### SU(1,1) parametrization

[latexpage]
In the last post, while getting hyperbolic, we have met the Lie group SU(1,1) – the group of $2\times 2$ complex matrices $A$ of determinant one, $\det A=1,$ and such that
$$AGA^*=G,\label{eq:su11df1}$$ where
$$G=\begin{bmatrix}1&0\\0&-1\end{bmatrix},$$
and star $A^*$ is Hermitian conjugated matrix to $A.$

It was our first meeting here with SU(1,1), and there was no enough time to decide whether we want to make friends with this group or not. Today we will look at SU(1,1) closer, in particular at her shape. I think we will like her … She has, as we will see below, the shape of the inside of a solid torus, donut-like shape:

How do we get this donut? One method is to visit a backery, another method is by doing some little algebra. We will choose the second method. In fact, we are going to get not only the shape of the group itself, but also the shape of the disk that the group is acting on! It will be the cross-section of the torus:

Let us take a matrix $A$ from SU(1,1). It is a complex $2\times 2$ matrix, so let us write it as
$$A=\begin{bmatrix}\lambda&\mu\\ \nu&\rho\end{bmatrix}.$$
From $\det A=1$ we get
$$\lambda\rho-\mu\nu=1.\label{eq:det}$$
On the other hand we have
$$AGA^*=\begin{bmatrix}\lambda&\mu\\ \nu&\rho\end{bmatrix}\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}\bar{\lambda}&\bar{\nu}\\ \bar{\mu}&\bar{\rho}\end{bmatrix}=\begin{bmatrix}\lambda\bar{\lambda}-\mu\bar{\mu}&\lambda\bar{\nu}-\mu\bar{\rho}\\\nu\bar{\lambda}-\rho\bar{\mu}&\nu\bar{\nu}-\rho\bar{\rho}\end{bmatrix},$$
and so from the determining Eq. (\ref{eq:su11df1}) we must have
\begin{align}
|\lambda|^2-|\mu|^2&=1,\tag{i}\\
|\rho|^2-|\nu|^2&=1,\tag{ii}\\
\lambda\bar{\nu}-\mu\bar{\rho}&=0,\tag{iii}\\
\nu\bar{\lambda}-\rho\bar{\mu}&=0.\tag{iv}
\end{align}
Eq. (iv) is just a complex conjugation of (iii), so it does not bring us any new information. The whole information is contained in the first three equations.

From the first two equations we deduce that, since $|\lambda|^2=1+|\mu|^2\geq 1$ and $|\rho|^2=1+|\nu|^2\geq 1,$ therefore $\lambda$ and $\rho$ are non-zero. From Eq. (iii), dividing by non-zero $\bar{\rho},$ we get
$$\mu=\frac{\lambda\bar{\nu}}{\bar{\rho}},\label{eq:mu}$$
which we substitute into Eq. (\ref{eq:det})
$\lambda\rho-\frac{\lambda|\nu|^2}{\bar{\rho}}=1,$
or, multiplying both sides with $\bar{\rho},$
$\lambda|\rho|^2-\lambda|\nu|^2=\bar{\rho}.$
Then, using Eq. (ii), we get $\lambda=\bar{\rho}$, thus
$$\rho=\bar{\lambda}.$$
Therefore, from Eq. (\ref{eq:mu}), we get $\mu=\bar{\nu},$ or
$$\nu=\bar{\mu}.$$

So, we obtain that every matrix $A$ of SU(1,1) is of the form
$$A=\begin{bmatrix}\lambda&\mu\\ \bar{\mu}&\bar{\lambda}\end{bmatrix}$$
where
$$|\lambda|^2-|\mu|^2=1.\label{eq:lm}$$
Conversely, every matrix of this form is in SU(1,1).

Now, we will determine the shape of our beauty.

