Compute the first three homology groups of the hollow sphere \(S^2\). You may use simplicial theory and a triangulation but do not simply say that \(H_1(S^2)\) is the abelianization of \(\pi_1(S^2)\).
Solution
Triangulate \(S^2\) as the boundary of a tetrahedron with vertices \[ v_0,v_1,v_2,v_3. \] Write \[ e_{ij}=[v_i,v_j] \qquad(i<j) \] for the oriented edges and \[ F_0=[v_1,v_2,v_3],\quad F_1=[v_0,v_2,v_3],\quad F_2=[v_0,v_1,v_3],\quad F_3=[v_0,v_1,v_2] \] for the oriented faces. The simplicial chain groups are \[ C_2\cong\mathbb Z^4, \qquad C_1\cong\mathbb Z^6, \qquad C_0\cong\mathbb Z^4, \] and \(C_3=0\).
1. The image of\[ \partial_1:C_1\to C_0 \]is exactly the subgroup of \(0\)-chains whose coefficients sum to zero.
For every \(i<j\), \[ \partial_1 e_{ij}=v_j-v_i, \] whose coefficients sum to zero. Thus \[ \operatorname{im}\partial_1 \subseteq \left\{\sum_{i=0}^3 a_i v_i:\sum_{i=0}^3a_i=0\right\}. \]
Conversely, if \[ c=a_0v_0+a_1v_1+a_2v_2+a_3v_3 \] has \[ a_0+a_1+a_2+a_3=0, \] then \[ c =a_1(v_1-v_0)+a_2(v_2-v_0)+a_3(v_3-v_0) =\partial_1(a_1e_{01}+a_2e_{02}+a_3e_{03}). \] Hence the reverse inclusion also holds.
2. The zeroth homology group is\[ H_0(S^2)\cong\mathbb Z. \]
Since \(\partial_0=0\), \[ H_0(S^2)=C_0/\operatorname{im}\partial_1. \] Define the augmentation \[ \varepsilon:C_0\to\mathbb Z, \qquad \varepsilon\left(\sum a_iv_i\right)=\sum a_i. \] This is surjective, and by 1 its kernel is precisely \(\operatorname{im}\partial_1\). Therefore the first isomorphism theorem gives \[ H_0(S^2) =C_0/\ker\varepsilon \cong\mathbb Z. \]
3. Every simplicial \(1\)-cycle is a simplicial boundary.
Write an arbitrary \(1\)-chain as \[ c =a_{01}e_{01}+a_{02}e_{02}+a_{03}e_{03} +a_{12}e_{12}+a_{13}e_{13}+a_{23}e_{23}. \] The equation \(\partial_1c=0\) is equivalent to \[ \begin{aligned} a_{01}+a_{02}+a_{03}&=0,\\ a_{01}-a_{12}-a_{13}&=0,\\ a_{02}+a_{12}-a_{23}&=0,\\ a_{03}+a_{13}+a_{23}&=0. \end{aligned} \] The last equation follows from the first three, and the first three give \[ a_{01}=a_{12}+a_{13}, \qquad a_{02}=-a_{12}+a_{23}, \qquad a_{03}=-a_{13}-a_{23}. \]
Now \[ \begin{aligned} \partial_2F_3&=e_{12}-e_{02}+e_{01},\\ \partial_2F_2&=e_{13}-e_{03}+e_{01},\\ \partial_2F_1&=e_{23}-e_{03}+e_{02}. \end{aligned} \] Substituting the cycle relations above yields \[ c =a_{12}\partial_2F_3 +a_{13}\partial_2F_2 +a_{23}\partial_2F_1. \] Hence \[ c\in\operatorname{im}\partial_2. \] Since every boundary is automatically a cycle because \[ \partial_1\partial_2=0, \] we conclude \[ \ker\partial_1=\operatorname{im}\partial_2. \]
4. The first homology group is\[ H_1(S^2)=0. \]
By definition, \[ H_1(S^2) =\ker\partial_1/\operatorname{im}\partial_2. \] The numerator and denominator are equal by 3, so the quotient is the zero group.
5. The kernel of\[ \partial_2:C_2\to C_1 \]is infinite cyclic, generated by\[ F_0-F_1+F_2-F_3. \]
The face boundaries are \[ \begin{aligned} \partial_2F_0&=e_{23}-e_{13}+e_{12},\\ \partial_2F_1&=e_{23}-e_{03}+e_{02},\\ \partial_2F_2&=e_{13}-e_{03}+e_{01},\\ \partial_2F_3&=e_{12}-e_{02}+e_{01}. \end{aligned} \] Let \[ z=aF_0+bF_1+cF_2+dF_3. \] If \(\partial_2z=0\), comparing respectively the coefficients of \[ e_{23},\quad e_{13},\quad e_{12} \] gives \[ a+b=0, \qquad -a+c=0, \qquad a+d=0. \] Thus \[ b=-a, \qquad c=a, \qquad d=-a. \] The remaining edge coefficients then vanish automatically. Hence every \(2\)-cycle has the form \[ z=a(F_0-F_1+F_2-F_3). \] Conversely, the displayed generator has zero boundary by the four formulas above. Therefore \[ \ker\partial_2 =\mathbb Z(F_0-F_1+F_2-F_3) \cong\mathbb Z. \]
6. The second homology group is\[ H_2(S^2)\cong\mathbb Z. \]
There are no \(3\)-simplices in the boundary of a tetrahedron, so \[ C_3=0 \qquad\text{and hence}\qquad \operatorname{im}\partial_3=0. \] Therefore, by 5, \[ H_2(S^2) =\ker\partial_2/\operatorname{im}\partial_3 =\ker\partial_2 \cong\mathbb Z. \]
7. Hence the first three homology groups of the hollow sphere are\[ \boxed{H_0(S^2)\cong\mathbb Z,\qquad H_1(S^2)=0,\qquad H_2(S^2)\cong\mathbb Z.} \]
This is the combination of 2, 4, and 6.