In an exercise for §1.2 we described a 3 dimensional CW complex obtained from the cube \(I^3\) by identifying opposite faces via a one-quarter twist. Compute the homology groups of this complex.
Solution
1. The complex \(X\) is the quaternionic \(3\)-manifold \(S^3/Q_8\): identifying opposite faces of the cube via a one-quarter twist gives the quotient of \(S^3\) by the quaternion group \(Q_8\).
this is the standard construction of the quaternionic manifold.
2. \(\pi_1(X) = Q_8\).
the fundamental group of this identification space is the quaternion group \(Q_8\) (the two edges give generators \(i, j\) with relations \(i^2 = j^2 = k^2 = ijk\)).
3. \(H_0(X) = \ZZ\).
\(X\) is connected.
4. \(H_1(X) = Q_8^{\text{ab}} = \ZZ/2 \times \ZZ/2\).
\(H_1 = \pi_1^{\text{ab}}\), and the abelianization of \(Q_8\) is \(\ZZ/2 \times \ZZ/2\) (the commutator subgroup of \(Q_8\) is \(\{\pm 1\}\), of order \(2\), so \(Q_8^{\text{ab}}\) has order \(4\) and is \(\ZZ/2 \times \ZZ/2\)).
5. \(H_2(X) = 0\).
by Poincaré duality (and the universal coefficient theorem), \(H_2(X) \cong H^1(X) \cong \operatorname{Hom}(H_1(X), \ZZ) = 0\), since \(H_1(X) = \ZZ/2 \times \ZZ/2\) is torsion.
6. \(H_3(X) = \ZZ\).
\(X\) is a closed orientable \(3\)-manifold, so \(H_3(X) = \ZZ\).
7. Q.E.D.
\(H_0 = \ZZ\), \(H_1 = \ZZ/2 \times \ZZ/2\), \(H_2 = 0\), \(H_3 = \ZZ\) (<1>3–<1>6).