indeed only be subgroups of the whole symmetry group for non normal MPS.
Moreover, we show that the fact that any possible SLOCC transformation among
normal MPS can be performed with a global transformation is no longer true for
non normal MPS.

Note that for any odd particle number $N$ we have $|\Psi_{A_{b}}\rangle=0$,
i.e., the MPS vanishes. Considering the definition of MPS [see Eq. (2)], this
can be easily seen as follows. If $\mathbf{v}=\mathbf{0}$, then both
$A_{b}^{0}$ and $A_{b}^{1}$ are matrices of the form
$\begin{pmatrix}0&0&\cdot\\\ 0&0&\cdot\\\ \cdot&\cdot&0\end{pmatrix}$, where
$\cdot$ indicates an arbitrary (vanishing or non-vanishing) entry. This form
is retained by any product of matrices of such a form, which is comprised of
odd factors (see also Appendix C). In particular, the trace vanishes and thus
$|\Psi_{A_{b}}\rangle=0$. Conversely, it may be straightforwardly verified
that $|\Psi_{A_{b}}\rangle\neq 0$ for any even $N\geq 4$ (unless $\det b=0$).
In the following, we hence consider even $N\geq 4$.

Let us first consider MPS associated to diagonalizable $T$ as in box (I) in
Fig. 4. and consider the particularly interesting case $m=N$. In standard
form, we then have
$T=\operatorname{diag}(e^{i\frac{r\pi}{N}},e^{-i\frac{r\pi}{N}})$ for
$r\in\\{0,\ldots,\frac{N}{2}\\}$. Considering only global $(b\rightarrow
c)$-cycles these different states appear to be inequivalent. Note, however,
that the premises of Lemmata 4 and 5 (which stated that this suffices to
conclude that the associated MPS are inequivalent) are not fulfilled, as
$\mathbf{v}=\mathbf{0}$. Indeed, additional equivalences become apparent
taking non-global $(b\rightarrow c)$-cycles into account. More precisely,
considering Eq. (32) and
$g_{k}=\operatorname{diag}(e^{-i\frac{rk\pi}{N}},e^{i\frac{rk\pi}{N}})$ and
$B_{11}^{(k)}=1$ for $k\in\\{0,\ldots,N-1\\}$, we find that all
$T=\operatorname{diag}(e^{i\frac{r\pi}{N}},e^{-i\frac{r\pi}{N}})$ for even $r$
are equivalent to $T=\text{1}$. For odd $r$ this construction does not work
due to a sign mismatch in $B_{11}^{(k)}B_{11}^{(k+1)}$ in Eq. (32). Instead,
those can be shown to be equivalent to
$T=\operatorname{diag}(i,-i)=i\sigma_{z}$ using
$g_{k}=\operatorname{diag}(e^{-i\frac{(r-N/2)k\pi}{N}},e^{i\frac{(r-N/2)k\pi}{N}})$.
Thus, there are (at most) two SLOCC classes within box (I) for $m=N$. The two
representative MPS associated to $T=\text{1}$ and $T=i\sigma_{z}$ are in fact
the Majumdar-Ghosh states [36]

$\displaystyle|\Psi_{A_{b}}\rangle\propto$
$\displaystyle|\psi^{-}\rangle_{01}\ldots|\psi^{-}\rangle_{N-2\,N-1}\pm|\psi^{-}\rangle_{12}\ldots|\psi^{-}\rangle_{N-1\,0},$

where $+$ (-) corresponds to $T=\text{1}$ ($T=i\sigma_{z}$), respectively.
Note that $g^{\otimes N}$ is a symmetry of $|\Psi_{A_{b}}\rangle$ for any $g$.
Thus, the study of cycles (see Figure 4) clearly has revealed only a subgroup
of the symmetry group, as may be expected for non-normal tensors.

##  VI Fiducial states for a bond dimension $D>3$ corresponding to diagonal
matrix pencils

We discuss here the generic case of fiducial states with bond dimension larger
than 3 (see also Sec. III). The fiducial states correspond to diagonal matrix
pencils [32], i.e., we have

$\displaystyle|A\rangle=|0\rangle_{A}|\Phi^{+}_{D}\rangle_{BC}+|1\rangle_{A}(D\otimes\text{1})|\Phi^{+}_{D}\rangle_{BC},$

where $D=\operatorname{diag}(x_{1},\ldots,x_{D})$, with $x_{1},\ldots,x_{D}$
being the eigenvalues of the corresponding matrix pencil. Its symmetries are
of the form

$\displaystyle g\otimes P_{\sigma}^{-1}D_{g}^{-1}\tilde{D}\otimes
P_{\sigma}^{-1}\tilde{D}^{-1},$

where $g=\begin{pmatrix}\alpha&\beta\\\ \gamma&\delta\end{pmatrix}$ is such
that the set of eigenvalues of the pencil is mapped into itself (see Sec.
III).

Depending on the eigenvalues of the diagonal matrix pencil, i.e., the entries
of $D$, we have either: (i) the eigenvalues are such that the linear
fractional transformation given in Eq. (13) exists, or (ii) no such
transformation exists (which is the generic case).

In case (ii), which is the generic case, the fiducial state has only the
trivial qubit symmetry, which implies that the corresponding TIMPS has only
the trivial symmetry. Moreover, any TIMPS which corresponds to a fiducial
state which is in such a SLOCC class has only the trivial symmetry. In case
(i) non–trivial symmetries exist. A simple example of such a state would be
the fiducial state given in Eq. (33) with $x_{k}=\omega^{k}$, where
$\omega=e^{i\frac{2\pi}{D}}$141414Using the theory of Möbius transformations
one might derive necessary and sufficient conditions on the fiducial states to
possess non-trivial symmetries.. In general we have that in this case, $g$ is
such that for all $i\in\\{1,\ldots,D\\}$, $x_{i}\mapsto x_{\sigma(i)}$ for
some permutation $\sigma$. Moreover, $D_{g}=\operatorname{diag}(\gamma
x_{1}+\delta,\ldots,\gamma x_{D}+\delta)$, and
$P_{\sigma}=\sum_{i}|\sigma(i)\rangle\langle\sigma|$, and $\tilde{D}$ is an
arbitrary invertible diagonal matrix (see equation above). Then, it is
immediate that $\text{1}\otimes b\otimes\text{1}|A\rangle$ has the symmetries

$\displaystyle g\otimes bP_{\sigma}^{-1}D_{g}^{-1}\tilde{D}b^{-1}\otimes
P_{\sigma}^{-1}\tilde{D}^{-1}=S\otimes bxb^{-1}\otimes y^{T},$

where the RHS is standard MPS notation for symmetries of the fiducial state.

We outline in the following how all the symmetries of the corresponding TIMPS
can be determined. As in Sec. IV one would start out by solving the
concatenation condition [see Eq. (III.2)]. More precisely, in order to
determine the physical symmetry of the MPS we have to identify the $N$-cycles
within the symmetry group of the fiducial state. That is, we solve the
concatenation rules:

$\displaystyle y_{k}bx_{k+1}b^{-1}$ $\displaystyle\propto\text{1}\text{ for
}k\in\\{0,\ldots,N-2\\},$ $\displaystyle y_{0}bx_{N-1}b^{-1}$
$\displaystyle\propto\text{1}.$

For $|A_{b}\rangle$ as fiducial state, this condition is equivalent to

$\displaystyle
bP_{\sigma_{k+1}}^{-1}D_{{g}_{k+1}}^{-1}\tilde{D}_{k+1}b^{-1}\propto
P_{\sigma_{k}}^{-1}\tilde{D}_{k}.$

Thus, we can derive as a necessary condition that
$bP_{\sigma_{k+1}}^{-1}D_{{g}_{k+1}}^{-1}\tilde{D}_{k+1}b^{-1}$ must be
similar to $P_{\sigma_{k}}^{-1}\tilde{D}_{k}$. This implies that these
matrices must have the same eigenvalues. Matrices of the form $P_{\sigma}D$,
where $P_{\sigma}$ is a permutation matrix and $D$ is a diagonal matrix, are
called _generalized permutation matrices_ [37]. It turns out that their
eigenvalues are easy to calculate, as the following lemma shows.

######  Lemma 6 (Eigenvalues of monomial matrices).

Let $P_{\sigma}$ be a permutation matrix and $D$ a diagonal matrix. Then the
eigenvalues of $P_{\sigma}D$ can be determined as follows. Assume $\sigma$
decomposes into $l$ distinct cycles $\pi_{1},\ldots,\pi_{l}$. Let $d_{i}$
denote the $\operatorname{length}(\pi_{i})$-th root of the product of the
entries of $D$ associated with the cycle $\pi_{i}$. Then the eigenvalues of
$P_{\sigma}D$ are
$\left(d_{1}e^{i\frac{2k\pi}{\operatorname{length}(\pi_{1})}}\right)_{k=0}^{\operatorname{length}(\pi_{1})-1}\cup\ldots\cup\left(d_{l}e^{i\frac{2k\pi}{\operatorname{length}(\pi_{l})}}\right)_{k=0}^{\operatorname{length}(\pi_{l})-1}$.

###### Proof.

Let us fix a cycle $\pi_{i}$ and restrict $P_{\sigma}D$ to the subspace
spanned by the basis elements that $\pi_{i}$ does not leave invariant. This
matrix then has characteristic polynomial
$\lambda^{\operatorname{length}(\pi_{i})}-d_{i}^{\operatorname{length}(\pi_{i})}=0$,
and thus its eigenvalues are as stated in the lemma. ∎

Using now these necessary conditions for the existence of a cycle, similar
tools as the ones presented in Sec IV can be utilized to determine all
symmetries of the corresponding TIMPS.

##  VII Entanglement and LOCC transformations

Before concluding, let us briefly discuss the implication of the results
derived here in the context of entanglement theory. As mentioned in the
introduction, if a state, $|\Psi\rangle$ can be transformed deterministically
via LOCC into some other state $|\Phi\rangle$, then $E(|\Psi\rangle)\geq
E(|\Phi\rangle)$ for any entanglement measure $E$. Hence, LOCC transformations
induce a partial order on the set of entangled states. As shown in [38, 39,
40] local symmetries play an important role in characterizing all possible
LOCC transformations among pure states. As we have characterized all the local
symmetries of the TIMPS, it is straightforward to determine possible LOCC
transformations (at least in case the number of symmetries is finite). To give
a simple example a TIMPS $|\Psi\rangle$ can be transformed deterministically
into a state $h_{1}\otimes\text{1}\otimes\text{1}\otimes\ldots|\Psi\rangle$,
where $h_{1}$ is determined by the symmetries of $|\Psi\rangle$. More
precisely, if the symmetries are unitary symmetries on (at least) all but one
system, system 1, then the above mentioned transformation is possible if and
only if there exists a finite set of probabilities $\\{p_{k}\\}$ and
symmetries $g^{k}$ such that
$\sum_{k}p_{k}\left(g^{k}_{1}\right)^{\dagger}h_{1}^{\dagger}h_{1}g^{k}_{1}\propto\text{1}$.
Recall that the symmetry group of an MPS may depend on the particle number
$N$. In drastic cases, an MPS may exhibit the trivial symmetry group for
certain $N$, while the symmetry group is non-trivial for other $N$. A simple
example would be the MPS $\Psi_{1\,D^{3}}$ from Section II.1. Thus, it can be
easily seen that whether $|\Psi\rangle$ can be transformed deterministically
into a state $h_{1}\otimes\text{1}\otimes\text{1}\otimes\ldots|\Psi\rangle$
via LOCC may depend on the particle number $N$. Another assertion concerning
reachability of states under LOCC is possible due to knowing the full symmetry
group of a TIMPS. Namely, TIMPS which possess non-trivial global symmetries,
but no local symmetries, such as, e.g., $\Psi_{G\,\text{fin}}$ from Section
II.1, are not reachable from any other state via an LOCC protocol involving a
finite number of rounds of classical communication [35].

In case a deterministic transformation is not posssible, one might study the
maximal success probability of transforming one TIMPS into another. We denote
by $P(\psi\rightarrow\phi)$ the maximal success probability for transforming
$\psi$ to $\phi$. It has been shown in [41] that
$P(\psi\rightarrow\phi)=\min_{\mu}\mu(\psi)/\mu(\phi)$, where $\mu$ denotes an
arbitrary entanglement monotone. For a generic set of states, this minimum can
be easily determined. This set is defined as the union of all SLOCC classes
which possess a representative whose single party reduced state is completely
mixed (critical state) and whose stabilizer is trivial. It has been shown that
this set is a full measure set in case of a homogeneous system, i.e. where all
local dimensions coincide [12]. For those generic multipartite states, we have
[11]

$P(\psi\rightarrow\phi)=\frac{||\phi||^{2}}{||\psi||^{2}}\frac{1}{\lambda_{max}(G^{-1}H)},$

where $G=g^{\dagger}g$, $H=h^{\dagger}h$, and $\psi=g\psi_{s}$,
$\phi=h\psi_{s}$, with $\psi_{s}$ the critical representative of the SLOCC
class and $g$ and $h$ are local operators. Here, $\lambda_{max}$ denotes the
maximal eigenvalue. Note that the maximal success probability can be easily
determined as $G$ and $H$ are local operators. Note that for these states it
is also possible to determine so–called SLOCC–paths along which on state can
be transformed optimally into the other [42]. Furthermore, for these states a
complete set of entanglement monotones, which can be easily computed, is known
[42]. Clearly all these results apply to TIMPS which belong to the above
mentioned full–measured set.

##  VIII Conclusion

We studied the symmetries of TIMPS with bond dimension $D=3$ and showed that
they are in strong contrast to TIMPS with bond dimension $D=2$. Depending on
the SLOCC class of the underlying fiducial state, very different symmetry and
entanglement properties (regarding SLOCC classes) occur. We illustrate the
rich variety of states by presenting TIMPS with particular symmetry groups.

In a future project it will be interesting to investigate how the stabilizer
groups and SLOCC classes presented here relate to the results on the
classification of phases of matter presented in [26, 25]. Furthermore, the
relaxation of locality in (S)LOCC as presented in [43] might reveal a more
coarse–grained structure of the SLOCC classes presented here.

##  IX Acknowledgments

M. H. and B. K. acknowledge financial support from the Austrian Science Fund
(FWF): the DK-ALM (W1259-N27), the stand alone project: P32273-N27, and the
SFB BeyondC (F7017). A.M. acknowledges funding from the European Research
Council (ERC) under the European Union’s Horizon 2020 research and innovation
programme through the ERC-CoG SEQUAM (grant agreement No. 863476) and ERC-CoG
GAPS (grant agreement No. 648913). J. I. Cirac acknowledges funding from the
ERC-AdG QUENOCOBA (No. 742102) and the DFG under Germany’s Excellence Strategy
(EXC2111-390814868), and the SFB BeyondC F7104 Project number 414325145.

