Is there a zero correlation length spin-$1$ chain in the Haldane phase? The ground state of the spin-$1$ AKLT model gives an example of a $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetry-protected topological (SPT) phase, the Haldane phase. This state is a nice example of the Haldane phase because it has an exact matrix product states (MPS) representation. This state has exponentially decaying connected correlation functions.
Is there an example of a spin-$1$ state that's in the Haldane phase but has zero correlation length?
In particular, I'd want the protecting symmetry to be the same $\mathbb{Z}_2 \times \mathbb{Z}_2$ that protects the AKLT state (i.e. $\pi$ rotations about $x$, $y$, and $z$ axes), or the inversion symmetry that protects the AKLT state. I'm currently learning about SPT phases, so it's possible I'm missing or misunderstanding something trivial.

For example, in $1$d, there's the spin-$1/2$ cluster state, which is again an exact matrix product state protected by a $\mathbb{Z}_2 \times \mathbb{Z}_2$  symmetry. The cluster state is even nicer than the AKLT state, in a sense, in that its correlation functions vanish identically after a finite distance. That is, unlike AKLT, its correlation length is zero.
My motivation is that I'm curious about whether there exists something similar to the cluster state but for spin-$1$ particles, and still in the Haldane phase and protected by the same symmetries (or a subset of them) as the AKLT state. I found a paper that mentions a generalization of cluster states to qudits, but it looks like the symmetry group there is larger, $\mathbb{Z}_3 \times \mathbb{Z}_3$ for spin-$1$. Further, it's not clear to me that breaking that symmetry down to some $\mathbb{Z}_2 \times \mathbb{Z}_2$ would yield the same $\mathbb{Z}_2 \times \mathbb{Z}_2$ ($\pi$ rotations about $x$, $y$, and $z$) that protects the AKLT state.
 A: The following gives a spin-1 chain which is $\mathrm{SO(3)}$ invariant, with a 2-site unit cell (i.e. breaking translation symmetry to translation by two sites):

*

*Start with two spin-$\tfrac12$ at each site.


*Place each of the spin-$\tfrac12$ in a maximally entangled singlet state with one of the spin-$\tfrac12$ at the adjacent site.


*The resulting state is $\mathrm{SO(3)}$ invariant, and in the Haldane phase, for the representation $\tfrac12\otimes \tfrac12=0\oplus1$ acting at each site.


*This representation can be naturally embedded in $1\otimes 1 = 0\oplus 1 \oplus 2$. Thus, the chain above can also be seen as an $\mathrm{SO(3)}$ invariant chain with representation $1\otimes 1$ per site, which is in the Haldane chain.


*We can now split the two spin-$1$s. This gives a chain with a spin 1 per site, but which breaks translational symmetry (it is translational invariant under translation by two sites).


*By construction, the chain (i) is in the Haldane phase and (ii) has correlation length 0.

One way to look at the resulting state is to note that within the 2-site unit cell, the resulting state cannot have spin 2, which is precisely the property the AKLT Hamiltonian enforces. Thus, on 2 sites, the state can be obtained by applying the AKLT construction on two sites to two dangling spin-1/2 (i.e., put a singlet inbetween the two spin-1/2, and project each pair onto spin 1). This map is then applied to a chain of singlets, but before that, one needs to apply the map which makes the 2-site AKLT construction an isometry. This map is precisely $\rho^{-1/2}$, where $\rho$ is the boundary state of the 2-site AKLT construction.
This should also provide the most tangible way to construct an MPS representation of the state.
In fact, this way one can construct a family which interpolates from the given state to the AKLT model: Note that $\rho$ must be a linear combination of the symmetric and antisymmetric projector, giving a one-parameter family which contains, among others, the AKLT state and the given model.
A: Here, I add some details based on Norbert Schuch's construction involving isometries replacing the AKLT projectors. Following a similar set of steps to that in Appendix A of this paper on AKLT scars, we immediately get that the two-site MPS matrices are
\begin{equation}
M^{[1,1]} = \frac{1}{\sqrt{2}}\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix}
\end{equation}
\begin{equation}
M^{[1,0]} = \frac{-1}{2}\begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}
\end{equation}
\begin{equation}
M^{[1,-1]} = \frac{-1}{\sqrt{2}}\begin{pmatrix}
0 & 0 \\
1 & 0
\end{pmatrix}
\end{equation}
\begin{equation}
M^{[0,0]} = \frac{-1}{2}\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
\end{equation}
where the superscripts stand for total spin $S$ and total spin-$z$ $S^z$ of the pair of spin-$1$s. The resulting transfer matrix is of the from $v v^\dagger$, where $v = \frac{1}{\sqrt{2}}(1,0,0,1)$.
This implies that two-spin connected correlation functions vanish as soon as the spins are in two different unit cells separated by at least one unit cell; such connected correlation functions vanish after the distance between the spins is $4$. Thus the correlation length is truly zero.

How about whether this state is an SPT protected by the correct $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetry; i.e. how does this state transform under $e^{i \pi S^x_{tot}}$ and $e^{i \pi S^z_{tot}}$?
Under $e^{i \pi S^x_{tot}}$, we have $|1,1\rangle \leftrightarrow -|1,-1\rangle$, while $|1,0\rangle \rightarrow -|1,0\rangle$ and $|0,0\rangle \rightarrow |0,0\rangle$. By inspection, this is realized through the virtual spin transformation of $M \rightarrow \sigma^x M \sigma^x$.
Under $e^{i \pi S^z_{tot}}$, we have $|1, \pm 1\rangle \rightarrow -|1,\pm 1\rangle$, while $|1,0\rangle \rightarrow |1,0\rangle$ and $|0,0\rangle \rightarrow |0,0\rangle$. By inspection, this is realized through the virtual spin transformation of $M \rightarrow \sigma^z M \sigma^z$.
This is important, as we have a minus sign on the right hand side of $\sigma^x \sigma^z = - \sigma^z \sigma^x$, which, by the arguments in this paper on SPT states, implies we have an entanglement spectrum degeneracy protected by the $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetry above, the hallmark of an SPT.
