I really don't understand how to calculate ramification points for a general map between Riemann Surfaces. If anyone has a good explanation of this, would they be prepared to share it? Disclaimer: I'd prefer an explanation that avoids talking about projective space!

I'll illustrate my problem with an example. The notion of degree for a holomorphic map is not well defined for non-compact spaces, such as algebraic curves in $\mathbb{C}^2$. I've had advice from colleagues not to worry about this and to use the notion of degree anyway, because it works in a different setting (I don't know which). In particular consider the algebraic curve defined by


and the first projection map


In order to find the ramification points of this we know that generically $v_f(x)=1$ and clearly when $z\neq0,\pm 1$ we have $|f^{-1}(z)|=3$ so the 'degree' should be $3$. Thus $z=0,\pm1$ are ramification points with branching order 3. I've had feedback that this is correct. Why did this work?

Now let's look at an extremely similar example. Consider the algebraic curve defined by


and the second projection map


Now again we see the 'degree' of $g$ should be $3$. Now $g^{-1}(i)=\{(1,i),(-1,i)\}$. So by the degree argument exactly one of these is a ramification point, of branching order 2. Is this correct? If so, how do I tell which one it is?

Finally in more generality, does this method work for the projection maps of all algebraic curves in $\mathbb{C}^2$? Sorry for the long exposition!

Edit: Here's an idea I just had. If our map $f$ is proper then we don't need $X$ to be compact for $\deg(f)$ to be well defined. Now the projection map is clearly proper (I think) so that's why this works. Am I right? This of course raises the natural question - 'what standard maps are proper'? I guess I should ask this in a separate question though!

  • $\begingroup$ You mean ramification index 3, for your first example. $\endgroup$
    – Zhen Lin
    May 29, 2012 at 9:35
  • $\begingroup$ Thanks - I've corrected it. $\endgroup$ May 29, 2012 at 9:52
  • $\begingroup$ Your algebraic curves are in $\mathbb C^2$, not $ \mathbb C$. $\endgroup$ May 29, 2012 at 10:02
  • $\begingroup$ Thanks - have corrected that too. $\endgroup$ May 29, 2012 at 10:36
  • $\begingroup$ The projection map $(z, w) \mapsto w$ is not always proper: consider the affine hyperbola $z w - 1 = 0$, and take any the preimage of any compact neighbourhood of $0$. $\endgroup$
    – Zhen Lin
    May 29, 2012 at 11:04

1 Answer 1


Let's look at your second example. Let $p(z, w) = w^2 - z^3 + z^2 + z$, and let $Y = \{ p(z, w) = 0 \}$. Then, $$p(z, i) = -z^3 + z^2 + z - 1 = -(z - 1)^2 (z+1)$$ so I claim $(1, i)$ has ramification index $2$ while $(-1, i)$ has ramification index $1$. Indeed, observe that \begin{align} \frac{\partial p}{\partial z} & = -3 z^2 + 2 z + 1 \\ \frac{\partial p}{\partial w} & = 2 w \end{align} so by an inverse function theorem argument, we find that $(z, w) \mapsto z$ is locally a chart near both $(-1, i)$ and $(1, i)$. In this chart, your function $g : Y \to \mathbb{C}$ is given by $z \mapsto \sqrt{z^3 - z^2 - z}$. Let us take Taylor expansions around $\pm 1$: \begin{align} g(z) - i & = -i (z-1)^2 + O((z-1)^3) \\ g(z) - i & = -2 i (z+1) + O((z+1)^2) \end{align} Hence, the ramification index at $(1, i)$ is indeed $2$ and at $(-1, i)$ it is $1$.

