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Hurwitz's encoding counts the number of branched self-coverings of a sphere, with prescribed ramification degrees at the critical points, as numbers of factorizations of the identity in a symmetric group with given cycle lengths. My question is:

Is there a classification of all "Hurwitz data" (namely, degrees at critical points) for which the covering is determined uniquely?

For example, if the branch data are {d,d}, then the map has to be $z^d$, up to Möbius transformations. A few other cases I found out are:

  • if the data are {d,m,(d+1-m)}, then the map has to be $\int z^{m-1}(1-z)^{d-m}$;
  • if the data are {d,m+(d-m),2}, then the map has to be $z^m(1-z)^{d-m}$;
  • if the data are {n+n,2+...+2,2+...+2}, then the map has to be $z^n/(1+z^n)^2$;
  • if the data are {m+n,m+n,3}, the map is $z^m((m-n)z-(m+n))^n/((m+n)z+(m-n))^n$.

On the other hand, if all critical values are simple (so the data is {2,2,...,2}), then there are exponentially many branched coverings (something like a Catalan number).

I'm aware of various combinatorial tools to compute the number of coverings, including "integration on the Deligne-Mumford stack", but all the literature I was able to google up was concerned about cases where there are no branched coverings, or lots.

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    $\begingroup$ The Hurwitz data as you have given them were not clear to me. I think they make sense to me now that I understand that "+" is serving as the character separating the parts in your partitions, and you are omitting parts equal to 1. $\endgroup$
    – Hugh Thomas
    Commented Nov 2, 2010 at 18:34
  • $\begingroup$ Yes, sorry, I should have been more explicit in my introduction. One starts with a list of partitions of d, and asks, using Hurwitz's reformulation, for the number of (lists of permutations, with cycle structure given by the partitions, whose product is 1 and which generate a transitive subgroup of Sym(d)) / diagonal action of Sym(d). $\endgroup$
    – grok
    Commented Nov 3, 2010 at 13:07
  • $\begingroup$ A simple linear algebra argument shows that, if there are 3 critical points, then the map f is determined by linear relations so is unique. One case I haven't been able to put into a family is the data {2+3,2+3,2+2}. $\endgroup$
    – grok
    Commented Nov 4, 2010 at 1:03

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