Notes:Homology

Definitions

1. Boundary operator: $\partial_n:C_n\rightarrow C_{n-1}$ given by $\partial_n:[a_0,\ldots a_n]\mapsto\sum^n_{i=0}(-1)^i[a_0,\ldots,\hat{a_i},\ldots,a_n]$
2. $\mathbf{n}$-cycles: $Z_n$ (a cycle is defined to have boundary 0, thus $Z_n=\text{Ker}(\partial_n)$ - todo - discussion)
3. $\mathbf{n}$-boundaries: $B_n$ (the image of $\partial_{n+1}$ - all boundaries)
   • Claim: $B_n\le Z_n$ (that is: $B_n$ is a subgroup of $Z_n$)
4. $\mathbf{n}$^th homology group: $H_n:=Z_n/B_n$

Examples 1: $G_1$

$\xymatrix{ & & x \ar@/_.25pc/[dll]_a \\ y \ar@/_.25pc/[drr]_b \\ & & z \ar@/_.5pc/[uu]^c \ar@/_2pc/[uu]_d }$
Our first example, $G_1$

Chain complex: $\xymatrix{ 0 \ar[r]^{\partial_2} & C_1 \ar[r]^{\partial_1} \ar@2{->}[d] & C_0 \ar[r]^{\partial_0=0} \ar@2{->}[d] & 0 \\ & {\langle a,b,c,d\rangle\cong\mathbb{Z}^4} & {\langle x,y,z\rangle\cong\mathbb{Z}^3} }$
$\partial_1:C_1\rightarrow C_0$ morphism:

• We have:
  1. $\partial_1(a)= y-x$,
  2. $\partial_1(b)= z-y$,
  3. $\partial_1(c)= x-z$ and
  4. $\partial_1(d)=x-z$ also
• We extend this to a group homomorphism by defining:
  • $\begin{array}{rcl}\partial_1(\alpha a+\beta b+\gamma c+\delta d)&:=&\alpha\partial_1(a)+\beta\partial_1(b)+\gamma\partial_1(c)+\delta\partial_1(d)\\ & =& \alpha(y-x)+\beta(z-y)+(\gamma+\delta)(x-z)\\& = &(-\alpha+\gamma+\delta)x+(\alpha-\beta)y+(-\gamma-\delta)z\end{array}$, we may write: $\begin{pmatrix}x\\y\\z\end{pmatrix}=\alpha\left(\begin{array}{c}-1\\ 1 \\ 0\end{array}\right)+\beta\begin{pmatrix}0\\-1\\1\end{pmatrix}+\gamma\begin{pmatrix}0\\-1\\-1\end{pmatrix}+\delta\begin{pmatrix}0\\-1\\-1\end{pmatrix}=\begin{pmatrix}-1 & 0 & 0 & 0 \\ 1 & -1 & -1 & -1 \\ 0 & 1 & -1 & -1\end{pmatrix}\begin{pmatrix}\alpha\\\beta\\\gamma\\\delta\end{pmatrix}$

Computing the homology groups:

• $H_0:=Z_0/B_0=\text{Ker}(\partial_0)/\text{Im}(\partial_1)$
  1. Computing $\text{Ker}(\partial_0)$ (result: $\text{Ker}(\partial_0)=C_0$)
     • By definition, $\partial_0:[a_0]\mapsto 0$, so everything in the domain of $\partial_0$ is in the kernel!
     • Thus $Z_0=C_0$
  2. Computing $\text{Im}(\partial_1)$