Description of SU(1, 1)

Question

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.

Answer 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 this:

$$ U = \left( \begin{array}{cc} \alpha & \beta \\ \gamma & \delta \end{array} \right) $$ 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 Bargmann'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 $Z$.

Answer 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 $$ g = \exp(\epsilon T) = \begin{pmatrix} \cosh(\epsilon\rho) + i\phi\frac{\sinh(\epsilon\rho)}{\rho} & A\frac{\sinh(\epsilon\rho)}{\rho} \\ A^*\frac{\sinh(\epsilon\rho)}{\rho} & \cosh(\epsilon\rho) - i\phi\frac{\sinh(\epsilon\rho)}{\rho} \\ \end{pmatrix} \quad\text{with}\quad \rho^2 = |A|^2-\phi^2 \,. $$

Answer 3

Given where you're starting at, particularly with $|α|^2 - |β|^2 = 1 $, your answer doesn't merely seem to work - it actually does work. Just define $θ = \sinh^{-1} |β| ≥ 0$. Remember: the "sinh" function is one-to-one over the real line (as well as over the non-negative reals), so it has an inverse. Then $|β| = \sinh θ$ and $$|α|^2 = 1 + \sinh^2 θ = \cosh^2 θ,$$ which implies the positive root $|α| = +\cosh θ$, since $|α| ≥ 0$. Moreover, it also follows from this that $|α| ≥ 1$.

Then $|α/|α|| = 1$. Thus we can define a unique $φ = \arg(α/|α|) ∈ [0,2π)$, from which it follows that $$α = \cosh θ e^{+iφ}, \hspace 1em α^* = \cosh θ e^{-iφ}.$$ A similar treatment applies to $β$ to yield the corresponding conclusions $$β = \sinh θ e^{+iω}, \hspace 1em β^* = \sinh θ e^{-iω},$$ where $ω = \arg(β/|β|) ∈ [0,2π)$ is uniquely determined ... provided that $|β| > 0$; i.e. $β ≠ 0$, or equivalently, $θ > 0$. In the case where $θ = 0$, similar formulae for $β$ and $β^*$ still follow, except that $ω ∈ [0,2π)$ is no longer unique and is arbitrary, since $\sinh θ = 0$.

One has the following equivalences $(θ,φ+2π,ω) ⇔ (θ,φ,ω) ⇔ (θ,φ,ω+2π)$, that allows one to extend the domain to all $(φ,ω) ∈ ℝ^2$, as a many-to-one map. Similarly, there is an equivalence transform $(+θ,φ,ω) ⇔ (-θ,φ,ω+π)$, that allows one to extend the domain to all $θ ∈ ℝ$, and in this way you can see how $ℝ^3$ is covering the manifold (technically: the "torsor") of $SU(2,1)$. It repeats every $2π$ for both of $φ$ and $ω$. It reflects on $θ$ for $(θ,ω)$ in the way just described, and it is degenerate for $ω$ at $θ = 0$.

A more fruitful decomposition can be found by, instead, writing $w + iy = \cosh θ e^{iφ}$ and $x + iz = \sinh θ e^{iω}$, for $(w,x,y,z) ∈ ℝ^4$, with $w^2 - x^2 + y^2 - z^2 = 1$. In contrast to the map just described, the correspondence $(α,β) ⇔ (w,x,y,z)$ is one-to-one over the $w-x-y-z$ hyperboloid. As already alluded to, in another reply, the correspondence to $SL(2,ℝ)$ is arrived at by: $$ {\left(\begin{array}{ll} w + i y & x + i z \\ x - i z & w - i y \end{array} \right) ∈ SU(2,1)} ⇔ {\left( \begin{array}{ll} w + z & x + y \\ x - y & w - z \end{array} \right) ∈ SL(2,ℝ)}. $$