# Is the spin-rotation symmetry of Kitaev model $D_2$ or $Q_8$?

It is known that the Kitaev Hamiltonian and its spin-liquid ground state both break the $SU(2)$ spin-rotation symmetry. So what's the spin-rotation-symmetry group for the Kitaev model?

It's obvious that the Kitaev Hamiltonian is invariant under $\pi$ rotation about the three spin axes, and in some recent papers, the authors give the "group"(see the Comments in the end) $G=\left \{1,e^{i\pi S_x}, e^{i\pi S_y},e^{i\pi S_z} \right \}$, where $(e^{i\pi S_x}, e^{i\pi S_y},e^{i\pi S_z})=(i\sigma_x,i\sigma_y,i\sigma_z )$, with $\mathbf{S}=\frac{1}{2}\mathbf{\sigma}$ and $\mathbf{\sigma}$ being the Pauli matrices.

But how about the quaternion group $Q_8=\left \{1,-1,e^{i\pi S_x}, e^{-i\pi S_x},e^{i\pi S_y},e^{-i\pi S_y},e^{i\pi S_z}, e^{-i\pi S_z}\right \}$, with $-1$ representing the $2\pi$ spin-rotation operator. On the other hand, consider the dihedral group $D_2=\left \{ \begin{pmatrix}1 & 0 &0 \\ 0& 1 & 0\\ 0&0 &1 \end{pmatrix},\begin{pmatrix}1 & 0 &0 \\ 0& -1 & 0\\ 0&0 &-1 \end{pmatrix},\begin{pmatrix}-1 & 0 &0 \\ 0& 1 & 0\\ 0&0 &-1 \end{pmatrix},\begin{pmatrix}-1 & 0 &0 \\ 0& -1 & 0\\ 0&0 &1 \end{pmatrix} \right \}$, and these $SO(3)$ matrices can also implement the $\pi$ spin rotation.

So, which one you choose, $G,Q_8$, or $D_2$ ? Notice that $Q_8$ is a subgroup of $SU(2)$, while $D_2$ is a subgroup of $SO(3)$. Furthermore, $D_2\cong Q_8/Z_2$, just like $SO(3)\cong SU(2)/Z_2$, where $Z_2=\left \{ \begin{pmatrix}1 & 0 \\ 0 &1\end{pmatrix} ,\begin{pmatrix}-1 & 0 \\ 0 &-1 \end{pmatrix} \right \}$.

Comments: The $G$ defined above is even not a group, since, e.g., $(e^{i\pi S_z})^2=-1\notin G$.

Remarks: Notice here that $D_2$ can not be viewed as a subgroup of $Q_8$, just like $SO(3)$ can not be viewed as a subgroup of $SU(2)$.

Supplementary: As an example, consider a two spin-1/2 system. We want to gain some insights that what kinds of wavefunctions preserves the $Q_8$ spin-rotation symmetry from this simplest model. For convenience, let $R_\alpha =e^{\pm i\pi S_\alpha}=-4S_1^\alpha S_2^\alpha$ represent the $\pi$ spin-rotation operators around spin axes $\alpha=x,y,z$, where $S_\alpha=S_1^\alpha+ S_2^\alpha$. Therefore, by saying a wavefunction $\psi$ has $Q_8$ spin-rotation symmetry, we mean $R_\alpha\psi=\lambda_ \alpha \psi$, with $\left |\lambda_ \alpha \right |^2=1$.

After a simple calculation, we find that a $Q_8$ spin-rotation symmetric wavefunction $\psi$ could only take one of the following 4 possible forms:

$(1) \left | \uparrow \downarrow \right \rangle-\left | \downarrow \uparrow \right \rangle$, with $(\lambda_x,\lambda_y,\lambda_z)=(1,1,1)$ (Singlet state with full $SU(2)$ spin-rotation symmetry), which is annihilated by $S_x,S_y,$ and $S_z$,

$(2) \left | \uparrow \downarrow \right \rangle+\left | \downarrow \uparrow \right \rangle$, with $(\lambda_x,\lambda_y,\lambda_z)=(-1,-1,1)$, which is annihilated by $S_z$,

$(3) \left | \uparrow \uparrow \right \rangle-\left | \downarrow \downarrow \right \rangle$, with $(\lambda_x,\lambda_y,\lambda_z)=(1,-1,-1)$, which is annihilated by $S_x$,

$(4) \left | \uparrow \uparrow \right \rangle+\left | \downarrow \downarrow \right \rangle$, with $(\lambda_x,\lambda_y,\lambda_z)=(-1,1,-1)$, which is annihilated by $S_y$.

Note that any kind of superposition of the above states would no longer be an eigenfunction of $R_\alpha$ and hence would break the $Q_8$ spin-rotation symmetry.

-
When talking the rotational symmetry, people tends to refer to the SO(3) group and its subgroups. So the symmetry group here is $\tilde{D}_2$. $D_4$ is its projective representation (or projective symmetry group). –  Everett You Dec 30 '13 at 15:49
To which papers are you referring? There's a notational irritation that always occurs when dealing with the Dihedral groups where some people write $D_n$ and some write $D_{m}$ - where $n$ is the number of edges and vertices of the n-gon and $m=2n$ is the number of group elements - for the same group. However, I don't think either of your $D_2$ or $D_4$ are groups. In the case of your $D_4$, $e^{i\pi S_x}e^{i\pi S_y}$ is not an element of your set. However, if both were taken to generate all elements of the group, it should be apparent that they would generate the same group here. –  Matthew Titsworth Dec 30 '13 at 15:56
@ Matthew TItsworth The notation I used here is obvious and it is not the key point of my question. –  K-boy Dec 30 '13 at 18:26
@K-boy : As you noticed, your "$D_2$" is not a group, so your notation is wrong, and your $\tilde D_2$ is the true $D_2$, the dihedral group of rank $4$. And your "$D_4$" is not the dihedral group of rank $8$, but the quaternion group $Q= Q_4$ of rank $8$ (the first in the family of the dicyclic groups $Q_{2n}$, of rank $4n$). And the true dihedral $D_4$ group is not isomorphic to $D_2 \times Z_2$ (while $D_2$ is isomorphic to $Z_2 \times Z_2$). Ref: Ramond, Group Theory, Cambridge, pages $13-17$. Finally, (abstract) groups, and representations of groups, are two different things. –  Trimok Dec 30 '13 at 20:18
@K-boy. Look at the order of the elements in your $D_4$. They are $\{1,2,4,4,4,4\}$. The The dihedral group of order $8$ has two elements of order $4$ and five elements of order $2$. See also here,here,and here. Trimok is right. –  Matthew Titsworth Dec 30 '13 at 20:43

The set $G$ gives the representation of the identity and generators of the abstract group of quaternions as elements in $SL(2,\mathbb C)$ which are also in $SU(2)$. Taking the completion of this yields the representation $Q_8$ of the quaternions presented in the question.
From the description of the symmetry group as coming from here, consider the composition of two $\pi$ rotations along the $\hat x$, $\hat y$, or $\hat z$ axis. This operation is not the identity operation on spins (that requires a $4\pi$ rotation). However, all elements of $D_2$ given above are of order 2.
This indicates that the symmetry group of the system should be isomorphic to the quaternions and $Q_8$ is the appropriate representation acting on spin states. The notation arising there for $D_2$ is probably from the dicyclic group of order $4\times 2=8$ which is isomorphic to the quaternions.
It probably worth mentioning that the quaternion group $Q$ is one of the two Schur covers of the Klein four-group $K$. The other one is $D_4$, the dihedral group of degree 4. –  Isidore Seville Dec 30 '13 at 22:26