# Why there is no meromorphic function of degree $d=1$ on any compact Riemann surface of positive genus?

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?

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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

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 $g(X)=0$

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@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.

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