What do the fixed points and symmetry of $f(x) = \frac {ax +b} {cx + d}$ tell us? Geometry, symmetry, and limits.

Question

Let $$f(x) = \frac {ax +b} {cx + d}$$ with $a,b,c,d \in \mathbb R, \neq 0$.

Suppose $f$ has fixed points $u,v, u \neq v$, and further suppose that $f(f(x)) = x$ wherever this is defined. Show that there exists a point $r = \frac {u +v } 2$ such that the domain and image of $f$ are both $\mathbb R - r$.

Source: Inspired by generalizing this problem. Note: Other solutions may exist. This question asks to verify or critique this solution below, which, after the lemma, uses geometrical symmetry.

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Lemma: The function $f(x) = \frac {ax +b} {cx + d}, c \neq 0$ is either a constant or a rectangular hyperbola with asymptotes parallel to the $x$ and $y$ axes.

Proof: Write $$f(x) = S \cdot \frac 1 {x - V_a} + H_a$$ with $H_a = \frac a c, V_a = \frac {-d} c$ and $S$ as needed below and note $$\begin{align*} f(x) &= \frac {ax + b} {cx + d} \\ &= \frac {a(x + \frac d c) - \frac {ad}{c} + b} {c(x + \frac d c)} \\ &= \frac a c + \frac {- \frac {ad} {c} + b}{c} \cdot \frac {1}{x + \frac {d} {c}}\\ &= S \cdot \frac 1 {x - V_a} + H_a. \end{align*}$$

If $f$ is non-constant, then $S \neq 0$, giving a rectangular hyperbola with asymptotes $y = H_a$ and $x = V_a$.

Main Proof: Since $u \neq v$, then $f$ is non-constant. By the lemma, $f$ is a hyperbola with asymptotes parallel to the $x$ and $y$ axes. Such a curve is defined everywhere but its vertical asymptote, and has image everywhere but its horizontal asymptote. The hyperbola's center is the unique point on both of these asymptotes; thus it suffices to show that its center is $(\frac {u+v} 2, \frac {u+v} 2)$.

Since $f(f(x)) = x$, then the line $y = x$ is an axis of symmetry of $f$. A hyperbola has only two axes of symmetry: its major (aka transverse) axis and its minor (aka conjugate) axis. Since $f$ has fixed points but a hyperbola never intersects its minor axis, then the line $y = x$ must be the major axis of $f$.

A hyperbola intersects its major axis in exactly two points, which for $f$ must be $(u,u)$ and $(v,v)$. Its center is their midpoint, which is consequently $(\frac {u+v} 2, \frac {u+v} 2)$. This completes the proof.

Answer 1

I read your proof and it seems correct, i like that it is really geometric. The only thing to me is that you use facts about hyperbolas which, although they might be well known, do not seem trivial to check from the analytic definition that you give. To you in the first lemma of the proof, an hyperbola is just the image of a function with the structure $\frac{A}{x-B} + C$, but the properties you use later (existence of major-minor axes of symmetries, their properties, number of intersections) are not really clear from this definition (at least to me).

This might be purely subjective, but since the original statement is written in analytic terms (fixed points, image, domain, the property $f(f(x)) = x$...) it calls to me that an analytic solution might be more suitable if possible.

I provide here another, maybe less geometric solution, but to me more elementary if you just know basic calculus. I know this is out of the scope of your question, as you said, but maybe it will be interesting to another one.

Take as requested $a,b,c,d\in\mathbb{R}-\lbrace 0\rbrace$. We assume $f(x) = \frac{ax+b}{cx+d}$ has two distinct fixed points. Since $c,d\neq 0$, this function is well defined for every $x \neq -\frac{d}{c}$, hence its domain is $\mathbb R -\lbrace -\frac{d}{c}\rbrace$.

The condition $f(f(x)) = x$ implies that for every $x\neq \frac c d$: $$x = \frac{a\frac{ax+b}{cx+d} + b}{c\frac{ax+b}{cx+d}+d}$$ which implies that, for every $x \neq \frac{c}{d}$: $$(ca+cd)x^2 + (d^2 - a^2)x-(ab+bd) = 0$$ This can only happen if each coefficient in this polynomial is zero, in particular $c(a+d) = 0$, from where we get $a=-d$ since $c\neq 0$.

