Hatcher Exercise 2.1.8 We compute the simplicial homology of the complex described in the text. Fix notation to refer to the vertices of each tetrahedron in the complex: v^i_j will refer to the j‘th vertex on the i‘th tetrahedron – throughout this problem, i used as an index ranging from 1 to n will be understood to be taken mod n. We perform the indicated identifications and examine equivalence classes of faces: We have n 3-simplices (the tetrahedra): \mathcal{T}_1 = \set{[v^1_0 v^1_1 v^1_2 v^1_3]}, \dots, \mathcal{T}_n = \set{[v^n_0 v^n_1 v^n_2 v^n_3]} For faces, we first identify the vertical faces with the next neighboring vertical face, in a cyclic fashion. This gives [v^1_1 v^1_2 v^1_3] \sim [v^2_0 v^2_2 v^2_3], [v^2_1 v^2_2 v^2_3] \sim [v^3_0 v^3_2 v^3_3], \dots, [v^n_1 v^n_2 v^n_3] \sim [v^1_0 v^1_2 v^1_3] Then we identify the bottom face of each tetrahedron with the top of the next tetrahedron around, giving: [v^1_0 v^1_1 v^1_2] \sim [v^2_0 v^2_1 v^2_3], [v^2_0 v^2_1 v^2_2] \sim [v^3_0 v^3_1 v^3_3], \dots, [v^n_0 v^n_1 v^n_2] \sim [v^1_0 v^1_1 v^1_3] So, we have 2n faces: % Examining carefully the identifications this induces, we have n+2 edges: % And only 2 vertices: % So we have \Delta_0 \cong \mathbb{Z}^2, \Delta_1 \cong \mathbb{Z}^{n+2}, \Delta_2 \cong \mathbb{Z}^{2n}, \Delta_3 \cong \mathbb{Z}^n. We calculate the homology groups: For \delta_0, we have \im \delta_0 = \set{0} and \ker \delta_0 = \Delta_0. For \delta_1, we have: % For \delta_2, we have: % For \delta_3, we have: % It’s clear that H_0 \cong \mathbb{Z}. Let’s represent the rest of the boundary maps as matrices, to compute the homology: % % % Where the matrices on the right are the corresponding Smith normal forms. Then, we compute the Betti numbers: % So, besides H_0, H_3 is the only homology group with non-trivial free rank, and H_1 has a non-trivial torsion coefficient coming from \mathsf{SNF}(\delta_2). We conclude: H_0 \cong \mathbb{Z} \qquad H_1 \cong \mathbb{Z}_n \qquad H_2 \cong 0 \qquad H_3 \cong \mathbb{Z}