From SU(1,1) to the Lorentz group

From the two-dimensional disk we are moving to three-dimensional space-time. We will meet Einstein-Poincare-Minkowski special relativity, though in a baby version, with x and y, but without z in space. It is not too bad, because the famous Lorentz transformations, with length contraction and time dilation happen already in two-dimensional space-time, with x and t alone. We will discover Lorentz transformations today. First in disguise, but then we will unmask them.

First we recall, from The disk and the hyperbolic model, the relation between the coordinates (x,y) on the Poincare disk x^2+y^2<1, and (X,Y,T) on the unit hyperboloid T^2-X^2-Y^2=1.

Space-time hyperboloid and the Poincare disk models

(1)   \begin{eqnarray*} X&=&\frac{2x}{1-x^2-y^2},\\ Y&=&\frac{2y}{1-x^2-y^2},\\ T&=&\frac{1+x^2+y^2}{1-x^2-y^2}. \end{eqnarray*}

(2)   \begin{eqnarray*} x&=&\frac{X}{1+T},\\ y&=&\frac{Y}{1+T}. \end{eqnarray*}

We have the group SU(1,1) acting on the disk with fractional linear transformations. With z=x+iy and A in SU(1,1)

(3)   \begin{equation*}A=\begin{bmatrix}\lambda&\mu\\ \nu&\rho\end{bmatrix},\end{equation*}

the fractional linear action is

(4)   \begin{equation*}A:z\mapsto z_1=\frac{\rho z+\nu}{\mu z+\lambda}.\end{equation*}

By the way, we know from previous notes that A is in SU(1,1) if and only if

(5)   \begin{equation*}\nu=\bar{\mu},\,\rho=\bar{\lambda},\quad|\lambda|^2-|\mu|^2=1.\end{equation*}

Having the new point on the disk, with coordinates (x_1,y_1) we can use Eq. (1) to calculate the new space-time point coordinates (X_1,Y_1,T_1). This is what we will do now. We will see that even if z_1 depends on z in a nonlinear way, the space-time coordinates transform linearly. We will calculate the transformation matrix L(A), and express it in terms of \lambda and \mu. We will also check that this is a matrix in the group SO(1,2).

The program above involves algebraic calculations. Doing them by hand is not a good idea. Let me recall a quote from Gottfried Leibniz, who, according to Wikipedia

He became one of the most prolific inventors in the field of mechanical calculators. While working on adding automatic multiplication and division to Pascal’s calculator, he was the first to describe a pinwheel calculator in 1685[13] and invented the Leibniz wheel, used in the arithmometer, the first mass-produced mechanical calculator. He also refined the binary number system, which is the foundation of virtually all digital computers.

“It is unworthy of excellent men to lose hours like slaves in the labor of calculation which could be relegated to anyone else if machines were used.”
— Gottfried Leibniz

I used Mathematica as my machine. The same calculations can be certainly done with Maple, or with free software like Reduce or Maxima. For those interested, the code that I used, and the results can be reviewed as a separate HTML document: From SU(1,1) to Lorentz.

Here I will provide only the results. It is important to notice that while the matrix A has complex entries, the matrix L(A) is real. The entries of L(A) depend on real and imaginary parts of \lambda and \mu

(6)   \begin{equation*}\lambda=\lambda_r+i\lambda_i,\, \mu=\mu_r+i\mu_i.\end{equation*}

Here is the calculated result for L(A):

(7)   \begin{equation*}L(A)=\begin{bmatrix}   -\lambda_i^2+\lambda_r^2-\mu_i^2+\mu_r^2 & 2 \lambda_i \lambda_r-2 \mu_i \mu_r & 2 \lambda_r \mu_r-2 \lambda_i \mu_i \\  -2 \lambda_i \lambda_r-2 \mu_i \mu_r & -\lambda_i^2+\lambda_r^2+\mu_i^2-\mu_r^2 & -2 \lambda_r \mu_i-2 \lambda_i \mu_r \\  2 \lambda_i \mu_i+2 \lambda_r \mu_r & 2 \lambda_i \mu_r-2 \lambda_r \mu_i & \lambda_i^2+\lambda_r^2+\mu_i^2+\mu_r^2  \end{bmatrix}\end{equation*}

In From SU(1,1) to Lorentz it is first verified that the matrix L(A) is of determinant 1. Then it is verified that it preserves the Minkowski space-time metric. With G defined as

(8)   \begin{equation*}G=\begin{bmatrix}-1&0&0\\0&-1&0\\0&0&1\end{bmatrix}\end{equation*}

we have

(9)   \begin{equation*}L(A)GL(A)^T=G.\end{equation*}

Since L(A)_{3,3}= \lambda_i^2+\lambda_r^2+\mu_i^2+\mu_r^2 =1+2|\mu|^2\geq 1>0, the transformation L(A) preserves the time direction. Thus L(A) is an element of the proper Lorentz group \mathrm{SO}^{+}(1,2).

Remark: Of course we could have chosen G with (+1,+1,-1) on the diagonal. We would have the group SO(2,1), and we would write the hyperboloid as X^2+Y^2-T^2=-1. It is a question of convention.

