# Which subgroup of $\mathrm{SL}(2,\mathbb{C})$ is this?

I am looking into sub-algebras of $\mathfrak{sl}_2(\mathbb{C})$ and the subgroups of $\mathrm{SL}(2,\mathbb{C})$ they generate. The basis of $\mathfrak{sl}_2(\mathbb{C})$ I am using consists of 3 anti-hermitian and 3 hermitian matrices, which for clarity I abbreviate by $A_i$ and $H_i$:

\begin{align} A_1&=\begin{pmatrix}i&0\\ 0&-i\end{pmatrix}& A_2&=\begin{pmatrix}0&1\\ -1&0\end{pmatrix}& A_3&=\begin{pmatrix}0&i\\ i&0\end{pmatrix} \\H_1&=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}& H_2&=\begin{pmatrix}0&-i\\ i&0\end{pmatrix}& H_3&=\begin{pmatrix}0&1\\ 1&0\end{pmatrix} \end{align}

Disregarding multipes (or halving all matrices first), the well-known commutation relations are

\begin{align} [A_i, A_j] &=\epsilon_{ijk}A_k& [A_i, H_j] &=\epsilon_{ijk}H_k& [H_i, H_j] &=-\epsilon_{ijk}A_k \end{align}

The sub-algebras and the sub-groups of $\mathrm{SL}(2,\mathbb{C})$ they generate that I comprehend are

(1) $\{A_1, A_2, A_3\} \rightarrow \mathrm{SU}(2)$ : $e^{A_j}$ is unitary as $A_j$ is anti-hermitian $(A_j = -A_j^\dagger)$; \begin{align} [A_1, A_2] &=A_3& [A_2, A_3] &=A_1& [A_3, A_1] &=A_2 \end{align} \begin{align} \\A_1 &\rightarrow \begin{pmatrix}e^{i\alpha}&0\\ 0&e^{-i\alpha}\end{pmatrix}& A_2 &\rightarrow \begin{pmatrix}\cos\alpha&\sin\alpha\\ -\sin\alpha&\cos\alpha\end{pmatrix}& A_3 &\rightarrow \begin{pmatrix}\cos\alpha&i\sin\alpha\\ i\sin\alpha&\cos\alpha\end{pmatrix} \end{align} general form: $\mathrm{SU}(2) = \{\begin{pmatrix}\alpha&-\overline{\beta}\\ \beta&\overline{\alpha}\end{pmatrix}: \left|\alpha\right|^2 + \left|\beta\right|^2 = 1\} = \{\begin{pmatrix}e^{i\phi}\cos\omega &-e^{-i\psi}\sin\omega \\ e^{i\psi}\sin\omega &e^{-i\phi}\cos\omega \end{pmatrix}: \omega, \phi, \psi \in \mathbb{R}\}$

(2) $\{A_1, H_2, H_3\} \rightarrow \mathrm{SU}(1,1)$ : $\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}X\begin{pmatrix}1&0\\ 0&-1\end{pmatrix} = -X^\dagger$ \begin{align} [A_1, H_2] &=H_3& [H_2, H_3] &=A_1& [H_3, A_1] &=-H_2 \end{align} \begin{align} \\A_1 &\rightarrow \begin{pmatrix}e^{i\alpha}&0\\ 0&e^{-i\alpha}\end{pmatrix}& H_2 &\rightarrow \begin{pmatrix}\cosh\alpha&-i\sinh\alpha\\ i\sinh\alpha&\cosh\alpha\end{pmatrix}& H_3 &\rightarrow \begin{pmatrix}\cosh\alpha&\sinh\alpha\\ \sinh\alpha&\cosh\alpha\end{pmatrix} \end{align} general form: $\mathrm{SU}(1,1) = \{\begin{pmatrix}\alpha&\overline{\beta}\\ \beta&\overline{\alpha}\end{pmatrix}: \left|\alpha\right|^2 - \left|\beta\right|^2 = 1\} = \{\begin{pmatrix}e^{i\phi}\cosh\omega &e^{-i\psi}\sinh\omega \\ e^{i\psi}\sinh\omega &e^{-i\phi}\cosh\omega \end{pmatrix}: \omega, \phi, \psi \in \mathbb{R}\}$

(3) $\{H_1, A_2, H_3\} \rightarrow \mathrm{SL}(2,\mathbb{R})$ : all 3 matrices are real; \begin{align} [H_1, A_2] &=H_3& [A_2, H_3] &=H_1& [H_3, H_1] &=-A_2 \end{align} \begin{align} \\H_1 &\rightarrow \begin{pmatrix}e^{\alpha}&0\\ 0&e^{-\alpha}\end{pmatrix}& A_2 &\rightarrow \begin{pmatrix}cos\alpha&sin\alpha\\ -sin\alpha&cos\alpha\end{pmatrix}& H_3 &\rightarrow \begin{pmatrix}cosh\alpha&sinh\alpha\\ sinh\alpha&cosh\alpha\end{pmatrix} \end{align} Obviously, $\mathrm{SL}(2,\mathbb{R})$ is isomorphic to $\mathrm{SU}(1,1)$, as can be seen from the exact similarity of their underlying Lie algebras.

But the subgroup resulting from a sub-algebra that I don't know how it is characterized is

(4) $\{H_2, A_3, H_1\} \rightarrow$ ??? \begin{align} [H_2, A_3] &=H_1& [A_3, H_1] &=H_2& [H_1, H_2] &=-A_3 \end{align} \begin{align} \\H_2 &\rightarrow \begin{pmatrix}cosh\alpha&-isinh\alpha\\ isinh\alpha&cosh\alpha\end{pmatrix}& A_3 &\rightarrow \begin{pmatrix}cos\alpha&isin\alpha\\ isin\alpha&cos\alpha\end{pmatrix}& H_1 &\rightarrow \begin{pmatrix}e^{\alpha}&0\\ 0&e^{-\alpha}\end{pmatrix} \end{align}

Obviously, the generated subgroup is again isomorphic to $\mathrm{SU}(1,1)$ (and to $\mathrm{SL}(2,\mathbb{R})$), because the underlying Lie-algebras have the same structure, but how is this subgroup best characterized in terms of the matrices it contains and their 4 components?

• That last statement is not obvious. It's in general false that having isomorphic Lie algebras implies being isomorphic as Lie groups. – Qiaochu Yuan Jun 13 '14 at 2:11

Looking further into the matter, I found that the members $g$ of the subgroup mentioned in (4) satisfy the relationship
$g^{-1}= -\begin{pmatrix}0&1\\ 1&0\end{pmatrix}g^\dagger\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$
$g=\begin{pmatrix}a&ib\\ ic&d\end{pmatrix}$, where $a,b,c,d \in \mathbb{R}$ and $ad+bc=1$.