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Let $\Gamma_g$ be a surface group of genus $g \geq 2$. A $2g$-tuple $(x_1,y_1, \dots,x_g,y_g) \in \Gamma_g^{2g}$ will be called a Surface Basis if we have the presentation $$\Gamma_g = \langle x_1, y_1, \dots, x_g, y_g \vert \prod_{i = 1}^{g}[x_i,y_i] = 1\rangle$$ Take some $1 \leq k \leq g$ and $H \leq \Gamma_g$ of finite index such that $x_1, \dots, x_k \in H$. One can show that $H$ is a surface group of genus $(g-1)[\Gamma_g : H] + 1$, and that it has a surface basis. My question is:

Is there a surface basis for $H$ containing $x_1, \dots, x_k$?

The analogous question for free groups has a positive answer as shown in Bases of free groups.

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    $\begingroup$ Is it possible to answer this using the abelianization? I think that your surface basis will induce a basis of the abelianization. The question would then reformulate to a question about the behaviour of $H_1$ of surfaces under coverings. $\endgroup$ – Matthias Wendt Sep 17 '14 at 10:01
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    $\begingroup$ A stronger (so it seems to me) statement which does not talk about finite index subgroups: given a genus $g$ surface and a wedge of circles on it (such that the classes of the circles in $H_1$ are linearly independent). Is it possible to complete this to a wedge of $2g$ circles whose complement is contractible? $\endgroup$ – Matthias Wendt Sep 17 '14 at 10:38
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    $\begingroup$ It's basically the same argument as in the free group case. You have a $\pi_1$-injective embedded rose in $\Gamma_g$ that lifts to the same thing in the cover corresponding to $H$. Now you just need to argue that this rose can be extended to a basis. $\endgroup$ – HJRW Sep 17 '14 at 13:19
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    $\begingroup$ (Hint: think about what the complement of the rose looks like.) $\endgroup$ – HJRW Sep 17 '14 at 13:22
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    $\begingroup$ @Pablo, on further reflection, if you want to do it computationally, perhaps you need to look up the 'Reidemeister--Schreier method'. $\endgroup$ – HJRW Sep 17 '14 at 14:15
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There is indeed a surface basis for $H$ containing $x_1,\ldots,x_k$. I'll give a topological proof, basically the same as the proof suggested in the comment of @HJRW. First, I'll give a topological re-interpretation of your problem, by formulating a topological property which is equivalent to the property "there is a surface basis of the fundamental group containing such-and-such a list of elements". Then I'll say why this topological property is preserved under the kind of covering maps in your question.


For each genus $g$ fix a ``standard'' surface $S_g$ with base point $p$ and an isomorphism $\pi_1(S_g,p) \approx \Gamma_g$, and with an embedded rank $2g$ rose consisting of loops $\xi_1,\eta_1,\ldots,\xi_g,\eta_g$ representing $x_1,y_1,\ldots,x_g,y_g$. When $S_g$ is cut open along the rose, the result is a $2g$-gon whose boundary is attached to the rose using the word given by the standard relator in your question. Let $N_k$ be a regular neighborhood of $\xi_1 \cup \cdots \cup \xi_k$. Note that $N_k$ is a connected planar surface with connected complement; it follows that $N_k$ has $k+1$ boundary components, and that $S_g - N_k$ is a connected surface of genus $g-k$ with $k+1$ boundary components.

Consider now an abstract surface $S'$ of some genus $p'$, and a rank $k$ rose embedded in $F$ consisting of loops $\gamma_1,\ldots,\gamma_k$ all with common base point $q$. Then the following are equivalent:

  1. there exists a surface basis for $\pi_1(S',p')$ containing $[\gamma_1],\ldots,[\gamma_k]$
  2. the regular neighborhood of the rose $\gamma_1 \cup \ldots \cup \gamma_k$ is a connected planar surface with connected complement.

I've already proved 1$\implies$2. For 2$\implies$1, it follows the regular neighborhood of the rose is a planar surface with $k+1$ boundary components and that its complement has genus $g'-k$ with $k+1$ boundary components, and this allows us to construct a homeomorphism between $S'$ and the standard surface $S_{g'}$ that takes $\gamma_1,\ldots,\gamma_k$ to $\xi_1,\ldots,\xi_k$.


Okay, so, consider a finite index subgroup of $\pi_1(S,p)$ containing $x_1,\ldots,x_k$. Let $f : S' \to S$ be the corresponding finite degree covering map with base point $p'$ lifting $p$. The loops $\xi_1,\ldots,\xi_k$ lift one-to-one to loops $\xi'_1,\ldots,\xi'_k$ based at $p'$. The regular neighborhood $N_k$ of $\xi_1\cup\cdots\cup\xi_k$ lifts one-to-one to a regular neighborhood $N'_k$ of $\xi'_1\cup\cdots\cup\xi'_k$. Since the restricted covering map $N'_k \to N_k$ is a homeomorphism, this proves that $N'_k$ is planar. It remains to prove that $S'-N'_k$ is connected.

Arguing by contradiction, assuming that $S' - N'_k$ is disconnected, consider a component $F$ of $S'-N'_k$ which meets $N'_k$. It follows that $\partial F = F \cap N'_k$ consists of a proper nonempty subset of the $k+1$ components of $\partial N'_k$. Let those components be enumerated $c'_1,\ldots,c'_j,c'_{j+1},\ldots,c'_{k+1}$ so that $\partial F = c'_1 \cup \cdots \cup c'_j$. Push this enumeration down to $N_k$ whose boundary circles are therefore enumerated $c_1,\ldots,c_j,c_{j+1},\ldots,c_{k+1}$.

Let the surface $F$ be decomposed as $F = F_1 \cup F_2$ where the covering map $S' \to S$ restricts to a covering map $F_1 \to S - N_k$ of some degree $\delta_1$, and it restricts to a covering map $F_2 \to N_k$ of some degree $\delta_2$. Let $\partial F_1$ be decomposed into two subsets $\partial F_1 = \partial_{\le j} F_1 \cup \partial_{>j} F_1$, with restricted covering maps $\partial_{\le j} F_1 \to c_1 \cup \cdots \cup c_j$ and $\partial_{>j} F_1 \to c_{j+1} \cup \cdots \cup c_{k+1}$ each of degree $\delta_1$ (the base spaces of these covering maps are not connected, nonetheless these covering maps each have constant degree $\delta_1$ over each component). It follows that the restricted map $\partial F_1 - (c'_1 \cup \cdots \cup c'_j) \to c_1 \cup \cdots \cup c_{k+1}$ is a covering map of unequal degrees over different components: it has degree $\delta_1-1$ over each component of $c_1,\ldots,c_j$ and it has degree degree $\delta_1$ over each component of $c_{j+1},\ldots,c_{k+1}$. However, we have an equation $$\partial F_1 - (c'_1 \cup \cdots \cup c'_j) = \partial F_2 $$ and the degree $\delta_2$ covering map $F_2 \to N_k$ restricts to a map $\partial F_2 \to c_1 \cup \cdots \cup c_{k+1}$ of equal degree $\delta_2$ over each component. This is a contradiction.

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  • $\begingroup$ @Mosher: That's great! As I have already commented, I would like to have formulas for elements in a surface basis of a finite index subgroup. Such formulas could give this result immediately, like in the free case, where the constructed basis from any Schreier transversal will contain the required elements. Here, this is not so clear because the relation will force us to remove some subset of a Schreier generating set. In view of that, I would very much like to see an algebraic argument for this, or a purely algebraic proof for the existence of surface bases for finite index subgroups. $\endgroup$ – Pablo Sep 18 '14 at 6:35

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