In the last post, while getting hyperbolic, we have met the Lie group SU(1,1) – the group of complex matrices of determinant one, and such that
and star is Hermitian conjugated matrix to
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 from SU(1,1). It is a complex matrix, so let us write it as
On the other hand we have
and so from the determining Eq. (1) we must have
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.
which we substitute into Eq. (4)
or, multiplying both sides with
Then, using Eq. (ii), we get , thus
Therefore, from Eq. (6), we get or
So, we obtain that every matrix of SU(1,1) is of the form
Conversely, every matrix of this form is in SU(1,1).
Now, we will determine the shape of our beauty.
where is the uniquely defined phase,
Then, using Eq. (10)
Thus is in the interior of the unit disk in the complex plane.
Conversely, given and with we can determine uniquely and From Eq. (13) we set
Finally, using Eq. (12), we set
Summarizing, the shape of SU(1,1) is that of the Cartesian product of the circle parametrized by and of the interior of the unit disk, parametrized by complex numbers with That is the interior of the solid torus, that is the donut without its skin.
Pretty and tasty. Not necessarily healthy, though….