3.3 Homology groups of simplicial complexes

3.3.1 Oriented simplexes

We may assign orientations to an $r$ -simplex for $r\ge 1$ . Instead of $⟨...⟩$ for an unoriented simplex, we will use $(...)$ to denote an oriented simplex. The symbol ${\sigma }_r$ is used to denote both types of simplex. An oriented 1-simplex ${\sigma }_1=(p_0{p}_1)$ is a directed line segment traversed in the direction $p_0\to p_1$ (figure 3.6(a)). Now

(a)

(b)

Figure 3.5. $(a)$ is a triangulation of a cylinder while $(b)$ is not.

Figure 3.6. An oriented 1-simplex $(a)$ and an oriented 2-simplex $(b)$ .

$(p_0{p}_1)$ should be distinguished from $(p_1{p}_0)$ . We require that

$$(p_0{p}_1)=-(p_1{p}_0).\text{\hspace{1em}(3.13)}$$

Here $-$ in front of $(p_1{p}_0)$ should be understood in the sense of a finitely generated Abelian group. In fact, $(p_1{p}_0)$ is regarded as the inverse of $(p_0{p}_1)$ . Going from $p_0$ to $p_1$ followed by going from $p_1$ to $p_0$ means going nowhere, $(p_0{p}_1)+(p_1{p}_0)=0$ , hence $-(p_1{p}_0)=(p_0{p}_1)$ .

Similarly, an oriented 2-simplex ${\sigma }_2=(p_0{p}_1{p}_2)$ is a triangular region $p_0{p}_1{p}_2$ with a prescribed orientation along the edges (figure 3.6(b)). Observe that the orientation given by $p_0{p}_1{p}_2$ is the same as that given by $p_2{p}_0{p}_1$ or $p_1{p}_2{p}_0$ but opposite to $p_0{p}_2{p}_1$ , $p_2{p}_1{p}_0$ or $p_1{p}_0{p}_2$ . We require that

$$\begin{array}{rl}(p_0{p}_1{p}_2)& =(p_2{p}_0{p}_1)=(p_1{p}_2{p}_0)\\ & =-\left(p_0{p}_2{p}_1\right)=-\left(p_2{p}_1{p}_0\right)=-\left(p_1{p}_0{p}_2\right).\end{array}$$

Let $P$ be a permutation of 0, 1, 2

$$P=\left(\begin{array}{ccc}0& 1& 2\\ i& j& k\end{array}\right).$$

These relations are summarized as

$$(p_i{p}_j{p}_k)=\mathrm{sgn}(P)(p_0{p}_1{p}_2)$$

where $\mathrm{sgn}(P)=+1$ ( $-1$ ) if $P$ is an even (odd) permutation.

An oriented 3-simplex ${\sigma }_3=(p_0{p}_1{p}_2{p}_3)$ is an ordered sequence of four vertices of a tetrahedron. Let

$$P=\left(\begin{array}{cccc}0& 1& 2& 3\\ i& j& k& l\end{array}\right)$$

be a permutation. We define

$$\left(p_i{p}_j{p}_k{p}_l\right)=\mathrm{sgn}(P)\left(p_0{p}_1{p}_2{p}_3\right).$$

It is now easy to construct an oriented $r$ -simplex for any $r\ge 1$ . The formal definition goes as follows. Take $r+1$ geometrically independent points $p_0,p_1,\dots ,p_r$ in ${\mathbb{R}}^m$ . Let $\{p_{i_0},p_{i_1},\dots ,p_{i_r}\}$ be a sequence of points obtained by a permutation of the points $p_0,\dots ,p_r$ . We define $\{p_0,\dots ,p_r\}$ and $\{p_{i_0},\dots ,p_{i_r}\}$ to be equivalent if

$$P=\left(\begin{array}{llll}0& 1& \dots & r\\ i_0& i_1& \dots & i_r\end{array}\right)$$

is an even permutation. Clearly this is an equivalence relation, the equivalence class of which is called an oriented $r$ -simplex. There are two equivalence classes, one consists of even permutations of $p_0,\dots ,p_r$ , the other of odd permutations. The equivalence class (oriented $r$ -simplex) which contains $\{p_0,\dots ,p_r\}$ is denoted by ${\sigma }_r=(p_0{p}_1\dots p_r)$ , while the other is denoted by $-{\sigma }_r=-(p_0{p}_1\dots p_r)$ . In other words,

$$\left(p_{i_0}{p}_{i_1}\dots p_{i_r}\right)=\mathrm{sgn}(P)\left(p_0{p}_1\dots p_r\right).\text{\hspace{1em}(3.14)}$$

For $r=0$ , we formally define an oriented 0-simplex to be just a point ${\sigma }_0=p_0$ .

3.3.2 Chain group, cycle group and boundary group

Let $K=\{{\sigma }_{\alpha }\}$ be an $n$ -dimensional simplicial complex. We regard the simplexes ${\sigma }_{\alpha }$ in $K$ as oriented simplexes and denote them by the same symbols ${\sigma }_{\alpha }$ as remarked before.

Definition 3.2

The $r$ -chain group $C_r(K)$ of a simplicial complex $K$ is a free Abelian group generated by the oriented $r$ -simplexes of $K$ . If $r>\dim K$ , $C_r(K)$ is defined to be 0. An element of $C_r(K)$ is called an $r$ -chain.

