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

Question

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 $SU(2)$.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy

Comments

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).

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Everett You

---

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.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Matthew Titsworth

---

@ Matthew TItsworth The notation I used here is obvious and it is not the key point of my question.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy

---

@ Matthew TItsworth I have added the expressions for $S_x,S_y,S_z$ to my question, and it's direct to show that $e^{i\pi S_x}e^{i\pi S_y}=e^{-i\pi S_z}\in D_4$ and hence $D_4$ is of course a group. While $D_2$ is not a group as I have mentioned in the Comments at the end of my question.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy

---

@ Everett You Thanks for your comments. But, for example, don't we usually say that the spin-1/2 Heisenberg model has global $SU(2)$ spin-rotation symmetry rather than saying $SO(3)$ symmetry?

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy

---

Of course it is. It helps if I pull the $i$ out. It also helps if I don't comment before I've had coffee. Apologies.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Matthew Titsworth

---

Take the closure of $D_2$ under multiplication. This obviously gives you the set $D_4$. So the sets $D_2$ and $D_4$ generate the same group. This was the reason for the statement about notation and asking which paper you are referring to when you say "in some recent papers, the authors give..." I searched through Kitaev's "Anyons..." paper, but there is no mention of it there. There's also no mention of $D_2$ in the Yao and Lee paper. I don't have a copy of the Baskaran paper immediately available. It would be helpful if you could clarify the context from which your question is drawn.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Matthew Titsworth

---

@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.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Trimok

---

@ Matthew TItsworth You are welcome. $D_2$ appears in the 3rd paragraph on page 1 of this paper.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy

---

@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.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Matthew Titsworth

---

@ Matthew TItsworth Thank you very much. I will rethink about it.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy

---

Maybe I made a mistake and my $D_4$ is not isomorphic to the dihedral group of rank 8.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy

---

@ Trimok Thanks for your corrections.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy

Answer 1

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.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Matthew Titsworth

Comments

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.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user Isidore Seville

---

@ Matthew TItsworth Thanks for your clear summary.

This post imported from StackExchange Physics at 2014-03-09 08:38 (UCT), posted by SE-user K-boy