After having seen some videos and read some articles, I am having some confusion about the symmetry group G of spin 1 Haldane chain. Being composed of spin 1 sites, it seems natural to consider SO(3) as the symmetry group of the bulk of the Haldane chain. But if we look at the ground state ( the valence bond state ) of the AKLT model, it also makes sense to treat each bond as having $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetry ( note that I am assuming Haldane chain and AKLT chains as having the same symmetry group. Also note that both candidates for G give me the correct cohomology: $H^2( SO(3),U(1)) = \mathbb{Z}_2$ as well as $H^2( \mathbb{Z}_2 \times \mathbb{Z}_2,U(1)) = \mathbb{Z}_2$.
Now, I am actually interested in calculating the group cohomology of Haldane chain, and then I would like to see the edge states ( i.e. two spin 1/2 sites at the ends of the chain ) as the projective representation of group G. But I am getting confused due to two different possibilities of G. I want to know what is the precise symmetry of Haldane spin 1 chain. Here are a few ideas that I am having:
My assumption in treating Haldane chain and AKLT chains as having same group G might be wrong.
Perhaps SO(3) is the group of Haldane chain only when we think in low energy limit. So I should substitue G = SO(3) if I am in low energy limit and should use G = $\mathbb{Z}_2 \times \mathbb{Z}_2$ otherwise.
I also read in a few articles that $\mathbb{Z}_2 \times \mathbb{Z}_2$ is the subgroup that remains after the symmetry of SO(3) is broken.
Any help/reference will be appreciated. In particular, if someone could point out what the correct groups are, and where my misunderstanding(s) lies, that would be great.
