Let \(G=\mathbb Z_2*\mathbb Z_2=\langle a,b\mid a^2=b^2=1\rangle\). Find the presentation complex \(X_G\) and the universal cover of \(X_G\), \(\widetilde X_G\). Find an index two subgroup of \(G\) and the corresponding covering space.
Solution
The presentation complex of \[ G=\langle a,b\mid a^2,b^2\rangle \] has one \(0\)-cell, two \(1\)-cells \(a,b\), and two \(2\)-cells attached by the degree-two words \(a^2\) and \(b^2\). Equivalently, \[ X_G\cong \mathbb{RP}^2\vee\mathbb{RP}^2. \]
Its universal cover can be described as a bi-infinite chain of \(2\)-spheres. Consecutive spheres meet in one point, and the sphere types alternate between the lifts of the two \(\mathbb{RP}^2\) summands. More explicitly, the lifted \(1\)-skeleton is the Cayley graph of \(\mathbb Z_2*\mathbb Z_2\), a bi-infinite line with alternating \(a\)- and \(b\)-steps; over each \(a\)-pair of adjacent vertices the two lifted \(a\)-edges and the two lifted \(a^2\)-cells form a copy of \(S^2\), and similarly for \(b\). This tree of simply connected pieces is simply connected, hence is \(\widetilde X_G\).
For an index-two subgroup, define \[ \varepsilon:G\to\mathbb Z/2, \qquad \varepsilon(a)=\varepsilon(b)=1. \] Then \[ H=\ker\varepsilon=\langle ab\rangle\cong\mathbb Z \] has index \(2\). The corresponding covering has two vertices \(v_0,v_1\). The lifts of the \(a\)-circle form a \(2\)-edge circle between these vertices, filled by two lifted \(a^2\)-cells to make a sphere; the \(b\)-cells form a second sphere with the same two vertices. Thus the index-two cover is two copies of \(S^2\) glued together at two distinct points. Its fundamental group is \(\mathbb Z\), agreeing with \(H\).