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In his "Iteration of Rational Functions" Beardon defines a critical point of a rational function (mapping the Riemann sphere to itself) as a point such that the function is not injective in any neighborhood of that point.

I'm trying to square Beardon's definition with the other definition, in which a critical point is a point at which the function is either not differentiable or the derivative is zero. Specifically, I'm trying to determine when/if/how I can use the second definition to evaluate whether or not $\infty$ is a critical point.

When is $\infty$ a critical point of a rational function mapping the Riemann sphere to itself?

A rule like, "a point $z\in\mathbb{C}$ is critical if $f'(z) = 0$ or $\infty$, and $\infty$ is critical if $1/f'(\infty) = 0$ or $\infty$," would fit nicely with Beardon's classification of fixed (and later periodic) points by multiplier. But I haven't been able to find this rule for critical points stated explicitly in Beardon or anywhere else.

Edit: Thinking about this a little more, I think the rule might be

Infinity is a critical point of $f$ if and only if zero is a critical point of $1/f$.

An argument goes: The valency of $f$ at any point is invariant under conjugation by a Möbius transformation, so instead of considering the valency of $f$ at $\infty$ we can consider the valency of the conjugate $g:z\to 1/f(1/z)$ at $\infty$. Now

$\begin{aligned} \frac{\partial}{\partial z}(1/f(1/z))(\infty) &= \lim_{a\to\infty}{\frac{\partial}{\partial z}}(1/f(1/z))(a)\\ &= \lim_{a\to 0}{\frac{\partial}{\partial z}}(1/f(z))(a)\\ &= \frac{\partial}{\partial z}(1/f(z))(0) \end{aligned}$.

And the second step there is supposed to be justified by the fact that everything in sight is continuous (though I'm afraid I might be forgetting some basic calculus gotcha). Anyway, assuming that the above steps are correct and assuming that knowledge of the derivative is all that's needed to determine the valency, then that should do it: The valency of $f$ at $\infty$ is the same as the valency of $1/f$ at zero.

azimut
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Aaron Golden
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1 Answers1

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Just choose an appropriate pair of charts. Write $f(z) = p(z) / q(z)$, where $p$ and $q$ are polynomial functions with no common zeros. We may assume $f$ is not a constant function – the constant case is trivial. There are a few cases:

  • $f(\infty) = \infty$ – look at the function $w \mapsto \frac{1}{f(1 / w)}$. The derivative is given by $$\frac{f' (1/w)}{w^2 f(1/w)^2} = \frac{p'(1/w) q(1/w) - p(1/w) q'(1/w)}{w^2 p(1/w)^2}$$ but it's not immediate how to extract anything useful from this expression. Instead, suppose the leading term of $p(z)$ is $a z^n$ while the leading term of $q(z)$ is $b z^m$. We must have $n > m$ if $f(\infty) = \infty$, and the numerator is of degree $n + m - 1$ (in terms of $1/w$) and the denominator is of degree $2 n - 2$. Now there are two subcases:

    • $m = n - 1$ – then the value of the expression at $w = 0$ is $\frac{b}{a}$; in particular, $\infty$ is not a critical point of $f$.
    • $m < n - 1$ – then the value of the expression at $w = 0$ is $0$, so $\infty$ is a critical point of $f$.

  • $f(\infty) \ne \infty$ – look at the function $w \mapsto f(1 / w)$. The derivative is given by $$-\frac{f' (1/w)}{w^2} = -\frac{p'(1/w) q(1/w) - p(1/w) q'(1/w)}{w^2 q(1/w)^2}$$ but again it's not clear what we can say from this. Let $a z^n$ be the leading term of $p(z)$ and let $b z^m$ be the leading term of $q(z)$. We must have $n \le m$ if $f(\infty) \ne \infty$, and the numerator is of degree $n + m - 1$ (if $n \ne m$) or $< n + m - 1$ (if $n = m$) and the denominator is of degree $2 m - 2$. Now there are four subcases:

    • $n = m = 1$ – then $f$ is a Möbius transformation and the value of the expression is non-zero; in particular, $\infty$ is not a critical point of $f$.
    • $n = m > 1$ – if the numerator is of degree $2 m - 2$ as well, then the value of the expression is non-zero; if the numerator is of degree $< 2 m - 2$, then the value of the expression is zero. (Both sub-subcases are possible, of course.)
    • $n = m - 1$ – then the value of the expression at $w = 0$ is $-\frac{a}{b}$; in particular, $\infty$ is not a critical point of $f$.
    • $n < m - 1$ – then the value of the expression at $w = 0$ is $0$, so $\infty$ is a critical point of $f$.

At any rate, the point is that there is no easy criterion purely in terms of the values of $f$ and $f'$.

Zhen Lin
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  • Thanks, but I'm not understanding why we can't use the $w\to1/f(1/w)$ function even when $f(\infty)\neq\infty$. If we do that everything works out the same in your first two cases, but we have an additional possibility: $m > n - 1$. In that case the expression is equal to $\infty$ at $w=0$, and $\infty$ is a critical point of $f$, right?

    As for your $f(\infty)\neq\infty$ case, I'm not sure that we can get much from considering $w\to f(1/w)$ because that function isn't conjugate to $f$. Maybe I'm missing something.

     – Aaron Golden Sep 25 '13 at 22:28
  • You can use either function if $f(\infty) \notin { 0, \infty }$. – Zhen Lin Sep 25 '13 at 22:35
  • I guess what I'm not understanding is why we can't use the first function, $w:1/f(1/w)$, when $f(\infty) = 0$. Then $m > n$, the degree of $p'(1/w)q(1/w) - p(1/w)q'(1/w)$ (in terms of $1/w$) is $n + m - 1$, and the degree of the denominator is still $2n - 2$, but since $m > n - 1$, $\frac{f'(1/w)}{w^2 f(1/w)^2}$ has a pole at $w = 0$, and $\infty$ is a critical point of $f$. – Aaron Golden Sep 26 '13 at 19:52
  • That's obviously not correct. The function $z \mapsto \frac{1}{z}$ vanishes at $\infty$, but $\infty$ is not a critical point. – Zhen Lin Sep 26 '13 at 20:00
  • Opps, I guess I never said it explicitly, but we don't have to think much when the degree of $f$ is less than two, as such functions have no critical points in $\mathbb{C}_\infty$. If we restrict our attention to rational functions of degree at least two, do we still need your $w\to f(1/w)$ case? – Aaron Golden Sep 26 '13 at 20:20
  • The function $z \mapsto \frac{z}{z^2 - 1}$ vanishes at $\infty$, but $\infty$ is not a critical point. It is simply not true that the derivative blowing up at $w = 0$ implies that there is a critical point at $w = 0$. – Zhen Lin Sep 26 '13 at 20:34
  • OK, that was my biggest misunderstanding, thinking that if the derivative had a pole then that indicated a critical point. Thank you so much for the thorough explanation. – Aaron Golden Oct 04 '13 at 05:40
  • Just to be clear, was "Infinity is a critical point of f if and only if zero is a critical point of 1/f". true? – mdave16 Mar 20 '17 at 16:24