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Is there an easy classification (and proof) of the possible branched covers between Seifert fibered integer homology spheres which are fiber-preserving and branched over fibers (or at least what the possible relations between the orders of the singular fibers should be)?

There exist branched covers of the form $f:\Sigma(p_1,\ldots,p_k) \to \Sigma(p_1,\ldots,p_{k-1})$ (branched cover over a regular fiber) or $\Sigma(p_1,\ldots,p_k \cdot n) \to \Sigma(p_1,\ldots,p_k)$ (branched cover over the singular fiber of order $p_k$). I think that everything should be something like a composition of maps of this form (or at least the Seifert invariants should be related by such moves). There is a very long and tedious brute force method given by applying the results of Huang (http://www.ams.org/journals/proc/2002-130-08/S0002-9939-02-06306-2/) that seems like it should work, but it also feels like there should be a quick way to prove this fact (or disprove it).

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To clarify, do you count $S^3\to \Sigma(2,3,5)$, the cover of the Poincare homology sphere? This map is fiber preserving, but the $S^1$ action on $S^3$ descends to an $S^1/\{\pm 1\}$ action on $\Sigma(2,3,5)$ since it factors through $\mathbb{RP}^3$. In other words, do you want generic fibers to map 1-1 to generic fibers? –  Ian Agol Jul 25 '13 at 20:36
    
I am only interested in the case that each has infinite fundamental group, but I don't want to require that generic fibers necessarily map 1-1 to generic fibers. In fact, I realized that I can only assume the maps to be fiber-preserving and branched over fibers (and not necessarily equivariant). Sorry for the mistake...will edit the question. –  Tye Lidman Jul 25 '13 at 21:04
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The class of homology sphere Seifert-fibered spaces fiber over 2-orbifolds $S^2(p_1,\ldots,p_k)$, where $\gcd(p_i,p_j)=1, i\neq j$. If one has a fiber-preserving map of one homology sphere to another, then there is a corresponding non-zero degree map between the base orbifolds obtained by quotienting the fibers.

Suppose that the map has degree $d >1$, so we have a degree $d$ map $\phi: S^2(p_1,\ldots,p_k)\to S^2(q_1,\ldots,q_l)$. I claim then that $l\leq 3$.

The maps between the underlying topological spaces of the orbifolds must be a branched cover $\varphi: S^2\to S^2$. Associate branching data to this map as follows. Let there be $r$ critical values of $\varphi$ (branch points). Associated to each branch point $y_i$ is branch data $A_i= (e_{i1},\ldots,e_{in_i})$ a partition of $d$, so $e_{i1}+\cdots+e_{in_i}=d$, and $n_i$ is the cardinality of $\varphi^{-1}(y_i)$. First, note that for each singular point of the orbifold $S^2(q_1,\ldots, q_l)$, the underlying point $y$ of $S^2$ must be a branch point, since otherwise $S^2(p_1,\ldots, p_k)$ would have at least $d$ orbifold points with branching multiple of $q_i$ for some $i$, contradicting that it underlies a homology sphere. So we have $l\leq r$, and we may assume that the first $l$ branch points of $\varphi$ correspond to orbifold points. Now, suppose $y_i$ is a branch point corresponding to an orbifold point of order $q_i$. Then the order of the orbifold singularity of the preimage with branch data $e_{ij}$ must be a multiple of $q_i/\gcd(e_{ij},q_i)$. Thus, we have that $\gcd(q_i/\gcd(e_{ij},q_i),q_i/\gcd(e_{im},q_i))=1$ for $1\leq j\leq m\leq n_i$. In particular, for at most one $j$ does $\gcd(e_{ij},q_i)=1$, and if so, all the rest must be multiples of $q_i$. For example, we might have $A_i=(1,q_i,\ldots,q_i)$, where $1+(n_i-1)q_i=d$. In this case, one may check either $q_i=2$ and $n_i\leq (d+1)/2$, or $n_i\leq d/2$. If there are no preimage branch points of order $1$, then $e_{ij}\geq 2$ for all $j$, and thus $n_i\leq d/2$. We also have from the euler characteristic computation (Hurwitz formula) $\sum_{i,j} (e_{ij}-1) = 2d-2$. Thus, we have $\sum_{i=1}^{l} \sum_{j=1}^{n_i} (e_{ij}-1) \leq 2d-2$. Since $\sum_j e_{ij}=d$, we have $ld-\sum_{i=1}^{l} n_i \leq 2d-2$. But $n_i\leq d/2$, except for one possible branch point which has $q_i=2$ and $n_i\leq (d+1)/2$. Then we get $(l/2-2)d\leq -1$, so $l \leq 3$. So if the degree is $>1$, then one gets stringent requirements on the image orbifold, namely that it has at most 3 branch points. I think some further analysis of the branch data for $l=3$ should rule out this possibility, except for the sort of example of $S^3\to \Sigma(2,3,5)$.

Otherwise, the map $\phi$ has degree 1. In this case, I think one can show that the map is a composition of the sorts of maps you suggest, since one can only decrease the orders of the orbifold points in a degree one branched cover between 2-orbifolds.

Addendum: I didn't work through the case $l=3$ completely, but here's a special case that works out cleanly. We have $\frac{1}{q_1}+\frac{1}{q_2}+\frac{1}{q_3}<1$ (the base orbifold can't be spherical or Euclidean). Suppose all of the branch points over these singularities have at least one branch datum of $1$. Then the branch data must look like $(1,k_2q_i,\ldots,k_{n_i}q_i)$, so $d\geq 1+(n_i-1)q_i$, or $n_i\leq 1+\frac{d-1}{q_i}$. By the Riemann-Hurwitz formula, $$d\leq n_1+n_2+n_3-2 \leq 1+ (d-1)(\frac{1}{q_1}+\frac{1}{q_2}+\frac{1}{q_3}) <d,$$ a contradiction. This case must hold, for example, when all the $q_i$ are prime powers.

The other cases should be even easier, since then the fractions $n_i/d$ should be much smaller, but I haven't gone through the analysis.

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I'm a bit confused. Let $l$ be at least 5. $\Sigma(p_1,\ldots,p_l)$ to $\Sigma(p_1,\ldots,p_{l-1})$ is a degree $p_l$ map. By Proposition 2.1 in Rong's sciencedirect.com/science/article/pii/016686419400108F, this map is a composition of a degree one map (thus $\pi_1$-surjective, and thus goes to a Seifert fibered integer homology sphere), followed by a degree $p_l$ fiber-preserving branched cover branched along fibers necessarily between Seifert integer homology spheres (these are never Euclidean manifolds). But, $l-1$ is at least 4. Sorry, I must be missing something. –  Tye Lidman Jul 26 '13 at 1:38
    
On the base orbifold, the induced map $S^2(p_1,\ldots,p_l)\to S^2(p_1,\ldots,p_{l-1})$ is degree 1. –  Ian Agol Jul 26 '13 at 3:33
    
Oh, I see. The case of degree 1 is actually easily dealt with in the paper of Huang (this is the case that the fiber degree equals the degree). The case of three singular fibers seems a little tedious, but I think it all works now. Thanks! This is much better than the approach I was taking. –  Tye Lidman Jul 26 '13 at 16:30
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