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##  Appendix A Characterization of cycles considering fiducial states
$\text{1}\otimes b\otimes\text{1}|M(\omega)\rangle$

In this appendix we present Table LABEL:tab:M10M11M1infcycleslong providing
details on normal TIMPS generated by fiducial states $\text{1}\otimes
b\otimes\text{1}|M(\omega)\rangle$ (see Section IV). The table lists all
possible cycles and provides parametrizations for all normal fiducial states
$\text{1}\otimes b\otimes\text{1}|M(\omega)\rangle$ s.t. $G_{b}$ exhibits the
respective cycles. See Section IV for the methods required to derive the
table, as well as an exemplary calculation for the cycle $C_{0}$.

Table 3: Characterization of all $b$ leading to non-trivial cycles for normal MPS generated by the fiducial states $\text{1}\otimes b\otimes\text{1}|M(\omega)\rangle$. The first four columns coincide with Table 2. The fifth column shows a parametrization (possible alternative parametrizations) of the set of all normal $b$’s s.t. $G_{b}$ exhibits the respective cycles. Parameter choices such that $b$ becomes a generalized permutation matrix must be excluded in order to have normality. The last column indicates the effective (complex) dimension of the respective sets ("-" indicates discrete sets). The matrix $D$ appearing in the fifth column is always an arbitrary diagonal matrix. As explained in Section IV, $D$ is actually irrelevant for the generated MPS. See Figure 2 in the main text for set-theoretic relations between all the displayed parametrized sets. Label | Subgroup(s) | Cycle | $N$ |  $b$ | #param.
---|---|---|---|---|---
$C_{0}$ | - | $S$ | 1 |  $D\operatorname{diag}(1,\omega,\omega^{2})^{l}\begin{pmatrix}b_{00}&b_{01}&b_{02}\\\ b_{02}&b_{00}&b_{01}\\\ b_{01}&b_{02}&b_{00}\end{pmatrix}D^{-1}$, where $b_{00},b_{01},b_{02}\in\mathbb{C}$, $l\in\\{0,1,2\\}$ | 2
$T_{0}^{\tau}$ | - | $\tau$ | 1 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}-b_{00}&-b_{01}&-b_{01}\\\ b_{01}&b_{11}&b_{12}\\\ b_{01}&b_{12}&b_{11}\end{pmatrix}D^{-1}$, or $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{01}&ib_{01}\\\ -b_{01}&-b_{11}&-b_{12}\\\ ib_{01}&b_{12}&-b_{11}\end{pmatrix}D^{-1}$, where $b_{00},b_{01},b_{11},b_{12}\in\mathbb{C}$ | 3
$T_{0}^{\epsilon}$ | - | $\epsilon$ | 1 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{01}&b_{02}\\\ b_{01}&b_{00}&b_{02}\\\ -b_{02}&-b_{02}&-b_{22}\end{pmatrix}D^{-1}$, or $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{01}&b_{02}\\\ -b_{01}&b_{00}&ib_{02}\\\ b_{02}&-ib_{02}&b_{22}\end{pmatrix}D^{-1}$, where $b_{00},b_{01},b_{02},b_{22}\in\mathbb{C}$ | 3
$T_{0}^{\kappa}$ | - | $\kappa$ | 1 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{01}&b_{02}\\\ -b_{01}&-b_{11}&-b_{01}\\\ b_{02}&b_{01}&b_{00}\end{pmatrix}D^{-1}$, where $b_{00},b_{01},b_{02},b_{11}\in\mathbb{C}$ | 3
$C_{1}$ | - | $S\otimes S^{2}$ | 2 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{01}&b_{02}\\\ b_{01}&b_{02}&b_{00}\\\ b_{02}&b_{00}&b_{01}\end{pmatrix}D^{-1}$, or $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{01}&b_{02}\\\ b_{01}&-b_{02}&-b_{00}\\\ b_{02}&-b_{00}&b_{01}\end{pmatrix}D^{-1}$, or $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{01}&b_{02}\\\ -b_{01}&-b_{02}&-ib_{00}\\\ b_{02}&ib_{00}&ib_{01}\end{pmatrix}D^{-1}$ where $b_{00},b_{01},b_{02}\in\mathbb{C}$ | 2
$T_{1}^{\tau}$ | $T_{0}^{\tau}$ | $\text{1}\otimes\tau$ | 2 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}0&b_{01}&b_{01}\\\ -b_{01}&-b_{11}&b_{11}\\\ -b_{01}&b_{11}&-b_{11}\end{pmatrix}D^{-1}$, where $b_{01},b_{11}\in\mathbb{C}$ | 1
$T_{1}^{\epsilon}$ | $T_{0}^{\epsilon}$ | $\text{1}\otimes\epsilon$ | 2 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}-b_{00}&b_{00}&-b_{02}\\\ b_{00}&-b_{00}&-b_{02}\\\ b_{02}&b_{02}&0\end{pmatrix}D^{-1}$, where $b_{00},b_{02}\in\mathbb{C}$ | 1
$T_{1}^{\kappa}$ | $T_{0}^{\kappa}$ | $\text{1}\otimes\kappa$ | 2 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{01}&-b_{00}\\\ -b_{01}&0&-b_{01}\\\ -b_{00}&b_{01}&b_{00}\end{pmatrix}D^{-1}$, where $b_{00},b_{01}\in\mathbb{C}$ | 1
$T_{2}^{\tau}$ |  $T_{0}^{\tau}$, $C_{1}$ | $\epsilon\otimes\kappa$ | 2 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{01}&b_{01}\\\ -b_{01}&-ib_{01}&b_{00}\\\ -b_{01}&b_{00}&-b_{01}\end{pmatrix}D^{-1}$, or $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{01}&-ib_{01}\\\ -b_{01}&ib_{01}&-ib_{00}\\\ -ib_{01}&ib_{00}&ib_{01}\end{pmatrix}D^{-1}$, where $b_{00},b_{01}\in\mathbb{C}$ | 1
$T_{2}^{\epsilon}$ |  $T_{0}^{\epsilon}$, $C_{1}$ | $\tau\otimes\kappa$ | 2 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}-b_{00}&-b_{01}&-ib_{00}\\\ -b_{01}&-b_{00}&-ib_{00}\\\ ib_{00}&ib_{00}&-b_{01}\end{pmatrix}D^{-1}$, or $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{01}&-b_{00}\\\ -b_{01}&b_{00}&-ib_{00}\\\ -b_{00}&ib_{00}&ib_{01}\end{pmatrix}D^{-1}$, where $b_{00},b_{01}\in\mathbb{C}$ | 1
$T_{2}^{\kappa}$ |  $T_{0}^{\kappa}$, $C_{1}$ | $\tau\otimes\epsilon$ | 2 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&ib_{00}&b_{02}\\\ -ib_{00}&b_{02}&-ib_{00}\\\ b_{02}&ib_{00}&b_{00}\end{pmatrix}D^{-1}$, or $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&-ib_{00}&b_{02}\\\ ib_{00}&-b_{02}&-ib_{00}\\\ b_{02}&ib_{00}&b_{00}\end{pmatrix}D^{-1}$, where $b_{00},b_{02}\in\mathbb{C}$ | 1
$C_{2}$ | $C_{0}$ | $\text{1}\otimes S\otimes S^{2}$ | 3 |  $D\operatorname{diag}(1,\omega,\omega^{2})^{l}\begin{pmatrix}1&1&\omega^{m}\\\ \omega^{m}&1&1\\\ 1&\omega^{m}&1\end{pmatrix}D^{-1}$, where $l\in\\{0,1,2\\}$, $m\in\\{1,2\\}$ | -
$T_{3}^{\tau}$ | - | $\text{1}\otimes\tau\otimes\tau$ | 3 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}0&b_{01}&b_{01}\\\ -b_{01}&-b_{11}&b_{11}\\\ b_{01}&-b_{11}&b_{11}\end{pmatrix}D^{-1}$, where $b_{01},b_{11}\in\mathbb{C}$ | 1
$T_{3}^{\epsilon}$ | - | $\text{1}\otimes\epsilon\otimes\epsilon$ | 3 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&-b_{00}&b_{02}\\\ b_{00}&-b_{00}&-b_{02}\\\ b_{02}&b_{02}&0\end{pmatrix}D^{-1}$, where $b_{00},b_{02}\in\mathbb{C}$ | 1
$T_{3}^{\kappa}$ | - | $\text{1}\otimes\kappa\otimes\kappa$ | 3 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{01}&b_{00}\\\ -b_{01}&0&b_{01}\\\ -b_{00}&b_{01}&-b_{00}\end{pmatrix}D^{-1}$, where $b_{00},b_{01}\in\mathbb{C}$ | 1
$T_{4}^{\circlearrowleft}$ | $C_{0}$ | $\tau\otimes\kappa\otimes\epsilon$ | 3 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{01}&b_{00}\omega^{m}\\\ b_{00}\omega^{m}&b_{00}&b_{01}\\\ b_{01}&b_{00}\omega^{m}&b_{00}\end{pmatrix}D^{-1}$, where $b_{00},b_{01}\in\mathbb{C}$, $m\in\\{0,1,2\\}$ | 1
$T_{4}^{\circlearrowright}$ | $C_{0}$ | $\tau\otimes\epsilon\otimes\kappa$ | 3 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}b_{00}&b_{00}\omega^{m}&b_{02}\\\ b_{02}&b_{00}&b_{00}\omega^{m}\\\ b_{00}\omega^{m}&b_{02}&b_{00}\end{pmatrix}D^{-1}$, where $b_{00},b_{01}\in\mathbb{C}$, $m\in\\{0,1,2\\}$ | 1
$T_{5}^{\circlearrowleft}$ |  $C_{2}$, $C_{0}$ | $\tau\otimes\tau\otimes\kappa$ | 3 |  $D\operatorname{diag}(1,\omega,\omega^{2})^{m+1}\begin{pmatrix}1&1&\omega^{m}\\\ \omega^{m}&1&1\\\ 1&\omega^{m}&1\end{pmatrix}D^{-1}$, where $m\in\\{1,2\\}$ | -
$T_{5}^{\circlearrowright}$ |  $C_{2}$, $C_{0}$ | $\tau\otimes\tau\otimes\epsilon$ | 3 |  $D\operatorname{diag}(1,\omega,\omega^{2})^{2m+1}\begin{pmatrix}1&1&\omega^{m}\\\ \omega^{m}&1&1\\\ 1&\omega^{m}&1\end{pmatrix}D^{-1}$, where $m\in\\{1,2\\}$ | -
$C_{3}$ | - | $\text{1}\otimes S\otimes\text{1}\otimes S^{2}$ | 4 |  $D\operatorname{diag}(1,\omega,\omega^{2})^{l}\begin{pmatrix}1&1&1\\\ 1&\omega^{m}&\omega^{2m}\\\ 1&\omega^{2m}&\omega^{m}\end{pmatrix}D^{-1}$, where $l\in\\{0,1,2\\}$, $m\in\\{1,2\\}$ | -
$T_{6}^{\tau}$ |  $C_{3}$, $T_{0}^{\tau}$ | $\kappa\otimes\tau\otimes\epsilon\otimes\tau$ | 4 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}1&1&1\\\ 1&\omega^{m}&\omega^{2m}\\\ 1&\omega^{2m}&\omega^{m}\end{pmatrix}D^{-1}$, where $m\in\\{1,2\\}$ | -
$T_{6}^{\epsilon}$ |  $C_{3}$, $T_{0}^{\epsilon}$ | $\tau\otimes\epsilon\otimes\kappa\otimes\epsilon$ | 4 |  $D\operatorname{diag}(1,\omega,\omega^{2})^{2m+1}\begin{pmatrix}1&1&1\\\ 1&\omega^{m}&\omega^{2m}\\\ 1&\omega^{2m}&\omega^{m}\end{pmatrix}D^{-1}$, where $m\in\\{1,2\\}$ | -
$T_{6}^{\kappa}$ |  $C_{3}$, $T_{0}^{\kappa}$ | $\epsilon\otimes\kappa\otimes\tau\otimes\kappa$ | 4 |  $D\operatorname{diag}(1,\omega,\omega^{2})^{m+1}\begin{pmatrix}1&1&1\\\ 1&\omega^{m}&\omega^{2m}\\\ 1&\omega^{2m}&\omega^{m}\end{pmatrix}D^{-1}$, where $m\in\\{1,2\\}$ | -
$C_{4}$ | $C_{1}$ | $\text{1}\otimes S\otimes S\otimes\text{1}\otimes S^{2}\otimes S^{2}$ | 6 |  $D\operatorname{diag}(1,\omega,\omega^{2})^{l}\begin{pmatrix}1&1&\omega^{m}\\\ 1&\omega^{m}&1\\\ \omega^{m}&1&1\end{pmatrix}D^{-1}$, where $l\in\\{0,1,2\\}$, $m\in\\{1,2\\}$ | -
$T_{7}^{\tau}$ |  $C_{4}$, $C_{1}$, $T_{2}^{\tau}$ , $T_{0}^{\tau}$ | $\kappa\otimes\tau\otimes\epsilon\otimes\epsilon\otimes\tau\otimes\kappa$ | 6 |  $D\operatorname{diag}(1,\omega,\omega^{2})^{m+1}\begin{pmatrix}1&1&\omega^{m}\\\ 1&\omega^{m}&1\\\ \omega^{m}&1&1\end{pmatrix}D^{-1}$, where $m\in\\{1,2\\}$ | -
$T_{7}^{\epsilon}$ |  $C_{4}$, $C_{1}$, $T_{2}^{\epsilon}$, $T_{0}^{\epsilon}$ | $\tau\otimes\epsilon\otimes\kappa\otimes\kappa\otimes\epsilon\otimes\tau$ | 6 |  $D\operatorname{diag}(1,\omega,\omega^{2})^{2m+1}\begin{pmatrix}1&1&\omega^{m}\\\ 1&\omega^{m}&1\\\ \omega^{m}&1&1\end{pmatrix}D^{-1}$, where $m\in\\{1,2\\}$ | -
$T_{7}^{\kappa}$ |  $C_{4}$, $C_{1}$, $T_{2}^{\kappa}$, $T_{0}^{\kappa}$ | $\epsilon\otimes\kappa\otimes\tau\otimes\tau\otimes\kappa\otimes\epsilon$ | 6 |  $D\operatorname{diag}(1,\omega,\omega^{2})\begin{pmatrix}1&1&\omega^{m}\\\ 1&\omega^{m}&1\\\ \omega^{m}&1&1\end{pmatrix}D^{-1}$, where $m\in\\{1,2\\}$ | -

##  Appendix B Proof of Observation 5 concerning the normality of MPS
generated by fiducial states $\text{1}\otimes
b\otimes\text{1}|M(\omega)\rangle$

In this appendix we prove Observation 5 from the main text, which concerns the
normality of fiducial states $\text{1}\otimes
b\otimes\text{1}|M(\omega)\rangle$. Before we do so, let us discuss a few
general remarks on the matter.

