2.1 Simplexes

Let $q\leq r$ a nonnegative integer. We choose $q+1$ points out of $p_{0},\dots,p_{r}$, define a $q$-simplex $\sigma_{q}=\left<p_{i_{0}}\dots p_{i_{q}}\right>$, which is called $q$-face of $\sigma_{r}$. Write $\sigma_{q}\leq\sigma_{r}$ if $\sigma_{q}$ is a face of $\sigma_{r}$, and moreover a proper face of $\sigma_{r}$ if $\sigma_{q}\neq\sigma_{r}$.

2.2 Simplicial complexes, polyhedra

Let $K$ be a set of finite number of simplexes in $\mathbb{R}^{m}$. If these simplexes are nicely fitted together, $K$ is a simplicial complex. By nicely we mean:

• if $\sigma\in K$, $\sigma'\leq\sigma$, then $\sigma'\in K$
• if $\sigma,\sigma'\in K$, then either $\sigma \cap \sigma'=\emptyset$ or $\sigma \cap\sigma'\leq\sigma,\sigma \cap\sigma'\leq\sigma'$
  $\text{dim}K$ is the largest dimension of its simplexes.
  \fcolorbox{green}{}{Example}
• $\sigma_{r}$ is a $r$-simplex and $K=\{\sigma'\mid \sigma'\leq\sigma_{r}\}$ be the set of faces of $\sigma_{r}$. $K$ is an $r$-dimensional simplical complex.
• 
  \fcolorbox{green}{}{End Example}

  Polyhedron: $\left| K \right|=\bigcup_{\sigma\in K}\sigma$ is called polyhedron of simplicial complex $K$. The dimension of $\left| K \right|$ is the same with $K$: $\text{dim} \left| K \right|=\text{dim}K$
  Let $K$ be a topological space. If there exists a simplicial complex K and a homeomorphism $f:|K| \to X$, $X$ is triangulable and the pair $(K,f)$ is triangulation of $X$.

  

Homology groups of simplicial complexes

3.1 Oriented Simplexes

Assign an orientation to an $r$-simplex for $r\geq 1$. Use $(\dots)$ for an oriented simplex. $(p_{0}p_{1})$ is a directed line segment from $p_{0}$ to $p_{1}$. $(p_{1}p_{0})=-(p_{0}p_{1})$. Similarly, for $\sigma_{2}=(p_{0}p_{1}p_{2})$ is a triangular rigion with a prescribed orientation along the edges. Any swap of the points produces a -1:

$$\begin{equation} (p_{i}p_{j}p_{k})=\text{sgn}(P)(p_{0}p_{1}p_{2}) \end{equation}$$

More generally, for $r$-simplex, select a permutation $P\in S_{r+1}$, we have that

$$\begin{equation} (p_{P(0)}\dots p_{P(r)}) = \text{sgn}(P)(p_{0}p_{1}\dots p_{r}) \end{equation}$$

3.2 Chain group, cycle group, boundary group

Let $K=\{\sigma_{\alpha}\}$ be an $n$-dimensional simplical complex. We regard $\sigma_{\alpha}$ as orinted, denote by $\sigma_{\alpha}$ as before.

Addition of r-chains:

$$c+c'=\sum_{i}(c_{i}+c_{i}')\sigma_{r,i}$$

We see that $C_{r}(K)$ is a free Abelian group of rank $I_{r}$, $C_{r}(K)\cong \mathbb{Z}^{I_{r}}$

It is easy to see the boundary operator is a homomorphism. There is a sequence of free Abelian groups and homomorphisms

$$0 \overset{ i }{ \to }C_{n}(K)\overset{ \partial_{n} }{ \to }C_{n-1}(K)\to\dots\overset{ \partial_{1} }{ \to }C_{0}(K)\overset{ \partial_{0} }{ \to }0$$

where $i:0\to C_{n}(K)$ is an inclusion map, 0 is regarded as unit element of $C_{n}(K)$

\fcolorbox{pink}{}{Lemma 3.3} $\partial_{r}\circ \partial_{r+1}$ is a zero map, $\partial_{r}\partial_{r+1}c=0$ for any $c\in C_{r+1}(K)$
\fcolorbox{pink}{}{End Lemma}
You can prove the lemma by calculating $\partial_{r}\partial_{r+1}c$ directly for some given $c$. This means that $B_{r}(K)\subset Z_{r}(K)$, or $\mathrm{Im} \partial_{r+1} \subset \ker \partial_{r}$.

3.3 Homology groups are topological invariants

We have defined 3 groups, how are they related to topological properties of K or to the topological space whose triangularization is K? Clearly $C_{r}(K)$ is not conserved under homeomorphism: If K is a triangle, $C_{1}(K)\cong \mathbb{Z}^{3}$; for square $C_{1}(K)\cong \mathbb{Z}^{4}$.

Since $B_{r}(K)$ is a subgroup of $Z_{r}(K)$,

$$H_{r}(K)\equiv \left\{[z]\mid z\in Z_{r}(K)\right\}$$

where $[z]$ denotes equivalence classes(homology class). $z\sim z'$ iff $z-z'\in B_{r}(K)$

3.4 Computation of $H_{0}$

3.5 More homology computations

The triangulation of Mobius strip. We can verify that $H_{1}(K)\cong \mathbb{Z}$, $H_{2}(K)=\{0\}$, $H_{0}(K)\cong\mathbb{Z}$ for being connected.

