# Conformal transformations and Möbius transformations

1. Oct 14, 2011

### mnb96

Hello,

I read somewhere that in 2D, the Möbius transformations do not represent all the possible conformal transformations, while according to Liouville's theorem, in spaces of dimension greater than 2 all the conformal transformation can be expressed as combinations of scaling/translation/rotation/inversion (=Möbius transformations).

Can anyone mention what are the conformal transformations of the plane that cannot be "captured" by Möbius transformations?

Thanks.

2. Oct 14, 2011

### Ben Niehoff

Every holomorphic function is a conformal transformation of the plane.

3. Oct 14, 2011

### mathwonk

4. Oct 15, 2011

### mnb96

Thanks.

Unfortunately I am not very familiar with holomorphic functions and Möbius transformations.
Could you please point out a straightforward explicit example of transformation that is conformal, but not Möbius?

5. Oct 15, 2011

### samalkhaiat

Conformal transformations of space-time are, by definition, those general coordinate transformations which preserve the angles between any two vectors, or what is the same, which scale the metric locally by an overall factor. In flat space-time of dimension $D > 2$, the infinitesimal conformal transformations generate a $(D+1)(D+2)/2$ dimensional Lie algebra isomorphic to that of the group $SO(2,D)$(*). In contrast, for D = 2 any ARBITRARY holomorphic mapping of the (compactified) complex plane IS angle-preserving. Thus, in complex coordinates $z, \bar{z}$, infinitesimal conformal transformations are generated by mappings which transform z;
$$z \rightarrow z + f(z),$$
by functions which do not depend on $\bar{z}$;
$$f(z) = - \sum_{n \in \mathbb{Z}} \epsilon_{n}z^{n+1},$$
and by analogous mappings of $\bar{z}$ with functions, $\bar{f}(\bar{z})$, which do not depend on z;
$$\bar{z}\rightarrow \bar{z} - \sum_{n \in \mathbb{Z}} \bar{\epsilon}_{n} \bar{z}^{n+1}.$$
It is easy to see that on functions of $z,\bar{z}$, these mappings are generated by differential operators;
$$\ell_{n}= - z^{n+1}\partial_{z} \ \mbox{and} \ \bar{\ell}_{n}= - \bar{z}^{n+1}\partial_{\bar{z}},$$
respectively. These operators satisfy the following algebra
$$[\ell_{m},\ell_{n}] = (m-n)\ell_{m+n},$$
$$[\bar{\ell}_{m},\bar{\ell}_{n}] = (m-n) \bar{\ell}_{m+n},$$
$$[\ell_{m}, \bar{\ell}_{n}] = 0.$$
In mathematical terms, this means that both the $\ell_{n}$ and the $\bar{\ell}_{n}$ span an infinite-dimensional Lie algebra known as the Witt algebra, moreover, these two algebras are combined as a direct sum. Thus the algebra of the local conformal group $\mathcal{D}^{\infty}(2)$ is a direct sum of two commuting Witt algebras. It is easy to show that the Witt algebra $\Span\{\ell_{n}\}$ is the same as the algebra of infinitesimal diffeomorphisms of the circle $S^{1}$ (try to prove this “easy to show” thing!). Thus, the local conformal group in 2D space-time is (essentially)
$$\mathcal{D}^{\infty}(2) = \Diff (S^{1}) \times \Diff (S^{1})$$

The Global Conformal Group in Two Dimensions; the Complex Mobius Group:

Let me remark that the general 2D conformal transformations are neither globally well defined nor invertible, even on the Riemann sphere $S^{2}= \mathbb{C}\cup \{\infty\}$ (think of it as the z plane plus a point at infinity). This is because the vector fields
$$V(z) = \sum_{n}a_{n}\ell_{n},$$
generating holomorphic transformations, are globally defined only if $a_{n}=0$ for $n < -1$ and $n > 1$. Namely, the action of $(-z^{n+1}\partial_{z})$ is non-singular at $z = 0$ only for $n \geq -1$, and non-singular at $z = \infty$ (where it acts as $(w^{1-n}\partial_{w})$ with $w = 1/ z$) only for $n \leq 1$. Thus, in order to obtain a global group of well-defined and invertible conformal transformations on the Riemann sphere, we must demand that the infinitesimal transformation $\delta z = \epsilon (z)$ does not have a pole at the “point at infinity”. In terms of a new coordinate system $w = 1/z$, the “point at infinity” is just an ordinary point, namely the origin $w = 0$.
Now,
$$\delta w = - \frac{\delta z}{z^{2}} = - \frac{\epsilon (z)}{z^{2}},$$
is non-singular at $w = 0$ if and only if $\epsilon / z^{2}$ is finite for $z \rightarrow \infty$. Thus, $\epsilon (z)$ must be quadratic polynomial in $z$. We therefore conclude that the global group of 2D infinitesimal conformal transformations have the form
$$\delta z = \alpha + \beta z + \gamma z^{2},$$
with $\alpha , \beta , \gamma$ being arbitrary complex parameters. These transformations integrate to the so-called complex Mobius group (or the complex projective group);
$$z \rightarrow \frac{az + b}{cz + d}, \ \ \ (1)$$
where $a,b,c,d$ are complex parameters and $ad – bc = 1$. Thus, the global conformal group in 2D is parametrized by 6 real parameters; being isomorphic to $SO(1,3) \approx SL(2,\mathbb{C}) / Z_{2}$. To understand why $SL(2,\mathbb{C})$ is the group of such transformations, note that a $2 \times 2$ complex matrix can act on a 2-dimensional complex vector space;
$$\left( \begin{array}{c} v_{1} \\ v_{2} \\ \end{array}\right) \rightarrow \left( \begin{array}{rr} a & b \\ c & d \end{array}\right) \left( \begin{array}{c} v_{1} \\ v_{2} \\ \end{array}\right) \ \ (2)$$
If we think of $v_{1}$ and $v_{2}$ as homogeneous coordinates for $CP^{1}$, then $v_{1}/v_{2} \equiv z$ is the stereographic projection from $CP^{1}$ to the z plane. From eq(2), we see that z transforms as in eq(1). Thus we learn that $SL(2,\mathbb{C})$ acts as the group of conformal mappings of the complex plane (including the point at infinity).

The Real Mobius Group $SL(2,\mathbb{R}) \approx SO(2,1)$:

Instead of the Riemann sphere, which is a closed surface of genus zero, we can consider an open Riemann surface given by an upper half-plane. In this case we obtain the real projective group $SL(2, \mathbb{R})$ as a subgroup of the full global conformal group
$$SO(2,2) \approx SL(2,\mathbb{R})_{1} \times SL(2,\mathbb{R})_{2}.$$
This is precisely the conformal group that one would get when “naively” extrapolating from $D > 2$ to $D = 2.$
The apparent similarity between the open case and the holomorphic part of the closed case extends to a similar relation in string theory between open and closed strings; the doubling of degrees of freedom in going from open strings to closed strings; $SL(2, \mathbb{C})$ has twice as many parameters as $SL(2,\mathbb{R})$.
Each infinite-dimensional Witt algebra $\Span\{\ell_{n}\}$ contains a finite-dimensional subalgebra, $\Span \{\ell_{-1},\ell_{0},\ell_{+1}\}$, isomorphic to $sl(2)$. When the field of scalars is restricted from $\mathbb{C}$ to $\mathbb{R}$, we obtain the real form $sl(2,\mathbb{R)}$. Exponentiating the action of $\{\ell_{-1},\ell_{0},\ell_{1}\}$ on the real axis y, gives the finite transformations:
Translation;
$$T(\lambda_{-1}) = e^{\lambda_{-1}\ell_{-1}}: \ y \rightarrow y + \lambda_{-1}$$
Scale;
$$S(\lambda_{0}) = e^{\lambda_{0}\ell_{0}}: \ y \rightarrow e^{\lambda_{0}}y$$
Special Conformal transformations;
$$e^{\lambda_{1}\ell_{1}}: \ y \rightarrow \frac{y}{1 - \lambda_{1}y}$$
Combining these transformations leads to the general form of an $SL(2, \mathbb{R})$ transformation;
$$y \rightarrow \frac{ay + b}{cy + d}, \ \ ad – bc = 1 \ \ (3)$$
This means that the $2 \times 2$ real matrices of the group $SL(2, \mathbb{R})$ have natural non-linear action on ONE real variable $y$! Does this surprise you? I hope not.
Since a, b, c and d are real, eq(3) clearly maps the real axis to itself. If we analytically continue to complex values of $y$, then $SL(2, \mathbb{R})$ transformations map the upper half y-plane to itself. Indeed, one can show that $SL(2, \mathbb{R})$ is the group of all one-to-one mappings of the upper half-plane to itself; $y \rightarrow \bar{y}(y)$, with $\bar{y}$ being an analytic function of y. In some region such transformation is conformal. Therefore, we may say that the mappings of the upper half-plane onto itself form a subgroup of the full conformal group, namely the real projective group $SL(2, \mathbb{R})$. And I should stop here.

Sam

(*) See

Last edited: Oct 15, 2011
6. Oct 16, 2011

"Could you please point out a straightforward explicit example of transformation that is conformal, but not Möbius?"
Yes, z^2 or z^3 or z^4... z^n.

7. Oct 17, 2011

### Bacle2

Any holomorphic map, i.e., any complex map satisfying Cauchy-Riemann in an open disk which also has non-zero derivative, is conformal. So , take f(z)=z2 in ℂ-{0} is conformal, but
it is not a Mobius map, and it is not conformal in ℂ, since f'(z)=2z is 0 at z=0.
The only entire conformal maps are the maps az+b, since there
is an equivalence between f(z) being 1-1 and f'(z) being 0 (standard proof is , if f(k+1)(z0) is the first non-zero power of the derivative, so that f(0)=f'(0)=...f(k)=0 for , then you can express, locally, f(z)=h(z)(z-z0)n , where h(z) is non-zero in a 'hood of z0 so that f(z)/h(z) admits a local n-th root, and n-th roots are n->1 functions, and so are not 1-1 *) and the only injective entire functions are the functions az+b.

*More technically, open balls are simply-connected, and logf(z) can be defined in simply-connected regions where f is non-zero --because the integral dz/z is independent of path in simply-connected regions not containing the origin. Once you can define a log, you can define an n-th root.

EDIT: If f'(z)=0 , then f is not 1-1, but f'(z)≠ 0 does not imply more than f being locally 1-1, by the inverse function theorem.

Maybe needs some tightening, but I think it is on the right track.

Last edited: Oct 17, 2011
8. Oct 17, 2011

### Bacle2

That should be: Any map f(z) :ℂ →ℂ is conformal, if/where it is analytic and f'(z)≠ 0. As Samalkhaiat posted in his thesis below (a real tour-de-france *) , the issue is different when you deal with maps between the 1-pt. compactification of ℂ to itself.