# Solving some constraints on a subgroup of a Lie group [closed]

Let $M$ be a rank-3 matrix, I am interested in searching all the group elements $g \in$ SU(3) Lie group, such that,

$$g^T M g =M.$$

Example 1. Let $$M[1]= \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right),$$ then we can find that there is at least a subgroup $g \in SU(2) \subset SU(3)$ that makes the $M[1]$ satisfies $$g^T M[1] g =M[1],$$ the $$g = \exp\left(\theta\sum_{k=1}^{3} i t_k \frac{\sigma_k}{2}\right) =\cos(\frac{\theta}{2})+i \sum_{k=1}^{3} t_k \sigma_k\sin(\frac{\theta}{2})$$

\begin{align} \sigma_1 = \sigma_x = \begin{pmatrix} 0&1 & 0\\ 1&0 & 0\\ 0&0 & 0 \end{pmatrix}, \sigma_2 = \sigma_y = \begin{pmatrix} 0&-i& 0\\ i&0& 0\\ 0 & 0& 0 \end{pmatrix}, \sigma_3 = \sigma_z = \begin{pmatrix} 1&0& 0\\ 0&-1& 0\\ 0 & 0 & 0 \end{pmatrix} \,. \end{align} Notice that any group element on $SU(2)$ can be parametrized by some $\theta$ and $(t_1,t_2,t_3)$. Also $\theta$ has a periodicity $[0,4 \pi)$.

Example 2. Let $$M[1]= \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right), M[2]= \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{array} \right), M[3]=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \\ \end{array} \right),$$

Can we find some subgroup $g \in G \subset$ SU(3) Lie group? such that $$g^T \{M[1], M[2], M[3]\} g =\{M[1], M[2], M[3]\}$$ This means that $g^TM[a]g=M[b]$ which may transform $a$ to a different value $b$. But overall the full set $\{M[1], M[2], M[3]\}$ is invariant under the transformation?

Example 3. . Let $$M[1]= \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right), M[2]= \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{array} \right), M[3]=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \\ \end{array} \right),$$ $$M[4]= -\left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right), M[5]= -\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \\ \end{array} \right), M[6]=-\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \\ \end{array} \right),$$

Can we find some subgroup $g \in G \subset$ SU(3) Lie group? such that $$g^T \{M[1], M[2], M[3],M[4], M[5], M[6]\} g =\{M[1], M[2], M[3],M[4], M[5], M[6]\}$$ This means that $g^TM[a]g=M[b]$ which may transform $a$ to a different value $b$. But overall the full set $\{M[1], M[2], M[3],M[4], M[5], M[6]\}$ is invariant under the transformation?

How can we write a Mathematica .nb file to solve this?

(Of course, there is always a trivial solution the identity $g=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$), but what are other solutions?

## closed as off-topic by Daniel Lichtblau, Henrik Schumacher, m_goldberg, José Antonio Díaz Navas, EdmundApr 10 '18 at 0:01

• The question does not concern the technical computing software Mathematica by Wolfram Research. Please see the help center to find out about the topics that can be asked here.
If this question can be reworded to fit the rules in the help center, please edit the question.