About the group of Mobius Transformations
dalthman
as
T(z) = (az+b)/(cz+d) ad-bc≠0
If we set M = the set of all Mobius transformations, then M forms a
group with the composition of functions.
I wonder whether it is known that
(1) Does all Mobius transformation have finite orders in the group M?
(2) Can we find all Abelian subgroups of M?
(3) What is known about the group structure of M?
José Carlos Santos
> The Mobius transformation in the book of complex function is a mapping
> as
>
> T(z) = (az+b)/(cz+d) ad-bc≠0
Please don't use HTML; use ASCII, as in "ad - bc is not zero" or
"ad - bc =/= 0" (I prefer the first one).
> If we set M = the set of all Mobius transformations, then M forms a
> group with the composition of functions.
>
> I wonder whether it is known that
>
> (1) Does all Mobius transformation have finite orders in the group M?
Of course not! Take f(z) = 2z, for instance (that is, take a = 2,
b = c = 0, and d = 1).
> (3) What is known about the group structure of M?
It's a simple group.
Best regards,
Jose Carlos Santos
Alain Verghote
***************************************
Dear friends,
Question 2) may receive a solution, we observe that T(z) {a,b,c,d} given
and T(T(z)) or T^[2](z) commute and also every integral power:
T^[i]U T^[j] will commute,
We must probably consider T^[r](z) any r real (or complex?)
built on {a,b,c,d} as an Abelian group...
Hope this helps,
Sincerely,Alain.
bijan
memed13
I need help on question 2, too. Is it possible to be explained it more detailed one more time.
Thanks in advance
David C. Ullrich
wrote:
>Hi,
>
>I need help on question 2, too. Is it possible to be explained it more detailed one more time.
Who knows whether that's possible? It would depend on what "question
2" is.
>Thanks in advance
David C. Ullrich
José Carlos Santos
>> I need help on question 2, too. Is it possible to be explained it more detailed one more time.
>
> Who knows whether that's possible? It would depend on what "question
> 2" is.
The subject of the post to which you were replying begins with "Re: ",
because the OP is replying to post from 2005! Question number 2 is:
Can we find all Abelian subgroups of the group of all Moebius
transformations of the complex plane?
Bill
I expect the answer is that we can. I think it probably helps if one
understands a little of the theory of elliptic, parabolic and hyperbolic
linear fractional maps. For starters, the cyclic subsgroups will provide
many Abelian subgroups, but clearly there are more. Interesting question!
--Bill
Bill
"Bill" <Bill_...@comcast.net> wrote in message
news:BMedncToRNxZna7V...@comcast.com...
In my reply, i had been thinking of the linear fractional maps which take
the unit disk into itself. The general question on the plane may be
slightly easier. It is no doubt useful to know that the entire group of
tranformations is generated by translations, dilations and inversions (i(z)
=1/z). How well do these commute? How many different subgroups could you
create just from the translations alone? Obviously the answer to this is
the same as the number of subgroups of the complex numbers (under the group
operaton +). There seem to be many,many subgroups in this category
alone--but probably not so many that you couldn't characterize them.
-Bill
David C. Ullrich
Oh.
This is more or less equivalent to finding all the abelian subgroups
of GL_2(C), although as often happens I suspect that the question
about GL_2 is easiest answered by looking at the representation
as Mobius transformations.
Say G is such a group and T is a non-trivial element of G.
Then T has one or two fixed points on the Riemann sphere S.
Now T has one or two fixed points on S. If T has two fixed points
then after a conjugation we can assume that T fixes 0 and infinity.
This shows that T(z) = Az, where A <> 1; one can calculate
that the only Mobius transformations commuting with this
are of the same form.
Suppose that T has only one fixed point. After a conjugation
we can assume that T fixes infinity. Since T has no finite
fized point it follows that T is a translation: T(z) = z + B,
where B <> 0. Now if T' commutes with T it follows
that T'(infinity) is fixed by T, hence T'(infinity) = infinity,
and hence T' is affine: T'(z) = az + b. Then the fact
that T' commutes with T shows that T' is a translation
as well.
So G must be conjugate to a subgroup of the translations
or a subgroup of the multiplication operators.
(Ie a subgroup of (C,+) or of C^*.)
>Best regards,
>
>Jose Carlos Santos
David C. Ullrich
Jack Schmidt
> If T has two fixed points then after a conjugation
> we can assume that T fixes 0 and infinity. This
> shows that T(z) = Az, where A <> 1; one can calculate
> that the only Mobius transformations commuting with
> this are of the same form.