Morally, what is going on is that your curves are dense open subsets of projective curves. Indeed, your first curve is given in homogeneous coordinates by $$w^3 - z (z^2 - u^2) = 0$$ and your second curve is given by $$w^2 u - z^3 + z^2 u + z u^2 = 0$$ and one can check by hand that these curves are smooth "at infinity", so we have the desired embedding of the original affine algebraic curves into projective (hence compact) algebraic curves. Degree is well-defined on the latter, so is well-defined on the former by restriction; the only trouble is that there may be "missing" preimages and so the equation relating degrees and ramification indices becomes an inequality: $$\text{deg}(g) \ge \sum_{x \in g^{-1} \{w\}} \nu_x (g)$$ For example, take the affine hyperbola $z w - 1 = 0$ and the projection $(z, w) \mapsto w$; this function has degree $1$ (once we embed it in the projective closure), but obviously there are no preimages of $0$ in the affine hyperbola.

Let's develop a generic method of dealing with affine plane curves. Let $p : \mathbb{C}^2 \to \mathbb{C}$ be a polynomial function in two variables, and suppose $Y = \{ p(z, w) = 0 \}$ is a smooth algebraic curve. Let $f : Y \to \mathbb{C}$ be the projection $(z, w) \mapsto w$. For each fixed complex number $b$, we get a polynomial function $p(-, b)$, say of degree $d$. Now, because $\mathbb{C}$ is algebraically closed, we can write $$p(z, b) = c (z - a_1)^{e_1} \cdots (z - a_n)^{e_n}$$ for some distinct complex numbers $a_1, \ldots, a_n$, $c \ne 0$, and positive integers $e_1, \ldots, e_n$, such that $e_1 + \cdots + e_n = d$. Suppose also that $$\frac{\partial p}{\partial w}(a_i, b) \ne 0$$ for all $a_i$; then an inverse function theorem argument shows that $(z, w) \mapsto z$ is a chart near each $(a_i, b)$. I claim that the ramification index of $f$ at $(a_i, b)$ is $e_i$ under these hypotheses. Indeed, when $z$ is a local parameter, we have $$0 = \frac{\partial p}{\partial z} + \frac{\mathrm{d} w}{\mathrm{d} z} \frac{\partial p }{\partial w}$$ so if $e_i > 1$, we have $\frac{\partial p}{\partial z} (a_i, b) = 0$, so we must have $\frac{\mathrm{d} w}{\mathrm{d} z} (a_i) = 0$ because $\frac{\partial p}{\partial w} (a_i, b) \ne 0 $ by hypothesis – implying $f(z) - b = O((z - a_i)^2)$. Playing around with total derivatives more, we eventually find that the first non-zero coefficient of $f(z) - b$ around $a_i$ is the coefficient of $(z - a_i)^{e_i}$, as required.

On the other hand, when we have $\frac{\partial p}{\partial w} (a_i, b) = 0$, then by non-degeneracy we must have $\frac{\partial p}{\partial z} (a_i, b) \ne 0$, and we must have $e_i = 1$ and $(z, w) \mapsto w$ is a chart near $(a_i, b)$. But then obviously the ramification index of $f$ at $(a_i, b)$ must be $1$. So in either case the ramification index of $f$ at $(a_i, b)$ is equal to $e_i$. Convenient, no?

  • $\begingroup$ Thank you very much - this is fantastic! Wish I could upvote it 10 times! $\endgroup$ May 29, 2012 at 11:06
  • $\begingroup$ To analyse (-1,i) could I simply observe that $\partial p/\partial z(-1,i)\neq 0$ so by an implicit function theorem argument $g$ is locally represented by $\textrm{id}$ and hence has ramification index $1$? $\endgroup$ May 29, 2012 at 11:15
  • $\begingroup$ Yes, exactly. But this doesn't happen often! $\endgroup$
    – Zhen Lin
    May 29, 2012 at 11:20
  • $\begingroup$ Okay, great. Think I understand now. For the purposes of explicit calculations (as I imagine will crop up in exams next week) it's a useful thing to spot! $\endgroup$ May 29, 2012 at 11:24
  • 8
    $\begingroup$ Since the OP wants to upvote Zhen several times, I have provided one (well-deserved) upvote. $\endgroup$ May 29, 2012 at 11:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .