Computing the infinitesimal generators for the Mobius transformation

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To compute the infinitesimal generators for the Möbius transformation, one must start with the constraint ad-bc=1, which reduces the number of parameters. The transformation can be expressed as t' = t'(t, a, b, c, ...), where a, b, and c are parameters. The generators will be functions rather than matrices, and there should be three generators corresponding to the three independent parameters. The general formula for the generator involves taking partial derivatives of the transformation with respect to each parameter at the identity transformation values. Understanding these steps is crucial for deriving the infinitesimal generators effectively.
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Homework Statement
The Mobius transformation is $$\frac{at+b}{ct+b},$$ with the constraint ##ad-bc=1##. Find the infinitesimal generators of its Lie algebra.
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I don't know where to start. I understand that the constraint ##ad-bc=1## gives us one less parameter since ##d=1+bc/a##. So we can rewrite our original function. I know how to compute the generators of matrix groups but in this case the generators will be functions. I also know there should be three of them since we have three independent parameters. However, I'm not sure what to do. I think we may have to take partial derivatives?
 
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[ I presume your denominator should be ##ct+d## ? ]

Suppose you have a transformation of the form $$ t' = t'(t,a,b,c,...)~,$$where ##a,b,c## are parameters of the transformation.

The general formula for the generator of such a coordinate transformation is $$X_a ~=~ \left. \frac{\partial t'}{\partial a} \right|_{a,b,c=\text{Id}} \; \frac{\partial}{\partial t}$$and similarly for ##b,c,...##

The "Id" notation denotes whatever value the parameter has at the identity transformation. E.g., in the Mobius case, for ##t' = t## we must have ##a=d=1##, and ##b = c = 0##.
 
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At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...