Let there be $I_r$ $r$ -simplexes in $K$ . We denote each of them by ${\sigma }_{r,i}$ ( $1\le i\le I_r$ ). Then $c\in C_r(K)$ is expressed as

$$c=\sum _{i=1}^{I_r}{c}_i{\sigma }_{r,i}{c}_i\in \mathbb{Z}.\text{\hspace{1em}(3.15)}$$

The integers $c_i$ are called the coefficients of $c$ . The group structure is given as follows. The addition of two $r$ -chains, $c={\sum }_i{c}_i{\sigma }_{r,i}$ and $c^{\prime }={\sum }_i{c}_i^{\prime }{\sigma }_{r,i}$ , is

$$c+c^{\prime }=\sum _i\left(c_i+c_i^{\prime }\right){\sigma }_{r,i}.\text{\hspace{1em}(3.16)}$$

The unit element is $0={\sum }_{i}0\cdot {\sigma }_{r,i}$ , while the inverse element of $c$ is $-c={\sum }_i(-c_i){\sigma }_{r,i}$ . [Remark: An oppositely oriented $r$ -simplex $-{\sigma }_r$ is identified with $(-1){\sigma }_r\in C_r(K)$ .] Thus, $C_r(K)$ is a free Abelian group of rank $I_r$ ,

$$C_r(K)\cong \underset{I_r}{\underbrace{\mathbb{Z}\oplus \mathbb{Z}\oplus \cdots \oplus \mathbb{Z}}}.\text{\hspace{1em}(3.17)}$$

Before we define the cycle group and the boundary group, we need to introduce the boundary operator. Let us denote the boundary of an $r$ -simplex ${\sigma }_r$ by ${\partial }_r{\sigma }_r$ . ${\partial }_r$ should be understood as an operator acting on ${\sigma }_r$ to produce its boundary. This point of view will be elaborated later. Let us look at the boundaries of lower-dimensional simplexes. Since a 0-simplex has no boundary, we define

$${\partial }_0{p}_0=0.\text{\hspace{1em}(3.18)}$$

For a 1-simplex $(p_0{p}_1)$ , we define

$${\partial }_1\left(p_0{p}_1\right)=p_1-p_0.\text{\hspace{1em}(3.19)}$$

The reader might wonder about the appearance of a minus sign in front of $p_0$ . This is again related to the orientation. The following examples will clarify this point. In figure 3.7(a), an oriented 1-simplex $(p_0{p}_2)$ is divided into two, $(p_0{p}_1)$ and $(p_1{p}_2)$ . We agree that the boundary of $(p_0{p}_2)$ is $\{p_0\}\cup \{p_2\}$ and so should be that of $(p_0{p}_1)+(p_1{p}_2)$ . If ${\partial }_1(p_0{p}_2)$ were defined to be $p_0+p_2$ , we would have ${\partial }_1(p_0{p}_1)+{\partial }_1(p_1{p}_2)=p_0+p_1+p_1+p_2$ . This is not desirable since $p_1$ is a fictitious boundary. If, instead, we take ${\partial }_1(p_0{p}_2)=p_2-p_0$ , we will have ${\partial }_1(p_0{p}_1)+{\partial }_1(p_1{p}_2)=p_1-p_0+p_2-p_1=p_2-p_0$ as expected. The next example is the triangle of figure 3.7(b). It is the sum of three oriented 1-simplexes,

Figure 3.7. (a) An oriented 1-simplex with a fictitious boundary $p_1$ . (b) A simplicial complex without a boundary.

$(p_0{p}_1)+(p_1{p}_2)+(p_2{p}_0)$ . We agree that it has no boundary. If we insisted on the rule ${\partial }_1(p_0{p}_1)=p_0+p_1$ , we would have

$${\partial }_1\left(p_0{p}_1\right)+{\partial }_1\left(p_1{p}_2\right)+{\partial }_1\left(p_2{p}_0\right)=p_0+p_1+p_1+p_2+p_2+p_0$$

which contradicts our intuition. If, on the other hand, we take ${\partial }_1(p_0{p}_1)=p_1-p_0$ , we have

$${\partial }_1\left(p_0{p}_1\right)+{\partial }_1\left(p_1{p}_2\right)+{\partial }_1\left(p_2{p}_0\right)=p_1-p_0+p_2-p_1+p_0-p_2=0$$

as expected. Hence, we put a plus sign if the first vertex is omitted and a minus sign if the second is omitted. We employ this fact to define the boundary of a general $r$ -simplex.

Let ${\sigma }_r(p_0\dots p_r)$ ( $r>0$ ) be an oriented $r$ -simplex. The boundary ${\partial }_r{\sigma }_r$ of ${\sigma }_r$ is an $(r-1)$ -chain defined by

$${\partial }_r{\sigma }_r\equiv \sum _{i=0}^r(-1{)}^i\left(p_0{p}_1\dots {\hat{p}}_i\dots p_r\right)\text{\hspace{1em}(3.20)}$$

where the point $p_i$ under $\hat{\mu }$ is omitted. For example,

$$\begin{array}{rl}& {\partial }_2\left(p_0{p}_1{p}_2\right)=\left(p_1{p}_2\right)-\left(p_0{p}_2\right)+\left(p_0{p}_1\right)\\ & {\partial }_3\left(p_0{p}_1{p}_2{p}_3\right)=\left(p_1{p}_2{p}_3\right)-\left(p_0{p}_2{p}_3\right)+\left(p_0{p}_1{p}_3\right)-\left(p_0{p}_1{p}_2\right).\end{array}$$

We formally define ${\partial }_0{\sigma }_0=0$ for $r=0$ .

The operator ${\partial }_r$ acts linearly on an element $c={\sum }_i{c}_i{\sigma }_{r,i}$ of $C_r(K)$ ,

$${\partial }_{r}c=\sum _i{c}_i{\partial }_r{\sigma }_{r,i}.\text{\hspace{1em}(3.21)}$$

The RHS of (3.21) is an element of $C_{r-1}(K)$ . Accordingly, ${\partial }_r$ defines a map

$${\partial }_r:C_r(K)\to C_{r-1}(K).\text{\hspace{1em}(3.22)}$$

${\partial }_r$ is called the boundary operator. It is easy to see that the boundary operator is a homomorphism.

