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.
Best Answer
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.