Skip to main content

Section 2.2 Tangles and Continued Fractions

Definition 2.2.1. Mobius Transformation.

A Mobius transformation is a function \(f \colon \mathbb{R} \to \mathbb{R}\) whose formula is given by

\begin{equation} f(x) = \frac{ax + b}{cx+d}\label{eq_mobius}\tag{2.2.1} \end{equation}

for some choice of the constants \(a,b,c,d \in \mathbb{R}\text{.}\)

Note that the real numbers in the above example are not the only domain of definition we could select. Mobius transformations also figure heavily in the transformational geometry of the complex plane (\(f \colon \mathbb{C} \to \mathbb{C}\) and \(a,b,c,d \in \mathbb{C}\)). And, crucially for our purposes, also can be considered to be functions on the rational number system.

Remark 2.2.2.

Convince yourself that if \(a,b,c,d\in \mathbb{Q}\) are all rational numbers, then so is \(\displaystyle\frac{ax+b}{cx+d}\text{.}\)

The Mobius transformations are particularly remarkable because they can also form a group under the operation of function composition.

The goal of our first activity is to identify our rational tangle group from Section 2.1 as a subgroup of \(M\text{.}\) That is, we'll find a collection of Mobius transformations that behave exactly like the elements of the rational tangle group we discovered:

Definition 2.2.4. Rational Tangle Group.

The rational tangle group \(\mathcal{C}\) is defined by the following presentation:

\begin{equation} \mathcal{C} = \bigl\langle T,R \; \mid \; R^2 = (TR)^3 = e \bigr\rangle.\label{eq_tanglegroup}\tag{2.2.2} \end{equation}

Recall that the (infinite-order) generator \(T\) was the “twist” movement in the rational tangle dance, while the (surprisingly order-two) generator \(R\) was the “rotation” movement.

Worksheet 2.2.1 Functional Fitness

The goal of this activity is to discover Mobius transformations, we'll call them \(t(x)\) and \(r(x)\text{,}\) which interact in the Mobius group exactly like the generators \(T\) and \(R\) respectively in the rational tangle group (2.2.2).

1.

Show that the function \(e(x) = x\) is in fact a Mobius transformation.

Hint

What values of \(a,b,c,d\) can we choose to obtain \(e(x)=x\text{?}\)

2.

Show that the function \(e(x) = x\) is the identity element of the Mobius group.

Hint

Use the definition of identity element found in Definition 1.1.3, remembering that the operation here is composition of functions!

3.

For the two Mobius transformations

\begin{align*} f(x) &= \frac{ax+b}{cx+d}\\ g(x) &= \frac{dx-b}{-cx+a} \end{align*}

compute and simplify the composition \(f\bigl(g(x)\bigr)\text{.}\) Use your answer to predict: what is the inverse of \(f\) in the Mobius group?

To discover the rational tangle group generators \(t(x)\) and \(r(x)\) as Mobius transformations, we'll use the two finite-order relations

\begin{align} R^2 &= e && r\bigl( r(x)\bigr) = x\label{eq_gentwo}\tag{2.2.3}\\ (TR)^3 &= e && t\bigl(r\bigl(t\bigl(r\bigl(t(r(x)\bigr)\bigr)\bigr)\bigr)\bigr) = x.\label{eq_genthree}\tag{2.2.4} \end{align}

The Desmos calculator below permits you to adjust through a range of values of \(a,b,c,d\) to view the Mobius transformation they define. The graph of the identity function (the diagonal line \(y=x\)) is also shown.

4.

Click to enable the display of the graph \(y=f\bigl(f(x)\bigr)\text{.}\)

Then, adjust the \(a,b,c,d\) sliders until you discover at least three different choices for which \(f\bigl(f(x)\bigr) = x\text{.}\) Write out their formulas.

  1. \(f(x) = \)

  2. \(f(x) = \)

  3. \(f(x) = \)

Try to find one that's “simplest” as possible (but not equal to \(e(x)=x\)). These functions are called functional square roots of the identity function.

5.

Refer to (2.2.3). Which rational tangle group element do you believe these functions might represent?

6.

Click to hide the display of the graph \(y=f\bigl(f(x)\bigr)\text{,}\) and enable instead the display of the graph \(y=f\bigl(f\bigl(f(x)\bigr)\bigr).\)

Then, adjust the \(a,b,c,d\) sliders until you discover at least three different choices for which \(f\bigl(f\bigl(f(x)\bigr)\bigr)) = x\text{.}\) Write out their formulas.

  1. \(f(x) = \)

  2. \(f(x) = \)

  3. \(f(x) = \)

Try to find one that's “simplest” as possible (but not equal to \(e(x)=x\)). These functions are called functional cube roots of the identity function.

7.

Refer to (2.2.4). Which rational tangle group element do you believe these functions might represent?

8.

Looking at your answers to #5 and #7, make and test a conjecture: What Mobius transformation \(t(x)\) could represent the rational tangle generator \(T\text{?}\) How do you know your choice is compatible with the results of #5 and #7?

One of the most important features of the functions \(t(x)\) and \(r(x)\) that you discovered in the previous activity is that the group of functions they generate is “fixed-point free”.

Definition 2.2.5. Fixed-Point Free.

Let \(G\) be a group of functions from \(\mathbb{R} \to \mathbb{R}\) with the operation of function composition.

We call \(G\) fixed-point free if the only function in \(G\) whose graph intersects the diagonal \(y=x\) is the identity function itself, \(e(x)=x\text{.}\)

In other words, for all \(f\in G\text{,}\) if there exists \(x_0\in \mathbb{R}\) such that \(f(x_0)=x_0\text{,}\) then we have \(f(x)=x\) for all \(x\text{.}\)

This vital property will permit us to trade the study of functions (which are hard) for the study of numbers (which are easier), according to the following theorem.

Definition 2.2.7.

Let \(Q\) be a word (in the alphabet \(T,R\)) in the rational tangle group. The fraction of \(Q\) is the rational number which is the value of the associated function word (in the alphabet of \(t(x),r(x)\)), evaluated at \(x=0\text{.}\)