Irrational Numbers and the Riemann Surface of a Multi-Valued Function - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T01:49:22Z http://mathoverflow.net/feeds/question/32697 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32697/irrational-numbers-and-the-riemann-surface-of-a-multi-valued-function Irrational Numbers and the Riemann Surface of a Multi-Valued Function David Corwin 2010-07-20T21:09:11Z 2011-01-06T04:41:46Z <p>Suppose a meromorphic function $f(z)$ has two poles, with residues $1$ and $\gamma$, respectively. Then the topology of the Riemann surface of the anti-derivative of $f(z)$ depends on whether or not $\gamma$ is irrational. More generally, the topology of a meromorphic function with $n$ poles with residues $\gamma_1,\gamma_2,\cdots,\gamma_n$ depends on the linearity of $\gamma_1,\cdots,\gamma_n$ over $\mathbb{Q}$. Has anyone considered this phenomenon? Are there relationships between the (co)homology groups of the covering and the residues? Could one attempt to prove the irrationality (or rationality) of a given complex number by considering the residues of the poles of a meromorphic function in this way? (In such a case, one would need other ways of extracting topological information about the given meromorphic function.)</p> http://mathoverflow.net/questions/32697/irrational-numbers-and-the-riemann-surface-of-a-multi-valued-function/37102#37102 Answer by Daniel Litt for Irrational Numbers and the Riemann Surface of a Multi-Valued Function Daniel Litt 2010-08-30T00:07:35Z 2011-01-06T04:41:46Z <p>EDIT 1/5/2010: I was a little dissatisfied with the informality of what I wrote below, so <a href="http://math.stanford.edu/~dlitt/exposnotes/IrrationalRiemann.pdf" rel="nofollow">here is a somewhat more formal writeup</a>. The statement is somehow geometrically obvious, but the proof is still a bit nice.</p> <p>EDIT: I initially claimed the sequence below is short exact, which is false; it is right exact. It is fixed below, with some explanation, and a bit of a geometric explanation of what's going on.</p> <p>I know little about the theory of linear independence over $\mathbb{Q}$, but I'll attempt an answer to this part of the question:</p> <blockquote> <p>Are there relationships between the (co)homology groups of the covering and the residues?</p> </blockquote> <p>The answer is "yes." Let $f$ be a meromorphic function on $\mathbb{C}$, and for convenience let's assume that it has poles $z_1, ..., z_n$ of order $1$, and no other singularities. Let $g$ be an antiderivative of $f$. Then the Riemann surface of $f$ is <code>$M_f=\mathbb{C}-\{z_1, ..., z_n\}$</code> and the Riemann surface of $g$, which we will denote by $M_g$, is a covering space of $M_f$ with covering map $\pi: M_g\to M_f$. Let $V\subset \mathbb{C}$ be the $\mathbb{Q}$-vector space spanned by $\operatorname{Res}_{z_1}(f), ..., \operatorname{Res}_{z_n}(f)$. Then there is a short exact sequence $$H_1(M_g, \mathbb{Q})\overset{H_1(\pi)}{\longrightarrow} H_1(M_f, \mathbb{Q})\overset{\int}{\longrightarrow} V\to 0,$$ where the map $\int$ is given as follows. Namely, $\int: H_1(M_f, \mathbb{Q})\to V$ is given by $[\gamma]\mapsto \frac{1}{2\pi i}\int_\gamma f~\operatorname{dz}$.</p> <p>Let's elucidate the connection to the linear independence of $\operatorname{Res}_{z_1}(f), ..., \operatorname{Res}_{z_n}(f)$ over $\mathbb{Q}$. $H_1(M_f, \mathbb{Q})$ is a $\mathbb{Q}$-vector space with a basis of cycles $[\lambda_1], ..., [\lambda_n]$ corresponding to the punctures $z_1, ..., z_n$. Then the map $\int$ sends $[\lambda_i]$ to $\operatorname{Res}_{z_i}(f)$. So the image of $H_1(M_g, \mathbb{Q})$ in $H_1(M_f, \mathbb{Q})$ is precisely the vector space of relations between the residues of $f$.</p> <p>Added: We can extend this right exact sequence into a longer sequence. In particular, by covering space theory we have that $\pi_1(M_g)\to \pi_1(M_f)$ is an injection. It is easy to see that the commutator subgroup of $\pi_1(M_f)$ is contained in the image of $\pi_1(M_g)$. By the Hurewicz theorem $$H_1(M_g, \mathbb{Q})\simeq \pi_1(M_g)^{Ab}\underset{\mathbb{Z}}{\otimes} \mathbb{Q}.$$ So the kernel of the map $H_1(M_g, \mathbb{Q})\to H_1(M_f, \mathbb{Q})$ is given by the image of $[\pi_1(M_f), \pi_1(M_f)]$ (which is contained in $\pi_1(M_g)$) in $H_1(M_g, \mathbb{Q})$. One can extend the exact sequence further back by looking at quotients of commutators in this manner.</p> <p>This first extension has a geometrical interpretation. Namely, let $h$ be a meromorphic function whose poles have the same locations as those of $f$, but whose residues are linearly independent over $\mathbb{Q}$. Then the antiderivative of $h$, denoted $s$ has Riemann surface $M_s$, which is a covering space over $M_f$, with covering map $\pi': M_s\to M_f$. By the properties of covering spaces, $\pi'$ factors through $\pi$, and it is not hard to see that $\pi_1(M_s)$ is exactly the commutator subgroup of $\pi_1(M_f)$. Then the sequence $$H_1(M_s, \mathbb{Q})\overset{H_1(\pi')}{\longrightarrow} H_1(M_g, \mathbb{Q})\overset{H_1(\pi)}{\longrightarrow} H_1(M_f, \mathbb{Q})\overset{\int}{\longrightarrow} V\to 0,$$ is exact, and coincides with the sequence described above.</p> <p>I don't know if the continuing left extensions of this sequence have similar geometric interpretations. Also, it would be nice to have a naturally arising description of this sequence, rather than the somewhat ad hoc one I've given.</p>