Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We have the Riemann-Hurwitz formula:

$$ 2g_X-2=d(2g_Y-2)+\sum_{x\in X}(e_x-1) $$

It is said that from this we can deduce that there is no meromorphic function of degree $d=1$ on any compact Riemann surface of positive genus.

I wonder how?

If I let $d=1$, I can get $$ 2(g_X-g_Y)=\sum_{x\in X}(e_x-1) $$

but what's next? Maybe I lack some knowledge about meromorphic function on Riemann surface, anyone can help?

share|cite|improve this question
Can't you also put $g_Y = 0$? – Michael Albanese Jan 9 '13 at 2:29
@MichaelAlbanese, eh, why can I? – hxhxhx88 Jan 9 '13 at 2:47
@hxhxhx88: a meromorphic function on a Riemann surface is the same as a holomorphic map to the Riemann sphere. This is the map you're applying the Riemann-Hurwitz formula to. – Qiaochu Yuan Jan 9 '13 at 3:12
@QiaochuYuan, got it, thank you very much! – hxhxhx88 Jan 9 '13 at 3:33
up vote 2 down vote accepted

Let $f:X\to \mathbb C$ a meromorphic function or equivalently a holomorphic function $f:X\to \mathbb P^1$ . Suppose d=1. Then f is bijective holomophic map and therefore a biholomorphic map. Follow that $g(X)=g(\mathbb P^1)$ because $g$ is a topological invariant.

another way:

Like you said : $$2( g(X)-g(\mathbb P^1))=\sum(e_x-1)$$

but $e_x=1$ for all x because $d=1$ and as we know $g(\mathbb P^1)=0$

Therefore $g(X)=0$

share|cite|improve this answer
@hxhxhx88 helped you? – user52188 Jan 9 '13 at 3:28
yes, your answer is quite helpful, but I'm still having problem: here I need to consider $f$ between two compact Riemann surface $X,Y$, so does such $f$ also equivalent to $X\rightarrow\mathbb{P}^1$? – hxhxhx88 Jan 9 '13 at 3:32
@hxhxhx88 exactly what I said. A meromorphic function take values in $ \mathbb C$ but you can see this like a holomorphic function $f:X\to \mathbb P^1$. – user52188 Jan 9 '13 at 3:58
@hxhxhx88 For me $\mathbb P^1=\mathbb C\cup{}\infty$ – user52188 Jan 9 '13 at 4:04
@hxhxhx88 "here I need to consider f between two compact Riemann surface $X,Y$, so does such f also equivalent to $f:X\to \mathbb P^1$?" The answer is yes! – user52188 Jan 9 '13 at 4:06

The degree of a map between compact Riemann surfaces is known to be a constant.

Recall that, locally around $P\in X$ and $f(P)=Q$, such maps look like $z\mapsto z^e$, where $e=e_P$ is unique.

The degree $d$ being constant is equivalent to the following: $$\sum_{P\mapsto Q} e_P=d.$$

Thus, since $e_P\geq 1$, one cannot have ramification at any point if $d=1$. On the other hand, this means that $f$ is bijective since $e_P=1$ for every point, and this shows that $f$ is biholomorphic (it is so locally since $e_P=1$, and it is globally since $f$ is bijective).

If you want to see a funny example where $d=1$ you may consider the compactification of the plane curve $$y^2=x^2(x+1),$$

which is not a Riemann surface since it has a node at the origin. The map $t\mapsto (t^2-1, t(t^2-1))$ is generically of degree $1$ (where $t=y/x$), but this fails to work at the origin.

Topologically, the target space looks like a sphere with two points identified, which is therefore not a topological manifold.

If $X,Y$ are as in the question, the degree-$1$ case is rather boring (biholomorphisms and that's that).

Thus, in the case where $g(X)$ and $g(Y)$ differ, no such map exists.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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