Finding infinitesimal Mobius transformations

Question

I'm working to understand a bit more about Mobius transformations. I understand that the Mobius group is isomorphic to $SL(2,\mathbb{C})$, which has as its Lie algebra $\mathfrak{sl}(2,\mathbb{C})$. However, Mobius transformations are complex functions, whereas the elements of $\mathfrak{sl}(2,\mathbb{C})$ are matrices with trace 0.

Can I bring back the connection to functions on the extended complex plane? In particular, I wanted to know whether elements of the algebra could correspond to infinitesimal conformal transformations. In physics, we often think of infinitesimal rotations and translations on a space. I wanted to think of infinitesimal Mobius transformations in the same way.

Answer 1

Suppose $A:[0,1]\to\mathrm{SL}_2(\mathbb{C})$ is a differentiable path, with $A(0)=I$, and $z\in\widehat{\mathbb{C}}$ is a constant. Then we may differentiate the transformation $A(t)z$ at $t=0$ to obtain $(\mathrm{d}A)(z)$:

$$ \left(\frac{az+b}{cz+d}\right)'=\frac{(\dot{a}z+\dot{b})(cz+d)-(az+b)(\dot{c}z+\dot{d})}{(cz+d)^2}=-\dot{c}z^2+(\dot{a}-\dot{d})z+\dot{b}. $$

Note $a=1,b=0,c=0,d=1$ at $t=0$.

The reason for the asymmetry (why $c$ goes with the quadratic term and $b$ with the constant term) is because of how we identify the complex projective line $\mathbb{CP}^1$ with $\widehat{\mathbb{C}}$ by identifying $(w,z)$ with $(w/z,1)$; if instead we used the second coordinate for $\widehat{\mathbb{C}}$ (i.e. $z/w$) it would have been the other way around.

Answer 2

I've continued to do some research online, and the closest I've found is a paper from 1968 titled "On generating subgroups of the Moebius group by pairs of infinitesimal transformations". In Introduction, the author states directly that

The infinitesimal transformations, i.e., elements of the Lie algebra, of the Moebius group (denoted by H— Lie algebra—h) are quadratics: $az^2 + 2bz + c$, $a,b,\text{ and } c$ complex constants.

I can see how this set makes sense as the Lie algebra, as it has complex dimension three just as we would expect from $\mathfrak{sl}(2, \mathbb{C})$. Without any other answers, I am going to take this and investigate it more. But for now, I'll close this question.