For a path-connected, locally path-connected, and semilocally simply-connected space \(X\), call a path-connected covering space \(\tilde{X} \to X\) abelian if it is normal and has abelian deck transformation group. Show that \(X\) has an abelian covering space that is a covering space of every other abelian covering space of \(X\), and that such a “universal” abelian covering space is unique up to isomorphism. Describe this covering space explicitly for \(X = S^1 \vee S^1\) and \(X = S^1 \vee S^1 \vee S^1\).
Solution
Let \[ G=\pi_1(X,x_0). \]
1. A connected normal covering corresponding to a normal subgroup \(H\triangleleft G\) is abelian exactly when\[ [G,G]\subseteq H. \]
For a connected normal covering with subgroup \(H\), the deck transformation group is \[ G/H. \] This quotient is abelian exactly when every commutator maps to the identity, equivalently when the commutator subgroup satisfies \[ [G,G]\subseteq H. \]
2. Let\[ p_{\mathrm{ab}}:X_{\mathrm{ab}}\to X \]be the connected covering corresponding to the subgroup\[ [G,G]\le G. \]Then \(p_{\mathrm{ab}}\) is an abelian covering with deck group\[ G/[G,G]=G_{\mathrm{ab}}. \]
The commutator subgroup is normal. By 1 its associated connected normal cover is abelian, and the standard normal-cover deck-group theorem identifies its deck group with the displayed quotient.
3. The cover \(X_{\mathrm{ab}}\) covers every other connected abelian covering of \(X\).
Let \[ q:Y\to X \] be any connected abelian covering, corresponding to a subgroup \(H\triangleleft G\). By 1, \[ [G,G]\subseteq H. \] The covering-space factorization criterion says that a connected cover for subgroup \(K\) covers the connected cover for subgroup \(H\) exactly when, after compatible basepoint choices, \[ K\subseteq H. \] Taking \(K=[G,G]\) gives a covering map \[ X_{\mathrm{ab}}\to Y \] over \(X\).
4. Any universal abelian covering is isomorphic to \(X_{\mathrm{ab}}\).
Suppose \[ q:U\to X \] is an abelian cover that covers every other abelian cover, and let its subgroup be \(K\). Since \(U\) is abelian, 1 gives \[ [G,G]\subseteq K. \] Since \(U\) covers \(X_{\mathrm{ab}}\), the factorization criterion gives \[ K\subseteq [G,G]. \] Thus \[ K=[G,G]. \] Connected coverings with the same subgroup are isomorphic over \(X\) after basepoint choice, hence unbased as well.
5. For\[ X=S^1\vee S^1, \qquad G=F(a,b), \]the universal abelian cover is the square lattice graph in \(\mathbb R^2\).
The abelianization is \[ F(a,b)_{\mathrm{ab}}\cong\mathbb Z^2. \] Use one vertex \[ v_{(m,n)} \] for each \((m,n)\in\mathbb Z^2\). Attach an oriented \(a\)-edge \[ v_{(m,n)}\to v_{(m+1,n)} \] and an oriented \(b\)-edge \[ v_{(m,n)}\to v_{(m,n+1)}. \] Map all \(a\)-edges to the first circle and all \(b\)-edges to the second. This is the labelled Cayley covering graph of \(\mathbb Z^2\). Its deck transformations are the translations \[ (m,n)\mapsto(m+p,n+q), \qquad(p,q)\in\mathbb Z^2, \] so the deck group is \(\mathbb Z^2\), and its subgroup is the kernel of \[ F(a,b)\to\mathbb Z^2, \] namely \([F,F]\).
6. For\[ X=S^1\vee S^1\vee S^1, \qquad G=F(a,b,c), \]the universal abelian cover is the cubic lattice graph in \(\mathbb R^3\).
Now \[ G_{\mathrm{ab}}\cong\mathbb Z^3. \] Put a vertex at every lattice point \[ (m,n,p)\in\mathbb Z^3 \] and join it by oriented \(a\)-, \(b\)-, and \(c\)-edges to the points obtained by adding the three standard basis vectors. Map these edge families to the three wedge circles. The deck group is the translation group \(\mathbb Z^3\), and the corresponding subgroup is the commutator subgroup of \(F(a,b,c)\).
7. Therefore the commutator-subgroup cover is the unique universal abelian covering, with the stated lattice models for the two examples.
Existence and universality are <1>2–<1>3, uniqueness is 4, and the explicit models are <1>5–<1>6.