Find all the connected \(2\)-sheeted covering spaces of the wedge product of two circles \(S^1\vee S^1\). Also, find all the index \(2\) subgroups of the free group on two generators, and explain the relationship between these problems.
Solution
Let \[ X=S^1\vee S^1, \qquad \pi_1(X,x_0)=F(a,b), \] where \(a\) and \(b\) are the two oriented circle generators.
1. A connected two-sheeted covering of \(X\) is determined, up to equivalence over \(X\), by a transitive monodromy homomorphism\[ \rho:F(a,b)\longrightarrow S_2. \]
Choose labels \(0,1\) for the two points in the fiber over \(x_0\). Lifting loops at \(x_0\) permutes this fiber, giving the monodromy homomorphism \(\rho\). The covering is connected exactly when the monodromy action on the fiber is transitive. Conversely, a permutation action of \(F(a,b)\) on two points constructs the corresponding covering graph by lifting the two loop edges according to the permutations \(\rho(a)\) and \(\rho(b)\). Changing the labeling of the fiber conjugates \(\rho\) in \(S_2\), which is exactly the equivalence ambiguity for the covering.
2. There are exactly three transitive monodromy homomorphisms \(F(a,b)\to S_2\) up to conjugacy.
Write \[ S_2=\{1,\tau\}, \qquad \tau=(01). \] Because \(F(a,b)\) is free, a homomorphism is determined arbitrarily by the pair \[ (\rho(a),\rho(b))\in S_2^2. \] The action on two points is transitive exactly when the image contains \(\tau\), so the pair cannot be \((1,1)\). Thus the three possibilities are \[ (\tau,1), \qquad (1,\tau), \qquad (\tau,\tau). \] Since \(S_2\) is abelian, conjugation does not identify any of these three homomorphisms.
3. The three connected double covers can be described as follows, with vertices \(v_0,v_1\) lying over the wedge point.
For each base loop labeled \(a\) or \(b\), there is one outgoing lifted edge of that label at each \(v_i\), and its terminal vertex is obtained by applying the corresponding monodromy permutation.
For \((\rho(a),\rho(b))=(\tau,1)\), the two lifted \(a\)-edges join \(v_0\) to \(v_1\), while there is a lifted \(b\)-loop at each vertex.
For \((\rho(a),\rho(b))=(1,\tau)\), there is a lifted \(a\)-loop at each vertex, while the two lifted \(b\)-edges join \(v_0\) to \(v_1\).
For \((\rho(a),\rho(b))=(\tau,\tau)\), both lifted \(a\)-edges and both lifted \(b\)-edges join \(v_0\) to \(v_1\).
Each graph is connected because at least one generator interchanges the two vertices, and 2 shows that there are no other connected two-sheeted covers.
4. Every index-\(2\) subgroup \(H\le F(a,b)\) is the kernel of a surjective homomorphism\[ F(a,b)\longrightarrow\ZZ/2\ZZ. \]
Every subgroup of index \(2\) is normal: left multiplication on the two cosets gives a nontrivial homomorphism \[ F(a,b)\longrightarrow S_2\cong\ZZ/2\ZZ \] whose kernel is \(H\). Equivalently, \[ F(a,b)/H\cong\ZZ/2\ZZ. \] Thus \(H\) is the kernel of the quotient map.
5. Hence \(F(a,b)\) has exactly three index-\(2\) subgroups:\[ \begin{aligned} H_a&=\ker\epsilon_a, &\epsilon_a(a)&=1,&\epsilon_a(b)&=0,\\ H_b&=\ker\epsilon_b, &\epsilon_b(a)&=0,&\epsilon_b(b)&=1,\\ H_{a+b}&=\ker\epsilon_{a+b}, &\epsilon_{a+b}(a)&=1,&\epsilon_{a+b}(b)&=1, \end{aligned} \]where the values lie in \(\ZZ/2\ZZ\).
A homomorphism from the free group \(F(a,b)\) to \(\ZZ/2\ZZ\) is determined by the images of \(a\) and \(b\). There are four choices, and surjectivity excludes only the choice \((0,0)\). By 4 every index-\(2\) subgroup occurs as the kernel of one of the remaining three maps. The three kernels are distinct because, for example, \[ b\in H_a\setminus H_b, \qquad a\in H_b\setminus H_a, \] and \[ a,b\notin H_{a+b}. \]
6. Explicit free bases for these three subgroups are\[ H_a=\langle b,a^2,aba^{-1}\rangle, \]\[ H_b=\langle a,b^2,bab^{-1}\rangle, \]and\[ H_{a+b}=\langle ba^{-1},a^2,ab\rangle. \]
Each subgroup has index \(2\), hence by the Schreier index formula has free rank \[ 1+2(2-1)=3. \] For \(H_a\), use the Schreier transversal \(\{1,a\}\). The nontrivial Schreier generators are \[ b, \qquad a^2, \qquad aba^{-1}. \] For \(H_b\), the symmetric computation with transversal \(\{1,b\}\) gives \[ a, \qquad b^2, \qquad bab^{-1}. \] For \(H_{a+b}\), using \(\{1,a\}\) gives \[ ba^{-1}, \qquad a^2, \qquad ab. \] Schreier’s theorem says that these are free bases of the respective kernels.
7. Under the classification of connected covering spaces, the three covers in 3 correspond exactly to the three subgroups in
5.
For a connected cover with chosen point \(\widetilde x_0\) over \(x_0\), the associated subgroup is \[ p_*\pi_1(\widetilde X,\widetilde x_0), \] equivalently the subgroup of \(F(a,b)\) whose monodromy fixes the chosen sheet. For a transitive action on two points, the image is all of \(S_2\), whose point stabilizer is trivial. Thus this subgroup is precisely the kernel of the monodromy homomorphism.
Consequently the monodromies \[ (\tau,1), \qquad (1,\tau), \qquad (\tau,\tau) \] correspond respectively to \[ H_a, \qquad H_b, \qquad H_{a+b}. \] Since index-\(2\) subgroups are normal, each is its own conjugacy class, so this is also the unbased covering-space classification.