Extending free group action on surfaces

Consider a principal $G$-bundle representing the given $G$-cobordism over the closed surface of genus one. It is enough to prove the case in which the total space of the bundle is a connected space. Moreover, the action of $G$ over the total space can be modified by an isotopy resulting in an action which depends completely on a pair of monodromies $(g,k)$, which are associated to two curves in the torus that intersect once. Denote by $P_{g}$ and $P_{k}$ the two principal $G$-bundles associated to these two curves. It follows that the action of $G$ over the total space is given by the product of the total spaces $P_{g}$ and $P_{k}$. Because of the assumptions, at least one of $P_{g}$ or $P_{k}$ is a connected space, let us assume that it is $P_{g}$. The extension of the action of $G$ is through the 3-dimensional handlebody constructed as follows. First, consider the disc $D$ as the union $(S^{1}\times(0,1])\cup\{0\}$, where $0$ is the center. For each circle $S^{1}\times\{r\}$, with $r\in(0,1]$, we take as monodromy the element $g$ so that the principal $G$-bundle is $P_{g}$, and we take the center $0$ as fixed point. Thus, over the disc we have the rotation by $2\pi/|g|$, with $|g|$ the order of $g\in G$. Taking the product of this disc together with the induced principal $G$-bundle $P_{k}$, we obtain the extension which makes the $G$-cobordism extendable. ∎

Consider a $G$-cobordism over a connected closed surface. By Proposition 5, we can write this $G$-cobordism as a connected sum of $G$-cobordisms with base space of genus one. This connected sum is done for the total space along trivial bundles over a circle. Since the connected sum is in the same bordism class as the disjoint union, we are decomposing this $G$-cobordism as a disjoint union of $G$-cobordisms with base space a closed surface of genus one. Because of Proposition 7, any $G$-cobordism over a closed surface of genus one is extendable, so the theorem follows. ∎

For $n=|G|$ and $n=p^{k}m$ with $p\not|m$, consider an element $f\in\mathcal{M}(G)_{p}$, hence the composition

$$F:=\operatorname{cor}\circ\operatorname{res}:\mathcal{M}(G)_{p}\longrightarrow\mathcal{M}(G)_{p}\,,$$

(19)

is given by $f\longmapsto f^{m}$, which is an automorphism of $\mathcal{M}(G)_{p}$. By the assumptions, the restriction $\operatorname{res}(f)$ is extendable by a $3$-manifold $M$ with an action of $Q$, therefore, applying the corestriction we obtain that $f^{m}$ is extendable by the $3$-manifold $\operatorname{cor}(M)$. Similarly, we can start with $F^{-1}(f)$ and we get that $f$ is extendable and the proposition follows. ∎

The relation (i) follows by (17). The use of (7), (17) and (11) implies

	$\displaystyle\left<c^{i},ac^{j}\right>\sim\left<c^{i},a\right>\left<c^{i},c^{j}\right>^{a}\sim\left<c^{i},a\right>\,,$		(22)
	$\displaystyle\left<c^{i},ac^{j}\right>=\left<cc^{i-1},ac^{j}\right>\sim\left<c^{i-1},ac^{j}\right>^{c}\left<c,ac^{j}\right>\sim\left<c^{i-1},a\right>\left<c,a\right>\,.$		(23)

Therefore, we obtain the relation

$$\left<c^{i},a\right>\sim\underbrace{\left<c,a\right>\left<c,a\right>\cdots\left<c,a\right>}_{i}\,,$$

(24)

which implies (ii). By (6) we obtain (iii) as follows

$$\left<ac^{i},c^{j}\right>\sim\left<c^{j},ac^{i}\right>^{-1}\sim\left<c,a\right>^{-j}\,.$$

(25)

Finally, we use (11) and (5),

$$\left<ac^{i},ac^{j}\right>\sim\left<c^{i},ac^{j}\right>^{a}\left<a,ac^{j}\right>\sim\left<c^{-1},a\right>^{i}\left<a,c^{j}\right>^{a}\,,$$