Let $K$ be an $n$ -dimensional simplicial complex. There exists a sequence of free Abelian groups and homomorphisms,

$$0\stackrel{i}{\to }{C}_n(K)\stackrel{{\partial }_n}{\to }{C}_{n-1}(K)\stackrel{{\partial }_{n-1}}{\to }\cdots \stackrel{{\partial }_2}{\to }{C}_1(K)\stackrel{{\partial }_1}{\to }{C}_0(K)\stackrel{{\partial }_0}{\to }0\text{\hspace{1em}(3.23)}$$

where $i:0\hookrightarrow C_n(K)$ is an inclusion map (0 is regarded as the unit element of $C_n(K)$ ). This sequence is called the chain complex associated with $K$ and is denoted by $C(K)$ . It is interesting to study the image and kernel of the homomorphisms ${\partial }_r$ .

Definition 3.3

If $c\in C_r(K)$ satisfies

$${\partial }_{r}c=0\text{\hspace{1em}(3.24)}$$

$c$ is called an $r$ -cycle. The set of $r$ -cycles $Z_r(K)$ is a subgroup of $C_r(K)$ and is called the $r$ -cycle group. Note that $Z_r(K)=\ker {\partial }_r$ . [Remark: If $r=0$ , ${\partial }_{0}c$ vanishes identically and $Z_0(K)=C_0(K)$ , see (3.23).]

Definition 3.4

Let $K$ be an $n$ -dimensional simplicial complex and let $c\in C_r(K)$ . If there exists an element $d\in C_{r+1}(K)$ such that

$$c={\partial }_{r+1}d\text{\hspace{1em}(3.25)}$$

then $c$ is called an $r$ -boundary. The set of $r$ -boundaries $B_r(K)$ is a subgroup of $C_r(K)$ and is called the $r$ -boundary group. Note that $B_r(K)=\mathrm{im}{\partial }_{r+1}$ . [Remark: $B_n(K)$ is defined to be 0.]

From lemma 3.1, it follows that $Z_r(K)$ and $B_r(K)$ are subgroups of $C_r(K)$ . We now prove an important relation between $Z_r(K)$ and $B_r(K)$ , which is crucial in the definition of homology groups.

Lemma 3.3

The composite map ${\partial }_r\circ {\partial }_{r+1}:C_{r+1}(K)\to C_{r-1}(K)$ is a zero map; that is, ${\partial }_r({\partial }_{r+1}c)=0$ for any $c\in C_{r+1}(K)$ .

Proof

Since ${\partial }_r$ is a linear operator on $C_r(K)$ , it is sufficient to prove the identity ${\partial }_r\circ {\partial }_{r+1}=0$ for the generators of $C_{r+1}(K)$ . If $r=0$ , ${\partial }_0\circ {\partial }_1=0$ since ${\partial }_0$ is a zero operator. Let us assume $r>0$ . Take $\sigma =(p_0\dots p_r{p}_{r+1})\in C_{r+1}(K)$ . We find

$$\begin{array}{rl}& {\partial }_r({\partial }_{r+1}\sigma )={\partial }_r\sum _{i=0}^{r+1}(-1{)}^i(p_0\dots \widehat{p_i}\dots p_{r+1})\\ & =\sum _{i=0}^{r+1}(-1{)}^i{\partial }_r\left(p_0\dots {\hat{p}}_i\dots p_{r+1}\right)\\ & =\sum _{i=0}^{r+1}(-1{)}^i(\sum _{j=0}^{i-1}(-1{)}^j\left(p_0\dots p_j\dots p_i\dots p_{r+1}\right)\\ & +\sum _{j=i+1}^{r+1}(-1{)}^{j-1}\left(p_0\dots {\hat{p}}_i\dots {\hat{p}}_j\dots p_{r+1}\right))\\ & =\sum _{j<i}(-1{)}^{i+j}\left(p_0\dots {\hat{p}}_j\dots {\hat{p}}_i\dots p_{r+1}\right)\\ & -\sum _{j>i}(-1{)}^{i+j}\left(p_0\dots {\hat{p}}_i\dots {\hat{p}}_j\dots p_{r+1}\right)=0\end{array}\text{\hspace{1em}(3.26)}$$

which proves the lemma.

Theorem 3.3

Let $Z_r(K)$ and $B_r(K)$ be the $r$ -cycle group and the $r$ -boundary group of $C_r(K)$ , then

$$B_r(K)\subset Z_r(K)(\subset C_r(K)).\text{\hspace{1em}(3.27)}$$

Proof

This is obvious from lemma 3.3. Any element $c$ of $B_r(K)$ is written as $c={\partial }_{r+1}d$ for some $d\in C_{r+1}(K)$ . Then we find ${\partial }_{r}c={\partial }_r({\partial }_{r+1}d)=0$ , that is, $c\in Z_r(K)$ . This implies $Z_r(K)\supset B_r(K)$ .

What are the geometrical pictures of $r$ -cycles and $r$ -boundaries? With our definitions, ${\partial }_r$ picks up the boundary of an $r$ -chain. If $c$ is an $r$ -cycle, ${\partial }_{r}c=0$ tells us that $c$ has no boundary. If $c={\partial }_{r+1}d$ is an $r$ -boundary, $c$ is the boundary of $d$ whose dimension is higher than $c$ by one. Our intuition tells us that a boundary has no boundary, hence $Z_r(K)\supset B_r(K)$ . Those elements of $Z_r(K)$ that are not boundaries play the central role in this chapter.

