2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ That last statement is not obvious. It's in general false that having isomorphic Lie algebras implies being isomorphic as Lie groups. $\endgroup$ – Qiaochu Yuan Jun 13 '14 at 2:11
0
$\begingroup$

Looking further into the matter, I found that the members $g$ of the subgroup mentioned in (4) satisfy the relationship

$g^{-1}= -\begin{pmatrix}0&1\\ 1&0\end{pmatrix}g^\dagger\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$

from which it can be derived that

$g=\begin{pmatrix}a&ib\\ ic&d\end{pmatrix}$, where $a,b,c,d \in \mathbb{R}$ and $ad+bc=1$.

This is the categorization I was looking for. I still don't know whether this subgroup has a name; I will post a separate question for that.

$\endgroup$

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.