Category: Geometry

Geometry, especially curvature and geodesics

Cayley transform for Easter

Posted on by arkajad

It is Easter Sunday when I writing this note. In Getting real we met the Cayley transform and its inverse.

Namely we have defined the unitary matrix \mathcal{C}

(1)   \begin{equation*}\mathcal{C}=\frac{1}{1+i}\begin{bmatrix}i&1\\-i&1\end{bmatrix}.\end{equation*}

Its inverse \mathcal{C}^{-1}=\mathcal{C}^* being given by

(2)   \begin{equation*}\mathcal{C}^{-1}=\frac{1}{1-i}\begin{bmatrix}-i&i\\1&1\end{bmatrix}.\end{equation*}

We will use the matrix \mathcal{C} in two ways: either for implementing a similarity transformation A\mapsto A'=\mathcal{C}^{-1}A\mathcal{C} resp. A'\mapsto \mathcal{C}A'\mathcal{C}^{-1}, or for implementing fractional linear transformation of the type

(3)   \begin{equation*}z\mapsto A\cdot{z}=\begin{bmatrix}\lambda&\mu\\ \nu&\rho\end{bmatrix}\cdot z= \frac{\rho z+\nu}{\mu z+\lambda}.\end{equation*}

In each case a factor, such as \frac{1}{1+i}, in front of the matrix is not important. It will cancel out. Two proportional matrices implement the same similarity transformation and the same fractional linear transformation. We have chosen the factor so as to have \mathcal{C} unitary of determinant 1, but that fact will play no role. What is important is the internal structure of \mathcal{C}.

In Getting real we have found that the similarity transformation

    \[ A\mapsto A'=\mathcal{C}^{-1}A\mathcal{C}\]

transforms complex SU(1,1) matrices into real SL(2,R) matrices. Let us now check the fractional linear transformation implemented by \mathcal{C}^{-1}. For a general case it is convenient to denote the fractional linear transformation (3) as A\cdot z. From Eq. (3) it can be easily verified that, whenever the results are finite, we have

(4)   \begin{equation*} B\cdot(A\cdot z) =(BA)\cdot z.\end{equation*}

One could extend the domain and the range of the transformation by replacing the complex plane by its one-point compactification, the Riemann sphere, but we will not need such an extension.

For us the crucial observation is that the transformation z\mapsto \mathcal{C}^{-1}\cdot z maps the unit disk D onto the upper half-plane \mathbb{H}, that is onto the set of complex numbers with positive imaginary part. To see this let us examine the properties of z' defined by

(5)   \begin{equation*}z'=\mathcal{C}^{-1}\cdot z=\frac{z+1}{iz-i}=\frac{(z+1)(-i\bar{z}+i)}{|z-1|^2}=\frac{i(1-|z|^2)+i(z-\bar{z})}{|z-1|^2}.\end{equation*}

The imaginary part of z' is evidently positive when |z|^2<1. It becomes zero when |z|^2=1. Conversely, the transformation z'\mapsto z=\mathcal{C}\cdot z'=\frac{z'-i}{z'+i} maps the upper half-plane \mathbb{H} in D. To see that this is indeed the case let us calculate |z|^2:

(6)   \begin{equation*}|z|^2=\frac{|z'-i|^2}{|z'+i|^2}=\frac{(z'-i)(\bar{z'}+i)}{(z'+i)(\bar{z'}-i)}=\frac{|z'|^2+1+i(z'-\bar{z'})}{|z'|^2+1-i(z'-\bar{z'})}.\end{equation*}

Now z'-\bar{z}'=2i\mathrm{Im}(z'), therefore if \mathrm{Im}(z')>0, then

(7)   \begin{equation*}|z|^2=\frac{|z'|^2+1-2\mathrm{Im}(z')}{|z'|^2+1+2\mathrm{Im}(z')}<1.\end{equation*}

Moreover, if \mathrm{Im}(z')=0, that is if z' is on the real axis, then its image z is on the unit circle.

We finish this Eater post with an ornament. Consider complex numbers z' of the form z'=(m+in)/(p+iq) where m,n,p,q are real integers with \mathrm{Im}(z')>0 and pq\neq0. Apply the Cayley transform to each such number and plot the number \matcal{C}\cdot z' as a point in the disk. Of course we have to restrict the size of m,n,p,q, say to \leq 9. The result is the following Easter Ornament:

Gaussian fractions on the disk.