3.3.3 Homology groups

So far we have defined three groups $C_r(K),Z_r(K)$ and $B_r(K)$ associated with a simplicial complex $K$ . How are they related to topological properties of $K$ or to the topological space whose triangulation is $K$ ? Is it possible for $C_r(K)$ to express any property which is conserved under homeomorphism? We all know that the edges of a triangle and those of a square are homeomorphic to each other. What about their chain groups? For example, the 1-chain group associated with a triangle is

$$\begin{array}{rl}{C}_1\left(K_1\right)& =\left\{i\left(p_0{p}_1\right)+j\left(p_1{p}_2\right)+k\left(p_2{p}_0\right)|i,j,k\in \mathbb{Z}\right\}\\ & \cong \mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z}\end{array}$$

while that associated with a square is

$$C_1(K_2)\cong \mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z}.$$

Clearly $C_1(K_1)$ is not isomorphic to $C_1(K_2)$ , hence $C_r(K)$ cannot be a candidate of a topological invariant. The same is true for $Z_r(K)$ and $B_r(K)$ . It turns out that the homology groups defined in the following provide the desired topological invariants.

Definition 3.5

Let $K$ be an $n$ -dimensional simplicial complex. The $r$ th homology group $H_r(K)$ , $0\le r\le n$ , associated with $K$ is defined by

$$H_r(K)\equiv Z_r(K)/B_r(K).\text{\hspace{1em}(3.28)}$$

[Remarks: If necessary, we define $H_r(K)=0$ for $r>n$ or $r<0$ . If we want to stress that the group structure is defined with integer coefficients, we

write $H_r(K;\mathbb{Z})$ . We may also define the homology groups with $\mathbb{R}$ -coefficients, $H_r(K;\mathbb{R})$ or those with ${\mathbb{Z}}_2$ -coefficients, $H_r(K;{\mathbb{Z}}_2)$ .]

Since $B_r(K)$ is a subgroup of $Z_r(K)$ , $H_r(K)$ is well defined. The group $H_r(K)$ is the set of equivalence classes of $r$ -cycles,

$$H_r(K)\equiv \{[z]|z\in Z_r(K)\}\text{\hspace{1em}(3.29)}$$

where each equivalence class $[z]$ is called a homology class. Two $r$ -cycles $z$ and $z^{\prime }$ are in the same equivalence class if and only if $z-z^{\prime }\in B_r(K)$ , in which case $z$ is said to be homologous to $z^{\prime }$ and denoted by $z\sim z^{\prime }$ or $[z]=[z^{\prime }]$ . Geometrically $z-z^{\prime }$ is a boundary of some space. By definition, any boundary $b\in B_r(K)$ is homologous to 0 since $b-0\in B_r(K)$ . We accept the following theorem without proof.

Theorem 3.4

Homology groups are topological invariants. Let $X$ be homeomorphic to $Y$ and let $(K,f)$ and $(L,g)$ be triangulations of $X$ and $Y$ respectively. Then we have

$$H_r(K)\cong H_r(L)r=0,1,2,\dots .\text{\hspace{1em}(3.30)}$$

In particular, if $(K,f)$ and $(L,g)$ are two triangulations of $X$ , then

$$H_r(K)\cong H_r(L)r=0,1,2,\dots .\text{\hspace{1em}(3.31)}$$

Accordingly, it makes sense to talk of homology groups of a topological space $X$ which is not necessarily a polyhedron but which is triangulable. For an arbitrary triangulation $(K,f)$ , $H_r(X)$ is defined to be

$$H_r(X)\equiv H_r(K)r=0,1,2,\dots .\text{\hspace{1em}(3.32)}$$

Theorem 3.4 tells us that this is independent of the choice of the triangulation $(K,f)$ .

Example 3.6

Let $K=\{p_0\}$ . The 0-chain is $C_0(K)=\{i{p}_0|i\in \mathbb{Z}\}\cong \mathbb{Z}$ . Clearly $Z_0(K)=C_0(K)$ and $B_0(K)=\{0\}({\partial }_0{p}_0=0$ and $p_0$ cannot be a boundary of anything). Thus

$$H_0(K)\equiv Z_0(K)/B_0(K)=C_0(K)\cong \mathbb{Z}.\text{\hspace{1em}(3.33)}$$

Exercise 3.1

Let $K=\{p_0,p_1\}$ be a simplicial complex consisting of two 0-simplexes. Show that

$$H_r(K)=\{\begin{array}{ll}\mathbb{Z}\oplus \mathbb{Z}& (r=0)\\ \{0\}& (r\ne 0).\end{array}\text{\hspace{1em}(3.34)}$$

Example 3.7

Let $K=\{p_0,p_1,(p_0{p}_1)\}$ . We have

$$C_0(K)=\{i{p}_0+j{p}_1|i,j\in \mathbb{Z}\}$$

$$C_1(K)=\{k(p_0{p}_1)|k\in \mathbb{Z}\}.$$

Since $(p_0{p}_1)$ is not a boundary of any simplex in $K$ , $B_1(K)=\{0\}$ and

$$H_1(K)=Z_1(K)/B_1(K)=Z_1(K).$$

If $z=m(p_0{p}_1)\in Z_1(K)$ , it satisfies

$${\partial }_{1}z=m{\partial }_1\left(p_0{p}_1\right)=m\left\{p_1-p_0\right\}=m{p}_1-m{p}_0=0.$$

Thus, $m$ has to vanish and $Z_1(K)=0$ , hence

$$H_1(K)=0.\text{\hspace{1em}(3.35)}$$

As for $H_0(K)$ , we have $Z_0(K)=C_0(K)=\{i{p}_0+j{p}_1\}$ and

$$B_0(K)=\mathrm{im}{\partial }_1=\left\{{\partial }_{1}i\left(p_0{p}_1\right)|i\in \mathbb{Z}\right\}=\left\{i\left(p_0-p_1\right)|i\in \mathbb{Z}\right\}.$$

