1 Result and method

As pointed out by Birman and Hilden [Reference Birman and Hilden2] (see also [Reference Atalan, Medetogullari and Ozan1, Reference Ghaswala and Winarski4, Reference Margalit and Winarski5]), an interesting problem is to find branched covers of a surface with the property that each homeomorphism of the base surface preserving the branching set can be lifted.

Fix a set $\mathfrak {B}=\{\mathfrak {x}_{1},\ldots ,\mathfrak {x}_{n}\}\subset S^{2}$ , with $n\ge 2$ . Let $\Sigma _{0,n}=S^{2}\setminus \mathfrak {B}$ . Let $x_{i}\in H_{1}(\Sigma _{0,n};\mathbb {Z})$ denote the homology class of a small loop enclosing $\mathfrak {x}_{i}$ . Given an abelian group A, a regular A-cover $\pi :\Sigma \to S^{2}$ with branching set $\mathfrak {B}$ is determined by an epimorphism $\phi _{\pi }: H_{1}(\Sigma _{0,n};\mathbb {Z})\twoheadrightarrow A$ , which satisfies $\phi _{\pi }(x_{1})+\cdots +\phi _{\pi }(x_{n})=0$ . Call two such covers $\phi :\Sigma \to S^{2}$ and $\phi ^{\prime }:\Sigma ^{\prime }\to S^{2}$ equivalent if there exist homeomorphisms $\tilde {g}:\Sigma \to \Sigma ^{\prime }$ , $g:S^{2}\to S^{2}$ such that $g(\mathfrak {B})=\mathfrak {B}$ and $g\circ \pi =\pi ^{\prime }\circ \tilde {g}$ .

Denote by $\mathbf {e}_{i}$ the vector with $1$ at the ith position and $0$ elsewhere.

We extend the results of [Reference Atalan, Medetogullari and Ozan1, Reference Ghaswala and Winarski4] by proving the following result.

According to [Reference Atalan, Medetogullari and Ozan1, Section 3], this essentially solves the problem for abelian covers.

Let G denote the group of automorphisms of $H_{1}(\Sigma _{0,n};\mathbb {Z})$ induced by all homeomorphisms of $\Sigma _{0,n}$ . Clearly, G is isomorphic to the permutation group on $\{x_{1},\ldots ,x_{n}\}$ . By [Reference Ghaswala and Winarski4, Lemma 2.1], a homeomorphism f of $\Sigma _{0,n}$ can be lifted if and only if there exists $\psi \in \text {Aut}(A)$ such that $\psi \circ \phi _{\pi }=\phi _{\pi }\circ \alpha $ , where $\alpha \in G$ is induced by f. This is equivalent to $\alpha (\ker \phi _{\pi })=\ker \phi _{\pi }$ , which in turn is equivalent to

$$ \begin{align*} \alpha(\ker\overline{\phi_{\pi}})=\ker\overline{\phi_{\pi}}, \end{align*} $$

where $\overline {\phi _{\pi }}: H_{1}(\Sigma _{0,n};\mathbb {Z}_{p^{k}}\kern-0.75pt)\twoheadrightarrow A$ is the map induced by $\phi _{\pi }$ . Instead of dealing with $\text {Aut}(A)$ , we work directly on $\ker \overline {\phi _{\pi }}$ , reducing the problem to finding all subgroups of $H_{1}(\Sigma _{0,n};\mathbb {Z}_{p^{k}}\kern-0.75pt)$ that are invariant under all $\alpha \in G$ . From this viewpoint, $\pi $ is equivalent to $\pi ^{\prime }$ if and only if $\beta (\ker \overline {\phi _{\pi ^{\prime }}})=\ker \overline {\phi _{\pi }}$ for some $\beta \in G$ .

For problems of this kind, a method was developed in [Reference Chen and Shen3].

