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)?



Degree of holomorphic maps between compact Riemann surfaces
The basic obstruction here is the [Riemann-Hurwitz formula](http://en.wikipedia.org/wiki/Riemann%25E2%2580%2593Hurwitz_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](http://en.wikipedia.org/wiki/Riemann_existence_theorem#Riemann.27s_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.


