?

Log in

No account? Create an account
Sigh. Have agreed to help one of my supervisees out with an… - Sally's Journal
January 13th, 2005
04:31 pm

[Link]

Previous Entry Share Next Entry
Sigh. Have agreed to help one of my supervisees out with an emergancy supervision. Unfortunately I can't answer the questions. This may be fun... Anyone out there know why three points define a mobius map completely, when a general mobius map clearly has *four* unknowns?

(13 comments | Leave a comment)

Comments
 
[User Picture]
From:atreic
Date:January 13th, 2005 04:35 pm (UTC)
(Link)
Is it something to do with two points spanning the complex plane, so there's another condition secretly built in?
[User Picture]
From:cartesiandaemon
Date:January 13th, 2005 04:35 pm (UTC)
(Link)
Isn't a moebius map az+b/cz+d? If so, you can in general cancel a parameter since all terms on numerator and denominator have a parameter in.

(But you can't *do* it, in case the missing parameter is zero.)
[User Picture]
From:fluffymark
Date:January 13th, 2005 04:38 pm (UTC)
(Link)
Yep, that's the reason. Beat me to it. :)
[User Picture]
From:atreic
Date:January 13th, 2005 04:41 pm (UTC)
(Link)
*hugs* Love you!
[User Picture]
From:cartesiandaemon
Date:January 13th, 2005 04:42 pm (UTC)
(Link)
*blush* Does that mean I was right? :)
[User Picture]
From:atreic
Date:January 13th, 2005 04:42 pm (UTC)
(Link)
And it *was* blindingly obvious. I feel so dumb :-)
[User Picture]
From:cartesiandaemon
Date:January 13th, 2005 04:46 pm (UTC)
(Link)
Well, I got it. Of course it was obvious :)
From:fluffymormegil
Date:January 13th, 2005 04:36 pm (UTC)
(Link)
Idle thought: Given that a three-unknowns three-equations simultaneous equation set is solved the instant you know two of the unknowns, does a similar effect apply here?
(But, IANAmathmo, so ignore me if I'm missing the point.)
[User Picture]
From:mhw
Date:January 13th, 2005 04:37 pm (UTC)
(Link)
I may be talking out of my bottom here, but isn't it three points and their images that define a Moebius transform?
[User Picture]
From:atreic
Date:January 13th, 2005 04:41 pm (UTC)
(Link)
Hmmm. Can show that if three points mapping to (0,1,inf) is a unique map, then three points mapping to any three points is a unique map.

Can find the (obviously going to be unique) map (z -z1)(z2-z3)/(z-z3)(z2-z1) But don't know *why* this is unique. I feel moronic. Ah well.
[User Picture]
From:cartesiandaemon
Date:January 13th, 2005 04:58 pm (UTC)
(Link)
Intuitively, because you have three unknowns and three constraints.

More rigorously? Um. *cudgels rusty complex plane bits of brain*. Prepend both your map and hypothetical other by one that takes (0,1,inf)->(z1,z2,z3). Then you have two maps taking (0,1,inf)->(0,1,inf). Show these are different by considering a point for which your and hypothetical maps differ. But the only map taking (0,1,inf) to (0,1,inf) is the identity (ummm... that must be easy to show, mustn't it?). Helps?
[User Picture]
From:atreic
Date:January 13th, 2005 05:10 pm (UTC)
(Link)
it's ok, I posted that before I read your comment. All is well. But I needed that map to do part 2 of the question, so I feel less thick now.
[User Picture]
From:cartesiandaemon
Date:January 13th, 2005 05:16 pm (UTC)
(Link)
I wasn't sure, and even compared time stamps with that post, and your first reply to me, but they were the same, so I couldn't tell which was first. But decided in the end all knowledge was good to share :)
Powered by LiveJournal.com