Is it possible to have topological degeneracy in 1D ?

I mean to have q-fold degenerate ground states on a ring which could not be lifted by local perturbation.

If the answer is no, then what is the physical (or mathematical) reason against having such a state in 1D ?

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Perhaps if you have $N$ segments of the wire which are topological superconductors ($p$-wave), all of these segments have non-topological segments between them, and the boundaries between the non-topological and topological sections are well separated then you would have a topologically protected $2^N$-fold degenerate ground state. However, it won't have topological order (i.e. long-range entanglement); I'm not 100% sure why. In a recent talk Kitaev mentioned that the 1-D $p$-wave chain does not have topological order. – NanoPhys May 2 '13 at 7:16

This may not be how people generally think about topological degeneracy, but if you define it to be any degeneracy or asymptotic degeneracy in the thermodynamic limit with the following two properties:

(1) it cannot be lifted by local perturbative Hamiltonians that respect the symmetry;

(2) it cannot be attributed to symmetry breaking, i.e. irreducible representation of dimension greater than 1;

then I suspect the answer to your question is affirmative. This definition makes it possible to speak of topological degeneracy between short-range entangled symmetric states.

My example is the Majumdar-Ghosh Hamiltonian for a spin-$\frac12$ chain (with periodic boundary condition and even number of sites), $$H = \sum_j \left( \boldsymbol S_j \cdot \boldsymbol S_{j+1} + \frac12 \boldsymbol S_j \cdot \boldsymbol S_{j+2}\right),$$ and my symmetry group is $SO(3)$. $H$ has two gapped, (exactly) degenerate ground states, the dimerized states with dimers on even and odd links, respectively. Both states are symmetric under $SO(3)$, so the degeneracy is not due to symmetry breaking. (By the way, both states are short-range entangled.) And I suspect that the degeneracy in the thermodynamic limit can survive any local perturbative Hamiltonian that respects $SO(3)$, on the following considerations.

First, any perturbations that respect not only $SO(3)$ but also the translation symmetry should leave the degeneracy intact, since the Lieb-Schultz-Mattis theorem is expected to apply. Secondly, arXiv:1208.0706 suggests that there is no quantum phase transition when one makes the nearest-neighbor coupling slightly different for even and odd links, which is one way to break the translation symmetry. So the degeneracy might in fact be robust against perturbations that respect just $SO(3)$ but not the translation symmetry.

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