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The Wiki article about Moebius transforms states:

A Möbius transformation can be obtained by first performing stereographic projection from the plane to the unit two-sphere, rotating and moving the sphere to a new location and orientation in space, and then performing stereographic projection (from the new position of the sphere) to the plane.

What's not clear to me:

At the very beginning the article states Moebius transforms are maps of the complex plane.

Further down (in the section ''Overview'') it states Moebius transforms are defined on the extended complex plane.

(1) Which is it?

And:

If a Moebius transform was indeed a continuous map on the complex plane and we wanted to define it on the extended complex plane, isn't there only one way it can be extended anyway?

(2) I mean, the extended map has to have $f(\infty) = \infty$ if we still want it to be continuous. Or does it not?

And if the extension of a continuous map on $\mathbb C$ to $\widehat{\mathbb C}$ is indeed unique:

(3) Why bother at all to consider maps on $\widehat{\mathbb C}$? It seems to me that there is no more information contained in such a map than in its restriction to $\mathbb C$.

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    $\begingroup$ A good example of a Möbius map is $f(z)=1/z$ ... to extend you make $f(0)=\infty$ and $f(\infty)=0$. $\endgroup$ – GEdgar Mar 22 '15 at 0:56
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Most Mobius transformations are not defined at some point in the complex plane. This is the point where the function takes the value $\infty$ in the extended plane.

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Including a point at infinity in the domain and target of a Möbius transformation (i.e., working on the Riemann sphere $\Sigma = \widehat{\mathbf{C}}$, a.k.a. the complex projective line, instead of on $\mathbf{C}$ itself) gains one a uniformity of treatment; a holomorphic diffeomorphism $T:\Sigma \to \Sigma$ is precisely a Möbius transformation, i.e., a fractional linear transformation $$ T(z) = \frac{az + b}{cz + d},\quad ad - bc \neq 0, $$ with the understandings that $T(-d/c) = \infty$ and $T(\infty) = a/c$ if $c \neq 0$, or $T(\infty) = \infty)$ if $c = 0$.

By contrast, if you restrict a fractional linear transformation to its natural domain in $\mathbf{C}$, you generally "lose a point", i.e., have to omit $-d/c$ from the domain and $a/c$ from the image (if $c \neq 0$). If you compose transformations, you can lose multiple points for similar reasons. This is more of a nuisance than a technical hurdle (because a rational function is uniquely determined by its values on a dense open set), but...it's a nuisance.

(A more subtle, analytic reason is that working on the sphere brings the advantages of a compact domain. But that's a motivation to tuck away for later.)

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