2
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
    Mar 5, 2017 at 19:04
  • $\begingroup$ Your expression for $m'$ is not correct. $\endgroup$ Mar 5, 2017 at 20:33
  • $\begingroup$ what should it be? (I am assuming ad-bc = 1 like i said) $\endgroup$
    – mdave16
    Mar 5, 2017 at 20:42
  • $\begingroup$ Sorry - I had missed that. You are correct. $\endgroup$ Mar 5, 2017 at 20:48
  • $\begingroup$ damn, the search continues $\endgroup$
    – mdave16
    Mar 5, 2017 at 20:49

2 Answers 2

1
$\begingroup$

$g\circ m = m\circ f$. 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 $f$.

$\endgroup$
5
  • $\begingroup$ 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
    Mar 5, 2017 at 21:09
  • 1
    $\begingroup$ 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 $\lim_{z\to\infty} f'(x) = \lim_{z\to 0} f'(1/z)$ $\endgroup$ 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
    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
    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
    Mar 14, 2017 at 22:19
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ This might be useful $\endgroup$
    – mdave16
    Mar 20, 2017 at 10:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.