Covers of surfaces

Let $\pi:\tilde{Y}\to S$ be the cover corresponding to $\pi_{1}(Y)$ and observe that the inclusion $Y\to S$ lifts to an embedding $Y\to\tilde{Y}$, which induces an isomorphism $\pi_{1}(Y)\to\pi_{1}(\tilde{Y})$. Therefore, any loop in $\tilde{Y}$ is a homotopic to one in the interior of $Y$. This implies that each boundary component of $Y$ is separating—otherwise, there is an essential loop of $\tilde{Y}$ which essentially intersects a component of $\partial Y$. Further, each component of $\tilde{Y}\smallsetminus Y$ is topologically a product. This shows $\tilde{Y}\cong Y\cup(\partial Y\times[0,\infty))$. ∎

Note that any finite-type surface can be identified as the interior of some compact surface $S$. Further, a geometrically characteristic cover of the interior of $S$ is the interior of a geometrically characteristic cover $\tilde{S}\to S$. Thus, the goal is to show that the interior of $\tilde{S}$ is one of the four surfaces listed.

Let $Q$ be the associated deck group of $\tilde{S}$. First assume that $\tilde{S}$ has no compact boundary components, so $\text{int}(\tilde{S})=\tilde{S}_{\circ}$. As $Q$ is infinite and acts cocompactly on $\tilde{S}$, the genus of $\text{int}(\tilde{S})$ is either $0$ or infinity, so the classification of surfaces (Theorem 2.2) implies that it is either the disc or the Loch Ness monster surface. Now, suppose that $\tilde{S}$ has a compact boundary component. As $Q$ is infinite, $\tilde{S}$ has infinitely many compact boundary components. Gluing discs to each compact boundary component, we obtain a surface without boundary, call it $Y$. By Proposition 3.1, $Y$ has a single end, and again the classification of surfaces (Theorem 2.2) implies that $Y$ is either the disc or the Loch Ness monster surface. To finish, observe that $\text{int}(\tilde{S})$ is obtained from $Y$ by puncturing all the infinitely many discs, so then $\text{int}(\tilde{S})$ is either the flute surface or the spotted Loch Ness monster surface. ∎

Let $\bar{\iota}:\mathcal{E}(\tilde{S})\to\mathcal{E}(\tilde{S}_{\circ})$ be defined as follows: given $\xi\in\mathcal{E}(\tilde{S})$, take a sequence of points $x_{i}\in\tilde{S}_{0}\subset\tilde{S}$ with $x_{i}\to\xi$, and define $\bar{\iota}(\xi)$ to be the limit in $\mathcal{E}(\tilde{S}_{\circ})$. It is readily verified that $\bar{\iota}$ is a well-defined continuous surjection.

As the action of $Q$ on $\tilde{S}$ is co-compact and proper, the end space of $Q$, denoted $\mathcal{E}(Q)$, is homeomorphic to $\mathcal{E}(\tilde{S})$. If $\tilde{S}_{\circ}$ has more than one end, then there is a non-constant continuous function $\phi:\mathcal{E}(\tilde{S}_{\circ})\to\mathbb{Z}$. Consider the composition

Pick cyclic subgroups of $\pi_{1}(S)$ representing the boundary components, and let $H_{j}$ denote their projections to $Q$. For any right coset $H_{j}g$, the intersection $\overline{H_{j}g}\cap\mathcal{E}(Q)$ is either empty (if $H_{j}$ is finite, which happens when that boundary component lifts to a closed curve in $\tilde{S}$) or projects to the endpoints in $\mathcal{E}(\tilde{S})$ of some non-compact boundary component of $\tilde{S}$, both of which (if they are distinct) map to the same element of $\mathcal{E}(\tilde{S}_{\circ})$.

We have established that $\phi$ is non-constant but that it is constant on $\overline{H_{j}g}\cap\mathcal{E}(Q)$ for each $j$ and each right coset $H_{j}g$. Therefore, by Theorem 2.6 there exists an action of $Q$ on a tree $T$, with no edge inversions and finite edge stabilizers, and no global fixed point, such that all the subgroups $H_{j}$ act elliptically. Every peripheral element of $\pi_{1}(S)$ projects to an element of $Q$ that is conjugate into one of the $H_{j}$’s and hence acts elliptically on $T$ as desired. ∎