In SU(1,1) straight lines on the disk we considered three one-parameter subgroups of SU(1,1):

(10)   \begin{eqnarray*} X_1&=&\begin{bmatrix}0&i\\-i&0\end{bmatrix},\\ X_2&=&\begin{bmatrix}0&1\\1&0\end{bmatrix},\\ X_3&=&\begin{bmatrix}i&0\\0&-i\end{bmatrix}. \end{eqnarray*}

(11)   \begin{eqnarray*} A_1(t)&=& \exp(tX_1)=\begin{bmatrix}\cosh(t)&i\sinh(t)\\-i\sinh(t)&\cosh(t)\end{bmatrix},\\ A_2(t)&=& \exp(tX_2)=\begin{bmatrix}\cosh(t)&\sinh(t)\\ \sinh(t)&\cosh(t)\end{bmatrix},\\ A_3(t)&=& \exp(tX_3)=\begin{bmatrix}e^{it}&0\\ 0&e^{-it}\end{bmatrix}. \end{eqnarray*}

We can now use Eq. (7) in order to see which space-time transformations they implement. Again I calculated it obeying Leibniz and using a machine (see From SU(1,1) to Lorentz).

A replica of the Stepped Reckoner of Leibniz form 1923 (original is in the Hannover Landesbibliothek)

Here are the results of the machine work:

(12)   \begin{equation*}L_1(t)=L(A_1(t))=\begin{bmatrix}  1 & 0 & 0 \\  0 & \cosh (2 t) & -\sinh (2 t) \\  0 & -\sinh (2 t) & \cosh (2 t)\end{bmatrix},\end{equation*}

(13)   \begin{equation*}L_2(t)=L(A_2(t))=\begin{bmatrix}  \cosh (2 t) & 0 & \sinh (2 t) \\  0 & 1 & 0 \\  \sinh (2 t) & 0 & \cosh (2 t) \end{bmatrix},\end{equation*}

(14)   \begin{equation*}L_3(\phi)=L(A_3(\phi))=\begin{bmatrix}  \cos (2 \phi ) & \sin (2 \phi ) & 0 \\  -\sin (2 \phi ) & \cos (2 \phi ) & 0 \\  0 & 0 & 1 \\ \end{bmatrix},\end{equation*}

The third family is a simple Euclidean rotation in the (X,Y) plane. That is why I denoted the parameter with the letter \phi. In order to “decode” the first two one-parameter subgroups it is convenient to introduce new variable v and set 2t=\mathrm{arctanh}(v). The group property L(t_1)L(t_2)=L(t_1+t_2) is then lost, but the matrices become evidently those of special Lorentz transformations, L_1(v) transforming Y and T, leaving X unchanged, and L_2(v) transforming (X,T) and leaving Y unchanged (though with a different sign of v). Taking into account the identities

(15)   \begin{eqnarray*} \cosh(\mathrm{arctanh} (v))&=&\frac{1}{\sqrt{1-v^2}},\\ \sinh(\mathrm{arctanh} (v))&=&\frac{v}{\sqrt{1-v^2}} \end{eqnarray*}

we get

(16)   \begin{equation*}L_1(v)=\begin{bmatrix}  1 & 0 & 0 \\  0 & \frac{1}{\sqrt{1-v^2}} & -\frac{v}{\sqrt{1-v^2}} \\  0 & -\frac{v}{\sqrt{1-v^2}} & \frac{1}{\sqrt{1-v^2}}\end{bmatrix},\end{equation*}

(17)   \begin{equation*}L_2(v)=\begin{bmatrix} \frac{1}{\sqrt{1-v^2}} & 0 & \frac{v}{\sqrt{1-v^2}} \\  0 & 1 & 0 \\ \frac{v}{\sqrt{1-v^2}} & 0 & \frac{1}{\sqrt{1-v^2}} \end{bmatrix},\end{equation*}

In the following posts we will use the relativistic Minkowski space distance on the hyperboloid for finding the distance formula on the Poincare disk.

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.

Ray Bradbury, The Martian Chroniques

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:

This time we take


The vector field comes as


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?

Our second field expedition

Our first field expedition was not without adventures. What kind of adventures they were – this can be now only guessed looking at some cryptic comments. They refer to the text that does not exist any more. I have changed the example. Last night Goddess talked to me in my dream, and She suggested a different avenue from that that I have had in my mind before. I obey. And so the second example follows. Very similar to the previous one, just a little been different.

Last time I considered the one parameter group of transformations defined, on the complex plane, by

    \[f_t:z\mapsto \frac{\cosh(t)z-i\sin(t)}{i\sinh(t) z+\cosh(t)},\]

where t\in\mathbf{R}.

Today I propose a variation:

    \[g_t:z\mapsto \frac{\cosh(t)z+\sinh(t)}{\sinh(t) z+\cosh(t)}.\]

Again, with some little effort (perhaps easier than it was before, because now there is no more imaginary i explicitly in the formula) we can verify that

g_{t+s}(z)=g_t(g_s(z)) and g_0(z)=z.

In order to obtain the vector field from this “flow of complex numbers” I am calculating the derivative:

    \[Y(z)=\frac{d}{dt} g_t(z)|_{t=0}.\]

I used Mathematica to get


To draw this vector field on the (x,y) plane I write z=x+iy, and calculate



    \[ Y(x,y)= (\mathrm{Re} (Y(z)),\mathrm{Im}(Y(z))=(1+y^2-x^2,-2xy).\]

Here is the vector field, and stream lines from this second field expedition

Of course there will also the third expedition. After that the mystery will slowly be be revealed.