Relationship between Möbius transformations and flows/vector fields I've noticed that the pictures illustrating the effect of Möbius transformations on the Riemann sphere (after stereographic projection to the plane) resemble the phase portrait of a vector field. For instance:

This is the representation of a hyperbolic transformation (from Wikipedia), and seems very similar to the phase portrait of a field generated by a positive and a negative charge located at the fixed points of the transformation (that is, of an irrotational and incompressible field). One could ever think of linking the fact that poles have integer order to the discretization of charges, using Gauss's divergence theorem. I guess the analogy arises from conformal mappings satisfying Laplace's equation, but I wonder how far can one push it.
This is where I'm doubting: in the example cited above, hyperbolic transformations are the stereographic projection of discrete rotations of the sphere. In general, Möbius transformations are discrete mappings from the Riemann sphere to itself, whereas a vector field suggests a continuous flow. Could we rigorously obtain a one-parameter subgroup of continuous transformations from those infinitesimal, discrete transformations, and could we define a (unique) vector field corresponding to this continuous subgroup (for example, writing each transformation as a differential equation)? How?
Please excuse me if my approach is hand-waving, I don't know much about this subject.
 A: I became curious about what that picture means. Here is my speculation: given a Mobius transformation as a matrix $ G \in SL(2,\mathbb{C}) $, it corresponds (by logarithm) to an element $ g \in \mathfrak{sl}_2 $, which then gives a vector field on $ \mathbb{R}^2 $ (by $ \frac{d}{dt} e^{g t} \big|_{t = 0}) $
Here are details for an easy example. Let
$$ g=
\begin{pmatrix}
0 & w \\
w & 0
\end{pmatrix}
$$
where $ w \in \mathbb{C} $.
He is in $ \mathfrak{sl}_2 $ because his trace is zero. Exponentiating:
$$
e^{ t g} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + t \begin{pmatrix} 0 & w \\ w & 0 \end{pmatrix} + O(t^2)
$$
This corresponds to the Mobius transformation:
\begin{align}
z &= \frac{z + t w }{z t w + 1} + O(t^2)\\
&= z + t \; w (1 - z^2) + O(t^2)
\end{align}
The $ O(t) $ term is the vector field. He has zeros at $ z = \pm 1 $, which means $ G = \exp(g) $ will have fixed points at $ z = \pm 1 $. With different choices of $ w $, $ g $ upon exponentation gives all Mobius transformations with those fixed points.
Writing $ w = a + b i $ and $ z = x + i y $, and taking the real and imaginary parts to get the $ \hat{x} $ and $ \hat{y} $ components:
$$
w ( 1 - z^2 ) \sim \left( 2 b x y + a( 1 - x^2 + y^2) \right) \hat{x} + \left( - 2 a x y + b(1 - x^2 + y^2) \right) \hat{y}
$$
If I set $ b = 0 $, then I get pictures like the one you posted. With $ a = 0 $, I get pictures that look like the B-field lines of two wires.
You can check that the divergence and curl of are nonzero, so it won't be an E field or B field on the nose. 
However, you can also check that after rescaling the vector field by a function, then it is the electric field of two point charges. So the stream lines are indeed the same!
EDIT: The electric field of one point charge (in 2d!) can be written as:
$$
\frac{( x - x_0) \hat{x} + (y - y_0) \hat{y} }{(x-x_0)^2+(y-y_0)^2} $$
Putting one charge at $ (1,0) $ and another charge at $ (-1,0) $, I get:
$$
\frac{( 2 - 2 x^2 + 2 y^2) \hat{x} + (4 x y ) \hat{y}
}{x^4 + 2 x^2 (-1 + y^2) + (1 + y^2)^2}$$
So you see, this is the same as the vector field from above with $ a = 0, b = 2 $ multiplied by the scalar function $  1 /  \; ( x^4 + 2 x^2 (-1 + y^2) + (1 + y^2)^2  ) $.
