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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:

enter image description here

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.

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  • $\begingroup$ The answers to this question might be related: math.stackexchange.com/questions/626465/… $\endgroup$ – dahemar May 2 '16 at 19:50
  • $\begingroup$ Not quite sure what you're after. There are many ways to construct one-parameter subgroups of Möbius transforms. For example $f(t):z \rightarrow 1/(t+1/z)$. $\endgroup$ – Dan Piponi May 9 '16 at 23:06
  • $\begingroup$ @DanPiponi Roughly, I'm looking for a way to write a vector field corresponding to a "flow" of continuos Möbius transformations. $\endgroup$ – dahemar May 10 '16 at 13:49
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    $\begingroup$ See this comment from @DanRust on the linked question: "It is true though that the invariant geodesics of a Mobius transformation can be generated from a geodesic flow which gives a canonical vector field [...]. The fixed points of the transformations correspond then to the zeros of the respective vector field." I'd like to find an expression for those vector fields. $\endgroup$ – dahemar May 10 '16 at 13:53
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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 ) $.

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  • $\begingroup$ Thank you! The analogy between the hyperbolic/elliptic transformations and the electric/magnetic field phase portraits is certainly interesting, although your remark on the divergence and curl being nonzero quite breaks it. Also, in the special case of transformations with one fixed point at infinity, you get a picture ressembling the electric/magnetic field phase portraits of a single wire. By the way, wrt your last point... could you give an example of what you mean by rescaling the vector field by a function to make it coincide with the electric field? $\endgroup$ – dahemar May 17 '16 at 12:49
  • $\begingroup$ Oh, and I thought conformal mappings always satisfy Laplace's equation, no? $\endgroup$ – dahemar May 17 '16 at 12:57
  • $\begingroup$ Sort of, but not in a way relevant to this problem I think. $\endgroup$ – user226970 May 17 '16 at 23:56

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