2
$\begingroup$

This is related to the post, but a simplified version of the problem. However, the form of $P_x,P_y,P_z$ are different and richer.

Let $$G=U(2),$$ be the unitary group. Here we consider $G$ in terms of the fundamental representation of U(2). Namely, all of $g \in G$ can be written as a rank-2 (2 by 2) matrices.

Can we find some subgroup of Lie group, $$k \in K \subset G= U(2) $$ such that

$$ k^T \{P_x, P_y, P_z, -P_x, - P_y, - P_z\} k =\{P_x, P_y, P_z, -P_x, - P_y, - P_z\}. $$ This means that set $\{P_x, P_y, P_z, -P_x, - P_y, - P_z\}$ is invariant under the transformation by $k$. Here $k^T$ is the transpose of $k$. What is the full subset (or subgroup) of $K$?

Here we define: $$ P_x = \left( \begin{array}{ccc} 0 & 1 \\ 1 & 0 \\ \end{array} \right),\;\;\;\; P_y = \left( \begin{array}{ccc} -1 & 0 \\ 0 & 1 \\ \end{array} \right),\;\;\;\; P_z = \left( \begin{array}{ccc} -i & 0 \\ 0 & -i \\ \end{array} \right).$$

This means that $k^T P_a k= \pm P_b$ which may transform $a$ to a different value $b$, where $a,b \in \{x,y,z \}$. But overall the full set $ \{P_x, P_y, P_z, -P_x, - P_y, - P_z\}$ is invariant under the transformation by $k$.

There must be a trivial element $k=$ the rank-2 identity matrix. But what else can it allow?

How could we determine the complete $K$?

$\endgroup$
5
  • $\begingroup$ I am guessing the answer is an order 8 quaternion then? $\endgroup$ – wonderich Apr 25 '18 at 4:23
  • $\begingroup$ see also qchu.wordpress.com/2011/02/12/su2-and-the-quaternions $\endgroup$ – wonderich Apr 25 '18 at 4:39
  • $\begingroup$ Sorry I mean the answer contain an order 8 quaternion (in a larger non-Abelian group), but then there is some additional cyclic group. $\endgroup$ – wonderich Apr 25 '18 at 4:42
  • $\begingroup$ $P_z$ is a scalar matrix, fixed by any conjugation. You probably have a typo. $\endgroup$ – orangeskid Apr 25 '18 at 4:49
  • $\begingroup$ This is correct -- I wrote what I mean, see also the previous question. The similar forms. $P_z$ is $-i$ times an identity. $\endgroup$ – annie marie heart Apr 25 '18 at 6:18
2
$\begingroup$

Answer. The subgroup $K$ of invariant matrices in $U(2)$ is isomorphic to the finite group $$ \operatorname{SL}(2,3)\to G\to\mathbb{Z}_2) \rtimes_\varphi \mathbb{Z}_2 \quad\simeq\quad \mathbb{Z}_2\to G\to S_4) \rtimes_\varphi \mathbb{Z}_2 $$ where $\operatorname{SL}(2,3)$ is the special linear group of degree 2 over a field of 3 elements (order 24), which is isomorphic to the binary von Dyck group with the parameters (2,3,3); it has the presentation: $\langle a,b,c \mid a^3=b^3=c^2=abc\rangle$.

In other words, $K\cong G\rtimes_\varphi\mathbb{Z}_2$ for some subgroup $G\subset K$ and $G/\operatorname{SL}(2,3)\cong\mathbb{Z}_2$ or $G/\mathbb{Z}_2\simeq S_4$.

For whom is interested, I list the all $96=24\times 4$ invariant matricies of $K$: $$ \begin{gather*} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \frac{(1+i)}{\sqrt{2}} \begin{pmatrix} 1 & 0 \\ 0 & -i \end{pmatrix}, \frac{(1-i)}{\sqrt{2}} \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}, \\ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, \frac{(1+i)}{\sqrt{2}} \begin{pmatrix} 0 & 1 \\ -i & 0 \end{pmatrix}, \frac{(1-i)}{\sqrt{2}} \begin{pmatrix} 0 & 1 \\ i & 0 \end{pmatrix}, \\ \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}, \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \frac{(1+i)}{2} \begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix}, \frac{(1-i)}{2} \begin{pmatrix} 1 & 1 \\ -i & i \end{pmatrix}, \\ \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ -1 & -1 \end{pmatrix}, \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}, \frac{(1+i)}{2} \begin{pmatrix} 1 & -1 \\ i & i \end{pmatrix}, \frac{(1-i)}{2} \begin{pmatrix} 1 & -1 \\ -i & -i \end{pmatrix}, \\ \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix}, \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & -i \\ i & -1 \end{pmatrix}, \frac{(1+i)}{2} \begin{pmatrix} 1 & -i \\ -1 & -i \end{pmatrix}, \frac{(1-i)}{2} \begin{pmatrix} 1 & -i \\ 1 & i \end{pmatrix}, \\ \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & i \\ -i & -1 \end{pmatrix}, \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}, \frac{(1+i)}{2} \begin{pmatrix} 1 & i \\ -1 & i \end{pmatrix}, \frac{(1-i)}{2} \begin{pmatrix} 1 & i \\ 1 & -i \end{pmatrix} \end{gather*} $$ product with $\langle i\rangle=\{\pm I,\pm iI\}$.

For given orders, the numbers of elements in $K$ and $\mathbb{Z}_4\times S_4$ are as follows:

order | K        | Z4 x S4
--------------------------
1     | 1        | 1
2     | 19       | 19
3     | 8        | 8
4     | 20       | 44
6     | 8        | 8
8     | 24       | 0
12    | 16       | 16
--------------------------
total | 96       | 96

This information shows that $K$ (GAP ID [96, 192]) in $U(2)$ is not isomorphic to the previous answer $\mathbb{Z}_4\times S_4$ (GAP ID [96,186]) in $U(3)$.

$\endgroup$
26
  • $\begingroup$ This is a surprising result to me again. $\endgroup$ – wonderich Apr 25 '18 at 13:56
  • $\begingroup$ If $\mathbb{Z}_4$ is just a cyclic group generated by $i$ times an identity matrix, why do you need a semi direct product? Why not just a direct product? $\endgroup$ – wonderich Apr 25 '18 at 13:56
  • $\begingroup$ @wonderich The difference between two groups $\mathbb{Z}_4\rtimes S_4$ and $\mathbb{Z}_4\times S_4$ were appended. Do you have any idea what this group is? $\endgroup$ – ChoF Apr 25 '18 at 14:02
  • $\begingroup$ I understand that one can generate some of your group elements by an subgroup of SU(2) rotations $$e^{i {\theta} \hat{n} \cdot \hat{\sigma}/2}=\cos( \frac{\theta}{2})+ i \hat{n} \cdot \hat{\sigma} \sin(\frac{\theta}{2})$$ with $ \hat{\sigma}$-Pauli matrices and $\hat{n}=(n_x,n_y,n_z)$ with $|\hat{n}|=1$ and choose $n_x=1$ or $n_y=1$ or $n_z=1$ at $\theta=\pi/2,\pi,3\pi/2, 2\pi,..$ or $\theta=m \pi/2$ with $m=0,1,2,3,4,5,6,7$ where $\theta$ is 4$\pi$ periodic. $\endgroup$ – wonderich Apr 25 '18 at 14:04
  • 1
    $\begingroup$ @wonderich I found $K$ from the list of GAP small groups, and there was only one group of order 96 having the element order list given above. $\endgroup$ – ChoF Apr 26 '18 at 11:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.