Showing that an involution of the projective line with a fixed point fixes exactly two points

Question

My idea to show this is to notice that a projective transformation is in general given by the form $\phi(z)=\frac{az+b}{cz+d}$, then $\phi(z)=z$ gives a second order equation with at most two different solutions. But I don’t see why there couldn’t be just one solution with multiciplity 2.

Answer 1

The question in the title does not say that the involution in question has to be a projective transformation, but let's assume that's part of the question statement. If $\phi$ is an involution then executing it twice is the identity, so you have

$$\phi(\phi(z))=\frac{(a^2+bc)z+b(a+d)}{c(a+d)z+(d^2+bc)}=z \;.$$

From this you can read

$$ b(a+d)=0\qquad c(a+d)=0\qquad a^2=d^2 \;.$$

Looking at the factors there is two cases to consider here. One is $a+d=0$ i.e. $d=-a$. Then all three equations are satisfied. The other case is $b=0,c=0,a=d$ which makes $\phi$ the identity. Let's exclude the identity (but note that it doesn't have exactly two fixed points). Looking at your question, you use

\begin{align*} \phi(z)=\frac{az+b}{cz+d}&=z\\ az+b&=(cz+d)z\\ cz^z + (d-a)z - b &= 0 \end{align*}

as a quadratic equation. Its solutions would be

$$z_{1,2}=\frac{(a-d)\pm\sqrt{(d-a)^2+4bc}}{2c}$$

which has a single solution of multiplicity $2$ if its discriminant is zero, i.e. if

$$(d-a)^2+4bc=0\;.$$

Substituting $d=-a$ you get

$$4a^2+4bc=0$$

as the condition for a double solution. Writing $c=-a^2/b$ you'd get

$$\phi(z)=\frac{az+b}{-\frac{a^2}{b}z-a}=-\frac{ab\,z+b^2}{a^2\,z+ab}=-\frac{b(az+b)}{a(az+b)}=-\frac ba$$

So your map would project all the points to a single point. There is a non-degeneracy assumption we should have stated explicitly, namely that in the equation at the top of my answer, the one for the involution property, we have

$$a^2+bc=d^2+bc\neq 0$$

which is violated by the “single solution of multiplicity two” choice of parameters. A geometrically similar constraint would be requiring your transformation to be invertible. You could express this in terms of the determinant of the projective transformation as

$$\begin{vmatrix}a&b\\c&d\end{vmatrix}\neq 0$$

which is a different equation in general, but for $d=-a$ the two conditions become equivalent.

So long answer short: yes, there is a condition where that quadratic equation you have mentioned looks as though would have a single solution of multiplicity two. But that choice of parameters corresponds to a degenerate projective transformation which would not be an involution.