Some notations and conventions. For a ring R, let $R^{\ell ,m}$ denote the set of $\ell \times m$ matrices over R. For $X\in R^{\ell ,m}$ , let $X_{i,j}$ denote its $(i,j)$ -entry and let $\langle X\rangle $ denote the subgroup of $R^{m}$ generated by the row vectors of X. For $Y\in \mathbb {Z}^{\ell ,m}$ , abusing the notation, we denote its image under the map $\mathbb {Z}^{\ell ,m}\rightarrow \mathbb {Z}_{p^{k}}^{\ell ,m}$ induced by the quotient map $\mathbb {Z}\twoheadrightarrow \mathbb {Z}_{p^{k}}$ also by Y. Let $S_{m}$ denote the permutation group on m elements. Embed $S_{m}$ as a subgroup of $\textrm{GL}(m,\mathbb {Z})$ by identifying $\sigma \in S_{m}$ with the matrix (denoted by the same notation) whose $(i,j)$ -entry is $\delta _{i,\sigma (j)}$ , where $\delta $ is the Kronecker symbol.

Clearly, $\mathbb {Z}_{p^{k}}^{m}/C\cong \mathbb {Z}_{p^{r_{\iota }}}\times \cdots \times \mathbb {Z}_{p^{r_{m}}}$ , with $\iota =\min \{i\colon r_{i}>0\}$ .

Let $b=n-1$ . Take $x_{1},\ldots ,x_{b}$ as generators for $H_{1}(\Sigma _{0,n};\mathbb {Z})\cong \mathbb {Z}^{b}$ . For each $\alpha \in G\cong S_{b+1}$ , let $T^{\alpha }\in \textrm {GL}(b,\mathbb {Z})$ denote the matrix determined by

$$ \begin{align*} \alpha(x_{i})=\sum_{j=1}^{b}(T^{\alpha})_{i,j}\cdot x_{j}, \quad i=1,\ldots,b. \end{align*} $$

Obviously, if $\alpha \in S_{b}$ , by which we mean $\alpha (x_{b+1})=x_{b+1}$ , then $T^{\alpha }=\alpha $ .

If $\ker \overline {\phi _{\pi }}=\langle PQ\omega \rangle $ , then $\alpha (\ker \overline {\phi _{\pi }})=\langle PQ\omega T^{\alpha }\rangle $ . Taking $\beta =\omega ^{-1}\in S_{b}$ , we have $\beta (\ker \overline {\phi _{\pi }})=\langle PQ\rangle $ . Hence, up to equivalence, we can assume $\ker \overline {\phi _{\pi }}=\langle PQ\rangle $ . Then $\alpha (\ker \overline {\phi _{\pi }})=\langle PQT^{\alpha }\rangle $ for each $\alpha \in G$ . By the criterion given by [Reference Chen and Shen3, Lemma 3.11], $\alpha (\ker \overline {\phi _{\pi }})=\ker \overline {\phi _{\pi }}$ is equivalent to

$$ \begin{align*}p^{r_{j}-r_{i}}\mid(QT^{\alpha}Q^{-1})_{i,j} \quad \text{for all } i<j.\end{align*} $$

Thus, it suffices to find tuples $(r_{1},\ldots ,r_{b};Q)$ consisting of integers $0\le r_{1}\le \cdots \le r_{b}>0$ and a matrix $Q\in \textrm{GL}(b,\mathbb {Z})$ , such that

(1.1) $$ \begin{align} Q_{i,i} = \,1, 1\le i\le b; \quad Q_{j,i}=0\le Q_{i,j}<p^{r_{j}-r_{i}} \quad \text{for all }i<j; \end{align} $$

(1.2) $$ \begin{align} & p^{r_{j}-r_{i}}\mid (QT^{\alpha}Q^{-1})_{i,j} \quad \text{for all }i<j \, \text{ and }\alpha\in S_{b+1}. \end{align} $$

Such a tuple determines a regular branched A-cover $\pi $ , with $A=\mathbb {Z}_{p^{r_{\iota }}}\times \cdots \times \mathbb {Z}_{p^{r_{b}}}$ , $\iota =\min \{i\colon r_{i}>0\}$ , and

(1.3) $$ \begin{align} \phi_{\pi}(x_{i})=\big((Q^{-1})_{i,\iota},\ldots,(Q^{-1})_{i,b}\big), \quad 1\le i\le b, \end{align} $$

with the understanding that $\phi _{\pi }(x_{b+1})=-\phi _{\pi }(x_{1})-\cdots -\phi _{\pi }(x_{b})$ .

2 Proof of Theorem 1.1

As a simple observation, (1.2) holds if and only if

(2.1) $$ \begin{align} p^{r_{j}-r_{i}}\mid(QXQ^{-1})_{i,j} \quad \text{for all }i<j, \end{align} $$

for all X in the subgroup of $\mathbb {Z}^{b,b}$ generated by $T^{\alpha }$ , $\alpha \in S_{b+1}$ .