Define a surjective (onto) homomorphism $f:Z_0(K)\to \mathbb{Z}$ by

$$f(i{p}_0+j{p}_1)=i+j.$$

Then we find

$$\ker f=f^{-1}(0)=B_0(K).$$

Theorem 3.1 states that $Z_0(K)/\ker f\cong \mathrm{im}f=\mathbb{Z}$ , or

$$H_0(K)=Z_0(K)/B_0(K)\cong \mathbb{Z}.\text{\hspace{1em}(3.36)}$$

Example 3.8

Let $K=\{p_0,p_1,p_2,(p_0{p}_1),(p_1{p}_2),(p_2{p}_0)\}$ , see figure 3.7(b). This is a triangulation of $S^1$ . Since there are no 2-simplexes in $K$ , we have $B_1(K)=0$ and $H_1(K)=Z_1(K)/B_1(K)=Z_1(K)$ . Let $z=i(p_0{p}_1)+j(p_1{p}_2)+k(p_2{p}_0)\in Z_1(K)$ where $i,j,k\in \mathbb{Z}$ . We require that

$$\begin{array}{rl}{\partial }_{1}z& =i\left(p_1-p_0\right)+j\left(p_2-p_1\right)+k\left(p_0-p_2\right)\\ & =(k-i)p_0+(i-j)p_1+(j-k)p_2=0.\end{array}$$

This is satisfied only when $i=j=k$ . Thus, we find that

$$Z_1(K)=\{i\{(p_0{p}_1)+(p_1{p}_2)+(p_2{p}_0)\}|i\in \mathbb{Z}\}.$$

This shows that $Z_1(K)$ is isomorphic to $\mathbb{Z}$ and

$$H_1(K)=Z_1(K)\cong \mathbb{Z}.\text{\hspace{1em}(3.37)}$$

Let us compute $H_0(K)$ . We have $Z_0(K)=C_0(K)$ and

$$\begin{array}{rl}{B}_0(K)& =\left\{{\partial }_1\left[l\left(p_0{p}_1\right)+m\left(p_1{p}_2\right)+n\left(p_2{p}_0\right)\right]|l,m,n\in \mathbb{Z}\right\}\\ & =\{(n-l)p_0+(l-m)p_1+(m-n)p_2\mid l,m,n\in \mathbb{Z}\}.\end{array}$$

Define a surjective homomorphism $f:Z_0(K)\to \mathbb{Z}$ by

$$f(i{p}_0+j{p}_1+k{p}_2)=i+j+k.$$

We verify that

$$\ker f=f^{-1}(0)=B_0(K).$$

From theorem 3.1 we find $Z_0(K)/\ker f\cong \mathrm{im}f=\mathbb{Z}$ , or

$$H_0(K)=Z_0(K)/B_0(K)\cong \mathbb{Z}.\text{\hspace{1em}(3.38)}$$

$K$ is a triangulation of a circle $S^1$ , and (3.37) and (3.38) are the homology groups of $S^1$ .

Exercise 3.2

Let $K=\{p_0,p_1,p_2,p_3,(p_0{p}_1),(p_1{p}_2),(p_2{p}_3),(p_3{p}_0)\}$ be a simplicial complex whose polyhedron is a square. Verify that the homology groups are the same as those of example 3.8 above.

Example 3.9

Let $K=\{p_0,p_1,p_2,(p_0{p}_1),(p_1{p}_2),(p_2{p}_0),(p_0{p}_1{p}_2)\}$ ; see figure 3.6(b). Since the structure of 0-simplexes and 1-simplexes is the same as that of example 3.8, we have

$$H_0(K)\cong \mathbb{Z}.\text{\hspace{1em}(3.39)}$$

Let us compute $H_1(K)=Z_1(K)/B_1(K)$ . From the previous example, we have

$$Z_1(K)=\{i\{(p_0{p}_1)+(p_1{p}_2)+(p_2{p}_0)\}|i\in \mathbb{Z}\}.$$

Let $c=m(p_0{p}_1{p}_2)\in C_2(K)$ . If $b={\partial }_{2}c\in B_1(K)$ , we have

$$\begin{array}{rl}b& =m\left\{\left(p_1{p}_2\right)-\left(p_0{p}_2\right)+\left(p_0{p}_1\right)\right\}\\ & =m\left\{\left(p_0{p}_1\right)+\left(p_1{p}_2\right)+\left(p_2{p}_0\right)\right\}m\in \mathbb{Z}.\end{array}$$

This shows that $Z_1(K)\cong B_1(K)$ , hence

$$H_1(K)=Z_1(K)/B_1(K)\cong \{0\}.\text{\hspace{1em}(3.40)}$$

Since there are no 3-simplexes in $K$ , we have $B_2(K)=\{0\}$ . Then $H_2(K)=Z_2(K)/B_2(K)=Z_2(K)$ . Let $z=m(p_0{p}_1{p}_2)\in Z_2(K)$ . Since ${\partial }_{2}z=m\{(p_1{p}_2)-(p_0{p}_2)+(p_0{p}_1)\}=0$ , $m$ must vanish. Hence, $Z_1(K)=\{0\}$ and we have

$$H_2(K)\cong \{0\}.\text{\hspace{1em}(3.41)}$$

Exercise 3.3

Let

$$\begin{array}{rl}& K=\{p_0,p_1,p_2,p_3,\left(p_0{p}_1\right),\left(p_0{p}_2\right),\left(p_0{p}_3\right),\left(p_1{p}_2\right),\left(p_1{p}_3\right),\left(p_2{p}_3\right),\\ & (p_0{p}_1{p}_2),(p_0{p}_1{p}_3),(p_0{p}_2{p}_3),(p_1{p}_2{p}_3)\}\end{array}$$

be a simplicial complex whose polyhedron is the surface of a tetrahedron. Verify that

$$H_0(K)\cong \mathbb{Z}{H}_1(K)\cong \{0\}{H}_2(K)\cong \mathbb{Z}.\text{\hspace{1em}(3.42)}$$

$K$ is a triangulation of the sphere $S^2$ and (3.42) gives the homology groups of $S^2$ .

