Homology: Homology Groups

Definition. Let $K$ be an $n$-dimensional simplicial complex. The $r$-th homology group $H_r(K)$, $0\leq r\leq n$, associated with $K$ is defined by
$$H_r(K)=Z_r(K)/B_r(K)$$

In the last theorem here, we have seen that $B_r(K)\subset Z_r(K)$, so the definition of a homology group makes sense.

The $r$-th homology group $H_r(K)$ can be written as the set of equivalence classes
$$H_r(K)=\{[z]: z\in Z_r(K)\}$$
where $[z]=z+B_r(K)$. Each equivalence class $[z]$ of an $r$-cycle $z\in Z_r(K)$ is called a homology class. Two $r$-cycles $z$ and $z'$ are in the same equivalent class if and only if $z-z'\in B_r(K)$. In this case, we say $z$ is homologous to $z'$ and write $z\sim z'$.

We are interested in studying homology groups because they are topological invariants as seen in the following theorem.

Theorem 1. Let $X$ be homeomorphic to $Y$ and let $(K,f)$ and $(L,g)$ be triangulations of $X$ and $Y$ respectively. Then we have
$$H_r(K)\cong H_r(L),\ r=0,1,2,\cdots$$
In particular, if $(K,f)$ and $(L,g)$ are two triangulations of $X$, then
$$H_r(K)\cong H_r(L),\ r=0,1,2,\cdots$$
Here, $\cong$ means "is isomorphic to" of course.

One can speak of homological groups of a topological space $X$ which is not necessarily a simplicial complex as long as it is triangulable. For an arbitrary triangulation $(K,f)$, $H_r(X)$ is defined to be
$$H_r(X):=H_r(K),\ r=0,1,2,\cdots$$

Example. Let $K=\{p_0\}$. Since $\dim K=0$, $H_r(K)=0$ for all $r\geq 1$. The $0$-chain is $C_0(K)=\{ip_0: i\in\mathbb{Z}\}\cong\mathbb{Z}$. $Z_0(K)=C_0(K)$ and $B_0(K)=0$ (since $K$ has no $1$-simplexes). Thus,
$$H_0(K)=Z_0(K)/B_0(K)\cong\mathbb{Z}$$

Theorem 2. If $K$ is a contractible space i.e. if it has the homotopy type of a single point then
$$
H_r(K)=\left\{\begin{array}{ccc}
0 & \mbox{if} & r\ne 0\\
\mathbb{Z} & \mbox{if} & r=0
\end{array}\right.
$$

If $K$ is a $2$-dimensional space, then $K$ is contractible if and only if it has no holes.

Exercise. Let $K=\{p_0,p_1\}$. Compute the homology groups of $K$ to confirm that $H_0(K)=\mathbb{Z}\oplus\mathbb{Z}$ and $H_r(K)=0$ for all $r\geq 1$.

Exercise. Let $K=\{p_0,p_1,(p_0p_1)\}$. Compute the homology groups of $K$ to confirm that $H_0(K)=\mathbb{Z}$ and $H_r(K)=0$ for all $r\geq 1$.

Example. Let $K=\{p_0,p_1,p_2,(p_0p_1),(p_1p_2),(p_2p_0)\}$. $K$ is a triangulation of $S^1$. Since $\dim K=1$, $H_r(K)=0$, $\forall r\geq 2$. $C_0(K)=Z_0(K)=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}$.
\begin{align*}
B_0(K)&=\{l\partial_1(p_0p_1)+m\partial_1(p_1p_2)+n\partial_1(p_2p_0): l,m,n\in\mathbb{Z}\}\\
&=\{l(p_1-p_0)+m(p_2-p_1)+n(p_0-p_2): l,m,n\in\mathbb{Z}\}\\
&=\{(n-l)p_0+(l-m)p_1+(m-n)p_2:l,m,n\in\mathbb{Z}\}
\end{align*}
Note that $(n-l)+(l-m)+(m-n)=0$. Define $f:Z_0(K)\longrightarrow\mathbb{Z}$ by $f(ip_0+jp_1+kp_3)=i+j+k$. Then $f$ is an onto homomorphism.
\begin{align*}
ip_0+jp_1+kp_3\in\ker f&\Leftrightarrow i+j+k=0\\
&\Leftrightarrow ip_0+jp_1+kp_3\in b_0(K)
\end{align*}
Hence, $\ker f=B_0(K)$ and $H_0(K)=Z_0(K)/B_0(K)\cong\mathbb{Z}$. Since there are no $2$-simplexes, $B_1(K)=0$. $C_1(K)=\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}$. Let $z\in Z_1(K)\subset C_1(K)$. Then
$z=i(p_0p_1)+j(p_1p_2)+k(p_2p_0)$ for some $i,j,k\in\mathbb{Z}$. Then
\begin{align*}
\partial_1z&=i(p_1-p_0)+j(p_2-p_1)+k(p_0-p_2)\\
&=(k-i)p_0+(i-j)p_1+(j-k)p_2=0\\
&\Rightarrow i=j=k
\end{align*}
Hence,
$$Z_1(K)=\{i[(p_0p_1)+(p_1p_2)+(p_2p_0)]: i\in\mathbb{Z}\}\cong\mathbb{Z}$$
since it is generated by a single $1$-chain $(p_0p_1)+(p_1p_2)+(p_2p_0)$, and $H_1(K)=Z_1(K)=\mathbb{Z}$.

Remark. $L=\{p_0,p_1,p_2,p_3,(p_0p_1),(p_1p_2),(p_2p_3),(p_3p_0)\}$ is a simplicial complex whose polyhedron is an oriented square. Since a square is homeomorphic to a circle, $L$ is also a triangulation of $S^1$. However, it is a less economical triangulation of $S^1$ than $K$ in the preceding example. By theorem 1, $H_r(L)\cong H_r(K)$ for all $r\geq 0$, i.e. $H_0(L)=H_1(L)=\mathbb{Z}$ and $H_r(L)=0$ for all $r\geq 2$.

Example. Let $K=\{p_0,p_1,p_2,(p_0p_1),(p_1p_2),(p_2p_0),(p_0p_1p_2)\}$.

Figure 1. A Contractible Simplicial Complex

 $K$ has no holes so it is contractible. By theorem 2, $H_0(K)=\mathbb{Z}$ and $H_r(K)=0$, $\forall r\geq 1$.