Recall that a fiducial state $\text{1}\otimes b\otimes\text{1}|A\rangle$ is
normal if and only if for some fixed $L$, it is possible to build products of
$A_{b}^{0}$ and $A_{b}^{1}$ comprised of $L$ factors, which form a basis for
all $D\times D$ matrices (see Section III). It suffices to consider $L$ up to
a certain upper bound depending on the bond dimension $D$ [44]. For any
concrete choice of $b$ it is simple to decide normality. To this end, one may
proceed as follows. First, one calculates all possible products of $A_{b}^{0}$
and $A_{b}^{1}$ of length $L$ leading to $2^{L}$ $D\times D$ matrices. Then,
one rearranges the matrix entries in order to form $D^{2}$-dimension vectors,
which are then used as columns of a $D^{2}\times 2^{L}$ matrix $M$. Obviously,
the tensor is normal if $\operatorname{rk}M=D^{2}$ and it is non-normal if
$\operatorname{rk}M<D^{2}$, hence, normality may be decided by computing the
rank151515In order to circumnavigate numeric imprecisions, one may e.g.
compute the singular values of $M$ and make sure that $D^{2}$ of them are
sufficiently different from 0. of $M$. Deciding normality of a continuous
family of $b$’s is more involved, though. Which products of $A_{b}^{0}$ and
$A_{b}^{1}$ one needs to consider in order to obtain a basis will typically
depend on the parameter choices in $b$. It is still comparably simple to show
that a family of $b$’s is normal for generic parameter choices. To this end,
one may construct the matrix $M$ as above and then consider the determinant of
a submatrix of $M$ obtained by selecting $D^{2}$ columns of $M$, which will be
some polynomial in the entries of $b$. In case the obtained polynomial is not
identically zero, this shows that the MPS is normal for generic parameter
choices. However, as mentioned above, typically there will be certain
particular parameter choices for which the polynomial vanishes. In order to
prove that the whole family leads to normal MPS, one thus often needs to
consider additional submatrices of $M$ (and their determinants) and show that
the obtained polynomials don’t have a common root (additionally assuming $\det
b\neq 0$).

We now make a small observation concerning the fact that an MPS is normal if
and only if any SLOCC equivalent MPS is, which is an immediate consequence of
the concepts introduced in [18].

######  Observation 9.

A (not necessarily TI) fiducial state
$g_{k}\otimes\text{1}\otimes\text{1}|A\rangle$ is normal with injectivity
length $L$ for any $g_{1}\ldots g_{N}$ if and only if $|A\rangle$ is.

Let us also recall Observation 4, which we have proven in the main text.
According to Observation 4, fiducial states $\text{1}\otimes
b\otimes\text{1}|M(\omega)\rangle$ are not normal if $b$ is a generalized
permutation matrix, or such that in any row or column $i$, $b_{ii}$ is the
only non-vanishing entry. Let us now restate and prove Observation 5.

Observation 5. All $b$ s.t. there exist non-trivial cycles in $G_{b}$ (see
Table LABEL:tab:M10M11M1infcycleslong) are normal with injectivty length
$L=4$, $5$, or $6$, unless $b$ fulfills the prerequisites of Observation 4.

###### Proof.

We use the parametrizations for $b$ as given in Table
LABEL:tab:M10M11M1infcycleslong. We first argue that in order to prove the
observation, it suffices to prove the statement for only some of the families
of $b$’s in Table LABEL:tab:M10M11M1infcycleslong. We will then argue that it
will be convenient to change the considered representative from
$\text{1}\otimes b\otimes\text{1}|M(\omega)\rangle$ to $\text{1}\otimes
b\otimes\text{1}|M^{1}(0)\oplus M^{1}(1)\oplus M^{1}(\infty)\rangle$. For each
of the relevant families, we then proceed as discussed above.

In order to prove the observation it suffices to consider $b(C_{0})$,
$b(C_{1})$, $b(C_{3})$, $b(T_{0}^{\tau,\epsilon,\kappa})$, and
$b(T_{3}^{\tau,\epsilon,\kappa})$, as all other relevant families are
subfamilies of the mentioned ones (see Figure 1). Moreover, it suffices to
prove the statement for $b(T_{0}^{\tau})$ and $b(T_{3}^{\tau})$ instead of all
the families $b(T_{0}^{\tau,\epsilon,\kappa})$ and
$b(T_{3}^{\tau,\epsilon,\kappa})$. The reason for this is that the families
are SLOCC equivalent to the families $b(T_{0}^{\tau})$ and $b(T_{3}^{\tau})$,
respectively, via a global operation. Due to Observation 9, normality is
retained under an SLOCC operation. Thus, we need to show the statement of the
observation for the families $b(C_{0})$,$b(C_{1})$, $b(C_{3})$,
$b(T_{0}^{\tau})$, and $b(T_{3}^{\tau})$ only. In order to do so, we proceed
as outlined in the discussion above Observation 9. Recall that the matrix $D$
within the parametrizations for $b$ within Table
LABEL:tab:M10M11M1infcycleslong does not alter the generated MPS, we will
hence disregard $D$ in the following. Note that the representative for the
fiducial state $|M(\omega)\rangle$ is SLOCC equivalent to $|M^{1}(0)\oplus
M^{1}(1)\oplus M^{1}(\infty)\rangle$. While using the former representative
was more convenient in the main text, as the physical operators constitute a
more natural representation of the symmetric group, the latter representative
will be more convenient here, as the tensor $A$ is more sparsely populated in
that case. Note that $|M(\omega)\rangle\propto\begin{pmatrix}1&\omega^{2}\\\
\omega&\omega^{2}\end{pmatrix}\otimes\operatorname{diag}(1,-\omega^{2},\omega)\otimes\text{1}|M^{1}(0)\oplus
M^{1}(1)\oplus M^{1}(\infty)\rangle$. Thus, $\text{1}\otimes
b\otimes\text{1}|M(\omega)\rangle$ is normal if and only if $\text{1}\otimes
b^{\prime}\otimes\text{1}|M^{1}(0)\oplus M^{1}(1)\oplus M^{1}(\infty)\rangle$
is for $b^{\prime}=\operatorname{diag}(1,-\omega^{2},\omega)b$. Moreover, note
that $b$ is of the form given in Observation 4 if and only if $b^{\prime}$ is.
Thus, it suffices to prove the statement of the observation considering the
tensor

$\displaystyle
A_{b^{\prime}}^{0}=\begin{pmatrix}0&b^{\prime}_{01}&b^{\prime}_{02}\\\
0&b^{\prime}_{11}&b^{\prime}_{12}\\\
0&b^{\prime}_{21}&b^{\prime}_{22}\end{pmatrix},\
A_{b^{\prime}}^{1}=\begin{pmatrix}b^{\prime}_{00}&b^{\prime}_{01}&0\\\
b^{\prime}_{10}&b^{\prime}_{11}&0\\\
b^{\prime}_{20}&b^{\prime}_{21}&0\end{pmatrix},$ (34)

where $b^{\prime}$ is given by the parametrizations of the families
$b(C_{0})$, $b(C_{1})$, $b(C_{3})$, $b(T_{0}^{\tau})$, and $b(T_{3}^{\tau})$
as in Table LABEL:tab:M10M11M1infcycleslong, multiplied with
$\operatorname{diag}(1,-\omega^{2},\omega)$ from the left. Let us now show the
statement for the individual families. In the following we simply write $b$
instead of $b^{\prime}$.

Let us start considering the family $b(C_{3})$. As this is a discrete family,
it is straightforward to show that the corresponding fiducial states are
normal with $L=4$. To this end, it suffices to construct the matrix $M$ as
described above Observation 9 and verify that it has rank $9$.

Let us now come to the continuous families. For each of these families, we
will provide several alternative choices of nine products of $A_{b}^{0}$ and
$A_{b}^{1}$ comprised of $L$ factors. Then, for each parameter choice within
the considered family of $b$’s, at least one of the alternatives will provide
a basis for all $3\times 3$ matrices (for generic parameter choices, all of
the alternatives do). Before we provide the concrete choices, a few remarks
are in order. First, note that the provided choices are by far not unique and,
moreover, there might exist choices such that less than the provided number of
alternatives suffice to show normality. Second, note that if nine operators
$\\{O_{i}\\}_{i}$ form a basis of $3\times 3$ matrices then so do the
operators $\\{XO_{i}\\}_{i}$ for any invertible $X$. Here it is advantageous
to consider products of $A_{b}^{0}$ and $A_{b}^{1}$ multiplied by $b^{-1}$
from the left instead of the mere products, such as, e.g.,
$b^{-1}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}$ or
$b^{-1}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1}$ instead of
$A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}$ or
$A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1}$. The reason for this is that due to the
form of $A^{0}=|1\rangle\langle 1|+|2\rangle\langle 2|$ and
$A^{1}=|0\rangle\langle 0|+|1\rangle\langle 1|$, we obtain matrices with only
four non-vanishing entries and it becomes much simpler to identify independent
ones.

Let us start with considering the family $b(T_{3}^{\tau})$. Here, it actually
suffices to consider a single choice of nine products of the operators
$A_{b}^{0}$, $A_{b}^{1}$ of length $L=4$, which turn out to be linearly
independent for all $b$ within the considered family. Considering
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}\\}$, it may be easily verified that the
determinant of the corresponding $9\times 9$ matrix reads
$-2048b_{01}^{16}b_{11}^{11}$. In order to have an invertible $b$ we have
$b_{01}\neq 0$ and $b_{11}^{11}\neq 0$. Hence, the given operator products are
linearly independent (and thus form a basis for all $3\times 3$ matrices) for
all $b$ within the considered family.

Let us now consider the family $b(C_{0})$. We consider the following
alternative sets of products with $L=4$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{0}\\}$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{0}\\}$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{0}\\}$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1}\\}$, as well as two with $L=6$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}\\}$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0}\\}$, which certify
normality as outlined above. As the corresponding determinants are lengthy
expressions, we abstain from displaying them here. Note that although $b$
within the considered family are generically normal with $L=4$, there indeed
exist examples that are normal with injectivity length $L=6$ (but not for a
smaller $L$).

Let us now consider the family $b(C_{1})$. We consider the following two
alternative sets of products with $L=4$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}\\}$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}\\}$, as well as the following three
alternative sets of products with $L=6$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}\\}$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1}\\}$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0}\\}$, which certify
normality as outlined above.

Let us now consider the family $b(C_{1})$. We consider the following two
alternative sets of products with $L=4$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{1}\\}$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0}\\}$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}\\}$, as well as the following three
alternative sets of products with $L=5$,
$\\{A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0}\\}$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}\\}$, as well as the following
three alternative sets of products with $L=6$,
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0}\\}$, and
$\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},\allowbreak
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{1}\\}$, which certify
normality as outlined above. Within the considered family there indeed exist
examples that are normal with injectivity lengths $L=4$, $L=5$, and $L=6$,
respectively.

∎

##  Appendix C Proof of Observation 7 concerning the normality of MPS
generated by fiducial states $\text{1}\otimes b\otimes\text{1}|LLT\rangle$

In this appendix we restate and prove Observation 7 from the main text, which
characterizes the normality of MPS generated by fiducial states
$\text{1}\otimes b\otimes\text{1}|LLT\rangle$.

Observation 7. $N$-qubit MPS associated to $\text{1}\otimes
b\otimes\text{1}|LLT\rangle$ are normal if and only if $b_{20}\neq 0$ and
$\mathbf{v}\neq\mathbf{0}$.

###### Proof.

We will first show that the tensor cannot be normal if $b_{20}=0$. We will
then parametrize $b$ in terms of $T$, $\mathbf{v}$, $b_{00}$ and $b_{10}$ as
in Eq. (22). Making use of Observation 9 we will argue that it suffices to
consider $b$ of a restricted form. We will then show that the tensor cannot be
normal if $\mathbf{v}=\mathbf{0}$. Finally, we will show that $b$ is normal
with injectivity length $L=4$ if $\mathbf{v}\neq\mathbf{0}$.

Let us now show that $b_{20}=0$ leads to non-normal MPS. For $b_{20}=0$, we
obtain

$\displaystyle\\{A_{b}^{0},A_{b}^{1}\\}=\\{\begin{pmatrix}0&b_{00}&b_{02}\\\
0&b_{10}&b_{12}\\\ 0&0&b_{22}\end{pmatrix},\begin{pmatrix}b_{00}&0&b_{01}\\\
b_{10}&0&b_{11}\\\ 0&0&b_{21}\end{pmatrix}\\}$ (35)

However, products of matrices of the form $\begin{pmatrix}\cdot&\cdot&\cdot\\\
\cdot&\cdot&\cdot\\\ 0&0&\cdot\end{pmatrix}$ remain of the same form (here,
$\cdot$ denots entries with arbitrary values). Hence, $A_{b}^{0}$ and
$A_{b}^{1}$ cannot generate all matrices.

Let us now consider $b$ such that $b_{20}\neq 0$. We choose wlog $b_{20}=1$.
As mentioned above, we now parametrize $b$ in terms of $T$, $\mathbf{v}$,
$b_{00}$, and $b_{10}$. We now make use of the fact that any obtained MPS is
SLOCC equivalent to an MPS associated to some $b$ with $T_{10}=0$, $b_{00}=0$,
and $b_{10}=0$ (see Section V.4). Thus, it suffices to consider the normality
of MPS for such $b$ in order characterize the normality for all remaining $b$
due to Observation 9.

Let us now consider the case $\mathbf{v}=\mathbf{0}$. Note that in this case
both $A_{b}^{0}$ and $A_{b}^{1}$ are matrices of the form
$\begin{pmatrix}0&0&\cdot\\\ 0&0&\cdot\\\ \cdot&\cdot&0\end{pmatrix}$. It may
be easily verified that any product of matrices of such a form comprised of an
odd number of factors is again of this form. Moreover, any product comprised
of an even number of factors is of the form $\begin{pmatrix}\cdot&\cdot&0\\\
\cdot&\cdot&0\\\ 0&0&\cdot\end{pmatrix}$. Hence, such $b$ cannot lead to
normal tensors/MPS.

