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I'm working on $S^2 \cong \mathbb{C} \cup \{\infty\}$. It's well known that mobius transformations are the only automorphisms of $S^2$.

Suppose I have a holomorphic function $f$ and a mobius transformation $m$, and I define $g = m\circ f \circ m^{-1}$. Then I want to show that if $\beta$ is a critical point of $f$, then $m(\beta)$ will be a critical point of $g$.

I just have one hitch in the (elementary) proof, so far I have this:

For later use let $m = (az + b)/(cz+d)$, then $m' = (cz + d)^{-2}$. Letting mobius transforms have determinant 1 so I'm assuming that $ad - bc = 1$.

I use chain rule on $g$: \begin{align} g'(z) &= (m'fm^{-1}(z)) \times (f'm^{-1}(z)) \times (m^{-1})'(z)\\ g'm &= (m'f(z)) \times (f'(z)) \times (m^{-1})'m(z)\\ &= (cf(z) + d)^{-2} \times f'(z) \times (-cm(z) + a)^{-2}\\ &= \frac{f'(z)(cz + d)^2}{(cf(z) + d)^{2}} \end{align} Now assuming that the denominator isn't $0$, this works. However, I can't see what to do when it is $0$. I've tried using L'hopital, but without much luck.

Alternatively, if someone knows a more insightful proof, I would be very glad.

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  • $\begingroup$ I've looked at the wiki page of critical points, and found that this is similar to the definition for charts, I wondered if there was a way to use them to solve this, but I don't know enough about them. $\endgroup$
    – mdave16
    Commented Mar 5, 2017 at 19:04
  • $\begingroup$ Your expression for $m'$ is not correct. $\endgroup$
    – Paul Sinclair
    Commented Mar 5, 2017 at 20:33
  • $\begingroup$ what should it be? (I am assuming ad-bc = 1 like i said) $\endgroup$
    – mdave16
    Commented Mar 5, 2017 at 20:42
  • $\begingroup$ Sorry - I had missed that. You are correct. $\endgroup$
    – Paul Sinclair
    Commented Mar 5, 2017 at 20:48
  • $\begingroup$ damn, the search continues $\endgroup$
    – mdave16
    Commented Mar 5, 2017 at 20:49

2 Answers 2

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$g\circ m = m\circ f$. Therefore $$(g'\circ m)m' = (m'\circ f)f'$$ If $f' = 0$, then either $g'\circ m = 0$, or $m' = 0$. The only place where $m' = 0$ is at $\infty$.

So this gets you that every critical point of $f$ except possibly for $z = \infty$ is also a critical point of $g\circ m$. To finish it, you need to examine what exactly it means for $\infty$ to be a critical point for $f$.

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  • $\begingroup$ I was also struggling for quite a while on what it meant for something to be critical at $\infty$, but i'll figure it out. Out of interest, why can't I finish my proof the way it is? $\endgroup$
    – mdave16
    Commented Mar 5, 2017 at 21:09
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    $\begingroup$ Your version divides through by $m'$. You got bogged down in the details of $m'\circ f/m'$. With great care, you could clean it up and eventually come to the same point I did above. But that does require care about when things are $0$ or $\infty$. Starting off with $g\circ m$ avoids those difficulties, so you only have to deal with the one remaining special case. As for the other: note that $\lim_{z\to\infty} f'(x) = \lim_{z\to 0} f'(1/z)$ $\endgroup$
    – Paul Sinclair
    Commented Mar 5, 2017 at 21:21
  • $\begingroup$ I slept on this, and i'm unsure again. What about when $m' \circ f = \infty$? We would have $0 \times \infty$ which is undefined, and if $f$ a polynomial, $f'/(cf + d)^2$ would eventually be $\infty$ under L'hopital. $\endgroup$
    – mdave16
    Commented Mar 6, 2017 at 12:49
  • $\begingroup$ I'm sorry for being slow, it might be correct, i'm just not seeing how it got rid of my original problem. $\endgroup$
    – mdave16
    Commented Mar 6, 2017 at 12:50
  • $\begingroup$ Found a neater proof -- and I think maybe more insightful, Will add it after writing it up better. $\endgroup$
    – mdave16
    Commented Mar 14, 2017 at 22:19
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Instead of viewing a critical at $x$ of $f$ as $f'(x) = 0$, view it geometrically. Let $x$ be a critical point of $f$ if $f$ is not injective in any neighbourhood of $x$. (As defined by Beardon, as well as differentiation textbooks). Now the claim follows trivially since $m$ is an automorphism.

The two statements $f'(\infty) = 0$ and $\infty$ critical point are unrelated.

Take $f = \frac {z} {z^2 - 1}$. Then $f' = \frac{-z^2 - 1}{(z^2 - 1)^2}$. Indeed $f'(\infty) = 0$, yet $\infty$ is not a critical point.

Take $f = z^2$. Then $f' = 2z$. Indeed $\infty$ is a critical point, but $f'(\infty) = \infty \neq 0$.

However, for a finite point, $f'(z) = 0$ is a still a valid test for critical points.

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  • $\begingroup$ This might be useful $\endgroup$
    – mdave16
    Commented Mar 20, 2017 at 10:19

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