Hurwitz Encoding - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:13:52Z http://mathoverflow.net/feeds/question/9415 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9415/hurwitz-encoding Hurwitz Encoding John Mangual 2009-12-20T07:55:29Z 2010-01-14T15:55:46Z <p>In "Random Matrices and Random Permutations" by Okounkov it says, "It is classically known that every problem about the combinatorics of a covering has a translation into a problem about permutations which arise as the monodromies around the ramification points." Apparently, this is called the "Hurwitz encoding" but I never got a clear explanation of how it works.</p> <p>I'd like to understand what it means for a covering of the sphere to be branched and how to identify its ramification points. Perhaps I should learn about how monodromy permutations are computed. Also, how to engineer branched covers which encode problems about the permutation group?</p> http://mathoverflow.net/questions/9415/hurwitz-encoding/9416#9416 Answer by Ryan Budney for Hurwitz Encoding Ryan Budney 2009-12-20T08:25:42Z 2009-12-20T08:32:19Z <p>What do you mean by a covering? </p> <p>It sounds like you're talking about the action of the fundamental group of the base space on $\pi_0$ of the fibre, for a covering space. Or some variant of it for a branched covering space. </p> <p>Take a look in Hatcher's Algebraic Topology textbook in Section 1.3, "Representating Covering Spaces By Permutations". </p> <p>Typically a branched covering space means a map $f : A \to B$ such that there is a co-dimension two subspace $A' \subset A$ such that $f$ when restricted to $A \setminus A' \to B \setminus f(A')$ is a covering space (locally trivial fibre bundle with discrete fibre), and $f$ in a regular neighbourhood of $A' \subset A$ usually satisfies additional constraints. Some authors can be pretty flexible on this. Usually you want a tubular neighbourhood, and the mapping on the level of normal bundles is modelled on the map $S^1 \ni z \longmapsto z^n \in S^1$. </p> http://mathoverflow.net/questions/9415/hurwitz-encoding/9424#9424 Answer by David Speyer for Hurwitz Encoding David Speyer 2009-12-20T15:14:49Z 2010-01-14T15:55:46Z <p><strong>What it means for a covering of a sphere to be branched:</strong> Let $f:X \to Y$ be a map of Riemann surfaces. We are particularly interested in the case that $Y$ is $\mathbb{CP}^1$; in this case, $Y$ has the topology of a sphere. At most points $y$ in $Y$, there will be a neighborhood $V$ of $y$ so that $f^{-1}(V)$ is just a union of $n$ disjoint copies of $U$, each mapping isomorphically to $V$. These are the points where there is no branching.</p> <p>At a few points of $y$, something different will happen. Let $x$ be a preimage of $y$, let $V$ be a small neighborhood of $y$ and let $U$ be the connected component of $f^{-1}(V)$ containing $x$. At these points, the map $f$ looks like $t \mapsto t^e$, as a map from the unit disc in $\mathbb{C}$ to itself. This is called branching. </p> <p>The generic situation is that, at finitely many points of $Y$, one of the preimages is branched with $e=2$ and the other $n-2$ preimages are unbranched.</p> <p>Algebraically, if $t$ is a local coordinate on $X$, then we have branching where $\partial f/\partial t$ vanishes. Even if we can't explicitly write $f$ as a function of $t$, if we can find a polynomial relation $P(f,t)=0$, then we have $(\partial P/\partial f)(\partial f/\partial t) = \partial P/\partial t$, so branching will occur when $\partial P/\partial t=0$.</p> <p><strong>The relation between branched covers and the symmetric group</strong>: Let $f: X \to \mathbb{CP}^1$ be a branched cover. Let $R \subset \mathbb{CP}^1$ be the points over which branching occurs. Then $f^{-1}(\mathbb{CP}^1 \setminus R) \to \mathbb{CP}^1 \setminus R$ is a cover, in the sense of algebraic topology. </p> <p>As you probably know, connected covers of a space $U$ are classified by subgroups of $\pi_1(U)$. In particular, degree $n$ covers are classified by index $n$ subgroups. I like to recast this and say that degree $n$ connected covers of $U$ are classified by transitive actions of $\pi_1(U)$ on an $n$-element set. This has the advantage that, more generally, we can say that degree $n$ covers of $U$ are classified by actions of $\pi_1(U)$ on an $n$-element set.