Let us now consider the case $\mathbf{v}\neq\mathbf{0}$. Here we distinguish
the two sub-cases $v_{1}\neq 0$ and $v_{1}=0$. In both cases we consider nine
products of $A_{b}^{0}$ and $A_{b}^{1}$ comprised of four factors and show
that the products are linearly independent. Let us first consider the case
$v_{1}\neq 0$. In this case, we find that the nine products

$\displaystyle\\{A_{b}^{0}A_{b}^{0}A_{b}^{1}A_{b}^{0},A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0},A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{1},$
$\displaystyle\qquad
A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1},A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0},A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{1},$
$\displaystyle\qquad
A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1},A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{0},A_{b}^{1}A_{b}^{1}A_{b}^{1}A_{b}^{1}\\}$
(36)

are linearly independent and thus form a basis for all $3\times 3$ matrices.
This may be easily verified by e.g. considering the determinant of a $9\times
9$ matrix $M$ whose columns are constructed by rewriting the nine matrices in
Eq. (36) as 9-dimensional vectors. We obtain $\det
M=-T_{00}^{7}T_{11}^{6}v_{1}^{10}$, which is non-vanishing for $v_{1}\neq 0$
(note that $T_{00},T_{11}\neq 0$ in order to have $\det b\neq 0$).

In case $v_{1}=0$ we instead consider the products

$\displaystyle\\{A_{b}^{0}A_{b}^{0}A_{b}^{0}A_{b}^{1},A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{0},A_{b}^{0}A_{b}^{1}A_{b}^{0}A_{b}^{1},$
$\displaystyle\qquad
A_{b}^{0}A_{b}^{1}A_{b}^{1}A_{b}^{0},A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{0},A_{b}^{1}A_{b}^{0}A_{b}^{0}A_{b}^{1},$
$\displaystyle\qquad
A_{b}^{1}A_{b}^{0}A_{b}^{1}A_{b}^{0},A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{0},A_{b}^{1}A_{b}^{1}A_{b}^{0}A_{b}^{1}\\}.$
(37)

Constructing $M$ as above and using $v_{1}=0$ we obtain $\det
M=-T_{00}^{11}T_{11}^{5}v_{0}^{4}$. As $\mathbf{v}\neq\mathbf{0}$ and thus
$v_{0}\neq 0$, the matrices in Eq. (37) form a basis for all $3\times 3$
matrices. This completes the proof. ∎

##  Appendix D Derivation of the symmetries of MPS generated by
$\text{1}\otimes b\otimes\text{1}|LLT\rangle$

In this appendix we provide details on the derivations of the symmetries of
MPS generated by $\text{1}\otimes b\otimes\text{1}|LLT\rangle$. In particular,
we provide the proof of Theorem 5 and we derive the procedure of deciding the
symmetries for a given $b$ as in the flowchart presented in Figure 4.

For readability we recite Theorem 5 here.

Theorem 5. ${g}_{0},\ldots,{g}_{N-1}$ is an $N$-cylce in $G_{b}$ 161616Recall
that we consider here $b_{20}=1$. We have dealt with the case $b_{20}=0$ in
Observation 6. if and only if there exist $B_{11}^{(k)}\in\mathbb{C}$ such
that for all $k\in\\{0,\ldots,N-1\\}$,

$\displaystyle{g}_{k+1}$
$\displaystyle=\frac{1}{B_{11}^{(k)}B_{11}^{(k+1)}}T{g}_{k}T^{-1},$ (38)
$\displaystyle\left[{g}_{k}-{B_{11}^{(k)}}\text{1}\right]\mathbf{v}$
$\displaystyle=\mathbf{0},$ (39) $\displaystyle B_{11}^{(k)}B_{11}^{(k+1)}$
$\displaystyle=\begin{cases}\pm 1&\text{if }x=i\\\
1&\text{otherwise}\end{cases}$ (40)

###### Proof.

In the main text, Eq. (40) has already been proven to be a necessary
condition, see Observation 8. Let us now prove that Eq. (38) is a necessary
condition. To this end, let us consider again the concatenation condition as
in Eq. (V.2) with the normalizations discussed in the main text. Note also
that, as discussed in the main text, due to $b_{20}\neq 0$, the
proportionality factor within Eq. (V.2) is fixed to $1/B_{11}^{(k)}$. It may
be easily seen that the condition is equivalent to the following vector
equation,

$\displaystyle\begin{pmatrix}-\delta_{k}&-\gamma_{k}&0&0\\\
-\delta_{k}b_{21}&-\gamma_{k}b_{21}&B_{11}^{(k)}b_{00}&0\\\
-\delta_{k}b_{22}&-\gamma_{k}b_{22}&0&B_{11}^{(k)}b_{00}\\\
-\beta_{k}&-\alpha_{k}&0&0\\\
-\beta_{k}b_{21}&-\alpha_{k}b_{21}&B_{11}^{(k)}b_{10}&0\\\
-\beta_{k}b_{22}&-\alpha_{k}b_{22}&0&B_{11}^{(k)}b_{10}\\\
0&0&B_{11}^{(k)}&0\\\
0&0&0&B_{11}^{(k)}\end{pmatrix}\begin{pmatrix}B_{01}^{(k)}\\\ B_{02}^{(k)}\\\
B_{01}^{(k+1)}\\\ B_{02}^{(k+1)}\\\
\end{pmatrix}=\begin{pmatrix}(B_{11}^{(k)}-\delta_{k})b_{00}-\gamma_{k}b_{10}\\\
B_{11}^{(k)}B_{11}^{(k+1)}(\alpha_{k+1}b_{01}-\gamma_{k+1}b_{02})-\delta_{k}b_{01}-\gamma_{k}b_{11}\\\
B_{11}^{(k)}B_{11}^{(k+1)}(-\beta_{k+1}b_{01}+\delta_{k+1}b_{02})-\delta_{k}b_{02}-\gamma_{k}b_{12}\\\
(B_{11}^{(k)}-\alpha_{k})b_{10}-\beta_{k}b_{00}\\\
B_{11}^{(k)}B_{11}^{(k+1)}(\alpha_{k+1}b_{11}-\gamma_{k+1}b_{12})-\beta_{k}b_{11}-\alpha_{k}b_{11}\\\
B_{11}^{(k)}B_{11}^{(k+1)}(-\beta_{k+1}b_{11}+\delta_{k+1}b_{12})-\beta_{k}b_{02}-\alpha_{k}b_{12}\\\
B_{11}^{(k)}\left((\alpha_{k+1}B_{11}^{(k+1)}-1)b_{21}-B_{11}^{(k+1)}\gamma_{k+1}b_{22}\right)\\\
B_{11}^{(k)}\left((\delta_{k+1}B_{11}^{(k+1)}-1)b_{22}-B_{11}^{(k+1)}\beta_{k+1}b_{21}\right)\end{pmatrix}.$
(41)

Considering this form of the equation, one easily obtains the following set of
necessary conditions that are independent of $B_{01}$ and $B_{02}$

$\displaystyle\gamma_{k}(b_{11}-b_{10}b_{21})+\delta_{k}(b_{01}-b_{00}b_{21})-B_{11}^{(k)}B_{11}^{(k+1)}\left(\alpha_{k+1}(b_{01}-b_{00}b_{21})+\gamma_{k+1}(b_{02}-b_{00}b_{22})\right)$
$\displaystyle=0$
$\displaystyle\gamma_{k}(b_{12}-b_{10}b_{22})+\delta_{k}(b_{02}-b_{00}b_{22})+B_{11}^{(k)}B_{11}^{(k+1)}\left(\beta_{k+1}(b_{01}-b_{00}b_{21})+\delta_{k+1}(b_{02}-b_{00}b_{22})\right)$
$\displaystyle=0$
$\displaystyle\alpha_{k}(b_{11}-b_{10}b_{21})+\beta_{k}(b_{01}-b_{00}b_{21})-B_{11}^{(k)}B_{11}^{(k+1)}\left(\alpha_{k+1}(b_{11}-b_{10}b_{21})+\gamma_{k+1}(b_{12}-b_{10}b_{22})\right)$
$\displaystyle=0$
$\displaystyle\alpha_{k}(b_{12}-b_{10}b_{22})+\beta_{k}(b_{02}-b_{00}b_{22})+B_{11}^{(k)}B_{11}^{(k+1)}\left(\beta_{k+1}(b_{11}-b_{10}b_{21})+\delta_{k+1}(b_{12}-b_{10}b_{22})\right)$
$\displaystyle=0,$ (42)

which are equivalent to Eq. (38) for $T$ as in Eq. (23), which proves that Eq.
(38) is a necessary condition.

Let us now also prove that Eq. (39) is a necessary condition. To this end, let
us again consider Eq. (41) and assume that ${g}_{0},\ldots,{g}_{N-1}$ satisfy
Eq. (38). It may be easily verified that then, the remaining conditions within
Eq. (41) read

$\displaystyle\begin{pmatrix}-\delta_{k}&-\gamma_{k}\\\
-\beta_{k}&-\alpha_{k}\\\ 1&0\\\
0&1\end{pmatrix}\begin{pmatrix}B_{01}^{(k)}\\\ B_{02}^{(k)}\end{pmatrix}=$
$\displaystyle\qquad\begin{pmatrix}-(B_{11}^{(k)}-\delta_{k})b_{00}+\gamma_{k}b_{10}\\\
\beta_{k}b_{00}-(B_{11}^{(k)}-\alpha_{k})b_{10}\\\
-(\alpha_{k}B_{11}^{(k)}-1)b_{21}+B_{11}^{(k)}\gamma_{k}b_{22}\\\
B_{11}^{(k)}\beta_{k}b_{21}-(\delta_{k}B_{11}^{(k)}-1)b_{22}\end{pmatrix}$
(43)

for all $k$. Eq. (43) has a solution $(B_{01}^{(k)},B_{02}^{(k)})^{T}$ for all
$k$ if and only if Eq. (39) is satisfied for all $k$.

Finally, note that if all of the conditions in the theorem are satisfied, then
the concatenation condition given in Eq. (V.2) is satisfied. This completes
the proof of the theorem. ∎

Let us briefly discuss the condition given in Eq. (38). Considering an
$N$-cycle, in particular, it implies that ${g}_{k}=T^{N}{g}_{k}T^{-N}$ if
$\text{tr}{{g}_{k}}\neq 0$ ($x\neq i$), or $N$ is even. To see this, we have
used Eq. (38) iteratively, moreover, we have used that $x\neq i$ implies that
$B_{11}^{(q)}B_{11}^{(q+1)}=1$ for all $q$, and that for an even $N$ we have
that $\prod_{q=0}^{N-1}\left(B_{11}^{(q)}\right)^{2}=1$. If $N$ is odd and
$\text{tr}{{g}_{k}}=0$ ($x=i$) instead, we obtain ${g}_{k}=\pm
T^{N}{g}_{k}T^{-N}$, i.e., an additional sign freedom emerges. In the former
case, we obtain equivalently $[T^{N},{g}_{k}]=0$. This condition is satisfied
if and only if either $[T,{g}_{k}]=0$, or $T\propto
R\operatorname{diag}(1,e^{i\frac{2r\pi}{N}})R^{-1}$ for some
$r\in\\{0,\ldots,N-1\\}$ and some matrix $R$ (as then $T^{N}=\text{1}$). Note
that $[T,{g}_{k}]=0$ implies a global symmetry, ${g}_{k}={g}$. Considering a
minus sign instead, ${g}_{0}=-T^{N}{g}_{0}T^{-N}$, we additionally obtain
solutions of the form
$T=S_{0}\tilde{D}H\operatorname{diag}(1,e^{i\frac{(2r+1)\pi}{N}})H\tilde{D}^{-1}S_{0}^{-1}$
for some $r\in\\{0,\ldots,N-2\\}$ and some diagonal matrix $\tilde{D}$, where
$H$ denotes the Hadamard matrix.

In the following discussion, let us now also take the condition in Eq. (39)
into account. One possiblility to satisfy this condition is to have
$b_{00}=-b_{21}$ as well as $b_{10}=-b_{22}$. Then, $\mathbf{v}=\mathbf{0}$
and Eq. (39) is obviously satisfied for all $k$. The second possibility is
that all ${g}_{k}$ share a common eigenvector, $\mathbf{v}$, corresponding to
the eigenvalue $B_{11}^{(k)}$, respectively. We will see that this second
option limits the symmetries of the MPS severely. To this end, first note that
$B_{11}^{(k)}$ must equal one of the eigenvalues of ${g}_{k}$, $x$ or $1/x$.
Moreover, using Eq. (39) for $k+1$, inserting ${g}_{k+1}$ as in Eq. (38) and
using $\left(B_{11}^{(k+1)}\right)^{2}=1/\left(B_{11}^{(k)}\right)^{2}$ yields
another useful condition,

$\displaystyle
T\left[{g}_{k}-\frac{1}{B_{11}^{(k)}}\text{1}\right]T^{-1}\mathbf{v}=\mathbf{0}.$
(44)

In other words, for each $k$, the eigenvectors of ${g}_{k}$ are given by
$\mathbf{v}$ and $T^{-1}\mathbf{v}$, corresponding to the eigenvalues
$B_{11}^{(k)}$ and $1/B_{11}^{(k)}$, respectively. One more notable
consequence is that considering once more Eq. (38) one obtains that either
$T\mathbf{v}\propto\mathbf{v}$, or $T\mathbf{v}\not\propto\mathbf{v}$ and
$T^{2}\mathbf{v}\propto\mathbf{v}$ (here we assumed
${g}_{k}\not\propto\text{1}$ to disregard trivial solutions). In the former
case we must have that $1/B_{11}^{(k)}=B_{11}^{(k)}=1/x=x=1$. Thus, only non-
diagonlizable $g_{k}$ are possible. In the latter case we have alternating
symmetries with ${g}_{k+1}={g}_{k}^{-1}$ for all $k$ (we obtain a global
symmetry in case $x=i$). Having established these useful facts, we are now in
the position to derive the process for determining all cycles in $G_{b}$ for a
given $b$.

###  D.1 Deriving the process depicted in Figure 4, which determines the
cycles in $G_{b}$ for a given $b$

We will now derive the process of determining the cycles for a given $b$ with
$b_{20}=1$ as depicted in Figure 4. The first step is to calculate $T$ and
$\mathbf{v}$ for the given $b$. We now consider the Jordan decomposition of
$T$, $T=RJR^{-1}$, where $J$ is the JNF of $T$. Depending on whether $T$ is
diagonalizable, we now distinguish two cases.

