### Getting real

It is better to get real instead of merely being right. But if so, how can we get real? Susan Campbell in her book “Getting Real” advocates Radical Honesty:

Honest communication is not only the quickest, most direct path to wholeness, it is also the least expensive. Without spending years in a therapist’s office, you can learn a set of communication practices that will lead you to the truth of your present experience and out of the morass of judgments, generalizations, shoulds, withholdings, assessments, and explanations about why you are the way you are. These practices are a way of using language to help you stay with your present felt experience — what you see, hear, smell, feel, remember, sense, and intuit. You can learn these practices in a relatively short time period, since so many other explorers have already charted the way. Buddhist meditation practice, Gestalt therapy, Jungian analysis, sensory awareness, Reichian and bioenergetic body work — these are the main underpinnings of the work that I call Getting Real.

[latexpage]
Today we will learn yet another way of getting real. Once we know how to be real, we will let ourselves to get complex. And then we will surf on the waves of complexity ….

Well, to some extent I am joking. But only to some extent. Indeed I want to get real. The group SU(1,1) that is at the basis of non-Euclidean hyperbolic geometry of the Poincare disk in the complex plane – that group is a group of complex $2\times 2$ matrices. That complicates calculations. It would be simpler to play with real matrices. Can it be done? Can complex be transformed into simple real? In this case the answer is yes. And the answer is provided by the magic of Cayley transform.

Let $\mathcal{C}$ denote the matrix
$$\mathcal{C}=\frac{1}{1+i}\begin{bmatrix}i&1\\-i&1\end{bmatrix}.$$
The Hermitian conjugated is
$$\mathcal{C}^*=\frac{1}{1-i}\begin{bmatrix}-i&i\\1&1\end{bmatrix}.$$
It is easy to verify that $\mathcal{C}\mathcal{C}^*=I,$ therefore $\mathcal{C}$ is unitary.
One can check that, moreover, $\det \mathcal{C}=1,$ therefore $\mathcal{C}$ is an element of the group SU(2) of complex unitary matrices of determinant one.

Now comes the magic. Let $A$ be a matrix from SU(1,1). We know it has the form
$$A=\begin{bmatrix}\lambda&\mu\\ \bar{\mu}&\bar{\lambda}\end{bmatrix}$$
where
$$|\lambda|^2-|\mu|^2=1.\label{eq:lm1}$$
Let us calculate $A’$ defined as
$$A’=\mathcal{C}^{-1}A\mathcal{C}.$$
It is clear that $A’$ has determinant one. Let us calculate $A’$ explicitly:

A’=\frac{1}{2}\begin{bmatrix}\lambda+\bar{\lambda}-\mu-\bar{\mu}&i(-\lambda+\bar{\lambda}-\mu+\bar{\mu})\\
i(\lambda-\bar{\lambda}-\mu+\bar{\mu})&\lambda+\bar{\lambda}+\mu+\bar{\mu}\end{bmatrix}

The magic is that the matrix $A’$ is real. One can easily check that conversely, for every $2\times 2$ real matrix $A’$ of determinant one, the matrix $A$ defined as $A=\mathcal{C}A’\mathcal{C}^{-1}$ is in the group SU(1,1). The group of real $2\times 2$ real matrices of determinant one is denoted SL(2,R) (“Special Linear group”). Thus we have established isomorphism between the complex group SU(1,1) and the real group SL(2,R). We became real. In the next post we will learn what becomes then of our hyperbolic geometry.
Being complex is fun. But being real is also fun – just another version of it.