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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 $K$?

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  • $\begingroup$ I am guessing the answer is an order 8 quaternion then? $\endgroup$
    – wonderich
    Apr 25, 2018 at 4:23
  • $\begingroup$ see also qchu.wordpress.com/2011/02/12/su2-and-the-quaternions $\endgroup$
    – wonderich
    Apr 25, 2018 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, 2018 at 4:42
  • $\begingroup$ $P_z$ is a scalar matrix, fixed by any conjugation. You probably have a typo. $\endgroup$
    – orangeskid
    Apr 25, 2018 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. $\endgroup$ Apr 25, 2018 at 6:18

1 Answer 1

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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 $U(3)$.

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  • $\begingroup$ This is a surprising result to me again. $\endgroup$
    – wonderich
    Apr 25, 2018 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, 2018 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, 2018 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, 2018 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, 2018 at 11:36

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