Let us first deal with the case that $T$ is not diagonalizable. Suppose we
haven an $N$-cycle ${g}_{0},\ldots,{g}_{N-1}$. Then, the only possibility to
fulfill Eq. (38) is that $[T,{g}_{k}]=0$. To see this, consider a consequence
of Eq. (38),
${g}_{k}T^{N}{g}_{k}^{-1}=\frac{1}{\prod_{k}\left(B_{11}^{(k)}\right)^{2}}T^{N}$,
which shows that $\prod_{k}\left(B_{11}^{(k)}\right)^{2}=1$. Thus we have
$[T^{N},{g}_{k}]=0$. From this, $[T,{g}_{k}]=0$ follows. It immediately
follows that only global symmetries are posssible, as
${g}_{k+1}\propto{g}_{k}$ due to Eq. (38). Moreover, using $[T,{g}]=0$ we
obtain that for all potential symmetries, ${g}=R\begin{pmatrix}1&y\\\
0&1\end{pmatrix}R^{-1}$ for any $y\in\mathbb{C}$. It yet remains to consider
Eq. (39). Recall that both $\mathbf{v}$ and $T\mathbf{v}$ are eigenvectors of
${g}$. As ${g}$ is not diagonalizable in the currently considered case, we
must have $T\mathbf{v}\propto\mathbf{v}$ (or $\mathbf{v}=0$). This leads to
another case distinction. If $T\mathbf{v}\not\propto\mathbf{v}$, then we have
no non-trivial cycles. However, if $T\mathbf{v}\propto\mathbf{v}$, then with
${g}$ as given above, all conditions in Theorem 5 are satisfied. Thus, in this
case we obtain the one-parametric family of gloabal symmetries given above.
This compeltes the case that $T$ is not diagonalizable and is shown in the
left branch of Figure 4.

Let us now discuss the case that $T$ is diagonalizable. We now additionally
distinguish the case that there exists an $m\in\mathbb{N}$ such that
$T^{m}\propto\text{1}$ from the case that there does not exist such an $m$.
Let us first discuss the latter case. Similarly to before, we suppose that we
have an $N$-cycle and obtain
${g}_{0}T^{N}{g}_{0}^{-1}=\frac{1}{\prod_{k}\left(B_{11}^{(k)}\right)^{2}}T^{N}$.
We must have $\prod_{k}\left(B_{11}^{(k)}\right)^{2}=1$, as otherwise
$T^{2N}\propto\text{1}$, contradicting the assumption. Thus, we obtain
$[{g}_{0},T^{N}]=0$, which implies $[{g}_{0},T]=0$, as
$T^{N}\not\propto\text{1}$. Hence, we have global symmetries only and
moreover, ${g}=R\operatorname{diag}(x,1/x)R^{-1}$ for any
$x\in\mathbb{C}\setminus\\{0\\}$. If $\mathbf{v}=\mathbf{0}$ the conditions in
Theorem 5 are satisfied and one indeed obtains the mentioned one-parametric
family of global symmetries. Let us now consider the case
$\mathbf{v}\neq\mathbf{0}$. Note that if $T\mathbf{v}\propto\mathbf{v}$ both
eigenvalues of ${g}$ coincide, which leads to trivial cycles only (recall that
${g}$ must be diagonalizable). Recall that $T^{2}\mathbf{v}\propto\mathbf{v}$
is a necessary condition to have non-trivial cycles. Thus, only the case
$T\mathbf{v}\not\propto\mathbf{v}$ and $T^{2}\mathbf{v}\propto\mathbf{v}$
remains. Note, however, that if $T^{2}\mathbf{v}=\lambda\mathbf{v}$ for some
$\lambda\in\mathbb{C}$, then also $T^{2}T\mathbf{v}=\lambda T\mathbf{v}$. As
$\mathbf{v}$ and $T\mathbf{v}$ are linearly independent, this implies that
$T^{2}=\lambda\text{1}$, which is contradicting the assumption that
$T^{N}\not\propto\text{1}$ for any $N\in\mathbb{N}$. Hence, in case
$\mathbf{v}\neq\mathbf{0}$ we have a trivial symmetry only.

Let us finally disucuss the case that $T$ is diagonalizable and moreover,
there is an $m\in\mathbb{N}$ such that $T^{m}\propto\text{1}$. In this case
one can write $T\propto R\begin{pmatrix}e^{i\frac{r\pi}{m}}&\\\
&e^{-i\frac{r\pi}{m}}\end{pmatrix}R^{-1}$ for some $r\in\mathbb{N}$. If
$\mathbf{v}=\mathbf{0}$, we first obtain the same global symmetries as in case
$T^{m}\not\propto\text{1}$, ${g}=R\operatorname{diag}(x,1/x)R^{-1}$ for any
$x\in\mathbb{C}\setminus\\{0\\}$. Moreover, we obtain $m$-cycles of the form
${g}_{k}=T^{k}{g}_{0}T^{-k}$ for any ${g}_{0}$, which (taking normalization
into account) effectively constitutes a three-parametric family of non-global
symmetries. Thus, in this case we have diagonalizable as well as non-
diagonalizable symmetries. In case $m$ is even, additionally certain
$m/2$-cycles emerge, which stem from the fact that in case $x=i$, we may have
${g}_{0}=-T^{m/2}{g}_{0}T^{m/2}$. As discussed earlier, this admits solutions
${g}_{0}=S_{0}\operatorname{diag}(i,-i)S_{0}^{-1}$ such that
$T=S_{0}\tilde{D}H\operatorname{diag}(e^{i\frac{(2r+1)\pi}{m}},e^{-i\frac{(2r+1)\pi}{m}})H\tilde{D}^{-1}S_{0}^{-1}$
for some $r\in\\{0,\ldots,m-2\\}$ and some diagonal matrix $\tilde{D}$. It may
be easily verified that this leads to the one-parametric family of
$m/2$-cycles with ${g}_{k}=R\begin{pmatrix}0&iye^{i\frac{2k(2r+1)\pi}{m}}\\\
i/ye^{-i\frac{2k(2r+1)\pi}{m}}&0\end{pmatrix}R^{-1}$ for
$y\in\mathbb{C}\setminus\\{0\\}$. Let uns now discuss the case that
$\mathbf{v}\neq 0$, in which additional restrictions must be satisfied. Let us
first discuss the subcase $T\mathbf{v}\propto\mathbf{v}$. Recall that in this
subcase, only cycles with non-diagonalizable ${g}_{k}$ are possible
(disregarding trivial cycles). We thus write
${g}_{0}=S_{0}\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}S_{0}^{-1}$, where
$S_{0}=\left(\mathbf{v},\mathbf{w}\right)$ with a freely choosable generalized
eigenvector $\mathbf{w}\in\mathbb{C}^{2}$. One then obtains the remaining
matrices forming an $m$-cycle
${g}_{k}=S_{0}\begin{pmatrix}1&e^{\frac{2kr\pi}{m}}\\\
0&1\end{pmatrix}S_{0}^{-1}$ for $k\in\\{0,\ldots,m-1\\}$. Thus, in this case
we obtain an effectively one-parametric family of non-diagonalizable, non-
global symmetries. Let us now discuss the remaining subcases. As $\mathbf{v}$
such that $T^{2}\mathbf{v}\not\propto\mathbf{v}$ do not allow for any non-
trivial cycle, actually the only subcase that remains open is the case that
$T\mathbf{v}\not\propto\mathbf{v}$, but $T^{2}\mathbf{v}\propto\mathbf{v}$.
Note that in this case we have $T^{2}\propto\text{1}$. Clearly, all ${g}_{k}$
must be diagonalizable, as each ${g}_{k}$ possesses the two linearly
independent vectors $\mathbf{v}$ and $T\mathbf{v}$ as eigenvectors. It may be
easily seen that a global symmetry with eigenvectors $\mathbf{v}$ and
$T\mathbf{v}$ and eigenvalues given by $\pm i$ satisfies the conditions given
in Theorem 5. Considering cycles with ${g}_{k}$ such that $x\neq i$, one
obtains that ${g}_{k+1}={g}_{k}^{-1}$, as the vectors $\mathbf{v}$ and
$T\mathbf{v}$ correspond to the eigenvalues $x$ and $1/x$ in an alternating
manner. Thus, one obtains a one-parametric family of two-cycles with a freely
choosable $x\in\mathbb{C}\setminus\\{0,\pm i\\}$. Clearly, however, for an odd
partilcle number only the global symmetry with $x=i$ remains.

##  Appendix E Details on the SLOCC classification of $\Psi(LLT)$ and proofs
of Lemmata 4 and 5

In this appendix, we first provide details on the characterization of
$(b\rightarrow c)$ cycles for fiducial states $|LLT\rangle$. We first state
and prove Observation 10. We then use the observation in order to prove
Lemmata 4 and 5, which lead to the SLOCC classification of normal MPS (Theorem
6), as explained in the main text.

From Eq. (33) it follows that $T_{b}$ and $T_{c}$ are either both
diagonalizable, or both non-diagonalizable and we may deal with SLOCC
equivalence for these two cases separately. By considering Eq. (32) one
straightforwardly obtains a few further necessary conditions for having an
$(b\rightarrow c)$ $N$-cycle. First, $\mathbf{v}_{b}=\mathbf{0}$ if and only
if $\mathbf{v}_{c}=\mathbf{0}$. Second, for any $k\in\mathbb{N}$ it holds that
$T_{b}^{k}\mathbf{v}_{b}\propto\mathbf{v}_{b}$ if and only if
$T_{c}^{k}\mathbf{v}_{c}\propto\mathbf{v}_{c}$. To see this, suppose that
$T_{c}^{k}\mathbf{v}_{c}\propto\mathbf{v}_{c}$ and consider

$\displaystyle T_{b}^{k}\mathbf{v}_{b}$ $\displaystyle\propto
g_{k}T_{c}g_{k-1}^{-1}g_{k-1}T_{c}\ldots
T_{c}g_{0}^{-1}g_{0}\mathbf{v}_{c}=g_{k}T_{c}^{k}\mathbf{v}_{c}$
$\displaystyle\propto g_{k}\mathbf{v}_{c}\propto\mathbf{v}_{b}.$ (45)

Third, there exists an $m_{b}$ such that $T_{b}^{m_{b}}\propto\text{1}$ if and
only if there exists an $m_{c}$ such that $T_{c}^{m_{c}}\propto\text{1}$. To
see this, suppose that there exists an $m_{c}$ such that
$T_{c}^{m_{c}}\propto\text{1}$. Then, consider $T_{b}^{Nm_{c}}$, where $N$ is
the particle number. Due to Eq. (33) we have $T_{b}^{Nm_{c}}\propto
g_{k}T_{c}^{Nm_{c}}g_{k}^{-1}\propto\text{1}$. Thus, we have
$T_{b}^{m_{b}}\propto\text{1}$ with $m_{b}=Nm_{c}$. Note, however, that we do
not necessarily have $T_{b}^{m_{b}}\propto\text{1}$ for $m_{b}=m_{c}$. We
present a simple counter example below. Finally, let us remark that if
$T_{c}\mathbf{v}_{c}\not\propto\mathbf{v}_{c}$, then we have that for any
$m\in\mathbb{N}$, $T_{c}^{m}\propto\text{1}$ if and only if
$T_{b}^{m}\propto\text{1}$ (with the same $m$). This can be seen as follows.
Suppose that $T_{c}^{m}\propto\text{1}$. Then we have
$T_{c}^{m}\mathbf{v}_{c}\propto\mathbf{v}_{c}$. As discussed above we thus
have $T_{b}^{m}\mathbf{v}_{b}=\lambda\mathbf{v}_{b}$ for some
$\lambda\in\mathbb{C}$. Furthermore, we have that
$T_{b}^{m}T_{b}\mathbf{v}_{b}=\lambda T_{b}\mathbf{v}_{b}$. As
$\mathbf{v}_{b}$ and $T_{b}\mathbf{v}_{b}$ are linearly independent by the
assumption, we have that $T_{b}^{m}=\lambda\text{1}$. Let us summarize all of
the discussed properties in the following observation.

######  Observation 10.

Consider $b$ and $c$ such that there exists an $(b\rightarrow c)$
$N$-cycle171717Recall that we are considering $b_{20}=c_{20}=1$.. Then we have

  1. (i)

$T_{b}$ is diagonalizable if and only if $T_{c}$ is.

  2. (ii)

$\mathbf{v}_{b}=\mathbf{0}$ if and only if $\mathbf{v}_{c}=\mathbf{0}$.

  3. (iii)

For any $k\in\mathbb{N}$ we have
$T_{b}^{k}\mathbf{v}_{b}\propto\mathbf{v}_{b}$ if and only if
$T_{c}^{k}\mathbf{v}_{c}\propto\mathbf{v}_{c}$.

  4. (iv)

There exists an $m_{b}\in\mathbb{N}$ such that $T_{b}^{m_{b}}\propto\text{1}$
if and only if there exists an $m_{c}\in\mathbb{N}$ such that
$T_{c}^{m_{c}}\propto\text{1}$.

  5. (v)

If $T_{c}\mathbf{v}_{c}\not\propto\mathbf{v}_{c}$, then for any
$m\in\mathbb{N}$ we have $T_{b}^{m}\propto\text{1}$ if and only if
$T_{c}^{m}\propto\text{1}$.

We formulate the observation in terms of cycles rather than SLOCC equivalence
of MPS as the considered family of $b,c$ encompasses non-normal instances, cf.
the discussion below Theorem 5. With this observation, we have established
that normal MPS belonging to different boxes within Figure 4 are SLOCC
inequivalent (even though they may have compatible symmetry group).

In Observation 10 we have shown that if there is a $(b\rightarrow c)$ cycle
and $T_{c}^{m_{c}}\propto\text{1}$ for some $m_{c}$, then
$T_{b}^{m_{b}}\propto\text{1}$ for some $m_{b}$. However, in the discussion
above the observation we have mentioned that we do not necessarily have
$T_{b}^{m_{b}}\propto\text{1}$ for $m_{b}=m_{c}$. Here we present a simple
counter example illustrating this. Consider $c$ such that
$\mathbf{v}_{c}=\mathbf{0}$ and
$T_{c}=\operatorname{diag}(e^{i\frac{\pi}{m_{c}}},e^{-i\frac{\pi}{m_{c}}})$,
i.e., $T_{c}^{m_{c}}\propto\text{1}$. Then, $b$ with
$\mathbf{v}_{b}=\mathbf{0}$ and
$T_{b}=\operatorname{diag}(e^{i(\frac{2}{N}+\frac{1}{m_{c}})\pi},e^{-i(\frac{2}{N}+\frac{1}{m_{c}})\pi})$
gives rise to an SLOCC equivalent MPS. This can be easily seen by writing
$T_{b}=g_{k+1}T_{c}g_{k}^{-1}$, where
$g_{k}=\operatorname{diag}(e^{i\frac{2k\pi}{N}},e^{-i\frac{2k\pi}{N}})$.
Suppose that $N$ is odd, moreover, $m_{c}$ and $N$ are coprime. Then,
$m_{b}=Nm_{c}$ is the smallest integer such that
$T_{b}^{m_{b}}\propto\text{1}$, in particular,
$T_{b}^{m_{c}}\not\propto\text{1}$.

