ for which $(\sigma(g),\beta)$ appears in the pattern $\zeta_{\omega}$.
The theorem demonstrates how this new pattern can be understood as expectation
values of evolved operators $\mathcal{E}^{\dagger}(s(g,\alpha,\alpha))$
evaluated on the original state. To be precise, the first half of the theorem
states that, on an SPT state,

If $\mathcal{E}$ is a generic locally $\sigma$-SS channel, then generically
$\langle\mathcal{E}^{\dagger}(s(g,\alpha,\alpha))\rangle\neq 0$ precisely for
the pairs $(g,\alpha)$ such that $\alpha=\sigma^{*}\beta$ for the (unique)
$\beta$ with $\langle s(\sigma(g),\beta,\beta)\rangle\neq 0$. (6.25)

To see why this is true, apply Lemma 3 to write
$\mathcal{E}^{\dagger}(s(g,\alpha,\alpha))$ as a sum of terms
$s(\sigma(g),\beta_{l},\beta_{r})$ with $\sigma^{*}\beta_{l,r}=\alpha$. Due to
the condition (6.24), these terms do not vanish (though the sum may be empty
if no $\beta_{l,r}$ exist). A pattern of zeros of an SPT state has a unique
entry $\beta$ per column, so the expectation values of the terms in the sum
vanish unless $\beta_{l}=\beta_{r}=\beta$; either zero or one terms do not
vanish. We have $\langle\mathcal{E}^{\dagger}(s(g,\alpha,\alpha))\rangle\neq
0$ when the nonvanishing term $\langle s(\sigma(g),\beta,\beta)\rangle$
appears in the sum. This can only happen when $\alpha=\sigma^{*}\beta$, and in
this case it generically happens, since generically the $\beta$-components of
the end operators are nonzero.

It remains to prove the second half of Theorem 2. The argument follows that of
section 4.1.1, except that $\sigma$ is no longer constrained to be connected
to the identity endomorphism. Now the condition that the transfer matrix (4.4)
has $\lambda_{\text{max}}=1$ implies that
$\mathcal{E}_{s}^{\dagger}(U_{g})=U_{h}$, which is to say that
$\mathcal{E}_{s}$ is $\sigma$-SS for some $\sigma$ with $\sigma(g)=h$. Then
apply the first half of Theorem 2 to see that the channel maps the phase
$\omega$ to the phase $\sigma^{*}\omega$, which by assumption is
$\omega^{\prime}$; therefore, $\sigma$ satisfies
$\sigma^{*}\omega=\omega^{\prime}$, as claimed.

As we mentioned earlier, the second half of Theorem 2 does not generalize from
uncorrelated noise to all causal channels. This is because there are causal
channels that are not locally $\sigma$-SS yet nevertheless transmute SPT
phases. For example, the phase $\omega$ is mapped to $\omega^{\prime}$ by
convex combinations of locally $\sigma_{i}$-SS channels where each
$\sigma_{i}^{*}\omega$ equals $\omega^{\prime}$. Additionally, one can add to
the convex combination a channel that is not $\sigma$-SS for any $\sigma$.
This extra factor annihilates string operator expectation values and so does
not alter the effect of the channel on string order. Finally, there are
channels that are $\sigma$-SS but not _locally_ $\sigma$-SS. These change the
bulk labels of strings from $g$ to $\sigma(g)$ and the end labels from
$\chi_{\alpha}$ to $\chi^{\nu}_{g}\chi_{\alpha}$, where $\nu$ is the
projectivity cocycle of $Q$. In doing so, they transform the pattern of zeros
in a way that endomorphism actions cannot; for example, if $\sigma=1$, the
cohomology invariant of the channel is simply added to that of the state:
$\omega\mapsto\omega+\nu$ [51].191919This is not surprising. In closed
systems, 1D SPT phases can be prepared by symmetric causal unitaries with
index $\nu=\omega$.

####  6.4.1 Edge modes perspective

The transformation of the SPT invariant $\omega$ under a $\sigma$-SS channel
can also be seen in terms of edge modes:

$\begin{split}\includegraphics[width=281.85034pt]{edgemodes2a.PNG}\end{split}$
(6.26)

The $\sigma$-SS condition means that $U_{g}$ is hit by $\sigma$ upon pulling
through $K_{i}$, and local realization means that $Q$ is linear. Then the edge
modes of the evolved MPS state transforms like $\sigma^{*}V\otimes Q$, which
has cocycle $\sigma^{*}\omega$.

####  6.4.2 Irrep probabilities perspective

Irrep probabilities were introduced in section 4.5, where it was shown that SS
Lindbladian evolution preserves them while non-SS Lindbladian evolution maps
them to the fully degenerate value $1/|G|$. In this section, we consider the
effects of (not necessarily Lindbladian) causal channels on irrep
probabilities.

