Find all the connected 2-sheeted and 3-sheeted covering spaces of \(S^1 \vee S^1\), up to isomorphism of covering spaces without basepoints.
Solution
Let the two oriented loops of \(S^1\vee S^1\) be labelled \(a\) and \(b\). An \(n\)-sheeted covering is determined by the monodromy permutations \[ (\sigma_a,\sigma_b)\in S_n\times S_n. \] It is connected exactly when the subgroup generated by \(\sigma_a\) and \(\sigma_b\) acts transitively on the \(n\) sheets. Changing the labelling of the sheets simultaneously conjugates both permutations, so unbased isomorphism classes are exactly simultaneous-conjugacy classes of transitive pairs.
1. There are exactly three connected \(2\)-sheeted coverings.
Write \(\tau=(12)\). The only transitive subgroup of \(S_2\) is \(S_2\) itself, so at least one of \(\sigma_a,\sigma_b\) must equal \(\tau\). Since \(S_2\) is abelian, simultaneous conjugacy changes nothing. Thus the three classes are \[ (\tau,1),\qquad (1,\tau),\qquad (\tau,\tau). \] Equivalently, in the corresponding two-vertex covering graphs, the \(a\)-edges switch vertices in the first case, the \(b\)-edges switch vertices in the second, and both colors switch vertices in the third; a trivial permutation gives one loop of that color at each vertex.
2. For \(3\) sheets, a transitive pair generates either \(A_3\) or \(S_3\).
The transitive subgroups of \(S_3\) are precisely the cyclic subgroup \(A_3=\langle(123)\rangle\) and \(S_3\). Hence a transitive pair either consists only of powers of one \(3\)-cycle and generates \(A_3\), or contains a transposition and generates \(S_3\) together with any element that makes the action transitive.
3. The four simultaneous-conjugacy classes generating \(A_3\) are represented by\[ (1,c),\qquad(c,1),\qquad(c,c),\qquad(c,c^{-1}), \]where \(c=(123)\).
If both entries lie in \(A_3\), transitivity requires that at least one be a nonidentity \(3\)-cycle. Simultaneous conjugation in \(S_3\) may replace \(c\) by \(c^{-1}\) in both coordinates at once, but it cannot change whether an entry is the identity nor whether two nonidentity entries are equal or inverse. These four possibilities are therefore distinct and exhaustive.
4. The three simultaneous-conjugacy classes generating \(S_3\) are represented by\[ (t,c),\qquad(c,t),\qquad(t,u), \]where\[ t=(12),\qquad u=(23),\qquad c=(123). \]
If one entry is a transposition and the other a \(3\)-cycle, simultaneous conjugacy acts transitively on such ordered pairs, giving one class of type \((t,c)\) and one with the coordinates reversed.
If both entries are transpositions, transitivity excludes the case in which they are equal, since one transposition fixes a sheet. Any ordered pair of distinct transpositions is simultaneously conjugate to \(((12),(23))\), giving one further class.
Two entries cannot generate a new transitive class with one transposition and the identity, since that subgroup fixes one sheet. Thus these three classes are exhaustive.
5. Hence there are exactly seven connected \(3\)-sheeted coverings, represented by\[ \boxed{ (1,c),\ (c,1),\ (c,c),\ (c,c^{-1}),\ (t,c),\ (c,t),\ (t,u). } \]
4.Each pair is transitive, and the simultaneous-conjugacy invariants used there show that no two listed pairs are equivalent.
6. Therefore the complete answer is three connected double covers and seven connected triple covers.
The classification of finite covers by monodromy pairs identifies every connected cover with exactly one of the classes listed in 1 and 5.