Lemma 4. Consider $b$ and $c$ in standard form which correspond to normal MPS
(i.e., $b_{20}=c_{20}=1$ and additionally
$\mathbf{v}_{b},\mathbf{v}_{c}\neq\mathbf{0}$). If there exists a
$(b\rightarrow c)$ $N$-cycle, then, $T_{b}=T_{c}$.

###### Proof.

In order to prove the lemma, we distinguish several cases, namely, non-
diagonalizable and diagonalizable $T_{c}$ and, moreover, $T_{c}$ such that
$T_{c}^{m}\propto\text{1}$ for some $m$ and $T_{c}$ such that
$T_{c}^{m}\not\propto\text{1}$ for all $m$. We show the statement of the lemma
for each of these cases separately.

Let us first consider non-diagonalizable $T_{c}$. Due to Observation 10 and
the chosen standard form for $T$ we have $T_{b}=T_{c}=\begin{pmatrix}1&1\\\
0&1\end{pmatrix}$. The statement is hence trivial for non-diagonalizable $T$.

Let us now consider diagonalizable $T_{b}$, $T_{c}$. Consider first the case
that there exists no $m_{c}$ such that $T_{c}^{m_{c}}\propto\text{1}$. Due to
Observation 10, the same must hold for $T_{b}$. Considering the standard form
for $T$ we may write $T_{c}=\operatorname{diag}(\sigma_{c},\sigma_{c}^{-1})$
and $T_{b}=\operatorname{diag}(\sigma_{b},\sigma_{b}^{-1})$. Using Eq. (33) we
necessarily have that either $\sigma_{b}^{N}=\pm\sigma_{c}^{N}$, or
$\sigma_{b}^{N}=\pm\sigma_{c}^{-N}$ (with a positive sign in case of even
$N$). Note that the latter case is only possible if $|\sigma_{c}|=1$. Thus,
$\sigma_{b}=e^{i\frac{q\pi}{N}}\sigma_{c}$ or
$\sigma_{b}=e^{i\frac{q\pi}{N}}\sigma_{c}^{-1}$ for some
$q\in\\{0,\ldots,2N-1\\}$ in case of odd $N$ and
$\sigma_{b}=e^{i\frac{2q\pi}{N}}\sigma_{c}$ or
$\sigma_{b}=e^{i\frac{2q\pi}{N}}\sigma_{c}^{-1}$ for some
$q\in\\{0,\ldots,N-1\\}$ in case of even $N$. Moreover, using Eq. (33) we
obtain that either $g_{k}=\operatorname{diag}(\alpha_{k},1/\alpha_{k})$ for
all $k$, or $g_{k}=\operatorname{diag}(\alpha_{k},1/\alpha_{k})\sigma_{x}$ for
all $k$, respectively181818Let us remark that in case
$\mathbf{v}_{c}=\mathbf{0}$ we also have $\mathbf{v}_{b}=\mathbf{0}$ and any
$\sigma_{b}$ as in the main text indeed leads to an SLOCC equivalent MPS,
which can be seen by considering
$g_{k}=\operatorname{diag}(e^{-i\frac{kq\pi}{N}},e^{i\frac{kq\pi}{N}})\sigma_{x}^{b}$
and $B_{11}^{(k)}=(-1)^{k}i$, where $b\in\\{0,1\\}$ for odd $N$ and
$g_{k}=\operatorname{diag}(e^{-i\frac{2kq\pi}{N}},e^{i\frac{2kq\pi}{N}})\sigma_{x}^{b}$
and $B_{11}^{(k)}=1$ for even $N$. Hence, there is a one-parametric family of
SLOCC classes corresponding to box (III) in Figure 4. . Due to the assumption
we have $\mathbf{v}_{c}\neq\mathbf{0}$. Due to Eq. (32) we have

$\displaystyle\frac{B_{11}^{(k)}}{B_{11}^{(k+1)}}g_{k}^{-1}g_{k+1}\mathbf{v}_{c}=\mathbf{v}_{c},$
(46)

and due to the considerations above we moreover have

$\displaystyle
g_{k}^{-1}g_{k+1}={B_{11}^{(k+1)}}{B_{11}^{(k)}}\operatorname{diag}(e^{i\frac{q\pi}{N}},e^{-i\frac{q\pi}{N}})$
(47)

for $q\in\\{0,\ldots,2N-1\\}$. Thus,
$\left(B_{11}^{(k)}\right)^{2}\operatorname{diag}(e^{i\frac{q\pi}{N}},e^{-i\frac{q\pi}{N}})\mathbf{v}_{c}=\mathbf{v}_{c}$.
Hence, either $q=0$, which implies $T_{b}=T_{c}$ ($T_{b}=T_{c}^{-1}$ is not
possible due to the chosen standard form), or $\mathbf{v}_{c}$ is proportional
to a standard basis vector, i.e., an eigenvector of $T_{c}$. In the latter
case we have $\left(B_{11}^{(k)}\right)^{2}=e^{\pm i\frac{q\pi}{N}}$ with
coinciding sign for all $k$. As we are considering $T$ in standard form ($\det
T_{b}=\det T_{c}=1$), we have
$\left(B_{11}^{(k)}\right)^{2}\left(B_{11}^{(k+1)}\right)^{2}=1$. We thus
obtain $e^{i\frac{2q\pi}{N}}=1$ and in further consequence
$\sigma_{b}=\pm\sigma_{c}$, or $\sigma_{b}=\pm\sigma_{c}^{-1}$. Hence,
$T_{b}=T_{c}$ due to the standard form.

Let us now consider the case that there exists an $m_{c}$ such that
$T_{c}^{m_{c}}\propto\text{1}$. Here we distinguish two subcases. First, if
$T_{c}^{N}\not\propto\text{1}$, then the same conclusions as for the case
$T_{c}^{m_{c}}\not\propto\text{1}$ for any $m_{c}\in\mathbb{N}$ can be drawn.
Second, if $T_{c}^{N}\propto\text{1}$, then Eq. (33) does not yield a
constraint. In particular, it does not imply that $g_{k}$ must be of either
diagonal or counter diagonal form. Thus, in the following, we deal with this
case separately. Due to Eq. (32) we have

$\displaystyle
g_{k}=\left(\prod_{i=0}^{k-1}B_{11}^{(i)}\right)\left(\prod_{i=1}^{k}B_{11}^{(i)}\right)T_{b}^{k}g_{0}T_{c}^{-k}$
(48)

for all $k$. We now consider
$\frac{1}{B_{11}^{(k)}}g_{k}\mathbf{v}_{c}=\frac{1}{B_{11}^{(0)}}g_{0}\mathbf{v}_{c}$
[an implication of Eq. (32)]. Inserting Eq. (48) we obtain the condition

$\displaystyle\underbrace{\left(\left(\prod_{i=0}^{k-1}B_{11}^{(i)}\right)^{2}T_{b}^{k}g_{0}T_{c}^{-k}-g_{0}\right)}_{M_{k}}\mathbf{v}_{c}=\mathbf{0}.$
(49)

As $\left(B_{11}^{(l)}B_{11}^{(l+1)}\right)^{2}=1$ for any $l$, the product in
Eq. (49) equals 1 for any even $k$. As $T_{c}^{N},T_{b}^{N}\propto\text{1}$
and due to the standard form we have
$T_{b}=\operatorname{diag}(e^{i\frac{q_{b}\pi}{N}},e^{-i\frac{q_{b}\pi}{N}})$
and
$T_{c}=\operatorname{diag}(e^{i\frac{q_{c}\pi}{N}},e^{-i\frac{q_{c}\pi}{N}})$
for some $q_{b},q_{c}\in\\{0,\ldots,\lfloor N/2\rfloor\\}$. Due to Eq. (49) we
have $\det M_{k}=0$ for any $k$. Moreover, due to the definition of $M_{k}$ we
have that

$\displaystyle\det M_{k}=2-2$
$\displaystyle\left(\alpha_{0}\delta_{0}\cos\frac{k\pi(q_{b}-q_{c})}{N}\right.$
$\displaystyle\quad\left.-\beta_{0}\gamma_{0}\cos\frac{k\pi(q_{b}+q_{c})}{N}\right)$
(50)

for any even $k$. We have

$\displaystyle\sum_{k\in\\{2,4,\ldots,2(N-1)\\}}\det
M_{k}=\begin{cases}2N&q_{b}\neq q_{c}\\\
0&q_{b}=q_{c}\in\\{0,\frac{N}{2}\\}\\\
-2N\beta_{0}\gamma_{0}&q_{b}=q_{c}\not\in\\{0,\frac{N}{2}\\}\end{cases}$ (51)

Note that we deliberately also sum over $k$ that are larger than $N$, we use
$N+l\equiv l$ for $l\in\\{0,\ldots,N-1\\}$. As $M_{k}$ must be singular for
any $k$ this shows that $q_{c}=q_{b}$ and hence $T_{c}=T_{b}$. This completes
the proof of the lemma. ∎

Lemma 5. Consider $b$ and $c$ which correspond to normal MPS (i.e.,
$b_{20}=c_{20}=1$ and additionally
$\mathbf{v}_{b},\mathbf{v}_{c}\neq\mathbf{0}$). If there exists a
$(b\rightarrow c)$ $N$-cycle, then, there also exists a $(b\rightarrow c)$
$1$-cycle.

###### Proof.

We prove the statement separately for the case of non-diagonalizable $T_{c}$,
diagonalizable $T_{c}$ such that $T_{c}^{N}\propto\text{1}$ and diagonalizable
$T_{c}$ such that $T_{c}^{N}\not\propto\text{1}$. We make use of the fact that
$T_{b}=T_{c}$ due to Lemma 4 in all of the cases.

Let us start by considering non-diagonalizable $T_{c}$. Due to Eq. (32) we
have $g_{k+1}\propto T_{c}g_{k}T_{c}^{-1}$. From the condition in Eq. (33) it
follows that $[g_{k},T_{c}]=0$. Thus, $g_{k+1}\propto g_{k}$ for all $k$ and
the statement of the lemma follows for non-diagonalizable $T_{c}$.

Let us now consider the case that $T_{c}$ is diagonalizable and
$T_{c}^{N}\not\propto\text{1}$. Recall that due to the considered standard
form of $c$, $T_{c}$ is diagonal. Due to Eq. (33) we have that $g_{k}$ is
either diagonal or counter-diagonal. Using $g_{k+1}\propto
T_{c}g_{k}T_{c}^{-1}$ we have that $g_{k+1}\propto g_{k}$ and the statement of
the lemma follows for the considered $T_{c}$.

Let us finally consider the case that $T_{c}$ is diagonalizable and
$T_{c}^{N}\propto\text{1}$. Note that if $T_{c}\propto\text{1}$ we have
$g_{k+1}\propto T_{c}g_{k}T_{c}^{-1}\propto g_{k}$ and the statement of the
lemma follows trivially. In the following we thus assume
$T_{c}\not\propto\text{1}$. We will make use of the relation

$\displaystyle\left(T_{c}^{2}g_{0}T_{c}^{-2}-g_{0}\right)\mathbf{v}_{c}=\mathbf{0},$
(52)

which may be derived as in the proof of Lemma 4. We now distinguish several
subcases. We first consider the subcase that
$T_{c}\mathbf{v}_{c}\propto\mathbf{v}_{c}$. We either have that
$\mathbf{v}_{c}\propto(1,0)^{T}$, or $\mathbf{v}_{c}\propto(0,1)^{T}$. Suppose
that $\mathbf{v}_{c}\propto(1,0)^{T}$. Note that we are considering $c$
corresponding to box (IV) in Figure 4. Considering $g_{k+1}\propto
T_{c}g_{k}T_{c}^{-1}$ and $g_{0}\mathbf{v}_{c}\propto g_{k}\mathbf{v}_{c}$ for
any $k$ we have that $g_{0}\mathbf{v}_{c}\propto
T_{c}^{k}g_{0}T_{c}^{-k}\mathbf{v}_{c}\propto T_{c}^{k}g_{0}\mathbf{v}_{c}$.
Considering $k=1$ and using $T_{c}\not\propto\text{1}$ we have that
$g_{0}\mathbf{v}_{c}$ must be proportional to a standard basis vector. Thus,
either $\alpha_{0}=0$ or $\gamma_{0}=0$. Unless $T_{c}^{2}\propto\text{1}$ we
must have $\gamma_{0}=0$ due to Eq. (52). Due to the fact that the symmetry
group of the considered fiducial states possesses an $N$-cycle yielding a
symmetry as displayed in box (IV) in Figure 4, there must as well exist an
$(b\rightarrow c)$ $N$-cycle with physical operators
$g^{\prime}_{0},g^{\prime}_{1},\ldots$, where
$g^{\prime}_{0}=g_{0}\begin{pmatrix}1&z\\\ 0&1\end{pmatrix}$ for any
$z\in\mathbb{C}$. Choosing $z=-\beta_{0}/\alpha_{0}$
($z=-\delta_{0}/\gamma_{0}$) if $\gamma_{0}=0$ ($\alpha_{0}=0$), we obtain a
$(b\rightarrow c)$ $N$-cycle with
$g^{\prime}_{0}=\begin{pmatrix}\alpha_{0}&0\\\ 0&\delta_{0}\end{pmatrix}$
($g^{\prime}_{0}=\begin{pmatrix}0&\beta_{0}\\\ \gamma_{0}&0\end{pmatrix}$),
respectively. Recall that the latter case can only occur if
$T_{c}^{2}\propto\text{1}$, which implies $T_{c}=\operatorname{diag}(i,-i)$
due to the chosen standard form. As $g^{\prime}_{k+1}\propto
T_{c}g^{\prime}_{k+1}T_{c}^{-1}$ we hence have that $g^{\prime}_{k+1}\propto
g^{\prime}_{0}$ for all $k$. Thus, we have an $(b\rightarrow c)$ 1-cycle. The
proof works similarly if $\mathbf{v}_{c}\propto(0,1)^{T}$, instead.