Is this true? If S takes infinity to 0 and 0 to infinity,
it still has a chance of commuting. For instance, Robert
Israel mentioned z -> -z and z -> 1/z as a pair of
commuting moebius transformations. The first fixes 0 and
infinity, and the latter exchanges them.
Mariano Suárez-Alvarez
wrote:
Well: if it exchanges them then it does not fix them :D
-- m
memed13
Can anyone helps me on showing it?
Thanks,
Mehmet
memed13
> This is more or less equivalent to finding all the
> abelian subgroups
> of GL_2(C), although as often happens I suspect that
> the question
> about GL_2 is easiest answered by looking at the
> representation
> as Mobius transformations.
>
My first question what do you mean by GL_2(C)?
> Say G is such a group and T is a non-trivial element
> of G.
> Then T has one or two fixed points on the Riemann
> sphere S.
>
That's right because cz^2 + (d-a)z-b =0 will be the fixed points of T. (Here T(z) = (az+b)/(cz+d) )
> Now T has one or two fixed points on S. If T has two
> fixed points
> then after a conjugation we can assume that T fixes 0
> and infinity.
> This shows that T(z) = Az, where A <> 1; one can
> calculate
> that the only Mobius transformations commuting with
> this
> are of the same form.
Here, I have question. A Mobius transformation has infinity as its only fixed point if and only if it is a translation. So if T(z)=Az, where A <>1 , T can only have infinity as fixed point, Is not it?
> Suppose that T has only one fixed point. After a
> conjugation
> we can assume that T fixes infinity. Since T has no
> finite
> fized point it follows that T is a translation: T(z)
> = z + B,
> where B <> 0. Now if T' commutes with T it follows
> that T'(infinity) is fixed by T, hence T'(infinity) =
> infinity,
> and hence T' is affine: T'(z) = az + b. Then the fact
> that T' commutes with T shows that T' is a
> translation
> as well.
My second question is about translation. A Mobius transformation has 0 and infinity as its only fixed points if and only if it is a dilation, is not it?
> So G must be conjugate to a subgroup of the
> translations
> or a subgroup of the multiplication operators.
> (Ie a subgroup of (C,+) or of C^*.)
Can you give me a detailed example on the conclusion part of your article, please. I need to see that at what forms of elements of group of Mobius transformation, can commutate with each other, this is to say that abelian subgroups of it. And How?
> >Best regards,
> >
> >Jose Carlos Santos
>
> David C. Ullrich
Thanks very much in advance...
Best regards,
Mehmet
memed13
> > This is more or less equivalent to finding all the
> > abelian subgroups
> > of GL_2(C), although as often happens I suspect
> that
> > the question
> > about GL_2 is easiest answered by looking at the
> > representation
> > as Mobius transformations.
> >
>
> My first question what do you mean by GL_2(C)?
>
OK, I understood what GL_2(C) was now. That is all invertible 2x2 matrices with entries in C
David C. Ullrich
<Jack.Schmi...@gmail.com> wrote:
>[ Let T be a moebius transformation ]
>> If T has two fixed points then after a conjugation
>> we can assume that T fixes 0 and infinity. This
>> shows that T(z) = Az, where A <> 1; one can calculate
>> that the only Mobius transformations commuting with
>> this are of the same form.
>
>Is this true?
No, sorry. I'm pretty sure it's true except for A = -1.
>If S takes infinity to 0 and 0 to infinity,
>it still has a chance of commuting. For instance, Robert
>Israel mentioned z -> -z and z -> 1/z as a pair of
>commuting moebius transformations. The first fixes 0 and
>infinity, and the latter exchanges them.
David C. Ullrich
David C. Ullrich
Erm, that was his point. If T(z) is -z then there is another
transformation commuting with T that's not of the form
I gave. (Being of that form is equivalent to fixing 0 an infinity.)
>-- m
David C. Ullrich
David C. Ullrich
wrote:
>Addition to translation and dilation, I think we have one more abelian subgroup o group of Mobius transformation is the inverse... that takes 0 to infinity and infinity to 0. But again I need to see how two elements of a subgroup of Mobius which consist of the elements of the form of Tz = Az^(-1) commutates with each other.
Find a keyboard with an Enter key!
What I said was not quite right. What you say you want to prove is far
from right - the maps 2/z and 3/z do not commutte.
>Can anyone helps me on showing it?
Take my proof and work out the details carefully. You'll find
an error where another example creeps in. I'm pretty sure
there's only one counterexample
>Thanks,
>
>Mehmet
David C. Ullrich