The triangulation of $\mathbb{R}P^{2}$, defined as $S^{2}$ where antipodal points identified.
By calculating $\partial_{2}z=0$ with $z=\sum_{i}m_{i}\Delta_{2i}$, where $\Delta_{2i}$ being all 2 simplexes, we find that $Z_{2}(K)$ is trivial, $H_{2}(K)=\{0\}$.
Consider $z=\sum_{i}m\Delta_{2,i}$ $\partial_{2}z=2m(p_{3}p_{5})+2m(p_{5}p_{4})+2m(p_{4}p_{3})$, so if $z\in Z_{2}(K)$, $m=0,$ so $Z_{2}(K)=\{0\}$. Note that any cycle is homologous to $nz$, with $z=(p_{3}p_{5})+(p_{5}p_{4})+(p_{4}p_{3})$. Furthermore, even multiple of $z$ is a boundary of 2-chain. Thus $z$ is a cycle and $z+z$ is homologous to 0. Hence we find that

$$\begin{equation} H_{1}(K)=\left\{[z]\mid [z]+[z] \sim [0]\right\}\cong \mathbb{Z}_{2} \end{equation}$$

This shows that a homology group is not necessarily free Abelian but may have the full structure of a finitely generated Abelian group.

Consider $T^{2}$ torus. r-th homology group is generated by boundaryless $r$-chains that are not boundaries of $r+1$ chains. For example, the surface of torus has no boundary but it is not boundary of some 3-chain. Thus $H_{2}(T^{2})$ is freely generated by 1 generator(the surface itself), $H_{2}(T^{2})\cong \mathbb{Z}$.

Look at $H_{1}(T^{2})$. Two loops $a,b$ have no boundaries but are not boundaries of some 2-chains. Take another loop $a'$, $a'$ is homologous to $a$ since $a'-a=\partial_{2}s$ for some $s$. So $H_{1}(T^{2})$ is freely generated by $a,b$, $H_{1}(T^{2})\cong \mathbb{Z}^{2}$. $H_{0}(T^{2})\cong \mathbb{Z}$ for $T^{2}$ connected.

Then generalizing the analysis to $\Sigma_{g}$. Since $\Sigma_{g}$ has no boundary and there are no 3-simplexes, the surface itself freely generates $H_{2}(\Sigma_{g})\cong \mathbb{Z}$. There are $2g$ independent paths, so $H_{1}(\Sigma_{g})\cong \mathbb{Z}^{2g}$. $H_{0}(\Sigma_{g})=\mathbb{Z}$ since it is connected.

For Klein bottles, let $z=\sum_{i}m\sigma_{2,i}\in Z_{2}(K)$, $\partial_{2}z=-2ma$, $a=(p_{0}p_{1})+(p_{1}p_{2})+(p_{2}p_{0})$. so $ker \partial_{2}=Z_{2}=H_{2}=\{0\}$; for $H_{1}$, we have that each 1-cycle is homologous to $ia+jb,i,j\in\mathbb{Z}$. For a 2-chain to have a boundary consisting $a,b$ only, all 2-simplex in $K$ must be added with the same coefficient. for such a 2-chain $z=\sum_{i}m\sigma_{2,i}$, $\partial z=2ma$. this shows that $2ma\sim 0$. Thus $H_{1}$ is generated by two cycles $a,b$ such that $a+a=0$, namely

$$\begin{equation} H_{1}(K)=\left\{i[a]+j[b]\mid i,j\in \mathbb{Z}\right\}\cong \mathbb{Z}_{2}\oplus \mathbb{Z} \end{equation}$$

3.4 General properties

1. Connectedness

So is true for $C_{r}(K)$, $Z_{r}(K)$ and $B_{r}(K)$

One corollary is that for disjoint union of $N$ connected components $H_{0}(K)\cong \mathbb{Z}^{N}$

2. Structure

The most general form of $H_{r}(K)$ is

$$\begin{equation} H_{r}(K)\cong \underbrace{ \mathbb{Z}\oplus \dots \oplus \mathbb{Z} }_{ f } \oplus \mathbb{Z}_{k_{1}}\oplus \dots \oplus \mathbb{Z}_{k_{p}} \end{equation}$$

From our experience $H_{r}(K)$ counts the number of $(r+1)$-d holes in $|K|$. The first part is free and the last p part are called torsion subgroup. Projective plane has $H_{1}(K)\cong \mathbb{Z}_{2}$, Klein bottle $H_{1}(K)\cong \mathbb{Z}\oplus \mathbb{Z}_{2}$. The torsion subgroup detects the twisting in polyhedron $\left| K \right|$ in a sense. We can now clarify why use $\mathbb{Z}$ coefficients for homology groups instead of $\mathbb{Z}2$ or $\mathbb{R}$. Since $\mathbb{Z}_{2}$ has no non-trivial subgroups, and $\mathbb{R}$ will make us not see the torsion subgroup, since $\mathbb{R}/ m\mathbb{R}\cong \{0\}$ for any $m\in\mathbb{Z}-\{0\}$. If $H_{r}(K;\mathbb{Z})$ is given by above, $H_{r}(K;\mathbb{R})\cong \mathbb{R}^{f}$.

3. Betti numbers, Euler-Poincare theorem

For example, $K=T^{2}$, $b_{i}(K)=1,2,1$ for $i=0,1,2$
$b_{i}(S^{2})=1,0,1$ for $i=0,1,2$

Since Betti numbers are topological invariants, $\chi(K)$ is conserved under homeomorphism. If $f:|K|\to X$, $g:|K'| \to X$ are two triangulations, $\chi(K)=\chi(K')$

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