Let us now consider the subcase that
$T_{c}\mathbf{v}_{c}\not\propto\mathbf{v}_{c}$ and $T_{c}^{2}\propto\text{1}$,
i.e., we are considering $c$ corresponding to box (VII) in Figure 4. It will
be convenient to introduce $S=(\mathbf{v}_{c},T\mathbf{v}_{c})$ and write
$g_{0}=S\begin{pmatrix}\alpha_{0,S}&\beta_{0,S}\\\
\gamma_{0,S}&\delta_{0,S}\end{pmatrix}S^{-1}$. Note that $T_{c}S\propto
S\sigma_{x}$. Thus, the condition $g_{0}\mathbf{v}_{c}\propto
g_{1}\mathbf{v}_{c}\propto T_{c}g_{0}T_{c}^{-1}\mathbf{v}_{c}$ yields
$\sigma_{x}(\beta_{0,S},\delta_{0,S})^{T}\propto(\alpha_{0,S},\gamma_{0,S})^{T}$.
Thus, we have $g_{0}=S\begin{pmatrix}\alpha_{0,S}&\lambda\gamma_{0,S}\\\
\gamma_{0,S}&\lambda\alpha_{0,S}\end{pmatrix}S^{-1}$ for some
$\lambda\in\mathbb{C}$. Note that the considered subcase can only occur in
case of even $N$. Thus, we may untilize the $2$-cycles displayed in box (VII)
in Figure 4 in order to conclude that there must exist another $(b\rightarrow
c)$ $N$-cycle with $g_{0}^{\prime},g_{1}^{\prime},\ldots$ such that
$g_{0}^{\prime}=S\begin{pmatrix}\alpha_{0,S}^{\prime}&\gamma_{0,S}^{\prime}\\\
\gamma_{0,S}^{\prime}&\alpha_{0,S}^{\prime}\end{pmatrix}S^{-1}$. Now, as
$g_{k+1}^{\prime}\propto
T_{c}g_{k}^{\prime}T_{c}^{-1}=S\sigma_{x}\begin{pmatrix}\alpha_{k,S}^{\prime}&\gamma_{k,S}^{\prime}\\\
\gamma_{k,S}^{\prime}&\alpha_{k,S}^{\prime}\end{pmatrix}\sigma_{x}S^{-1}$, we
have $g_{k}^{\prime}\propto g_{0}^{\prime}$ for all $k$. Thus, there exists a
$(b\rightarrow c)$ $1$-cycle.

Let us finally consider the subcase that
$T_{c}\mathbf{v}_{c}\not\propto\mathbf{v}_{c}$ and
$T_{c}^{2}\not\propto\text{1}$, i.e., we are considering $c$ corresponding to
box (VIII) in Figure 4. As $T_{c}^{2}\not\propto\text{1}$, Eq. (52) yields
$\begin{pmatrix}0&\beta_{0}\\\
\gamma_{0}&0\end{pmatrix}\mathbf{v}_{c}=\mathbf{0}$. As $\mathbf{v}_{c}$ is
not an eigenvector of $T_{c}$, i.e., not proportional to a standard basis
vector, this yields $\beta_{0}=0$ and $\gamma_{0}=0$, i.e., $g_{0}$ is
diagonal. As $g_{k+1}\propto T_{c}g_{k}T_{c}^{-1}$ and $T_{c}$ is diagonal
too, we have $g_{k}\propto g_{0}$ for all $k$. Thus, there exists a
$(b\rightarrow c)$ $1$-cycle. This completes the proof of the lemma. ∎

##  Appendix F Symmetries of the TIMPS corresponding to the remaining 4 SLOCC
classes of the fiducial state

We briefly discuss here the four remaining SLOCC classes of the fiducial
states. The subsequent subsections are all structured in the same way. We
present the symmetries of the fiducial states and the concatenation rules,
which would then, analogously to the derivation in the main text, allow to
determine the symmetries, SLOCC transformations, and SLOCC classes of the
corresponding TIMPS. Instead of completely characterizing here all the
symmetries (and SLOCC classes), we just highlight some particularly
interesting symmetries, which do not occur in the cases studied in the main
text. The notation used for the representatives refer to the value and the
degeneracy of the eigenvalues of the corresponding matrix pencil (see also Sec
III).

###  F.1 TIMPS corresponding to the fiducial states represented by $M^{3}(0)$

We consider here the SLOCC class of the fiducial state which is represented by
the state by
$|M^{3}(0)\rangle=|0\rangle(|01\rangle+|12\rangle)+|1\rangle(|00\rangle+|11\rangle+|22\rangle)$.

The matrix pencil reads

$\displaystyle\mathcal{P}=M^{3}(0)=\begin{pmatrix}\lambda&\mu&0\\\
0&\lambda&\mu\\\ 0&0&\lambda\end{pmatrix}.$ (53)

####  F.1.1 Symmetries of the fiducial state

As before, the symmetries of the fiducial state can be straight forwardly
determined using the corresponding matrix pencil (or directly from the state).
Symmetries of the form $\text{1}\otimes B\otimes C$ read

$\displaystyle B$ $\displaystyle=\begin{pmatrix}1&B_{01}&B_{02}\\\
0&1&B_{01}\\\ 0&0&1\end{pmatrix}$ $\displaystyle C^{T}$
$\displaystyle=\begin{pmatrix}1&-B_{01}&B_{01}^{2}-B_{02}\\\ 0&1&-B_{01}\\\
0&0&1\end{pmatrix}.$ (54)

The symmetries involving the physical symmetry ${g}$ can be easily determined
using Eq. (13). As the matrix pencil possesses the eigenvalue 0, we have
$\beta=0$. Moreover, w.l.o.g. we choose the normalization $\delta=1$. Hence,
any physical symmetry can be chosen to be lower triangular. For any given $g$
the corresponding matrices $B_{g},C_{g}$ can then easily be determined and we
otbain

$\displaystyle{g}$ $\displaystyle=\begin{pmatrix}\alpha&0\\\
\gamma&1\end{pmatrix}$ $\displaystyle B_{g}$
$\displaystyle=\begin{pmatrix}1/\alpha&0&0\\\ 0&1&-\gamma\\\
0&0&\alpha\end{pmatrix}$ $\displaystyle C_{g}^{T}$
$\displaystyle=\begin{pmatrix}\alpha&-\gamma&0\\\ 0&1&0\\\
0&0&1/\alpha\end{pmatrix}.$ (55)

The symmetry group of the state $|M^{3}(0)\rangle$ is thus given by

$\displaystyle{g}\otimes B_{g}B\otimes C_{g}C.$ (56)

As shown in [18] (see also Sec. III), symmetries and SLOCC transformations
between TIMPS generated by any fiducial state within the SLOCC class of the
representativ $|M^{3}(0)\rangle$ can be determined by focusing on the states
of the form $\text{1}\otimes b\otimes\text{1}|M^{3}(0)\rangle$. The tensor $A$
associated to $\text{1}\otimes b\otimes\text{1}|M^{3}(0)\rangle$ reads

$\displaystyle A_{0}=\begin{pmatrix}0&b_{00}&b_{01}\\\ 0&b_{10}&b_{11}\\\
0&b_{20}&b_{21}\end{pmatrix},\ A_{1}=\begin{pmatrix}b_{00}&b_{01}&b_{02}\\\
b_{10}&b_{11}&b_{12}\\\ b_{20}&b_{21}&b_{22}\end{pmatrix}.$ (57)

####  F.1.2 Concatenation rules

As explained in the main text, the symmetries of the MPS corresponding to a
particular fiducial state can be determined by the corresponding concatenation
rules. In case of the fiducial state $\text{1}\otimes
b\otimes\text{1}|M^{3}(0)\rangle$ the concatenation rules read

$\displaystyle
b\underbrace{\begin{pmatrix}1/\alpha_{k+1}&B_{01}^{(k+1)}/\alpha_{k+1}&B_{02}^{(k+1)}/\alpha_{k+1}\\\
0&1&B_{01}^{(k+1)}-\gamma_{k+1}\\\
0&0&\alpha_{k+1}\end{pmatrix}}_{LHS_{k+1}}b^{-1}\propto$
$\displaystyle\underbrace{\begin{pmatrix}1/\alpha_{k}&(B_{01}^{(k)}+\gamma_{k})/\alpha_{k}&(B_{02}^{(k)}+B_{01}^{(k)}\gamma_{k})/\alpha_{k}\\\
0&1&B_{01}^{(k)}\\\ 0&0&\alpha_{k}\end{pmatrix}}_{RHS_{k}}$ (58)

Observe that the matrices involved in Eq. (F.1.2) are upper triangular. Thus,
the diagonal elements are the eigenvalues of the matrices. As the determinant
of the matrices on both sides equals 1, the proportionality factor must be a
thrid root of unity and it follows that either $\alpha_{k+1}=\alpha_{k}$, or
$\alpha_{k+1}=1/\alpha_{k}$. Note that all eigenvalues are distinct as long as
$\alpha_{k}\neq\pm 1$. Similar tools as presented in the main text can be used
to determine all the symmetries and to identify the normal MPS corresponding
to fiducial states in this SLOCC class. One finds, apart from global
symmetries for instance two-cycles.

###  F.2 TIMPS corresponding to the fiducial states represented by
$M^{2}(0)\oplus M^{1}(0)$

The SLOCC class we consider here is represented by the fiducial state by
$|M^{2}(0)\oplus
M^{1}(0)\rangle=|0\rangle(|01\rangle)+|1\rangle(|00\rangle+|11\rangle+|22\rangle)$.
The corresponding matrix pencil

$\displaystyle\mathcal{P}=M^{2}(0)\oplus
M^{1}(0)=\begin{pmatrix}\lambda&\mu&0\\\ 0&\lambda&0\\\
0&0&\lambda\end{pmatrix}$ (59)

has an eingenvalue zero.

####  F.2.1 Symmetries of the fiducial state

Analogously to before we determine the symmetries by symmetries of the form
$\text{1}\otimes B\otimes C$, which are given by (choosing a convenient
normalization)

$\displaystyle B$ $\displaystyle=\begin{pmatrix}1&B_{01}&B_{02}\\\ 0&1&0\\\
0&B_{21}&B_{22}\end{pmatrix}$ $\displaystyle C^{T}$
$\displaystyle=\begin{pmatrix}1&-B_{01}+\frac{B_{02}B_{21}}{B_{22}}&-\frac{B_{02}}{B_{22}}\\\
0&1&0\\\ 0&-\frac{B_{21}}{B_{22}}&1/B_{22}\end{pmatrix}.$ (60)

and symmetries of the form $g\otimes B_{g}\otimes C_{g}$. In the case
considered here they are given by

$\displaystyle{g}$ $\displaystyle=\begin{pmatrix}\alpha&0\\\
\gamma&1\end{pmatrix}$ $\displaystyle B_{g}$
$\displaystyle=\begin{pmatrix}1&0&0\\\ 0&\alpha&0\\\ 0&0&1\end{pmatrix}$
$\displaystyle C_{g}^{T}$ $\displaystyle=\begin{pmatrix}1&-\gamma/\alpha&0\\\
0&1/\alpha&0\\\ 0&0&1\end{pmatrix}.$ (61)

As before we chose w.l.o.g. the normalization $\delta=1$. Note that, as in the
previously considered case, the matrix pencil has a single eigenvalue, 0, and
thus we have $\beta=0$ in ${g}$. The symmetry group of the state
$|M^{2}(0)\oplus M^{1}(0)\rangle$ is thus given by

$\displaystyle{g}\otimes B_{g}B\otimes C_{g}C.$ (62)

The tensor $A$ associated to $\text{1}\otimes b\otimes\text{1}|M^{2}(0)\oplus
M^{1}(0)\rangle$ reads

$\displaystyle A_{0}=\begin{pmatrix}0&b_{00}&0\\\ 0&b_{10}&0\\\
0&b_{20}&0\end{pmatrix},\ A_{1}=\begin{pmatrix}b_{00}&b_{01}&b_{02}\\\
b_{10}&b_{11}&b_{12}\\\ b_{20}&b_{21}&b_{22}\end{pmatrix}.$ (63)

####  F.2.2 Concatenation rules

For the fiducial state $\text{1}\otimes b\otimes\text{1}|M^{2}(0)\oplus
M^{1}(0)\rangle$ the concatenation rules read

$\displaystyle b\underbrace{\begin{pmatrix}1&B_{01}^{(k+1)}&B_{02}^{(k+1)}\\\
0&\alpha_{k+1}&0\\\
0&B_{21}^{(k+1)}&B_{22}^{(k+1)}\end{pmatrix}}_{LHS_{k+1}}b^{-1}\propto\underbrace{\begin{pmatrix}1&B_{01}^{(k)}+\gamma_{k}&B_{02}^{(k)}\\\
0&\alpha_{k}&0\\\ 0&B_{21}^{(k)}&B_{22}^{(k)}\end{pmatrix}}_{RHS_{k}}$ (64)

The symmetries can be worked out similarly to the main text. To give an
example, one finds for $b$ given by

$\displaystyle b=\begin{pmatrix}b_{00}&b_{01}&b_{02}\\\ 0&b_{11}&b_{12}\\\
1&b_{21}&-b_{00}-b_{11}\end{pmatrix}\mbox{ with }b_{12}\neq 0$ (65)

3-cycles with $\alpha_{2}=1/(\alpha_{0}\alpha_{1})$ for arbitrary
$\alpha_{0},\alpha_{1}\in\mathbb{C}\setminus\\{0\\}$. Moreover,
$\gamma_{k}=\frac{\text{tr}\\{b^{-1}\det
b\\}}{b_{12}}(\alpha_{k}-1)=\frac{b_{00}^{2}+b_{02}+b_{00}b_{11}+b_{11}^{2}+b_{12}b_{21}}{b_{12}}(\alpha_{k}-1)$.

###  F.3 TIMPS corresponding to the fiducial states represented by
$M^{2}(0)\oplus M^{1}(\infty)$

Here, we consider the SLOCC class of fiducial states represented by
$|M^{2}(0)\oplus
M^{1}(\infty)\rangle=|0\rangle(|01\rangle+|22\rangle)+|1\rangle(|00\rangle+|11\rangle)$.

The corresponding matrix pencil reads

$\displaystyle\mathcal{P}=M^{2}(0)\oplus
M^{1}(\infty)=\begin{pmatrix}\lambda&\mu&0\\\ 0&\lambda&0\\\
0&0&\mu\end{pmatrix}.$ (66)

It has two distinct eigenvalues, 0 and $\infty$.

