# A Fractional Linear Transformation Class Property

Let $\mathcal{S}$ be the class of Fractional Linear Transformations (or FLT's) consisting of $f: [-1,0] \rightarrow [-1,0]$ such that $f(x) = \frac{ax+b}{cx+d}$ where $a,b,c,d \in R$, and $f'(x)>0$ for all $x \in [-1,0]$.

Notice that given a fixed number $c>0$ (and you may suppose that $c \neq 1$ if necessary), there is a unique FLT $F \in \mathcal{S}$ such that the following three conditions hold:

$F(-1)=-1$, $F(0)=0$, and $\dfrac{F'(0)}{F'(-1)}=c^2$

I would like to ask for examples of other classes of maps $\mathcal{C}$ (that is other than the class of FLT's $\mathcal{S}$) such that:

1. $f'(x)>0$ for $x \in [-1,0]$ for all $f \in \mathcal{C}$.

2. The same property as above of there existing a unique element in the class $\mathcal{C}$ satisfying the three conditions above holds.

3. Additionally, the class of maps $\mathcal{C}$ satisfies the property of being closed under (functional) composition of its members.

Thank you in advance to all those who respond,

E(up)lio M.

• In light of Robert Israel's answer, did you mean to restrict $\mathcal{S}$ to include only orientation-preserving FLTs? Apr 18, 2012 at 1:41
• @Vaughn: All fractional linear transformation (and all meromorphic functions) preserve orientation. Apr 18, 2012 at 2:04
• @Andreas: I see I've fallen into my usual trap of only thinking in terms of real numbers and neglecting the complex world. I interpreted the question to be dealing with fractional linear transformations $x\mapsto \frac{ax+b}{cx+d}$ on $\mathbb{R}\cup\{\infty\}$, where $a,b,c,d$ are all real, and so one can distinguish between FLTs with $f'>0$ (which I called orientation-preserving) and those with $f'<0$. Of course since rotation by $\pi$ in $\mathbb{C}$ is orientation-preserving the distinction vanishes if we think in $\mathbb{C}$, exactly as you say. Apr 18, 2012 at 3:03
• I suppose we need the OP to clarify whether $\mathcal{S}$ refers to transformations on $\mathbb{R}\cup\{\infty\}$ or on $\hat{\mathbb{C}}$. That is, is this a question in real analysis or complex analysis? Apr 18, 2012 at 3:06
• We also need a few other points to be resolved. What kinds of maps do you want, e.g. in the case of $\hat C$ do you want holomorphic bijections, or just holomorphic self-maps? You have restricted your class of maps $\mathcal{C}$ to be closed under composition; do you want them to be invertible and closed under inverse as well? The answers will have very different characters depending on how these points are resolved. Apr 18, 2012 at 12:26

You're asking for a one-parameter family $\mathcal{C}$ of maps on a closed interval $I$ such that

1. the endpoints of $I$ are fixed by every map in $\mathcal{C}$;
2. the maps in $\mathcal{C}$ are strictly increasing;
3. $\mathcal{C}$ is closed under composition;
4. every positive real number appears exactly once as the ratio of the derivatives of $\mathcal{C}$ at the right and left endpoints of $I$.

The fourth condition means that $\mathcal{C}$ should be parametrised by $\mathbb{R}^+$, so let's write $\mathcal{C} = \{ \psi_c \mid c>0 \}$, where $c$ represents the square root of the ratio between the derivates at the endpoints, as in your question. The chain rule together with the third condition means that we should require $$\psi_{c_1 c_2} = \psi_{c_1} \circ \psi_{c_2}$$ for all $c_1,c_2>0$. We can reparametrise $\mathcal{C}$ by putting $t=\log c$ (or $t=-\log c$) and writing $\phi_t = \psi_c$, this becomes $$\phi_{t_1+t_2} = \phi_{t_1} \circ \phi_{t_2}.$$ In other words, the family $\mathcal{C}$ defines an action of $\mathbb{R}$ on $I$, or in the language of dynamical systems/ODEs, a flow on $I$. This flow comes from integrating a vector field $V(x)$, or if you prefer, solving the ODE $$(*) \qquad \qquad \frac d{dt} \phi_t(x) = V(\phi_t(x)).\qquad\qquad\qquad$$ Conditions 1 and 2 mean that $V$ should vanish at the endpoints of $I$ and should be positive everywhere else. But modulo some regularity concerns, that's all you need. If you let $V\colon I\to [0,\infty)$ be any Lipschitz continuous function that vanishes at the endpoints of $I$ and is positive on its interior, then solving $(*)$ will give you a family of smooth maps (the time-$t$ maps of the flow) that satisfy your conditions.