Let I denote the identity matrix. Let $E_{u}^{v}$ denote the matrix whose $(u,v)$ -entry is $1$ and the other entries are all $0$ .

For the permutation $\eta _{i}\in S_{b+1}$ switching i and $b+1$ ,

$$ \begin{align*}T^{\eta_{i}}=\left(\begin{array}{ccccc} 1 & \ & \ & \ & \ \\ \ & \ddots & \ & \ & \ \\ -1 & \cdots & -1 & \cdots & -1 \\ \ & \ & \ & \ddots & \ \\ \ & \ & \ & \ & 1 \end{array}\right)=I-E_{i}^{i}-\sum\limits_{j=1}^{b}E_{i}^{j}.\end{align*} $$

Clearly, for each $\alpha \in S_{b+1}\setminus S_{b}$ , there uniquely exists $u\in \{1,\ldots ,b\}$ , $\sigma \in S_{b}$ , such that $\alpha =\eta _{u}\sigma $ . Then

(2.2) $$ \begin{align} T^{\sigma}-T^{\alpha}=(I-T^{\eta_{u}})T^{\sigma}=E_{u}^{\sigma^{-1}(u)}+\sum_{j=1}^{b}E_{u}^{j}. \end{align} $$

The difference of two such matrices can give rise to $E_{u}^{v}-E_{u}^{w}$ for any $v\ne w$ .

Taking $i=1$ and $X=E_{1}^{v}-E_{1}^{w}$ in (2.1), we obtain

(2.3) $$ \begin{align} (Q^{-1})_{v,j}\equiv (Q^{-1})_{w,j}\pmod{p^{r_{j}-r_{1}}}. \end{align} $$

In particular, setting $w=j=b-1$ and $v=b$ leads to $r_{b-1}=r_{1}$ (so that $r_{i}=r_{1}$ for all $i<b$ ). By (1.1), $Q_{i,j}=0$ for all $i,j$ with $1\le i<j<b$ .

If $k=r_{b}=r_{1}$ , then $Q_{i,b}=0$ for $1\le i<b$ , so that $Q=I$ . In this case, $A=\mathbb {Z}_{p^{k}}^{b}$ , and by (1.3), $\phi (x_{i})=\mathbf {e}_{i}$ .

Now suppose $k>r_{1}$ . Setting $w=j=b$ in (2.3) leads to

$$ \begin{align*}(Q^{-1})_{v,b}\equiv 1\pmod{p^{r_{b}-r_{1}}}.\end{align*} $$

Hence $Q_{v,b}\equiv -1\pmod {p^{r_{b}-r_{1}}}$ for all $v<b$ . From (2.2), we see that for $i<b$ ,

$$ \begin{align*}(Q(T^{\sigma}-T^{\alpha})Q^{-1})_{i,b}\equiv (\delta_{u,i}-\delta_{u,b})(b+1)\pmod{p^{r_{b}-r_{1}}},\end{align*} $$

so $p^{r_{b}-r_{1}}\mid (Q(T^{\sigma }-T^{\alpha })Q^{-1})_{i,b}$ is equivalent to

$$ \begin{align*} p^{r_{b}-r_{1}}\mid b+1. \end{align*} $$

Once this holds, for each $\sigma \in S_{b}$ and all $i<b$ ,

$$ \begin{align*}(QT^{\sigma} Q^{-1})_{i,b}\equiv\sum_{v=1}^{b}(Q\sigma)_{i,v}\equiv 0\pmod{p^{r_{b}-r_{1}}}.\end{align*} $$

Therefore, (1.2) is fulfilled.

There are two possible cases.

• • If $r_{1}>0$ , then $A=\mathbb {Z}_{p^{r_{1}}}^{b-1}\times \mathbb {Z}_{p^{k}}$ and, by (1.3),

  $$ \begin{align*}\phi(x_{i})=(\mathbf{e}_{i},1), 1\le i\le b-1; \quad \phi(x_{b})=(\mathbf{0},1).\end{align*} $$

• • If $r_{1}=0$ , then $A=\mathbb {Z}_{p^{k}}$ and $\phi (x_{i})=1$ for each i.