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For a homework question, I am required to "describe the Lie group SU(1, 1)". This is a bit ambiguous, but I think what that means is I need to find a parametrisation of the elements of the group. I know that the general form of a matrix in $\mathrm{SU}(1, 1)$ is given by $$\left( \begin{array}{ccc} \alpha & \beta \\ \beta^* & \alpha^* \end{array} \right), $$ where $|\alpha|^2 - |\beta|^2 = 1$. So, I'm trying to get a "parametrisation" of the entries of this matrix, in the same way that $\left( \begin{array}{ccc} a & -b \\ b & a \end{array} \right) \in \mathrm{SO}(2)$ can be parametrised by $g (\theta) = \left( \begin{array}{ccc} \mathrm{cos}(\theta) & -\mathrm{sin}(\theta) \\ \mathrm{sin}(\theta) & \mathrm{cos}(\theta) \end{array} \right) $, with $\theta \in (-\pi, \pi] $. I'm pretty sure that such a parametrisation for $\mathrm{SU}(1, 1)$ would involve 3 parameters, and the exponential function and hyperbolic trigonometric functions. I can't quite see how to get it based on the intrinsic form of the group elements though. My guess would be something like $$g(\omega, \phi, \theta) = \left( \begin{array}{ccc} e^{i\phi}\mathrm{cosh}(\theta) & e^{i\omega}\mathrm{sinh}(\theta) \\ e^{-i\omega}\mathrm{sinh}(\theta) & e^{-i\phi}\mathrm{cosh}(\theta) \end{array} \right)$$ This seems to work, but presumably I need to justify it. In other words, I suppose I would need to prove that this does generate all elements of $\mathrm{SU}(1, 1)$ and that at least three parameters are needed. So my question is: starting with the generic description of the elements of $\mathrm{SU}(1, 1)$, how can we derive a parametrisation like the one above? And how can we be sure that this "works" (in the sense that it generates all the elements of the group and uses the least possible number of parameters)?

Also, as part of the description, I think I need to show that $\mathrm{SU}(1, 1)$ is isomorphic to $\mathrm{SL}(2, \mathbb{R})$, and I'm not entirely sure how to do this. I've read somewhere that the "Cayley transform" gives an isomorphism, but I don't really know what that is. Is there an "easy" way to see that these two groups are isomorphic? Apologies if these questions sound too simple. Any help would be appreciated.

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There is nice wikipedia article on Cayley transform which might be of help. – Sasha Aug 25 '11 at 18:53

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 $Z$.

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I'm not sure this actually solves the second question - it does prove that the space is 3-dimensional, but wouldn't he still have to show that his formula doesn't produce a discrete subgroup of $SU(1,1)$ instead of the group itself (for instance, $PSU(1,1)$)? – Steven Stadnicki Aug 25 '11 at 17:14
Thanks for this, but I agree with Steven: I still need to show that my expression produces all of $\mathrm{SU}(1, 1)$ (or equivalently that your expression for elements in $\mathrm{SL}(2, \mathbb{R})$ does in fact produce all of $\mathrm{SL}(2, \mathbb{R})$). Also, as a follow up question, what would be the minimum domain necessary for the parameters? Presumably we would have to take $\theta \geq 0$, but I'm not sure about $\phi$ and $\psi$ (just $(0, 2\pi]$?). – saurs Aug 25 '11 at 17:59
@saurs Please see my updated answer with proof of isomorphism between $SL(2,R)$ and $SU(1,1)$. – Sasha Aug 26 '11 at 5:31
+1; (i) To help the interested reader, the relevant pages in the Bargmann paper is mostly p.589-591. (ii) Note that the symplectic group $Sp(2,\mathbb{R})=SL(2,\mathbb{R})$. (iii) The subgroup $H_1\subseteq SL(2,\mathbb{C}) $ that leaves the $x_1$-variable invariant is in detail $$H_1~=~\left\{ \begin{bmatrix} a & ib \\ ic & d\end{bmatrix} | a,b,c,d\in \mathbb{R}, ad+bc=1\right\}.$$ – Qmechanic Jun 20 '13 at 23:09
@LeoSpencer For a general 2 by 2 matrix $U$, $U^\dagger J U = \begin{pmatrix} u_{1,1}^\ast u_{1,1} - u_{2,1}^\ast u_{2,1} & u_{1,1}^\ast u_{1,2} - u_{2,1}^\ast u_{2,2} \cr u_{1,2}^\ast u_{1,1} - u_{2,2}^\ast u_{2,1} & u_{1,2}^\ast u_{1,2} - u_{2,2}^\ast u_{2,2} \end{pmatrix}$. Let $u_{1,1}$ be $\alpha$, $u_{1,2}$ be $\beta$, $u_{2,1}$ be $\gamma$ and $u_{2,2}$ be $\delta$. Equations stated in my post simply express equality of the just computed matrix to $J$ element-wise. – Sasha Nov 25 '13 at 13:34

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