# Degree of holomorphic maps between compact Riemann surfaces

Can all nonzero degree map between compact Riemann surfaces (both genus >1 ) be deformed to holomorphic maps, if we can change the conformal structures on them? The simplest case: does there exist holomorphic map of degree one from \Sigma_m to \Sigma_n (m>n>1)?

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The genus is invariant for biholomorphisms... –  Francesco Polizzi Sep 2 '11 at 12:32
The map, if exists, is ofc not biholomorphic but has some branch points:) –  zalver Sep 2 '11 at 13:01
What is "a map"? A continuous function? –  Igor Rivin Sep 2 '11 at 13:50
Re the first question: if $f$ is a smooth map between two 2-manifolds such that all singular points (= points where the differential is not bijective) of $f$ are isolated, then, given a complex structure on the target there is a complex structure on the source that makes $f$ holomorphic. –  algori Sep 2 '11 at 15:15
I think the answer by Francesco is correct. The answer for is there a holomorphic map of degree $1$ $f:\Sigma _m\rightarrow \Sigma _n$ is No when $m\not= n$, to see this just apply the universal property of blowups: By the assumptions, the preimage of a point is just a point (even though we may have critical points), and thus the universal property of blowups says that $\Sigma _m$ is biholomorphic to a blowup of $\Sigma _n$, but the latter is just $\Sigma _n$. –  anonymous Sep 2 '11 at 16:04

The basic obstruction here is the Riemann-Hurwitz formula: If there is a degree $d$ map from a Riemann surface of genus $g$ to one of genus $h$, with branch points of orders $e_1$, $e_2$, ..., $e_r$, then $$2(g-1) = 2d (h-1) + \sum (e_i -1).$$ As a corollary, $2(g-1) \geq 2d (h-1)$. This has the standard corollaries: One always has $g \geq h$. If $g=h \geq 2$ then $d=1$. And, combined with the fact that a map of degree $1$ can have no braching, if $d=1$ then $g=h$.
In the positive direction, one has the Riemann existence theorem. Given a Riemann surface $X$ of genus $g$, fix a finite number of points $x_1$, ..., $x_r$ on $X$. Give a $d$-fold covering of $X \setminus \{ x_1, \ldots, x_r \}$ which completes topologically to a branched covering of $X$ by a surface $Y$ of genus $h$. Then there is a Riemann structure on $Y$ such that $Y \to X$ is a map of Riemann surfaces. So, if you can make your map look like a branched covering, then it will come from an actual holomorphic map, and you can even specify the downstairs holomorphic stucture in advance.