Let \(S\) be the closed orientable surface of genus 2 and let \(C\) be the commutator subgroup of \(\pi_1 (S, \ast)\). Let \(\tilde S\) be the cover corresponding to \(C\). Is the covering map \(\tilde S \to S\) regular?
The term “normal” is sometimes used as a synonym for regular in this context.
What is the group of deck transformations?
Give an example of a nontrivial element of \(\pi_1 (S, \ast)\) which lifts to a trivial deck transformation.
Solution
Set \[ G=\pi_1(S,*), \qquad C=[G,G]. \]
1. The subgroup \(C\) is normal in \(G\).
The commutator subgroup of any group is characteristic. Indeed, if \[ \varphi:G\longrightarrow G \] is an automorphism, then \[ \varphi([x,y])=[\varphi(x),\varphi(y)] \] for every \(x,y\in G\). Thus \(\varphi(C)\subseteq C\); applying the same argument to \(\varphi^{-1}\) gives equality. Every characteristic subgroup is normal, so \(C\trianglelefteq G\).
2. The covering\[ p:\widetilde S\longrightarrow S \]corresponding to \(C\) is regular.
For a connected covering of a connected, locally path-connected space, the covering is regular exactly when the corresponding subgroup of the fundamental group is normal. By 1, the corresponding subgroup \(C\) is normal in \(G\). Hence \(p\) is regular.
3. The deck transformation group is\[ \operatorname{Deck}(p)\cong G/C. \]
For a regular covering corresponding to a normal subgroup \(H\trianglelefteq\pi_1(S,*)\), the deck transformation group is canonically isomorphic to \[ \pi_1(S,*)/H. \] Apply this with \(H=C\).
4. One has\[ G/C\cong\ZZ^4. \]
For a closed orientable surface of genus \(2\), \[ G \cong \left\langle a_1,b_1,a_2,b_2 \ \middle|\ [a_1,b_1][a_2,b_2]=1 \right\rangle. \] The quotient by \(C=[G,G]\) is the abelianization \(G_{\mathrm{ab}}\). After abelianization, every commutator is trivial, so the displayed relator imposes no relation among the four generator classes. Therefore \[ G/C=G_{\mathrm{ab}}\cong\ZZ^4. \]
5. The element\[ \gamma=[a_1,b_1] \]is nontrivial in \(G\).
Let \(F(a,b)\) be the free group on \(a,b\). Define a map on the generators of the surface presentation by \[ a_1\longmapsto a, \qquad b_1\longmapsto b, \qquad a_2\longmapsto b, \qquad b_2\longmapsto a. \] The surface relator maps to \[ [a,b][b,a] =[a,b][a,b]^{-1} =1, \] so this assignment induces a homomorphism \[ \psi:G\longrightarrow F(a,b). \] But \[ \psi(\gamma)=[a,b]=aba^{-1}b^{-1}, \] which is a nonempty reduced word in the free group and hence is not the identity. Thus \(\gamma\ne1\) in \(G\).
6. The nontrivial element \(\gamma=[a_1,b_1]\) induces the trivial deck transformation of \(\widetilde S\).
By definition, \[ \gamma\in[G,G]=C. \] Under the isomorphism in 3, the deck transformation associated to an element \(g\in G\) depends only on its coset \(gC\). Thus \[ \gamma C=C, \] the identity element of \(G/C\), so \(\gamma\) induces the identity deck transformation.
Equivalently, the lift of a loop representing \(\gamma\) from a chosen point of the fiber is closed. The associated deck transformation fixes that point, and a deck transformation of a connected cover that fixes one point is the identity.
Therefore \[ \boxed{p\text{ is regular},\qquad \operatorname{Deck}(p)\cong\ZZ^4,\qquad [a_1,b_1]\ne1\text{ acts trivially}.} \]