We begin by constructing a prototype for the map $\tilde{f}$, which we call $\tilde{f}_{1}$. Following Shalen [14], we construct the $Q$-equivariant map $\tilde{f}_{1}\mskip 0.5mu\colon\thinspace\tilde{S}\to T$ as follows. Fix a triangulation $\tau$ of $S$, and consider its associated $Q$-invariant triangulation $\tilde{\tau}$ of $\tilde{S}$. Let $\tilde{X}_{0}\subset T^{(0)}$ contain a single representative for each orbit equivalence class of $\tilde{\tau}^{(0)}$, and pick a map

that is arbitrary, except that for every component $\beta\subset\partial\tilde{S}$, we require that $\tilde{h}^{(0)}$ takes $\tilde{X}_{0}\cap\beta$ to a single vertex of $T$ that is fixed by the stabilizer $Q_{\beta}\subset Q$; this can be done as $Q_{\beta}$ is the projection to $Q$ of a peripheral $\mathbb{Z}$-subgroup of $\pi_{1}(S)$, so acts with a global fixed point on $T$ by assumption. (Note that $Q_{\beta}$ must fix a vertex, since it acts simplicially without edge inversions.)

There is a unique $Q$-equivariant extension of $\tilde{h}^{(0)}$ to a map

The map $\tilde{f}_{1}^{(0)}$ is readily extended to a piecewise linear map $\tilde{f}_{1}\mskip 0.5mu\colon\thinspace\tilde{S}\to T$ (see Figure 1). Observe that, under this piecewise linear extension, each boundary component $\beta$ of $\partial\tilde{S}$ is mapped to a point by $\tilde{f}_{1}$, as its vertices are all mapped to the same point by construction.

Let $Z$ be the set of all edge midpoints of $T$, let $\tilde{C}_{1}=\tilde{f}_{1}^{-1}(Z)$, and let $C_{1}=\pi(C_{1})$. We claim that each component $\tilde{C}_{1}$ (resp., $C_{1}$) is a simple closed curve. By the construction of $\tilde{f}_{1}$, we know that the components of $\tilde{C}_{1}$ form a locally finite family of pairwise-disjoint 1-submanifolds of $\tilde{S}$. As $\pi$ is a local homeomorphism and $\tilde{C}_{1}$ is invariant under the action of $Q$, we have that $C_{1}$ is a compact 1-submanifold of $S$; in particular, each component of $C_{1}$ is a simple closed curve. Now, let $\tilde{c}$ be a component of $\tilde{C}_{1}$. Then $\pi(\tilde{c})$ is compact, and as the action of $Q$ on $T$ has finite edge stabilizers, the stabilizer of $\tilde{c}$ is finite, implying that $\tilde{c}$ is compact. Therefore, $\tilde{C}_{1}$ is a union of simple closed curves. We now need to edit $\tilde{f}_{1}$ so that no component of $\tilde{C}_{1}$ nor $C_{1}$ is peripheral or bounds a disk.

First, suppose that there is a component of $C_{1}$ that bounds a disk. Let $\Delta_{1},\ldots,\Delta_{k}$ be the outermost disks bounded by components of $C_{1}$, so that if $c$ is a component of $C_{1}$ bounding a disk, then there exists a unique $j\in\{1,\ldots,k\}$ such that $c\subset\Delta_{j}$. Choose a lift $\tilde{\Delta}_{j}$ in $\tilde{S}$ for each of the $\Delta_{j}$, and for $q\in Q$, let $\tilde{\Delta}_{j}^{q}=q\cdot\tilde{\Delta}_{j}$. Let $e_{j}^{q}$ be the edge with vertices $v_{j}^{q},w_{j}^{q}\in T$ whose midpoint is $\tilde{f_{1}}(\tilde{\partial}\Delta_{j}^{q})$. By the construction of $C_{1}$, all the $1$-cells in $\tilde{\tau}$ that intersect $\tilde{\partial}\Delta_{j}^{q}$ are mapped to segments in $T$ that all contain the edge $e_{j}^{q}$. If the $0$-cells in $\tilde{\Delta}_{j}^{q}$ are mapped into the $v_{j}^{q}$ side of $T\smallsetminus\{e_{j}^{q}\}$, we can define a new map

that maps the 0-cells in $\Delta_{j}^{q}$ to $w_{j}^{q}$, but agrees with $\tilde{f}_{1}^{(0)}$ otherwise. We can now extend $\tilde{f}_{2}^{(0)}$ to a $Q$-invariant piecewise linear map $\tilde{f}_{2}\mskip 0.5mu\colon\thinspace\tilde{S}\to T$ in such a way that $\tilde{f}_{2}^{-1}(z)=\tilde{f}_{1}^{-1}(z)$ for any $z\in Z\smallsetminus(\bigcup\tilde{\Delta}_{j}^{q})$. Setting $\tilde{C}_{2}=\tilde{f}_{2}^{-1}(Z)$ and $C_{2}=\pi(\tilde{C}_{2})$, we have that $C_{2}\subset C_{1}$ and no component of $C_{2}$ bounds a disk.

