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linear fractional transformation fixing a line

pantboio

Member
Nov 20, 2012
45
I'm trying to find the set $\mathscr{F}$ of all linear fractional transformations (l.f.t.) of the unit disc D in itself which map 1 in 1, -1 in -1 and i in -i. By l.f.t. i mean a function$$f(z)=\frac{az+b}{cz+d}$$with $a,b,c,d\in\mathbb C$, $ad-bc\neq0$.I know that this kind of maps sends lines to lines and circles to circles. In this particular case, $f$ fixes the real axis. The only functions i found are $f(z)=\frac{1}{z}$ and $f(z)=\overline z$ , but the first doesn't map the unit disc in itself and the second is not a l.f.m. So should i conclude that $\mathscr{F}=\varnothing$?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,706
I'm trying to find the set $\mathscr{F}$ of all linear fractional transformations (l.f.t.) of the unit disc D in itself which map 1 in 1, -1 in -1 and i in -i. By l.f.t. i mean a function$$f(z)=\frac{az+b}{cz+d}$$with $a,b,c,d\in\mathbb C$, $ad-bc\neq0$.I know that this kind of maps sends lines to lines and circles to circles. In this particular case, $f$ fixes the real axis. The only functions i found are $f(z)=\frac{1}{z}$ and $f(z)=\overline z$ , but the first doesn't map the unit disc in itself and the second is not a l.f.m. So should i conclude that $\mathscr{F}=\varnothing$?
A linear fractional transformation is uniquely specified by three points and their images. You have already found the (unique) LFT taking $1$ to $1$, $-1$ to $-1$ and $i$ to $-i$, namely $f(z) = 1/z$. Since this does not take the unit disc to itself, you are right to conclude that $\mathscr{F}=\varnothing.$
 

chisigma

Well-known member
Feb 13, 2012
1,704
I'm trying to find the set $\mathscr{F}$ of all linear fractional transformations (l.f.t.) of the unit disc D in itself which map 1 in 1, -1 in -1 and i in -i. By l.f.t. i mean a function$$f(z)=\frac{az+b}{cz+d}$$with $a,b,c,d\in\mathbb C$, $ad-bc\neq0$.I know that this kind of maps sends lines to lines and circles to circles. In this particular case, $f$ fixes the real axis. The only functions i found are $f(z)=\frac{1}{z}$ and $f(z)=\overline z$ , but the first doesn't map the unit disc in itself and the second is not a l.f.m. So should i conclude that $\mathscr{F}=\varnothing$?
Why $f(z)=\frac{1}{z}$ don't map the unit disk in itself?... Setting $z= e^{i\ \theta}$ is $f(z)=e^{-i\ \theta}$... right?... and, of course, $f(z)= \frac{1}{z}$ when is a=0, b=1, c=1, d=0...


Kind regards


$\chi$ $\sigma$
 
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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,706
Why $f(z)=\frac{1}{z}$ don't map the unit disk in itself?... Setting $z= e^{i\ \theta}$ is $f(z)=e^{-i\ \theta}$... right?...
In English, "disc" means "circle together with its interior". Maybe Italian distinguishes between these two concepts in a different way? :)

Kind regards,

Opalg (Handshake)
 

pantboio

Member
Nov 20, 2012
45
Why $f(z)=\frac{1}{z}$ don't map the unit disk in itself?... Setting $z= e^{i\ \theta}$ is $f(z)=e^{-i\ \theta}$... right?...


Kind regards


$\chi$ $\sigma$
By unit disc i mean the set of complex numbers with modulus less than 1, not only the unit circle
 

pantboio

Member
Nov 20, 2012
45
I'm trying to find the set $\mathscr{F}$ of all linear fractional transformations (l.f.t.) of the unit disc D in itself which map 1 in 1, -1 in -1 and i in -i. By l.f.t. i mean a function$$f(z)=\frac{az+b}{cz+d}$$with $a,b,c,d\in\mathbb C$, $ad-bc\neq0$.I know that this kind of maps sends lines to lines and circles to circles. In this particular case, $f$ fixes the real axis.
i'm no longer sure if what i said is correct. We have seen that such a function doesn't exist. But suppose it exists, with this property:
1) $f$ is a fractional linear transformation
2)$f$ map the unit disc onto itself
3)$f$ fixes 1 and -1, and maps i to -i

Can i conclude that $f$ fixes the real axis? Couldn't it happen that $f$ maps the real axis into a circle?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,706
i'm no longer sure if what i said is correct. We have seen that such a function doesn't exist. But suppose it exists, with this property:
1) $f$ is a fractional linear transformation
2)$f$ map the unit disc onto itself
3)$f$ fixes 1 and -1, and maps i to -i

Can i conclude that $f$ fixes the real axis? Couldn't it happen that $f$ maps the real axis into a circle?
You have already seen that there is no function satisfying all three of those conditions. So forget (2) and part of (3), and suppose that $f$ is a fractional linear transformation that fixes 1 and -1. Then yes, it could happen that $f$ maps the real axis into a circle. If $L$ denotes the set of lines and $C$ denotes the set of circles, then a fractional linear transformation always maps $L\cup C$ to $L\cup C$. But it does not need to map $L$ to $L$, or $C$ to $C$.
 