Getting real

Posted on by arkajad

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.

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

(1)   \begin{equation*}\mathcal{C}=\frac{1}{1+i}\begin{bmatrix}i&1\\-i&1\end{bmatrix}.\end{equation*}

The Hermitian conjugated is

(2)   \begin{equation*}\mathcal{C}^*=\frac{1}{1-i}\begin{bmatrix}-i&i\\1&1\end{bmatrix}.\end{equation*}

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

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

where

(4)   \begin{equation*}|\lambda|^2-|\mu|^2=1.\end{equation*}

Let us calculate A' defined as

(5)   \begin{equation*}A'=\mathcal{C}^{-1}A\mathcal{C}.\end{equation*}

It is clear that A' has determinant one. Let us calculate A' explicitly:

(6)   \begin{equation*}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}\end{equation*}

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.

Circumference to radius ratio

Posted on by arkajad

In Texas kids learn about the number “pi” in 6th grade. I do not know how it is in Kansas. But we are not in Kansas anyway, so we can learn it now.

We are not in Texas, and we are not in Kansas. We are on the unit disk with hyperbolic geometry defined by the transitive action of SU(1,1) – as it was discussed in the recent series of posts. We know the line element:

(1)   \begin{equation*}ds^2=\frac{4(dx^2+dy^2)}{(1-x^2-y^2)^2}.\end{equation*}

We can now ask this pressing question: what is the ratio of the circumference of a circle to its radius?

Any point of the disk is as good as any other, but in the standard coordinates (x,y) that we are using the origin of the disk is most convenient. Let us introduce polar coordinates

(2)   \begin{eqnarray*} x&=&\rho\cos\phi,\\ y&=&\rho\sin\phi. \end{eqnarray*}

We need to express our line element in terms of (\rho,\phi). We calculate

    \[ dx =d\rho\cos \phi-\rho\sin\phi d\phi,\]

    \[ dy =d\rho \sin\phi+\rho\cos\phi d\phi,\]

    \[dx^2= d\rho^2\cos^2\phi-2\rho\sin\phi\cos\phi d\rho d\phi+\rho^2\sin^2\phi d\phi^2,\]

    \[dy^2= d\rho^2\sin^2\phi+2\rho\sin\phi\cos\phi d\rho d\phi+\rho^2\cos^2\phi d\phi^2,\]

    \[dx^2+dy^2=d\rho^2+\rho^2 d\phi^2,\]

    \[ds^2=4\frac{d\rho^2+\rho^2d\phi^2}{(1-\rho^2)^2}.\]

Consider now a circle with \rho=\rho_0 and \phi running from 0 to 2\pi. What is the length C of its circumference? We have to integrate ds. Going around circumference \rho is constant, so d\rho=0. Therefore

    \[C=\int_0^{2\pi}ds=\int_0^{2\pi}\frac{2\rho_0 d\phi}{1-\rho_0^2}=\frac{4\pi \rho_0}{1-\rho_0^2}.\]

What is the length r of its radius? Along the radius \phi is constant, so d\phi=0, so

    \[r=\int_0^{\rho_0} ds=\int_0^{\rho_0}\frac{2d\rho}{1-\rho^2}=2\mathrm{arctanh}\rho_0.\]

From the last equation we get

    \[\rho_0=\tanh\frac{r}{2}.\]

Substituting into the formula for C:

    \[C=4\pi\frac{\sinh(r/2)/\cosh(r/2)}{1-\sinh^2(r/2)/\cosh^2(r/2)}=2\pi\frac{2\sinh(r/2)\cosh(r/2)}{\cosh^2(r/2)-\sinh^2(r/2)}.\]

Thus

(3)   \begin{equation*}C=2\pi \sinh r,\end{equation*}

and

(4)   \begin{equation*}\frac{\mathrm{circumference}}{\mathrm{radius}}=\frac{C}{r}=2\pi\frac{\sinh r}{r}.\end{equation*}

We can find this formula in Wikipedia article Hyperbolic geometry, in the section Circles and disks (putting there the curvature R=-1).

Looking at the graph of the function \sinh r/r

we see that the ratio circumference to radius on the disk is always greater than 2\pi. In fact the ratio is growing faster and faster with increasing r.