Now, suppose that $C_{2}$ contains a peripheral component. Then each peripheral component of $C_{2}$ bounds an annulus with a component of $\partial S$. Let $A_{1},\ldots,A_{m}$ be the outermost annuli, so that if $c$ is a component of $C_{2}$ bounding an annulus with a component of $\partial S$, then there exists a unique $j\in\{1,\ldots,m\}$ such that $c\subset A_{j}$. Each lift of an $A_{j}$ is again an annulus in $\tilde{S}$ as the component of $\tilde{C}_{2}$ are compact. We now proceed identically as we did above with the disks $\Delta_{j}$ replaced with the annuli $A_{j}$. The result is a $Q$-equivariant map $\tilde{f}\mskip 0.5mu\colon\thinspace\tilde{S}\to T$ so that $\tilde{C}:=\tilde{f}^{-1}(Z)$ and $C:=\pi(\tilde{C})$ have neither peripheral nor inessential components. ∎

Since the action of the deck group $Q$ produced by Lemma 3.2 has no global fixed point, the image of the map in Proposition 3.3 has to contain an edge midpoint, the $1$-submanifold $C\subset S$ it produces is non-empty. Take $\gamma$ to be any component of $C$. Any component of $\pi^{-1}(\gamma)$ is a component of $\tilde{C}$, and hence is compact. ∎

It suffices to show that if some component of $\tilde{S}_{\circ}\mathbin{{\smallsetminus}\mspace{-5.0mu}{\smallsetminus}}\pi^{-1}(C)$ has more than one end, then there is an essential, nonperipheral simple closed curve $\gamma\subset S\mathbin{{\smallsetminus}\mspace{-5.0mu}{\smallsetminus}}C$ such that every component of $\pi^{-1}(\gamma)\subset\tilde{S}$ is compact. For then we can add such curves recursively to $C$ to produce $D$ as required. Since the number of isotopy classes in a collection of pairwise disjoint curves on $S$ is bounded, this process terminates.

So, let $\Sigma$ be a component of $S\mathbin{{\smallsetminus}\mspace{-5.0mu}{\smallsetminus}}C$, and assume that we have a component $\tilde{\Sigma}\subset\tilde{S}\mathbin{{\smallsetminus}\mspace{-5.0mu}{\smallsetminus}}\tilde{C}$ such that $\tilde{\Sigma}_{\circ}$ has more than one end. We claim that the restriction of $\pi$ to $\tilde{\Sigma}$ is a regular covering map $\tilde{\Sigma}\to\Sigma.$ Indeed, if two points $p,q\in\tilde{\Sigma}$ have the same projection to $\Sigma$, then there is a deck transformation $g$ of $\tilde{S}\to S$ with $g(p)=q$, and since $\tilde{C}$ is invariant under the deck group, we have $g(\tilde{\Sigma})=\tilde{\Sigma}$, implying $g$ restricts to a deck transformation of $\tilde{\Sigma}\to\Sigma$ taking $p$ to $q$.

Corollary 3.4 then says that there is some essential, non-peripheral curve in $\gamma\subset\Sigma$ such that every component of $\pi|_{\tilde{\Sigma}}^{-1}(\gamma)$ is compact. But since the cover $\pi:\tilde{S}\to S$ is regular, this implies that every component of $\pi^{-1}(\gamma)\subset\tilde{S}$ is compact, so we are done. ∎

We observe that if $\alpha$ is a separating curve in $S$, such that both sides of $A$ and $B$ of $S\smallsetminus\alpha$ are not precompact, then $\alpha$ can be extended to a free basis for $\pi_{1}(A)$ and $\pi_{1}(B)$. By Van Kampen’s theorem, $\pi_{1}(S)=\pi_{1}(A)*\pi_{1}(B)$. It is not hard to see from this that any basis for $\pi_{1}(Y)$ can be extended to a basis for $\pi_{1}(S)$. ∎