3.3.4 Computation of $H_0(K)$

Examples 3.6-3.9 and exercises 3.2, 3.3 share the same zeroth homology group, $H_0(K)\cong \mathbb{Z}$ . What is common to these simplicial complexes? We have the following answer.

Theorem 3.5

Let $K$ be a connected simplicial complex. Then

$$H_0(K)\cong \mathbb{Z}.\text{\hspace{1em}(3.43)}$$

Proof

Since $K$ is connected, for any pair of 0-simplexes $p_i$ and $p_j$ , there exists a sequence of 1-simplexes $(p_i{p}_k),(p_k{p}_l),\dots ,(p_m{p}_j)$ such that ${\partial }_1((p_i{p}_k)+(p_k{p}_l)+\cdots +(p_m{p}_j))=p_j-p_i$ . Then it follows that $p_i$ is homologous to $p_j$ , namely $[p_i]=[p_j]$ . Thus, any 0-simplex in $K$ is homologous to $p_1$ say. Suppose

$$z=\sum _{i=1}^{I_0}{n}_i{p}_i\in Z_0(K)$$

where $I_0$ is the number of 0-simplexes in $K$ . Then the homology class $[z]$ is generated by a single point,

$$[z]=\left[\sum _i{n}_i{p}_i\right]=\sum _i{n}_i[p_i]=\sum _i{n}_i[p_1].$$

It is clear that $[z]=0$ , namely $z\in B_0(K)$ , if $\sum n_i=0$ .

Let ${\sigma }_j=(p_{j,1}{p}_{j,2})$ ( $1\le j\le I_1$ ) be 1-simplexes in $K$ , $I_1$ being the number of 1-simplexes in $K$ , then

$$\begin{array}{rl}{B}_0(K)& =\mathrm{im}{\partial }_1\\ & =\left\{{\partial }_1\left(n_1{\sigma }_1+\cdots +n_{I_1}{\sigma }_{I_1}\right)|n_1,\dots ,n_{I_1}\in \mathbb{Z}\right\}\\ & =\left\{n_1\left(p_{1,2}-p_{1,1}\right)+\cdots +n_{I_1}\left(p_{I_1,2}-p_{I_1,1}\right)\mid n_1,\dots ,n_{I_1}\in \mathbb{Z}\right\}.\end{array}$$

Note that $n_j$ ( $1\le j\le I_1$ ) always appears as a pair $+n_j$ and $-n_j$ in an element of $B_0(K)$ . Thus, if

$$z=\sum _j{n}_j{p}_j\in B_0(K)\text{t h e n}\sum _j{n}_j=0.$$

Now we have proved for a connected complex $K$ that $z=\sum n_i{p}_i\in B_0(K)$ if and only if $\sum n_i=0$ .

Define a surjective homomorphism $f:Z_0(K)\to \mathbb{Z}$ by

$$f\left(n_1{p}_1+\cdots +n_{I_0}{p}_{I_0}\right)=\sum _{i=1}^{I_0}{n}_i.$$

We then have $\ker f=f^{-1}(0)=B_0(K)$ . It follows from theorem 3.1 that $H_0(K)=Z_0(K)/B_0(K)=Z_0(K)/\ker f\cong \mathrm{im}f=\mathbb{Z}$ .

3.3.5 More homology computations

Example 3.10

This and the next example deal with homology groups of nonorientable spaces. Figure 3.8 is a triangulation of the Möbius strip. Clearly $B_2(K)=0$ . Let us take a cycle $z\in Z_2(K)$ ,

Figure 3.8. A triangulation of the Möbius strip.

$$\begin{array}{rl}& z=i\left(p_0{p}_1{p}_2\right)+j\left(p_2{p}_1{p}_4\right)+k\left(p_2{p}_4{p}_3\right)\\ & +l\left(p_3{p}_4{p}_5\right)+m\left(p_3{p}_5{p}_1\right)+n\left(p_1{p}_5{p}_0\right).\end{array}$$

$z$ satisfies

$$\begin{array}{rl}& {\partial }_{2}z=i\left\{\left(p_1{p}_2\right)-\left(p_0{p}_2\right)+\left(p_0{p}_1\right)\right\}\\ & +j\left\{\left(p_1{p}_4\right)-\left(p_2{p}_4\right)+\left(p_2{p}_1\right)\right\}\\ & +k\left\{\left(p_4{p}_3\right)-\left(p_2{p}_3\right)+\left(p_2{p}_4\right)\right\}\\ & +l\left\{\left(p_4{p}_5\right)-\left(p_3{p}_5\right)+\left(p_3{p}_4\right)\right\}\\ & +m\left\{\left(p_5{p}_1\right)-\left(p_3{p}_1\right)+\left(p_3{p}_5\right)\right\}\\ & +n\left\{\left(p_5{p}_0\right)-\left(p_1{p}_0\right)+\left(p_1{p}_5\right)\right\}=0.\end{array}$$

Since each of $(p_0{p}_2)$ , $(p_1{p}_4)$ , $(p_2{p}_3)$ , $(p_4{p}_5)$ , $(p_3{p}_1)$ and $(p_5{p}_0)$ appears once and only once in ${\partial }_{2}z$ , all the coefficients must vanish, $i=j=k=l=m=n=$

0. Thus, $Z_2(K)=\{0\}$ and

$$H_2(K)=Z_2(K)/B_2(K)\cong \{0\}.\text{\hspace{1em}(3.44)}$$

To find $H_1(K)$ , we use our intuition rather than doing tedious computations. Let us find the loops which make complete circuits. One such loop is

$$z=\left(p_0{p}_1\right)+\left(p_1{p}_4\right)+\left(p_4{p}_5\right)+\left(p_5{p}_0\right).$$

Then all the other complete circuits are homologous to multiples of $z$ . For example, let us take

$$z^{\prime }=\left(p_1{p}_2\right)+\left(p_2{p}_3\right)+\left(p_3{p}_5\right)+\left(p_5{p}_1\right).$$