pantboio

Member
Nov 20, 2012
45
You have already seen that there is no function satisfying all three of those conditions. So forget (2) and part of (3), and suppose that $f$ is a fractional linear transformation that fixes 1 and -1. Then yes, it could happen that $f$ maps the real axis into a circle. If $L$ denotes the set of lines and $C$ denotes the set of circles, then a fractional linear transformation always maps $L\cup C$ to $L\cup C$. But it does not need to map $L$ to $L$, or $C$ to $C$.
tell me if i'm right: substitute condition 3) with the following:3')$f$ fixes 1. So
1)$f$ is fractional linear map
2)$f$ maps the unit disc in itself
3') $f$ fixes 1
I want to prove that, in this particular case, $f$ maps real axis in real axis (so a line into a line, non into a circle). I remember that all fractional linear of unit disc have the form (up to a rotation)

$$f(z)=\frac{z-a}{1-\overline a z}$$
for $|a|<1$. Condition 3') implies that $a$ must be real. Hence $f$ maps real numbers to real numbers
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,706
tell me if i'm right: substitute condition 3) with the following:3')$f$ fixes 1. So
1)$f$ is fractional linear map
2)$f$ maps the unit disc in itself
3') $f$ fixes 1
I want to prove that, in this particular case, $f$ maps real axis in real axis (so a line into a line, non into a circle). I remember that all fractional linear of unit disc have the form (up to a rotation)

$$f(z)=\frac{z-a}{1-\overline a z}$$
for $|a|<1$. Condition 3') implies that $a$ must be real. Hence $f$ maps real numbers to real numbers
That argument is incorrect, because you are neglecting the rotation. The general form for an FLT is $$f(z)=\lambda\frac{z-a}{1-\overline a z},$$ where $|a|<1$ and $|\lambda|=1.$ It is quite possible for the fraction $\frac{z-a}{1-\overline a z}$ to shift the point 1 to some other point on the circle, and then for the rotation by $\lambda$ to shift it back to 1.

For example, if $a = \frac12(1+i)$ and $\lambda = i$ then the map becomes $$f(z)=\frac{i(2z-1-i)}{2-(1-i)z}.$$ You can check that this fixes $1$ but not $-1$, and it takes the real axis to a circle.
 
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pantboio

Member
Nov 20, 2012
45
That argument is incorrect, because you are neglecting the rotation. The general form for an FLT is $$f(z)=\lambda\frac{z-a}{1-\overline a z},$$ where $|a|<1$ and $|\lambda|=1.$ It is quite possible for the fraction $\frac{z-a}{1-\overline a z}$ to shift the point 1 to some other point on the circle, and then for the rotation by $\lambda$ to shift it back to 1.

For example, if $a = \frac12(1+i)$ and $\lambda = i$ then the map becomes $$f(z)=\frac{2z-1-i}{2-(1-i)z}.$$ You can check that this fixes $1$ but not $-1$, and it takes the real axis to a circle.
Thank you very much. Let's see if i can state something correct. Let $f$ satisfie the following conditions:
1)$f$ is a linear fractional map;
2)$f$ maps the unit disc in itself;
3)$f$ fixes 1 and -1.
Hence $f$ is the identity.
proof: $f$ has the form
$$f(z)=\lambda\frac{z-a}{1-\overline a z}$$
for some $|\lambda|=1$. Condition 3) implies
$$\lambda=1$$
Thus
$$f(z)=\frac{z-a}{1-\overline a z}$$
and again by 3) i get $a$ real. Hence $f$ has infinite fixed points in the unit disc ( the real segment (-1,1) ). By identity principle for holomorphic function, i conclude $f=id$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,706
Thank you very much. Let's see if i can state something correct. Let $f$ satisfie the following conditions:
1)$f$ is a linear fractional map;
2)$f$ maps the unit disc in itself;
3)$f$ fixes 1 and -1.
Hence $f$ is the identity.
proof: $f$ has the form
$$f(z)=\lambda\frac{z-a}{1-\overline a z}$$
for some $|\lambda|=1$. Condition 3) implies
$$\lambda=1$$
Thus
$$f(z)=\frac{z-a}{1-\overline a z}$$
and again by 3) i get $a$ real. Hence $f$ has infinite fixed points in the unit disc ( the real segment (-1,1) ). By identity principle for holomorphic function, i conclude $f=id$
Sorry to disappoint you, but this is still not right. (Crying)

You are correct that those conditions imply that $f(z)=\frac{z-a}{1-a z}$ for some real $a$ with $|a|<1$. But it does not follow that $f$ fixes the segment (-1,1), unless $a=0$. More precisely, it maps that segment to itself, but it does not fix each point in the segment. In fact, it stretches part of the segment and shrinks another part.

You can see for example that $f(0)=-a$. The only value of $a$ for which $f$ fixes the whole segment is $a=0$. In that case only, $f$ becomes the identity map.
 

pantboio

Member
Nov 20, 2012
45
what i'm trying to do is to generalize the original situation, where $f$ fixes 1 and -1 and maps i to -i. But i probably considered the wrong way, since f doesn't fix a line, it fixes the unit circle. So what i'm looking for is some result that states something as: if $f$ is a l.f.t. that fixes a circle (not necessarily the unit one) then........i don't know, then it's is a rotation, or it is an inversion......and by this carachterization i would like to deduce that there are no such maps sending i to -i and fixing 1 and -1 (even if i know there is a very easy way to see this, as we noticed in previous posts)