(26)

and by (6) we obtain (iv). ∎

It suffices to note that for pairs $\langle a,b\rangle$ and $\langle x,y\rangle$, with $a,b,x,y\in S_{n}$, such that $a$ and $b$ fix $k$ there is the relation

	$\displaystyle\left<x,y\right>\left<a,b\right>$	$\displaystyle\sim\left<a,b\right>\left<b,a\right>\left<x,y\right>\left<a,b\right>$
		$\displaystyle\sim\left<a,b\right>\left<x^{[b,a]},y^{[b,a]}\right>\,,$

where we have used (12). An iterative application of this process, allows us to put all terms fixing $k$ to the left in the sequence. ∎

First, we observe that from (7) and (11), we can assume that all the elements $\sigma_{i},\tau_{i},\sigma^{\prime}_{j},\tau^{\prime}_{j}\in S_{n}$ are transpositions. By (8), every pair $\langle\sigma,\tau\rangle$ with $\sigma$ and $\tau$ disjoint transpositions is in the same class as the pair $u:=\left<(1,2),(3,4)\right>$. Therefore, we can assume that the pairs are of the form $\left<(i,j),(j,k)\right>$, with $i,j$ and $k$ different numbers.

By exhaustion, the proposition follows for the symmetric group $S_{n}$, with $4\leq n\leq 6$. We proceed by induction for $n\geq 7$ and we suppose that for $k<n$, the generator of the Schur multiplier $\mathcal{M}(S_{n})$ is given by the element $u:=\left<(1,2),(3,4)\right>$. Set by $m$ the maximum of $r$ and $s$, for the sequences $\langle\sigma_{1},\tau_{1}\rangle\cdots\langle\sigma_{r},\tau_{r}\rangle$ and $\langle\sigma^{\prime}_{1},\tau^{\prime}_{1}\rangle\cdots\langle\sigma^{\prime}_{s},\tau^{\prime}_{s}\rangle$. For $m=1$, the proposition follows from the triviality of $\mathcal{M}(S_{3})$. Suppose that our proposition follows for sequences with length $l<m$. We consider the sequence

$$\langle\sigma_{1},\tau_{1}\rangle\cdots\langle\sigma_{r},\tau_{r}\rangle\left(\langle\sigma^{\prime}_{1},\tau^{\prime}_{1}\rangle\cdots\langle\sigma^{\prime}_{s},\tau^{\prime}_{s}\rangle\right)^{-1}\sim\langle\sigma_{1},\tau_{1}\rangle\cdots\langle\sigma_{r},\tau_{r}\rangle\langle\tau^{\prime}_{s},\sigma^{\prime}_{s}\rangle\cdots\langle\tau^{\prime}_{1},\sigma^{\prime}_{1}\rangle\,,$$

(31)

which has trivial commutator and length given by $M:=r+s\leq 2m$. Let $x\in\{1,\cdots,n\}$ be the number that is fixed by the most terms of the sequence (31). Given that the sequences have non trivial terms, each term permutes $3$ different numbers in $\{1,2,\cdots,n\}$. Therefore, the number $x$ is not fixed by at most $\frac{3(r+s)}{n}$ terms. Given that $\frac{3(r+s)}{n}\leq\frac{3(2m)}{7}<m$, hence $x$ is not fixed by at most $m-1$ terms. By Lemma 17, we can find an equivalent sequence for (31), with the following form

$$\underbrace{\left<\alpha_{1},\beta_{1}\right>\cdots\left<\alpha_{t},\beta_{t}\right>}_{\text{fix number x}}\underbrace{\left<\alpha_{t+1},\beta_{t+1}\right>\cdots\left<\alpha_{M},\beta_{M}\right>}_{\text{do not fix number x}}\,,$$

(32)

where:

• i)

  $M-t<m$;

• ii)

  the $\alpha_{i},\beta_{i}\in S_{n}$, with $0\leq i\leq t$, fix $x$; and

• iii)

  the $\alpha_{j},\beta_{j}\in S_{n}$, with $t+1\leq j\leq M$, at least one does not fix $x$.