Incidentally, you can use $(*)$ to see how both the family $\mathcal{S}$ of FLTs and the family in Robert Israel's (second) answer fit into this scheme. If you write $$\phi_t(x) = \frac{e^{-t} x}{1+(1-e^{-t})x}$$ for the FLT satisfying your conditions (with $c=e^{-t}$ and $t\in\mathbb{R}$ arbitrary), then an easy calculation shows that $\phi_{s+t} = \phi_s \circ \phi_t$, so this defines a flow, and we can compute $$V(x) = \frac{d\phi_t(x)}{dt}|_{t=0} = -x(1+x).$$ In other words, these FLTs are just the time-$t$ maps of the logistic flow on the interval $[-1,0]$. You can play a similar game with the arctan maps, but I haven't worked it out to see if the formula comes out cleanly.

Actually, the family of arctan maps in Robert Israel's answer illustrates a (superficially) different way of approaching the question, which is to fix a homeomorphism $h$ between the interior of $I$ and the real line (or the half-line) and then define the family of maps $\mathcal{C}$ to be those that are conjugated to translations (or homotheties) by $h$. So in his example, you conjugate to the negative half-line by $h(x) = \tan(\pi x/2)$ and then let $g_c = h^{-1} \circ (y\mapsto cy) \circ h$.

Of course, logs turn the half-line into the line and multiplication into addition, so this turns into translation on the line under another change of coordinates. And finding a change of coordinates that turns a map into translation on the line is just another way of saying that you find that map as a solution of an ODE, so it boils down to what I described above. Your choice... you can choose the vector field $V$, or you can choose the homeomorphism $h$. Either will give you quite a large number of examples.

• @Vaughn: Thanks very much. I'm going to read it carefully and get back to you. Apr 21, 2012 at 17:05
• @Vaugn: Thanks again for your answer. I appreciate especially you answering the question in a general context. Apr 21, 2012 at 17:26

$F \in {\cal S}$ is not unique. Both $f(z) = \frac{cz}{1+(1-c) z}$ and $f(z) = \frac{-cz}{1+(1+c)z}$ satisfy the conditions.

• I have edited the question. Now there is a unique element satisfying the updated conditions. Apr 19, 2012 at 15:57

I'm going to answer the question under the assumption that the domain is the extended complex plane and the maps are holomorphic bijections. This simplifies the question considerably, because these maps are exactly the fractional linear transformations of the extended complex plane. The class of maps in the OP's question is simply the subgroup of the full group of fractional linear transformations that fixes both $-1$ and $0$. This group is isomorphic to the multiplicative group of nonzero complex numbers; the same isomorphism is given either by $f \mapsto f'(0)$ or $f \mapsto 1 / f'(-1)$. Robert Israel's objection can be addressed by observing that there is a unique fractional linear transformation fixing $-1$, $0$, and having a given value of $f'(0)$. In this case, the question boils down to: what are the sub-semigroups of the nonzero complex numbers? The discrete subgroups can be easily listed, they are all finite cyclic subgroups of the unit complex numbers and infinite cyclic subgroups generated by nonunit complex numbers. There are many indiscrete subgroups as well, of arbitrary finite rank, of infinite rank, with various torsion properties. If you do not insist on closure under inverse then you are asking for all subsemigroups of the nonzero complex numbers...

But if you require only that that the maps be holomorphic but not bijections, I'm not sure what to say.

Try the functions $$g_c(x) = \frac{2}{\pi} \arctan(c \tan(\pi x/2))$$ for $c > 0$, where $g_c(-1)$ is defined as $-1$.

• That is an interesting example for sure. Thanks for that! (I'm going to take the risk of thinking out loud: it seems like a fairly generalizable construction--I wonder precisely what property of `tan' makes it all work out) Apr 20, 2012 at 4:24