Since $|\lambda|\geq 1,$ we can write
$$\lambda=|\lambda|e^{i\theta},\label{eq:ml0}$$
where $\theta$ is the uniquely defined phase, $0\leq\theta<2\pi.$ Let $$z=\frac{\bar{\mu}}{\lambda}.\label{eq:z}$$ Then, using Eq. (\ref{eq:lm}) $$|z|^2=\frac{|\mu|^2}{|\lambda|^2}=\frac{|\lambda|^2-1}{|\lambda|^2}=1-\frac{1}{|\lambda|^2}<1.\label{eq:ll}$$ Thus $z$ is in the interior of the unit disk in the complex plane. Conversely, given $\theta$ and $z,$ with $|z|<1$ we can determine uniquely $\lambda$ and $\mu.$ From Eq. (\ref{eq:ll}) we set $$|\lambda|=\frac{1}{\sqrt{1-|z|^2}},\label{eq:ml1}$$ then Finally, using Eq. (\ref{eq:z}), we set $$\mu=\bar{\lambda} \bar{z}.\label{eq:ml2}$$ Summarizing, the shape of SU(1,1) is that of the Cartesian product of the circle parametrized by $\theta,$ $0\leq\theta<2\pi,$ and of the interior of the unit disk, parametrized by complex numbers $z$ with $|z|<1.$ That is the interior of the solid torus, that is the donut without its skin.

Pretty and tasty. Not necessarily healthy, though….

### Getting hyperbolic

Without knowing it, during the last three posts (Our first field expedition, Our second field expedition, The Third Expedition) we became hyperbolic. Hyperbolic and conformal. Conformal and relativistic, relativistic and non-Euclidean.

How so?

It will take us a while to get accustomed to these terms and, moreover, there is no good systematic way of progressing. So, we will progress in a somewhat haphazard way – that is the natural way.

The three streams that we had seen in three recent notes have one thing in common: they all map the unit disk of the complex plane into itself. Here are again these streams, but now, with the unit disk marked up:

I do like disks. Throwing a disk so that it would fly really far,and in the intended direction, is an art.

[latexpage]
The three one-parameter transformation groups and their vector fields that we have met are

$$f_t:z\mapsto \frac{\cosh(t)z-i\sinh(t)}{i\sinh(t) z+\cosh(t)},$$
$$X(x,y)=(2xy,y^2-x^2-1).$$
$$g_t:z\mapsto \frac{\cosh(t)z+\sinh(t)}{\sinh(t) z+\cosh(t)}.$$
$$Y(x,y)=(1+y^2-x^2,-2xy).$$
$$h_t:z\mapsto e^{-2it}z.$$
$$Z(x,y)=(2y,-2x).$$

Looking at the pictures of flows it is not completely clear that $f_t$ and $g_t$ indeed map the disk into itself. For $f_t$ it is not clear from the picture what happens to $z=-i$ at the bottom of the disk. and for $g_t$ it is not clear what happens to $z=1$ at the right. But we can easily calculate from the equations that $f_t(-i)=-i$ for all $t,$ and, similarly, $g_t(1)=1$ for all $t.$
So $-i$ is the fixed point of $f_t,$ and $1$ is a fixed point of $g_t.$ (The disk origin $z=0$ is a fixed point of $h_t.$ ) Yet the fact that $f_t$ and $g_t$ map the unit disk onto itself (and the unit circle onto itself) is not completely evident and needs a proof, that is if one wants to know rather than to believe. We will return to this problem later on, in the future posts. For now there are more urgent questions:

What it has to do with hyperbolic, conformal, relativistic and non-Euclidean? Hyperbolic – we can partly guess: there are hyperbolic functions in the formula. But what they have to do with hyperbolas? Conformal? Our transformations preserve angles, but that is not evident. Relativistic? Once we know why hyperbolic, we will start to understand about being relativistic and non-Euclidean.

To start with, let me tell you right away how I got these formulas. First, I will tell it without explaining and/or elaborating. I will explain the details and steps later on.

