Degree of holomorphic maps between compact Riemann surfaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:59:09Z http://mathoverflow.net/feeds/question/74349 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74349/degree-of-holomorphic-maps-between-compact-riemann-surfaces Degree of holomorphic maps between compact Riemann surfaces zalver 2011-09-02T12:03:08Z 2011-09-02T20:27:10Z <p>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)?</p> http://mathoverflow.net/questions/74349/degree-of-holomorphic-maps-between-compact-riemann-surfaces/74375#74375 Answer by David Speyer for Degree of holomorphic maps between compact Riemann surfaces David Speyer 2011-09-02T16:37:00Z 2011-09-02T20:27:10Z <p>The basic obstruction here is the <a href="http://en.wikipedia.org/wiki/Riemann%25E2%2580%2593Hurwitz_formula" rel="nofollow">Riemann-Hurwitz formula</a>: 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 <code>$$2(g-1) = 2d (h-1) + \sum (e_i -1).$$</code> 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 $g=h$.</p> <p>In the positive direction, one has the <a href="http://en.wikipedia.org/wiki/Riemann_existence_theorem#Riemann.27s_existence_theorem" rel="nofollow">Riemann existence theorem</a>. 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 <code>$X \setminus \{ x_1, \ldots, x_r \}$</code> 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.</p>