####  F.3.1 Symmetries of the fiducial state

Symmetries of the form $\text{1}\otimes B\otimes C$ (choosing a convenient
normalization) read

$\displaystyle B$ $\displaystyle=\begin{pmatrix}1&B_{01}&0\\\ 0&1&0\\\
0&0&B_{22}\end{pmatrix}$ $\displaystyle C^{T}$
$\displaystyle=\begin{pmatrix}1&-B_{01}&0\\\ 0&1&0\\\
0&0&1/B_{22}\end{pmatrix}.$ (67)

The matrix pencil has two distinct eigenvalues, $0$ and $\infty$. Thus, in
order to satiesfy Eq. (13) we have $\beta=\gamma=0$ for any symmetry on the
physical system, ${g}$. W.l.o.g. we normalize $g$ such that $\delta=1$. For a
particular $g$ one can then easily determine a particular $B_{g},C_{g}$, such
that ${g}\otimes B_{g}B\otimes C_{g}$ is a symmetry. The operators are given
by

$\displaystyle{g}$ $\displaystyle=\begin{pmatrix}\alpha&0\\\ 0&1\end{pmatrix}$
$\displaystyle B_{g}$ $\displaystyle=\begin{pmatrix}1/\alpha&0&0\\\ 0&1&0\\\
0&0&1/\alpha\end{pmatrix}$ $\displaystyle C_{g}^{T}$
$\displaystyle=\begin{pmatrix}\alpha&0&0\\\ 0&1&0\\\ 0&0&1\end{pmatrix}.$ (68)

The symmetry group of the state $|M^{2}(0)\oplus M^{1}(\infty)\rangle$ is thus
given by

$\displaystyle{g}\otimes B_{g}B\otimes C_{g}C.$ (69)

The tensor $A$ associated to $\text{1}\otimes b\otimes\text{1}|M^{2}(0)\oplus
M^{1}(\infty)\rangle$ reads

$\displaystyle A_{0}=\begin{pmatrix}0&b_{00}&b_{02}\\\ 0&b_{10}&b_{12}\\\
0&b_{20}&b_{22}\end{pmatrix},\ A_{1}=\begin{pmatrix}b_{00}&b_{01}&0\\\
b_{10}&b_{11}&0\\\ b_{20}&b_{21}&0\end{pmatrix}.$ (70)

####  F.3.2 Concatenation rules

The concatenation rules are given by

$\displaystyle
b\underbrace{\begin{pmatrix}1/\alpha_{k+1}&B_{01}^{(k+1)}/\alpha_{k+1}&0\\\
0&1&0\\\
0&0&B_{22}^{(k+1)}/\alpha_{k+1}\end{pmatrix}}_{LHS_{k+1}}b^{-1}\propto$
$\displaystyle\underbrace{\begin{pmatrix}1/\alpha_{k}&B_{01}^{(k)}/\alpha_{k}&0\\\
0&1&0\\\ 0&0&B_{22}^{(k)}\end{pmatrix}}_{RHS_{k}}.$ (71)

####  F.3.3 Symmetries of the MPS

In this class particularly interesting new symmetries occur, which we will
exemplary present here.

Considering for instance $b=\begin{pmatrix}0&b_{01}&-b_{12}b_{21}\\\
0&0&b_{12}\\\ 1&b_{21}&0\end{pmatrix}$ leads to $N$-cycles such that
$\alpha_{k+1}\alpha_{k}^{2}\alpha_{k-1}=1$ for all $k$. Now, $\alpha_{k}$ may
be expressed in terms of $\alpha_{0}$ and $\alpha_{1}$ as follows. For even
$k$ we have $\alpha_{k}=\frac{1}{\alpha_{0}^{k-1}\alpha_{1}^{k}}$ and for odd
$k$ we have $\alpha_{k}=\alpha_{0}^{k-1}\alpha_{1}^{k}$. It may be easily
verified that for odd $N$, one obtains only three non-trivial global
symmetries with $\alpha_{k}=\alpha$ such that $\alpha^{4}=1$, i.e.,
$\alpha\in\\{\pm 1,\pm i\\}$. In contrast to that, for even $N$ we obtain only
a less stringent condition, $(\alpha_{0}\alpha_{1})^{N}=1$. Thus, we have
$N$-cycles with arbitrary $\alpha_{0}\in\mathbb{C}$,
$\alpha_{1}=1/\alpha_{0}e^{i\frac{2r\pi}{N}}$ for some
$r\in\\{0,\ldots,N-1\\}$, and then
$\alpha_{k}=\begin{cases}\alpha_{0}e^{-i\frac{2rk\pi}{N}}&\text{for even }k\\\
\frac{1}{\alpha_{0}}e^{i\frac{2rk\pi}{N}}&\text{for odd }k\end{cases}$.
Interestingly, the eigenvalues of ${g}_{k}$ do not coincide (up to rescaling)
for all $k$.

Another interesting symmetry occurs for instance for
$b=\begin{pmatrix}b_{00}&-b_{00}^{2}&b_{02}\\\ 1&-b_{00}&0\\\
0&b_{21}&0\end{pmatrix}.$

We find 4-cycles with $\alpha_{2}=1/\alpha_{0}$ and $\alpha_{3}=1/\alpha_{1}$
and also 2-cycles with $\alpha_{1}=1/\alpha_{0}$ and 1-cycles with
$\alpha_{0}=-1$.

###  F.4 TIMPS corresponding to the fiducial states represented by
$M^{1}(0)\oplus M^{1}(0)\oplus M^{1}(\infty)$

The representing fiducial state of this SLOCC class is $|M^{1}(0)\oplus
M^{1}(0)\oplus
M^{1}(\infty)\rangle=|0\rangle|22\rangle+|1\rangle(|00\rangle+|11\rangle)$.
The corresponding matrix pencil reads

$\displaystyle\mathcal{P}=M^{1}(0)\oplus M^{1}(0)\oplus
M^{1}(\infty)=\begin{pmatrix}\lambda&0&0\\\ 0&\lambda&0\\\
0&0&\mu\end{pmatrix}.$ (72)

The matrix pencil has (as in the previous case) two distinct eigenvalues, $0$
and $\infty$. However, the degeneracy is different compared to the case
studied before.

####  F.4.1 Symmetries of the fiducial state

Symmetries of the form $\text{1}\otimes B\otimes C$ are given by

$\displaystyle B$ $\displaystyle=\begin{pmatrix}B_{00}&B_{01}&0\\\
B_{10}&B_{11}&0\\\ 0&0&B_{22}\end{pmatrix}$ $\displaystyle C^{T}$
$\displaystyle=B^{-1}$ (73)

In the following we use the normalization $B_{22}=1$.

Symmetries involving ${g}$ (and choosing a proper normalization) are of the
following simple form

$\displaystyle{g}$ $\displaystyle=\begin{pmatrix}\alpha&0\\\ 0&1\end{pmatrix}$
$\displaystyle B_{g}$ $\displaystyle=\begin{pmatrix}1&0&0\\\ 0&1&0\\\
0&0&1/\alpha\end{pmatrix}$ $\displaystyle C_{g}$ $\displaystyle=\text{1}.$
(74)

As the matrix pencil now has two distinct eigenvalues, $0$ and $\infty$, we
have $\beta=\gamma=0$ in ${g}$. The symmetry group of the state
$|M^{1}(0)\oplus M^{1}(0)\oplus M^{1}(\infty)\rangle$ is thus given by

$\displaystyle{g}\otimes B_{g}B\otimes C.$ (75)

The tensor $A$ associated to $\text{1}\otimes b\otimes\text{1}|M^{1}(0)\oplus
M^{1}(0)\oplus M^{1}(\infty)\rangle$ reads

$\displaystyle A_{0}=\begin{pmatrix}0&0&b_{02}\\\ 0&0&b_{12}\\\
0&0&b_{22}\end{pmatrix},\ A_{1}=\begin{pmatrix}b_{00}&b_{01}&0\\\
b_{10}&b_{11}&0\\\ b_{20}&b_{21}&0\end{pmatrix}.$ (76)

####  F.4.2 Concatenation rules

As in all the previous cases the symmetries of the correspongind MPS can be
determined by solving the concatatation rules, which are given in this case by

$\displaystyle b\underbrace{\begin{pmatrix}B_{00}^{(k+1)}&B_{01}^{(k+1)}&0\\\
B_{10}^{(k+1)}&B_{11}^{(k+1)}&0\\\
0&0&1/\alpha_{k+1}\end{pmatrix}}_{LHS_{k+1}}b^{-1}\propto$
$\displaystyle\underbrace{\begin{pmatrix}B_{00}^{(k)}&B_{01}^{(k)}&0\\\
B_{10}^{(k)}&B_{11}^{(k)}&0\\\ 0&0&1\end{pmatrix}}_{RHS_{k}}.$ (77)

##  Appendix G Table combining results on the symmetry group and the SLOCC
classification of the TIMPS $\Psi(LLT)$

type | normal | # symm. | cycles |  # SLOCC classes | representatives
---|---|---|---|---|---
IIb | yes |  $1$-param. |  1-cycle: ${g}=R\begin{pmatrix}1&y\\\ 0&1\end{pmatrix}R^{-1}$ for any $y\in\mathbb{C}$ |  $1$-param. |  $T=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}$, $\mathbf{v}=(x,0)^{T}$, $x\in\mathbb{C}\setminus\\{0\\}$.
IV | yes |  $1$-param. ($m$ divides $N$), 1 (else) |  $m$-cycle: ${g}_{k}=S_{0}\begin{pmatrix}1&e^{i\frac{2kr\pi}{m}}\\\ 0&1\end{pmatrix}S_{0}^{-1}$, where $S_{0}=(\mathbf{v},\,\mathbf{w})$ for any $\mathbf{w}\in\mathbb{C}^{2}$ |  $m$ considering a fixed $m$ 191919Note that there exist exactly $m$ different pairs of $r$ and $\mathbf{v}$ in the most right column. However, the SLOCC classes corresponding to e.g. $m=2$ would also be counted in the case of $m=4$ (as $T^{2}\propto\text{1}$ implies $T^{4}\propto\text{1}$). |  $T=\begin{pmatrix}e^{i\frac{r\pi}{m}}&0\\\ 0&e^{-i\frac{r\pi}{m}}\end{pmatrix}$ for $r\in\\{1,\ldots,\lfloor\frac{m-1}{2}\rfloor\\}$, $\mathbf{v}\in\\{(1,0)^{T},(0,1)^{T}\\}$. Moreover, $T\in\\{\text{1},i\sigma_{z}\\}$ ($T=\text{1}$) for even $m$ (odd $m$), respectively, and $\mathbf{v}=(1,0)^{T}$.
V | yes |  $1$ (trivial) |  none |  $1$-param. |  $T=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}$, $\mathbf{v}=(0,x)^{T}$, $x\in\mathbb{C}\setminus\\{0\\}$.
VI | yes |  $1$ (trivial) |  none |  $2$-param., generic |  $T=\begin{pmatrix}\sigma&0\\\ 0&1/\sigma\end{pmatrix}$, $\sigma\in\mathbb{D}$ s.t. $\sigma^{n}\neq 1$ for any $n\in\mathbb{N}$, $\mathbf{v}=(1,0)^{T}$, $\mathbf{v}=(0,1)^{T}$, or$\mathbf{v}=(v_{1},1/v_{1})^{T}$, $v_{1}\in\mathbb{C}\setminus\\{0\\}$.
VII | yes |  $1$-param. (even $N$), 2 (odd $N$) |  2-cycle: ${g}_{0}$ s.t. ${g}_{0}\mathbf{v}\propto\mathbf{v}$, ${g}_{0}T\mathbf{v}\propto T\mathbf{v}$ with any $x\in\mathbb{C}\setminus\\{0,i\\}$, ${g}_{1}={g}_{0}^{-1}$.
1-cylce: ${g}_{0}$ as above, but with $x=i$. |  $1$-param. |  $T=\begin{pmatrix}i&0\\\ 0&-i\end{pmatrix}$, $\mathbf{v}=(v_{1},1/v_{1})^{T}$, $v_{1}\in\mathbb{C}\setminus\\{0\\}$.
VIII | yes |  $1$ (trivial) |  none |  $1$-param. |  $T=\begin{pmatrix}e^{i\frac{s\pi}{m}}&0\\\ 0&e^{-i\frac{s\pi}{m}}\end{pmatrix}$ for $s\in\\{0,\ldots,\lfloor m/2\rfloor\\}$, $\mathbf{v}=(v_{1},1/v_{1})^{T}$, $v_{1}\in\mathbb{C}\setminus\\{0\\}$.
I | no |  $3$-param. |  1-cycle: ${g}=R\begin{pmatrix}x&\\\ &1/x\end{pmatrix}R^{-1}$ for any $x\in\mathbb{C}\setminus\\{0\\}$.
$m$-cycle: ${g}_{k}=T^{k}g_{0}T^{-k}$ for any ${g}_{0}$.
$m/2$-cycle: ${g}_{k}=R\begin{pmatrix}0&iz\\\ i/z&0\end{pmatrix}R^{-1}$, $z=ye^{i\frac{2k(2r+1)\pi}{m}}$ for any $y\in\mathbb{C}\setminus\\{0\\}$ | 2 (even $N$) |  $T\in\\{\text{1},i\sigma_{z}\\}$, $\mathbf{v}=\mathbf{0}$
IIa | no |  $1$-param. |  1-cycle: ${g}=R\begin{pmatrix}1&y\\\ 0&1\end{pmatrix}R^{-1}$ for any $y\in\mathbb{C}$ | 1 (even $N$) |  $T=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}$, $\mathbf{v}=\mathbf{0}$
III | no |  $1$-param. |  1-cycle: ${g}=R\begin{pmatrix}x&\\\ &1/x\end{pmatrix}R^{-1}$ for any $x\in\mathbb{C}\setminus\\{0\\}$ |  $1$-param. (even $N$) |  $T=\begin{pmatrix}\sigma&0\\\ 0&1/\sigma\end{pmatrix}$, $\mathbf{v}=\mathbf{0}$, $\sigma\in\mathbb{C}\setminus\\{0\\}$ s.t. there exists no $m\in\mathbb{N}$ s.t. $\sigma^{m}=\pm 1$.
Table 4: Summary of the symmetries (cycles in $G_{b}$) and SLOCC
classification of all normal MPS generated by $\text{1}\otimes
b\otimes\text{1}|LLT\rangle$ (first part) plus partial results on SLOCC
classes of some non-normal MPS (second part of the table). The “type”
corresponds to the labels displayed in Figure 4. Note that for non-normal MPS
the symmetry group might be larger than displayed, as the utilized methods may
fail to identify the full symmetry group, but yield a subgroup instead.
Similarly, additional non-normal MPS that have not been identified as SLOCC
equivalent might in fact be equivalent. The displayed non-normal MPS vanish in
case of odd $N$. As in Figure 4, by $m$ we denote a number such that
$T^{m}\propto\text{1}$ (if such a number exists) and write $T\propto
R\begin{pmatrix}e^{i\frac{r\pi}{m}}&\\\
&e^{-i\frac{r\pi}{m}}\end{pmatrix}R^{-1}$, $r\in\\{0,\ldots,N-1\\}$ for some
matrix $R$ and $r\in\\{0,\ldots,\lfloor m/2\rfloor\\}$.

In this appendix we show Table 4, which summarizes the results on the symmetry
groups and SLOCC classes of MPS corresponding to fiducial states in the $LLT$
class.