</p> <p>Now, in this case, $\pi_1(\mathbb{CP}^1 \setminus R)$ is isomorphic to the group generated by $R$, modulo the relation $\prod{r \in R} [r]=1$. You should be warned that this isomorphism depends on some choices. First of all, we need to choose a base point $y$ in $\mathbb{CP}^1 \setminus R$! Even once we've done that, we need to choose loops based at $y$, circling each of the elements of $R$, and disjoint from each other away from $y$. These will then be the classes $[r]$. The order that the product above is taken is related to the circular order at which these loops come in to $y$.</p> <p>So, to covers of $\mathbb{CP}^1 \setminus R$ correspond to maps from this group to $S_n$. To give such a map, we choose an element $b(r)$ of $S_n$ for each $r$ in $R$; these must obey $\prod_{r \in R} b(r)=1$. (I am being very sloppy about when two such maps give isomorphic covers, and, indeed, what it means to say two covers are isomorphic.)</p> <p>The relation between the geometry of the cover, and the map $b(r)$ is the following: If the permutation $b(r)$ has cycles of lengths $e_1$, $e_2$, ..., $e_k$, then $f^{-1}(r)$ contains $k$ points, which are branched with degrees $e_1$, $e_2$, ..., $e_k$.</p> <p><strong>Useful facts to know:</strong> The cover is connected if and only if the action of the $b(r)$ on $[n]$ is transitive.</p> <p>The genus of $X$ is given by, the Riemmann-Hurwitz formula: <code>$$2g-2 = -2n+\sum_{r \in R} (n-\#\mbox{cycles of r}).$$</code></p> <p>It is very difficult to obtain an explicit equation for $X$ from the data of $R$ and $b:R \to S_n$.</p> http://mathoverflow.net/questions/9415/hurwitz-encoding/9426#9426 Answer by Alberto García-Raboso for Hurwitz Encoding Alberto García-Raboso 2009-12-20T15:39:03Z 2009-12-20T15:39:03Z <p>To make things concrete, let me consider a holomorphic map of compact Riemann surfaces $f: S' \to s$. That $f$ is a covering of degree $d$ means that every point $p \in s$ has an open neighborhood $\mathcal{U}_p$ such that <code>$$f^{-1}\big( \mathcal{U}_p \big) = \coprod_{i = 1}^d \mathcal{V}_{p,i}$$</code> where the <code>$\mathcal{V}_{p,i}$</code> are disjoint and homeomorphic to $\mathcal{U}_p$. There is a strong restriction to existence of these coverings: the Euler characteristic of $S'$ has to be $d$ times that of $S$; e.g., if $S = \mathbb{CP}^1$, then $\chi(S') = -2d$, i.e., $S'$ consists of $d$ copies of $\mathbb{CP}^1$ itself.</p> <p>You can get a lot more mileage allowing for ramification. What this means is that you only require your map $f$ to be a covering away from a finite set of points. Over these so-called ramification points there are less than $d$ points. The restriction on Euler characteristics now includes a term that accounts for the behavior of your map around these points, and it is known as the Riemann-Hurwitz formula: $$\chi(S') = d \chi(S) + \mathrm{deg} R$$ Here $R$ is the ramification divisor and you can associate to it a nonnegative integer, its degree.</p> <p>The example that you want to keep in mind is $z \mapsto z^2$ on $\mathbb{CP}^1$, which is known as the Riemann surface of the square root function: over any point $z \neq 0$ there are two points, while over the origin there is only one! Actually, the same happens over the point at infinity: the ramification locus consists of the points $0$ and $\infty$; their inverse images are called branching points. You can google for images of "Riemann surface square root" to see a picture of this.</p> <p>Now consider a loop going once around the origin in the base $\mathbb{CP}^1$, fix a lift of the basepoint $p$ and lift the whole loop to the $\mathbb{CP}^1$ upstairs. You'll see that the endpoint of this lift is precisely the other point in the inverse image of $p$. This gives you a homomorphism of $\pi_1(\mathbb{CP}^1 - \lbrace 0, \infty \rbrace) \cong \mathbb{Z}$ to the symmetric group in 2 letters, that takes the generator to the transposition $(12)$.