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Question about trivializing an SPT phase via group extension.

+ 4 like - 0 dislike
167 views

Consider a d (spatial) dimensional SPT phase with an on-site symmetry G, classified by some non-trivial cocycle  $\omega^{d+1}(\{g_i\}) \neq \delta \mu^d(\{g_i\})$, $[\omega^{d+1}(\{g_i\})] \in H^{d+1}(G,U(1))$, . In a recent paper, Wang, Wen and Witten construct gapped boundaries via a suitable group extension $1 \longrightarrow H  \overset{i}{\longrightarrow} K  \overset{r}{\longrightarrow} G  \longrightarrow 1 $ such that the cocycle fro $H$ defined via pullback is trivial $r^*\omega^{d+1}(\{h_i\}) = \omega^{d+1}(\{r(h_i)\})= \delta \mu^d(\{h_i\})$. The gapped boundary corresponds to an $H$ invariant theory but with $K$ gauged so that the global symmetry is $H/K \cong G$ as required. The following sentence they say however confuses me: 

" By definition, two states in two different $G$-SPT phases cannot smoothly deform into each other via deformation paths that preserve the $G$-symmetry. However, two such $G$-SPT states may be able to smoothly deform into each other if we view them as systems with the extended $H$-symmetry and deform them along the paths that preserve the $H$-symmetry. " 

To me, it sounds like the two sentences contradict each other.
Q1) If there is an $H$ invariant deformation path to connect the system to a trivial state, then does that not automatically give us a $G$ invariant deformation path?
Q2) Does sentence 2 somehow only apply to the boundary rather than the bulk? 

asked Sep 14 in Theoretical Physics by abhishodh (35 points) [ no revision ]

Please give a background reference. What is SPT? 

What is SPT?

I apologize for the delayed response. The response by Xiao-Gang Wen in this post in physics stack exchange gives a nice explanation on what a Symmetry Protected Topological (SPT) phase is and is not. https://physics.stackexchange.com/questions/135398/definition-of-short-range-entanglement

1 Answer

+ 1 like - 0 dislike

The point is indeed a subtle one. If you begin with a G SPT and consider it as a K SPT, then there is still no local unitary commuting with the K action which can take you to a product state, since the generators of K are just the generators of G. However, if you allow the introduction of new ancilla bits to your sites, then it matters whether we consider it a G or a K action since we must extend our operators to act on the ancillas as well. If it's to be a G action, this is a stricter set of relations that the new operators need to satisfy than if it's to be merely a K action.

answered Sep 22 by Ryan Thorngren (1,655 points) [ no revision ]

Thanks for your answer Ryan. I understand this better after working through a few examples. For the benefit of anyone else who might stumble upon this post, I would also like to add that  within K, the original generators of G no longer form a subgroup. This means that when we add ancillary spins that transform faithfully under the extended group, K, the system is no longer G invariant. Hence, any K-invariant circuit that trivializes the G SPT state is also not G invariant which is consistent with standard SPT assertions. 

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