# Fractionalization and the structure of spin rotation group?

As we know, the phenomena of fractionalizations in condensed matter physics is fantastic, like fractional spin, fractional charge , fractional statistics, .... And one key point is that the quasiparticals must be created or annihilated by pair.

On the other hand, consider the groups $SU(2)$ and $SO(3)$, they are the rotation groups for half-integer and integer spins, respectively. And we know that $SU(2)/\mathbb{Z}_2=SO(3)$, which means that each element in $SO(3)$ can be viewed as one pair $(U,-U)$, where $U\in SU(2)$ (otherwise put: the coset $\left\{U, -U\right\} \subset SU(2)$ in the quotient group $SU(2)/\mathbb{Z}_2$ is our element in $SO(3)$).

So I wonder that whether is there any underlying connection between the pair nature of quasiparticals in topological phase in physics side and the pair structure relating $SU(2)$ and $SO(3)$ in mathematics side?

Thank you very much.

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Dear K-boy: I added a phrase about the coset $\left\{U, -U\right\}$ - just another way to say the "one pair in SU(2)" thing, which some readers might find a little obtuse. Group theorists will likely know what you are talking about, but spelling it out might make it clearer for the rest of us. Delete it if you feel it changes the sense, but it's nice if a wider audience can understand the question, even if they can't answer it. – WetSavannaAnimal aka Rod Vance Aug 8 '13 at 9:47
The phenomenon you mentioned is called "symmetry fractionalization" (arxiv.org/abs/1012.4470), i.e. the symmetry of the system is $SO(3)$, but the quasiparticle carries fractionalized symmetry (or projective representation) $SU(2)$. This phenomenon belongs to the symmetry enriched topological (SET) order, classified (partly) by the group cohomology $H^2(SO(3),\mathbb{Z}_2)=\mathbb{Z}_2$, meaning that the non-trivial quasiparticles can be trivialized in pairs. – Everett You Aug 9 '13 at 1:23
@K-boy Yes, the physics is very much the same as AKLT chain (and AKLT chain is a SPT not a SET phase). But when you talk about "fractionalization", you mean something is broken apart and its pieces must be DECONFINED. In the AKLT chain, the spin-1/2 objects are confined (as 1+1D gauge theory always confining) to the end of the chain and can not move freely in the system. Moreover, if you close the AKLT chain, then no spin-1/2 excitations actually exist in the bulk. So you seem to break spin-1 into spin-1/2's, but they then confined back to spin-1, and we should not call it a fractionalization. – Everett You Aug 11 '13 at 2:42
@K-boy So "symmetry fractionalization" actually means you can not only break spin-1 into spin-1/2's, but also the spin-1/2 excitations are defined in the bulk. Only in this case, we call it a "successful" fractionalization. And such senario can only be achieved in 2+1D or higher dimensions. In 2+1D, Yao, Fu and Qi (arxiv.org/abs/1012.4470) constructed a AKLT loop liquid state with exact solvable models. In that state, the deconfined spin-1/2 excitations arise in the spin-1 system, and the $\mathbb{Z}_2$ topo. order coexists with the fractionalization. So it is actually a SET phase. – Everett You Aug 11 '13 at 2:50
@K-boy In fact, all "successful" fractionalization must come along with topological order, otherwise you can not gauge away the unphysical degrees of freedom arise in the fractionalization. Then it is not hard to understand why symmetry fractionalization is actually SET. – Everett You Aug 11 '13 at 2:53

The group of rotations of an $N$-dimensional space is $SO(N)$. Being a symmetry of nature, classical systems transform according to representations of $SO(N)$.
Quantum mechanics, on the other hand, allows systems which transform according to the universal covering groups of classical symmetries. This is the reason why we get in three dimensional quantum theory representations of $SU(2)$ which are not true representations of $SO(3)$, (the half integer spin representations). More generally, we have, in quantum theory, representations of $Spin(N) = SO(N) \ltimes \mathbb{Z}_2$.
However in the case of a two spatial dimensions, $SO(2) \cong U(1)$, and the universal covering of $U(1)$ is not $Spin(2)$ but rather $\mathbb{R}$.
In contrast to $SO(2)$ or $U(1)$ which allow discrete values of the two dimensional spin: $u = e^{i n \theta}$ $n \in \mathbb{Z}$, $0\le \theta <2 \pi$, the universal covering $\mathbb{R}$ allows a continuum of spin values.