Now, the two possible fixed points are given by the formula $x = \frac{-dx+b}{cx+d}$, therefore both fixed points must satisfy $cx^2 +2dx -b = 0$ Hence:

$$x_{\pm} = \frac{-2d\pm\sqrt{4d^2 +4bc}}{2c}$$

Call $u = x_+, v = x_-$. Notice that we have $\frac{u+ v}{2} = -\frac{d}{c}$ as requested. It only rests to see that the image of $f$ is $\mathbb R - \lbrace -\frac d c \rbrace$. But setting $f(x) = y$ and isolating $x$ we can see that the inverse function of $f$ is: $$g(y) = \frac{-dy+b}{cy-a}$$ which is only not defined at $y=\frac a c = -\frac d c$.

Answer 2

A fractional linear transformation $f(z)=\frac{az+b}{cz+d}$ (assume $ad-bc=1$, although your question is about specific elements with determinant $-1$) will have two fixed points, the solutions of $cz^2+(d-a)z-b=0$, $$ z=\frac{a-d\pm\sqrt{d^2-2ad+a^2+4cb}}{2c}=\frac{a-d\pm\sqrt{(a+d)^2\pm4}}{2c}, $$ determined by the squared trace of the associated matrix (we're working in $PSL_2(\mathbb{R})$). There are three classes of transformations:

• elliptic ($\text{tr}^2 < 4$) with a fixed point in the upper half-plane, conjugate to something in $SO_2(\mathbb{R})$,
• parabolic ($\text{tr}^2 = 4$) with a single fixed point on the real line (including the possibility of the point at infinity), conjugate to an upper triangular matrix,
• hyperbolic ($\text{tr}^2 > 4$) with two real roots, conjugate to a diagonal matrix.

Geometrically, these are orientatin-preserving isometries of the hyperbolic plane (upper half-plane model). The three classes correspond to

• rotations about a point in the hyperbolic plane
• movement along horocycles (sort of a rotation about a point on the ideal boundary)
• translation along the geodesic connecting the two fixed points.

You can extend the geometric picture by defining $f(z)=\frac{a\bar{z}+b}{c\bar{z}+d}$ if $ad-bc=-1$, allowing hyperbolic reflections.

The inverse of $f$ is $f^{-1}(z)=\frac{dz-b}{-cz+a}$, corresponding to the inverse of the associated matrix. These rational functions are the automorphisms of the projective line $\mathbb{R}\cup\{\infty\}=P^1(\mathbb{R})$ (viewed as the ideal boundary of the upper half-plane model of the hyperbolic plane above).

The value of $f$ at $\infty$ is $\frac{a}{c}$ (which is $\infty$ if $c=0$, these are the parabolics fixing $\infty$) and the point that gets mapped to $\infty$ is $\frac{-d}{c}$ when the denominator is zero (again an upper triangular parabolic fixing $\infty$ if $c=0$).

You ask specifically about $f^2(z)=z$, i.e. order two elements. These are either reflections about some geodesic, or a rotation of $\pi$ about some point in the upper half-plane. These are conjugate to $$ \left( \begin{array}{cc} 0&-1\\ 1&0\\ \end{array} \right)=-1/z,\quad \left( \begin{array}{cc} -1&0\\ 0&1\\ \end{array} \right)=-\bar{z} $$ which has fixed points $\pm i$ (rotation about $i$) or fixes a whole line $f(z)=-\bar{z}$. Require two real fixed points indicates the second case (a reflection).

Let's say the two real (or $\infty$), distinct, fixed points are $a/c$, $b/d$. Then $\frac{dz-b}{-cz+a}$ maps $b/d$ to zero and $a/c$ to infinity, $-\bar{z}$ reflects about the line with ideal boundary $\{0,\infty\}$, and then $\frac{az+b}{cz+d}$ maps $\{0,\infty\}$ back to $\{a/c, b/d\}$. We get $$ \left( \begin{array}{cc} a&b\\ c&d\\ \end{array} \right) \left( \begin{array}{cc} -1&0\\ 0&1\\ \end{array} \right) \left( \begin{array}{cc} d&-b\\ -c&a\\ \end{array} \right) $$ which will fix $\{a/c, b/d\}$ (and the geodesic connecting them).

If we want to single out infinity for whatever reason, and assume $\infty\not\in\{a/c,b/d\}$, then there will be one point that gets reflected to infinity, and the inverse will reflect infinity back to that point. This point is the zero of the denominator of the above $$ \left( \begin{array}{cc} -1&2ab\\ -2cd&ad+bc\\ \end{array} \right), \quad x=\frac{ad+bc}{2cd}=\frac{1}{2}\left(\frac{a}{c}+\frac{b}{d}\right), $$ i.e. the point getting flip-flopped to infinity is the average of the fixed points.