The surface \(M_g\) of genus \(g\), embedded in \(\mathbb{R}^3\) in the standard way, bounds a compact region \(R\). Two copies of \(R\), glued together by the identity map between their boundary surfaces \(M_g\), form a closed 3-manifold \(X\). Compute the homology groups of \(X\) via the Mayer–Vietoris sequence for this decomposition of \(X\) into two copies of \(R\). Also compute the relative groups \(H_i(R, M_g)\).
Solution
The region \(R\) bounded by the standard genus-\(g\) surface is a genus-\(g\) handlebody. Hence \[ R\simeq\bigvee^g S^1, \] so \[ H_0(R)=\mathbb Z, \qquad H_1(R)=\mathbb Z^g, \qquad H_i(R)=0\quad(i\ge2). \] The inclusion \[ i:M_g=\partial R\hookrightarrow R \] induces a surjection \[ i_*:H_1(M_g)=\mathbb Z^{2g}\twoheadrightarrow H_1(R)=\mathbb Z^g \] whose kernel has rank \(g\): one may choose the usual symplectic basis so that the meridians die in \(R\) and the longitudes form a basis of \(H_1(R)\).
Let \[ X=R_1\cup_{M_g}R_2 \] be the double.
1. The homology of \(X\) is\[ H_i(X)\cong \begin{cases} \mathbb Z,&i=0,3,\\ \mathbb Z^g,&i=1,2,\\ 0,&\text{otherwise}. \end{cases} \]
Mayer–Vietoris gives in high degrees \[ 0\to H_3(X)\to H_2(M_g)=\mathbb Z \to H_2(R_1)\oplus H_2(R_2)=0, \] so \(H_3(X)\cong\mathbb Z\). Next, \[ 0\to H_2(X)\to H_1(M_g) \xrightarrow{(i_*,-i_*)} H_1(R_1)\oplus H_1(R_2) \to H_1(X)\to0. \] The kernel of \((i_*,-i_*)\) is exactly \(\ker i_*\cong\mathbb Z^g\), hence \[ H_2(X)\cong\mathbb Z^g. \] Since \(i_*\) is onto, the image of \((i_*,-i_*)\) is the anti-diagonal copy of \(\mathbb Z^g\) in \(\mathbb Z^g\oplus\mathbb Z^g\). Its cokernel is \(\mathbb Z^g\), so \[ H_1(X)\cong\mathbb Z^g. \] Finally \(X\) is connected, giving \(H_0(X)=\mathbb Z\).
2. The relative homology of the handlebody pair is\[ H_i(R,M_g)\cong \begin{cases} \mathbb Z,&i=3,\\ \mathbb Z^g,&i=2,\\ 0,&\text{otherwise}. \end{cases} \]
The long exact sequence of the pair begins \[ 0=H_3(R)\to H_3(R,M_g)\to H_2(M_g)=\mathbb Z\to H_2(R)=0, \] so \(H_3(R,M_g)\cong\mathbb Z\). The next portion is \[ 0\to H_2(R,M_g)\to H_1(M_g) \xrightarrow{i_*}H_1(R) \to H_1(R,M_g) \to H_0(M_g)\xrightarrow{\cong}H_0(R). \] Since \(i_*\) is surjective with kernel \(\mathbb Z^g\), exactness gives \[ H_2(R,M_g)\cong\mathbb Z^g, \qquad H_1(R,M_g)=0, \qquad H_0(R,M_g)=0. \]