We find that $z\sim z^{\prime }$ since

$$z-z^{\prime }={\partial }_2\left\{\left(p_2{p}_1{p}_4\right)+\left(p_2{p}_4{p}_3\right)+\left(p_3{p}_4{p}_5\right)+\left(p_1{p}_5{p}_0\right)\right\}.$$

If, however, we take

$$z^{\prime \prime }=\left(p_1{p}_4\right)+\left(p_4{p}_5\right)+\left(p_5{p}_0\right)+\left(p_0{p}_2\right)+\left(p_2{p}_3\right)+\left(p_3{p}_1\right)$$

we find that $z^{\prime \prime }\sim 2z$ since

$$\begin{array}{rl}& 2z-z^{\prime \prime }=2\left(p_0{p}_1\right)+\left(p_1{p}_4\right)+\left(p_4{p}_5\right)+\left(p_5{p}_0\right)-\left(p_0{p}_2\right)\\ & -\left(p_2{p}_3\right)-\left(p_3{p}_1\right)\\ & ={\partial }_2\{\left(p_0{p}_1{p}_2\right)+\left(p_1{p}_4{p}_2\right)+\left(p_2{p}_4{p}_3\right)+\left(p_3{p}_4{p}_5\right)\\ & +\left(p_3{p}_5{p}_1\right)+\left(p_0{p}_1{p}_5\right)\}.\end{array}$$

We easily verify that all the closed circuits are homologous to $nz$ , $n\in \mathbb{Z}$ . $H_1(K)$ is generated by just one element $[z]$ ,

$$H_1(K)=\{i[z]|i\in \mathbb{Z}\}\cong \mathbb{Z}.\text{\hspace{1em}(3.45)}$$

Since $K$ is connected, it follows from theorem 3.5 that $H_0(K)=\{i[p_a]|i\in \mathbb{Z}\}\cong \mathbb{Z}$ , $p_a$ being any 0-simplex of $K$ .

Example 3.11

The projective plane $\mathbb{R}{P}^2$ has been defined in example 2.5(c) as the sphere $S^2$ whose antipodal points are identified. As a coset space, we may take the hemisphere (or the disc $D^2$ ) whose opposite points on the boundary $S^1$ are identified, see figure 2.5(b). Figure 3.9 is a triangulation of the projective plane. Clearly $B_2(K)=\{0\}$ . Take a cycle $z\in Z_2(K)$ ,

Figure 3.9. A triangulation of the projective plane.

$$\begin{array}{rl}& z=m_1\left(p_0{p}_1{p}_2\right)+m_2\left(p_0{p}_4{p}_1\right)+m_3\left(p_0{p}_5{p}_4\right)\\ & +m_4\left(p_0{p}_3{p}_5\right)+m_5\left(p_0{p}_2{p}_3\right)+m_6\left(p_2{p}_4{p}_3\right)\\ & +m_7\left(p_2{p}_5{p}_4\right)+m_8\left(p_2{p}_1{p}_5\right)+m_9\left(p_1{p}_3{p}_5\right)+m_{10}\left(p_1{p}_4{p}_3\right).\end{array}$$

The boundary of $z$ is

$$\begin{array}{rl}& {\partial }_{2}z=m_1\left\{\left(p_1{p}_2\right)-\left(p_0{p}_2\right)+\left(p_0{p}_1\right)\right\}\\ & +m_2\left\{\left(p_4{p}_1\right)-\left(p_0{p}_1\right)+\left(p_0{p}_4\right)\right\}\\ & +m_3\left\{\left(p_5{p}_4\right)-\left(p_0{p}_4\right)+\left(p_0{p}_5\right)\right\}\\ & +m_4\left\{\left(p_3{p}_5\right)-\left(p_0{p}_5\right)+\left(p_0{p}_3\right)\right\}\\ & +m_5\left\{\left(p_2{p}_3\right)-\left(p_0{p}_3\right)+\left(p_0{p}_2\right)\right\}\\ & +m_6\left\{\left(p_4{p}_3\right)-\left(p_2{p}_3\right)+\left(p_2{p}_4\right)\right\}\\ & +m_7\left\{\left(p_5{p}_4\right)-\left(p_2{p}_4\right)+\left(p_2{p}_5\right)\right\}\\ & +m_8\left\{\left(p_1{p}_5\right)-\left(p_2{p}_5\right)+\left(p_2{p}_1\right)\right\}\\ & +m_9\left\{\left(p_3{p}_5\right)-\left(p_1{p}_5\right)+\left(p_1{p}_3\right)\right\}\\ & +m_{10}\left\{\left(p_4{p}_3\right)-\left(p_1{p}_3\right)+\left(p_1{p}_4\right)\right\}=0.\end{array}$$

Let us look at the coefficient of each 1-simplex. For example, we have $(m_1-m_2)(p_0{p}_1)$ , hence $m_1-m_2=0$ . Similarly,

$$\begin{array}{rl}& -m_1+m_5=0,m_4-m_5=0,m_2-m_3=0,m_1-m_8=0,\\ & m_9-m_{10}=0,-m_2+m_{10}=0,m_5-m_6=0,m_6-m_7=0,\\ & m_6+m_{10}=0.\end{array}$$

These ten conditions are satisfied if and only if $m_i=0$ , $1\le i\le 10$ . This means that the cycle group $Z_2(K)$ is trivial and we have

$$H_2(K)=Z_2(K)/B_2(K)\cong \{0\}.\text{\hspace{1em}(3.46)}$$

Before we calculate $H_1(K)$ , we examine $H_2(K)$ from a slightly different viewpoint. Let us add all the 2-simplexes in $K$ with the same coefficient,

$$z\equiv \sum _{i=1}^{10}m{\sigma }_{2,i}m\in \mathbb{Z}.$$

Observe that each 1-simplex of $K$ is a common face of exactly two 2-simplexes. As a consequence, the boundary of $z$ is

$${\partial }_{2}z=2m\left(p_3{p}_5\right)+2m\left(p_5{p}_4\right)+2m\left(p_4{p}_3\right).\text{\hspace{1em}(3.47)}$$