We find that strongly symmetric channels twisted by automorphisms preserve the
degeneracies of irrep probabilities (though permute the irrep probabilities
themselves) for all SPT phases, while those twisted by noninvertible
endomorphisms reduce the degeneracy if the initial state is in an MNC phase
and either reduce or preserve it (depending on the phase) for non-MNC phases.
In light of the claim of Ref. [45] that the degeneracy of irrep probabilities
measures SPT complexity, this result reflects the behaviour of complexity that
we observed in section 6.1; in particular, in Figure 2. Meanwhile, channels
that are not $\sigma$-SS for any endormorphism $\sigma$ send the degeneracy to
the maximal value $|G|$, regardless of the initial state. This result means
that irrep probabilities and their degeneracy detect whether a channel is
twisted strongly symmetric.

To see that automorphism-twisted SS channels permute irrep probabilities, use
what we learned in section 6.4 about the transformation of the string
operators $\langle s(U_{g},\mathds{1},\mathds{1})\rangle$ to compute

$p_{\alpha}\longmapsto\frac{1}{|G|}\sum_{g}\chi_{\alpha}(g)\langle
s(U_{\sigma(g)},\mathds{1},\mathds{1})\rangle=\frac{1}{|G|}\sum_{g}\chi_{\alpha}(\sigma^{-1}(g))\langle
s(U_{g},\mathds{1},\mathds{1})\rangle=p_{(\sigma^{-1})^{*}\alpha}~{}.$ (6.27)

A similar computation can be performed for endomorphisms. Fix a set of
elements $h\in G$ that represent the cosets of the quotient $G/\ker\sigma$.
Then

$\displaystyle\begin{split}p_{\alpha}\longmapsto\,&\frac{1}{|G|}\sum_{g}\chi_{\alpha}(g)\langle
s(U_{\sigma(g)},\mathds{1},\mathds{1})\rangle\\\
=&\left(\frac{|\ker\sigma|}{|G|}\sum_{h}\chi_{\alpha}(h)\langle
s(U_{\sigma(h)},\mathds{1},\mathds{1})\rangle\right)\left(\frac{1}{|\ker\sigma|}\sum_{k\in\ker\sigma}\chi_{\alpha}(k)\right)~{}.\end{split}$
(6.28)

Orthogonality of characters means that the sum over $k\in\ker\sigma$ enforces
the constraint that $\alpha$ restricted to $\ker\sigma$ is trivial. For the
characters $\alpha$ without this property (of which there is at least one if
$\sigma$ is noninvertible), the corresponding irrep probability $p_{\alpha}$
must vanish. When at least one of the $p_{\alpha}$’s vanishes, they cannot be
fully degenerate, so there is less degeneracy than for a MNC state. We
conclude that SS channels twisted by noninvertible endomorphisms reduce the
degeneracy of MNC phases. Finally, channels that are not $\sigma$-SS for any
$\sigma$ annihilate the string order parameters without $g=1$, so we get again
the result (4.27) that the irrep probabilities become maximally degenerate,
regardless of the initial state.

Figure 4: A random state in the trivial phase of
${\mathbb{Z}}_{4}\times{\mathbb{Z}}_{4}$, which has nondegenerate irrep
probabilities. Under non-$\sigma$-SS channels the irrep probabilities become
exactly fully degenerate. For $\sigma$-SS channels, the nondegeneracy of the
trivial phase is preserved. The irrep numbered by $i=n+4m$ is given by
$(a,b)\mapsto\exp(\pi i(na+mb))$. The channels are defined in the text below.

Figure 5: A close-up of the irrep probabilities of a random state in the
trivial phase evolved by a $\sigma$-SS channel (in this case, the $1$-SS
dephasing channel) shows that they have no enforced degeneracy.

To illustrate these results, we perform numerical checks on random example SPT
states with symmetry $G={\mathbb{Z}}_{4}\times{\mathbb{Z}}_{4}$. We generate
random, injective, symmetric MPS of bond dimension $D=16$ in a particular SPT
phase, with support in all irrep sectors. Then we act on them with channels
satisfying various symmetry conditions. The resulting irrep probabilities are
depicted in Figures 4, 5 (trivial phase) and Figures 6, 7 (nontrivial phases).
For $Z_{4}\times{\mathbb{Z}}_{4}$, the automorphisms are those $\sigma$ with
$\det\sigma=1,3$, while the noninvertible endomorphisms are those with
$\det\sigma=0,2$.

We now give the explicit constructions for the example channels used for this
numerical investigation. We denote these twisted strongly symmetric channels