Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

EDIT: To make my question more precise i think we can narrow it down to this. Say you have a simple polygon that includes the origin, that is completely contained in the unit disk, we can 'blow up' (description by Patrick) the polygon until it completely covers the disk. This 'blowing up' map can explicitly be given by the Schwarz–Christoffel mapping which is conformal. Now the similarity transform that scales (and maybe translates) the polygon so that it still is contained in the disk, but has maximal area, is trivially conformal. Can we somehow 'upgrade' the Schwarz–Christoffel map all the way to this similarity transform - maybe by intermediate maps that are all conformal and distort the lengths less but maybe fill less and less of the disk?

I am looking for an algorithm/theorem that helps me with the following: Given a convex set in the plane, I want to map it to the unit disk so that the image (of the convex set) has maximal area. With 'map to the unit disk' I mean that the image of the convex set is completely contained in the unit disk.

I know that for example for the Riemann mapping theorem you want a biholomorphic map - so I guess that also means you fill the whole disk. Whereas here the whole disk doesn't need to be filled. Does the Riemann-mapping theorem imply there will always be such a map so that the image has area 1? I would then be interested in restricting the map maybe all the way down to a homogeneous Euclidean transform. I am specifically interested in the case where the convex set is what you get from the intersection of half-planes (convex polygon). If there is a conformal such mapping does it mean that the image is a scaled/rotated version of the polygon maybe?

Anyway I am interested in the gritty details of accomplishing such a mapping (or of a similar sort).

share|cite|improve this question
I notice that you have flagged a couple of your own questions as "not proper questions." Why? – mixedmath Dec 15 '12 at 23:55
up vote 2 down vote accepted

Yes. There is always such a mapping by the Riemann mapping theorem, because your convex set is a non-empty simply connected open set (if you need the border, simply consider the continuation of the map from the convex set open to the open unit disk and "add the border" afterwards. You're not losing any properties there.)

I think the most natural way of seeing this is to fix a point in the convex set and map it to the origin, and then "blow your set up", as if you were blowing air in a balloon. I don't think such a map will end up naturally conformal (I mean, "blowing up" your convex set is probably not a good way to see a conformal map) but it might give you an idea of how it can happen. I don't have any explicit construction in mind though. It's just my intuition of conformal maps behaving "nicely".

Note that there is no explicit function that maps conformally the square into the disk, so that you're not hoping for something explicit all the time.

share|cite|improve this answer
Yeah that's what i thought - and thanks for thinking it through. But as i state in my question i am especially interessted in finding such a mapping expicitly. Since i wrote the question i have become aware of the so called Schwarz–Christoffel mapping - this was what i wanted! (in part) - but it is too complicatd to use for me (although i want to try more). Since it maps from the upper half plane to a polygon you can compose that with a caylay transform (or exp). – Peter Sheldrick Jul 30 '11 at 21:50
You comment a lot, do you. XD I'll think it through too. – Patrick Da Silva Jul 31 '11 at 2:42
If you're willing to approximate the conformal mapping then it would be useful to know what are you willing to do with those mappings. Knowing that may lead to different tools and might involve other MSE users to tell you more about your problem. =) – Patrick Da Silva Jul 31 '11 at 2:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.