Description of SU(1, 1)

Solution 1:

Since $SU(1,1)$ is defined as a set of 2 by 2 matrices $U$ with unit determinant such that $U^\dagger J U = J$, where $J = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right)$. You precisely get that the general form of such matrix is

$$ U = \left( \begin{array}{cc} \alpha & \beta \\ \gamma & \delta \end{array} \right) $$ where $ \vert \alpha \vert^2 - \vert \beta \vert^2 = 1$, $ \vert \delta \vert^2 - \vert \gamma \vert^2 = 1$ and $(\alpha \, \beta^\ast)^\ast = \gamma \, \delta^\ast$. This implies $\delta = \alpha^\ast$ and $\gamma = \beta^\ast$ as you surely know.

Now $ \vert \alpha \vert^2 - \vert \beta \vert^2 = 1$ is one real equation for 4 real parameters in 2 complex numbers. The equation leaves you with 3 free parameters, and your solution is the most general one for $SU(1,1)$.

For the second part of your question. Elements of $SL(2, \mathbb{R})$ are 2 by 2 matrices with real coefficients and unit determinant. Both $SL(2, \mathbb{R})$ and $SU(1,1)$ are real forms of $SL(2, \mathbb{C})$.

The isomorphism is established as follows (see Bargamann's article). Define 2 by 2 Hermitian matrix $Z$ as follows: $$ Z = \left( \begin{array}{ll} x_0 + x_3 & x_1 + i x_2 \\ x_1 - i x_2 & x_0 - x_3 \end{array} \right) $$ Notice that $\det Z = x_0^2 - x_1^2-x_2^2 - x_3^2$. Mappings $Z \to g^\dagger Z g$, where $g$ is complex 2 by 2 matrices with unit determinant, i.e. $SL(2, \mathbb{C})$ preserve the determinant.

It is not hard to see that $SU(1,1)$ correspond to those transformations that fix hyper-plane $x_3 = 0$, while $SL(2,\mathbb{R})$ correspond to those transformations that fix hyperplane $x_2=0$, while transformations that fix $x_1=0$ correspond to $Sp(2, \mathbb{R})$. They are all isomorphic via rotation in the ambient space formed by $Z$.

Solution 2:

Another way to parametrise elements of $SU(1,1)$ is through its Lie algebra. In the vicinity of the identity element, the elements of $g\in SU(1,1)$ can be written as the exponential of an element $T$ of its Lie algebra: $$ g = \exp(\epsilon T) \,. $$ The condition $$ g^\dagger J g = J \quad\text{for}\quad J= \begin{pmatrix} 1 & 0 \\ 0 & - 1 \end{pmatrix} $$ can be expanded for $\epsilon\ll 1$ and is equivalent to $$ T^\dagger J + J T = 0 $$ which then implies $$ T = \begin{pmatrix} i\phi & A \\ A^* & -i\phi \end{pmatrix} \qquad \phi\in\mathbb{R},\quad A\in\mathbb{C}\,. $$ As you can see, $\phi$ and $A$ contain three real degrees of freedom, as required. Also because $\det g = 1$ we get $\operatorname{Tr} T = 0$. You can then exponentiate $T$ to get $$ g = \exp(\epsilon T) = \begin{pmatrix} \cosh(\epsilon\rho) + i\phi\frac{\sinh(\epsilon\rho)}{\rho} & A\frac{\sinh(\epsilon\rho)}{\rho} \\ A^*\frac{\sinh(\epsilon\rho)}{\rho} & \cosh(\epsilon\rho) - i\phi\frac{\sinh(\epsilon\rho)}{\rho} \\ \end{pmatrix} \quad\text{with}\quad \rho^2 = |A|^2-\phi^2 \,. $$