Thus, if $z\in Z_2(K)$ , $m$ must vanish and we find $Z_2(K)=\{0\}$ as before. This observation remarkably simplifies the computation of $H_1(K)$ . Note that any 1-cycle is homologous to a multiple of

$$z=\left(p_3{p}_5\right)+\left(p_5{p}_4\right)+\left(p_4{p}_3\right)$$

cf example 3.10. Furthermore, equation (3.47) shows that an even multiple of $z$ is a boundary of a 2-chain. Thus, $z$ is a cycle and $z+z$ is homologous to 0. Hence, we find that

$$H_1(K)=\{[z]|[z]+[z]\sim [0]\}\cong {\mathbb{Z}}_2.\text{\hspace{1em}(3.48)}$$

This example shows that a homology group is not necessarily free Abelian but may have the full structure of a finitely generated Abelian group. Since $K$ is connected, we have $H_0(K)\cong \mathbb{Z}$ .

It is interesting to compare example 3.11 with the following examples. In these examples, we shall use the intuition developed in this section on boundaries and cycles to obtain results rather than giving straightforward but tedious computations.

Example 3.12

Let us consider the torus $T^2$ . A formal derivation of the homology groups of $T^2$ is left as an exercise to the reader: see Fraleigh (1976), for example. This is an appropriate place to recall the intuitive meaning of the homology groups. The $r$ th homology group is generated by those boundaryless $r$ -chains that are not, by themselves, boundaries of some $(r+1)$ -chains. For example, the surface of the torus has no boundary but it is not a boundary of some 3-chain. Thus, $H_2(T^2)$ is freely generated by one generator, the surface itself, $H_2(T^2)\cong \mathbb{Z}$ . Let us look at $H_1(T^2)$ next. Clearly the loops $a$ and $b$ in figure 3.10 have no boundaries but are not boundaries of some 2-chains. Take another loop $a^{\prime }$ . $a^{\prime }$ is homologous to $a$ since $a^{\prime }-a$ bounds the shaded area of figure 3.10.

Figure 3.10. $a^{\prime }$ is homologous to $a$ but $b$ is not. $a$ and $b$ generate $H_1(T^2)$ .

Hence, $H_1(T^2)$ is freely generated by $a$ and $b$ and $H_1(T^2)\cong \mathbb{Z}\oplus \mathbb{Z}$ . Since $T^2$ is connected, we have $H_0(T^2)\cong \mathbb{Z}$ .

Now it is easy to extend our analysis to the torus ${\Sigma }_g$ of genus $g$ . Since ${\Sigma }_g$ has no boundary and there are no 3-simplexes, the surface ${\Sigma }_g$ itself freely generates $H_2(T^2)\cong \mathbb{Z}$ . The first homology group $H_1({\Sigma }_g)$ is generated by those loops which are not boundaries of some area. Figure 3.11 shows the standard choice for the generators. We find

Figure 3.11. $a_i$ and $b_i$ $(1\le i\le g)$ generate ${\mathrm{H}}_1({\Sigma }_g)$

$$\begin{array}{rl}{H}_1\left({\Sigma }_g\right)& =\left\{i_1\left[a_1\right]+j_1\left[b_1\right]+\cdots +i_g\left[a_g\right]+j_g\left[b_g\right]\right\}\\ & \cong \underset{2g}{\underbrace{\mathbb{Z}\oplus \mathbb{Z}\oplus \cdots \oplus \mathbb{Z}}}.\end{array}\text{\hspace{1em}(3.49)}$$

Since ${\Sigma }_g$ is connected, $H_0({\Sigma }_g)\cong \mathbb{Z}$ . Observe that $a_i(b_i)$ is homologous to the edge $a_i(b_i)$ of figure 2.12. The $2g$ curves $\{a_i,b_i\}$ are called the canonical system of curves on ${\Sigma }_g$ .

Example 3.13

Figure 3.12 is a triangulation of the Klein bottle. Computations of the homology groups are much the same as those of the projective plane. Since $B_2(K)=0$ , we have $H_2(K)=Z_2(K)$ . Let $z\in Z_2(K)$ . If $z$ is a combination of all the 2-simplexes of $K$ with the same coefficient, $z=\sum m{\sigma }_{2,i}$ , the inner 1-simplexes cancel out to leave only the outer 1-simplexes

Figure 3.12. A triangulation of the Klein bottle.

$${\partial }_{2}z=-2ma$$

where $a=(p_0{p}_1)+(p_1{p}_2)+(p_2{p}_0)$ . For ${\partial }_{2}z$ to be 0, the integer $m$ must vanish and we have

$$H_2(K)=Z_2(K)\cong \{0\}.\text{\hspace{1em}(3.50)}$$

To compute $H_1(K)$ we first note, from our experience with the torus, that every 1-cycle is homologous to $ia+jb$ for some $i,j\in \mathbb{Z}$ . For a 2-chain to have a boundary consisting of $a$ and $b$ only, all the 2-simplexes in $K$ must be added with the same coefficient. As a result, for such a 2-chain $z=\sum m{\sigma }_{2,i}$ , we have $\partial z=2ma$ . This shows that $2ma\sim 0$ . Thus, $H_1(K)$ is generated by two cycles $a$ and $b$ such that $a+a=0$ , namely

$$H_1(K)=\{i[a]+j[b]|i,j\in \mathbb{Z}\}\cong {\mathbb{Z}}_2\oplus \mathbb{Z}.\text{\hspace{1em}(3.51)}$$

We obtain $H_0(K)\cong \mathbb{Z}$ since $K$ is connected.

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