Moreover, by the proof of Lemma 17, the elements $\alpha_{i},\beta_{i},\alpha^{\prime}_{j},\beta^{\prime}_{j}\in S_{n}$ are again transpositions. Now we consider the sequences

$$A:=\left<\alpha_{1},\beta_{1}\right>\cdots\left<\alpha_{t},\beta_{t}\right>$$

(33)

and

$$B:=\left(\left<\alpha_{t+1},\beta_{t+1}\right>\cdots\left<\alpha_{M},\beta_{M}\right>\right)^{-1}=\left<\beta_{M},\alpha_{M}\right>\cdots\left<\beta_{t+1},\alpha_{t+1}\right>\,,$$

(34)

where both sequences have the same commutator. Furthermore, the sequence $A$ has pairs composed by elements in $S_{n-1}$ because they fix $x$. By our induction hypothesis, for $n$, we conclude that the Schur multiplier $\mathcal{M}(S_{n-1})$ is generated by $u=\langle(1,2),(3,4)\rangle$. Therefore, $A\sim u^{i}C$ for $i\in\{0,1\}$ and $C$ is a sequence of pairs with elements in $S_{n-1}$. We can take $C$ to be of length $<m$, as it is has the same commutator as the chain $B$ of length $<m$. By the other induction hypothesis, for $m$, since $B$ and $C$ have length less than $m$, then there is $j\in\{0,1\}$ such that $B\sim u^{j}C$. This shows that the product of our initial sequences in (31) is equivalent to $u^{i-j}$ and the proof of the proposition follows. ∎

Because of the relations (7) and (11) in $\mathcal{M}(S_{n})$, we have the following

	$\displaystyle\langle(1,2)(3,4),(1,3)(2,4)\rangle$	$\displaystyle\sim\langle(3,4),(2,3)(1,4)\rangle\langle(1,2),(1,3)(2,4)\rangle$
		$\displaystyle\sim\langle(3,4),(2,3)\rangle\langle(2,4),(1,4)\rangle\langle(1,2),(1,3)\rangle\langle(2,3),(2,4)\rangle$

We also have from (8), (11) and $\langle(2,4),(1,4)\rangle=\langle(2,3),(1,3)\rangle^{(3,4)}$ that

	$\displaystyle\langle(2,4),(1,4)\rangle$	$\displaystyle\sim\langle(3,4),[(2,3),(1,3)]\rangle\langle(2,3),(1,3)\rangle=\langle(3,4),(1,2)(1,3)\rangle\langle(2,3),(1,3)\rangle$
		$\displaystyle\sim\langle(3,4),(1,2)\rangle\langle(3,4),(2,3)\rangle\langle(2,3),(1,3)\rangle=u\langle(3,4),(2,3)\rangle\langle(2,3),(1,3)\rangle$

As $[(3,4),(2,3)]=[(2,3),(2,4)]$, $[(2,3),(1,3)]=[(1,3),(1,2)]$ and $\mathcal{M}(S_{3})=0$, we have that $\langle(3,4),(2,3)\rangle\sim\langle(2,3),(2,4)\rangle$ and $\langle(2,3),(1,3)\rangle\sim\langle(1,3),(1,2)\rangle$. Therefore,

	$\displaystyle\langle(1,2)(3,4),(1,3)(2,4)\rangle$	$\displaystyle\sim\langle(2,3),(2,4)\rangle\langle(2,4),(1,4)\rangle\langle(1,2),(1,3)\rangle\langle(2,3),(2,4)\rangle$
		$\displaystyle\sim u\langle(2,3),(2,4)\rangle^{2}\langle(1,3),(1,2)\rangle\langle(1,2),(1,3)\rangle\langle(2,3),(2,4)\rangle$
		$\displaystyle\sim u\langle(2,3),(2,4)\rangle^{3}\sim u=\langle(1,2),(3,4)\rangle\,,$

where $\langle(2,3),(2,4)\rangle^{3}$ vanishes since $[(2,3),(2,4)]^{3}=1$ and $\mathcal{M}(S_{3})=0$. As a consequence, the element $\langle(1,2)(3,4),(1,3)(2,4)\rangle$ is nontrivial in $\mathcal{M}(S_{n})$, and hence it is also nontrivial in $\mathcal{M}(A_{n})$ for $n\geq 4$. ∎

