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A non-constant holomorphic function $f:X\to Y$ between Riemann Surfaces has a branch point at $p\in X$ if there is no open neighbourhood around p on which $f$ is injective.

I looked up examples and saw that sometimes people only look at the zeroes of the derivative and say that these are the branch points. Why is this the case?

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  • $\begingroup$ How does the local normal form of a holomorphic function look? $\endgroup$
    – Daniel Fischer
    Commented Jul 30, 2017 at 18:34
  • $\begingroup$ The normal form around $p$ is $z\mapsto z^k$. If its derivative has a zero, this means that $k>1$, so $p$ is a branch point. But why is it then enough to check the derivatives of $f$? $\endgroup$
    – user451226
    Commented Jul 30, 2017 at 18:43
  • $\begingroup$ Because if the derivative at $x$ is nonzero, then $k = 1$ at $x$, and $f$ is injective on every small enough neighbourhood of $x$. $\endgroup$
    – Daniel Fischer
    Commented Jul 30, 2017 at 18:45
  • $\begingroup$ Ok, this means I find the points that are no branch points by looking at the points where the derivative does not vanish. But what about the points where the derivative vanish? Are those actually branch points or are they just candidates and one needs to check if they really are branch points by using a different method? $\endgroup$
    – user451226
    Commented Jul 30, 2017 at 18:55
  • $\begingroup$ The derivative vanishing means $k > 1$, so a point is a branch point if and only if the derivative vanishes there. $\endgroup$
    – Daniel Fischer
    Commented Jul 30, 2017 at 18:57

1 Answer 1

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For an analytic function in a neighbourhood of $p$, if $f'(p) \ne 0$ then $f$ is injective in a neighbourhood of $p$, while if $f'(p) = 0$ it is not. Indeed,
$f - f(p)$ has a zero of order $m$ at $p$ if and only if $f'$ has a zero of order $m-1$ there (no zero at all in the case $m=1$). For $w$ in a deleted neighbourhood of $f(p)$, the zeros of $f - w$ in a neighburhood of $p$ are all simple, and the number of them is equal to $m$ by the Argument Principle.

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  • $\begingroup$ No, but $f'(p) = 0$ does. $\endgroup$
    – Robert Israel
    Commented Jul 31, 2017 at 2:49

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