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I am having a hard time understanding how we find mobius maps from circles, discs to half planes etc. I know how to find maps that take a set of points to another but not sets. I know about cross ratios, orientation principle, etc but I might be skipping some other important concept here. For example the below image is something I found enter image description here

My doubts in this are -

  1. How were the specific points chosen. Even though 1, i, -1 can be thought of as the boundary points of the disk in the half plane Im(z) > 0 but I don't understand the selection of points 0, 1, infinity. Why take points in the real axis?

  2. I didn't understand anything achieved or done in the last line. Why is f(0) important and how the final inference?

I am afraid I might be missing some fundamental theorems. Any help is appreciated.

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2 Answers 2

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  1. They chose three points on the circle $|z|=1$ and three points on the line $Im(z)=0$. Moebius transformations get determined when you choose the images of three points. They chose the points on the boundaries of the disc and half plane, respectively, because Moebius transformations are open maps. Therefore, they should send boundaries to boundaries. Not only the sets of points are important in the choice, but also their ordering. Moebius transformations preserve angles. Therefore, if the points $1,i,-1$ are chosen in that order, such that $|z|<1$ is to the left when you move from one to that other along $|z|=1$, you also want to choose their corresponding images, in such a way that the half-plane $Im(z)>0$ is to the left of the line $Im(z)=0$, when you travel it following the order of the points $0,1,\infty$.

  2. The last step is actually redundant if the argument above about the order has been given. An alternative argument, which they are using, would be that since Moebius transformations are open maps, they send open sets to open sets. Therefore, if you have already ensured that the boundary of the unit disc $|z|=1$ is sent to the boundary of the upper half plane $Im(z)=0$, then if you take a point in the interior of $|z|<0$ and find that its image is in the inetior of the upper half plane, then all other points of those interiors are being mapped to the other's interior.

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  • $\begingroup$ Nice explanation. thanks. But why choose only the positive real axis and not the negative part i.e. negative infinity to zero on the real axis $\endgroup$
    – deadcode
    Jul 16, 2018 at 23:28
  • $\begingroup$ @deadcode No reason. You could have chosen any three points. The only conditions, are that if the three in the domain are on the circle $|z|=1$, then the three of the image need to be on the line $Im(z)=0$. Then it comes the thing about the ordering, both sets keeping the interior to the left, or both to the right. Notice that there are many Moebius transformations that map the unit disc onto the upper half plane. $\endgroup$
    – user574889
    Jul 16, 2018 at 23:29
  • $\begingroup$ So am I correct in understanding that if we choose three arbitrary points in one set and three arbitrary points in the other set keeping in mind orientation, we can find a mobius transform mapping them to one another and this reflects the whole set if chosen in the boundary? And is this MT unique for the sets or only for the chosen points? $\endgroup$
    – deadcode
    Jul 16, 2018 at 23:34
  • $\begingroup$ @deadcode Yes, fixing the two ordered triples of different points determines uniquely the Moebius transformation. See the first link above. $\endgroup$
    – user574889
    Jul 16, 2018 at 23:37
  • $\begingroup$ I think I get it now. thanks $\endgroup$
    – deadcode
    Jul 16, 2018 at 23:37
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The map will take the boundary of the domain to the boundary of the image. The real line is the boundary of the upper half plane. That is why points on the real line were chose.

Now we know that it maps a region bounded by the unit circle to a region bound by the real line. By the first could be the interior or exterior of the circle, and the second could be the upper or lower half plane. Checking the value of $f(0)$ resolves this question.

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  • $\begingroup$ Doesn't the upper half plane Im(z)>0 have boundary in the negative real axis as well? negative infinity to zero to infinity $\endgroup$
    – deadcode
    Jul 16, 2018 at 23:24
  • $\begingroup$ The "real line" is the whole real number line, positive, negative, and zero. $\endgroup$
    – Lee Mosher
    Jul 16, 2018 at 23:28
  • $\begingroup$ It's the whole axis; we could have chosen two negative reals, or one positive and one negative. But we're looking at the Riemann sphere; there's only one $\infty$ and all lines pass through it. $\endgroup$
    – saulspatz
    Jul 16, 2018 at 23:28
  • $\begingroup$ oh, I get the idea. This is so non intuitive for people living in the real world $\endgroup$
    – deadcode
    Jul 16, 2018 at 23:30

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