</p> <p>The latter is called the monodromy representation, and generalizing this example you can see that a branched covering of degree $d$ gives you a homomorphism from the fundamental group of the base Riemann surface with the ramification locus deleted to the symmetric group in $d$ letters. Conversely, it can be proved that to such a representation you can associate a branched covering, and that this correspondence is bijective.</p> <p>The book from which I learned all this stuff is Miranda's <em>Algebraic Curves and Riemann Surfaces</em> (chapter III, section 4). You might want to read it and then fish out further references.</p> http://mathoverflow.net/questions/9415/hurwitz-encoding/9437#9437 Answer by Matheus for Hurwitz Encoding Matheus 2009-12-20T18:58:33Z 2009-12-20T18:58:33Z <p>Although it is probably only tangentially related, I believe that the study of square-tiled surfaces has something to do with your question: such surfaces are defined by a finite collection of unit squares together with identifications of parallel sides by translations (under the constraint that the surface we get in this way is connected). Note that square-tiled surfaces are branched corverings of the torus (i.e., the standard unit square with identification of its sides by horizontal and vertical translations) and they can be encoded by two permutations in the following way: after numbering arbitrarily our squares, we can form a permutation h associating to the ith square its right neighbor and a permutation v associating to the ith square its top neighbor. Of course, the combinatorial data (h,v) completely determine your surface and the cycles of the commutator [h,v] give the monodromies around the branching points. </p> <p>Actually, this point of view was used by Eskin and Okounkov in their computation of the volume of moduli spaces of Abelian differentials and you can find further details in their paper. </p> http://mathoverflow.net/questions/9415/hurwitz-encoding/9464#9464 Answer by Greg Kuperberg for Hurwitz Encoding Greg Kuperberg 2009-12-21T07:59:58Z 2009-12-21T07:59:58Z <p>I'd like to add two remarks to David Speyer's great explanation of the Hurwitz encoding of a branched covering. A key question first studied by Hurwitz is when two branched coverings are "the same", in the sense that there is some way to braid the branch points to send one covering to the other covering. To address this question, you start with the obvious invariants, which are called "Hurwitz data". Obviously, each branch point is labelled by the cycle structure of its permutation. Obviously, the permutations of all of the branch points multiply to 1. (At least, these are both obvious once you understand what David said.) Slightly less obviously, the group $G$ generated by the permutations $g_1, g_2, \ldots, g_n$ is another invariant. It is exactly the Galois group of the branched covering, if you either interpret the branched covering as a field extension of $\mathbb{C}(z)$, or if you interpret the branched coverings as a Galois category. In one definition, the Hurwitz data is exactly the choice of $G$ and abstract monodromy elements assigned to the branch points that generate $G$. A geometrical branched covering with sheets is then defined by this data and a choice of a subgroup $H \subset G$ representing an intermediate, irregular covering. (Well, with the extra condition that the normal core of $H$ is trivial, so that the homomorphism from $G$ to permutations is injective.)</p> <p>After you identify these obvious invariants, you might think that it is time to show that the Hurwitz action of the braid group is transitive on coverings with the same Hurwitz data. However, there is another invariant that was anything but obvious in Hurwitz' time, but fairly standard these days. It is easier to see this invariant if you have an unbranched covering with Galois group $G$ of a closed surface $S$. The covering comes from a homomorphism $f:\pi_1(S) \to G$, and the invariant the homology class <code>$f_*([S]) \in H_2(G)$</code>. A branched covering of the sphere gives you a relative version of this invariant; the homology group that it lives in depends on the monodromy elements as well as on the group $G$. In any case it is often a non-trivial invariant.</p>