Find the Euler characteristic of \(X = (\Delta^5)^{(2)}\), the 2-skeleton of the 5-simplex. Show that \(H_1(X) = 0\), \(H_2(X)\) is free abelian, and compute the rank of \(H_2(X)\).
Solution
1. The complex \(X=(\Delta^5)^{(2)}\) has \(6\) vertices, \(15\) edges, and \(20\) two-simplices.Hence\[ \chi(X)=6-15+20=11. \]
The \(k\)-simplices of \(\Delta^5\) correspond to the \((k+1)\)-element subsets of its \(6\) vertices. Therefore \[ c_0=\binom61=6, \qquad c_1=\binom62=15, \qquad c_2=\binom63=20. \] Since \(X\) has no cells above dimension \(2\), its Euler characteristic is the alternating sum shown.
2. The complex \(X\) is simply connected, so\[ H_1(X;\mathbb Z)=0. \]
The full simplex \(\Delta^5\) is contractible and therefore has trivial fundamental group. The complex \(\Delta^5\) is obtained from its \(2\)-skeleton \(X\) by attaching cells of dimensions \(3,4,\) and \(5\). Attaching cells of dimension at least \(3\) does not change the fundamental group, since the fundamental group of a CW complex depends only on its \(2\)-skeleton. Thus \[ \pi_1(X)\cong\pi_1(\Delta^5)=0. \] The first homology group is the abelianization of the fundamental group, hence \[ H_1(X;\mathbb Z)=0. \]
3. The group \(H_2(X;\mathbb Z)\) is free abelian.
Because \(X\) is \(2\)-dimensional, its cellular chain complex has \(C_3(X)=0\). Therefore \[ H_2(X;\mathbb Z)=\ker\bigl(\partial_2:C_2(X)\to C_1(X)\bigr). \] The cellular group \(C_2(X)\) is free abelian on the \(20\) two-simplices, so \[ C_2(X)\cong\mathbb Z^{20}. \] Every subgroup of a finitely generated free abelian group is free abelian, hence so is \(H_2(X;\mathbb Z)\).
4. The rank of \(H_2(X;\mathbb Z)\) is \(10\).
The complex \(X\) is connected, so \[ H_0(X;\mathbb Z)\cong\mathbb Z. \] There are no homology groups above degree \(2\), and 2 gives \(H_1(X)=0\). The Euler-characteristic formula therefore yields \[ 11 =\chi(X) =\operatorname{rank}H_0(X)-\operatorname{rank}H_1(X)+\operatorname{rank}H_2(X) =1+\operatorname{rank}H_2(X). \] Consequently \[ \operatorname{rank}H_2(X;\mathbb Z)=10. \]
5. Thus\[ \chi(X)=11, \qquad H_1(X;\mathbb Z)=0, \qquad H_2(X;\mathbb Z)\cong\mathbb Z^{10}. \]
The Euler characteristic is 1, the vanishing of \(H_1\) is 2, and <1>3–<1>4 show that \(H_2\) is free abelian of rank \(10\).