I do like the groups SU(2,2) and its baby version SU(1,1). Here I am using SU(1,1) – the special unitary group for signature $(1,1)$. This is the group of all $2\times 2$ complex matrices $A$ of determinant one, $\det A=1,$ that have the property that
$$AGA^*=G,\label{eq:su11df}$$, where
$$G=\begin{bmatrix}1&0\\0&-1\end{bmatrix},$$
and star $A^*$ is Hermitian conjugated matrix to $A.$

If $t\mapsto A(t)$ is a path in SU(1,1) through identity, $A(0)=I,$ then, by differentiating the defining equation
$$A(t)GA(t)^*=G,$$
at $t=0,$ and denoting $X=dA(t)/dt|_{t=0}$ we obtain
$$XG+GX^*=0.\label{eq:su11l}$$
Now, if
$X=\begin{bmatrix}a&b\\c&d\end{bmatrix},$
then, from the condition $\det A(t)=1$ it follows that $\mathrm{Tr}(X)=0,$ that is $d=-a.$ Then for $X^*$ we get
$X^*=\begin{bmatrix}a^*&c^*\\b^*&-a^*\end{bmatrix},$
and from Eq. (\ref{eq:su11l}) we get $a^*=-a,$ and $c=b^*.$ Thus $X$ is of the form
$$X=\begin{bmatrix}i\alpha&b\\b^*&-i\alpha\end{bmatrix},$$
where $\alpha$ is real and $b$ is complex. This is the Lie algebra of the group SU(1,1). I first take its generator
$$X=\begin{bmatrix}0&i\\-i&0.\end{bmatrix}$$
I exponentiate (using, for instance, Mathematica, but by hand it is also easy, once you notice that $X^2=I$:

$$F(t)=\exp(t X)=\begin{bmatrix}\cosh t&i\sinh t\\-i\sinh t&\cosh t\end{bmatrix}$$

My prescription for acting with $2\times 2$ complex matrices on complex numbers is via linear fractional transformations. Thus if $A$ is of the form
$$A=\begin{bmatrix}\lambda&\mu\\ \nu&\rho\end{bmatrix},\quad \lambda,\mu,\nu,\rho\in \mathbf{C},$$
then it acts on complex numbers as
$$A:z\mapsto \frac{\rho z+\nu}{\mu z + \lambda}.$$
That is how I arrived at $f_t.$ Doing the same for $X=\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$ I got $g_t$, and with $X=\left(\begin{smallmatrix}i&0\\0&-i\end{smallmatrix}\right)$ I obtained $h_t.$

That was a short summary. With time it will all become clear. Little by little it will also becoming clear what is, or can be, the relation of these
games with matrices to spinning tops and Dzhanibekov’s T-handles. In short: I am interested in relativistic T-handles. That is when not only space but also time are set in motion. It is kinda futuristic, but that is why it is fun to do it.

### The Third Expedition

April 2000:  THE THIRD EXPEDITION

The ship came down from space. It came from the stars and the black velocities, and the shining movements, and the silent gulfs of space.
….
“Mars!” cried Navigator Lustig.
“Good old Mars!” said Samuel Hinkston, archaeologist.
“Well,” said Captain John Black.
The rocket landed on a lawn of green grass.
….
“I’ll be damned,” whispered Lustig, rubbing his face with his numb fingers. “I’ll be damned.”
“It just can’t be,” said Samuel Hinkston.

And we are here on our third field expedition. During the first expedition we had seen the stream moving down:

Then, in our second expedition the stream was flowing to the right.

What will happen in our third expedition? Will it flow up now? No. It will be something else. You will say “I’ll be damned, It just can’t be.” And yet here it is:
[latexpage]
This time we take

$h(t,z)=e^{-2it}z.$

The vector field comes as

$Z(x,y)=(2y,-2x).$

Its plot is:

And the streamlines:

Oh, no, sorry, these are from Mars. Mathematica gives these:

Now the mystery is this: Combining 1 and 2 one can get 3, combining 2 and 3 one can get 1, combining 3 and 1 one can get 2. Perhaps. It resonates, doesn’t it?

But how exactly this happens? Where are the details that devil dwells in?