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When given the mobius transformation and three points $f(z)=\frac{az+b}{cz+d}$ for $z,a,b,c,d\in \mathbb{C},ad-bc\neq0, f(z_1)=f1,f(z_2)=f_2,f(z_3)=f_3$,

what is the approach to determine the unique transformation? Because during the substitutions I always end up with a quadratic equation involving one of the variables. How do I then definitively disprove the other solution?

smaillis
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  • Is it possible that this leads to a sign ambiguity for $a$ (say), but that the same sign ambiguity occurs in $b,c,d$, so that it cancels itself out of the fraction $\frac{az+b}{cz+d}$? (Remember that a single Möbius transformation is represented by infinitely many quadruples $a,b,c,d$, since one can scale by a nonzero constant.) – Greg Martin Mar 17 '21 at 05:28
  • yes, theres is a sign ambiguity, but what do you mean it cancels? what happens is when i find the two $a$(say), i then obtain 2 set of $b,c,d$ accordingly. However, when substituting the second set back into $ad-bc = C$ where C is a constant given by the question, it produces not C, and when the transformation is plotted, the second set doesn't map all three points, but I don't understand why it shouldn't equals to C. – smaillis Mar 17 '21 at 05:36
  • "...What is the approach to determine the unique transformation? ...How do I then definitively disprove the other solution?". The direction of the implication, when resolving the corresponding generated quadratic equation is this direction: $\implies.~~$ This means that the roots of the quadratic equation are candidate solutions, that each may or may not satisfy the original constraints. Therefore, you must automatically check each individual candidate solution against the original constraints, to see if the candidate solution is successful. – user2661923 Mar 17 '21 at 06:43
  • It's not clear we're going to be able to provide further assistance without details of computations in the OP. – Greg Martin Mar 17 '21 at 06:49

1 Answers1

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A good idea is to use $(0,\infty,1)$ as a "way station" (midpoint) between two triples.

To go from $(z_1,z_2,z_3)$ to $(0,\infty,1)$, we may put $z-z_1$ in a numerator to ensure $f(z_1)=0$, we may put $z-z_2$ in a denominator to ensure $f(z_2)=\infty$, and then normalize to ensure $f(z_3)=1$:

$$ f(z)=\frac{z-z_1}{z-z_2}\cdot\frac{z_3-z_2}{z_3-z_1}, $$

the "cross-ratio." The corresponding matrix for this transformation is the product

$$ \begin{bmatrix} z_3-z_2 & 0 \\ 0 & z_3-z_1 \end{bmatrix} \begin{bmatrix} 1 & -z_1 \\ 1 & -z_2 \end{bmatrix}. $$

To go from $(z_1,z_2,z_3)$ to $(f_1,f_2,f_3)$, then, we compose with the inverse of the transformation that goes from $(f_1,f_2,f_3)$ to $(0,\infty,1)$, which (using $(AB)^{-1}=B^{-1}A^{-1}$) would yield

$$ \begin{bmatrix} 1 & -f_1 \\ 1 & -f_2 \end{bmatrix}^{-1} \begin{bmatrix} f_3-f_2 & 0 \\ 0 & f_3-f_1 \end{bmatrix}^{-1} \begin{bmatrix} z_3-z_2 & 0 \\ 0 & z_3-z_1 \end{bmatrix} \begin{bmatrix} 1 & -z_1 \\ 1 & -z_2 \end{bmatrix}. $$

The title suggests you also want to understand why there is a unique transformation between triples. Suppose there were two transformations between a pair of triples. We may compose both with a map that turns the second triple into $(0,\infty,1)$, so we end up with two transformations that turn a triple into $(0,\infty,1)$. But then we may precompose with a map that turns $(0,\infty,1)$ into the first triple, and we end up with two transformations from $(0,\infty,1)$ to itself. But it is an easier exercise to show there is only one such transformation.

anon
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