Can a symmetry-preserving unitary transformation that goes from a trivial SPT to a non-trivial SPT be local? This question concerns the very interesting paper: ''Symmetry protected topological (SPT) orders
and the group cohomology of their symmetry group'' by Chen et al., http://arxiv.org/abs/1106.4772
In this paper, in Section II.F. (left column of page 7), it is said that the unitary transformation U that goes from the trivial SPT state to a non-trivial SPT state whose wavefunction is written as a product of group cocycles is local. Indeed, citing the paper, ``Then, using the local unitary transformation $U=\prod_{\vartriangle} \nu_3(1,g_i,g_j,g_k) \prod_{\triangledown} \nu_3^{-1}(1,g_i,g_j,g_k)$, we find that the above ideal ground state wave function is given by $\Phi = U \Phi_0$ and...''
where $\Phi_0$ is the trivial SPT state and $\Phi$ the non-trivial one.
Given the fact that $U$ is symmetry-preserving, my question is: how can U be local, since, by definition, two different SPT phases cannot be linked by a symmetry-preserving local unitary transformation (as stated in the paper)?
 A: The trivial and non-trivial SPT states both are symmetric under on-site unitary symmetry transformations. The trivial and non-trivial SPT states can be mapped into each other by local unitary transformations (the $U$ in you question). Although such a $U$ is a local unitary transformation, 1. it is not on-site, 2. it is not the on-site unitary symmetry transformations, 3. it is not symmetric under the on-site symmetry transformations.
If two states can be mapped into each other by symmetric local unitary transformations, then the two states have the same SPT order.
A: There is an important distinction here which I feel the other answers have not addressed. One can check that the unitary $U$ as given in the question is indeed symmetric in the sense that $[U,W(g)] = 0$, where $W(g)$ is the representation of the symmetry. It is also a local unitary, in the sense that one can find a local (possibly time-dependent) Hamiltonian $H(t)$ such that $U$ is the time evolution of $H$, $$U = \mathcal{T} \exp\left(-i\int_0^1 H(t) dt\right).$$ However, one can prove that the Hamiltonian $H(t)$ cannot be chosen to be symmetric, i.e. necessarily $[H(t),W(g)] \neq 0$ for some $t$. Only local unitaries that can be generated by symmetric Hamiltonians have a right to be called symmetric local unitaries.
(If we instead define a local unitary to be a finite-depth quantum circuit, then the corresponding statement is that $U$ cannot be written as a finite-depth quantum circuit in which each layer is individually symmetric.)
A slightly different (but equivalent) way to think about it is that $U$ is only symmetric on a system with no boundaries. If we try to implement it on a system with boundary we find that it must break the symmetry. A truly symmetric local unitary should respect the symmetry regardless of boundary conditions.
A: It seems that you thought $\Phi_0$ is a trivial SPT while $\Phi$ is nontrivial. This does not make sense without defining the symmetry transformation. The fact that $\Phi=U\Phi_0$ means both are product state ($U$ is a honest local unitary transformation). However, symmetry transformation is defined differently in the two states: for $\Phi$ the symmetry action is simply $|g_i\rangle\rightarrow |gg_i\rangle$, while on $\Phi_0$, the symmetry action is much more complicated (pull back the simple one on  $\Phi$ by $U$). So both are nontrivial SPT phases if $\nu_3$ is a nontrivial cocycle, but with different definitions of symmetry transformations. In other words, if we start from $\Phi_0$ with "trivial" symmetry transformation $|g_i\rangle\rightarrow |gg_i\rangle$, transforming to $\Phi_0$ by $U$, one still gets a trivial SPT phase.