By Theorem 9 we know that any free action of a finite abelian group is extendable. For dihedral groups $D_{2n}$, we have two cases to consider. One is for $n=2k+1$, but since $\mathcal{M}(D_{4k+2})=0$, then any free action is extendable. The other is for $n=2k$, where by Corollary 16, the generator of the Schur multiplier is represented by a $G$-cobordism over a closed surface of genus one, therefore, by Proposition 7 these free actions are extendable. Now consider free actions of the symmetric groups $S_{n}$, since $\mathcal{M}(S_{n})=0$ for $n\leq 3$, it remains to prove the extension for $n\geq 4$. Similar as for dihedral groups, by Corollary 19 these free actions are extendable. For the alternating groups $A_{n}$ the free actions are extendable for $n\leq 3$. Again, for $n\geq 4$ and $n\neq 6,7$, by Corollary 21 these actions are extendable. In the case of free actions of $A_{n}$ for $n=6,7$, we notice that the Sylow subgroups of $A_{6}$ and $A_{7}$ have the following isomorphic types $\{D_{8},\mathbb{Z}_{3}\times\mathbb{Z}_{3}\times\mathbb{Z}_{3},\mathbb{Z}_{5},\mathbb{Z}_{7}\}$ and because of Proposition 14, we obtain that these free actions are extendable. ∎

The extension is performed in some steps. First, we consider the quotient of the surface by the hyperelliptic involutions in some order and smooth the corners with the aim to obtain a smooth closed oriented surface. The hyperelliptic involutions act in the set of ramification points with signature $>2$ and in the quotient we still have ramifications points grouped into pairs. Then we connect the complementary monodromies by cylinders in order to have a free action over a closed oriented surface. From Theorem 9, we have that this free action is extendable and by the proof of Proposition 15, this extension is by means of a 3-dimensional handlebody. Now we disconnect the complementary monodromies by cutting in each of the cylinders that we glued. Each cylinder has only one fixed point by the proof of Proposition 7. Finally, we extend the action to the original surface by an unfolding process using the hyperelliptic involutions in the reverse order in which we constructed the initial quotient surface. ∎

By Proposition 15, the extension problem reduces to a finite product of the same $D_{2n}$-cobordism induced by the pair $\langle c,a\rangle\sim\langle ab,a\rangle\sim\langle b,a\rangle^{a}$. Thus, it is enough to solve the extension problem for the pair $\langle b,a\rangle$ and for the $D_{2n}$-cobordism over the pair of pants with entries in $[D_{2n},D_{2n}]=\langle c^{2}\rangle$. The last reduces to the extension of $G$-cobordisms over pair of pants where $G=[D_{2n},D_{2n}]$, which follows because the group is cyclic. For the pair $\langle b,a\rangle$ we construct a representative $D_{2n}$-cobordism and there are two cases to consider:

• (i)

  For $n=2k$, we consider the disjoint union of two spheres, where each one is the gluing of two $n$-gons by the boundary. Denote by $T$ the operation of switching from one sphere to the other and by $S$ the operation of switching from one $n$-gon to the other in the same sphere. The action of $a$ lifts to the composition $Sa=aS$ and the action of $b$ lifts to the composition $Tb=bT$. We obtain $n=2k$ fixed points over each sphere plus the north a south poles. Then for each fixed point (except the north and south pole), we remove a small disc around it and we connect the holes for opposite fixed points with a cylinder. In a similar way as for abelian groups, this action extends with the north and south pole as unique ramification points. For $k=2$, in Figure 7, we draw an illustrator of this construction.

  

• (ii)

  For $n=2k+1$, we consider one sphere as the gluing of two $n$-gons by the boundary. Denote by $S$ the operation of switching from one $n$-gon to the other. Thus the action of $a$ lifts to the composition $Sa=aS$ and the action of $b$ lifts to the composition $Sb=bS$. We obtain $2n$ fixed points plus the north a south poles. Then we perform the same procedure to construct the extension of the action, as in the even case.

∎