# Symmetry fractionalization of symmetry enriched topological phases

I have been working on symmetry-protected topological (SPT) phases, a.k.a symmetry-protected trivial phases. The classification of SPT relies on the projective representation of a symmetry group (at least for 1D), i.e.

$$U(g_1) U(g_2) = e^{i \phi}U(g_1 g_2)$$

(which differs from the normal representation theory up to a U(1) phase factor).

Thus symmetry fractionalization arises, e.g. in AKLT chain, there are two SU(2) degrees of freedom at the edges of the chain, the combination of which give the original SO(3) symmetry.

There are some new papers about symmetry enriched topological (SET) phases, which also rely on projective representation of a symmetry group (as far as I understand). As a consequences, symmetry fractionalization also appears and carries fractional charges, etc.

I'm wondering if the classification for SPT and SET is essentially the same, but of course SPT distinguishes different "trivial phases", while SET distinguishes different "topological phases".

• Why was this downvoted? – Ruben Verresen Feb 27 '17 at 10:09
• I have no clue... It seems an interesting question about the difference between symmetry fractionalization in SPT and SET. In addition, the presence of fractional charges (more precisely, excitations) is an indication of topological order but not all topological orders would have this feature. Perhaps SET would be the condition of the presence of fractional charges. (I'm not sure) – Exhaustive Feb 27 '17 at 10:56

## 1 Answer

The notion of the "fractional charge" carried by anyons in an SET phases is actually subtly different from the projective representations that appear at the boundary of a one-dimensional SPT. To illustrate this distinction, consder the case of a $U(1)$ symmetry. One can attempt to define a "projective representation" corresponding to, say, 1/3 charge as observed in the quasiparticles of the fractional quantum Hall effect: $$V(e^{i\theta}) = e^{i\theta/3}$$ Observe, however, that this "projective representation" differs from the trivial representation only be a phase factor. Given that the action of a symmetry on the boundary of an SPT is only determined up to a phase factor, there is no well-defined sense in which an SPT boundary can be said to carry fractional $U(1)$ charge.

The reason why anyons in an SET are different and it is meaningful for them to carry fractional $U(1)$ charge is a rather subtle. We have to go back and understand why, exactly, the symmetry action on the boundary of an SPT is only defined up to a phase factor. Basically, one consider a one-dimensional chain with open boundary conditions. The system acts linearly (non-projectively) on the whole chain. By reducing to the low-energy modes at the boundary, we conclude that the system still acts linearly on the boundary (which comprises two points $a$ and $b$). Locality implies that we can decompose the symmetry action on the boundary as $U(g) = V_a(g) \otimes V_b(g)$, where $V_a(g)$ acts on the left edge and $V_b(g)$ acts on the right edge. But this decomposition is not quite unique, since we can multiply $V_a(g)$ by any phase factor $\beta(g)$ so long as we also multiply $V_b(g)$ by $\beta(g)^{-1}$. Hence the phase factor ambiguity.

It might seem that the situation for anyons is similar. Consider a state $|a,\overline{a}\rangle$ containing an anyon $a$ and its antiparticle $\overline{a}$, well separated from each other. Then we can decompose the symmetry action on this state as a product $V_a(g) \otimes V_{\overline{a}}(g)$. By a similar argument as before, it would seem that $V_a(g)$ is only defined up to a phase factor transformation.

But, there is something else we can do! Suppose that $n$ copies of $a$ fuse to the vacuum. (For simplicitly, I'll assume $n=3$). This allows us to consider the state $|a,a,a\rangle$ and decompose the symmetry action on this state as $U(g) = V_a(g) \otimes V_a(g) \otimes V_a(g)$. Now observe that the only phase factor transformations we can do on $V_a(g)$ are of the form $V_a(g) \to \beta(g) V_a(g)$, where $\beta(g)^3 = 1$. This does not allow us to eliminate the charge 1/3 projective representation described above, hence charge 1/3 is actually distinct from charge 0.

The upshot is that, whereas projective representation of an SPT boundary is classified by the cohomology group $H^2(G, U(1))$, for an anyon of which $n$ copies fuse to the vacuum it is actually $H^2(G, Z_n)$ instead.

There are also more general things that can happen in an SET, for example the symmetry can change an anyon into a different anyon type.

• Thanks so much for pointing out the difference between SPT and SET with respect of "projective representation". As far as I understand, the fusion rule of anyons in SET phase is also an important feature to define the phase in projective representation, which doesn't appear in symmetry fractionalization of SPT. – Exhaustive Mar 3 '17 at 10:35
• Actually, I have a follow-up question. Since the ground state of SPT is symmetry-fractionalized (e.g. AKLT chain), we could consider the symmetry in the Hamiltonian (SO(3)) has broken simultaneously. In fact, if we take the both edges of AKLT chain and put them together, we still recover SO(3) symmetry in the Hamiltonian. Is the symmetry in the Hamiltonian really broken, or in another word, how to understand the phenomenon? (this holds also for SET I suppose) Thanks so much, Dominic! – Exhaustive Mar 3 '17 at 10:40
• I would not say that the symmetry is spontaneously broken since there is no order parameter in the bulk. But there are non local transformations which map SPT phases in 1-D to spontaneous symmetry breaking phases. I wrote about them in arxiv.org/abs/1304.0783 – Dominic Else Mar 3 '17 at 15:32
• Yes, I agree that there is no local order parameter in the bulk. But the fact is symmetry of ground state has changed. There are some SPT phases that have non-local order parameter (e.g. string order parameter in Haldane phase), but there is no general way to find them for arbitrary SPT phases as far as I understand. Maybe when symmetry fractionalization happens, there must be some "hiden" order parameter that will appear. (I guess) – Exhaustive Mar 4 '17 at 14:46