Let $N=\pi_{*}(\pi_{1}(\tilde{S}))$. As $N$ fails to be locally finite index in $\pi_{1}(S)$, there exists a finitely generated group $H<\pi_{1}(S)$ such that $[H:H\cap N]$ is infinite. Using the fact that $H$ is finitely generated, we can find a free factor $A$ of $\pi_{1}(S)$ containing $H$; let $R$ denote the rank of $A$. Let $Y\subset S$ be a $\pi_{1}$-injective subsurface so that $\pi_{1}(Y)$ contains a free factor $B$ of $\pi_{1}(S)$ of rank $R$. Then there exists an automorphism $\varphi$ of $\pi_{1}(S)$ such that $\varphi(A)=B$. As $N$ is characteristic, $\varphi(N)=N$, implying that

In particular, as $\varphi(H)<\pi_{1}(Y)$, we have that the index of $\pi_{1}(Y)\cap N$ in $\pi_{1}(Y)$ is infinite. Given a component $\tilde{Y}$ of $\pi^{-1}(Y)$, the restriction of $\pi$ to $\tilde{Y}$ is a covering $\tilde{Y}\to Y$ with corresponding subgroup $\pi_{1}(Y)\cap N$, and hence $\tilde{Y}$ is an infinite-sheeted cover of $Y$, implying it is noncompact. ∎

Let $\varphi\in\mathrm{Aut}(\pi_{1}(Y))$. By Lemma 3.6, $\pi_{1}(Y)$ is a free factor of $\pi_{1}(S)$, and so there exists $B<\pi_{1}(S)$ such that $\pi_{1}(S)$ is isomorphic to $\pi_{1}(Y)*B$. We can therefore extend $\varphi$ to an automorphism $\hat{\varphi}$ of $\pi_{1}(S)$ by declaring $\hat{\varphi}(b)=b$ for all $b\in B$ and $\hat{\varphi}|_{\pi_{1}(Y)}=\varphi$. So,

Hence, $N\cap\pi_{1}(Y)$ is characteristic in $\pi_{1}(Y)$. ∎

Let $S$ be an infinite-type surface, and let $\bar{S}$ denote the surface obtained by removing an open annular neighborhood of each isolated planar end of $S$. That is, $\bar{S}$ is obtained by replacing the isolated planar ends of $S$ with compact boundary components.

Let $\pi\mskip 0.5mu\colon\thinspace\tilde{S}\to\bar{S}$ be an infinite-sheeted characteristic cover. We will prove that $\tilde{S}_{\circ}$ is one ended (recall that $\tilde{S}_{\circ}$ is obtained by deleting the non-compact boundary components of $\tilde{S}$). Note that we are actually interested in the topology of $int(\tilde{S})$; however, if $\tilde{S}_{\circ}$ is one ended, then the result is a consequence of the classification of surfaces.

Let $K\subset\tilde{S}_{\circ}$ be a compact subset. By possibly enlarging $K$, we can assume that no component of $\tilde{S}_{\circ}\smallsetminus K$ is precompact. Our goal is to show $\tilde{S}_{\circ}\smallsetminus K$ is connected.

Let $R$ be the constant of Lemma 3.7. We can exhaust $\bar{S}$ by compact subsurfaces whose boundary components are separating and whose complementary components are not precompact. Thus, we may find two nested compact subsurfaces $Y\subset X\subset\bar{S}$ with the following properties.

• •

  The interior of $Y$ contains $\pi(K)$.

• •

  Each boundary component of $X$ is separating, and each component of $\bar{S}\smallsetminus int(X)$ is noncompact,

• •

  Each component of $X\smallsetminus int(Y)$ has a fundamental group containing a free factor of rank $R$.

Now let $\tilde{Y}$ be the component of $\pi^{-1}(Y)$ containing $K$, and let $\tilde{X}\supset\tilde{Y}$ be the associated component of $\pi^{-1}(X)$. By Lemma 3.8, the restriction $\pi$ to $\tilde{X}$ gives a characteristic cover $\tilde{X}\to X$, so it follows from Proposition 3.1 that $\tilde{X}_{\circ}$ is one ended. In particular, $\tilde{X}_{\circ}\smallsetminus K$ has only one component which is not precompact. On the other hand, by Lemma 3.7, no component of $\tilde{X}\smallsetminus\tilde{Y}$ is precompact, so the same is true of $\tilde{X}_{\circ}\smallsetminus K$. This shows the complement of $K$ in $\tilde{X}_{\circ}$, and hence in $\tilde{S}_{\circ}$, is connected. ∎