Construct a 3 dimensional \(\Delta\)-complex \(X\) from \(n\) tetrahedra \(T_1, \cdots, T_n\) by the following two steps. First arrange the tetrahedra in a cyclic pattern as in the figure, so that each \(T_i\) shares a common vertical face with its two neighbors \(T_{i-1}\) and \(T_{i+1}\), subscripts being taken mod \(n\). Then identify the bottom face of \(T_i\) with the top face of \(T_{i+1}\) for each \(i\). Show the simplicial homology groups of \(X\) in dimensions 0, 1, 2, 3 are \(\mathbb{Z}\), \(\mathbb{Z}_n\), \(0\), \(\mathbb{Z}\), respectively. [The space \(X\) is an example of a lens space; see Example 2.43 for the general case.]
Solution
Use the notation in Hatcher’s figure. After all identifications there are two vertices \(v,w\), \(n+2\) edges, \(2n\) faces, and \(n\) tetrahedra. Choose generators \[ p,b,f_1,\dots,f_n\in C_1, \qquad A_1,\dots,A_n,B_1,\dots,B_n\in C_2, \qquad T_1,\dots,T_n\in C_3, \] with subscripts modulo \(n\).
1. The boundary map \(\partial_1:C_1\to C_0\) has rank \(1\).
The edges \(p\) and \(b\) are loops, while each \(f_i\) joins \(v\) to \(w\). Thus \[ \partial_1(p)=\partial_1(b)=0, \qquad \partial_1(f_i)=w-v. \] Hence \(\operatorname{im}\partial_1=\mathbb Z(w-v)\) has rank one. Therefore \[ H_0(X)\cong\mathbb Z. \]
2. The boundaries of the tetrahedra can be oriented so that\[ \partial_3(T_i)=A_i-A_{i-1}+B_{i-1}-B_i. \]Consequently \(\ker\partial_3\) is generated by\[ T_1+\cdots+T_n. \]
The displayed formula follows directly by taking the alternating sum of the four oriented faces of \(T_i\) and applying the prescribed face identifications. If \[ \sum_i c_i\partial_3(T_i)=0, \] then the coefficient of \(A_i\) is \(c_i-c_{i+1}\), hence all \(c_i\) are equal. Conversely, the sum of all \(\partial_3(T_i)\) telescopes to zero. Thus \[ \ker\partial_3\cong\mathbb Z. \] Since there are no \(4\)-simplices, \[ H_3(X)\cong\mathbb Z. \]
3. After elementary integral row and column operations, the matrix of \(\partial_2:C_2\to C_1\) has Smith normal form with nonzero diagonal entries\[ 1,\dots,1,n, \]where there are \(n\) entries equal to \(1\) and one entry equal to \(n\).
For the chosen orientations, the \(A_i\)- and \(B_i\)-boundaries are \[ \partial_2(A_i)=p+f_i-f_{i+1}, \qquad \partial_2(B_i)=b+f_i-f_{i+1}, \] with indices modulo \(n\). Subtract the \(A_i\)-columns successively from \(A_{i+1}\) and likewise for the \(B_i\)-columns. The difference columns isolate the \(f_i-f_{i+1}\) relations. Next subtract one \(A\)-column from the corresponding \(B\)-column to isolate \(b-p\). These operations split off \(n\) unit invariant factors. The sole remaining relation is the sum of the cyclic differences, which contributes the invariant factor \(n\). Equivalently, the quotient of \(\ker\partial_1\) by \(\operatorname{im}\partial_2\) is generated by one class with relation \(n\) times that class equal to zero.
4. Hence\[ H_1(X)\cong\mathbb Z/n\mathbb Z. \]
Since \(\ker\partial_1\) has rank \[ (n+2)-1=n+1, \] and the Smith form in 3 has rank \(n+1\), the quotient has no free part. Its only nontrivial invariant factor is \(n\), giving \[ H_1(X)=\ker\partial_1/\operatorname{im}\partial_2\cong\mathbb Z/n\mathbb Z. \]
5. One has\[ H_2(X)=0. \]
Write a general \(2\)-chain as \[ z=\sum_i a_iA_i+\sum_i b_iB_i. \] Using the formulas from 3, the coefficients of \(p\) and \(b\) in \(\partial_2z\) are respectively \[ \sum_i a_i, \qquad \sum_i b_i, \] and the coefficient of \(f_i\) is \[ (a_i+b_i)-(a_{i-1}+b_{i-1}). \] Hence \(\partial_2z=0\) exactly when all \(a_i+b_i\) are equal and both displayed sums vanish. If the common value is \(c\), then \[ 0=\sum_i(a_i+b_i)=nc, \] so \(c=0\) in \(\mathbb Z\). Thus \(b_i=-a_i\) for all \(i\) and \(\sum_i a_i=0\).
2. Hence\[ \ker\partial_2=\operatorname{im}\partial_3, \]so \(H_2(X)=0\).
6. Therefore\[ \boxed{ H_0(X)\cong\mathbb Z, \quad H_1(X)\cong\mathbb Z/n\mathbb Z, \quad H_2(X)=0, \quad H_3(X)\cong\mathbb Z. } \]
Combine <1>1–<1>5.