MathOverflow is a question and answer site for professional mathematicians. 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

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)?

share|cite|improve this question
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
up vote 2 down vote accepted

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.

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.